Definitions, Rotation Axis and Linear Momentum
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2. Angular Momentum Of A Rigid Body
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3. Rotational Energy Of A Rigid Body
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4. Analysis Through Forces And Moments, Rolling Without Slipping
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5. Rolling With Slipping
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Exercises
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[{"Name":"Definitions, Rotation Axis and Linear Momentum","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro","Duration":"6m 54s","ChapterTopicVideoID":9121,"CourseChapterTopicPlaylistID":5405,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9121.jpeg","UploadDate":"2017-03-23T08:06:24.1770000","DurationForVideoObject":"PT6M54S","Description":null,"MetaTitle":"Intro: Video + Workbook | Proprep","MetaDescription":"Rigid Body - Definitions, Rotation Axis and Linear Momentum. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/rigid-body/definitions%2c-rotation-axis-and-linear-momentum/vid9394","VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:03.360","Text":"Hello. In this section of the course,"},{"Start":"00:03.360 ","End":"00:09.210","Text":"we\u0027re going to be speaking about what rigid bodies are and what are the axis of rotation."},{"Start":"00:09.210 ","End":"00:12.360","Text":"Let\u0027s start with what is a rigid body?"},{"Start":"00:12.360 ","End":"00:16.125","Text":"A rigid body is something that is solid,"},{"Start":"00:16.125 ","End":"00:18.240","Text":"not a liquid or a gas,"},{"Start":"00:18.240 ","End":"00:20.220","Text":"but a solid thing,"},{"Start":"00:20.220 ","End":"00:24.720","Text":"a table, your laptop whatever it might be."},{"Start":"00:24.720 ","End":"00:29.500","Text":"The exact definition is"},{"Start":"00:29.540 ","End":"00:36.729","Text":"that a rigid body that the distance between 2 consecutive points is constant."},{"Start":"00:36.830 ","End":"00:39.480","Text":"Don\u0027t get too confused with what that means,"},{"Start":"00:39.480 ","End":"00:42.300","Text":"it means that we\u0027re dealing with a solid,"},{"Start":"00:42.300 ","End":"00:44.895","Text":"and so not a liquid, and a gas."},{"Start":"00:44.895 ","End":"00:48.950","Text":"The distance between each particle is going to be"},{"Start":"00:48.950 ","End":"00:54.300","Text":"constant because the particles aren\u0027t moving within the body."},{"Start":"00:54.830 ","End":"00:59.270","Text":"The main difference that we\u0027re speaking about is up until now in the course,"},{"Start":"00:59.270 ","End":"01:03.515","Text":"we\u0027ve spoken about either point masses or the center of mass."},{"Start":"01:03.515 ","End":"01:05.345","Text":"Taking a large objects,"},{"Start":"01:05.345 ","End":"01:07.370","Text":"finding where its center of mass is and then"},{"Start":"01:07.370 ","End":"01:11.105","Text":"considering that large object as a point mass, again."},{"Start":"01:11.105 ","End":"01:15.840","Text":"We\u0027ve been speaking about what\u0027s happening outside the body"},{"Start":"01:15.840 ","End":"01:20.005","Text":"and making our calculation slightly easier by finding the center of mass."},{"Start":"01:20.005 ","End":"01:22.610","Text":"Right now, with the rigid body,"},{"Start":"01:22.610 ","End":"01:28.135","Text":"we\u0027re going to start speaking about what\u0027s happening within the body itself."},{"Start":"01:28.135 ","End":"01:32.765","Text":"Instead of changing a large body into a point-mass,"},{"Start":"01:32.765 ","End":"01:36.394","Text":"we\u0027re going to be keeping it a large body and considering the forces,"},{"Start":"01:36.394 ","End":"01:41.770","Text":"how they affect every individual part of the bodies."},{"Start":"01:41.770 ","End":"01:43.950","Text":"That is a rigid body."},{"Start":"01:43.950 ","End":"01:48.260","Text":"Now, what we can do is we can imagine that we have some kind of axis."},{"Start":"01:48.260 ","End":"01:51.160","Text":"This is going to be our axis of rotation."},{"Start":"01:51.160 ","End":"02:01.085","Text":"It\u0027s some invisible line around which our rigid body is going to rotate around."},{"Start":"02:01.085 ","End":"02:05.355","Text":"The axis of rotation can be either inside,"},{"Start":"02:05.355 ","End":"02:08.270","Text":"going through a certain point in the body."},{"Start":"02:08.270 ","End":"02:11.775","Text":"Now, I\u0027m drawing the weirdest body I can think."},{"Start":"02:11.775 ","End":"02:14.315","Text":"This is some body and it\u0027s rigid."},{"Start":"02:14.315 ","End":"02:16.355","Text":"It\u0027s some solid structure."},{"Start":"02:16.355 ","End":"02:23.630","Text":"The axis of rotation can go right through the body at any random point."},{"Start":"02:23.630 ","End":"02:26.210","Text":"It can also go through its center of mass,"},{"Start":"02:26.210 ","End":"02:27.770","Text":"if this is CM,"},{"Start":"02:27.770 ","End":"02:29.540","Text":"it can also go through our CM,"},{"Start":"02:29.540 ","End":"02:33.805","Text":"and it can also be on a point outside."},{"Start":"02:33.805 ","End":"02:38.480","Text":"Here\u0027s the axis of rotation that could go through a center of mass"},{"Start":"02:38.480 ","End":"02:42.905","Text":"and it can also go at any point outside the body as well."},{"Start":"02:42.905 ","End":"02:46.325","Text":"The only thing that this axis represents,"},{"Start":"02:46.325 ","End":"02:50.405","Text":"it means that the rigid body is rotating around it,"},{"Start":"02:50.405 ","End":"02:52.015","Text":"just around one point."},{"Start":"02:52.015 ","End":"02:54.310","Text":"I\u0027ll rub out these."},{"Start":"02:57.770 ","End":"03:05.240","Text":"It\u0027s rotating around 1 point and every single molecule in"},{"Start":"03:05.240 ","End":"03:12.550","Text":"this body is rotating around the same point at the exact same angular velocity."},{"Start":"03:12.550 ","End":"03:15.190","Text":"That\u0027s the definition."},{"Start":"03:15.950 ","End":"03:22.460","Text":"Where the whole body is rotating around this axis of rotation,"},{"Start":"03:22.460 ","End":"03:28.835","Text":"which is going through a certain point or relative to a certain point of the body and"},{"Start":"03:28.835 ","End":"03:36.925","Text":"every single molecule is rotating at the exact same angular velocity around this axis."},{"Start":"03:36.925 ","End":"03:43.040","Text":"Now, these definitions and using this idea of a rigid body and the axis"},{"Start":"03:43.040 ","End":"03:49.150","Text":"of rotation is going to help us in future to solve lots of different questions."},{"Start":"03:49.150 ","End":"03:52.210","Text":"Let\u0027s exactly see what this means."},{"Start":"03:52.210 ","End":"03:55.010","Text":"This means that every single point,"},{"Start":"03:55.010 ","End":"03:57.515","Text":"say, this is a point on this body,"},{"Start":"03:57.515 ","End":"04:04.710","Text":"is rotating around the axis in circles."},{"Start":"04:05.750 ","End":"04:11.990","Text":"The plane of the circle is at 90 degrees to the axis of rotation."},{"Start":"04:11.990 ","End":"04:14.555","Text":"That\u0027s this angle over here."},{"Start":"04:14.555 ","End":"04:16.550","Text":"Every single point also,"},{"Start":"04:16.550 ","End":"04:19.315","Text":"this point drawn in a different color,"},{"Start":"04:19.315 ","End":"04:24.675","Text":"is also rotating around this point."},{"Start":"04:24.675 ","End":"04:27.280","Text":"Each point, let\u0027s say,"},{"Start":"04:27.280 ","End":"04:30.995","Text":"this point over here going in this direction has"},{"Start":"04:30.995 ","End":"04:35.360","Text":"angular velocity Omega and this point over"},{"Start":"04:35.360 ","End":"04:39.110","Text":"here going in this direction also has"},{"Start":"04:39.110 ","End":"04:46.385","Text":"angular velocity Omega and this point as well as angular velocity of Omega."},{"Start":"04:46.385 ","End":"04:49.730","Text":"Now, the only difference between each point,"},{"Start":"04:49.730 ","End":"04:53.105","Text":"as we know, they\u0027re all moving with the same angular velocity."},{"Start":"04:53.105 ","End":"04:54.725","Text":"However, as we know,"},{"Start":"04:54.725 ","End":"04:57.920","Text":"the equation of Omega goes that"},{"Start":"04:57.920 ","End":"05:07.380","Text":"the line velocity is equal to Omega multiplied by its radius from its axis of rotation."},{"Start":"05:09.110 ","End":"05:14.955","Text":"Obviously, this point is at a radius, let\u0027s say,"},{"Start":"05:14.955 ","End":"05:20.165","Text":"r1 and this point is at a different radius,"},{"Start":"05:20.165 ","End":"05:23.240","Text":"r2 from the axis of rotation."},{"Start":"05:23.240 ","End":"05:31.820","Text":"The line velocities of these 2 points will be v_1 will equal to omega,"},{"Start":"05:31.820 ","End":"05:33.500","Text":"which is the same in both points,"},{"Start":"05:33.500 ","End":"05:41.125","Text":"multiplied by r1 because it\u0027s distance from the axis of rotation is different and v_2,"},{"Start":"05:41.125 ","End":"05:44.080","Text":"this is point number 1 and this is point number 2,"},{"Start":"05:44.080 ","End":"05:49.470","Text":"will equal to the same Omega but r_2."},{"Start":"05:49.520 ","End":"05:56.280","Text":"This is the main difference between these 2 points on the rigid body."},{"Start":"05:56.410 ","End":"06:01.160","Text":"The most important things to remember from what we\u0027ve learned is first of all,"},{"Start":"06:01.160 ","End":"06:07.415","Text":"the equation, v for straight line velocity is equal to omega r,"},{"Start":"06:07.415 ","End":"06:10.955","Text":"where omega is the angular velocity and r is the radius,"},{"Start":"06:10.955 ","End":"06:16.550","Text":"which is the distance from the axis of rotation which means that every point on"},{"Start":"06:16.550 ","End":"06:23.630","Text":"the rigid body will have the same Omega but will have a different v. Also,"},{"Start":"06:23.630 ","End":"06:27.740","Text":"that a rigid body rotates around some axis of rotation,"},{"Start":"06:27.740 ","End":"06:31.970","Text":"which will go through either a single-point through the body,"},{"Start":"06:31.970 ","End":"06:34.850","Text":"which can be anywhere from a random point or the center of"},{"Start":"06:34.850 ","End":"06:38.600","Text":"mass or it can be at a distance from"},{"Start":"06:38.600 ","End":"06:46.595","Text":"the rigid body and the whole body rotates around that point and that point doesn\u0027t move,"},{"Start":"06:46.595 ","End":"06:50.385","Text":"it remains constant and that is it."},{"Start":"06:50.385 ","End":"06:54.630","Text":"That is what is important to remember from this lesson."}],"ID":9394},{"Watched":false,"Name":"Linear Momentum of a Rigid Body","Duration":"5m 52s","ChapterTopicVideoID":9122,"CourseChapterTopicPlaylistID":5405,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:06.015","Text":"we\u0027re going to be speaking about the momentum of a rigid body."},{"Start":"00:06.015 ","End":"00:09.480","Text":"For instance, if I have any shape,"},{"Start":"00:09.480 ","End":"00:15.915","Text":"the weirder shape that I could possibly have, something like this."},{"Start":"00:15.915 ","End":"00:20.280","Text":"If it\u0027s moving along in this direction"},{"Start":"00:20.280 ","End":"00:24.120","Text":"and simultaneously it\u0027s rotating and then it changes direction,"},{"Start":"00:24.120 ","End":"00:27.340","Text":"rotates this way and it moves up and down and all around."},{"Start":"00:27.340 ","End":"00:29.115","Text":"If you\u0027re following my arrow,"},{"Start":"00:29.115 ","End":"00:31.670","Text":"all of this, you might be thinking,"},{"Start":"00:31.670 ","End":"00:37.530","Text":"how on earth am I meant to work out what its momentum is going to be?"},{"Start":"00:37.530 ","End":"00:41.070","Text":"I\u0027m reminding you that momentum is this,"},{"Start":"00:41.070 ","End":"00:43.905","Text":"my P. What is this going to be?"},{"Start":"00:43.905 ","End":"00:47.120","Text":"It\u0027s going to be the easiest thing ever. It\u0027s very simple."},{"Start":"00:47.120 ","End":"00:51.095","Text":"It\u0027s simply going to be the total mass of the body"},{"Start":"00:51.095 ","End":"00:56.340","Text":"multiplied by the velocity of its center of mass."},{"Start":"00:56.340 ","End":"00:59.285","Text":"Then let\u0027s say its center of mass is somewhere here,"},{"Start":"00:59.285 ","End":"01:04.355","Text":"then it doesn\u0027t matter if the body is rotating this way or this way, whatever."},{"Start":"01:04.355 ","End":"01:07.790","Text":"We\u0027re just taking the center of mass as a point mass and then we\u0027re just"},{"Start":"01:07.790 ","End":"01:12.270","Text":"seeing how it moves and the velocity of it as it moves."},{"Start":"01:14.240 ","End":"01:18.190","Text":"Of course, we also have to remember that our momentum is"},{"Start":"01:18.190 ","End":"01:22.345","Text":"a vector quantity and so is our velocity."},{"Start":"01:22.345 ","End":"01:26.450","Text":"Let\u0027s give an example."},{"Start":"01:26.450 ","End":"01:31.350","Text":"If here we have a ceiling, this is a ceiling,"},{"Start":"01:31.350 ","End":"01:33.150","Text":"and from the ceiling,"},{"Start":"01:33.150 ","End":"01:36.974","Text":"we have some rod attached."},{"Start":"01:36.974 ","End":"01:44.755","Text":"The rod is rigid and it\u0027s of mass m. I\u0027m going to say it\u0027s of mass m,"},{"Start":"01:44.755 ","End":"01:46.945","Text":"it\u0027s of length l,"},{"Start":"01:46.945 ","End":"01:50.570","Text":"and it\u0027s moving like a pendulum."},{"Start":"01:50.570 ","End":"01:54.510","Text":"Appear down, up to this side."},{"Start":"01:54.510 ","End":"01:58.395","Text":"Let\u0027s say it\u0027s going at Omega 0."},{"Start":"01:58.395 ","End":"02:02.615","Text":"Also, Omega 0 is given to us and we know this and we\u0027re being"},{"Start":"02:02.615 ","End":"02:08.340","Text":"asked to find what its momentum is equal to."},{"Start":"02:09.680 ","End":"02:14.315","Text":"You\u0027ll notice that we have here circular motion."},{"Start":"02:14.315 ","End":"02:18.455","Text":"That means that every section on the rod will be traveling"},{"Start":"02:18.455 ","End":"02:21.980","Text":"at a different velocity to a different section,"},{"Start":"02:21.980 ","End":"02:25.595","Text":"for instance, because this section is closer to the origin."},{"Start":"02:25.595 ","End":"02:30.140","Text":"So its Omega will be"},{"Start":"02:30.140 ","End":"02:36.315","Text":"significantly slower than the Omega that is over here for this section."},{"Start":"02:36.315 ","End":"02:38.750","Text":"Now the trick here when dealing with"},{"Start":"02:38.750 ","End":"02:43.085","Text":"a rigid body is that you don\u0027t have to consider this at all."},{"Start":"02:43.085 ","End":"02:46.310","Text":"We can just rub that out."},{"Start":"02:46.310 ","End":"02:50.120","Text":"All we have to do is look at the center of"},{"Start":"02:50.120 ","End":"02:55.025","Text":"mass because that\u0027s the only velocity that we have to take into account."},{"Start":"02:55.025 ","End":"02:57.995","Text":"If this is the center of mass,"},{"Start":"02:57.995 ","End":"03:06.950","Text":"then we know that bearing in mind that this rod has a uniform density."},{"Start":"03:06.950 ","End":"03:10.025","Text":"That means that its center of mass is going to be right in its center,"},{"Start":"03:10.025 ","End":"03:14.700","Text":"which will be at L/2."},{"Start":"03:15.170 ","End":"03:23.735","Text":"Then all we have to do in order to work our P is write down our mass,"},{"Start":"03:23.735 ","End":"03:25.010","Text":"which we know it\u0027s M,"},{"Start":"03:25.010 ","End":"03:28.160","Text":"and then multiply by my V_CM."},{"Start":"03:28.160 ","End":"03:30.125","Text":"The velocity of the center of mass."},{"Start":"03:30.125 ","End":"03:34.010","Text":"Now we know that velocity when dealing with"},{"Start":"03:34.010 ","End":"03:41.985","Text":"circular motion is equal to Omega multiplied by R,"},{"Start":"03:41.985 ","End":"03:43.770","Text":"R being the radius."},{"Start":"03:43.770 ","End":"03:53.530","Text":"Here, our Omega is Omega 0 and our R is going to be this distance L/2."},{"Start":"03:54.140 ","End":"03:59.435","Text":"Now the last thing that we have to do is just include in our direction."},{"Start":"03:59.435 ","End":"04:02.150","Text":"We can just say that it\u0027s either moving in"},{"Start":"04:02.150 ","End":"04:06.875","Text":"the Theta direction if we\u0027re going according to polar coordinates,"},{"Start":"04:06.875 ","End":"04:10.280","Text":"or we can say that at this moment in time, specifically,"},{"Start":"04:10.280 ","End":"04:16.110","Text":"it\u0027s going in this direction and we can say that that is the x direction."},{"Start":"04:17.740 ","End":"04:25.895","Text":"Let\u0027s take a look for 1 second to see how we derive this equation over here. Let\u0027s begin."},{"Start":"04:25.895 ","End":"04:30.025","Text":"We know that our momentum for an entire shape,"},{"Start":"04:30.025 ","End":"04:39.800","Text":"let\u0027s say this rod over here is going to be the sum of each individual mass m_i,"},{"Start":"04:39.800 ","End":"04:45.065","Text":"multiplied by its velocity, multiplied by v_i."},{"Start":"04:45.065 ","End":"04:51.015","Text":"We\u0027re going to sum all the momentums of all of this over here."},{"Start":"04:51.015 ","End":"04:53.645","Text":"We\u0027re summing all of the momentums."},{"Start":"04:53.645 ","End":"04:55.325","Text":"Now, what you\u0027ll notice,"},{"Start":"04:55.325 ","End":"05:01.955","Text":"if you remember the equation for V_CM is going to be,"},{"Start":"05:01.955 ","End":"05:04.130","Text":"actually won\u0027t be a vector,"},{"Start":"05:04.130 ","End":"05:14.385","Text":"is equal to the sum of m_i v_i divided by the total mass."},{"Start":"05:14.385 ","End":"05:17.870","Text":"This is the velocity for the center of mass if you remember."},{"Start":"05:17.870 ","End":"05:21.890","Text":"That means if I want to get this expression,"},{"Start":"05:21.890 ","End":"05:25.430","Text":"so I need to isolate it because that\u0027s what appears over here."},{"Start":"05:25.430 ","End":"05:30.500","Text":"All I have to do is multiply my total mass by my V_CM."},{"Start":"05:30.720 ","End":"05:34.540","Text":"That actually is a vector quantity."},{"Start":"05:34.540 ","End":"05:37.280","Text":"Then I isolate this out."},{"Start":"05:37.280 ","End":"05:44.585","Text":"It\u0027s just going to be mass multiplied by my velocity of the center of mass."},{"Start":"05:44.585 ","End":"05:46.955","Text":"That\u0027s how you derive this."},{"Start":"05:46.955 ","End":"05:52.930","Text":"Super easy. That\u0027s the end of the lesson."}],"ID":9395}],"Thumbnail":null,"ID":5405},{"Name":"2. Angular Momentum Of A Rigid Body","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Angular Momentum Of A Rigid Body","Duration":"9m 2s","ChapterTopicVideoID":9126,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.165","Text":"Hello. In the last lesson,"},{"Start":"00:03.165 ","End":"00:05.715","Text":"we spoke about momentum of a rigid body."},{"Start":"00:05.715 ","End":"00:07.995","Text":"We were speaking about linear momentum."},{"Start":"00:07.995 ","End":"00:12.134","Text":"Right now we\u0027re going to speak about angular momentum."},{"Start":"00:12.134 ","End":"00:15.750","Text":"Let\u0027s remind ourselves of the equation."},{"Start":"00:15.750 ","End":"00:19.680","Text":"Angular momentum is a vector quantity denoted by"},{"Start":"00:19.680 ","End":"00:24.600","Text":"a capital L. It equals to the position vector,"},{"Start":"00:24.600 ","End":"00:30.255","Text":"cross multiplied with the momentum vector, okay?"},{"Start":"00:30.255 ","End":"00:33.250","Text":"The linear momentum vector."},{"Start":"00:34.640 ","End":"00:38.085","Text":"Let\u0027s deal with our first case."},{"Start":"00:38.085 ","End":"00:39.455","Text":"Our first case."},{"Start":"00:39.455 ","End":"00:43.370","Text":"Let\u0027s say we\u0027re dealing with some disc."},{"Start":"00:43.370 ","End":"00:48.070","Text":"It\u0027s moving in a straight line."},{"Start":"00:48.800 ","End":"00:54.410","Text":"Currently, we\u0027re ignoring whether it\u0027s rotating about itself or what it\u0027s doing."},{"Start":"00:54.410 ","End":"00:59.165","Text":"Its total movement is in a straight line direction like this."},{"Start":"00:59.165 ","End":"01:01.359","Text":"No internal movement."},{"Start":"01:01.359 ","End":"01:04.670","Text":"Just exterior movement in straight line."},{"Start":"01:04.670 ","End":"01:10.550","Text":"We\u0027ll remember that when we\u0027re dealing with a large disc or some large shape,"},{"Start":"01:10.550 ","End":"01:15.450","Text":"then in order to find the momentum, for instance."},{"Start":"01:15.450 ","End":"01:19.175","Text":"We\u0027re actually finding the momentum of the center of mass."},{"Start":"01:19.175 ","End":"01:22.100","Text":"Which as we know from our previous video,"},{"Start":"01:22.100 ","End":"01:25.459","Text":"it\u0027s going to be the total mass of the shape multiplied"},{"Start":"01:25.459 ","End":"01:30.035","Text":"by the velocity of the center of mass."},{"Start":"01:30.035 ","End":"01:35.570","Text":"That means that in order to work out the angular momentum,"},{"Start":"01:35.570 ","End":"01:41.630","Text":"we\u0027re also going to be working out the angular momentum of the center of mass."},{"Start":"01:41.630 ","End":"01:43.745","Text":"Because when we\u0027re dealing with rigid bodies,"},{"Start":"01:43.745 ","End":"01:46.760","Text":"then they\u0027re large bodies."},{"Start":"01:46.760 ","End":"01:50.750","Text":"There\u0027s lots of different pieces to the body."},{"Start":"01:50.750 ","End":"01:56.095","Text":"It makes it very complicated sometimes to do the working out in a different way to this."},{"Start":"01:56.095 ","End":"01:59.055","Text":"Rather we work out the center of mass."},{"Start":"01:59.055 ","End":"02:03.860","Text":"That is going to then be the position vector of the center of"},{"Start":"02:03.860 ","End":"02:10.240","Text":"mass cross multiplied with our momentum vector."},{"Start":"02:10.240 ","End":"02:14.320","Text":"Again, for the center of mass."},{"Start":"02:15.620 ","End":"02:20.290","Text":"Now let\u0027s give some worked example."},{"Start":"02:20.660 ","End":"02:23.750","Text":"We\u0027re trying to find the angular momentum of"},{"Start":"02:23.750 ","End":"02:26.255","Text":"this disk which is traveling in a straight line."},{"Start":"02:26.255 ","End":"02:27.560","Text":"Okay, as we said,"},{"Start":"02:27.560 ","End":"02:31.740","Text":"we have to find where its center of mass is. Let\u0027s say from here."},{"Start":"02:31.740 ","End":"02:38.305","Text":"We\u0027re doing all the working out based on this point mass in the center of the body."},{"Start":"02:38.305 ","End":"02:45.560","Text":"This arrow here is the velocity of the center of mass."},{"Start":"02:45.560 ","End":"02:50.735","Text":"Okay? Now we have our position vector to our center of mass."},{"Start":"02:50.735 ","End":"02:52.580","Text":"Now, if you remember,"},{"Start":"02:52.580 ","End":"02:57.860","Text":"if we go back to the chapter where we speak about angular momentum."},{"Start":"02:57.860 ","End":"03:02.495","Text":"You\u0027ll remember that the angular momentum of"},{"Start":"03:02.495 ","End":"03:07.790","Text":"a body traveling in a straight line is equal to its mass,"},{"Start":"03:07.790 ","End":"03:11.790","Text":"multiplied by its velocity of the center of mass."},{"Start":"03:15.350 ","End":"03:20.925","Text":"Then this is multiplied by the r effective."},{"Start":"03:20.925 ","End":"03:23.880","Text":"Just to remind you what the r effective is."},{"Start":"03:23.880 ","End":"03:31.895","Text":"If I carry on drawing a parallel line from the line of my velocity,"},{"Start":"03:31.895 ","End":"03:36.650","Text":"and then all I have to do is draw an arrow"},{"Start":"03:36.650 ","End":"03:42.960","Text":"perpendicular to this parallel line that I\u0027ve drawn with my velocity to the origins."},{"Start":"03:42.960 ","End":"03:46.770","Text":"Let\u0027s just move it like so."},{"Start":"03:46.770 ","End":"03:49.180","Text":"Okay, this angle,"},{"Start":"03:49.180 ","End":"03:58.360","Text":"the angle between this is 90 degrees and then this is my r effective."},{"Start":"03:58.360 ","End":"04:01.625","Text":"This was case number 1, okay,"},{"Start":"04:01.625 ","End":"04:07.190","Text":"of shape moving in a straight line and working out its angular momentum."},{"Start":"04:07.190 ","End":"04:11.250","Text":"Now, this was the easiest example."},{"Start":"04:11.250 ","End":"04:15.410","Text":"Now let\u0027s move on to case number 2."},{"Start":"04:15.410 ","End":"04:17.855","Text":"Case number 2,"},{"Start":"04:17.855 ","End":"04:24.090","Text":"we are speaking about a rigid body which rotates around a fixed axis of rotation."},{"Start":"04:24.590 ","End":"04:27.895","Text":"Again, we\u0027re going to take this disc,"},{"Start":"04:27.895 ","End":"04:33.460","Text":"and let\u0027s say that its fixed axis of rotation is right in the center."},{"Start":"04:33.460 ","End":"04:40.055","Text":"Now the equation for angular momentum is going to not be this,"},{"Start":"04:40.055 ","End":"04:43.625","Text":"but it will rather equal L,"},{"Start":"04:43.625 ","End":"04:50.115","Text":"angular momentum is equal to I multiplied by Omega."},{"Start":"04:50.115 ","End":"04:58.670","Text":"Our I, you\u0027ll remember is the moment of inertia and Omega is our angular velocity."},{"Start":"04:58.670 ","End":"05:03.680","Text":"Let\u0027s say this is a fixed axis of rotation and it\u0027s rotating,"},{"Start":"05:03.680 ","End":"05:08.820","Text":"let\u0027s say in this direction at an angular velocity of Omega."},{"Start":"05:09.920 ","End":"05:13.145","Text":"Again, if this disc is large,"},{"Start":"05:13.145 ","End":"05:20.180","Text":"so we\u0027re going to take the moment of inertia from the center."},{"Start":"05:20.180 ","End":"05:26.420","Text":"That would mean that our L is going to be our L of our center of mass,"},{"Start":"05:26.420 ","End":"05:29.585","Text":"which is going to be our moment of"},{"Start":"05:29.585 ","End":"05:34.620","Text":"inertia of the center of mass multiplied by our angular velocity."},{"Start":"05:34.620 ","End":"05:37.430","Text":"To remind you, if you don\u0027t remember how to work this out,"},{"Start":"05:37.430 ","End":"05:40.130","Text":"then please go back to the chapter where we discussed moment of"},{"Start":"05:40.130 ","End":"05:45.245","Text":"inertia and how to work it out for different shapes including that of a disc."},{"Start":"05:45.245 ","End":"05:47.480","Text":"I\u0027ll just remind you that the moment of inertia of"},{"Start":"05:47.480 ","End":"05:50.510","Text":"a disc when its axis of rotation is in the center,"},{"Start":"05:50.510 ","End":"05:52.800","Text":"is going to be 1/2m*R^2,"},{"Start":"05:56.060 ","End":"05:59.265","Text":"and then multiplied by Omega."},{"Start":"05:59.265 ","End":"06:05.680","Text":"This is the angular momentum of a rigid body disc."},{"Start":"06:05.680 ","End":"06:09.220","Text":"Then just as another example,"},{"Start":"06:09.220 ","End":"06:11.770","Text":"what if our axes of rotation, okay,"},{"Start":"06:11.770 ","End":"06:13.870","Text":"so now this isn\u0027t the axes of rotation,"},{"Start":"06:13.870 ","End":"06:15.250","Text":"this is just our center of mass."},{"Start":"06:15.250 ","End":"06:18.640","Text":"Let\u0027s say that our axes of rotation is over here."},{"Start":"06:18.640 ","End":"06:21.235","Text":"Again it\u0027s rotating and an Omega,"},{"Start":"06:21.235 ","End":"06:26.970","Text":"which means that it will be rotating around this axis of rotation over here."},{"Start":"06:26.970 ","End":"06:29.175","Text":"Let\u0027s call the axis of rotation A."},{"Start":"06:29.175 ","End":"06:35.955","Text":"Then in order to find our angular momentum about point A."},{"Start":"06:35.955 ","End":"06:45.570","Text":"It will be the moment of I_A multiplied by the angular velocity Omega."},{"Start":"06:45.570 ","End":"06:48.300","Text":"What does that going to be?"},{"Start":"06:48.300 ","End":"06:51.480","Text":"We can say that I_A is going to be"},{"Start":"06:51.480 ","End":"06:59.940","Text":"equal to our I_cm+mR^2."},{"Start":"06:59.940 ","End":"07:04.940","Text":"Then this is going to be multiplied by Omega."},{"Start":"07:04.940 ","End":"07:08.030","Text":"Remember this is Steiner additive."},{"Start":"07:08.030 ","End":"07:13.340","Text":"So we\u0027re adding the mass multiplied by the distance squared."},{"Start":"07:13.340 ","End":"07:17.240","Text":"The distance from here is our R."},{"Start":"07:17.240 ","End":"07:22.025","Text":"Our center of mass is a distance R from the axis of rotation."},{"Start":"07:22.025 ","End":"07:26.840","Text":"Then just if you want to work it out on your own,"},{"Start":"07:26.840 ","End":"07:28.490","Text":"but just as the answer,"},{"Start":"07:28.490 ","End":"07:37.420","Text":"it will work out to 3/2mR^2 multiplied by Omega."},{"Start":"07:37.420 ","End":"07:40.240","Text":"For this specifically."},{"Start":"07:40.480 ","End":"07:43.430","Text":"Another example to bear in mind"},{"Start":"07:43.430 ","End":"07:49.130","Text":"is that my axes"},{"Start":"07:49.130 ","End":"07:53.810","Text":"of rotation can also be not even on any point on the disc,"},{"Start":"07:53.810 ","End":"07:56.075","Text":"for instance, it could be here."},{"Start":"07:56.075 ","End":"07:59.660","Text":"Then I can say that my shape is moving in"},{"Start":"07:59.660 ","End":"08:04.505","Text":"circular motion around this axis at a velocity of Omega."},{"Start":"08:04.505 ","End":"08:08.075","Text":"Then let\u0027s say that this is point B,"},{"Start":"08:08.075 ","End":"08:09.455","Text":"and then it\u0027s the same thing."},{"Start":"08:09.455 ","End":"08:12.800","Text":"My angular momentum here of"},{"Start":"08:12.800 ","End":"08:17.000","Text":"the rigid body is just going to be a moment of inertia about point B."},{"Start":"08:17.000 ","End":"08:18.950","Text":"Okay, then I have to add,"},{"Start":"08:18.950 ","End":"08:21.305","Text":"so it will be my I_cm plus"},{"Start":"08:21.305 ","End":"08:26.510","Text":"my specific Steiner additive for this distance over here multiplied by Omega."},{"Start":"08:26.510 ","End":"08:30.650","Text":"The same over here, just my Steiner additive will change."},{"Start":"08:30.650 ","End":"08:36.395","Text":"Now, another note that\u0027s super important is that right now in this question,"},{"Start":"08:36.395 ","End":"08:37.700","Text":"right now at the basic,"},{"Start":"08:37.700 ","End":"08:40.910","Text":"we\u0027re dealing with a shape that\u0027s moving completely"},{"Start":"08:40.910 ","End":"08:46.145","Text":"radially about our axes of rotation wherever it might be."},{"Start":"08:46.145 ","End":"08:50.405","Text":"Okay, if the shape is moving in some direction relative to B."},{"Start":"08:50.405 ","End":"08:53.495","Text":"It\u0027s a different case and we\u0027ll speak about it later in this chapter."},{"Start":"08:53.495 ","End":"08:56.165","Text":"But for now, these equations are correct for"},{"Start":"08:56.165 ","End":"08:59.935","Text":"a complete radial movement around our axis of rotation."},{"Start":"08:59.935 ","End":"09:03.640","Text":"Okay, that\u0027s the end of this lesson."}],"ID":9399},{"Watched":false,"Name":"Exercise 1","Duration":"14m 58s","ChapterTopicVideoID":9127,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.945","Text":"Hello. In this question that we\u0027re being told,"},{"Start":"00:03.945 ","End":"00:07.620","Text":"that a disk of mass M and a radius R"},{"Start":"00:07.620 ","End":"00:11.970","Text":"is at rest and attached to a frictionless axis at its center."},{"Start":"00:11.970 ","End":"00:14.430","Text":"This is its axis of rotation,"},{"Start":"00:14.430 ","End":"00:16.695","Text":"but currently, the disk is at rest."},{"Start":"00:16.695 ","End":"00:20.085","Text":"Then we\u0027re being told that a small ball of mass"},{"Start":"00:20.085 ","End":"00:25.665","Text":"m moves at a velocity of v_0 towards the disk."},{"Start":"00:25.665 ","End":"00:28.470","Text":"The ball hits the disk from the left,"},{"Start":"00:28.470 ","End":"00:32.550","Text":"a distance d from its center. What does that mean?"},{"Start":"00:32.550 ","End":"00:34.710","Text":"The ball is traveling in a straight line,"},{"Start":"00:34.710 ","End":"00:37.410","Text":"so will hit here."},{"Start":"00:37.410 ","End":"00:44.045","Text":"This distance, if we take a parallel line,"},{"Start":"00:44.045 ","End":"00:48.890","Text":"so this distance over here is d,"},{"Start":"00:48.890 ","End":"00:56.615","Text":"and this is a 90-degree angle."},{"Start":"00:56.615 ","End":"01:00.545","Text":"We\u0027re told that the ball sticks onto the disk."},{"Start":"01:00.545 ","End":"01:03.245","Text":"It will attach itself over here,"},{"Start":"01:03.245 ","End":"01:09.080","Text":"and then the ball and the disk begin rotating about the axis."},{"Start":"01:09.080 ","End":"01:14.920","Text":"We\u0027re being asked, once the ball has stuck onto the disk and everything is rotating."},{"Start":"01:14.920 ","End":"01:16.850","Text":"At the initial moment,"},{"Start":"01:16.850 ","End":"01:21.569","Text":"what is the angular velocity of the system?"},{"Start":"01:21.700 ","End":"01:26.300","Text":"We can tell from the question that our angular velocity,"},{"Start":"01:26.300 ","End":"01:30.260","Text":"our Omega will be in this direction and this is our unknown,"},{"Start":"01:30.260 ","End":"01:31.880","Text":"this is what we need to find out."},{"Start":"01:31.880 ","End":"01:38.120","Text":"Now, gravity doesn\u0027t play a part in this question because we\u0027re dealing on the x,"},{"Start":"01:38.120 ","End":"01:40.455","Text":"y plane and not into the page and"},{"Start":"01:40.455 ","End":"01:43.460","Text":"our gravity is at 90 degrees to the plane that we\u0027re working on."},{"Start":"01:43.460 ","End":"01:49.170","Text":"We\u0027re also being told that the system is frictionless."},{"Start":"01:50.030 ","End":"01:53.735","Text":"Now let\u0027s begin answering the question."},{"Start":"01:53.735 ","End":"02:01.165","Text":"Whenever you\u0027re dealing with a rigid body and questions in general, but definitely,"},{"Start":"02:01.165 ","End":"02:03.290","Text":"when dealing with rigid bodies and when"},{"Start":"02:03.290 ","End":"02:07.725","Text":"dealing with some body colliding with another body,"},{"Start":"02:07.725 ","End":"02:10.065","Text":"so collision questions,"},{"Start":"02:10.065 ","End":"02:14.710","Text":"you\u0027re going to want to look at these 3 options."},{"Start":"02:14.770 ","End":"02:18.380","Text":"They are conservation of momentum,"},{"Start":"02:18.380 ","End":"02:24.005","Text":"conservation of energy, and conservation of angular momentum."},{"Start":"02:24.005 ","End":"02:26.695","Text":"Conservation of momentum."},{"Start":"02:26.695 ","End":"02:30.000","Text":"How do we know that momentum is being conserved?"},{"Start":"02:30.000 ","End":"02:37.115","Text":"Conservation of momentum occurs when the sum of all of the forces is equal to 0."},{"Start":"02:37.115 ","End":"02:45.180","Text":"Conservation of energy means that our forces are conservative."},{"Start":"02:46.850 ","End":"02:51.740","Text":"When we\u0027re speaking about conservation of angular momentum,"},{"Start":"02:51.740 ","End":"03:01.425","Text":"we\u0027re saying the sum of all of the external torques is equal to 0."},{"Start":"03:01.425 ","End":"03:04.860","Text":"These are the things that we have to check."},{"Start":"03:05.150 ","End":"03:08.240","Text":"The first thing that we\u0027re going to check is"},{"Start":"03:08.240 ","End":"03:11.750","Text":"whether the sum of the external forces is going to be equal to 0."},{"Start":"03:11.750 ","End":"03:15.485","Text":"We\u0027re checking if there\u0027s conservation of momentum."},{"Start":"03:15.485 ","End":"03:20.570","Text":"It\u0027s very easy to look at this system over here and say, yeah,"},{"Start":"03:20.570 ","End":"03:24.230","Text":"there is conservation of momentum because there\u0027s only 2 bodies,"},{"Start":"03:24.230 ","End":"03:25.880","Text":"and once this body hits this body,"},{"Start":"03:25.880 ","End":"03:27.320","Text":"they\u0027ll just carry on moving."},{"Start":"03:27.320 ","End":"03:30.800","Text":"However, no. In this question and in all questions where there\u0027s"},{"Start":"03:30.800 ","End":"03:37.245","Text":"some kind of axis that our disk is attached to, something concrete,"},{"Start":"03:37.245 ","End":"03:42.270","Text":"here we can say that the disk is attached with the axis to"},{"Start":"03:42.270 ","End":"03:48.800","Text":"some table because we\u0027re being told that this is a frictionless axis,"},{"Start":"03:48.800 ","End":"03:54.195","Text":"so we know that it\u0027s some real physical objects here."},{"Start":"03:54.195 ","End":"03:56.265","Text":"We have a third body."},{"Start":"03:56.265 ","End":"03:58.370","Text":"If we look at what happens with this third body,"},{"Start":"03:58.370 ","End":"03:59.810","Text":"the axis over here."},{"Start":"03:59.810 ","End":"04:01.985","Text":"Our ball is going to travel,"},{"Start":"04:01.985 ","End":"04:03.590","Text":"hit our disk over here,"},{"Start":"04:03.590 ","End":"04:06.350","Text":"and then our disk is going to want to carry on"},{"Start":"04:06.350 ","End":"04:09.890","Text":"moving in this direction from the impact of the force."},{"Start":"04:09.890 ","End":"04:13.790","Text":"However, because we have this solid axis over here,"},{"Start":"04:13.790 ","End":"04:18.095","Text":"some normal force over here,"},{"Start":"04:18.095 ","End":"04:22.325","Text":"a reaction force is going to be working from the axis,"},{"Start":"04:22.325 ","End":"04:25.580","Text":"which means that we have an external force."},{"Start":"04:25.580 ","End":"04:27.755","Text":"Our normal is an external force."},{"Start":"04:27.755 ","End":"04:31.580","Text":"Therefore, the sum of our external forces does not equal 0,"},{"Start":"04:31.580 ","End":"04:35.730","Text":"which means that there\u0027s no conservation of momentum."},{"Start":"04:36.190 ","End":"04:42.665","Text":"The thing to remember is that if somebody is held in place,"},{"Start":"04:42.665 ","End":"04:45.200","Text":"that means that if there\u0027s a collision,"},{"Start":"04:45.200 ","End":"04:47.750","Text":"then there will be an external force"},{"Start":"04:47.750 ","End":"04:51.360","Text":"which means that there\u0027s no conservation of momentum."},{"Start":"04:51.800 ","End":"04:56.705","Text":"Now we\u0027re going to move on to conservation of energy."},{"Start":"04:56.705 ","End":"05:00.520","Text":"What we\u0027re dealing with here in this collision is plastic collision."},{"Start":"05:00.520 ","End":"05:05.779","Text":"The first thing that we have to know is what is plastic collision?"},{"Start":"05:05.779 ","End":"05:14.100","Text":"Plastic collision is collision that is completely inelastic. What does that mean?"},{"Start":"05:14.100 ","End":"05:16.550","Text":"In a perfect plastic collision,"},{"Start":"05:16.550 ","End":"05:21.755","Text":"the colliding particles will attach to each other and stick together,"},{"Start":"05:21.755 ","End":"05:23.750","Text":"which is precisely what\u0027s happening here."},{"Start":"05:23.750 ","End":"05:26.300","Text":"We\u0027re being told in the question that the ball"},{"Start":"05:26.300 ","End":"05:29.090","Text":"is hitting the disk and they stick together and rotate."},{"Start":"05:29.090 ","End":"05:31.190","Text":"Which means that this is plastic collision."},{"Start":"05:31.190 ","End":"05:35.070","Text":"In elastic collision, the 2 shapes,"},{"Start":"05:35.070 ","End":"05:37.860","Text":"the 2 bodies will bounce off each other."},{"Start":"05:39.140 ","End":"05:41.880","Text":"This is plastic collision."},{"Start":"05:41.880 ","End":"05:43.560","Text":"In plastic collision,"},{"Start":"05:43.560 ","End":"05:49.160","Text":"you have to know as a rule that there\u0027s no energy conservation ever."},{"Start":"05:49.160 ","End":"05:56.405","Text":"First of all, we see the word that the ball hits the disk and they stick together."},{"Start":"05:56.405 ","End":"05:59.225","Text":"Plastic collision, no energy conservation."},{"Start":"05:59.225 ","End":"06:02.765","Text":"The second thing that you could use if you forget this,"},{"Start":"06:02.765 ","End":"06:07.945","Text":"is that because we don\u0027t know what\u0027s happening inside,"},{"Start":"06:07.945 ","End":"06:11.430","Text":"we don\u0027t really know what\u0027s going on with the energy."},{"Start":"06:11.430 ","End":"06:15.485","Text":"If I didn\u0027t have enough information,"},{"Start":"06:15.485 ","End":"06:20.650","Text":"then I wouldn\u0027t even go about this route in this specific question,"},{"Start":"06:20.650 ","End":"06:22.380","Text":"in a question that was similar."},{"Start":"06:22.380 ","End":"06:27.750","Text":"Here we specifically know that there\u0027s no conservation of energy."},{"Start":"06:28.580 ","End":"06:33.140","Text":"Now the third thing is conservation of angular momentum,"},{"Start":"06:33.140 ","End":"06:38.590","Text":"is the sum of all of the external torques equal to 0. Let\u0027s take a look."},{"Start":"06:38.590 ","End":"06:40.860","Text":"We\u0027ll look at the size."},{"Start":"06:40.860 ","End":"06:45.455","Text":"As we know, the size of the torque is going to be equal to"},{"Start":"06:45.455 ","End":"06:50.465","Text":"the size of our radius vector, the distance,"},{"Start":"06:50.465 ","End":"06:55.619","Text":"the force is from the axis of rotation multiplied"},{"Start":"06:55.619 ","End":"07:02.750","Text":"by the size of the force and then multiplied by sine of the angle."},{"Start":"07:02.750 ","End":"07:06.170","Text":"Now we can see here that our force is the normal force."},{"Start":"07:06.170 ","End":"07:10.400","Text":"But seeing it here, it\u0027s acting from the axis of rotation,"},{"Start":"07:10.400 ","End":"07:12.410","Text":"meaning its distance, its r,"},{"Start":"07:12.410 ","End":"07:16.225","Text":"is 0 because it\u0027s at the axis of rotation."},{"Start":"07:16.225 ","End":"07:22.040","Text":"If this is 0, then that means that this whole thing is equal to 0."},{"Start":"07:22.040 ","End":"07:26.615","Text":"So that means that the sum of the external torques is equal to 0."},{"Start":"07:26.615 ","End":"07:31.588","Text":"Great. We have conservation of angular momentum."},{"Start":"07:31.588 ","End":"07:34.705","Text":"Now if you\u0027re in a test situation,"},{"Start":"07:34.705 ","End":"07:38.035","Text":"in order to make sure that you get the first couple of points,"},{"Start":"07:38.035 ","End":"07:40.330","Text":"then I would write in big letters,"},{"Start":"07:40.330 ","End":"07:46.165","Text":"there\u0027s conservation of angular momentum and then you\u0027ll already get a few points."},{"Start":"07:46.165 ","End":"07:49.825","Text":"Now onto the working out."},{"Start":"07:49.825 ","End":"07:53.125","Text":"What we know is that the angular momentum"},{"Start":"07:53.125 ","End":"07:57.950","Text":"before is going to equal the angular momentum after."},{"Start":"07:57.960 ","End":"08:04.030","Text":"We can say that our total angular momentum initially,"},{"Start":"08:04.030 ","End":"08:08.110","Text":"so this is before the collision,"},{"Start":"08:08.110 ","End":"08:12.820","Text":"is going to be equal to the angular momentum of the ball,"},{"Start":"08:12.820 ","End":"08:14.500","Text":"so I\u0027ll write b,"},{"Start":"08:14.500 ","End":"08:21.010","Text":"plus the angular momentum of the disk, L_d."},{"Start":"08:21.010 ","End":"08:30.200","Text":"Then this is going to be equal to the total angular momentum after final."},{"Start":"08:30.570 ","End":"08:37.330","Text":"Let\u0027s begin by seeing what our angular momentum of the disk is."},{"Start":"08:37.330 ","End":"08:40.585","Text":"Initially, we\u0027re going to solve from the beginning."},{"Start":"08:40.585 ","End":"08:44.965","Text":"We\u0027re told in the question that the disk is at rest."},{"Start":"08:44.965 ","End":"08:51.070","Text":"We know that our angular momentum is going to be I multiply by Omega,"},{"Start":"08:51.070 ","End":"08:53.920","Text":"but seeing as my Omega is equal to 0,"},{"Start":"08:53.920 ","End":"08:56.845","Text":"so this is going to be equal to 0."},{"Start":"08:56.845 ","End":"09:02.680","Text":"Then let\u0027s work out the angular momentum of the ball initially."},{"Start":"09:02.680 ","End":"09:07.870","Text":"This is going to be r cross multiplied"},{"Start":"09:07.870 ","End":"09:13.495","Text":"by our p. Now this looks like it\u0027s going to be really difficult."},{"Start":"09:13.495 ","End":"09:18.745","Text":"However, we know that our ball is moving in a straight line at a velocity of v is 0."},{"Start":"09:18.745 ","End":"09:21.310","Text":"In one of the previous lessons,"},{"Start":"09:21.310 ","End":"09:24.535","Text":"we said that the equation for"},{"Start":"09:24.535 ","End":"09:29.440","Text":"the angular momentum for a body moving in a straight line is going to"},{"Start":"09:29.440 ","End":"09:39.290","Text":"be equal to its mass multiplied by its velocity multiplied by its r effective."},{"Start":"09:39.360 ","End":"09:44.150","Text":"Let\u0027s take a look at what r effective is."},{"Start":"09:44.910 ","End":"09:53.139","Text":"If we draw, this will be our r over here."},{"Start":"09:53.139 ","End":"09:55.855","Text":"So this is our r vector."},{"Start":"09:55.855 ","End":"10:04.915","Text":"However, we know that r effective is the perpendicular component to our velocity vector."},{"Start":"10:04.915 ","End":"10:09.220","Text":"Our velocity vector is going along this line over here and"},{"Start":"10:09.220 ","End":"10:13.540","Text":"the perpendicular component we can see is this over here."},{"Start":"10:13.540 ","End":"10:20.485","Text":"We know already from the question that this distance here is d. Alternatively,"},{"Start":"10:20.485 ","End":"10:24.760","Text":"we can also say that r vector,"},{"Start":"10:24.760 ","End":"10:26.080","Text":"we can even take it from here."},{"Start":"10:26.080 ","End":"10:27.895","Text":"It makes no difference."},{"Start":"10:27.895 ","End":"10:30.250","Text":"This has meant to be a straight line."},{"Start":"10:30.250 ","End":"10:32.814","Text":"So if this was our r vector,"},{"Start":"10:32.814 ","End":"10:36.205","Text":"so we can continue this line."},{"Start":"10:36.205 ","End":"10:38.020","Text":"This is a parallel line,"},{"Start":"10:38.020 ","End":"10:40.120","Text":"and the component,"},{"Start":"10:40.120 ","End":"10:44.004","Text":"which is perpendicular to our velocity vector,"},{"Start":"10:44.004 ","End":"10:52.480","Text":"is this, and its size is d. Now we can simply just rewrite this."},{"Start":"10:52.480 ","End":"10:55.300","Text":"We know that our mass of the ball is m,"},{"Start":"10:55.300 ","End":"10:57.415","Text":"our velocity is v_0,"},{"Start":"10:57.415 ","End":"11:07.011","Text":"and r effective is d. Now we\u0027re moving back to this equation."},{"Start":"11:07.011 ","End":"11:09.490","Text":"So we know that at the end of the question,"},{"Start":"11:09.490 ","End":"11:14.545","Text":"the ball is sticking onto the disk and they\u0027re both rotating about an axis."},{"Start":"11:14.545 ","End":"11:18.265","Text":"If we remember from one of the previous lessons,"},{"Start":"11:18.265 ","End":"11:24.505","Text":"that the equation for the angular momentum of a shape rotating about"},{"Start":"11:24.505 ","End":"11:32.770","Text":"an axis in perfect radial motion is going to be I multiplied by Omega."},{"Start":"11:32.770 ","End":"11:41.110","Text":"Now we know that our total angular momentum at"},{"Start":"11:41.110 ","End":"11:48.955","Text":"the beginning is going to be equal to 0 plus m(v_0)d,"},{"Start":"11:48.955 ","End":"11:55.380","Text":"which is going to be equal to our L total final,"},{"Start":"11:55.380 ","End":"11:59.980","Text":"which is also equal to I Omega."},{"Start":"11:59.980 ","End":"12:04.375","Text":"Now I remind you that we\u0027re trying to find out what our Omega is."},{"Start":"12:04.375 ","End":"12:10.760","Text":"Now what we have to do is we have to work out the moment of inertia of this system."},{"Start":"12:12.390 ","End":"12:15.215","Text":"Let\u0027s work out where my I is."},{"Start":"12:15.215 ","End":"12:20.830","Text":"My I is, if you remember back to chapter in moments of inertia,"},{"Start":"12:20.830 ","End":"12:23.500","Text":"we speak about the additivity of I."},{"Start":"12:23.500 ","End":"12:34.180","Text":"That means that my total I is going to be my I of the disk plus my I of the ball."},{"Start":"12:34.180 ","End":"12:39.010","Text":"Let\u0027s see. The moment of inertia of the disk,"},{"Start":"12:39.010 ","End":"12:41.425","Text":"so we work it out in that chapter."},{"Start":"12:41.425 ","End":"12:49.495","Text":"It\u0027s half times the mass of the disk multiplied by the radius squared."},{"Start":"12:49.495 ","End":"12:52.060","Text":"This is going to be r^2."},{"Start":"12:52.060 ","End":"12:56.620","Text":"This is of course its moment of inertia when the axis of rotation is at its center."},{"Start":"12:56.620 ","End":"12:57.790","Text":"If it isn\u0027t at its center,"},{"Start":"12:57.790 ","End":"12:59.560","Text":"then it\u0027s a slightly different case and"},{"Start":"12:59.560 ","End":"13:02.420","Text":"you can go back to the chapter if you don\u0027t remember."},{"Start":"13:02.790 ","End":"13:06.700","Text":"Then my I of the ball,"},{"Start":"13:06.700 ","End":"13:10.720","Text":"so seeing this in the question,"},{"Start":"13:10.720 ","End":"13:13.075","Text":"we\u0027re not told the radius of the ball."},{"Start":"13:13.075 ","End":"13:16.555","Text":"We can consider our ball as some point mass."},{"Start":"13:16.555 ","End":"13:21.055","Text":"Which means that all we have to do is write down,"},{"Start":"13:21.055 ","End":"13:24.355","Text":"we multiply its mass, which is m,"},{"Start":"13:24.355 ","End":"13:28.555","Text":"multiplied by its distance from the axis of rotation."},{"Start":"13:28.555 ","End":"13:31.330","Text":"After the collision, its distance from the axis of"},{"Start":"13:31.330 ","End":"13:34.360","Text":"rotation is just the radius of our disk,"},{"Start":"13:34.360 ","End":"13:38.545","Text":"which is R. It\u0027s going to be m multiplied by R"},{"Start":"13:38.545 ","End":"13:44.260","Text":"squared because we\u0027re considering the ball as a point mass."},{"Start":"13:44.260 ","End":"13:50.035","Text":"That means that our total I is going to be,"},{"Start":"13:50.035 ","End":"13:53.650","Text":"so we have a common factor, R^2,"},{"Start":"13:53.650 ","End":"14:03.940","Text":"and then we have a half of M plus m. Therefore,"},{"Start":"14:03.940 ","End":"14:13.360","Text":"we can say that we have m(v_0)d is equal to I Omega,"},{"Start":"14:13.360 ","End":"14:15.505","Text":"and we have our I over here,"},{"Start":"14:15.505 ","End":"14:25.630","Text":"which is R squared half of capital M plus m multiplied by Omega."},{"Start":"14:25.630 ","End":"14:27.415","Text":"Now because in the question,"},{"Start":"14:27.415 ","End":"14:30.370","Text":"we\u0027re finding the angular velocity of the system,"},{"Start":"14:30.370 ","End":"14:34.504","Text":"so I just have to isolate out this Omega."},{"Start":"14:34.504 ","End":"14:43.990","Text":"Therefore, I\u0027ll get the Omega is equal to m(v_0)d divided by"},{"Start":"14:43.990 ","End":"14:50.169","Text":"R^2 half of M plus"},{"Start":"14:50.169 ","End":"14:58.760","Text":"m. That is the end of our question."}],"ID":9400},{"Watched":false,"Name":"Exercise 2","Duration":"10m 13s","ChapterTopicVideoID":9128,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:03.015","Text":"Hello. In this question,"},{"Start":"00:03.015 ","End":"00:06.060","Text":"we\u0027re being told that we have a disc of radius capital"},{"Start":"00:06.060 ","End":"00:10.785","Text":"R. This over here and have mass capital M,"},{"Start":"00:10.785 ","End":"00:17.805","Text":"which is rotating about an axis at the center at a constant angular velocity of Omega 0."},{"Start":"00:17.805 ","End":"00:19.275","Text":"A man of mass,"},{"Start":"00:19.275 ","End":"00:22.755","Text":"m over here stands at the edge of the disk."},{"Start":"00:22.755 ","End":"00:26.910","Text":"The man jumps off the disk and his velocity at this moment is"},{"Start":"00:26.910 ","End":"00:31.980","Text":"v_0 in the radial direction relative to the ground."},{"Start":"00:31.980 ","End":"00:35.295","Text":"Relative to the ground is very important here in this question."},{"Start":"00:35.295 ","End":"00:38.175","Text":"That means that when he jumps off,"},{"Start":"00:38.175 ","End":"00:46.340","Text":"his velocity is such what is the discs angular velocity after the jump,"},{"Start":"00:46.340 ","End":"00:51.570","Text":"so we\u0027re being asked to find Omega after the jump."},{"Start":"00:52.340 ","End":"00:58.805","Text":"How can we do this? We know that we have to look out for the conservation of energy,"},{"Start":"00:58.805 ","End":"01:00.350","Text":"the conservation of momentum,"},{"Start":"01:00.350 ","End":"01:03.755","Text":"and conservation of angular momentum."},{"Start":"01:03.755 ","End":"01:10.320","Text":"Energy momentum, angular momentum, EMA."},{"Start":"01:10.540 ","End":"01:14.165","Text":"Now, just like in the previous question, where,"},{"Start":"01:14.165 ","End":"01:19.100","Text":"remember we had a disk and a ball hitting the disk and then the ball would"},{"Start":"01:19.100 ","End":"01:24.570","Text":"stick to the disk and we said that that was a plastic collision."},{"Start":"01:24.570 ","End":"01:29.970","Text":"Which meant that there was no conservation of momentum."},{"Start":"01:30.620 ","End":"01:35.135","Text":"There\u0027s no conservation of momentum because this is the exact same question,"},{"Start":"01:35.135 ","End":"01:37.019","Text":"but the opposite way around."},{"Start":"01:37.019 ","End":"01:40.520","Text":"Now the man and the disk are connected and when he jumps off,"},{"Start":"01:40.520 ","End":"01:42.070","Text":"it\u0027s like he breaks off the disk."},{"Start":"01:42.070 ","End":"01:45.530","Text":"That\u0027s the exact opposite in the opposite direction,"},{"Start":"01:45.530 ","End":"01:49.100","Text":"I mean, but it means that there\u0027s no conservation of momentum."},{"Start":"01:49.100 ","End":"01:50.810","Text":"What about of energy?"},{"Start":"01:50.810 ","End":"01:59.255","Text":"Now again, in the last question we spoke about if something is rotating about some axis,"},{"Start":"01:59.255 ","End":"02:03.500","Text":"then that means that the axes is acting as the third body,"},{"Start":"02:03.500 ","End":"02:07.290","Text":"which is applying some reaction for us."},{"Start":"02:07.290 ","End":"02:10.430","Text":"Let\u0027s say it\u0027s in this direction, the normal force."},{"Start":"02:10.430 ","End":"02:13.100","Text":"He said that if we have this reaction for us,"},{"Start":"02:13.100 ","End":"02:17.585","Text":"that means that the sum of the external forces does not equal 0,"},{"Start":"02:17.585 ","End":"02:21.830","Text":"which means that there\u0027s no conservation of energy."},{"Start":"02:21.830 ","End":"02:24.860","Text":"The only thing that we have is conservation of"},{"Start":"02:24.860 ","End":"02:29.795","Text":"angular momentum and let\u0027s just say why we have this,"},{"Start":"02:29.795 ","End":"02:32.480","Text":"because we know that the sum of"},{"Start":"02:32.480 ","End":"02:40.250","Text":"the external torques is equal to 0 and this is because the normal force,"},{"Start":"02:40.250 ","End":"02:43.280","Text":"which is the external force which we\u0027re working out the torque for,"},{"Start":"02:43.280 ","End":"02:45.875","Text":"is working from the axis of rotation,"},{"Start":"02:45.875 ","End":"02:48.890","Text":"which means that it\u0027s position vector,"},{"Start":"02:48.890 ","End":"02:50.915","Text":"it\u0027s a vector is equal to 0,"},{"Start":"02:50.915 ","End":"02:56.610","Text":"which means that its torque is equal to 0."},{"Start":"02:57.040 ","End":"03:00.755","Text":"Now let\u0027s look at how we work this out."},{"Start":"03:00.755 ","End":"03:04.625","Text":"I\u0027m going to start with my initial angular momentum."},{"Start":"03:04.625 ","End":"03:07.490","Text":"My L_i, which we know is going to be equal"},{"Start":"03:07.490 ","End":"03:10.955","Text":"to my moment of inertia of the entire system,"},{"Start":"03:10.955 ","End":"03:15.605","Text":"so that\u0027s the disk and the man multiplied by its angular velocity,"},{"Start":"03:15.605 ","End":"03:17.740","Text":"which here is Omega 0,"},{"Start":"03:17.740 ","End":"03:19.595","Text":"so that\u0027s my initial."},{"Start":"03:19.595 ","End":"03:22.465","Text":"Now I\u0027m going to look at what my I is,"},{"Start":"03:22.465 ","End":"03:24.150","Text":"my moment of inertia."},{"Start":"03:24.150 ","End":"03:33.368","Text":"My moment of inertia is going to be equal to my I of the disk plus my I of the man, my."},{"Start":"03:33.368 ","End":"03:36.530","Text":"To I of the disk if we go back to what"},{"Start":"03:36.530 ","End":"03:40.190","Text":"we learned in our chapter on moment of inertia is going to"},{"Start":"03:40.190 ","End":"03:47.755","Text":"be 1/2 mass of the disk multiplied by its radius squired then the inertia of the man."},{"Start":"03:47.755 ","End":"03:54.520","Text":"We can see because he has a mass of M and we\u0027re not given a radius."},{"Start":"03:54.520 ","End":"03:58.010","Text":"We just know that his distance from the axis of rotation,"},{"Start":"03:58.010 ","End":"03:59.795","Text":"because he\u0027s standing at the end of the disk,"},{"Start":"03:59.795 ","End":"04:01.490","Text":"must be capital I."},{"Start":"04:01.490 ","End":"04:06.210","Text":"We\u0027re going to look at him and consider him as a point-mass."},{"Start":"04:06.210 ","End":"04:11.180","Text":"Which means that his moment of inertia is simply going to be m is"},{"Start":"04:11.180 ","End":"04:17.160","Text":"mass multiplied by his distance from the axis of rotation squared."},{"Start":"04:17.630 ","End":"04:21.260","Text":"That means that my L_i is going to be,"},{"Start":"04:21.260 ","End":"04:25.845","Text":"if I just substitute this in my lifetime as R^2,"},{"Start":"04:25.845 ","End":"04:29.870","Text":"and then we have 1/2 capital M plus small"},{"Start":"04:29.870 ","End":"04:34.775","Text":"m. I just wrote this up here multiplied by Omega 0."},{"Start":"04:34.775 ","End":"04:39.385","Text":"What is my L final."},{"Start":"04:39.385 ","End":"04:47.610","Text":"At the end, my angular momentum is going to be my I(d),"},{"Start":"04:51.370 ","End":"04:58.955","Text":"but at the end, so I\u0027ll put a tag over here multiplied by its angular momentum,"},{"Start":"04:58.955 ","End":"05:01.580","Text":"which is what I\u0027m trying to find out this Omega."},{"Start":"05:01.580 ","End":"05:03.960","Text":"This is unknown"},{"Start":"05:05.390 ","End":"05:13.440","Text":"plus my L_m."},{"Start":"05:13.440 ","End":"05:18.500","Text":"What is the L_m? So we know that it\u0027s going to be his position vector cross"},{"Start":"05:18.500 ","End":"05:24.750","Text":"multiplied with his momentum vector so we can see that,"},{"Start":"05:24.750 ","End":"05:27.135","Text":"there\u0027s 2 ways of solving this as you know,"},{"Start":"05:27.135 ","End":"05:31.130","Text":"1 of the ways is by working out its size and direction."},{"Start":"05:31.130 ","End":"05:33.560","Text":"If we first work out its size,"},{"Start":"05:33.560 ","End":"05:39.360","Text":"without the direction in the meantime so we can see that it\u0027s going to be"},{"Start":"05:40.270 ","End":"05:44.705","Text":"the size of I times the size of p multiplied by"},{"Start":"05:44.705 ","End":"05:49.475","Text":"sine of the angle between and then we\u0027ll add in the direction in a second."},{"Start":"05:49.475 ","End":"05:53.240","Text":"As we can see, the I vector is going to be"},{"Start":"05:53.240 ","End":"06:00.455","Text":"going here to the man in a straight line from the origin."},{"Start":"06:00.455 ","End":"06:05.315","Text":"My P, as we know so let\u0027s write this down,"},{"Start":"06:05.315 ","End":"06:12.565","Text":"my P is equal to my mass times my velocity."},{"Start":"06:12.565 ","End":"06:13.915","Text":"Now as we can see,"},{"Start":"06:13.915 ","End":"06:17.155","Text":"my velocity is in the exact same direction,"},{"Start":"06:17.155 ","End":"06:23.270","Text":"so it\u0027s parallel to my I vector, which means that,"},{"Start":"06:23.270 ","End":"06:31.660","Text":"well, I\u0027ll have here will be equal to the size of my I vector multiplied by m,"},{"Start":"06:31.660 ","End":"06:33.220","Text":"which is a scalar vector,"},{"Start":"06:33.220 ","End":"06:39.115","Text":"multiplied by the size of my velocity times sine of the angle."},{"Start":"06:39.115 ","End":"06:41.215","Text":"Now because they\u0027re both parallel,"},{"Start":"06:41.215 ","End":"06:46.170","Text":"there\u0027s 0 angle between them and Sine of 0 is equal to 0,"},{"Start":"06:46.170 ","End":"06:49.565","Text":"so this whole thing is going to be equal to 0."},{"Start":"06:49.565 ","End":"06:54.500","Text":"All of this is equal to 0,"},{"Start":"06:54.500 ","End":"07:01.805","Text":"which means that my final angular momentum is simply going to be."},{"Start":"07:01.805 ","End":"07:07.220","Text":"Now my I_d. Sorry, this was not going to have"},{"Start":"07:07.220 ","End":"07:09.860","Text":"a tag here because it\u0027s going to be"},{"Start":"07:09.860 ","End":"07:12.740","Text":"the exact same thing as what I wrote here just without this,"},{"Start":"07:12.740 ","End":"07:13.880","Text":"because this is not the total I_d."},{"Start":"07:13.880 ","End":"07:18.545","Text":"It\u0027s simply going to be equal to half capital"},{"Start":"07:18.545 ","End":"07:24.140","Text":"M R^2 multiplied by this Omega, which is an unknown."},{"Start":"07:24.140 ","End":"07:27.545","Text":"Now because we have conservation of angular momentum."},{"Start":"07:27.545 ","End":"07:34.545","Text":"I\u0027m reminding you over here so I can say that my L_f equal to my L_i,"},{"Start":"07:34.545 ","End":"07:39.830","Text":"so let\u0027s scroll down a little bit to make some more space so that means that my L_f"},{"Start":"07:39.830 ","End":"07:46.200","Text":"which is 1/2 m^2 multiplied by my unknown Omega,"},{"Start":"07:46.200 ","End":"07:48.540","Text":"is equal to my L_i,"},{"Start":"07:48.540 ","End":"07:58.100","Text":"which is equal to R^2 1/2 M plus m multiplied by Omega 0."},{"Start":"07:58.100 ","End":"08:01.220","Text":"As we can see, we can divide both sides by R^2."},{"Start":"08:01.220 ","End":"08:03.410","Text":"Now all we have to do is isolate out for"},{"Start":"08:03.410 ","End":"08:06.800","Text":"our Omega and then we\u0027ve answered the question so then we\u0027ll get that"},{"Start":"08:06.800 ","End":"08:13.740","Text":"our Omega is equal to 1/2 m plus m multiplied by"},{"Start":"08:13.740 ","End":"08:22.770","Text":"Omega 0 and then all of this is going to be divided by 1/2 M Omega."},{"Start":"08:23.570 ","End":"08:27.180","Text":"This is the final answer."},{"Start":"08:27.180 ","End":"08:33.305","Text":"Now let\u0027s give a little bit more of an intuitive explanation as to what is happening."},{"Start":"08:33.305 ","End":"08:38.570","Text":"What\u0027s happening is that when the man is standing on the disk,"},{"Start":"08:38.570 ","End":"08:44.600","Text":"as we can see, everything is rotating in this direction at a angular velocity of Omega 0."},{"Start":"08:44.600 ","End":"08:48.485","Text":"Which means that at this exact moment when he\u0027s standing at the edge of the disk,"},{"Start":"08:48.485 ","End":"08:52.180","Text":"his velocity is in fact into the page,"},{"Start":"08:52.180 ","End":"08:54.650","Text":"but then from the moment he jumps,"},{"Start":"08:54.650 ","End":"08:58.520","Text":"he only has this component so he\u0027s"},{"Start":"08:58.520 ","End":"09:02.945","Text":"missing this component over here just disappears so how does that happen?"},{"Start":"09:02.945 ","End":"09:06.485","Text":"That means that he\u0027s jumping at some diagonal,"},{"Start":"09:06.485 ","End":"09:10.640","Text":"which is what\u0027s going to cancel out the component going in the other direction,"},{"Start":"09:10.640 ","End":"09:13.090","Text":"in the other diagonal direction."},{"Start":"09:13.090 ","End":"09:15.170","Text":"That will be canceled out,"},{"Start":"09:15.170 ","End":"09:19.690","Text":"leaving just this velocity in this component over here."},{"Start":"09:19.690 ","End":"09:24.380","Text":"Which is what was crucial to us solving this and"},{"Start":"09:24.380 ","End":"09:31.739","Text":"finding the angular momentum of the man himself."},{"Start":"09:31.739 ","End":"09:34.005","Text":"That\u0027s number 1, number 2,"},{"Start":"09:34.005 ","End":"09:38.630","Text":"if he\u0027s jumping at this angle so that means that as he jumps,"},{"Start":"09:38.630 ","End":"09:43.100","Text":"he\u0027s pushing backwards with his feet so let\u0027s draw it over here."},{"Start":"09:43.100 ","End":"09:45.020","Text":"He\u0027s pushing backwards with his feet in"},{"Start":"09:45.020 ","End":"09:48.810","Text":"this direction in order to push himself off the disk."},{"Start":"09:48.810 ","End":"09:52.670","Text":"This force over here, this blue arrow,"},{"Start":"09:52.670 ","End":"09:56.630","Text":"is in fact what is causing this change in angular velocity."},{"Start":"09:56.630 ","End":"10:01.790","Text":"The angular velocity doesn\u0027t just randomly changed from the fact that the man isn\u0027t on"},{"Start":"10:01.790 ","End":"10:07.400","Text":"here it\u0027s changing because there\u0027s external force is pushing against the disk,"},{"Start":"10:07.400 ","End":"10:10.814","Text":"which is changing the velocity."},{"Start":"10:10.814 ","End":"10:13.710","Text":"That\u0027s the end of the lesson."}],"ID":9401},{"Watched":false,"Name":"Angular Momentum In a Complex Movement","Duration":"6m 11s","ChapterTopicVideoID":9129,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:06.405","Text":"Hello, in this lesson we\u0027re going to continue speaking about angular momentum."},{"Start":"00:06.405 ","End":"00:08.295","Text":"In the previous lesson,"},{"Start":"00:08.295 ","End":"00:11.010","Text":"we learned about how to work out the angular momentum of"},{"Start":"00:11.010 ","End":"00:14.620","Text":"some disk which is traveling in a straight line."},{"Start":"00:14.620 ","End":"00:20.160","Text":"We said in that lesson that sometimes we\u0027ll have more complicated movement."},{"Start":"00:20.160 ","End":"00:24.585","Text":"In this lesson, we\u0027re going to be speaking about finding the angular momentum"},{"Start":"00:24.585 ","End":"00:29.655","Text":"of a shape moving in a more complex form."},{"Start":"00:29.655 ","End":"00:33.570","Text":"Here we have a disc with its center over here."},{"Start":"00:33.570 ","End":"00:35.940","Text":"The disk is of mass m,"},{"Start":"00:35.940 ","End":"00:37.840","Text":"it\u0027s a radius R,"},{"Start":"00:37.840 ","End":"00:41.510","Text":"and it has a velocity going in a straight line,"},{"Start":"00:41.510 ","End":"00:44.120","Text":"which is starting from its center of mass,"},{"Start":"00:44.120 ","End":"00:48.050","Text":"which is over here, at a velocity of v, center of mass."},{"Start":"00:48.050 ","End":"00:50.365","Text":"That\u0027s the velocity of the center of mass."},{"Start":"00:50.365 ","End":"00:54.395","Text":"Simultaneously, as this desk is moving in straight line,"},{"Start":"00:54.395 ","End":"00:58.745","Text":"it\u0027s also rotating at an angular velocity of Omega."},{"Start":"00:58.745 ","End":"01:02.120","Text":"You can imagine it like the wheels of a car,"},{"Start":"01:02.120 ","End":"01:04.490","Text":"so one of the car wheels."},{"Start":"01:04.490 ","End":"01:09.190","Text":"Now, this is where our origin is so this is our axis."},{"Start":"01:09.190 ","End":"01:15.790","Text":"We can say that the distance that the center of masses from the origin is"},{"Start":"01:15.790 ","End":"01:24.690","Text":"a distance d. Now we\u0027re going to see how to work out the angular momentum of this shape."},{"Start":"01:24.940 ","End":"01:28.850","Text":"The first thing that we\u0027re going to do is we\u0027re going to work out"},{"Start":"01:28.850 ","End":"01:32.930","Text":"the angular momentum when it\u0027s moving just in a straight line."},{"Start":"01:32.930 ","End":"01:36.110","Text":"We\u0027re going to ignore the fact that it\u0027s rotating."},{"Start":"01:36.110 ","End":"01:39.350","Text":"As we know, when we\u0027re dealing with"},{"Start":"01:39.350 ","End":"01:44.725","Text":"some rigid bodies so we can look at it as a point-mass."},{"Start":"01:44.725 ","End":"01:53.285","Text":"We can multiply it by its mass multiplied by its velocity in the center of mass, so V_cm."},{"Start":"01:53.285 ","End":"01:54.665","Text":"Then as we know,"},{"Start":"01:54.665 ","End":"01:57.230","Text":"for the angular momentum of a shape moving in"},{"Start":"01:57.230 ","End":"02:01.655","Text":"a straight line is going to be multiplied by its R effective,"},{"Start":"02:01.655 ","End":"02:03.050","Text":"which means the distance,"},{"Start":"02:03.050 ","End":"02:10.695","Text":"the center of masses from the origin when we take it at 90 degrees to the velocity."},{"Start":"02:10.695 ","End":"02:13.490","Text":"Here the distance is d,"},{"Start":"02:13.490 ","End":"02:16.355","Text":"because here we have an angle of 90 degrees."},{"Start":"02:16.355 ","End":"02:20.060","Text":"Similarly, if our shape moved over here,"},{"Start":"02:20.060 ","End":"02:24.785","Text":"so this is the center of mass and our desk is around it, still,"},{"Start":"02:24.785 ","End":"02:30.870","Text":"the velocity is going in a straight line, like so."},{"Start":"02:30.870 ","End":"02:35.135","Text":"As we know, if we carry on"},{"Start":"02:35.135 ","End":"02:42.600","Text":"the parallel line and we take down the component which is 90 degrees to it,"},{"Start":"02:42.700 ","End":"02:46.880","Text":"to where the origin is to over here,"},{"Start":"02:46.880 ","End":"02:49.385","Text":"then we\u0027ll see that this distance is the same as this,"},{"Start":"02:49.385 ","End":"02:55.140","Text":"which is a distance d. Because as you remember,"},{"Start":"02:55.520 ","End":"03:00.490","Text":"this would be our r vector."},{"Start":"03:00.950 ","End":"03:04.825","Text":"Then when we multiply it by sine of the angle,"},{"Start":"03:04.825 ","End":"03:06.430","Text":"because we have to take the size,"},{"Start":"03:06.430 ","End":"03:11.860","Text":"which would be mv times the size of r times sine of the angle,"},{"Start":"03:11.860 ","End":"03:18.070","Text":"so that will give us this component over here on the y-axis,"},{"Start":"03:18.070 ","End":"03:19.405","Text":"if we want to call it that,"},{"Start":"03:19.405 ","End":"03:22.250","Text":"which will be d again."},{"Start":"03:22.490 ","End":"03:29.890","Text":"At whichever point the shape is here on its line of travel,"},{"Start":"03:29.890 ","End":"03:35.210","Text":"it\u0027s angular momentum is always going to be this."},{"Start":"03:36.770 ","End":"03:43.985","Text":"Now this is the angular momentum of the center of mass."},{"Start":"03:43.985 ","End":"03:49.910","Text":"I wrote this here, the angular momentum of the center of mass. What does that mean?"},{"Start":"03:49.910 ","End":"03:51.980","Text":"Instead of looking at the entire shape,"},{"Start":"03:51.980 ","End":"03:54.710","Text":"we\u0027re looking at the center of mass and"},{"Start":"03:54.710 ","End":"03:59.120","Text":"its angular momentum relative to the origin. That\u0027s what that is."},{"Start":"03:59.120 ","End":"04:06.655","Text":"Now what we have to do is we have to add on the internal angular momentum,"},{"Start":"04:06.655 ","End":"04:10.140","Text":"or the angular momentum around the center of mass."},{"Start":"04:10.140 ","End":"04:15.330","Text":"Because we also have the angular momentum caused by this rotation."},{"Start":"04:15.580 ","End":"04:18.700","Text":"Let\u0027s see what this is."},{"Start":"04:18.700 ","End":"04:22.855","Text":"This is simply going to be the I,"},{"Start":"04:22.855 ","End":"04:25.280","Text":"the moment of inertia of the center of mass,"},{"Start":"04:25.280 ","End":"04:29.490","Text":"multiplied by its angular velocity omega."},{"Start":"04:30.320 ","End":"04:36.020","Text":"This term over here has 3 names which refer to this."},{"Start":"04:36.020 ","End":"04:38.900","Text":"The angular momentum about the center of mass,"},{"Start":"04:38.900 ","End":"04:41.045","Text":"which is different to the center of mass."},{"Start":"04:41.045 ","End":"04:43.850","Text":"This means the angular momentum going around when"},{"Start":"04:43.850 ","End":"04:47.440","Text":"the axis of rotation is the center of mass itself."},{"Start":"04:47.440 ","End":"04:50.850","Text":"That\u0027s this, so that refers to the rotation."},{"Start":"04:50.850 ","End":"04:55.265","Text":"We can call it also the internal angular momentum."},{"Start":"04:55.265 ","End":"05:00.280","Text":"L is angular momentum and also L_CM,"},{"Start":"05:00.280 ","End":"05:05.390","Text":"which is the same as angular momentum about center of mass."},{"Start":"05:05.720 ","End":"05:10.115","Text":"Let\u0027s remind you of an equation that we\u0027ve come across before,"},{"Start":"05:10.115 ","End":"05:14.730","Text":"which is basically this just written out in a more formal way."},{"Start":"05:15.560 ","End":"05:19.970","Text":"This is the equation that we saw in the chapter on"},{"Start":"05:19.970 ","End":"05:25.300","Text":"angular momentum and we also saw how to derive this equation."},{"Start":"05:25.300 ","End":"05:28.025","Text":"You\u0027ll notice that over here,"},{"Start":"05:28.025 ","End":"05:34.360","Text":"this refers to these two,"},{"Start":"05:34.940 ","End":"05:42.280","Text":"and then this I_CM Omega refers to this to the L_CM."},{"Start":"05:43.010 ","End":"05:48.200","Text":"The only difference over here is that when"},{"Start":"05:48.200 ","End":"05:52.759","Text":"I\u0027m working out the angular momentum about the axis of rotation,"},{"Start":"05:52.759 ","End":"05:56.840","Text":"which here for the I_CM Omega,"},{"Start":"05:56.840 ","End":"06:00.410","Text":"which I\u0027m looking at this as a point-mass."},{"Start":"06:00.410 ","End":"06:04.530","Text":"That\u0027s why I write I_CM Omega."},{"Start":"06:04.530 ","End":"06:06.770","Text":"We\u0027ve already seen this, if you can\u0027t remember,"},{"Start":"06:06.770 ","End":"06:09.110","Text":"please go back to that lesson."},{"Start":"06:09.110 ","End":"06:11.670","Text":"That\u0027s the end of this lesson."}],"ID":9402},{"Watched":false,"Name":"Example","Duration":"14m 59s","ChapterTopicVideoID":9130,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:03.690","Text":"Hello. In the last couple of lessons,"},{"Start":"00:03.690 ","End":"00:07.520","Text":"we\u0027ve spoken about different ways to work out the angular momentum."},{"Start":"00:07.520 ","End":"00:09.780","Text":"Right now, we\u0027re going to give"},{"Start":"00:09.780 ","End":"00:13.860","Text":"a worked example where we\u0027re going to work out through different ways."},{"Start":"00:13.860 ","End":"00:17.530","Text":"The angular momentum of this rod over here."},{"Start":"00:19.340 ","End":"00:30.335","Text":"Let\u0027s say that the length of a rod is L and it has mass M. Then"},{"Start":"00:30.335 ","End":"00:36.980","Text":"let\u0027s say that the distance the rod is from the origin is"},{"Start":"00:36.980 ","End":"00:44.615","Text":"d. The rod is going to rotate at a radius of d around this origin."},{"Start":"00:44.615 ","End":"00:52.630","Text":"This is the axis of rotation and the rod is going to rotate at this distance."},{"Start":"00:52.760 ","End":"00:59.730","Text":"Now we\u0027re trying to find our angular momentum."},{"Start":"01:01.520 ","End":"01:05.030","Text":"Let\u0027s call the origin point O,"},{"Start":"01:05.030 ","End":"01:11.005","Text":"which means that we\u0027re trying to find our angular momentum about this point O."},{"Start":"01:11.005 ","End":"01:13.385","Text":"As we know the equation,"},{"Start":"01:13.385 ","End":"01:16.940","Text":"you can write this out on your equation sheet."},{"Start":"01:16.940 ","End":"01:21.155","Text":"The equation for the angular momentum about a fixed axis."},{"Start":"01:21.155 ","End":"01:23.195","Text":"This is important,"},{"Start":"01:23.195 ","End":"01:25.505","Text":"so that\u0027s L is equal to I,"},{"Start":"01:25.505 ","End":"01:26.555","Text":"the moment of inertia,"},{"Start":"01:26.555 ","End":"01:30.160","Text":"multiplied by Omega, its angular velocity."},{"Start":"01:30.160 ","End":"01:37.235","Text":"Over here, our L is going to be our moment of inertia about the origin."},{"Start":"01:37.235 ","End":"01:42.870","Text":"O multiplied by its angular velocity."},{"Start":"01:42.870 ","End":"01:47.120","Text":"Notice that this is specifically if every single piece of"},{"Start":"01:47.120 ","End":"01:51.405","Text":"this rod is also rotating about the origin."},{"Start":"01:51.405 ","End":"01:53.945","Text":"Soon we\u0027ll see an example where this doesn\u0027t happen,"},{"Start":"01:53.945 ","End":"01:56.330","Text":"in which case we cannot use this equation."},{"Start":"01:56.330 ","End":"02:00.125","Text":"But right now this equation is fine because every single piece of the rod"},{"Start":"02:00.125 ","End":"02:04.950","Text":"is moving around our origin."},{"Start":"02:05.090 ","End":"02:07.220","Text":"Let\u0027s see how we do this now."},{"Start":"02:07.220 ","End":"02:09.330","Text":"What is our I_o?"},{"Start":"02:09.590 ","End":"02:16.490","Text":"Our I around the origin is going to be equal to our I center of mass."},{"Start":"02:16.490 ","End":"02:19.115","Text":"The center of mass of the rod, which is going to be,"},{"Start":"02:19.115 ","End":"02:22.149","Text":"if I mass is distributed uniformly."},{"Start":"02:22.149 ","End":"02:25.075","Text":"It\u0027ll be right in the center of the rod."},{"Start":"02:25.075 ","End":"02:29.600","Text":"At a distance of L/2 plus,"},{"Start":"02:29.600 ","End":"02:34.175","Text":"and then we have to add in because our axis of rotation isn\u0027t the center of mass."},{"Start":"02:34.175 ","End":"02:37.655","Text":"We have to add in our Steiner,"},{"Start":"02:37.655 ","End":"02:40.095","Text":"which is our mass."},{"Start":"02:40.095 ","End":"02:47.630","Text":"The mass of the rod multiplied by the distance squared from the axis of rotation."},{"Start":"02:47.630 ","End":"02:51.540","Text":"That means this distance,"},{"Start":"02:54.130 ","End":"03:03.025","Text":"this distance is going to be L/2 for that plus our d for this section."},{"Start":"03:03.025 ","End":"03:06.435","Text":"That\u0027s this black distance over here."},{"Start":"03:06.435 ","End":"03:15.390","Text":"This is going to be m multiplied by the distance from the axis of rotation squared."},{"Start":"03:15.390 ","End":"03:17.820","Text":"Our I_CM."},{"Start":"03:17.820 ","End":"03:21.740","Text":"You could write this on your equation sheet and also in the chapter on moment of inertia,"},{"Start":"03:21.740 ","End":"03:27.560","Text":"we really discuss how to figure out the moment of inertia of different shapes."},{"Start":"03:27.560 ","End":"03:29.645","Text":"Specifically for a rod,"},{"Start":"03:29.645 ","End":"03:34.160","Text":"it\u0027s ML^2 divided by 12."},{"Start":"03:34.160 ","End":"03:38.000","Text":"If you want to see how to get this or you can\u0027t remember how,"},{"Start":"03:38.000 ","End":"03:40.535","Text":"go back to these lessons."},{"Start":"03:40.535 ","End":"03:42.770","Text":"That means that my I is zero,"},{"Start":"03:42.770 ","End":"03:44.405","Text":"I\u0027m going to substitute this in here."},{"Start":"03:44.405 ","End":"03:54.540","Text":"It\u0027s going to be ML^2 divided by 12 plus ML/2 plus d^2."},{"Start":"03:54.540 ","End":"03:56.900","Text":"Then if I know my Omega,"},{"Start":"03:56.900 ","End":"04:01.020","Text":"so it simple, I just have to multiply."},{"Start":"04:03.200 ","End":"04:05.603","Text":"I just substituted this in."},{"Start":"04:05.603 ","End":"04:07.895","Text":"I saw that there was a common factor M over here,"},{"Start":"04:07.895 ","End":"04:10.740","Text":"so I took that out and then multiplied."},{"Start":"04:10.740 ","End":"04:15.610","Text":"That was my I_0 I substituted in and then multiplied by Omega."},{"Start":"04:15.860 ","End":"04:19.600","Text":"That is the answer."},{"Start":"04:24.110 ","End":"04:30.340","Text":"Another way that we can work this out is by using this equation over here."},{"Start":"04:30.340 ","End":"04:34.610","Text":"That is the total angular momentum is going to be equal"},{"Start":"04:34.610 ","End":"04:39.370","Text":"to I_CM across multiplied by our momentum,"},{"Start":"04:39.370 ","End":"04:46.940","Text":"P_CM of the center of mass plus the angular momentum of the center of mass."},{"Start":"04:46.940 ","End":"04:52.230","Text":"Let\u0027s see what this is. I\u0027m just going to rub all of this out and also this."},{"Start":"04:55.060 ","End":"04:58.590","Text":"Let\u0027s see how this works out."},{"Start":"04:58.590 ","End":"05:02.900","Text":"From our I_CM and I also suggest you write this equation out."},{"Start":"05:02.900 ","End":"05:05.195","Text":"Our I_CM is going to be."},{"Start":"05:05.195 ","End":"05:06.710","Text":"our distance from the origin,"},{"Start":"05:06.710 ","End":"05:08.735","Text":"which just like the same over here,"},{"Start":"05:08.735 ","End":"05:15.050","Text":"are black line over here will be d plus L/2 the origin until the center of mass is"},{"Start":"05:15.050 ","End":"05:23.340","Text":"a distance of L/2 Plus d. That\u0027s our I_CM."},{"Start":"05:24.490 ","End":"05:31.200","Text":"Our P_CM. We know that our mass is"},{"Start":"05:31.200 ","End":"05:37.640","Text":"our capital M. Then we have to multiply it by velocity by our V_CM."},{"Start":"05:37.640 ","End":"05:40.410","Text":"The velocity of the center of mass."},{"Start":"05:41.330 ","End":"05:44.610","Text":"Let\u0027s see what our velocity of center of mass is."},{"Start":"05:44.610 ","End":"05:46.950","Text":"Here\u0027s our center of mass."},{"Start":"05:46.950 ","End":"05:50.235","Text":"This is, it\u0027s V_CM over here."},{"Start":"05:50.235 ","End":"05:51.680","Text":"What is its V_CM?"},{"Start":"05:51.680 ","End":"05:55.385","Text":"Because it\u0027s doing angular circular motion,"},{"Start":"05:55.385 ","End":"05:59.875","Text":"we know that it\u0027s going to be Omega multiplied by"},{"Start":"05:59.875 ","End":"06:09.870","Text":"the radius from the axis of rotation, I_CM."},{"Start":"06:09.870 ","End":"06:13.800","Text":"What is our I _CM? Let\u0027s try this here."},{"Start":"06:13.800 ","End":"06:21.630","Text":"It\u0027s going to be our V_CM is equal to Omega multiplied by I_CM."},{"Start":"06:21.630 ","End":"06:23.310","Text":"What is this equal to?"},{"Start":"06:23.310 ","End":"06:29.150","Text":"Our I_CM is equal to this distance over here again,"},{"Start":"06:29.150 ","End":"06:32.915","Text":"d up until this point over here, plus L/2."},{"Start":"06:32.915 ","End":"06:38.110","Text":"d plus L/2."},{"Start":"06:39.080 ","End":"06:43.810","Text":"When we multiply these 2 together."},{"Start":"06:44.630 ","End":"06:47.010","Text":"Let\u0027s see how we do this."},{"Start":"06:47.010 ","End":"06:52.640","Text":"We\u0027ll get that our r_CM cross multiplied with"},{"Start":"06:52.640 ","End":"06:59.375","Text":"a P_CM is going to be equal to the mass multiplied by Omega."},{"Start":"06:59.375 ","End":"07:03.815","Text":"Then we have d plus L/2 times d plus L/2,"},{"Start":"07:03.815 ","End":"07:09.670","Text":"so d plus L/2^2."},{"Start":"07:10.490 ","End":"07:14.850","Text":"That we have to add on is our L_CM."},{"Start":"07:14.850 ","End":"07:16.310","Text":"What is our L_CM?"},{"Start":"07:16.310 ","End":"07:20.750","Text":"We know that it\u0027s going to be our moment of inertia of the center of"},{"Start":"07:20.750 ","End":"07:26.855","Text":"mass multiplied by the Omega of the center of mass."},{"Start":"07:26.855 ","End":"07:30.790","Text":"Notice this."},{"Start":"07:36.210 ","End":"07:39.175","Text":"Here we have our Omega_CM,"},{"Start":"07:39.175 ","End":"07:43.820","Text":"which is different to this Omega over here."},{"Start":"07:44.400 ","End":"07:49.390","Text":"Now, specifically in this example over here,"},{"Start":"07:49.390 ","End":"07:52.615","Text":"our Omega_CM and our Omega the same,"},{"Start":"07:52.615 ","End":"07:54.850","Text":"so I could have just written Omega."},{"Start":"07:54.850 ","End":"08:01.779","Text":"However, you will have some questions where your angular velocity"},{"Start":"08:01.779 ","End":"08:04.720","Text":"around the center of mass will not be the"},{"Start":"08:04.720 ","End":"08:09.350","Text":"same as your angular velocity around the origin."},{"Start":"08:09.350 ","End":"08:12.060","Text":"Let\u0027s see why in this example"},{"Start":"08:12.060 ","End":"08:18.520","Text":"specifically our Omega_CM and this Omega are actually the same."},{"Start":"08:19.470 ","End":"08:22.300","Text":"As our rod is rotating,"},{"Start":"08:22.300 ","End":"08:25.850","Text":"it will be going like this."},{"Start":"08:26.010 ","End":"08:31.270","Text":"This is what the rod will look"},{"Start":"08:31.270 ","End":"08:41.515","Text":"like as it rotates around the origin over here."},{"Start":"08:41.515 ","End":"08:48.655","Text":"Bearing in mind the radial distance until the minimum point."},{"Start":"08:48.655 ","End":"08:51.640","Text":"Let\u0027s label that in red,"},{"Start":"08:51.640 ","End":"08:57.460","Text":"so the distance between the origin and the red point is always going to"},{"Start":"08:57.460 ","End":"09:03.995","Text":"be this distance d. Just to put that in,"},{"Start":"09:03.995 ","End":"09:13.345","Text":"so this distance will be d and similarly with this distance,"},{"Start":"09:13.345 ","End":"09:22.525","Text":"it will also be d. This minimum point is this red dot and it\u0027s going around in a circle."},{"Start":"09:22.525 ","End":"09:24.895","Text":"Then if we label in,"},{"Start":"09:24.895 ","End":"09:26.305","Text":"let\u0027s say green,"},{"Start":"09:26.305 ","End":"09:27.850","Text":"our center of mass,"},{"Start":"09:27.850 ","End":"09:34.429","Text":"so this is our center of mass and our center of masses all the way like this."},{"Start":"09:36.540 ","End":"09:40.000","Text":"Imagine that all of these rods are of equal length."},{"Start":"09:40.000 ","End":"09:42.580","Text":"It\u0027s the same rod just rotating around."},{"Start":"09:42.580 ","End":"09:48.135","Text":"If we take a look, we know that our center of mass,"},{"Start":"09:48.135 ","End":"09:54.120","Text":"this is going around the origin at angular velocity of Omega."},{"Start":"09:54.120 ","End":"10:04.480","Text":"Now, if you notice that means that to do this entire circle, it was Omega."},{"Start":"10:05.210 ","End":"10:09.225","Text":"Now, if you notice the time taken for"},{"Start":"10:09.225 ","End":"10:15.345","Text":"our green dots to rotate the whole way round and get back to where they started from,"},{"Start":"10:15.345 ","End":"10:19.980","Text":"is the same time that it takes for all of our red dots or for this red dot"},{"Start":"10:19.980 ","End":"10:25.110","Text":"to do this circle and also return to its original point."},{"Start":"10:25.110 ","End":"10:32.605","Text":"Because our center of mass and our minimum point arrive at their final destination,"},{"Start":"10:32.605 ","End":"10:36.850","Text":"which is the exact same point that they started at the exact same time,"},{"Start":"10:36.850 ","End":"10:40.810","Text":"so we know that this Omega and this Omega will be the same."},{"Start":"10:40.810 ","End":"10:42.520","Text":"If this didn\u0027t happen,"},{"Start":"10:42.520 ","End":"10:46.990","Text":"if our center of mass arrived later than our minimum point,"},{"Start":"10:46.990 ","End":"10:54.260","Text":"or vice versa, then our Omega_CM and our Omega would not be the same."},{"Start":"10:55.560 ","End":"11:01.195","Text":"Then I could add onto here my I_CM and as I know,"},{"Start":"11:01.195 ","End":"11:04.525","Text":"my I_CM is going to be of a rod,"},{"Start":"11:04.525 ","End":"11:06.160","Text":"which is going to be the same as here,"},{"Start":"11:06.160 ","End":"11:11.260","Text":"so it will be ML^2 divided by 12 and"},{"Start":"11:11.260 ","End":"11:13.780","Text":"then multiply by Omega and we don\u0027t have"},{"Start":"11:13.780 ","End":"11:16.780","Text":"to add in the CM because it\u0027s the same Omega is over here."},{"Start":"11:16.780 ","End":"11:25.130","Text":"This will be plus ML^2 divided by 12 multiplied by Omega,"},{"Start":"11:25.230 ","End":"11:31.480","Text":"so this is our L. Then this is equal to, if we take a look,"},{"Start":"11:31.480 ","End":"11:35.410","Text":"we can take my Omega out as a common factor and then my M."},{"Start":"11:35.410 ","End":"11:41.270","Text":"Then I can write over here M and then I have,"},{"Start":"11:41.310 ","End":"11:43.420","Text":"let\u0027s do over here,"},{"Start":"11:43.420 ","End":"11:51.025","Text":"L^2 divided by 12 plus this d"},{"Start":"11:51.025 ","End":"12:00.010","Text":"plus L/2^2 and then all of this multiplied by Omega."},{"Start":"12:00.010 ","End":"12:07.010","Text":"Now let\u0027s scroll up and we have the exact same equation over here."},{"Start":"12:09.060 ","End":"12:14.605","Text":"You\u0027ll see that we got the exact same answer by working it out"},{"Start":"12:14.605 ","End":"12:20.240","Text":"in 2 separate ways with 2 different equations,"},{"Start":"12:20.730 ","End":"12:23.410","Text":"so these are the 2 ways."},{"Start":"12:23.410 ","End":"12:26.560","Text":"Now, this equation is always correct."},{"Start":"12:26.560 ","End":"12:28.630","Text":"If you use it correctly,"},{"Start":"12:28.630 ","End":"12:30.910","Text":"then it will always work."},{"Start":"12:30.910 ","End":"12:34.900","Text":"This equation you have to be a bit careful with."},{"Start":"12:34.900 ","End":"12:38.740","Text":"Because sometimes it won\u0027t always work."},{"Start":"12:38.740 ","End":"12:40.645","Text":"When will it not always work?"},{"Start":"12:40.645 ","End":"12:43.340","Text":"I\u0027m going to give an example right now."},{"Start":"12:43.830 ","End":"12:48.685","Text":"In this example, specifically the 1 that we worked out over here,"},{"Start":"12:48.685 ","End":"12:53.515","Text":"it does work because every single point is rotating around,"},{"Start":"12:53.515 ","End":"12:58.590","Text":"as we said, like the rays coming out of a sun."},{"Start":"12:58.590 ","End":"13:04.065","Text":"It\u0027s just rotating around in a uniform circle."},{"Start":"13:04.065 ","End":"13:07.035","Text":"However, when will it not work?"},{"Start":"13:07.035 ","End":"13:13.704","Text":"Is if the rod is not only rotating around the origin,"},{"Start":"13:13.704 ","End":"13:18.745","Text":"but also the rod itself is rotating around its center of mass."},{"Start":"13:18.745 ","End":"13:25.420","Text":"That would mean if the rod is rotating around itself,"},{"Start":"13:25.420 ","End":"13:30.880","Text":"whilst it as a shape is also rotating around here,"},{"Start":"13:30.880 ","End":"13:37.390","Text":"so it would look something like if I take slow snapshots at each moment,"},{"Start":"13:37.390 ","End":"13:41.620","Text":"it would look something like this."},{"Start":"13:41.620 ","End":"13:43.450","Text":"Let\u0027s say the rod itself,"},{"Start":"13:43.450 ","End":"13:46.990","Text":"this is the center of mass and it\u0027s rotating in"},{"Start":"13:46.990 ","End":"13:52.430","Text":"this direction and then this is going in this direction,"},{"Start":"13:54.420 ","End":"13:57.805","Text":"so it\u0027s rotating around,"},{"Start":"13:57.805 ","End":"14:00.220","Text":"it\u0027s going around in a spiral type of way."},{"Start":"14:00.220 ","End":"14:02.785","Text":"In this case,"},{"Start":"14:02.785 ","End":"14:06.025","Text":"this equation will not work."},{"Start":"14:06.025 ","End":"14:10.555","Text":"It\u0027s only if every single point on the shape"},{"Start":"14:10.555 ","End":"14:15.475","Text":"is doing perfect circular motion around the origin."},{"Start":"14:15.475 ","End":"14:20.470","Text":"If that isn\u0027t happening and there\u0027s some spiral thing going along as we\u0027re going around,"},{"Start":"14:20.470 ","End":"14:24.955","Text":"this equation will not work and you have to use this equation and of course,"},{"Start":"14:24.955 ","End":"14:28.825","Text":"you have to note when using this equation and that type of situation,"},{"Start":"14:28.825 ","End":"14:35.830","Text":"your Omega_CM, this over here will not be the same as this Omega."},{"Start":"14:35.830 ","End":"14:39.580","Text":"Because this Omega is the angular velocity of the center of"},{"Start":"14:39.580 ","End":"14:43.465","Text":"mass around the origin and this would be"},{"Start":"14:43.465 ","End":"14:47.425","Text":"the angular velocity that the actual shape is moving at"},{"Start":"14:47.425 ","End":"14:52.585","Text":"around the axis at its center of mass,"},{"Start":"14:52.585 ","End":"14:54.100","Text":"so you would have to change that,"},{"Start":"14:54.100 ","End":"14:56.425","Text":"but this equation as a whole would work."},{"Start":"14:56.425 ","End":"14:59.300","Text":"That\u0027s the end of the lesson."}],"ID":9403},{"Watched":false,"Name":"Exercise 3","Duration":"32m 16s","ChapterTopicVideoID":9131,"CourseChapterTopicPlaylistID":5406,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:02.460","Text":"Hello. In this question,"},{"Start":"00:02.460 ","End":"00:06.329","Text":"we\u0027re being given an equilateral triangle where each side is"},{"Start":"00:06.329 ","End":"00:10.275","Text":"of length L. At each corner of the triangle,"},{"Start":"00:10.275 ","End":"00:19.694","Text":"a mass of m is placed and we\u0027re told that the rods of length L are massless."},{"Start":"00:19.694 ","End":"00:21.960","Text":"Now, in our first question,"},{"Start":"00:21.960 ","End":"00:26.530","Text":"we\u0027re being asked to find the center of mass."},{"Start":"00:27.950 ","End":"00:32.745","Text":"Let\u0027s see how we do this."},{"Start":"00:32.745 ","End":"00:38.940","Text":"The first thing that we have to do is we have to put down our axes."},{"Start":"00:38.940 ","End":"00:42.660","Text":"Let\u0027s say that our origin will be over here."},{"Start":"00:42.660 ","End":"00:44.310","Text":"Let\u0027s label each ball."},{"Start":"00:44.310 ","End":"00:45.975","Text":"We\u0027ll say that this is Ball Number 1,"},{"Start":"00:45.975 ","End":"00:48.640","Text":"this is 2, and this is 3."},{"Start":"00:49.100 ","End":"00:51.159","Text":"Here\u0027s our axes,"},{"Start":"00:51.159 ","End":"00:57.330","Text":"and let\u0027s say that this is our x-axis and that this is our y-axis."},{"Start":"00:57.330 ","End":"01:03.503","Text":"The first thing that we\u0027re going to do is we can work out where our x center of mass is,"},{"Start":"01:03.503 ","End":"01:04.590","Text":"because that\u0027s really easy."},{"Start":"01:04.590 ","End":"01:08.764","Text":"Our x_cm, from symmetry,"},{"Start":"01:08.764 ","End":"01:10.970","Text":"we can see that it\u0027s going to be right in"},{"Start":"01:10.970 ","End":"01:14.120","Text":"the center because this is an equilateral triangle."},{"Start":"01:14.120 ","End":"01:17.350","Text":"We know that L divided by 2,"},{"Start":"01:17.350 ","End":"01:21.600","Text":"we\u0027ll have our x center of mass."},{"Start":"01:21.600 ","End":"01:27.169","Text":"Now our y center of mass will be slightly more difficult. Why is that?"},{"Start":"01:27.169 ","End":"01:31.490","Text":"Because our height that we need to"},{"Start":"01:31.490 ","End":"01:38.790","Text":"find is this length over here and we have to find the center of it."},{"Start":"01:39.920 ","End":"01:42.345","Text":"Let\u0027s see how we do this."},{"Start":"01:42.345 ","End":"01:45.949","Text":"Now because we know that in an equilateral triangle,"},{"Start":"01:45.949 ","End":"01:50.520","Text":"each angle in the triangle is going to be 60 degrees,"},{"Start":"01:50.520 ","End":"01:54.775","Text":"so what we can do because we want to find out what this length is,"},{"Start":"01:54.775 ","End":"01:57.540","Text":"and this is the opposite side,"},{"Start":"01:57.540 ","End":"02:00.013","Text":"it\u0027s the opposite side to the angle,"},{"Start":"02:00.013 ","End":"02:05.100","Text":"so what we can do is we can use our trigonometric identities."},{"Start":"02:05.500 ","End":"02:11.044","Text":"We know that if we do sine of the angle,"},{"Start":"02:11.044 ","End":"02:12.664","Text":"which is 60 degrees,"},{"Start":"02:12.664 ","End":"02:17.720","Text":"it\u0027s going to equal the opposite side divided by the hypotenuse."},{"Start":"02:17.720 ","End":"02:19.595","Text":"The opposite side,"},{"Start":"02:19.595 ","End":"02:24.150","Text":"so let\u0027s call this height y_3."},{"Start":"02:24.170 ","End":"02:29.435","Text":"The opposite side is y_3 divided by the hypotenuse."},{"Start":"02:29.435 ","End":"02:30.559","Text":"This is the hypotenuse."},{"Start":"02:30.559 ","End":"02:37.550","Text":"It\u0027s of length L. Sine 60 is root 3/2."},{"Start":"02:37.550 ","End":"02:40.640","Text":"Then we\u0027re going to isolate to get our y_3."},{"Start":"02:40.640 ","End":"02:47.285","Text":"Our y_3 is going to be root 3/2"},{"Start":"02:47.285 ","End":"02:54.350","Text":"multiplied by L. Now we\u0027re going to write out the equation for our y_cm."},{"Start":"02:54.350 ","End":"02:58.145","Text":"Our y center of mass is going to be equal to,"},{"Start":"02:58.145 ","End":"03:02.964","Text":"if you remember, it is the sum of all of the heights,"},{"Start":"03:02.964 ","End":"03:11.075","Text":"y_i, multiplied by each mass divided by the sum of the masses,"},{"Start":"03:11.075 ","End":"03:13.820","Text":"which is also equal to the total mass."},{"Start":"03:13.820 ","End":"03:18.195","Text":"Let\u0027s do this. Y_1 is going to be this ball."},{"Start":"03:18.195 ","End":"03:23.029","Text":"We can see that it\u0027s at y equals 0 because it\u0027s at the origin."},{"Start":"03:23.029 ","End":"03:26.424","Text":"0 times m is going to be 0."},{"Start":"03:26.424 ","End":"03:32.535","Text":"Then our next y is our y position for Ball Number 2,"},{"Start":"03:32.535 ","End":"03:34.905","Text":"which again it\u0027s at y equals 0,"},{"Start":"03:34.905 ","End":"03:37.429","Text":"so this is again going to be plus 0."},{"Start":"03:37.429 ","End":"03:39.315","Text":"Then for Ball Number 3,"},{"Start":"03:39.315 ","End":"03:41.400","Text":"it\u0027s going to be this height y_3,"},{"Start":"03:41.400 ","End":"03:42.825","Text":"which we just worked out,"},{"Start":"03:42.825 ","End":"03:44.700","Text":"multiplied by its mass,"},{"Start":"03:44.700 ","End":"03:51.030","Text":"m. It\u0027s going to be plus root 3 divided by"},{"Start":"03:51.030 ","End":"04:00.634","Text":"2L multiplied by m. Then this is going to be divided by the total mass of the system,"},{"Start":"04:00.634 ","End":"04:02.645","Text":"which is m plus m plus m,"},{"Start":"04:02.645 ","End":"04:04.855","Text":"so divided by 3m."},{"Start":"04:04.855 ","End":"04:08.969","Text":"Then we can cross out the 2 m\u0027s because they cancel"},{"Start":"04:08.969 ","End":"04:12.515","Text":"out and we\u0027ll get that our y center of mass is"},{"Start":"04:12.515 ","End":"04:21.690","Text":"equal to root 3L divided by 6,"},{"Start":"04:21.690 ","End":"04:30.115","Text":"which is also equal to L/2 root 3."},{"Start":"04:30.115 ","End":"04:32.469","Text":"Now on to the next question."},{"Start":"04:32.720 ","End":"04:36.019","Text":"In this question, we\u0027re being told that the system is"},{"Start":"04:36.019 ","End":"04:40.205","Text":"rotating about the center of mass at an angular velocity of Omega."},{"Start":"04:40.205 ","End":"04:46.550","Text":"The whole system is doing circular motion,"},{"Start":"04:46.550 ","End":"04:48.530","Text":"Omega, about the center of mass,"},{"Start":"04:48.530 ","End":"04:52.780","Text":"which is around about here."},{"Start":"04:54.560 ","End":"04:58.305","Text":"Now we\u0027re also being told that Ball Number 3,"},{"Start":"04:58.305 ","End":"05:01.425","Text":"this one, is released from the current position."},{"Start":"05:01.425 ","End":"05:04.270","Text":"Current position right now and it\u0027s at the bottom,"},{"Start":"05:04.270 ","End":"05:09.710","Text":"the glue which is sticking it to the rods comes loose and the ball is released."},{"Start":"05:09.710 ","End":"05:14.670","Text":"Then we\u0027re being asked what is the velocity of the ball after release?"},{"Start":"05:14.670 ","End":"05:16.590","Text":"Let\u0027s see what this is."},{"Start":"05:16.590 ","End":"05:20.490","Text":"We\u0027re going to call this velocity v_3 because it\u0027s of Ball Number 3,"},{"Start":"05:20.490 ","End":"05:21.799","Text":"and as we know,"},{"Start":"05:21.799 ","End":"05:26.910","Text":"it\u0027s equal to Omega multiplied by its radius."},{"Start":"05:27.230 ","End":"05:30.259","Text":"What is its radius? Let\u0027s see."},{"Start":"05:30.259 ","End":"05:34.674","Text":"Its radius is referring to this distance from the ball itself,"},{"Start":"05:34.674 ","End":"05:37.275","Text":"until the center of mass."},{"Start":"05:37.275 ","End":"05:39.315","Text":"This is its radius."},{"Start":"05:39.315 ","End":"05:46.360","Text":"Now we know what our center of mass is and we know what our entire length is,"},{"Start":"05:46.360 ","End":"05:48.900","Text":"y_3, because we worked it out."},{"Start":"05:48.900 ","End":"05:55.535","Text":"We know that our radius is going to be equal to y_3,"},{"Start":"05:55.535 ","End":"05:58.045","Text":"this total length over here,"},{"Start":"05:58.045 ","End":"06:01.860","Text":"minus our y_cm,"},{"Start":"06:01.860 ","End":"06:04.840","Text":"which is this over here."},{"Start":"06:05.390 ","End":"06:09.274","Text":"We\u0027re taking the whole length minus this length over here,"},{"Start":"06:09.274 ","End":"06:11.780","Text":"and then we\u0027ll be left with a radius."},{"Start":"06:11.780 ","End":"06:14.300","Text":"What is our y_3?"},{"Start":"06:14.300 ","End":"06:21.540","Text":"It\u0027s going to be root 3 divided by 2 multiplied by L minus our y_cm,"},{"Start":"06:21.540 ","End":"06:26.295","Text":"which is L/2 root 3."},{"Start":"06:26.295 ","End":"06:36.584","Text":"Then what we will get is that it\u0027s 1 over root 3 multiplied by L. Then we\u0027ll get that"},{"Start":"06:36.584 ","End":"06:41.190","Text":"our v of the ball is going to be Omega"},{"Start":"06:41.190 ","End":"06:48.460","Text":"multiplied by L multiplied by 1 over root 3."},{"Start":"06:49.160 ","End":"06:53.329","Text":"Of course not forgetting that this is a vector quantity,"},{"Start":"06:53.329 ","End":"06:55.805","Text":"so now we have to decide its direction."},{"Start":"06:55.805 ","End":"07:01.604","Text":"Because our angular velocity is going in this clockwise direction,"},{"Start":"07:01.604 ","End":"07:05.310","Text":"so we know that our velocity, our v_3,"},{"Start":"07:05.310 ","End":"07:10.470","Text":"is going to be in this direction,"},{"Start":"07:10.470 ","End":"07:12.325","Text":"which as we can see,"},{"Start":"07:12.325 ","End":"07:15.040","Text":"is in the negative x direction because this is"},{"Start":"07:15.040 ","End":"07:18.235","Text":"the positive x direction and it\u0027s pointing in this direction."},{"Start":"07:18.235 ","End":"07:23.160","Text":"It\u0027s going to be in the negative x direction."},{"Start":"07:23.160 ","End":"07:27.545","Text":"Then we just have to add that in over here as well."},{"Start":"07:27.545 ","End":"07:32.955","Text":"Now onto question c. What is the remaining section\u0027s V_cm?"},{"Start":"07:32.955 ","End":"07:36.225","Text":"We\u0027re trying to find the remaining section,"},{"Start":"07:36.225 ","End":"07:38.685","Text":"the 2 balls connected by a rod,"},{"Start":"07:38.685 ","End":"07:42.745","Text":"so their velocity for the center of mass."},{"Start":"07:42.745 ","End":"07:44.529","Text":"Let\u0027s go back to here."},{"Start":"07:44.529 ","End":"07:46.375","Text":"Now, in question a,"},{"Start":"07:46.375 ","End":"07:49.000","Text":"we were asked to find the center of mass,"},{"Start":"07:49.000 ","End":"07:52.545","Text":"but that was of the 3 balls,"},{"Start":"07:52.545 ","End":"07:54.410","Text":"but now we\u0027re being asked of"},{"Start":"07:54.410 ","End":"07:57.940","Text":"the remaining section and to find where the center of mass is,"},{"Start":"07:57.940 ","End":"08:00.875","Text":"and then to find the velocity of that center of mass."},{"Start":"08:00.875 ","End":"08:04.590","Text":"The center of mass of"},{"Start":"08:04.590 ","End":"08:09.230","Text":"the remaining section is going to be the center of mass between these 2 balls,"},{"Start":"08:09.230 ","End":"08:14.180","Text":"which is of course, because of the symmetry going to be right in the center."},{"Start":"08:14.180 ","End":"08:17.665","Text":"Let\u0027s call this cm_1,"},{"Start":"08:17.665 ","End":"08:21.180","Text":"2, between Ball 1 and Ball 2."},{"Start":"08:21.180 ","End":"08:27.645","Text":"This center of mass between Ball 1 and Ball 2 always existed also in question a,"},{"Start":"08:27.645 ","End":"08:31.145","Text":"but in question a, we were asked to find the center of mass of the entire system,"},{"Start":"08:31.145 ","End":"08:33.140","Text":"including this ball over here."},{"Start":"08:33.140 ","End":"08:35.815","Text":"Now we\u0027re just looking at this."},{"Start":"08:35.815 ","End":"08:40.935","Text":"Now we\u0027re trying to find the velocity of this."},{"Start":"08:40.935 ","End":"08:44.150","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"08:44.150 ","End":"08:50.670","Text":"see whether there\u0027s conservation of momentum."},{"Start":"08:50.670 ","End":"08:55.330","Text":"As we know, in order to check if there\u0027s conservation of momentum,"},{"Start":"08:55.330 ","End":"09:03.750","Text":"so we have to see if the sum of the external forces is equal to 0."},{"Start":"09:04.460 ","End":"09:10.730","Text":"Now let\u0027s take a look and see if there is conservation of momentum."},{"Start":"09:10.730 ","End":"09:13.684","Text":"Right now we\u0027re looking at this system,"},{"Start":"09:13.684 ","End":"09:19.370","Text":"and it doesn\u0027t matter that it\u0027s rotating at an angular velocity of Omega."},{"Start":"09:19.370 ","End":"09:21.814","Text":"If we\u0027re going to look at this shape from very far away,"},{"Start":"09:21.814 ","End":"09:28.665","Text":"it\u0027s going to look like some point mass and we\u0027re going to be looking at it like this,"},{"Start":"09:28.665 ","End":"09:30.235","Text":"at the whole entire shape,"},{"Start":"09:30.235 ","End":"09:33.930","Text":"because right now we\u0027re looking at it right from the beginning."},{"Start":"09:34.100 ","End":"09:37.970","Text":"We can see that if we\u0027re looking at it from a far,"},{"Start":"09:37.970 ","End":"09:40.420","Text":"it will look something like this for instance."},{"Start":"09:40.420 ","End":"09:43.679","Text":"It looks like 1 point mass in the distance."},{"Start":"09:43.679 ","End":"09:47.670","Text":"We can see that there\u0027s no external forces acting on it."},{"Start":"09:47.670 ","End":"09:50.165","Text":"We\u0027re not told about any gravity acting."},{"Start":"09:50.165 ","End":"09:52.220","Text":"It\u0027s just this system rotating,"},{"Start":"09:52.220 ","End":"09:55.474","Text":"which means that because there\u0027s no external forces acting,"},{"Start":"09:55.474 ","End":"09:59.480","Text":"then there is conservation of momentum."},{"Start":"10:00.260 ","End":"10:02.915","Text":"We can say, yes,"},{"Start":"10:02.915 ","End":"10:05.360","Text":"there\u0027s conservation of momentum."},{"Start":"10:05.360 ","End":"10:09.350","Text":"Now let\u0027s write out the equation."},{"Start":"10:09.350 ","End":"10:13.069","Text":"As we can see, our center of mass"},{"Start":"10:13.069 ","End":"10:19.470","Text":"is located in this area over here with our x and y coordinates."},{"Start":"10:20.620 ","End":"10:24.260","Text":"Then from a second section,"},{"Start":"10:24.260 ","End":"10:32.770","Text":"we can see that the velocity of ball number 3 is in the negative x-direction."},{"Start":"10:32.770 ","End":"10:35.664","Text":"Ball number 3 is going to be flying in this direction."},{"Start":"10:35.664 ","End":"10:38.200","Text":"Now if ball number 3 is flying in this direction,"},{"Start":"10:38.200 ","End":"10:42.640","Text":"so we can know that once ball number 3 has been separated,"},{"Start":"10:42.640 ","End":"10:49.314","Text":"our body of ball number 1 and ball number 2 will be moving in the positive x-direction."},{"Start":"10:49.314 ","End":"10:50.836","Text":"Because if you look at it,"},{"Start":"10:50.836 ","End":"10:55.235","Text":"it doesn\u0027t make sense that if after the section separate."},{"Start":"10:55.235 ","End":"10:57.590","Text":"This section is moving over here."},{"Start":"10:57.590 ","End":"11:00.950","Text":"It doesn\u0027t make sense that this section would suddenly move over here."},{"Start":"11:00.950 ","End":"11:03.260","Text":"Because we are speaking about conservation of momentum,"},{"Start":"11:03.260 ","End":"11:06.020","Text":"so this situation can happen."},{"Start":"11:06.020 ","End":"11:12.830","Text":"Instead, we\u0027ll have that ball 1 and 2 are moving in the positive x-direction."},{"Start":"11:12.830 ","End":"11:16.055","Text":"If we look at this as the center of mass,"},{"Start":"11:16.055 ","End":"11:19.430","Text":"so once ball number 3 is separated,"},{"Start":"11:19.430 ","End":"11:22.490","Text":"so the center of mass will just move here and then will be"},{"Start":"11:22.490 ","End":"11:26.945","Text":"located here and here and so on and so forth along the x-axis."},{"Start":"11:26.945 ","End":"11:32.229","Text":"Now that we\u0027ve seen that our movement of our center of"},{"Start":"11:32.229 ","End":"11:37.720","Text":"mass of balls number 1 and 2 is going to be in the positive x-direction."},{"Start":"11:37.720 ","End":"11:43.250","Text":"Let\u0027s write the equation for the momentum."},{"Start":"11:43.250 ","End":"11:44.690","Text":"The equation for the momentum,"},{"Start":"11:44.690 ","End":"11:47.315","Text":"as we just saw, it\u0027s going to be on the x-axis."},{"Start":"11:47.315 ","End":"11:50.015","Text":"I\u0027m going to write momentum for the x-axis."},{"Start":"11:50.015 ","End":"11:52.850","Text":"Before. Let\u0027s right here before."},{"Start":"11:52.850 ","End":"11:56.015","Text":"This is before our ball 3 separates."},{"Start":"11:56.015 ","End":"12:02.420","Text":"Momentum is going to be equal to 0 before. Why is this?"},{"Start":"12:02.420 ","End":"12:04.774","Text":"As discussed before, when we\u0027re looking at"},{"Start":"12:04.774 ","End":"12:07.670","Text":"our system of 3 balls from very very far away,"},{"Start":"12:07.670 ","End":"12:10.760","Text":"we\u0027re just looking at its center of mass."},{"Start":"12:10.760 ","End":"12:16.475","Text":"Even though the shape itself is rotating at an angular velocity of Omega,"},{"Start":"12:16.475 ","End":"12:18.350","Text":"around, it doesn\u0027t matter,"},{"Start":"12:18.350 ","End":"12:20.329","Text":"because we\u0027re just looking at the center of mass and"},{"Start":"12:20.329 ","End":"12:23.285","Text":"the center of mass isn\u0027t moving, it\u0027s stationary."},{"Start":"12:23.285 ","End":"12:27.545","Text":"Because the 3 balls are moving around the center of mass,"},{"Start":"12:27.545 ","End":"12:30.785","Text":"but the center of mass is stationary."},{"Start":"12:30.785 ","End":"12:36.590","Text":"That means that if we\u0027re working out our momentum for the center of mass,"},{"Start":"12:36.590 ","End":"12:42.395","Text":"it\u0027s going to be mass multiplied by its velocity and its velocity is 0."},{"Start":"12:42.395 ","End":"12:46.069","Text":"It\u0027s going to be 0. Then afterwards,"},{"Start":"12:46.069 ","End":"12:53.600","Text":"we have to write down the momentum of our third body, ball number 3."},{"Start":"12:53.600 ","End":"12:56.734","Text":"We have this, so this is 1 body."},{"Start":"12:56.734 ","End":"13:01.550","Text":"Then we have ball number 3."},{"Start":"13:02.050 ","End":"13:05.345","Text":"Let\u0027s write the momentum for ball number 3."},{"Start":"13:05.345 ","End":"13:06.769","Text":"It\u0027s going to be mass,"},{"Start":"13:06.769 ","End":"13:08.420","Text":"which we know is m,"},{"Start":"13:08.420 ","End":"13:10.760","Text":"and then multiply it by its velocity."},{"Start":"13:10.760 ","End":"13:14.390","Text":"Its velocity we worked out in B, it\u0027s V_3."},{"Start":"13:14.390 ","End":"13:17.090","Text":"Right now, we\u0027ll substitute everything in afterwards,"},{"Start":"13:17.090 ","End":"13:18.815","Text":"but right now it\u0027s V_3."},{"Start":"13:18.815 ","End":"13:23.705","Text":"Then we\u0027re going to add in this,"},{"Start":"13:23.705 ","End":"13:27.125","Text":"the momentum of body 1, 2."},{"Start":"13:27.125 ","End":"13:29.330","Text":"What is this momentum?"},{"Start":"13:29.330 ","End":"13:35.495","Text":"It\u0027s plus, and then we\u0027re going to write 2m because there\u0027s 2 masses here and each one is"},{"Start":"13:35.495 ","End":"13:42.185","Text":"of mass m. So 2m multiplied by the velocity that we\u0027re trying to find."},{"Start":"13:42.185 ","End":"13:49.550","Text":"We\u0027re just going to label it V_cm. That\u0027s that."},{"Start":"13:49.550 ","End":"13:53.120","Text":"This is after."},{"Start":"13:53.120 ","End":"13:58.100","Text":"This is after ball number 3 has separated from the bodies."},{"Start":"13:58.100 ","End":"14:00.545","Text":"Now all we have to do,"},{"Start":"14:00.545 ","End":"14:04.115","Text":"is we just have to isolate out this V_cm."},{"Start":"14:04.115 ","End":"14:10.430","Text":"We\u0027ll have negative mv_3"},{"Start":"14:10.430 ","End":"14:15.890","Text":"divided by 2m,"},{"Start":"14:15.890 ","End":"14:19.925","Text":"is going to equal our V_cm."},{"Start":"14:19.925 ","End":"14:24.590","Text":"We can divide both sides by m. Then let\u0027s look at what our V_3 is,"},{"Start":"14:24.590 ","End":"14:27.155","Text":"so our V_3 is a negative."},{"Start":"14:27.155 ","End":"14:29.390","Text":"Negative, negative is positive."},{"Start":"14:29.390 ","End":"14:37.440","Text":"Then we have Omega L divided by root 3."},{"Start":"14:37.750 ","End":"14:40.490","Text":"Then here we have a 2."},{"Start":"14:40.490 ","End":"14:45.890","Text":"We know that this is going to be in the negative x-direction and this is going"},{"Start":"14:45.890 ","End":"14:51.500","Text":"to equal our V_cm."},{"Start":"14:51.500 ","End":"14:53.824","Text":"Now for question d,"},{"Start":"14:53.824 ","End":"14:56.239","Text":"sorry, there\u0027s a mistake here."},{"Start":"14:56.239 ","End":"14:58.789","Text":"There\u0027s not meant to be a minus over here."},{"Start":"14:58.789 ","End":"15:02.239","Text":"As we said, it was negative and then V_3 has a negative,"},{"Start":"15:02.239 ","End":"15:03.830","Text":"so it was negative, negative."},{"Start":"15:03.830 ","End":"15:06.530","Text":"This is going in the positive x-direction,"},{"Start":"15:06.530 ","End":"15:09.754","Text":"which as we also saw over here,"},{"Start":"15:09.754 ","End":"15:10.955","Text":"we already said,"},{"Start":"15:10.955 ","End":"15:14.340","Text":"it would be moving in the positive x-direction."},{"Start":"15:14.380 ","End":"15:17.465","Text":"Question d, we\u0027re being asked,"},{"Start":"15:17.465 ","End":"15:23.220","Text":"what is the angular velocity of this section about its center of mass?"},{"Start":"15:23.730 ","End":"15:29.185","Text":"Now what we understand is that our center of mass is also moving."},{"Start":"15:29.185 ","End":"15:31.840","Text":"Of our ball number 1 and 2,"},{"Start":"15:31.840 ","End":"15:35.125","Text":"is also moving in the positive direction x-direction."},{"Start":"15:35.125 ","End":"15:40.900","Text":"But we know now from the question that it\u0027s also rotating around itself."},{"Start":"15:40.900 ","End":"15:44.855","Text":"It\u0027s going to be traveling like this,"},{"Start":"15:44.855 ","End":"15:48.185","Text":"in some kind of spiral."},{"Start":"15:48.185 ","End":"15:52.100","Text":"Let\u0027s see how we go about answering this question."},{"Start":"15:52.100 ","End":"15:53.720","Text":"In the previous section,"},{"Start":"15:53.720 ","End":"15:57.470","Text":"we already saw that the momentum is"},{"Start":"15:57.470 ","End":"16:01.550","Text":"conserved because the sum of the external forces was equal to 0."},{"Start":"16:01.550 ","End":"16:03.019","Text":"We can\u0027t use that now,"},{"Start":"16:03.019 ","End":"16:06.455","Text":"the next thing to look at is if there\u0027s conservation of energy."},{"Start":"16:06.455 ","End":"16:10.355","Text":"Now there\u0027s no conservation of energy because we said that"},{"Start":"16:10.355 ","End":"16:15.590","Text":"ball number 3 breaks off from balls 1 and 2."},{"Start":"16:15.590 ","End":"16:18.259","Text":"As we said in the previous question,"},{"Start":"16:18.259 ","End":"16:22.400","Text":"is similar to plastic collision,"},{"Start":"16:22.400 ","End":"16:24.380","Text":"except the other way round."},{"Start":"16:24.380 ","End":"16:27.935","Text":"As we said, in cases like this with plastic collision,"},{"Start":"16:27.935 ","End":"16:30.695","Text":"then energy is never conserved."},{"Start":"16:30.695 ","End":"16:36.845","Text":"The last thing that we have to check is conservation of angular momentum."},{"Start":"16:36.845 ","End":"16:42.800","Text":"What we\u0027re trying to check is if the sum of all of"},{"Start":"16:42.800 ","End":"16:49.800","Text":"the external torques is equal to 0."},{"Start":"16:49.960 ","End":"16:54.919","Text":"As we saw, the sum of our external forces is equal to 0."},{"Start":"16:54.919 ","End":"16:57.484","Text":"If the sum of all external forces is equal to 0,"},{"Start":"16:57.484 ","End":"17:02.420","Text":"that means that the sum of the external torques is equal to 0."},{"Start":"17:02.420 ","End":"17:08.345","Text":"That means that there is conservation of angular momentum."},{"Start":"17:08.345 ","End":"17:12.950","Text":"Now what we\u0027re going to do is we\u0027re going to write out the equation."},{"Start":"17:12.950 ","End":"17:22.520","Text":"Before, our equation for angular momentum is going to be our I of body 1,"},{"Start":"17:22.520 ","End":"17:24.080","Text":"2, and 3."},{"Start":"17:24.080 ","End":"17:29.165","Text":"Of the whole thing before it split up, multiplied by Omega."},{"Start":"17:29.165 ","End":"17:32.430","Text":"Omega was given to us in the question."},{"Start":"17:32.440 ","End":"17:37.640","Text":"Now let\u0027s work out what our I_123 is."},{"Start":"17:37.640 ","End":"17:39.589","Text":"Let\u0027s just write this."},{"Start":"17:39.589 ","End":"17:43.260","Text":"This is before the split."},{"Start":"17:44.740 ","End":"17:49.770","Text":"What is our I_123?"},{"Start":"17:54.160 ","End":"17:59.190","Text":"Right now we\u0027re in the original diagram."},{"Start":"17:59.230 ","End":"18:02.689","Text":"What we\u0027re going to do is we have to find out"},{"Start":"18:02.689 ","End":"18:06.170","Text":"what our moment of inertia is for this entire body."},{"Start":"18:06.170 ","End":"18:10.025","Text":"We know that it\u0027s rotating around its center of mass."},{"Start":"18:10.025 ","End":"18:12.200","Text":"This is axis of rotation."},{"Start":"18:12.200 ","End":"18:16.639","Text":"We have to find also for now when we\u0027re working"},{"Start":"18:16.639 ","End":"18:21.275","Text":"out the angular momentum and also for afterwards around these exact same axis,"},{"Start":"18:21.275 ","End":"18:25.880","Text":"our moments of inertia and angular momentums."},{"Start":"18:25.880 ","End":"18:28.490","Text":"For body 1, 2,"},{"Start":"18:28.490 ","End":"18:30.740","Text":"3 our I,"},{"Start":"18:30.740 ","End":"18:33.140","Text":"we can say as we know,"},{"Start":"18:33.140 ","End":"18:40.670","Text":"its mass times its radius squared from the axis of rotation."},{"Start":"18:40.670 ","End":"18:43.504","Text":"Mass times distance squared from axis of rotation."},{"Start":"18:43.504 ","End":"18:46.625","Text":"Now because of the symmetry between the 3 points,"},{"Start":"18:46.625 ","End":"18:49.295","Text":"and we know that each of them is a distance R"},{"Start":"18:49.295 ","End":"18:52.490","Text":"away because this is an equilateral triangle."},{"Start":"18:52.490 ","End":"18:54.860","Text":"We already worked out what our R is."},{"Start":"18:54.860 ","End":"18:56.510","Text":"It\u0027s this over here,"},{"Start":"18:56.510 ","End":"18:59.600","Text":"1 over root 3L."},{"Start":"18:59.600 ","End":"19:10.084","Text":"We can say that it\u0027s just equal to 3 multiplied by R radius,"},{"Start":"19:10.084 ","End":"19:13.569","Text":"which is 1 over"},{"Start":"19:13.569 ","End":"19:19.885","Text":"root 3L^2 multiplied by"},{"Start":"19:19.885 ","End":"19:26.540","Text":"the mass m. This is our I before."},{"Start":"19:27.790 ","End":"19:32.360","Text":"Now because of conservation of angular momentum."},{"Start":"19:32.360 ","End":"19:36.320","Text":"We can say that our angular momentum before,"},{"Start":"19:36.320 ","End":"19:39.110","Text":"is equal to our angular momentum after."},{"Start":"19:39.110 ","End":"19:41.839","Text":"Here we\u0027ll write down after."},{"Start":"19:41.839 ","End":"19:46.160","Text":"Now, let\u0027s work out what our angular momentum after is."},{"Start":"19:46.160 ","End":"19:50.970","Text":"Fast, let\u0027s work out our angular momentum of ball number 3."},{"Start":"19:51.370 ","End":"19:57.650","Text":"Again, because before we worked out our angular momentum about this axis of rotation,"},{"Start":"19:57.650 ","End":"20:00.965","Text":"again, we\u0027re going to have to work it out from here."},{"Start":"20:00.965 ","End":"20:04.474","Text":"As we saw, our ball number 3 flies off"},{"Start":"20:04.474 ","End":"20:07.865","Text":"in the negative x-direction going in a straight line."},{"Start":"20:07.865 ","End":"20:11.480","Text":"As we know from our chapter on angular momentum,"},{"Start":"20:11.480 ","End":"20:18.785","Text":"the equation for the angular momentum of a body moving in a straight line,"},{"Start":"20:18.785 ","End":"20:23.270","Text":"is its mass multiplied by its velocity."},{"Start":"20:23.270 ","End":"20:32.855","Text":"Here it\u0027s going to be V_3 and then multiplied by its distance from the axis of rotation."},{"Start":"20:32.855 ","End":"20:39.410","Text":"Now, the distance from the axis of rotation has to be the shortest distance."},{"Start":"20:39.410 ","End":"20:41.615","Text":"The perpendicular components."},{"Start":"20:41.615 ","End":"20:45.349","Text":"Now because our velocity is in the negative x-direction"},{"Start":"20:45.349 ","End":"20:50.585","Text":"and our axis of rotation is in the negative y-direction."},{"Start":"20:50.585 ","End":"20:54.890","Text":"We can see that there\u0027s a 90-degree angle between these 2,"},{"Start":"20:54.890 ","End":"20:56.599","Text":"which means that we don\u0027t have to work out"},{"Start":"20:56.599 ","End":"21:01.640","Text":"the perpendicular component is already given to us and it\u0027s simply our I."},{"Start":"21:01.810 ","End":"21:05.045","Text":"Then we can write down our I,"},{"Start":"21:05.045 ","End":"21:07.100","Text":"which again, we know what it is."},{"Start":"21:07.100 ","End":"21:12.440","Text":"It\u0027s this times 1 over root 3L."},{"Start":"21:14.740 ","End":"21:20.060","Text":"Then in order to check the sign if there\u0027s meant to be a positive or negative over here."},{"Start":"21:20.060 ","End":"21:23.150","Text":"Again, we can go back to our diagram and we"},{"Start":"21:23.150 ","End":"21:27.410","Text":"see that our arrow is pointing in the negative x direction,"},{"Start":"21:27.410 ","End":"21:31.940","Text":"which is pushing our body in this direction."},{"Start":"21:31.940 ","End":"21:34.039","Text":"This direction, as we can see,"},{"Start":"21:34.039 ","End":"21:37.950","Text":"is in the positive direction that we were given for our Omega."},{"Start":"21:38.620 ","End":"21:43.115","Text":"We can just leave this expression as a positive."},{"Start":"21:43.115 ","End":"21:51.150","Text":"Now we\u0027re going to work out the angular momentum of body 1 and 2, our second shape."},{"Start":"21:51.580 ","End":"21:55.520","Text":"This is now the complicated section of the question,"},{"Start":"21:55.520 ","End":"21:58.410","Text":"which isn\u0027t even that complicated."},{"Start":"22:00.490 ","End":"22:04.580","Text":"When dealing with our shape over here,"},{"Start":"22:04.580 ","End":"22:09.395","Text":"so we know that our center of mass is moving and also simultaneously the whole thing,"},{"Start":"22:09.395 ","End":"22:13.025","Text":"the 2 balls are rotating around the center of mass."},{"Start":"22:13.025 ","End":"22:17.345","Text":"What we\u0027re dealing with is a complex angular momentum."},{"Start":"22:17.345 ","End":"22:23.030","Text":"We know from our equation that we learned before that our L,"},{"Start":"22:23.030 ","End":"22:25.789","Text":"when dealing with complex angular momentum,"},{"Start":"22:25.789 ","End":"22:30.604","Text":"is going to be L around our center of mass."},{"Start":"22:30.604 ","End":"22:32.439","Text":"This one of 1,"},{"Start":"22:32.439 ","End":"22:39.980","Text":"2 plus the angular momentum of the shape moving"},{"Start":"22:39.980 ","End":"22:47.600","Text":"relative to this point over here, so L_CM."},{"Start":"22:47.600 ","End":"22:52.160","Text":"Again, this L_CM_12 is going to be"},{"Start":"22:52.160 ","End":"22:57.283","Text":"the angular momentum of the balls rotating about this axis of rotation,"},{"Start":"22:57.283 ","End":"23:01.205","Text":"the center of mass between these 2 shapes, 2 bodies,"},{"Start":"23:01.205 ","End":"23:05.809","Text":"plus the angular momentum of this center of"},{"Start":"23:05.809 ","End":"23:12.260","Text":"mass relative to the total center of mass of the 3 shapes."},{"Start":"23:12.260 ","End":"23:18.875","Text":"Because this is our axis of rotation by which we\u0027re writing all of our equations."},{"Start":"23:18.875 ","End":"23:23.348","Text":"As we can see, this distance over here,"},{"Start":"23:23.348 ","End":"23:32.915","Text":"our y_CM, which is this where we worked out over here,"},{"Start":"23:32.915 ","End":"23:35.190","Text":"which is this number over here."},{"Start":"23:35.380 ","End":"23:38.460","Text":"Let\u0027s scroll back."},{"Start":"23:39.360 ","End":"23:42.385","Text":"Just like we did with ball Number 3,"},{"Start":"23:42.385 ","End":"23:45.370","Text":"when dealing with working out"},{"Start":"23:45.370 ","End":"23:49.230","Text":"the angular momentum of a shape which is moving in a straight line."},{"Start":"23:49.230 ","End":"23:51.395","Text":"We can put here a positive."},{"Start":"23:51.395 ","End":"23:54.740","Text":"When dealing with the shape moving in a straight line,"},{"Start":"23:54.740 ","End":"23:58.325","Text":"we have to write down this equation."},{"Start":"23:58.325 ","End":"24:04.490","Text":"It\u0027s going to be the total mass of our now 2 body system."},{"Start":"24:04.490 ","End":"24:10.850","Text":"It\u0027s going to be 2M multiplied by the velocity."},{"Start":"24:10.850 ","End":"24:19.775","Text":"Our velocity is going to be this what we worked out in our previous question."},{"Start":"24:19.775 ","End":"24:23.930","Text":"This over here multiplied by the distance."},{"Start":"24:23.930 ","End":"24:27.080","Text":"It is from the axis of rotation,"},{"Start":"24:27.080 ","End":"24:31.830","Text":"which is our y_cm."},{"Start":"24:32.200 ","End":"24:34.924","Text":"It\u0027s distance is this, and remember,"},{"Start":"24:34.924 ","End":"24:37.850","Text":"it has to be the perpendicular component."},{"Start":"24:37.850 ","End":"24:42.919","Text":"As we can see, because it\u0027s moving in this direction, it\u0027s 90,"},{"Start":"24:42.919 ","End":"24:47.015","Text":"it\u0027s perpendicular component is going to be this y_CM,"},{"Start":"24:47.015 ","End":"24:49.429","Text":"because there\u0027s a 90-degree angle over here."},{"Start":"24:49.429 ","End":"24:51.125","Text":"If it wasn\u0027t perpendicular,"},{"Start":"24:51.125 ","End":"24:54.695","Text":"then we would multiply it by sine of the angle."},{"Start":"24:54.695 ","End":"24:56.480","Text":"Here, sine of the angle is 90,"},{"Start":"24:56.480 ","End":"24:58.920","Text":"so it\u0027s just multiplied by 1."},{"Start":"24:59.290 ","End":"25:05.720","Text":"Multiply by y_cm so we can go back to here and as we see,"},{"Start":"25:05.720 ","End":"25:15.710","Text":"our y_cm is this so L divided by 2 root 3."},{"Start":"25:15.710 ","End":"25:18.170","Text":"This is our y_cm."},{"Start":"25:18.170 ","End":"25:20.225","Text":"Up until now,"},{"Start":"25:20.225 ","End":"25:24.515","Text":"we worked out this section, our L_CM."},{"Start":"25:24.515 ","End":"25:27.740","Text":"But now what we want to do is we want to work out"},{"Start":"25:27.740 ","End":"25:32.580","Text":"the angular momentum of these 2 shapes around their center of mass."},{"Start":"25:32.710 ","End":"25:39.995","Text":"Till now we did this and now we\u0027re going to do this."},{"Start":"25:39.995 ","End":"25:43.999","Text":"Our equation that we\u0027re going to use for our body, 1,"},{"Start":"25:43.999 ","End":"25:50.249","Text":"2 rotating around itself is going to be our I of bodies 1,"},{"Start":"25:50.249 ","End":"25:54.070","Text":"2 multiplied by our Omega."},{"Start":"25:54.070 ","End":"25:58.495","Text":"Our Omega are Omega tag and this is our unknown."},{"Start":"25:58.495 ","End":"26:02.900","Text":"This is what we\u0027re trying to find, the angular velocity."},{"Start":"26:02.910 ","End":"26:06.170","Text":"Let\u0027s scroll down a little bit."},{"Start":"26:06.420 ","End":"26:09.520","Text":"Let\u0027s see what our I_1,2 is."},{"Start":"26:09.520 ","End":"26:11.440","Text":"We\u0027ll write it over here."},{"Start":"26:11.440 ","End":"26:14.780","Text":"Our I_1,2 is equal to,"},{"Start":"26:14.780 ","End":"26:15.994","Text":"as we can see over here,"},{"Start":"26:15.994 ","End":"26:19.415","Text":"it\u0027s going to be the total mass of the system."},{"Start":"26:19.415 ","End":"26:21.169","Text":"We know that there are 2 masses."},{"Start":"26:21.169 ","End":"26:23.300","Text":"There\u0027s ball Number 1 and ball Number 2."},{"Start":"26:23.300 ","End":"26:25.085","Text":"Each one is of mass M,"},{"Start":"26:25.085 ","End":"26:29.020","Text":"so 2M and then we have to multiply all"},{"Start":"26:29.020 ","End":"26:33.115","Text":"of this by the distance squared to the center of mass."},{"Start":"26:33.115 ","End":"26:35.500","Text":"As we can see, our distance,"},{"Start":"26:35.500 ","End":"26:38.050","Text":"each ball is from the center of mass,"},{"Start":"26:38.050 ","End":"26:40.975","Text":"is L divided by 2."},{"Start":"26:40.975 ","End":"26:48.640","Text":"Now we can scroll back over here and back down and then we\u0027re going to multiply"},{"Start":"26:48.640 ","End":"26:56.165","Text":"it by L over 2 squared because that is the distance from the center of mass."},{"Start":"26:56.165 ","End":"26:58.669","Text":"Again, this section,"},{"Start":"26:58.669 ","End":"27:01.654","Text":"we\u0027re working out the angular momentum"},{"Start":"27:01.654 ","End":"27:05.150","Text":"about the center of mass of the 2 balls rotating around each"},{"Start":"27:05.150 ","End":"27:08.330","Text":"other and then this section was the correction in"},{"Start":"27:08.330 ","End":"27:12.900","Text":"order to make it relative to our original axes of rotation."},{"Start":"27:13.240 ","End":"27:15.710","Text":"Let\u0027s do this."},{"Start":"27:15.710 ","End":"27:18.860","Text":"I\u0027m literally just going to substitute this in."},{"Start":"27:18.860 ","End":"27:20.954","Text":"Before we have I of 1,"},{"Start":"27:20.954 ","End":"27:23.210","Text":"2, 3 multiplied by Omega."},{"Start":"27:23.210 ","End":"27:33.769","Text":"We have 3M multiplied by L divided by root 3 squared,"},{"Start":"27:33.769 ","End":"27:43.445","Text":"which is going to equal to M L divided by root 3 multiplied by V_3."},{"Start":"27:43.445 ","End":"27:46.280","Text":"Let\u0027s scroll a bit to the side."},{"Start":"27:46.280 ","End":"27:48.845","Text":"This was our V_3."},{"Start":"27:48.845 ","End":"27:54.545","Text":"We\u0027ll put this here because we just want to take the positive value."},{"Start":"27:54.545 ","End":"27:57.829","Text":"Here we\u0027re going to multiply this by Omega and another L"},{"Start":"27:57.829 ","End":"28:01.520","Text":"so it\u0027s squared and another root 3."},{"Start":"28:01.520 ","End":"28:10.279","Text":"We can just write root 3 squared is just 3 multiplied by Omega and then we\u0027re going"},{"Start":"28:10.279 ","End":"28:19.535","Text":"to add to this I_2 multiplied by M and then it\u0027s divided by 2."},{"Start":"28:19.535 ","End":"28:21.215","Text":"We can take that away."},{"Start":"28:21.215 ","End":"28:25.475","Text":"It\u0027s going to be M multiplied by our V_cm, which is this."},{"Start":"28:25.475 ","End":"28:29.510","Text":"M Omega L divided"},{"Start":"28:29.510 ","End":"28:36.455","Text":"by 2 root 3 multiplied over here by this root 3."},{"Start":"28:36.455 ","End":"28:38.975","Text":"It\u0027s just going to be 2 times 3,"},{"Start":"28:38.975 ","End":"28:43.205","Text":"so 6 and then"},{"Start":"28:43.205 ","End":"28:50.435","Text":"plus our I_1,2 multiplied by our unknown Omega."},{"Start":"28:50.435 ","End":"28:58.220","Text":"Then plus 2ML squared divided by 4."},{"Start":"28:58.220 ","End":"29:06.660","Text":"I just squared this so I can rub this out and rub this out and have a 2 over here."},{"Start":"29:06.790 ","End":"29:10.595","Text":"Now we and then multiply it over here by"},{"Start":"29:10.595 ","End":"29:14.570","Text":"Omega and also over here I forgot to multiply by Omega."},{"Start":"29:14.570 ","End":"29:20.390","Text":"Perfect. Now, we can see that in every single one of our terms we have M. We can divide"},{"Start":"29:20.390 ","End":"29:26.405","Text":"everything by M to simplify and we could also see that in every single one of our terms,"},{"Start":"29:26.405 ","End":"29:34.380","Text":"we have at least one L. Let\u0027s just rewrite this without the square roots."},{"Start":"29:34.570 ","End":"29:38.105","Text":"Instead of having l,"},{"Start":"29:38.105 ","End":"29:43.380","Text":"we\u0027ll have L squared and instead of root 3 squared we\u0027ll just be left with I_3."},{"Start":"29:43.630 ","End":"29:50.270","Text":"Now we can see that we can also divide each side by L. Here I can take off my squared."},{"Start":"29:50.270 ","End":"29:52.129","Text":"Here, I can take off my squared,"},{"Start":"29:52.129 ","End":"29:54.005","Text":"and here I can rub off my L,"},{"Start":"29:54.005 ","End":"29:56.970","Text":"and here I can take off the squared."},{"Start":"29:57.910 ","End":"30:02.405","Text":"This Omega, of course is our unknown so here there\u0027s a tag."},{"Start":"30:02.405 ","End":"30:11.090","Text":"Now, we can just take all of these terms and switch them to this side of the equal side,"},{"Start":"30:11.090 ","End":"30:15.455","Text":"and then divide both sides by L divided by 2."},{"Start":"30:15.455 ","End":"30:17.029","Text":"Let\u0027s see how I do this."},{"Start":"30:17.029 ","End":"30:22.430","Text":"Here, I can get my like terms together because it will help me simplify."},{"Start":"30:22.430 ","End":"30:24.620","Text":"This is going to equal to,"},{"Start":"30:24.620 ","End":"30:32.230","Text":"so I know that in both sides I have Omega divided by 3 and then we"},{"Start":"30:32.230 ","End":"30:41.920","Text":"have over here L multiplied by that and over here we have plus 1/2."},{"Start":"30:41.920 ","End":"30:46.055","Text":"Then we have plus, what is this?"},{"Start":"30:46.055 ","End":"30:47.885","Text":"L over 2,"},{"Start":"30:47.885 ","End":"30:50.915","Text":"and this is our Omega tag that we don\u0027t know."},{"Start":"30:50.915 ","End":"30:56.300","Text":"Over here we have 3L divided by 3."},{"Start":"30:56.300 ","End":"30:59.105","Text":"We can cross off this as well."},{"Start":"30:59.105 ","End":"31:07.685","Text":"Here we just have Omega L. Now,"},{"Start":"31:07.685 ","End":"31:10.400","Text":"we can take our like terms to one side,"},{"Start":"31:10.400 ","End":"31:12.455","Text":"so everything has omega."},{"Start":"31:12.455 ","End":"31:21.185","Text":"We have Omega and then I have L and then here I have negative and this will be L over 3."},{"Start":"31:21.185 ","End":"31:24.155","Text":"Negative L over 3,"},{"Start":"31:24.155 ","End":"31:29.670","Text":"and then negative 1 over 6,"},{"Start":"31:30.070 ","End":"31:36.845","Text":"and this is going to equal L over 2 multiplied by Omega tag."},{"Start":"31:36.845 ","End":"31:44.495","Text":"Then L minus L over 3 will be 2L over 3."},{"Start":"31:44.495 ","End":"31:55.099","Text":"What we\u0027ll have is Omega multiplied by 2 divided by L and then this is going to"},{"Start":"31:55.099 ","End":"31:59.780","Text":"be multiplied by 2L divided by 3"},{"Start":"31:59.780 ","End":"32:06.750","Text":"minus 1/6 which is going to equal our Omega tag."},{"Start":"32:07.450 ","End":"32:10.309","Text":"This is our final answer."},{"Start":"32:10.309 ","End":"32:13.400","Text":"Just did some algebra over here."},{"Start":"32:13.400 ","End":"32:17.279","Text":"That is the end of our question."}],"ID":9404}],"Thumbnail":null,"ID":5406},{"Name":"3. Rotational Energy Of A Rigid Body","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Circular KE","Duration":"33m 20s","ChapterTopicVideoID":9137,"CourseChapterTopicPlaylistID":5407,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:02.745","Text":"Hello. In this lesson,"},{"Start":"00:02.745 ","End":"00:07.230","Text":"we\u0027re going to be speaking about rotational kinetic energy."},{"Start":"00:07.230 ","End":"00:15.165","Text":"Let\u0027s say that we have some rod and it\u0027s rotating around some axis,"},{"Start":"00:15.165 ","End":"00:19.170","Text":"so let\u0027s draw that over here."},{"Start":"00:19.170 ","End":"00:24.400","Text":"It\u0027s rotating around like so."},{"Start":"00:24.620 ","End":"00:31.245","Text":"The general equation for kinetic energy, as everyone knows,"},{"Start":"00:31.245 ","End":"00:36.179","Text":"is equal to 1/2 multiplied by the mass,"},{"Start":"00:36.179 ","End":"00:40.510","Text":"multiplied by the velocity^2."},{"Start":"00:40.630 ","End":"00:45.560","Text":"How would we find the kinetic energy of this rod?"},{"Start":"00:45.560 ","End":"00:50.494","Text":"What some people might say is to say,"},{"Start":"00:50.494 ","End":"00:54.410","Text":"we have to find some velocity for this rod."},{"Start":"00:54.410 ","End":"00:57.229","Text":"Let\u0027s write the VCM,"},{"Start":"00:57.229 ","End":"00:59.870","Text":"the velocity of the center of mass and then we will"},{"Start":"00:59.870 ","End":"01:02.720","Text":"find out the kinetic energy of the shape."},{"Start":"01:02.720 ","End":"01:06.605","Text":"But no, this is incorrect,"},{"Start":"01:06.605 ","End":"01:13.835","Text":"we can take the velocity of the center of mass when working out these questions."},{"Start":"01:13.835 ","End":"01:17.550","Text":"Let\u0027s see what we can do."},{"Start":"01:18.260 ","End":"01:25.804","Text":"What we will do is we\u0027ll say that the kinetic energy of any body,"},{"Start":"01:25.804 ","End":"01:27.095","Text":"not just this rod,"},{"Start":"01:27.095 ","End":"01:29.135","Text":"but this is an example."},{"Start":"01:29.135 ","End":"01:31.929","Text":"What we\u0027ll do is we\u0027ll sum it up,"},{"Start":"01:31.929 ","End":"01:38.680","Text":"so we\u0027ll take some small area and it has some mass Mi,"},{"Start":"01:38.680 ","End":"01:44.250","Text":"and it has some velocity in this direction, V_i."},{"Start":"01:44.250 ","End":"01:47.370","Text":"Then we\u0027ll put it into this format so we\u0027ll say we"},{"Start":"01:47.370 ","End":"01:50.325","Text":"have 1/2 multiplied by the mass of this piece,"},{"Start":"01:50.325 ","End":"01:56.490","Text":"so multiply by Mi multiplied by the velocity of this V_i^2."},{"Start":"01:56.490 ","End":"02:00.185","Text":"Then in order to get for the whole body,"},{"Start":"02:00.185 ","End":"02:04.020","Text":"will sum up all of the I\u0027s."},{"Start":"02:05.020 ","End":"02:09.470","Text":"I\u0027m going to work this out and I\u0027m going to show"},{"Start":"02:09.470 ","End":"02:15.680","Text":"you that it doesn\u0027t end up being the same as multiplying by VCM."},{"Start":"02:15.680 ","End":"02:20.309","Text":"I\u0027m going to show you why this is incorrect and why this is correct."},{"Start":"02:20.900 ","End":"02:25.594","Text":"I\u0027ve written a very important sentence for you over here."},{"Start":"02:25.594 ","End":"02:33.620","Text":"Every point on the body on this undergoes circular motion with the same angular velocity."},{"Start":"02:33.620 ","End":"02:41.884","Text":"If we say that the whole rod is moving with angular velocity Omega."},{"Start":"02:41.884 ","End":"02:44.465","Text":"Then every single point,"},{"Start":"02:44.465 ","End":"02:45.874","Text":"this point over here,"},{"Start":"02:45.874 ","End":"02:47.329","Text":"this point over here,"},{"Start":"02:47.329 ","End":"02:48.814","Text":"this point over here,"},{"Start":"02:48.814 ","End":"02:56.550","Text":"they\u0027re all rotating around the axis of rotation with this Omega."},{"Start":"02:57.140 ","End":"03:01.710","Text":"It might be useful to write this sentence down somewhere."},{"Start":"03:01.990 ","End":"03:05.854","Text":"Let\u0027s work this out, I labeled this in red,"},{"Start":"03:05.854 ","End":"03:10.230","Text":"you should write this on your formula sheets."},{"Start":"03:12.100 ","End":"03:16.109","Text":"As we know, the lever will be our V_i,"},{"Start":"03:16.109 ","End":"03:22.700","Text":"let\u0027s see our V_i is going to be equal to when we\u0027re working with Omega,"},{"Start":"03:22.700 ","End":"03:26.470","Text":"so it will be equal to our Omega,"},{"Start":"03:26.470 ","End":"03:31.730","Text":"which is known multiplied by the radius of each piece."},{"Start":"03:31.730 ","End":"03:35.569","Text":"Obviously, the radius is different for every piece,"},{"Start":"03:35.569 ","End":"03:40.759","Text":"so for instance, the radius here is this short."},{"Start":"03:40.759 ","End":"03:43.830","Text":"The radius to this point,"},{"Start":"03:45.260 ","End":"03:49.545","Text":"is like this and also to this point."},{"Start":"03:49.545 ","End":"03:52.530","Text":"Every radius is different."},{"Start":"03:52.530 ","End":"04:00.885","Text":"Then when I have that, I can rewrite my E_k as the sum on Is"},{"Start":"04:00.885 ","End":"04:05.760","Text":"of 1/2mi and then instead of V_i I can"},{"Start":"04:05.760 ","End":"04:12.265","Text":"write Omega r_i and all of this squared because my V_i is squared."},{"Start":"04:12.265 ","End":"04:14.639","Text":"When this is written like this,"},{"Start":"04:14.639 ","End":"04:17.889","Text":"my Omega and my 1/2 is the same for everyone,"},{"Start":"04:17.889 ","End":"04:21.055","Text":"so I can move them outside of my summation mark."},{"Start":"04:21.055 ","End":"04:26.075","Text":"I can write 1/2 Omega and then sum,"},{"Start":"04:26.075 ","End":"04:28.829","Text":"and all the i\u0027s, sorry,"},{"Start":"04:28.829 ","End":"04:33.075","Text":"my Omega will be squared and then this will be multiplied by"},{"Start":"04:33.075 ","End":"04:40.090","Text":"m_i and then r_i^2 because it was in the brackets being squared."},{"Start":"04:41.000 ","End":"04:45.364","Text":"For those of you that notice and those that haven\u0027t it doesn\u0027t matter."},{"Start":"04:45.364 ","End":"04:52.065","Text":"This what\u0027s inside our summation is equal to our I,"},{"Start":"04:52.065 ","End":"04:54.180","Text":"our moment of inertia."},{"Start":"04:54.180 ","End":"05:02.530","Text":"Then we can rewrite this as 1/2I Omega."},{"Start":"05:03.470 ","End":"05:08.489","Text":"We have 1/2I Omega^2 and as you can see,"},{"Start":"05:08.489 ","End":"05:12.610","Text":"this definitely does not equal this."},{"Start":"05:14.090 ","End":"05:20.480","Text":"This is the basic equation for our kinetic energy when working in circular motion,"},{"Start":"05:20.480 ","End":"05:23.705","Text":"for some body rotating around its axis."},{"Start":"05:23.705 ","End":"05:27.424","Text":"I suggest writing this in your formula sheets."},{"Start":"05:27.424 ","End":"05:30.985","Text":"There\u0027s also another equation."},{"Start":"05:30.985 ","End":"05:33.595","Text":"I\u0027m going to write it."},{"Start":"05:33.595 ","End":"05:36.275","Text":"I don\u0027t know if you\u0027ve seen it already,"},{"Start":"05:36.275 ","End":"05:45.179","Text":"but it is equal to 1/2 the total mass multiplied by V_center of"},{"Start":"05:45.179 ","End":"05:52.319","Text":"mass^2 plus 1/2 multiplied"},{"Start":"05:52.319 ","End":"05:58.720","Text":"by I_cm multiplied by Omega^2."},{"Start":"06:00.080 ","End":"06:07.950","Text":"This equation is slightly different so let\u0027s see when we use each one."},{"Start":"06:07.950 ","End":"06:15.660","Text":"This equation we\u0027re going to use when our axis is fixed."},{"Start":"06:16.040 ","End":"06:19.714","Text":"When we\u0027re working on a fixed axis,"},{"Start":"06:19.714 ","End":"06:21.320","Text":"so it\u0027s not moving."},{"Start":"06:21.320 ","End":"06:25.565","Text":"In this case, it would be if our axis of rotation was,"},{"Start":"06:25.565 ","End":"06:28.939","Text":"for instance, fixed onto a table and it wasn\u0027t moving."},{"Start":"06:28.939 ","End":"06:35.740","Text":"The rod is just rotating around the axis and that\u0027s the only movement that\u0027s going on."},{"Start":"06:36.550 ","End":"06:42.665","Text":"When is this used?"},{"Start":"06:42.665 ","End":"06:46.260","Text":"It\u0027s when axis is moving."},{"Start":"06:48.740 ","End":"06:53.774","Text":"In this case, if we had our rod,"},{"Start":"06:53.774 ","End":"06:58.580","Text":"and then if we had our axes of rotation over here."},{"Start":"06:58.580 ","End":"07:02.300","Text":"It\u0027s not fixed to a table so the axis,"},{"Start":"07:02.300 ","End":"07:07.919","Text":"let\u0027s say that this"},{"Start":"07:07.919 ","End":"07:13.475","Text":"is the x-axis and this is our axis of rotation and it\u0027s the z axes."},{"Start":"07:13.475 ","End":"07:18.634","Text":"Our rod can simultaneously spin around the axis."},{"Start":"07:18.634 ","End":"07:25.805","Text":"The axes itself can move along the x-axis, for instance."},{"Start":"07:25.805 ","End":"07:29.870","Text":"Simultaneously was the rod is going"},{"Start":"07:29.870 ","End":"07:34.310","Text":"around in circles and we\u0027re going forwards along the page,"},{"Start":"07:34.310 ","End":"07:38.250","Text":"so that is for this equation."},{"Start":"07:38.600 ","End":"07:41.460","Text":"Where does this equation come from?"},{"Start":"07:41.460 ","End":"07:43.264","Text":"If we look at this section,"},{"Start":"07:43.264 ","End":"07:45.909","Text":"It\u0027s very similar to this equation."},{"Start":"07:45.909 ","End":"07:49.669","Text":"What we\u0027re doing is this is the internal energy,"},{"Start":"07:49.669 ","End":"07:55.614","Text":"so this is the kinetic energy caused by the rod going around the axis of rotation."},{"Start":"07:55.614 ","End":"08:01.114","Text":"Just like over here but then we have to add the external kinetic energy,"},{"Start":"08:01.114 ","End":"08:07.510","Text":"which is as a result of the rod of the axis of rotation moving forward."},{"Start":"08:07.510 ","End":"08:12.000","Text":"Every piece over here, just like over here,"},{"Start":"08:12.000 ","End":"08:17.029","Text":"so it\u0027s going to have the same V_i going in this direction,"},{"Start":"08:17.029 ","End":"08:21.380","Text":"but it\u0027s also going to have due to the axis moving every piece is"},{"Start":"08:21.380 ","End":"08:26.535","Text":"also going to have the V_cm."},{"Start":"08:26.535 ","End":"08:35.025","Text":"Also over here, there will be VI and in this direction V_cm."},{"Start":"08:35.025 ","End":"08:37.910","Text":"That is where this comes from."},{"Start":"08:37.910 ","End":"08:42.874","Text":"We can write out like here how we wrote our V_i,"},{"Start":"08:42.874 ","End":"08:49.865","Text":"for this equation,"},{"Start":"08:49.865 ","End":"08:59.322","Text":"these are vector quantities of course."},{"Start":"08:59.322 ","End":"09:06.830","Text":"It will also equal Omega multiplied by r_i."},{"Start":"09:07.100 ","End":"09:10.125","Text":"But then because of the external motion,"},{"Start":"09:10.125 ","End":"09:16.180","Text":"we\u0027ll also add in our V of the center of mass."},{"Start":"09:16.230 ","End":"09:23.030","Text":"Notice the blue line I didn\u0027t extend it until the end."},{"Start":"09:26.070 ","End":"09:31.060","Text":"This equation is also for an axis moving"},{"Start":"09:31.060 ","End":"09:37.309","Text":"and also when the rod is rotating about its center of mass."},{"Start":"09:37.309 ","End":"09:43.740","Text":"To recap, this equation is when the body is moving"},{"Start":"09:43.740 ","End":"09:50.974","Text":"at any place on the rod but the important thing is that the axis is fixed."},{"Start":"09:50.974 ","End":"09:53.139","Text":"The axis is not moving."},{"Start":"09:53.139 ","End":"09:56.920","Text":"This equation refers to the axis"},{"Start":"09:56.920 ","End":"10:05.845","Text":"moving and also that the body is rotating about the center of mass."},{"Start":"10:05.845 ","End":"10:10.884","Text":"Our body is rotating around it and only then we can use this equation."},{"Start":"10:10.884 ","End":"10:18.834","Text":"Most of the questions that you will encounter will be with these 2 examples."},{"Start":"10:18.834 ","End":"10:24.820","Text":"These 2 equations will be relevant for the case where we have"},{"Start":"10:24.820 ","End":"10:30.924","Text":"some kind of body that is rotating not about its center of mass and the axis is moving."},{"Start":"10:30.924 ","End":"10:32.664","Text":"We\u0027re going to speak about now,"},{"Start":"10:32.664 ","End":"10:34.999","Text":"it\u0027s a lot rarer though."},{"Start":"10:36.450 ","End":"10:39.535","Text":"I\u0027m going to scroll down a little bit."},{"Start":"10:39.535 ","End":"10:43.659","Text":"We\u0027re going to speak about the example when"},{"Start":"10:43.659 ","End":"10:48.100","Text":"the axis is moving and the body is rotating about a different point,"},{"Start":"10:48.100 ","End":"10:51.230","Text":"not necessarily the center of mass."},{"Start":"10:51.630 ","End":"10:56.420","Text":"When we have our axis of rotation,"},{"Start":"10:56.640 ","End":"11:02.125","Text":"and we have our rod,"},{"Start":"11:02.125 ","End":"11:05.545","Text":"which is rotating, let\u0027s say,"},{"Start":"11:05.545 ","End":"11:09.625","Text":"around this point, right at the edge."},{"Start":"11:09.625 ","End":"11:18.621","Text":"We can say that this point specifically over here has some velocity V_0."},{"Start":"11:18.621 ","End":"11:21.985","Text":"That means that simultaneously the point over here,"},{"Start":"11:21.985 ","End":"11:27.591","Text":"which is rotating, is going to have the velocity V_i."},{"Start":"11:27.591 ","End":"11:29.934","Text":"Also, going in this direction,"},{"Start":"11:29.934 ","End":"11:33.520","Text":"it\u0027s going to have V_o,"},{"Start":"11:33.520 ","End":"11:37.889","Text":"so then just like I wrote over here,"},{"Start":"11:37.889 ","End":"11:45.700","Text":"I could write that my V_i is going to be equal to Omega r_i."},{"Start":"11:45.700 ","End":"11:51.804","Text":"Then instead of here plus V_cm which is what was going on here,"},{"Start":"11:51.804 ","End":"11:55.975","Text":"so I\u0027ll write plus V_o."},{"Start":"11:55.975 ","End":"11:59.320","Text":"Just because it\u0027s not clear,"},{"Start":"11:59.320 ","End":"12:04.051","Text":"this V_o is what\u0027s pushing the axis,"},{"Start":"12:04.051 ","End":"12:06.235","Text":"I can draw it at a bit of a different angle,"},{"Start":"12:06.235 ","End":"12:08.009","Text":"let\u0027s say like this,"},{"Start":"12:08.009 ","End":"12:13.225","Text":"it\u0027s what\u0027s pushing our body to move in this direction as well,"},{"Start":"12:13.225 ","End":"12:15.325","Text":"because the axis is also moving."},{"Start":"12:15.325 ","End":"12:20.890","Text":"You can also look at this equation as similar to relative motion."},{"Start":"12:20.890 ","End":"12:27.940","Text":"This is the motion of the rod around the axis of rotation."},{"Start":"12:27.940 ","End":"12:30.025","Text":"Then we add in this,"},{"Start":"12:30.025 ","End":"12:34.015","Text":"and then this v_i is relative to the ground,"},{"Start":"12:34.015 ","End":"12:38.575","Text":"so it has the circular motion of the section"},{"Start":"12:38.575 ","End":"12:45.620","Text":"plus the added motion of it moving in a direction as a whole."},{"Start":"12:45.870 ","End":"12:48.009","Text":"If you can\u0027t remember this,"},{"Start":"12:48.009 ","End":"12:55.310","Text":"go back to the lessons on relative velocities and relative motion."},{"Start":"12:55.890 ","End":"13:05.100","Text":"In order to make this the most general way of writing this formula,"},{"Start":"13:05.100 ","End":"13:07.780","Text":"this Omega r_i,"},{"Start":"13:07.780 ","End":"13:11.980","Text":"so this Omega is in fact a vector itself."},{"Start":"13:11.980 ","End":"13:19.300","Text":"We can see that if our Omega is in this direction,"},{"Start":"13:19.300 ","End":"13:22.525","Text":"so the direction of its vector will be"},{"Start":"13:22.525 ","End":"13:27.249","Text":"upwards through the right-hand rule, so it\u0027s a vector."},{"Start":"13:27.249 ","End":"13:32.365","Text":"Then if we\u0027re multiplying these 2 vectors together,"},{"Start":"13:32.365 ","End":"13:36.010","Text":"then we\u0027re going to end up with a cross-product."},{"Start":"13:36.010 ","End":"13:44.830","Text":"We\u0027re going to have Omega cross multiplied by r_i and then plus our V_0."},{"Start":"13:46.560 ","End":"13:51.355","Text":"This is the more general form because in this way,"},{"Start":"13:51.355 ","End":"13:53.379","Text":"you\u0027ll see that you\u0027ll anyway,"},{"Start":"13:53.379 ","End":"13:58.795","Text":"get the size of Omega multiplied by r_i."},{"Start":"13:58.795 ","End":"14:02.410","Text":"Because here, if you\u0027re doing the cross multiplication specifically here,"},{"Start":"14:02.410 ","End":"14:05.955","Text":"the degree between them is 90."},{"Start":"14:05.955 ","End":"14:07.499","Text":"It\u0027s at a right angle,"},{"Start":"14:07.499 ","End":"14:11.039","Text":"which means that we\u0027ll have Omega multiplied by r,"},{"Start":"14:11.039 ","End":"14:12.990","Text":"the size is multiplied by sin of"},{"Start":"14:12.990 ","End":"14:16.734","Text":"the angle so we would have gotten the same as what we have here,"},{"Start":"14:16.734 ","End":"14:19.510","Text":"but this is in the more general form."},{"Start":"14:19.510 ","End":"14:23.810","Text":"This is what we\u0027re going to use in order to derive this equation."},{"Start":"14:23.940 ","End":"14:29.755","Text":"Lets substitute this into our equation over here."},{"Start":"14:29.755 ","End":"14:37.045","Text":"What we will get is our E_k is going to be equal to"},{"Start":"14:37.045 ","End":"14:45.175","Text":"the sum and all of the i\u0027s of 1/2 multiplied by m_i multiplied by V_i^2,"},{"Start":"14:45.175 ","End":"14:49.150","Text":"so I\u0027ll put this in everything."},{"Start":"14:49.150 ","End":"14:54.145","Text":"Omega cross multiplied with r_i"},{"Start":"14:54.145 ","End":"15:00.715","Text":"plus V_o and all of this squared."},{"Start":"15:00.715 ","End":"15:04.675","Text":"Let\u0027s see how to solve this."},{"Start":"15:04.675 ","End":"15:10.285","Text":"What we\u0027re going to do is we\u0027re going to expand out the brackets. Let\u0027s see."},{"Start":"15:10.285 ","End":"15:17.630","Text":"We write the sum on i multiplied by 1/2, multiplied by m_i."},{"Start":"15:18.630 ","End":"15:24.500","Text":"Then this expression squared is just going to be the size."},{"Start":"15:24.900 ","End":"15:29.170","Text":"Whether there\u0027s this angle or they\u0027re perpendicular to each other,"},{"Start":"15:29.170 ","End":"15:30.834","Text":"it doesn\u0027t really matter."},{"Start":"15:30.834 ","End":"15:32.859","Text":"You\u0027re going to end up,"},{"Start":"15:32.859 ","End":"15:37.540","Text":"no matter which way you look at it with Omega^2 multiplied"},{"Start":"15:37.540 ","End":"15:42.925","Text":"by r_i^2 and it\u0027s going to be the size of them without the vector quantities."},{"Start":"15:42.925 ","End":"15:53.410","Text":"Then we\u0027re going to have plus 2 times Omega cross multiplied by r vector."},{"Start":"15:53.410 ","End":"15:55.765","Text":"Then because we\u0027re working with vectors,"},{"Start":"15:55.765 ","End":"16:04.010","Text":"this is going to be scalar multiplication with our V_o vector."},{"Start":"16:04.010 ","End":"16:09.580","Text":"Then we\u0027re going to add in our V_0,"},{"Start":"16:09.580 ","End":"16:13.735","Text":"and again it\u0027s just the size without a vector because it\u0027s squared,"},{"Start":"16:13.735 ","End":"16:19.190","Text":"and then it\u0027s going to be squared."},{"Start":"16:22.590 ","End":"16:30.790","Text":"Let\u0027s carry on. Let\u0027s write this out."},{"Start":"16:30.790 ","End":"16:33.460","Text":"I\u0027ll take my 1/2 out,"},{"Start":"16:33.460 ","End":"16:37.660","Text":"and then I\u0027ll write the sum on our is,"},{"Start":"16:37.660 ","End":"16:44.544","Text":"and then we\u0027ll write Omega squared,"},{"Start":"16:44.544 ","End":"16:48.205","Text":"then multiplied by my m_i,"},{"Start":"16:48.205 ","End":"16:53.740","Text":"multiplied by my r_i^2."},{"Start":"16:53.740 ","End":"16:57.474","Text":"Now, what this works out to be,"},{"Start":"16:57.474 ","End":"17:04.870","Text":"it\u0027s going to be our 1/2 multiplied by a moment of inertia our I around the origin,"},{"Start":"17:04.870 ","End":"17:07.809","Text":"so our I_0,"},{"Start":"17:07.809 ","End":"17:11.496","Text":"multiplied by Omega squared,"},{"Start":"17:11.496 ","End":"17:16.809","Text":"because this makes up our I_0 and then multiplied by Omega squared,"},{"Start":"17:16.809 ","End":"17:21.350","Text":"and this is around the origin."},{"Start":"17:21.710 ","End":"17:28.260","Text":"In other words, this is the normal energy that we would have had if our axis was fixed."},{"Start":"17:28.260 ","End":"17:30.165","Text":"See over here."},{"Start":"17:30.165 ","End":"17:31.920","Text":"If our axis was fixed,"},{"Start":"17:31.920 ","End":"17:34.360","Text":"this is the energy that we would get."},{"Start":"17:34.410 ","End":"17:38.695","Text":"Now I\u0027m going to write out the third expression,"},{"Start":"17:38.695 ","End":"17:42.880","Text":"I\u0027m not yet looking at this with our V_0^2."},{"Start":"17:42.880 ","End":"17:48.130","Text":"Plus and then we\u0027re going to have again 1/2 on the outside,"},{"Start":"17:48.130 ","End":"17:57.115","Text":"and then sum on all the is of m_i multiplied by v_0^2."},{"Start":"17:57.115 ","End":"17:59.680","Text":"Let\u0027s take a look at what this means."},{"Start":"17:59.680 ","End":"18:02.799","Text":"Our V_0^2 is going to be the same for all of"},{"Start":"18:02.799 ","End":"18:06.160","Text":"the pieces because it\u0027s just the motion of the axes,"},{"Start":"18:06.160 ","End":"18:10.105","Text":"so all of the pieces have this v_0 as well."},{"Start":"18:10.105 ","End":"18:17.919","Text":"That means I can take it out and write it out before my Sigma."},{"Start":"18:17.919 ","End":"18:23.754","Text":"Then that means that I\u0027m just summing up all of the small masses,"},{"Start":"18:23.754 ","End":"18:25.600","Text":"all of the masses of the different sections,"},{"Start":"18:25.600 ","End":"18:29.815","Text":"which means that I\u0027m dealing with the total mass of the body,"},{"Start":"18:29.815 ","End":"18:32.349","Text":"which means that I\u0027m going to be left"},{"Start":"18:32.349 ","End":"18:40.540","Text":"with 1/2m total multiplied by V_0^2."},{"Start":"18:40.540 ","End":"18:44.104","Text":"This is just the normal equation for kinetic energy."},{"Start":"18:44.104 ","End":"18:48.269","Text":"Notice that when we were dealing with our center of mass,"},{"Start":"18:48.269 ","End":"18:54.660","Text":"so here was written just plus 1/2m total V_cm^2."},{"Start":"18:54.660 ","End":"18:57.880","Text":"Let\u0027s take a look."},{"Start":"19:00.840 ","End":"19:05.630","Text":"What\u0027s written in this section is the same as here."},{"Start":"19:08.310 ","End":"19:12.279","Text":"The only reason that we\u0027re not doing with V_cm^2 is"},{"Start":"19:12.279 ","End":"19:15.640","Text":"because here it was moving with its v center of mass,"},{"Start":"19:15.640 ","End":"19:19.869","Text":"and here it\u0027s moving at some other point on the origin."},{"Start":"19:19.869 ","End":"19:25.030","Text":"We have the 3rd expression to add onto here,"},{"Start":"19:25.030 ","End":"19:27.145","Text":"and soon we\u0027ll speak about why"},{"Start":"19:27.145 ","End":"19:31.165","Text":"this 3rd expression doesn\u0027t appear when we\u0027re dealing with the center of mass."},{"Start":"19:31.165 ","End":"19:34.839","Text":"When the object is rotating around its center of mass,"},{"Start":"19:34.839 ","End":"19:39.489","Text":"this expression doesn\u0027t need to be added,"},{"Start":"19:39.489 ","End":"19:41.155","Text":"but here it does,"},{"Start":"19:41.155 ","End":"19:44.510","Text":"so let\u0027s add it."},{"Start":"19:47.310 ","End":"19:53.185","Text":"What I\u0027m adding is the sum."},{"Start":"19:53.185 ","End":"19:56.635","Text":"As we can see, I have 1/2 and then multiplied by 2,"},{"Start":"19:56.635 ","End":"19:58.450","Text":"so that cancels up."},{"Start":"19:58.450 ","End":"20:03.990","Text":"I have the sum of m_i and then multiplied"},{"Start":"20:03.990 ","End":"20:10.750","Text":"by my Omega cross multiplied by my r_i,"},{"Start":"20:10.750 ","End":"20:17.270","Text":"and then this scalar multiplication with my V_0."},{"Start":"20:18.390 ","End":"20:25.840","Text":"There\u0027s a little trick when we have this formation because as you can see,"},{"Start":"20:25.840 ","End":"20:28.810","Text":"my r_i and my m_i are dependent on i,"},{"Start":"20:28.810 ","End":"20:32.785","Text":"so I have to keep them after my Sigma,"},{"Start":"20:32.785 ","End":"20:39.550","Text":"but my Omega and my V_0 are the same for every single piece in the body,"},{"Start":"20:39.550 ","End":"20:41.260","Text":"so I can move them out of my Sigma,"},{"Start":"20:41.260 ","End":"20:42.745","Text":"so that\u0027s what I want to do."},{"Start":"20:42.745 ","End":"20:47.110","Text":"Let\u0027s see how I do that. If I have a case of a cross"},{"Start":"20:47.110 ","End":"20:53.965","Text":"multiplied by b and then scalar multiplied by c,"},{"Start":"20:53.965 ","End":"20:59.080","Text":"I can rewrite this as c cross multiplied"},{"Start":"20:59.080 ","End":"21:05.870","Text":"by a and then scalar multiplied by b."},{"Start":"21:06.900 ","End":"21:10.630","Text":"If I use this little trick,"},{"Start":"21:10.630 ","End":"21:12.609","Text":"then I can rewrite"},{"Start":"21:12.609 ","End":"21:14.449","Text":"this"},{"Start":"21:22.080 ","End":"21:25.330","Text":"as the sum"},{"Start":"21:25.330 ","End":"21:30.384","Text":"on my i\u0027s of m_i multiplied by,"},{"Start":"21:30.384 ","End":"21:40.975","Text":"then I just have V_0 instead of my c cross multiplied by my Omega,"},{"Start":"21:40.975 ","End":"21:46.280","Text":"and then scalar multiplied by my r_i."},{"Start":"21:47.010 ","End":"21:54.189","Text":"Then that means that I can take out this."},{"Start":"21:54.189 ","End":"22:00.625","Text":"I can have V_0 cross multiplied by my Omega,"},{"Start":"22:00.625 ","End":"22:10.149","Text":"and then Sigma i m_i multiplied by r_i."},{"Start":"22:10.149 ","End":"22:17.425","Text":"Then with my Sigma m_i multiplied by r_i vector."},{"Start":"22:17.425 ","End":"22:24.175","Text":"If you remember what the equation is for my r_cm,"},{"Start":"22:24.175 ","End":"22:30.235","Text":"it equals to the sum of m_i multiplied by"},{"Start":"22:30.235 ","End":"22:37.540","Text":"r_i vector divided by the sum of all of my masses."},{"Start":"22:37.540 ","End":"22:41.089","Text":"This is equal to obviously my total mass."},{"Start":"22:43.350 ","End":"22:52.970","Text":"As we can see, this then is going to equal to my total mass multiplied by my r_cm."},{"Start":"22:53.730 ","End":"22:55.915","Text":"Let\u0027s write this here."},{"Start":"22:55.915 ","End":"23:05.499","Text":"This will equal to V_0 cross multiplied by Omega."},{"Start":"23:05.499 ","End":"23:12.149","Text":"Then it\u0027s going to be multiplied by my m total"},{"Start":"23:12.149 ","End":"23:19.995","Text":"multiplied by my r_cm."},{"Start":"23:19.995 ","End":"23:23.910","Text":"Let\u0027s write down the whole expression for"},{"Start":"23:23.910 ","End":"23:28.720","Text":"the energy in a neat way so that we can understand what\u0027s happening."},{"Start":"23:29.480 ","End":"23:36.540","Text":"Finally, we\u0027ll get that our kinetic energy is equal to"},{"Start":"23:36.540 ","End":"23:43.934","Text":"1/2 I_o Omega^2 plus"},{"Start":"23:43.934 ","End":"23:53.085","Text":"1\\2 m total V_0^2 plus this weird thing,"},{"Start":"23:53.085 ","End":"24:01.080","Text":"which is m total multiplied by our r_cm scalar multiplication"},{"Start":"24:01.080 ","End":"24:10.270","Text":"with a V_o vector cross multiplied by r Omega vector."},{"Start":"24:10.790 ","End":"24:14.744","Text":"Here we have our equation for kinetic energy."},{"Start":"24:14.744 ","End":"24:19.701","Text":"Notice that in my 3rd expression,"},{"Start":"24:19.701 ","End":"24:21.690","Text":"so while we were working out over here,"},{"Start":"24:21.690 ","End":"24:24.704","Text":"my r_cm vector, what does it mean?"},{"Start":"24:24.704 ","End":"24:29.310","Text":"It denotes the location of the center of mass relative to the axis."},{"Start":"24:29.310 ","End":"24:34.140","Text":"That doesn\u0027t mean relative to somewhere in the lab or"},{"Start":"24:34.140 ","End":"24:39.225","Text":"to the ground or to some random axis."},{"Start":"24:39.225 ","End":"24:42.464","Text":"It\u0027s specifically the location of the center of"},{"Start":"24:42.464 ","End":"24:46.155","Text":"mass of the body relative to the axis of rotation."},{"Start":"24:46.155 ","End":"24:55.485","Text":"That\u0027s r_cm and this equation is to be used when the body is rotating about an axis."},{"Start":"24:55.485 ","End":"24:57.450","Text":"At any point, the center of mass,"},{"Start":"24:57.450 ","End":"24:59.789","Text":"the edge just off the center of mass,"},{"Start":"24:59.789 ","End":"25:04.556","Text":"anywhere and simultaneously the axis is moving,"},{"Start":"25:04.556 ","End":"25:06.225","Text":"so the axis isn\u0027t fixed."},{"Start":"25:06.225 ","End":"25:14.940","Text":"Notice that this equation is pretty much the same aside from this weird thing here."},{"Start":"25:14.940 ","End":"25:18.960","Text":"This is only because we are saying that"},{"Start":"25:18.960 ","End":"25:22.574","Text":"our center of mass is not where our axis of rotation is."},{"Start":"25:22.574 ","End":"25:25.950","Text":"If our center of mass was our axis of rotation,"},{"Start":"25:25.950 ","End":"25:29.115","Text":"then our r_cm would equal to 0."},{"Start":"25:29.115 ","End":"25:33.885","Text":"Because the distance between the axis of rotation and the center of mass will be 0,"},{"Start":"25:33.885 ","End":"25:38.580","Text":"which means that this whole thing over here would disappear."},{"Start":"25:38.580 ","End":"25:45.659","Text":"Then obviously instead of r_i origin and our v origin,"},{"Start":"25:45.659 ","End":"25:48.150","Text":"so we\u0027ll have our r_cm and our V_cm."},{"Start":"25:48.150 ","End":"25:52.980","Text":"Because our origin is our axis,"},{"Start":"25:52.980 ","End":"25:55.169","Text":"so around our axis of rotation."},{"Start":"25:55.169 ","End":"25:59.339","Text":"Obviously, if our axis of rotation is our center of mass,"},{"Start":"25:59.339 ","End":"26:02.265","Text":"so we can just replace our o with cm."},{"Start":"26:02.265 ","End":"26:05.159","Text":"We get the exact same expression that we"},{"Start":"26:05.159 ","End":"26:13.180","Text":"got over here when speaking about solely the center of mass."},{"Start":"26:13.220 ","End":"26:17.549","Text":"Go through this video again and write"},{"Start":"26:17.549 ","End":"26:23.740","Text":"down everything that\u0027s in blue and whatever I squared in red."},{"Start":"26:24.170 ","End":"26:32.380","Text":"What I\u0027m going to do is I\u0027m going to do a little example of how to use this equation."},{"Start":"26:33.050 ","End":"26:40.350","Text":"Imagine that we have some rod, like so,"},{"Start":"26:40.350 ","End":"26:43.979","Text":"and its center of mass is over here"},{"Start":"26:43.979 ","End":"26:49.976","Text":"and it\u0027s origin which is its axis of rotation, is over here."},{"Start":"26:49.976 ","End":"26:59.979","Text":"Of course, the axis of rotation is moving forwards at some V_0 in this direction."},{"Start":"27:00.860 ","End":"27:09.970","Text":"The rod is of course also rotating around this axis of rotation in this direction."},{"Start":"27:11.570 ","End":"27:14.655","Text":"Then what we\u0027re going to get as"},{"Start":"27:14.655 ","End":"27:19.390","Text":"the motion continues is something looking a bit like this."},{"Start":"27:19.640 ","End":"27:24.104","Text":"What\u0027s happening is that as the rod moves forwards,"},{"Start":"27:24.104 ","End":"27:26.519","Text":"it\u0027s also simultaneously rotating."},{"Start":"27:26.519 ","End":"27:28.530","Text":"The important things to notice is that"},{"Start":"27:28.530 ","End":"27:31.800","Text":"the axis of rotation carries on in a straight line."},{"Start":"27:31.800 ","End":"27:33.795","Text":"It\u0027s not moving."},{"Start":"27:33.795 ","End":"27:37.150","Text":"Still with V_0."},{"Start":"27:39.110 ","End":"27:42.255","Text":"It\u0027s always in the same position."},{"Start":"27:42.255 ","End":"27:48.015","Text":"It\u0027s never moving. Then the center of mass is still in the same place."},{"Start":"27:48.015 ","End":"27:52.184","Text":"This is the center of mass and just like all the rest of the points,"},{"Start":"27:52.184 ","End":"27:57.765","Text":"it\u0027s moving, because it\u0027s not the axis of rotation."},{"Start":"27:57.765 ","End":"28:04.935","Text":"We\u0027re going to assume that our V_0 is given and that\u0027s so as our Omega."},{"Start":"28:04.935 ","End":"28:10.679","Text":"Let\u0027s say that the distance our center of masses from our axis of"},{"Start":"28:10.679 ","End":"28:18.345","Text":"rotation is d and obviously our Omega our rotation is into the page."},{"Start":"28:18.345 ","End":"28:24.495","Text":"Let\u0027s see now how we work out the kinetic energy."},{"Start":"28:24.495 ","End":"28:27.330","Text":"First what I\u0027m going to do is I want to work out"},{"Start":"28:27.330 ","End":"28:30.750","Text":"the kinetic energy at every single moment in time."},{"Start":"28:30.750 ","End":"28:35.145","Text":"That means that as my rod rotates."},{"Start":"28:35.145 ","End":"28:39.120","Text":"If I say that this angle here is Theta,"},{"Start":"28:39.120 ","End":"28:44.355","Text":"so I\u0027ll workout my kinetic energy as a function of Theta."},{"Start":"28:44.355 ","End":"28:46.620","Text":"Let\u0027s see how I do this."},{"Start":"28:46.620 ","End":"28:48.840","Text":"Going from this equation up here,"},{"Start":"28:48.840 ","End":"28:54.254","Text":"so I\u0027ll have 1/2 multiplied by I_o,"},{"Start":"28:54.254 ","End":"28:59.880","Text":"which I can work out easily and also through Steiner multiplied by"},{"Start":"28:59.880 ","End":"29:06.225","Text":"my Omega^2 plus 1/2 the total mass."},{"Start":"29:06.225 ","End":"29:10.829","Text":"Let\u0027s say that that\u0027s M. The total mass of the rod is"},{"Start":"29:10.829 ","End":"29:17.114","Text":"M multiplied by V_0 ^2 plus."},{"Start":"29:17.114 ","End":"29:19.875","Text":"We\u0027re doing this section."},{"Start":"29:19.875 ","End":"29:24.479","Text":"My total mass which is capital M and"},{"Start":"29:24.479 ","End":"29:30.250","Text":"my r_cm which is the location of the center of mass relative to the axis of rotation."},{"Start":"29:30.590 ","End":"29:33.870","Text":"We know that the size of this is going to be"},{"Start":"29:33.870 ","End":"29:42.300","Text":"d. My V_0 cross, my Omega."},{"Start":"29:42.300 ","End":"29:45.720","Text":"You can see that my V_0 is pointing in the right direction,"},{"Start":"29:45.720 ","End":"29:50.969","Text":"say on the x-axis and my Omega is into the page which"},{"Start":"29:50.969 ","End":"29:56.849","Text":"means that they\u0027re perpendicular 90 degrees to one another."},{"Start":"29:56.849 ","End":"30:04.000","Text":"What I\u0027ll get is multiplied by V_0 Omega."},{"Start":"30:04.550 ","End":"30:10.590","Text":"The direction of this vector is going to be upwards."},{"Start":"30:10.590 ","End":"30:19.845","Text":"Our vector of V_0 cross multiplied by Omega is in the upwards direction."},{"Start":"30:19.845 ","End":"30:24.870","Text":"You can see that by using the right-hand rule."},{"Start":"30:24.870 ","End":"30:27.809","Text":"We have that in the upwards direction."},{"Start":"30:27.809 ","End":"30:34.035","Text":"What we have to do is we have to find the direction of all of this."},{"Start":"30:34.035 ","End":"30:37.260","Text":"Here we know that the direction is upwards,"},{"Start":"30:37.260 ","End":"30:41.290","Text":"our direction of r_cm."},{"Start":"30:41.360 ","End":"30:48.975","Text":"Because I know the direction of this vector and I know that my r_cm is this d,"},{"Start":"30:48.975 ","End":"30:55.575","Text":"which means that my r_cm is this vector over here."},{"Start":"30:55.575 ","End":"30:58.875","Text":"This is r_cm."},{"Start":"30:58.875 ","End":"31:00.944","Text":"You can\u0027t really see it."},{"Start":"31:00.944 ","End":"31:04.469","Text":"But this blue arrow over here is my r_cm vector."},{"Start":"31:04.469 ","End":"31:10.275","Text":"Because I have a scalar multiplication between my r_cm and this vector."},{"Start":"31:10.275 ","End":"31:13.620","Text":"What I have to do is I have to multiply this expression,"},{"Start":"31:13.620 ","End":"31:15.389","Text":"which is the size of the vector,"},{"Start":"31:15.389 ","End":"31:19.155","Text":"multiplied by cosine of the angle"},{"Start":"31:19.155 ","End":"31:26.099","Text":"between my r_cm and my v_0 cross Omega."},{"Start":"31:26.099 ","End":"31:30.704","Text":"The angle between this blue arrow and this red arrow,"},{"Start":"31:30.704 ","End":"31:33.689","Text":"cosine of that angle and I multiply"},{"Start":"31:33.689 ","End":"31:37.869","Text":"that here in order to get the direction of the vector."},{"Start":"31:38.240 ","End":"31:43.259","Text":"If we have a look over here,"},{"Start":"31:43.259 ","End":"31:49.965","Text":"V_0 cross Omega is still be in this direction over here."},{"Start":"31:49.965 ","End":"31:53.400","Text":"This is also this, the red arrow."},{"Start":"31:53.400 ","End":"31:59.104","Text":"The blue arrow denoting my r_cm is going to be just this over here."},{"Start":"31:59.104 ","End":"32:04.740","Text":"From my axes of rotation until my center of mass."},{"Start":"32:04.740 ","End":"32:11.175","Text":"We can see that the angle between my red arrow and my blue arrow is exactly Theta."},{"Start":"32:11.175 ","End":"32:14.475","Text":"All I have to do is multiply by this,"},{"Start":"32:14.475 ","End":"32:18.495","Text":"by cosine of the angle, which is cos(Theta)."},{"Start":"32:18.495 ","End":"32:26.080","Text":"This is my kinetic energy at every single point in this system."},{"Start":"32:26.330 ","End":"32:32.189","Text":"The trick here was to just write this out."},{"Start":"32:32.189 ","End":"32:38.615","Text":"Then in the last section to write out the size of each of the vectors."},{"Start":"32:38.615 ","End":"32:44.239","Text":"Our r_cm we knew that the size was d and then to work out the size of this."},{"Start":"32:44.239 ","End":"32:46.174","Text":"Because they were perpendicular,"},{"Start":"32:46.174 ","End":"32:50.129","Text":"it was just a V_0 multiplied by Omega."},{"Start":"32:50.129 ","End":"32:52.739","Text":"Then to remember that because this is"},{"Start":"32:52.739 ","End":"32:56.430","Text":"a vector quantity that we have to also write the direction and"},{"Start":"32:56.430 ","End":"33:03.204","Text":"to remember that when we have a vector dot product with another vector,"},{"Start":"33:03.204 ","End":"33:07.174","Text":"so we have to multiply it by cosine of the angle."},{"Start":"33:07.174 ","End":"33:12.920","Text":"Then the trick is to see where our 2 vectors are pointing and"},{"Start":"33:12.920 ","End":"33:18.458","Text":"finding the angle between them and then multiplying by cosine of that angle."},{"Start":"33:18.458 ","End":"33:20.709","Text":"That\u0027s the end of this lesson."}],"ID":9407},{"Watched":false,"Name":"Exercise 1","Duration":"15m 50s","ChapterTopicVideoID":9138,"CourseChapterTopicPlaylistID":5407,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:03.030","Text":"Hello. In this question,"},{"Start":"00:03.030 ","End":"00:06.030","Text":"we\u0027re being told that a rod of length capital L"},{"Start":"00:06.030 ","End":"00:09.480","Text":"and mass capital M is attached to the ceiling."},{"Start":"00:09.480 ","End":"00:14.505","Text":"The rod rotates at an initial angular velocity of Omega 0."},{"Start":"00:14.505 ","End":"00:18.225","Text":"What is the maximum angle which the rod will reach?"},{"Start":"00:18.225 ","End":"00:20.745","Text":"As we can see, here\u0027s our rod,"},{"Start":"00:20.745 ","End":"00:26.340","Text":"and it has a length L and mass capital M,"},{"Start":"00:26.340 ","End":"00:29.370","Text":"and here is its axis of rotation."},{"Start":"00:29.370 ","End":"00:32.010","Text":"This point here doesn\u0027t move,"},{"Start":"00:32.010 ","End":"00:35.575","Text":"and the rod rotates around this."},{"Start":"00:35.575 ","End":"00:41.240","Text":"Its angular velocity is Omega 0 at the beginning,"},{"Start":"00:41.240 ","End":"00:47.700","Text":"and we\u0027re being asked what is the maximum angle that the rod will reach."},{"Start":"00:47.700 ","End":"00:51.245","Text":"Let\u0027s take a look at how we\u0027re going to do this."},{"Start":"00:51.245 ","End":"00:55.070","Text":"The first thing that we\u0027re going to do is we\u0027re going to label"},{"Start":"00:55.070 ","End":"01:00.870","Text":"some arbitrary angle, over here,"},{"Start":"01:00.870 ","End":"01:06.080","Text":"and we\u0027re going to imagine that this angle is"},{"Start":"01:06.080 ","End":"01:12.240","Text":"Theta and this is the maximum angle which the rod reaches."},{"Start":"01:13.340 ","End":"01:16.865","Text":"This is the angle that we\u0027re trying to find."},{"Start":"01:16.865 ","End":"01:22.550","Text":"Now the way that we\u0027re going to solve this question is by using the fact that all of"},{"Start":"01:22.550 ","End":"01:28.530","Text":"the forces here are conservation forces."},{"Start":"01:28.530 ","End":"01:32.970","Text":"We\u0027re going to use the idea of the conservation of energy."},{"Start":"01:33.020 ","End":"01:35.220","Text":"What forces do we have?"},{"Start":"01:35.220 ","End":"01:36.540","Text":"We have mg,"},{"Start":"01:36.540 ","End":"01:42.035","Text":"okay and then we have some force acting over here at the axis of rotation."},{"Start":"01:42.035 ","End":"01:45.754","Text":"Now, because this point isn\u0027t moving,"},{"Start":"01:45.754 ","End":"01:49.350","Text":"so no work is being done here,"},{"Start":"01:49.350 ","End":"01:51.240","Text":"so it\u0027s still okay,"},{"Start":"01:51.240 ","End":"01:55.890","Text":"so we can use the conservation of energy in order to solve this question."},{"Start":"01:57.070 ","End":"02:02.040","Text":"Let\u0027s begin by solving this."},{"Start":"02:02.750 ","End":"02:07.220","Text":"We\u0027re going to say that our energy at the beginning,"},{"Start":"02:07.220 ","End":"02:11.255","Text":"so E_i is going to be equal to,"},{"Start":"02:11.255 ","End":"02:17.790","Text":"as per usual, our kinetic energy plus our potential energy."},{"Start":"02:17.790 ","End":"02:22.250","Text":"The kinetic energy of the system plus the potential energy of the system."},{"Start":"02:22.250 ","End":"02:26.330","Text":"Now, kinetic energy at the beginning,"},{"Start":"02:26.330 ","End":"02:28.205","Text":"so let\u0027s take a look."},{"Start":"02:28.205 ","End":"02:32.465","Text":"A rod is rotating about a fixed axis,"},{"Start":"02:32.465 ","End":"02:37.490","Text":"so that means that we have to use the equation that we looked at from"},{"Start":"02:37.490 ","End":"02:45.950","Text":"our previous lesson for something with angular velocity and a fixed axis."},{"Start":"02:45.950 ","End":"02:53.585","Text":"The equation for that is equal to 1/2I_0, so not ICM,"},{"Start":"02:53.585 ","End":"02:56.720","Text":"because the axis of rotation isn\u0027t at the center of mass,"},{"Start":"02:56.720 ","End":"02:59.240","Text":"but rather over here at the origin,"},{"Start":"02:59.240 ","End":"03:06.419","Text":"we can say, multiplied by the angular velocity squared."},{"Start":"03:07.040 ","End":"03:12.910","Text":"Then our I_0 is going to be equal to"},{"Start":"03:12.910 ","End":"03:18.700","Text":"now because the axis of rotation is at the edge and not at the center of mass."},{"Start":"03:18.700 ","End":"03:25.215","Text":"It\u0027s going to be 1/3ML^2."},{"Start":"03:25.215 ","End":"03:28.600","Text":"We mean to know this equation off by heart, and of course,"},{"Start":"03:28.600 ","End":"03:31.375","Text":"if the axis of rotation was in the center,"},{"Start":"03:31.375 ","End":"03:34.390","Text":"then here it would be 1 divided by 12."},{"Start":"03:34.390 ","End":"03:37.960","Text":"In order to know how to work this out,"},{"Start":"03:37.960 ","End":"03:42.560","Text":"you can go back to the chapter on moments of inertia."},{"Start":"03:42.650 ","End":"03:45.725","Text":"Now we have our E_k,"},{"Start":"03:45.725 ","End":"03:49.760","Text":"so now we have to see what our potential energy is."},{"Start":"03:51.440 ","End":"03:55.125","Text":"Let\u0027s write out, our U."},{"Start":"03:55.125 ","End":"04:04.690","Text":"As we know, our U is equal to Mg multiplied by the height h. Now,"},{"Start":"04:04.690 ","End":"04:06.880","Text":"an important thing over here to note,"},{"Start":"04:06.880 ","End":"04:08.740","Text":"is that when we\u0027re speaking about our hL,"},{"Start":"04:08.740 ","End":"04:12.490","Text":"we\u0027re actually speaking about the height of the center of mass."},{"Start":"04:12.490 ","End":"04:14.470","Text":"This is really important."},{"Start":"04:14.470 ","End":"04:16.655","Text":"So I\u0027m going to label over here,"},{"Start":"04:16.655 ","End":"04:18.850","Text":"here\u0027s our center of mass,"},{"Start":"04:18.850 ","End":"04:21.800","Text":"right in the center of the rod."},{"Start":"04:23.030 ","End":"04:26.950","Text":"The reason for it being the center of mass,"},{"Start":"04:26.950 ","End":"04:29.860","Text":"you have to derive the equation, it doesn\u0027t really matter,"},{"Start":"04:29.860 ","End":"04:34.430","Text":"but you\u0027ll see that when you derive the equation,"},{"Start":"04:34.430 ","End":"04:42.160","Text":"it\u0027s going to end up that the potential energy is located where the center of masses."},{"Start":"04:42.160 ","End":"04:44.540","Text":"If we do this,"},{"Start":"04:44.540 ","End":"04:47.140","Text":"so right at the beginning,"},{"Start":"04:47.140 ","End":"04:50.850","Text":"our center of mass is over here,"},{"Start":"04:50.850 ","End":"04:54.700","Text":"so it\u0027s located at L over 2."},{"Start":"04:54.800 ","End":"05:01.895","Text":"Because it\u0027s right in the center and our rod is of length L. Right at the beginning,"},{"Start":"05:01.895 ","End":"05:05.300","Text":"our initial potential energy,"},{"Start":"05:05.300 ","End":"05:11.360","Text":"is going to equal to mass times gravity times our h_cm,"},{"Start":"05:11.360 ","End":"05:13.920","Text":"which is L over 2."},{"Start":"05:13.920 ","End":"05:16.380","Text":"This is very important,"},{"Start":"05:16.380 ","End":"05:18.480","Text":"because this changes everything."},{"Start":"05:18.480 ","End":"05:25.025","Text":"Then we can say that this height over here right at the bottom is where H is equal to 0."},{"Start":"05:25.025 ","End":"05:27.155","Text":"Our center of mass is at L over 2."},{"Start":"05:27.155 ","End":"05:31.320","Text":"This is really critical to answering this question."},{"Start":"05:32.030 ","End":"05:37.715","Text":"Now we can substitute all of this in to our initial energy."},{"Start":"05:37.715 ","End":"05:41.440","Text":"Now we\u0027re going to find our energy at the end."},{"Start":"05:41.440 ","End":"05:43.744","Text":"Our E final."},{"Start":"05:43.744 ","End":"05:50.370","Text":"So this is again going to equal to our E_k so here\u0027s the initial,"},{"Start":"05:50.900 ","End":"05:57.400","Text":"our final kinetic energy plus our final potential energy."},{"Start":"05:57.400 ","End":"06:01.190","Text":"Let\u0027s take a look at our final kinetic energy."},{"Start":"06:01.190 ","End":"06:04.805","Text":"At the end, it\u0027s going to be at its maximum angle,"},{"Start":"06:04.805 ","End":"06:09.275","Text":"which means that it\u0027s going to rotate to the maximum angle,"},{"Start":"06:09.275 ","End":"06:11.675","Text":"stop here for a second and go back down."},{"Start":"06:11.675 ","End":"06:13.805","Text":"Which means that right at the end,"},{"Start":"06:13.805 ","End":"06:16.610","Text":"our kinetic energy is 0 because there\u0027s"},{"Start":"06:16.610 ","End":"06:20.410","Text":"no movement because it\u0027s stationary and then it goes back down."},{"Start":"06:20.410 ","End":"06:23.505","Text":"That was easy, and then"},{"Start":"06:23.505 ","End":"06:28.385","Text":"our final potential energy is again"},{"Start":"06:28.385 ","End":"06:34.270","Text":"going to be mass times gravity times the height of the center of mass."},{"Start":"06:34.270 ","End":"06:37.490","Text":"Obviously, the highest of our center of mass is going to"},{"Start":"06:37.490 ","End":"06:41.740","Text":"be different this time. Let\u0027s take a look."},{"Start":"06:41.740 ","End":"06:43.530","Text":"Right at the beginning,"},{"Start":"06:43.530 ","End":"06:47.405","Text":"the height of our center of mass was all of this."},{"Start":"06:47.405 ","End":"06:50.225","Text":"However, once it\u0027s rotated,"},{"Start":"06:50.225 ","End":"06:53.850","Text":"our center of mass is going to be over here."},{"Start":"06:54.320 ","End":"07:00.635","Text":"Now what we have to do is we have to find this height over here,"},{"Start":"07:00.635 ","End":"07:01.715","Text":"and as you can see,"},{"Start":"07:01.715 ","End":"07:03.685","Text":"it\u0027s a different height."},{"Start":"07:03.685 ","End":"07:10.595","Text":"Now what we want to do is we want to find this difference over here."},{"Start":"07:10.595 ","End":"07:14.015","Text":"Because up until this dotted line,"},{"Start":"07:14.015 ","End":"07:16.975","Text":"we know that the height is L divided by 2."},{"Start":"07:16.975 ","End":"07:20.990","Text":"Now we want to find what this height is over here,"},{"Start":"07:20.990 ","End":"07:22.775","Text":"what this difference is over here,"},{"Start":"07:22.775 ","End":"07:24.770","Text":"then we\u0027ll add it to L over 2,"},{"Start":"07:24.770 ","End":"07:29.430","Text":"and then we\u0027ll have our h_cm for our U_f."},{"Start":"07:29.930 ","End":"07:33.270","Text":"Now I\u0027m going to work out this section,"},{"Start":"07:33.270 ","End":"07:36.785","Text":"and this is a calculation that is very repetitive,"},{"Start":"07:36.785 ","End":"07:39.530","Text":"so if you just learn how to do this,"},{"Start":"07:39.530 ","End":"07:43.560","Text":"it will really help in an exam situation."},{"Start":"07:44.270 ","End":"07:49.010","Text":"What I\u0027ve done is I\u0027ve just marked this height over here, which is known,"},{"Start":"07:49.010 ","End":"07:50.090","Text":"which is L over 2,"},{"Start":"07:50.090 ","End":"07:54.570","Text":"and now we\u0027re trying to find this height over here."},{"Start":"07:55.670 ","End":"07:59.470","Text":"What I\u0027m going to do is from this point,"},{"Start":"07:59.470 ","End":"08:03.675","Text":"I\u0027m going to draw a perpendicular line to the rod."},{"Start":"08:03.675 ","End":"08:07.005","Text":"It\u0027s going to look something like this,"},{"Start":"08:07.005 ","End":"08:11.190","Text":"and the angle over here is 90 degrees."},{"Start":"08:11.190 ","End":"08:16.195","Text":"From my point of center of mass at the maximum point to the rose,"},{"Start":"08:16.195 ","End":"08:18.805","Text":"I draw a perpendicular line,"},{"Start":"08:18.805 ","End":"08:27.078","Text":"and now what I want to do is I want to find what this length is."},{"Start":"08:27.078 ","End":"08:29.340","Text":"I enlarge the picture."},{"Start":"08:29.340 ","End":"08:31.020","Text":"Now, as we can see,"},{"Start":"08:31.020 ","End":"08:35.670","Text":"this black line over here is obviously the rod just at its maximum point,"},{"Start":"08:35.670 ","End":"08:38.440","Text":"which means that it\u0027s the same length."},{"Start":"08:38.510 ","End":"08:43.890","Text":"That means that this length over"},{"Start":"08:43.890 ","End":"08:51.150","Text":"here is also L over 2 to the center of mass."},{"Start":"08:51.150 ","End":"08:59.050","Text":"That also means that this length over here is also L over 2."},{"Start":"08:59.090 ","End":"09:02.970","Text":"What I want to do is I want to find this length,"},{"Start":"09:02.970 ","End":"09:05.430","Text":"where the black spiral is."},{"Start":"09:05.430 ","End":"09:10.995","Text":"What I\u0027m going to do is I\u0027m going to look at this triangle over here."},{"Start":"09:10.995 ","End":"09:17.910","Text":"I\u0027m drawing this red triangle over here."},{"Start":"09:17.910 ","End":"09:23.220","Text":"What I\u0027m doing is I\u0027m looking at this triangle and obviously here there\u0027s 90 degrees,"},{"Start":"09:23.220 ","End":"09:26.170","Text":"so it\u0027s a right angle triangle."},{"Start":"09:26.360 ","End":"09:30.102","Text":"Let\u0027s call what we\u0027re looking for x."},{"Start":"09:30.102 ","End":"09:32.655","Text":"Now we can look over here."},{"Start":"09:32.655 ","End":"09:35.595","Text":"We can see that this is the right angle triangle,"},{"Start":"09:35.595 ","End":"09:41.270","Text":"the red triangle, and we can see that the hypotenuse is L divided by 2."},{"Start":"09:41.270 ","End":"09:45.510","Text":"Then we also have this angle over here, Theta."},{"Start":"09:45.510 ","End":"09:51.360","Text":"Then we can write down that cosine of"},{"Start":"09:51.360 ","End":"09:57.705","Text":"Theta is equal to adjacent over hypotenuse."},{"Start":"09:57.705 ","End":"09:59.715","Text":"Our adjacent is our x,"},{"Start":"09:59.715 ","End":"10:01.290","Text":"which is what we\u0027re trying to find,"},{"Start":"10:01.290 ","End":"10:06.315","Text":"divided by the hypotenuse which is L over 2,"},{"Start":"10:06.315 ","End":"10:09.760","Text":"so divided by L over 2."},{"Start":"10:10.610 ","End":"10:16.095","Text":"Now, we can rearrange to find our x because this is what we want to find."},{"Start":"10:16.095 ","End":"10:17.730","Text":"We isolate out our x,"},{"Start":"10:17.730 ","End":"10:26.980","Text":"which means that our x is equal to L divided by 2 cosine of Theta."},{"Start":"10:27.980 ","End":"10:30.690","Text":"Now I have my x,"},{"Start":"10:30.690 ","End":"10:33.960","Text":"and now what I want to find is this length over here."},{"Start":"10:33.960 ","End":"10:36.360","Text":"Now because here there is a right angle,"},{"Start":"10:36.360 ","End":"10:41.085","Text":"then that means that this length over here,"},{"Start":"10:41.085 ","End":"10:43.320","Text":"we can call this y,"},{"Start":"10:43.320 ","End":"10:47.930","Text":"is also equal to this y over here."},{"Start":"10:47.930 ","End":"10:51.155","Text":"How am I going to find out what that is?"},{"Start":"10:51.155 ","End":"10:55.250","Text":"I\u0027ll just write down that my height here, so my y,"},{"Start":"10:55.250 ","End":"10:57.950","Text":"which is also equal to my Delta H,"},{"Start":"10:57.950 ","End":"10:59.485","Text":"my change in height."},{"Start":"10:59.485 ","End":"11:02.370","Text":"That is equal to this,"},{"Start":"11:02.370 ","End":"11:06.090","Text":"my L over 2, minus my x."},{"Start":"11:06.090 ","End":"11:09.915","Text":"Then I know what my x is so I have L over 2,"},{"Start":"11:09.915 ","End":"11:11.789","Text":"which is a common factor,"},{"Start":"11:11.789 ","End":"11:18.330","Text":"1 minus cosine of Theta."},{"Start":"11:18.330 ","End":"11:20.745","Text":"I just substituted that in."},{"Start":"11:20.745 ","End":"11:23.820","Text":"This is my change in height."},{"Start":"11:23.820 ","End":"11:26.490","Text":"This is what this arrow is equal to."},{"Start":"11:26.490 ","End":"11:29.355","Text":"In the more general form,"},{"Start":"11:29.355 ","End":"11:32.764","Text":"so I\u0027ll write it down in blue."},{"Start":"11:32.764 ","End":"11:35.940","Text":"Now I showed you how to get to this equation but you should"},{"Start":"11:35.940 ","End":"11:39.465","Text":"write this down in your equation sheets."},{"Start":"11:39.465 ","End":"11:46.785","Text":"My Delta H, the height that I\u0027ve risen if we have some circular motion."},{"Start":"11:46.785 ","End":"11:50.910","Text":"Because here there\u0027s circular motion to get to this section,"},{"Start":"11:50.910 ","End":"11:54.135","Text":"so it equals to the radius."},{"Start":"11:54.135 ","End":"11:58.830","Text":"Here specifically we\u0027re going according to the center of mass."},{"Start":"11:58.830 ","End":"12:04.410","Text":"We know that our center of mass is at a distance of L over"},{"Start":"12:04.410 ","End":"12:09.880","Text":"2 here away from the axis of rotation so that\u0027s this, the L over 2."},{"Start":"12:09.880 ","End":"12:16.200","Text":"Then multiply by 1 minus cosine of Theta."},{"Start":"12:16.200 ","End":"12:20.550","Text":"If we were going according to an end point over here right at the end,"},{"Start":"12:20.550 ","End":"12:24.840","Text":"so the radius of rotation will"},{"Start":"12:24.840 ","End":"12:29.055","Text":"be L. Then we would just substitute an L over here and again,"},{"Start":"12:29.055 ","End":"12:31.980","Text":"1 minus cosine of Theta."},{"Start":"12:31.980 ","End":"12:37.060","Text":"This is useful to write in your formula sheets."},{"Start":"12:37.760 ","End":"12:40.290","Text":"Now I have my Delta h,"},{"Start":"12:40.290 ","End":"12:46.346","Text":"so now I can substitute all of this in to find out my new h_cm."},{"Start":"12:46.346 ","End":"12:54.640","Text":"We have mg and then remember my h_cm is L over 2 plus this section over here."},{"Start":"12:55.280 ","End":"13:02.060","Text":"I\u0027ll have L over 2 plus my y,"},{"Start":"13:02.060 ","End":"13:05.990","Text":"which is going to be L over 2,"},{"Start":"13:05.990 ","End":"13:13.490","Text":"1 minus cosine of Theta."},{"Start":"13:15.650 ","End":"13:19.290","Text":"Now I have all of my ingredients,"},{"Start":"13:19.290 ","End":"13:24.045","Text":"so now let\u0027s work out what our maximum angle is."},{"Start":"13:24.045 ","End":"13:26.680","Text":"Because that\u0027s what we\u0027re looking for."},{"Start":"13:34.520 ","End":"13:40.935","Text":"What we\u0027re going to do is we\u0027re going to use our conservation of energy."},{"Start":"13:40.935 ","End":"13:44.190","Text":"We know that our E_ i is equal to our E_f,"},{"Start":"13:44.190 ","End":"13:48.020","Text":"so let\u0027s substitute everything in."},{"Start":"13:48.020 ","End":"13:51.440","Text":"My E_i is going to be my E_ki,"},{"Start":"13:51.440 ","End":"13:55.860","Text":"which is going to be 1/2 my I_0."},{"Start":"13:55.970 ","End":"13:58.875","Text":"I\u0027ll substitute that in a second."},{"Start":"13:58.875 ","End":"14:03.210","Text":"Omega squared plus my U_i,"},{"Start":"14:03.210 ","End":"14:09.530","Text":"which is plus mg, l over 2."},{"Start":"14:09.530 ","End":"14:13.260","Text":"Then that\u0027s going to equal to my E_f,"},{"Start":"14:13.260 ","End":"14:16.200","Text":"which is my kinetic energy is equal to 0,"},{"Start":"14:16.200 ","End":"14:18.300","Text":"so it\u0027s just going to be equal to this."},{"Start":"14:18.300 ","End":"14:27.810","Text":"Mg, and then multiplied by L divided by 2,"},{"Start":"14:27.810 ","End":"14:35.440","Text":"and then inside the brackets it will be 2 minus cosine of Theta."},{"Start":"14:35.480 ","End":"14:43.155","Text":"Then we can say that this is equal to 1/2 I_ 0 Omega."},{"Start":"14:43.155 ","End":"14:47.265","Text":"This is obviously Omega 0^2,"},{"Start":"14:47.265 ","End":"14:55.575","Text":"which is equal to mg multiplied by L over 2."},{"Start":"14:55.575 ","End":"14:57.285","Text":"Then inside the brackets,"},{"Start":"14:57.285 ","End":"15:02.085","Text":"1 minus cosine of Theta."},{"Start":"15:02.085 ","End":"15:05.310","Text":"Then all I have to do now is isolate out"},{"Start":"15:05.310 ","End":"15:11.190","Text":"my cosine Theta and substitute in what my I_0 is equal to, which is this."},{"Start":"15:11.190 ","End":"15:15.570","Text":"With simple algebra you can do this on a pen and paper,"},{"Start":"15:15.570 ","End":"15:21.000","Text":"but you\u0027ll come to a final answer of cosine of Theta is equal to 1"},{"Start":"15:21.000 ","End":"15:29.260","Text":"minus L Omega 0^2 divided by 3g."},{"Start":"15:30.650 ","End":"15:34.089","Text":"This is the final answer."},{"Start":"15:34.089 ","End":"15:37.290","Text":"Obviously to get Theta the maximum angle,"},{"Start":"15:37.290 ","End":"15:45.600","Text":"you\u0027ll just do Theta is equal to cosine to the negative 1 of this expression over here."},{"Start":"15:45.600 ","End":"15:48.360","Text":"That\u0027s how you solve this question."},{"Start":"15:48.360 ","End":"15:50.680","Text":"That\u0027s the end of the lesson."}],"ID":9408},{"Watched":false,"Name":"Exercise 2","Duration":"35m 44s","ChapterTopicVideoID":9139,"CourseChapterTopicPlaylistID":5407,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Hello. In this question,"},{"Start":"00:03.045 ","End":"00:12.059","Text":"we have a ceiling with a rod of length L and of mass M attached to it."},{"Start":"00:12.059 ","End":"00:16.035","Text":"This dot over here is its axis of rotation."},{"Start":"00:16.035 ","End":"00:22.709","Text":"The rod can move up and down like so about this axis of rotation."},{"Start":"00:22.709 ","End":"00:27.449","Text":"We\u0027re told that a mass m is traveling in"},{"Start":"00:27.449 ","End":"00:32.669","Text":"a straight line with a velocity of u_0 towards the rod."},{"Start":"00:32.669 ","End":"00:38.280","Text":"It eventually hits the rod and an angle of 90 degrees to the rod,"},{"Start":"00:38.280 ","End":"00:44.180","Text":"its velocity and a distance of x down the rod."},{"Start":"00:44.180 ","End":"00:48.244","Text":"The collision over here is plastic collision."},{"Start":"00:48.244 ","End":"00:50.779","Text":"Now our first question is,"},{"Start":"00:50.779 ","End":"00:56.370","Text":"what is the angular velocity of the system right after collision?"},{"Start":"00:57.110 ","End":"01:00.990","Text":"Let\u0027s see how we can answer this."},{"Start":"01:00.990 ","End":"01:02.959","Text":"What we\u0027re going to do is,"},{"Start":"01:02.959 ","End":"01:06.199","Text":"we\u0027re going to do it in the way that we always answer"},{"Start":"01:06.199 ","End":"01:11.040","Text":"questions of this type and we\u0027re going to see what is conserved."},{"Start":"01:11.900 ","End":"01:16.550","Text":"We can have either momentum is conserved,"},{"Start":"01:16.550 ","End":"01:18.650","Text":"or energy is conserved,"},{"Start":"01:18.650 ","End":"01:22.410","Text":"or angular momentum is conserved."},{"Start":"01:22.820 ","End":"01:25.655","Text":"For a momentum to be conserved,"},{"Start":"01:25.655 ","End":"01:33.250","Text":"we have to say that the sum of the external forces is equal to 0."},{"Start":"01:33.250 ","End":"01:35.645","Text":"Let\u0027s take a look if this is happening."},{"Start":"01:35.645 ","End":"01:38.540","Text":"Now, when our ball hits our rod,"},{"Start":"01:38.540 ","End":"01:42.020","Text":"so because the rod is attached to the ceiling"},{"Start":"01:42.020 ","End":"01:45.710","Text":"somehow with a nail or a wire, it doesn\u0027t matter."},{"Start":"01:45.710 ","End":"01:47.220","Text":"It\u0027s attached to the ceiling,"},{"Start":"01:47.220 ","End":"01:50.674","Text":"so that means that when the ball hits the rod,"},{"Start":"01:50.674 ","End":"01:54.859","Text":"this nail, where the axis of rotation is,"},{"Start":"01:54.859 ","End":"01:58.114","Text":"is applying some force."},{"Start":"01:58.114 ","End":"02:01.705","Text":"A reaction force to this."},{"Start":"02:01.705 ","End":"02:04.135","Text":"If there is a reaction force,"},{"Start":"02:04.135 ","End":"02:07.730","Text":"so we know that is going to be an external force."},{"Start":"02:07.730 ","End":"02:13.530","Text":"Here we can say that our external forces are not equal to 0."},{"Start":"02:14.510 ","End":"02:20.225","Text":"This is what we\u0027re checking, and that means that momentum is not conserved."},{"Start":"02:20.225 ","End":"02:27.915","Text":"As a rule, if we have something attached to some axis,"},{"Start":"02:27.915 ","End":"02:30.855","Text":"and it can rotate about this axis,"},{"Start":"02:30.855 ","End":"02:34.610","Text":"so we can know that momentum will never be conserved"},{"Start":"02:34.610 ","End":"02:39.450","Text":"because the axis will always apply an external force."},{"Start":"02:41.000 ","End":"02:43.869","Text":"Now let\u0027s take a look at our energy."},{"Start":"02:43.869 ","End":"02:47.075","Text":"Because we know that we have plastic collision,"},{"Start":"02:47.075 ","End":"02:50.770","Text":"so let\u0027s write plastic collision."},{"Start":"02:50.770 ","End":"02:55.370","Text":"That means that there\u0027s no conservation of energy."},{"Start":"02:55.370 ","End":"02:57.545","Text":"Remember we spoke about this."},{"Start":"02:57.545 ","End":"03:00.769","Text":"Anytime that you know that there\u0027s plastic collision,"},{"Start":"03:00.769 ","End":"03:04.965","Text":"so that means inelastic collision."},{"Start":"03:04.965 ","End":"03:09.425","Text":"The 2 bodies join and attach themselves to one another,"},{"Start":"03:09.425 ","End":"03:12.725","Text":"or a body breaks into 2,"},{"Start":"03:12.725 ","End":"03:16.280","Text":"which is the opposite version of plastic collision,"},{"Start":"03:16.280 ","End":"03:21.180","Text":"so energy is never conserved in this type of situation."},{"Start":"03:21.890 ","End":"03:24.695","Text":"Now for our angular momentum."},{"Start":"03:24.695 ","End":"03:28.249","Text":"We know that if angular momentum is conserved,"},{"Start":"03:28.249 ","End":"03:33.175","Text":"so that will mean that the sum of the external torque,"},{"Start":"03:33.175 ","End":"03:37.985","Text":"or moment is going to be equal to 0."},{"Start":"03:37.985 ","End":"03:40.264","Text":"That\u0027s what we need to check."},{"Start":"03:40.264 ","End":"03:42.709","Text":"If we take a look over here,"},{"Start":"03:42.709 ","End":"03:48.259","Text":"the only place where we can have an external torque is where we have our external force."},{"Start":"03:48.259 ","End":"03:50.720","Text":"We saw our external force was when we were looking over"},{"Start":"03:50.720 ","End":"03:54.810","Text":"here and that was from our axis of rotation."},{"Start":"03:55.100 ","End":"03:59.134","Text":"This is the only place that we can have an external torque."},{"Start":"03:59.134 ","End":"04:05.779","Text":"Our reaction force is acting from the axis of rotation,"},{"Start":"04:05.779 ","End":"04:07.865","Text":"which means that its radius,"},{"Start":"04:07.865 ","End":"04:15.800","Text":"it\u0027s r the distance the force is from the axis of rotation is going to be equal to 0,"},{"Start":"04:15.800 ","End":"04:20.785","Text":"the radius of the force."},{"Start":"04:20.785 ","End":"04:26.810","Text":"That means that its torque is also going to be equal to 0."},{"Start":"04:27.650 ","End":"04:33.715","Text":"That means that the sum of our external force is equal to 0."},{"Start":"04:33.715 ","End":"04:37.055","Text":"The way that we\u0027re going to work this out is by using"},{"Start":"04:37.055 ","End":"04:41.580","Text":"the idea of conservation of angular momentum."},{"Start":"04:41.900 ","End":"04:46.374","Text":"Now in order for us to work this out,"},{"Start":"04:46.374 ","End":"04:48.999","Text":"so what we\u0027re going to do is we\u0027re going to write"},{"Start":"04:48.999 ","End":"04:53.435","Text":"an equation for our angular momentum before,"},{"Start":"04:53.435 ","End":"04:55.554","Text":"so our initial angular momentum."},{"Start":"04:55.554 ","End":"05:02.785","Text":"Then we\u0027re going to say that it equals to our final angular momentum after the collision."},{"Start":"05:02.785 ","End":"05:05.979","Text":"Then through that, we\u0027re going to find what"},{"Start":"05:05.979 ","End":"05:10.550","Text":"our angular velocity or our Omega is equal to."},{"Start":"05:11.720 ","End":"05:17.155","Text":"Let\u0027s start with our L_i before the collision."},{"Start":"05:17.155 ","End":"05:25.590","Text":"This is going to equal to the angular momentum of this body."},{"Start":"05:25.590 ","End":"05:30.844","Text":"As we know from our previous lessons when we\u0027re dealing with"},{"Start":"05:30.844 ","End":"05:38.275","Text":"some body or some mass traveling in a straight line relative to an axis,"},{"Start":"05:38.275 ","End":"05:47.630","Text":"or an origin, its angular momentum is going to be equal to mv_0d."},{"Start":"05:47.630 ","End":"05:49.430","Text":"We\u0027ve seen this equation before,"},{"Start":"05:49.430 ","End":"05:55.800","Text":"this is the angular momentum of an object traveling in a straight line."},{"Start":"05:56.470 ","End":"06:00.200","Text":"Now let\u0027s take a look at what our v_0 is."},{"Start":"06:00.200 ","End":"06:04.595","Text":"Our v_ 0 is obviously the velocity of the mass,"},{"Start":"06:04.595 ","End":"06:07.880","Text":"so here we said that it equals to u_0."},{"Start":"06:07.880 ","End":"06:17.400","Text":"Our d is the perpendicular distance it is from the axis of rotation,"},{"Start":"06:17.400 ","End":"06:19.480","Text":"so the perpendicular component."},{"Start":"06:19.480 ","End":"06:22.160","Text":"We can see that that is this x."},{"Start":"06:22.160 ","End":"06:24.215","Text":"Remember how we work it out."},{"Start":"06:24.215 ","End":"06:26.560","Text":"We draw a line,"},{"Start":"06:26.560 ","End":"06:31.355","Text":"carrying on from the arrow where its velocity is in a straight line."},{"Start":"06:31.355 ","End":"06:35.194","Text":"Then we draw a perpendicular line to that,"},{"Start":"06:35.194 ","End":"06:39.170","Text":"to where axis of rotation is and that is our d,"},{"Start":"06:39.170 ","End":"06:41.470","Text":"so in this case, it\u0027s x."},{"Start":"06:41.470 ","End":"06:44.389","Text":"Now, we\u0027re going to add onto this,"},{"Start":"06:44.389 ","End":"06:47.164","Text":"our angular momentum of the rod."},{"Start":"06:47.164 ","End":"06:51.695","Text":"Now, our angular momentum of the rod is going to be equal to, of course,"},{"Start":"06:51.695 ","End":"06:54.575","Text":"right now 0 because it\u0027s not moving,"},{"Start":"06:54.575 ","End":"06:56.795","Text":"it\u0027s stationary, It\u0027s just hanging here."},{"Start":"06:56.795 ","End":"06:59.420","Text":"Now if I substitute all of this in,"},{"Start":"06:59.420 ","End":"07:04.970","Text":"so we\u0027ll have mu_0x."},{"Start":"07:04.970 ","End":"07:07.325","Text":"That\u0027s our L_i."},{"Start":"07:07.325 ","End":"07:11.510","Text":"Also, I wrote here in brackets that the equation"},{"Start":"07:11.510 ","End":"07:16.414","Text":"for our angular momentum of the rod would be I Omega."},{"Start":"07:16.414 ","End":"07:20.490","Text":"But because it\u0027s Omega is equal to 0, it\u0027s equal to 0."},{"Start":"07:21.580 ","End":"07:28.127","Text":"Now we\u0027re going to work out our L_final."},{"Start":"07:28.127 ","End":"07:31.335","Text":"This is going to equal."},{"Start":"07:31.335 ","End":"07:33.539","Text":"Now we can work this out in 2 ways."},{"Start":"07:33.539 ","End":"07:39.060","Text":"Either we can work it out as working out the final angular momentum of the ball and"},{"Start":"07:39.060 ","End":"07:41.505","Text":"the final angular momentum of the rod"},{"Start":"07:41.505 ","End":"07:46.184","Text":"or we can look at the rod and the ball because there\u0027s plastic collision."},{"Start":"07:46.184 ","End":"07:51.519","Text":"We know that the 2 shapes are going to join together."},{"Start":"07:51.530 ","End":"07:57.285","Text":"That means that we can also look at them as just 1 body."},{"Start":"07:57.285 ","End":"07:59.475","Text":"Let\u0027s look at it at 1 body."},{"Start":"07:59.475 ","End":"08:03.990","Text":"We know that the equation for this is going to equal I Omega."},{"Start":"08:03.990 ","End":"08:06.105","Text":"I total of the 2 shapes."},{"Start":"08:06.105 ","End":"08:14.805","Text":"Now, in order to work out what our I total is we can use the idea of the additivity of I."},{"Start":"08:14.805 ","End":"08:21.810","Text":"That means that we can say that our I total is equal to our I ball,"},{"Start":"08:21.810 ","End":"08:30.490","Text":"so the ball is this plus IR whether R is the rod."},{"Start":"08:31.310 ","End":"08:35.429","Text":"Let\u0027s see what our I of the ball is."},{"Start":"08:35.429 ","End":"08:39.180","Text":"We can see that our ball is like a point mass."},{"Start":"08:39.180 ","End":"08:45.735","Text":"The equation for the I of a point mass is going to be mr^2."},{"Start":"08:45.735 ","End":"08:48.739","Text":"Here we know that its I,"},{"Start":"08:48.739 ","End":"08:52.145","Text":"its distance from the axis of rotation is x."},{"Start":"08:52.145 ","End":"08:56.404","Text":"This is just going to equal to mx^2."},{"Start":"08:56.404 ","End":"09:02.010","Text":"Then our I of the rod is just of a normal rod,"},{"Start":"09:02.010 ","End":"09:04.605","Text":"where its axis of rotation is at the end."},{"Start":"09:04.605 ","End":"09:08.910","Text":"As we know, that is going to be 1/3 the mass of"},{"Start":"09:08.910 ","End":"09:14.744","Text":"the rod multiplied by L^2, its length squared."},{"Start":"09:14.744 ","End":"09:22.635","Text":"Now I total is going to be this plus this and then we can just substitute it in here."},{"Start":"09:22.635 ","End":"09:26.655","Text":"I\u0027m going back, therefore,"},{"Start":"09:26.655 ","End":"09:30.360","Text":"our L_f is going to be I total,"},{"Start":"09:30.360 ","End":"09:31.815","Text":"which is the I of the ball,"},{"Start":"09:31.815 ","End":"09:37.830","Text":"which is going to be mx^2 plus I of the rod,"},{"Start":"09:37.830 ","End":"09:41.799","Text":"plus 1/3 ML^2,"},{"Start":"09:41.799 ","End":"09:47.020","Text":"and all of this multiplied by Omega."},{"Start":"09:47.020 ","End":"09:49.789","Text":"Now, all we have to do is we have to say that"},{"Start":"09:49.789 ","End":"09:53.014","Text":"our L_f is equal to L_i like we said over here,"},{"Start":"09:53.014 ","End":"09:56.030","Text":"and then isolate out r Omega."},{"Start":"09:56.030 ","End":"10:02.360","Text":"Now we can say that our mu_0x,"},{"Start":"10:02.360 ","End":"10:03.635","Text":"so that\u0027s our L_i,"},{"Start":"10:03.635 ","End":"10:07.865","Text":"is equal to mx^2"},{"Start":"10:07.865 ","End":"10:15.055","Text":"plus 1/3 ML^2 Omega."},{"Start":"10:15.055 ","End":"10:18.165","Text":"Then to isolate out the Omega,"},{"Start":"10:18.165 ","End":"10:21.959","Text":"we can just divide both sides by what\u0027s inside the bracket."},{"Start":"10:21.959 ","End":"10:27.735","Text":"We\u0027ll get mu_0x divided by"},{"Start":"10:27.735 ","End":"10:36.279","Text":"mx^2 plus 1/3 ML^2."},{"Start":"10:38.090 ","End":"10:43.410","Text":"Now we\u0027ve found our angular velocity of the system after collision,"},{"Start":"10:43.410 ","End":"10:47.020","Text":"so let\u0027s go on to question Number 2."},{"Start":"10:47.090 ","End":"10:53.250","Text":"In question Number 2, we\u0027re being asked what is the maximum angle the rod will reach?"},{"Start":"10:53.250 ","End":"10:56.354","Text":"Obviously, we\u0027re meaning the rod and the ball."},{"Start":"10:56.354 ","End":"10:58.185","Text":"Because of the plastic collision,"},{"Start":"10:58.185 ","End":"11:00.705","Text":"they\u0027re both attached to each other."},{"Start":"11:00.705 ","End":"11:05.260","Text":"Let\u0027s see what that means it\u0027s going to look like."},{"Start":"11:05.990 ","End":"11:10.229","Text":"What\u0027s going to happen is that the ball will collide"},{"Start":"11:10.229 ","End":"11:14.740","Text":"with a rod and it will swing to this position over here."},{"Start":"11:14.960 ","End":"11:20.849","Text":"Let\u0027s say that this angle over here is equal"},{"Start":"11:20.849 ","End":"11:27.100","Text":"to Theta and this is the angle that we\u0027re trying to find."},{"Start":"11:27.830 ","End":"11:29.970","Text":"How do we solve this?"},{"Start":"11:29.970 ","End":"11:33.704","Text":"Now, as usual, we\u0027re going to look at what is being conserved."},{"Start":"11:33.704 ","End":"11:37.035","Text":"As we saw, momentum isn\u0027t being conserved"},{"Start":"11:37.035 ","End":"11:43.020","Text":"because we still have the forces acting from this axis of rotation."},{"Start":"11:43.020 ","End":"11:48.704","Text":"Also, which is different too in question Number 1,"},{"Start":"11:48.704 ","End":"11:57.854","Text":"our momentum isn\u0027t conserved and our angular momentum also isn\u0027t conserved."},{"Start":"11:57.854 ","End":"11:59.295","Text":"Why isn\u0027t it conserved?"},{"Start":"11:59.295 ","End":"12:05.570","Text":"Because over here we can see that our mg, so our m,"},{"Start":"12:05.570 ","End":"12:10.769","Text":"our total mass is pointing in this direction,"},{"Start":"12:10.769 ","End":"12:17.894","Text":"which means that it is in itself causing a moment of force some torque."},{"Start":"12:17.894 ","End":"12:21.090","Text":"The reason why it wasn\u0027t acting over here in question Number"},{"Start":"12:21.090 ","End":"12:24.329","Text":"1 when the rod was hanging straight down was"},{"Start":"12:24.329 ","End":"12:31.019","Text":"because our mg was acting parallel and on the same line as our radius,"},{"Start":"12:31.019 ","End":"12:33.659","Text":"which means that when you work it out,"},{"Start":"12:33.659 ","End":"12:35.504","Text":"it just equals 0."},{"Start":"12:35.504 ","End":"12:38.940","Text":"We don\u0027t have to count it because it\u0027s parallel, but here,"},{"Start":"12:38.940 ","End":"12:44.729","Text":"because it isn\u0027t parallel and it has some component so there will be an external torque,"},{"Start":"12:44.729 ","End":"12:48.299","Text":"which means that our L isn\u0027t conserved."},{"Start":"12:48.299 ","End":"12:54.849","Text":"That means that we can look at our energy as being conserved."},{"Start":"12:55.130 ","End":"12:58.590","Text":"What we can say about the conservation of energy"},{"Start":"12:58.590 ","End":"13:02.355","Text":"is that when the collision itself happened,"},{"Start":"13:02.355 ","End":"13:04.560","Text":"energy wasn\u0027t conserved, however,"},{"Start":"13:04.560 ","End":"13:12.220","Text":"from the slightest millisecond after the collision itself."},{"Start":"13:12.800 ","End":"13:16.725","Text":"Onwards our energy is going to be conserved."},{"Start":"13:16.725 ","End":"13:19.934","Text":"We\u0027re going to use this in order to solve the question."},{"Start":"13:19.934 ","End":"13:25.650","Text":"Now another quick note on why our angular momentum isn\u0027t being conserved is just"},{"Start":"13:25.650 ","End":"13:27.359","Text":"because at the collision we have"},{"Start":"13:27.359 ","End":"13:31.664","Text":"some angular momentum because it\u0027s being pushed over here."},{"Start":"13:31.664 ","End":"13:34.034","Text":"However, in its final stage,"},{"Start":"13:34.034 ","End":"13:37.394","Text":"we can see that at its maximum angle,"},{"Start":"13:37.394 ","End":"13:40.335","Text":"it\u0027s going to have no angular momentum because"},{"Start":"13:40.335 ","End":"13:43.680","Text":"its angular velocity is going to equal to 0."},{"Start":"13:43.680 ","End":"13:46.170","Text":"Because it\u0027s going to swing up for a second,"},{"Start":"13:46.170 ","End":"13:49.020","Text":"stop for another second some moments of time"},{"Start":"13:49.020 ","End":"13:53.799","Text":"stop at the maximum height and then swing back down."},{"Start":"13:54.500 ","End":"13:58.170","Text":"That\u0027s also why our L isn\u0027t being conserved."},{"Start":"13:58.170 ","End":"14:01.664","Text":"Using the idea of conservation of energy,"},{"Start":"14:01.664 ","End":"14:04.030","Text":"let\u0027s see how we solve this."},{"Start":"14:04.910 ","End":"14:11.110","Text":"Now let\u0027s write out our equation for our energy."},{"Start":"14:12.080 ","End":"14:14.894","Text":"Our energy, in general,"},{"Start":"14:14.894 ","End":"14:17.429","Text":"is going to be our kinetic energy but because we\u0027re"},{"Start":"14:17.429 ","End":"14:20.640","Text":"dealing with circular motion so the equation"},{"Start":"14:20.640 ","End":"14:27.855","Text":"for kinetic energy when dealing with circular motion is 1/2I total,"},{"Start":"14:27.855 ","End":"14:34.964","Text":"of the rod and of the ball multiplied by Omega^2,"},{"Start":"14:34.964 ","End":"14:38.684","Text":"That\u0027s the kinetic energy plus potential energy."},{"Start":"14:38.684 ","End":"14:45.510","Text":"Now, potential energy will be our mass so mass,"},{"Start":"14:45.510 ","End":"14:49.125","Text":"I\u0027m doing a tilde over here to differentiate it from"},{"Start":"14:49.125 ","End":"14:53.835","Text":"the mass solely of the rod because it\u0027s the mass of the rod and of the ball,"},{"Start":"14:53.835 ","End":"15:00.275","Text":"multiplied by g and then multiplied by h. Now notice that this isn\u0027t regular h,"},{"Start":"15:00.275 ","End":"15:04.955","Text":"it\u0027s h_cm, the heights of the center of mass."},{"Start":"15:04.955 ","End":"15:08.645","Text":"Now notice that once our ball is attached to the rod,"},{"Start":"15:08.645 ","End":"15:11.820","Text":"our center of mass isn\u0027t going to be at the edge of the rod,"},{"Start":"15:11.820 ","End":"15:13.519","Text":"it\u0027s not going to be where the ball is"},{"Start":"15:13.519 ","End":"15:15.380","Text":"and it\u0027s not going to be in the center of the rod,"},{"Start":"15:15.380 ","End":"15:17.874","Text":"it\u0027s going to be somewhere in-between."},{"Start":"15:17.874 ","End":"15:24.920","Text":"Let\u0027s say it\u0027s here. Now what we have to do is we have to work out where"},{"Start":"15:24.920 ","End":"15:28.080","Text":"the center of mass is and I\u0027m"},{"Start":"15:28.080 ","End":"15:32.450","Text":"reminding us that I total we worked out for question Number 1."},{"Start":"15:32.450 ","End":"15:37.444","Text":"Our I_T was equal to"},{"Start":"15:37.444 ","End":"15:45.479","Text":"the I of the ball plus I of the rod and we\u0027ll soon substitute this in."},{"Start":"15:45.479 ","End":"15:50.939","Text":"What we\u0027re going to do now is work out where our center of mass is."},{"Start":"15:52.120 ","End":"15:55.505","Text":"The way I\u0027m going to find my h_cm,"},{"Start":"15:55.505 ","End":"15:57.845","Text":"so let\u0027s write this here is,"},{"Start":"15:57.845 ","End":"16:00.455","Text":"I\u0027m going to say that right at the beginning,"},{"Start":"16:00.455 ","End":"16:01.970","Text":"my h_cm at the beginning."},{"Start":"16:01.970 ","End":"16:05.120","Text":"I\u0027m going to say that going up in this direction is"},{"Start":"16:05.120 ","End":"16:09.980","Text":"my x-axis and I\u0027m going to say"},{"Start":"16:09.980 ","End":"16:15.625","Text":"that this place over here is at h equals 0, where the ceiling is."},{"Start":"16:15.625 ","End":"16:20.294","Text":"The way to solve this because I have 1 mass and then this is a rod."},{"Start":"16:20.294 ","End":"16:22.935","Text":"Assuming that the rod is uniform,"},{"Start":"16:22.935 ","End":"16:27.179","Text":"I\u0027m going to have its center of mass right in the center."},{"Start":"16:27.700 ","End":"16:30.920","Text":"Then I have here this mass."},{"Start":"16:30.920 ","End":"16:34.054","Text":"What I\u0027m going to do is I\u0027m going to find the center of mass,"},{"Start":"16:34.054 ","End":"16:36.780","Text":"of the center of masses."},{"Start":"16:37.040 ","End":"16:41.300","Text":"I\u0027m finding the center of mass of the 2 center masses so I"},{"Start":"16:41.300 ","End":"16:45.759","Text":"have this point and this point and then finding the center of mass between them."},{"Start":"16:45.759 ","End":"16:50.300","Text":"That\u0027s going to be somewhere around here but let\u0027s work it out."},{"Start":"16:50.300 ","End":"16:53.900","Text":"Because I\u0027m working along the x-axis so I\u0027m going to say that"},{"Start":"16:53.900 ","End":"17:00.665","Text":"my h_cm is equal to my x_cm."},{"Start":"17:00.665 ","End":"17:07.384","Text":"What this is equal to is we have a uniform rod so"},{"Start":"17:07.384 ","End":"17:13.909","Text":"this is going to be its mass multiplied by its distance from the axis of rotation."},{"Start":"17:13.909 ","End":"17:18.365","Text":"Its distance from the axis of rotation is negative L over 2,"},{"Start":"17:18.365 ","End":"17:25.709","Text":"multiplied by negative L over 2 plus my point mass,"},{"Start":"17:25.709 ","End":"17:31.489","Text":"which is of mass m and its distance because it\u0027s a point mass from the axis of rotation,"},{"Start":"17:31.489 ","End":"17:32.630","Text":"which as we can see,"},{"Start":"17:32.630 ","End":"17:36.196","Text":"is a distance of negative x."},{"Start":"17:36.196 ","End":"17:42.220","Text":"Then we\u0027re going to divide this by the total mass,"},{"Start":"17:42.220 ","End":"17:48.460","Text":"which is m plus m. Now,"},{"Start":"17:48.460 ","End":"17:49.600","Text":"I have my h_c_m,"},{"Start":"17:49.600 ","End":"17:56.800","Text":"so now I can substitute this in and I can work out what my initial energy is."},{"Start":"17:56.800 ","End":"18:01.644","Text":"My initial energy is going to be 0.5 of our IT,"},{"Start":"18:01.644 ","End":"18:03.400","Text":"which we\u0027ll substitute in later,"},{"Start":"18:03.400 ","End":"18:10.135","Text":"but it\u0027s this multiplied by r Omega squared when our Omega we found in question number 1."},{"Start":"18:10.135 ","End":"18:13.839","Text":"Then we\u0027re going to add plus our total mass,"},{"Start":"18:13.839 ","End":"18:19.270","Text":"which is going to be m plus m multiplied by gravity,"},{"Start":"18:19.270 ","End":"18:21.895","Text":"multiplied by our h_cm."},{"Start":"18:21.895 ","End":"18:27.430","Text":"Then we\u0027re going to have a negative over here to take out this and then we\u0027re"},{"Start":"18:27.430 ","End":"18:34.615","Text":"going to have multiplied by ml over 2 plus mx,"},{"Start":"18:34.615 ","End":"18:41.065","Text":"and then it\u0027s going to be divided by m plus m. Now,"},{"Start":"18:41.065 ","End":"18:44.720","Text":"we can cross out this in a second."},{"Start":"18:44.820 ","End":"18:48.080","Text":"Then my E final."},{"Start":"18:48.330 ","End":"18:52.000","Text":"It\u0027s kinetic energy is going to be equal to"},{"Start":"18:52.000 ","End":"18:55.914","Text":"0 because when it gets to its final destination,"},{"Start":"18:55.914 ","End":"18:59.135","Text":"it\u0027s stopped rotating, it\u0027s stationary."},{"Start":"18:59.135 ","End":"19:01.515","Text":"Its Omega will be 0, which means that,"},{"Start":"19:01.515 ","End":"19:03.675","Text":"this whole expression here will be 0,"},{"Start":"19:03.675 ","End":"19:07.480","Text":"and then I have to add on my mgh."},{"Start":"19:09.300 ","End":"19:11.800","Text":"My mg and then, again,"},{"Start":"19:11.800 ","End":"19:14.080","Text":"it\u0027s going to be h_cm, but now,"},{"Start":"19:14.080 ","End":"19:17.139","Text":"we\u0027re going to do a tag over here because"},{"Start":"19:17.139 ","End":"19:21.850","Text":"its height is different to right at the beginning."},{"Start":"19:21.850 ","End":"19:23.935","Text":"Let\u0027s take a look."},{"Start":"19:23.935 ","End":"19:27.099","Text":"We can see that the center of the mass of the rod goes"},{"Start":"19:27.099 ","End":"19:30.310","Text":"in some kind of circular formation till here."},{"Start":"19:30.310 ","End":"19:33.010","Text":"This is CM of the rod."},{"Start":"19:33.010 ","End":"19:39.310","Text":"Then my center of mass is going to go like this."},{"Start":"19:39.310 ","End":"19:43.099","Text":"Here\u0027s my center of mass afterwards."},{"Start":"19:43.320 ","End":"19:50.854","Text":"Now, what we\u0027re trying to find is"},{"Start":"19:50.854 ","End":"19:59.580","Text":"the height from h equals 0 until this point over here, the perpendicular component."},{"Start":"19:59.580 ","End":"20:01.439","Text":"Now, what I\u0027m going to draw,"},{"Start":"20:01.439 ","End":"20:05.949","Text":"is I\u0027m going to take at 90 degrees to the rod,"},{"Start":"20:05.949 ","End":"20:10.120","Text":"a line that connects to my center of mass."},{"Start":"20:10.120 ","End":"20:12.625","Text":"Now, what I want to find,"},{"Start":"20:12.625 ","End":"20:16.435","Text":"is this height until the blue line."},{"Start":"20:16.435 ","End":"20:25.370","Text":"This is my h_cm tag to this blue line which corresponds to this point."},{"Start":"20:25.890 ","End":"20:33.775","Text":"Now, we know that the distance from here to my center of mass."},{"Start":"20:33.775 ","End":"20:40.630","Text":"We saw before that\u0027s going to be my x_cm or my h_cm."},{"Start":"20:40.630 ","End":"20:45.610","Text":"This length over here is"},{"Start":"20:45.610 ","End":"20:54.699","Text":"my h_c_m originally and because this angle is Theta and this is hypotenuse,"},{"Start":"20:54.699 ","End":"20:57.804","Text":"if we\u0027re looking at the triangle,"},{"Start":"20:57.804 ","End":"21:01.520","Text":"if we\u0027re looking at this triangle."},{"Start":"21:04.110 ","End":"21:10.419","Text":"We can see that it\u0027s a right angle triangle because this angle over here is 90 degrees."},{"Start":"21:10.419 ","End":"21:11.995","Text":"Here we have our Theta,"},{"Start":"21:11.995 ","End":"21:18.894","Text":"and here we have hypotenuse so that means that our h_cm,"},{"Start":"21:18.894 ","End":"21:25.360","Text":"this height over here is simply going to be our x_cm,"},{"Start":"21:25.360 ","End":"21:28.520","Text":"which is our normal h_cm,"},{"Start":"21:28.950 ","End":"21:35.540","Text":"over here, multiplied by cosine of Theta."},{"Start":"21:37.260 ","End":"21:43.280","Text":"Because we\u0027re finding this component of this length."},{"Start":"21:43.920 ","End":"21:46.509","Text":"Now, all I have to do,"},{"Start":"21:46.509 ","End":"21:51.770","Text":"is I have to say that my E_i is equal to my E_f,"},{"Start":"21:55.260 ","End":"22:02.410","Text":"and then I can substitute in my values and then solve to find my Theta."},{"Start":"22:02.410 ","End":"22:04.239","Text":"Let\u0027s write this out."},{"Start":"22:04.239 ","End":"22:11.920","Text":"I have my 0.5 IT Omega squared minus and then this and this can"},{"Start":"22:11.920 ","End":"22:21.110","Text":"cross off so minus g_ml over 2 plus mx."},{"Start":"22:25.290 ","End":"22:30.639","Text":"Then all of this is going to be equal to my total mass so it\u0027s m"},{"Start":"22:30.639 ","End":"22:37.164","Text":"plus m multiplied by g,"},{"Start":"22:37.164 ","End":"22:40.869","Text":"multiplied by my h_cm,"},{"Start":"22:40.869 ","End":"22:43.960","Text":"which is my h_cm cosine of Theta,"},{"Start":"22:43.960 ","End":"22:51.109","Text":"which is going to be multiplied by negative L over 2 m"},{"Start":"22:52.800 ","End":"22:59.140","Text":"plus negative mx divided"},{"Start":"22:59.140 ","End":"23:04.390","Text":"by m plus m. Again,"},{"Start":"23:04.390 ","End":"23:06.850","Text":"these terms can cross off and again,"},{"Start":"23:06.850 ","End":"23:07.900","Text":"I have a minus here,"},{"Start":"23:07.900 ","End":"23:12.769","Text":"so I can take out the common factor and put in a minus at the beginning."},{"Start":"23:13.500 ","End":"23:20.079","Text":"Then I\u0027m going to be left with 1/2i total Omega squared"},{"Start":"23:20.079 ","End":"23:28.075","Text":"negative g_ml over 2 plus mx,"},{"Start":"23:28.075 ","End":"23:34.960","Text":"which is going to be equal to negative g and then"},{"Start":"23:34.960 ","End":"23:42.070","Text":"ml over 2 plus mx."},{"Start":"23:42.070 ","End":"23:47.140","Text":"Then I can add this term to both sides."},{"Start":"23:47.140 ","End":"23:50.360","Text":"Sorry, this is an x over here."},{"Start":"23:52.290 ","End":"23:57.830","Text":"This was the problem, I forgot to multiply over here by cosine of Theta."},{"Start":"24:00.450 ","End":"24:02.890","Text":"I was missing a cosine of Theta,"},{"Start":"24:02.890 ","End":"24:05.365","Text":"so that explains what was happening."},{"Start":"24:05.365 ","End":"24:09.235","Text":"Then I can group together my like terms."},{"Start":"24:09.235 ","End":"24:17.199","Text":"Then I\u0027ll have a 0.5_i total Omega squared is going to be equal to,"},{"Start":"24:17.199 ","End":"24:20.949","Text":"then I\u0027m going to add to both sides this,"},{"Start":"24:20.949 ","End":"24:31.370","Text":"so I\u0027ll have g_ml over 2 plus m_ x,"},{"Start":"24:31.500 ","End":"24:36.715","Text":"and then I will have 1"},{"Start":"24:36.715 ","End":"24:44.302","Text":"minus cosine of Theta."},{"Start":"24:44.302 ","End":"24:52.090","Text":"Now all I have to do is to try and get my cosine of Theta out."},{"Start":"24:52.100 ","End":"24:56.160","Text":"Then you just rearrange this by"},{"Start":"24:56.160 ","End":"24:59.715","Text":"doing a little bit of algebra to isolate out your cosine Theta."},{"Start":"24:59.715 ","End":"25:03.014","Text":"We\u0027ll get negative 1/2I total,"},{"Start":"25:03.014 ","End":"25:07.515","Text":"and you can substitute in what your I total is multiplied by Omega squared,"},{"Start":"25:07.515 ","End":"25:10.470","Text":"and the Omega is what we found over here in question number 1,"},{"Start":"25:10.470 ","End":"25:15.165","Text":"divided by g multiplied by ML over 2 plus mx,"},{"Start":"25:15.165 ","End":"25:16.965","Text":"and then all of this plus 1,"},{"Start":"25:16.965 ","End":"25:19.440","Text":"and this is equal to cosine of Theta."},{"Start":"25:19.440 ","End":"25:23.355","Text":"That is how you can work out the maximum angle."},{"Start":"25:23.355 ","End":"25:27.340","Text":"I\u0027m going to go back to the diagram."},{"Start":"25:29.450 ","End":"25:34.620","Text":"What separated it out so that we could find where our maximum angle is,"},{"Start":"25:34.620 ","End":"25:37.904","Text":"is that we knew that at that maximum angle,"},{"Start":"25:37.904 ","End":"25:43.154","Text":"Omega, our angular velocity will be there equal to 0,"},{"Start":"25:43.154 ","End":"25:46.815","Text":"which means that our kinetic energy over here would be equal to 0."},{"Start":"25:46.815 ","End":"25:48.885","Text":"Then that\u0027s how we could solve it."},{"Start":"25:48.885 ","End":"25:51.930","Text":"That was the very important thing to remember."},{"Start":"25:51.930 ","End":"25:57.690","Text":"That at the maximum angle or the maximum height,"},{"Start":"25:57.690 ","End":"26:00.225","Text":"depending on which question you\u0027re being asked,"},{"Start":"26:00.225 ","End":"26:03.769","Text":"then always your kinetic energy,"},{"Start":"26:03.769 ","End":"26:05.719","Text":"or specifically in these type of questions,"},{"Start":"26:05.719 ","End":"26:08.520","Text":"will be equal to 0."},{"Start":"26:08.770 ","End":"26:16.705","Text":"Now we can go on and look at question number 3."},{"Start":"26:16.705 ","End":"26:22.574","Text":"Question 3 is asking us to find which length of x will cause"},{"Start":"26:22.574 ","End":"26:28.630","Text":"the force being applied by the ceiling to the rod to be equal to 0."},{"Start":"26:28.670 ","End":"26:32.370","Text":"This is a question that doesn\u0027t appear very often."},{"Start":"26:32.370 ","End":"26:36.075","Text":"It\u0027s also got nothing to do with questions number 1 and 2."},{"Start":"26:36.075 ","End":"26:39.015","Text":"Questions 1 and 2 are very, very common."},{"Start":"26:39.015 ","End":"26:42.059","Text":"This is a slightly rarer question."},{"Start":"26:42.059 ","End":"26:44.760","Text":"But let\u0027s see how we solve this."},{"Start":"26:44.760 ","End":"26:48.434","Text":"The question is giving us a bit of a problem"},{"Start":"26:48.434 ","End":"26:53.325","Text":"because in question 1 or 2, I can\u0027t remember,"},{"Start":"26:53.325 ","End":"27:01.770","Text":"we said that the sum of the external forces did not equal to 0,"},{"Start":"27:01.770 ","End":"27:07.390","Text":"and therefore, we didn\u0027t have conservation of momentum."},{"Start":"27:07.400 ","End":"27:10.619","Text":"Here suddenly, we\u0027re saying that"},{"Start":"27:10.619 ","End":"27:13.949","Text":"the external force that we said which wasn\u0027t equal to 0,"},{"Start":"27:13.949 ","End":"27:16.560","Text":"we\u0027re being asked to find the length of x."},{"Start":"27:16.560 ","End":"27:18.060","Text":"Here x is an unknown,"},{"Start":"27:18.060 ","End":"27:20.145","Text":"so some length over here,"},{"Start":"27:20.145 ","End":"27:24.255","Text":"which will cause that external force to equal 0."},{"Start":"27:24.255 ","End":"27:27.460","Text":"Let\u0027s see how we do this."},{"Start":"27:28.430 ","End":"27:32.040","Text":"In actual fact, what we\u0027re saying is that there\u0027s"},{"Start":"27:32.040 ","End":"27:37.185","Text":"some length x at which our ball will hit."},{"Start":"27:37.185 ","End":"27:42.959","Text":"If we\u0027re using that length then the sum of our external forces will equal 0,"},{"Start":"27:42.959 ","End":"27:46.965","Text":"which means that there will be conservation of momentum."},{"Start":"27:46.965 ","End":"27:52.240","Text":"Now we\u0027re being told to try and find that length."},{"Start":"27:53.390 ","End":"27:59.384","Text":"There\u0027s a very specific x for which there is conservation of momentum."},{"Start":"27:59.384 ","End":"28:02.969","Text":"But for all the other x\u0027s aside from that one value,"},{"Start":"28:02.969 ","End":"28:04.709","Text":"there isn\u0027t conservation of momentum,"},{"Start":"28:04.709 ","End":"28:10.590","Text":"which means that we were still correct in what we were saying in the previous questions."},{"Start":"28:10.590 ","End":"28:12.645","Text":"In order to answer this question,"},{"Start":"28:12.645 ","End":"28:15.615","Text":"I\u0027m going to assume that there is conservation of momentum."},{"Start":"28:15.615 ","End":"28:20.800","Text":"I\u0027m going to say that my momentum is equal to a constant."},{"Start":"28:22.220 ","End":"28:26.280","Text":"Let\u0027s see how we do this. If there\u0027s conservation of momentum,"},{"Start":"28:26.280 ","End":"28:32.685","Text":"that means that our initial momentum is equal to our final momentum."},{"Start":"28:32.685 ","End":"28:34.920","Text":"Let\u0027s write down the equation."},{"Start":"28:34.920 ","End":"28:37.769","Text":"Our initial momentum is obviously just going to be"},{"Start":"28:37.769 ","End":"28:41.460","Text":"the momentum of the ball because that\u0027s the only thing which is working."},{"Start":"28:41.460 ","End":"28:48.240","Text":"That means that our mass of the ball multiplied by its velocity, which is u_0."},{"Start":"28:48.240 ","End":"28:50.415","Text":"That\u0027s the initial momentum,"},{"Start":"28:50.415 ","End":"28:53.700","Text":"and this is equal to the final momentum."},{"Start":"28:53.700 ","End":"28:55.995","Text":"How do we work this out?"},{"Start":"28:55.995 ","End":"28:58.875","Text":"This is the trick of the question."},{"Start":"28:58.875 ","End":"29:05.410","Text":"It\u0027s going to be something to do with its angular movement."},{"Start":"29:06.770 ","End":"29:12.370","Text":"In order to work out P final,"},{"Start":"29:13.040 ","End":"29:19.965","Text":"the reason why it\u0027s difficult is because we take the mass multiplied by the velocity."},{"Start":"29:19.965 ","End":"29:24.480","Text":"Now, because we\u0027re dealing with the rod and with a ball together,"},{"Start":"29:24.480 ","End":"29:26.700","Text":"every single piece and the rod,"},{"Start":"29:26.700 ","End":"29:28.320","Text":"it isn\u0027t a point-mass."},{"Start":"29:28.320 ","End":"29:31.980","Text":"Every piece has a different velocity."},{"Start":"29:31.980 ","End":"29:37.664","Text":"How do I know which velocity to multiply by the mass?"},{"Start":"29:37.664 ","End":"29:45.135","Text":"The way to do it, so P total will be the total momentum of a body,"},{"Start":"29:45.135 ","End":"29:52.150","Text":"is equal to its mass multiplied by the velocity of the center of mass."},{"Start":"29:52.150 ","End":"29:56.389","Text":"Obviously this mass is the total mass of the system."},{"Start":"29:56.389 ","End":"29:59.060","Text":"I write here a total as well."},{"Start":"29:59.060 ","End":"30:02.850","Text":"You can write this in your formula sheets."},{"Start":"30:03.080 ","End":"30:06.060","Text":"In one of our previous questions,"},{"Start":"30:06.060 ","End":"30:09.720","Text":"we already found that our center of mass for the ball and"},{"Start":"30:09.720 ","End":"30:13.815","Text":"the rod together is equal to x_cm."},{"Start":"30:13.815 ","End":"30:15.975","Text":"That\u0027s where our center of mass is."},{"Start":"30:15.975 ","End":"30:18.990","Text":"Let\u0027s say that we know this right now."},{"Start":"30:18.990 ","End":"30:22.755","Text":"Now, what I have to do is find the velocity of this point."},{"Start":"30:22.755 ","End":"30:25.515","Text":"How do I find the velocity of this point?"},{"Start":"30:25.515 ","End":"30:30.885","Text":"I have to remember that it\u0027s moving in angular motion."},{"Start":"30:30.885 ","End":"30:35.475","Text":"It\u0027s angular motion is around this axis of rotation."},{"Start":"30:35.475 ","End":"30:44.220","Text":"I already know that for a body moving with angular motion around some axis,"},{"Start":"30:44.220 ","End":"30:52.780","Text":"every single point in the body is also moving with angular motion about this axis."},{"Start":"30:53.330 ","End":"30:56.460","Text":"Because of that sentence that I just said,"},{"Start":"30:56.460 ","End":"31:02.279","Text":"so I know that my V center of mass is going to be equal to, as we know,"},{"Start":"31:02.279 ","End":"31:04.770","Text":"our equation, r Omega,"},{"Start":"31:04.770 ","End":"31:07.335","Text":"which we found in our first questions,"},{"Start":"31:07.335 ","End":"31:09.660","Text":"that\u0027s this from my first question,"},{"Start":"31:09.660 ","End":"31:12.165","Text":"multiplied by the radius."},{"Start":"31:12.165 ","End":"31:13.635","Text":"What is the radius?"},{"Start":"31:13.635 ","End":"31:18.250","Text":"Our radius is going to be our x center of mass."},{"Start":"31:20.060 ","End":"31:22.545","Text":"This is our radius,"},{"Start":"31:22.545 ","End":"31:27.195","Text":"its distance from the axis of rotation."},{"Start":"31:27.195 ","End":"31:32.309","Text":"In that case we can say that our P final is going to be equal"},{"Start":"31:32.309 ","End":"31:39.915","Text":"to total mass multiplied by our V,"},{"Start":"31:39.915 ","End":"31:43.305","Text":"which is our Omega from our first question,"},{"Start":"31:43.305 ","End":"31:45.000","Text":"multiplied by our radius,"},{"Start":"31:45.000 ","End":"31:47.560","Text":"which is x center of mass."},{"Start":"31:48.290 ","End":"31:53.579","Text":"That is going to be equal to our mass of"},{"Start":"31:53.579 ","End":"31:59.580","Text":"the rod plus our mass of the point mass multiplied by r Omega from question 1,"},{"Start":"31:59.580 ","End":"32:03.255","Text":"multiplied by our x center of mass."},{"Start":"32:03.255 ","End":"32:09.240","Text":"Now what I have to do is I have to plug this into my equation over here."},{"Start":"32:09.240 ","End":"32:12.539","Text":"I have my P initial is equal to my P final."},{"Start":"32:12.539 ","End":"32:16.410","Text":"My P initial is mu_0,"},{"Start":"32:16.410 ","End":"32:19.169","Text":"which is equal to my P final,"},{"Start":"32:19.169 ","End":"32:26.040","Text":"which is going to be equal to m plus m multiplied by my Omega from question 1,"},{"Start":"32:26.040 ","End":"32:29.415","Text":"multiplied by my x_ cm."},{"Start":"32:29.415 ","End":"32:34.360","Text":"Now I\u0027m just going to isolate out my x_cm."},{"Start":"32:34.580 ","End":"32:40.229","Text":"What I\u0027m going to get is I\u0027m going to have mu_0 divided"},{"Start":"32:40.229 ","End":"32:47.320","Text":"by M plus m multiplied by Omega."},{"Start":"32:48.050 ","End":"32:51.525","Text":"Now we have where x center of masses is."},{"Start":"32:51.525 ","End":"32:53.939","Text":"But we\u0027re not trying to find our x center of mass,"},{"Start":"32:53.939 ","End":"32:57.100","Text":"we\u0027re trying to find this length of x."},{"Start":"33:04.790 ","End":"33:07.364","Text":"What we\u0027re trying to find in fact,"},{"Start":"33:07.364 ","End":"33:10.665","Text":"is at which height can the ball hit."},{"Start":"33:10.665 ","End":"33:13.274","Text":"Our height that the ball will hit is our x,"},{"Start":"33:13.274 ","End":"33:16.935","Text":"such that the force here will be equal to 0."},{"Start":"33:16.935 ","End":"33:19.470","Text":"That means our external force will be equal to 0,"},{"Start":"33:19.470 ","End":"33:22.600","Text":"and then we\u0027ll have conservation of momentum."},{"Start":"33:23.170 ","End":"33:25.530","Text":"Right now we have our x_cm,"},{"Start":"33:25.530 ","End":"33:27.855","Text":"but we\u0027re trying to find the height of the ball."},{"Start":"33:27.855 ","End":"33:31.289","Text":"If you remember the equation that links our x_cm with the height of"},{"Start":"33:31.289 ","End":"33:35.205","Text":"the ball is the equation for to find our x center of mass."},{"Start":"33:35.205 ","End":"33:43.095","Text":"That\u0027s the mass of the rod multiplied by its position from the axis of rotation."},{"Start":"33:43.095 ","End":"33:47.385","Text":"The center of mass plus the point mass and"},{"Start":"33:47.385 ","End":"33:51.900","Text":"its distance from the axis of rotation, which is x."},{"Start":"33:51.900 ","End":"33:54.850","Text":"This is what we\u0027re trying to find."},{"Start":"33:55.250 ","End":"33:58.410","Text":"That\u0027s a question mark on top."},{"Start":"33:58.410 ","End":"34:01.545","Text":"Then divide it by the total mass,"},{"Start":"34:01.545 ","End":"34:09.105","Text":"which is M plus m. This is exactly what we\u0027re trying to find. This over here."},{"Start":"34:09.105 ","End":"34:12.134","Text":"I\u0027ll make this a bit clear."},{"Start":"34:12.134 ","End":"34:15.465","Text":"This is what we\u0027re trying to find."},{"Start":"34:15.465 ","End":"34:19.875","Text":"Now we can say that this x_cm is equal to this x_cm."},{"Start":"34:19.875 ","End":"34:30.620","Text":"We\u0027ll say that our mu_0 divided by M plus m multiplied by Omega is equal to"},{"Start":"34:30.620 ","End":"34:40.860","Text":"ml over 2 plus mx divided by M plus m. I can cross out"},{"Start":"34:40.860 ","End":"34:49.590","Text":"this M plus m and this M plus m. Now I have to isolate out to get this x. I\u0027ll"},{"Start":"34:49.590 ","End":"34:54.434","Text":"have mu_0 divided by"},{"Start":"34:54.434 ","End":"35:00.400","Text":"Omega minus ml over 2."},{"Start":"35:00.400 ","End":"35:03.245","Text":"That\u0027s going to equal mx."},{"Start":"35:03.245 ","End":"35:11.950","Text":"Then I can divide all of this by m. Then I\u0027ll have my x that we were looking for."},{"Start":"35:12.050 ","End":"35:17.294","Text":"Now I just rearrange this a bit to simplify the expression."},{"Start":"35:17.294 ","End":"35:23.010","Text":"Then we get that at this specific x and only at this x,"},{"Start":"35:23.010 ","End":"35:25.725","Text":"we have a conservation of momentum."},{"Start":"35:25.725 ","End":"35:32.166","Text":"It\u0027s 2mu_0 minus ML Omega divided by 2 Omega m,"},{"Start":"35:32.166 ","End":"35:34.586","Text":"and at every single other x,"},{"Start":"35:34.586 ","End":"35:38.084","Text":"at every single other length that the ball can hit,"},{"Start":"35:38.084 ","End":"35:41.280","Text":"there will not be conservation of momentum."},{"Start":"35:41.280 ","End":"35:44.589","Text":"That\u0027s the end of this question."}],"ID":9409}],"Thumbnail":null,"ID":5407},{"Name":"4. Analysis Through Forces And Moments, Rolling Without Slipping","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Comparing Angular and Linear Motion","Duration":"13m 59s","ChapterTopicVideoID":9151,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"Hello. In this lesson,"},{"Start":"00:02.610 ","End":"00:05.730","Text":"we\u0027re going to be speaking about motion in a straight line,"},{"Start":"00:05.730 ","End":"00:09.645","Text":"what we\u0027ve learned in high school,"},{"Start":"00:09.645 ","End":"00:11.700","Text":"and compare it to,"},{"Start":"00:11.700 ","End":"00:16.630","Text":"and find all the similarities between circular motion."},{"Start":"00:16.790 ","End":"00:22.275","Text":"What we\u0027re going to do in this lesson is write out a table"},{"Start":"00:22.275 ","End":"00:28.300","Text":"where we\u0027re going to see the similarities and their comparisons."},{"Start":"00:29.780 ","End":"00:34.550","Text":"The first thing that we\u0027re going to speak about is our position."},{"Start":"00:34.550 ","End":"00:37.325","Text":"If here in our beginning point,"},{"Start":"00:37.325 ","End":"00:46.080","Text":"we say that our position is at x=0 and then this is some different x;"},{"Start":"00:46.080 ","End":"00:47.975","Text":"after a certain amount of time,"},{"Start":"00:47.975 ","End":"00:52.620","Text":"our body would have moved to a new position."},{"Start":"00:53.840 ","End":"00:56.375","Text":"In motion in a straight line,"},{"Start":"00:56.375 ","End":"00:59.580","Text":"we see that our position is x."},{"Start":"00:59.580 ","End":"01:01.455","Text":"In circular motion however,"},{"Start":"01:01.455 ","End":"01:05.315","Text":"if we draw a line up until here,"},{"Start":"01:05.315 ","End":"01:07.325","Text":"and this is our first position,"},{"Start":"01:07.325 ","End":"01:10.205","Text":"so this is at Theta is equal to 0,"},{"Start":"01:10.205 ","End":"01:14.405","Text":"and then if our shape was to move to here,"},{"Start":"01:14.405 ","End":"01:20.120","Text":"if this angle is Theta,"},{"Start":"01:20.120 ","End":"01:24.148","Text":"its new position will now be at Theta."},{"Start":"01:24.148 ","End":"01:29.275","Text":"The position in circular motion is denoted by the angle Theta."},{"Start":"01:29.275 ","End":"01:34.920","Text":"The next thing we\u0027re going to deal with is velocity."},{"Start":"01:36.110 ","End":"01:39.873","Text":"When we\u0027re dealing with motion in a straight line,"},{"Start":"01:39.873 ","End":"01:44.030","Text":"we know that the velocity that the body would have traveled at,"},{"Start":"01:44.030 ","End":"01:52.715","Text":"so we label it with a V and we say that it\u0027s equal to the derivative of our x-dot,"},{"Start":"01:52.715 ","End":"01:53.945","Text":"which is also, of course,"},{"Start":"01:53.945 ","End":"01:59.030","Text":"equal to the change in x over time,"},{"Start":"01:59.030 ","End":"02:01.450","Text":"so dx by dt."},{"Start":"02:01.450 ","End":"02:04.420","Text":"As time changes, our position changes."},{"Start":"02:04.420 ","End":"02:07.129","Text":"Similarly with circular motion,"},{"Start":"02:07.129 ","End":"02:10.415","Text":"our velocity when dealing with circular motion is our Omega,"},{"Start":"02:10.415 ","End":"02:12.125","Text":"it\u0027s our angular velocity,"},{"Start":"02:12.125 ","End":"02:17.010","Text":"and this is equal to Theta dot, and of course,"},{"Start":"02:17.010 ","End":"02:22.520","Text":"Theta dot is equal to d Theta by dt,"},{"Start":"02:22.520 ","End":"02:26.610","Text":"the change in the angle as time goes by."},{"Start":"02:28.100 ","End":"02:32.920","Text":"The next thing that we have is acceleration."},{"Start":"02:34.760 ","End":"02:39.650","Text":"That is how fast our velocity is changing."},{"Start":"02:39.650 ","End":"02:41.945","Text":"In motion in straight line,"},{"Start":"02:41.945 ","End":"02:46.370","Text":"we have a for acceleration is equal to our change in"},{"Start":"02:46.370 ","End":"02:52.620","Text":"velocity and that equals to x double dot."},{"Start":"02:53.920 ","End":"02:58.385","Text":"That\u0027s the second derivative of our x."},{"Start":"02:58.385 ","End":"03:01.490","Text":"In circular motion for acceleration,"},{"Start":"03:01.490 ","End":"03:05.945","Text":"we have Alpha, which is equal to Omega dot,"},{"Start":"03:05.945 ","End":"03:10.330","Text":"which is how fast our angular velocity is changing,"},{"Start":"03:10.330 ","End":"03:12.815","Text":"and that, of course, is equal to"},{"Start":"03:12.815 ","End":"03:17.760","Text":"the second derivative of our Theta, so Theta double-dot."},{"Start":"03:18.830 ","End":"03:23.345","Text":"Now, if we\u0027re being told that our body is of mass m,"},{"Start":"03:23.345 ","End":"03:25.310","Text":"in a straight line,"},{"Start":"03:25.310 ","End":"03:29.810","Text":"we\u0027ll say that our body has a mass m. It\u0027s"},{"Start":"03:29.810 ","End":"03:35.164","Text":"parallel in circular motion is our moment of inertia."},{"Start":"03:35.164 ","End":"03:37.870","Text":"When we\u0027re speaking about mass in a straight line,"},{"Start":"03:37.870 ","End":"03:39.620","Text":"speaking about I,"},{"Start":"03:39.620 ","End":"03:42.125","Text":"a moment of inertia in circular motion,"},{"Start":"03:42.125 ","End":"03:44.135","Text":"soon we\u0027ll understand why,"},{"Start":"03:44.135 ","End":"03:47.835","Text":"but it\u0027s good to know."},{"Start":"03:47.835 ","End":"03:52.684","Text":"Then we have momentum,"},{"Start":"03:52.684 ","End":"03:55.560","Text":"mom is for momentum."},{"Start":"03:55.560 ","End":"03:57.620","Text":"Motion in a straight line,"},{"Start":"03:57.620 ","End":"03:59.840","Text":"we have regular momentum, which is P,"},{"Start":"03:59.840 ","End":"04:05.495","Text":"and we know that that equals to our mass multiplied by our velocity."},{"Start":"04:05.495 ","End":"04:08.675","Text":"In circular motion, we have angular momentum,"},{"Start":"04:08.675 ","End":"04:10.055","Text":"which is denoted L,"},{"Start":"04:10.055 ","End":"04:16.395","Text":"angular momentum, and it equals to I multiplied by Omega."},{"Start":"04:16.395 ","End":"04:18.176","Text":"Now, see the similarities."},{"Start":"04:18.176 ","End":"04:23.235","Text":"Here, we have our momentum is equal to mass times our velocity,"},{"Start":"04:23.235 ","End":"04:25.340","Text":"and here in our angular momentum,"},{"Start":"04:25.340 ","End":"04:26.810","Text":"we have the equivalent of mass,"},{"Start":"04:26.810 ","End":"04:31.620","Text":"which is our I, and the equivalent of our velocity, which is Omega."},{"Start":"04:33.110 ","End":"04:37.115","Text":"Lastly, with motion in a straight line,"},{"Start":"04:37.115 ","End":"04:40.040","Text":"we have our forces working."},{"Start":"04:40.040 ","End":"04:41.180","Text":"In circular motion,"},{"Start":"04:41.180 ","End":"04:44.641","Text":"instead of forces, we speak about torques."},{"Start":"04:44.641 ","End":"04:49.910","Text":"Again, this is either torque or our moment of force or our moments,"},{"Start":"04:49.910 ","End":"04:52.620","Text":"they all mean the same thing."},{"Start":"04:54.410 ","End":"04:58.325","Text":"Now, by working out the similarities and"},{"Start":"04:58.325 ","End":"05:04.386","Text":"our parallel terms when dealing with motion in a straight line and with circular motion,"},{"Start":"05:04.386 ","End":"05:07.730","Text":"it\u0027s going to be a lot easier for us when we\u0027re working out"},{"Start":"05:07.730 ","End":"05:11.630","Text":"questions with circular motion to understand exactly what we\u0027re"},{"Start":"05:11.630 ","End":"05:15.530","Text":"doing and what we\u0027re speaking about and to understand the problem at"},{"Start":"05:15.530 ","End":"05:20.150","Text":"hand because when we\u0027re given a question with circular motion,"},{"Start":"05:20.150 ","End":"05:22.475","Text":"as you can see, it\u0027s the exact same thing"},{"Start":"05:22.475 ","End":"05:25.000","Text":"as when you\u0027re dealing with motion in a straight line,"},{"Start":"05:25.000 ","End":"05:27.710","Text":"just we\u0027re dealing with slightly different letters."},{"Start":"05:27.710 ","End":"05:30.980","Text":"So instead of m, we have I, instead of F,"},{"Start":"05:30.980 ","End":"05:36.900","Text":"we have Tau, but it\u0027s exactly the same type of working out."},{"Start":"05:37.550 ","End":"05:42.320","Text":"Now, we\u0027re going to give an example of how our equations for motion in"},{"Start":"05:42.320 ","End":"05:47.020","Text":"a straight line are exactly the same as our equations for circular motion."},{"Start":"05:47.020 ","End":"05:52.610","Text":"For instance, we know from basic kinematics the first thing that we learned is"},{"Start":"05:52.610 ","End":"05:55.880","Text":"the equation for our position with motion in"},{"Start":"05:55.880 ","End":"05:59.825","Text":"a straight line is equal to our starting position x_0,"},{"Start":"05:59.825 ","End":"06:07.250","Text":"plus our initial velocity or our constant velocity,"},{"Start":"06:07.250 ","End":"06:09.920","Text":"multiplied by the time."},{"Start":"06:09.920 ","End":"06:12.605","Text":"How would we do this in circular motion?"},{"Start":"06:12.605 ","End":"06:14.375","Text":"It\u0027s the exact same thing."},{"Start":"06:14.375 ","End":"06:18.010","Text":"Instead of x, we have Theta,"},{"Start":"06:18.010 ","End":"06:21.270","Text":"so its angular position,"},{"Start":"06:21.270 ","End":"06:22.810","Text":"and then instead of x_0,"},{"Start":"06:22.810 ","End":"06:26.636","Text":"we\u0027ll have its initial angular position,"},{"Start":"06:26.636 ","End":"06:31.145","Text":"plus our initial velocity; instead of velocity,"},{"Start":"06:31.145 ","End":"06:36.620","Text":"we have our Omega or our constant of angular velocity,"},{"Start":"06:36.620 ","End":"06:40.070","Text":"and then multiplied by t. This is"},{"Start":"06:40.070 ","End":"06:44.450","Text":"our equation for our position when dealing with circular motion,"},{"Start":"06:44.450 ","End":"06:46.055","Text":"which is the exact same thing,"},{"Start":"06:46.055 ","End":"06:51.810","Text":"just denote it with different letterings as our motion in a straight line."},{"Start":"06:51.810 ","End":"06:54.480","Text":"Then if we have some acceleration,"},{"Start":"06:54.480 ","End":"07:01.253","Text":"we remember that our equation for x will be plus 1/2 at^2."},{"Start":"07:01.253 ","End":"07:04.880","Text":"Over here, we\u0027ll also have the exact same thing plus"},{"Start":"07:04.880 ","End":"07:09.500","Text":"1/2 and then instead of acceleration for motion in a straight line,"},{"Start":"07:09.500 ","End":"07:11.780","Text":"we have acceleration in circular motion,"},{"Start":"07:11.780 ","End":"07:14.865","Text":"so it\u0027s Alpha t^2,"},{"Start":"07:14.865 ","End":"07:18.050","Text":"the exact same equation."},{"Start":"07:18.050 ","End":"07:21.530","Text":"Now, let\u0027s prove it with more equations."},{"Start":"07:21.530 ","End":"07:23.300","Text":"If we have our velocity,"},{"Start":"07:23.300 ","End":"07:25.700","Text":"again, of course it\u0027s as a function of time,"},{"Start":"07:25.700 ","End":"07:32.520","Text":"so we know that that is equal to v_0 plus at."},{"Start":"07:32.560 ","End":"07:35.750","Text":"Now in circular motion,"},{"Start":"07:35.750 ","End":"07:38.780","Text":"so instead of velocity we have Omega and again,"},{"Start":"07:38.780 ","End":"07:40.220","Text":"as a function of t,"},{"Start":"07:40.220 ","End":"07:42.725","Text":"so that will equal to instead of v_0,"},{"Start":"07:42.725 ","End":"07:45.620","Text":"it will be Omega 0 plus instead of a,"},{"Start":"07:45.620 ","End":"07:52.680","Text":"it will be Alpha multiplied by t. The exact same thing."},{"Start":"07:52.780 ","End":"07:59.075","Text":"More. Say we have in motion in a straight line,"},{"Start":"07:59.075 ","End":"08:02.030","Text":"F is equal to ma,"},{"Start":"08:02.030 ","End":"08:06.020","Text":"I most used almost equation."},{"Start":"08:06.020 ","End":"08:10.175","Text":"In circular motion, our F is our Tau,"},{"Start":"08:10.175 ","End":"08:15.590","Text":"so Tau is equal to m. We say we speak about our I,"},{"Start":"08:15.590 ","End":"08:17.135","Text":"our moment of inertia,"},{"Start":"08:17.135 ","End":"08:21.530","Text":"and then a is our Alpha, I Alpha."},{"Start":"08:21.530 ","End":"08:24.875","Text":"Now, remember this equation because it\u0027s very useful and"},{"Start":"08:24.875 ","End":"08:28.790","Text":"we will use it a lot so remember this."},{"Start":"08:28.790 ","End":"08:30.620","Text":"Our next equation is,"},{"Start":"08:30.620 ","End":"08:32.240","Text":"of course, our momentum."},{"Start":"08:32.240 ","End":"08:34.640","Text":"P is equal to mv."},{"Start":"08:34.640 ","End":"08:41.285","Text":"Our m is our I and our v is our Omega so our angular momentum is I omega."},{"Start":"08:41.285 ","End":"08:44.480","Text":"Another example is our work,"},{"Start":"08:44.480 ","End":"08:49.475","Text":"so our w. It\u0027s equal to the integral on"},{"Start":"08:49.475 ","End":"08:56.180","Text":"our force multiplied by our dr. Now,"},{"Start":"08:56.180 ","End":"09:02.150","Text":"in circular motion, so we\u0027ll be doing the integral on instead of force,"},{"Start":"09:02.150 ","End":"09:05.360","Text":"on our Tau, on our torque."},{"Start":"09:05.360 ","End":"09:08.000","Text":"Then instead of our dr,"},{"Start":"09:08.000 ","End":"09:11.210","Text":"so dr denotes our position."},{"Start":"09:11.210 ","End":"09:16.235","Text":"Instead of our dr, so our I can also be our x,"},{"Start":"09:16.235 ","End":"09:18.540","Text":"we\u0027ll be doing d Theta."},{"Start":"09:19.710 ","End":"09:24.820","Text":"I, remember, is just your position when you\u0027re dealing with integrals usually,"},{"Start":"09:24.820 ","End":"09:28.150","Text":"if you don\u0027t know which axes you\u0027re integrating on."},{"Start":"09:28.150 ","End":"09:31.790","Text":"Here, it\u0027s d Theta, it\u0027s the exact same thing."},{"Start":"09:32.320 ","End":"09:38.150","Text":"Now, let\u0027s give a short little example of how to solve a question."},{"Start":"09:38.150 ","End":"09:41.660","Text":"Say that we have this angle over here,"},{"Start":"09:41.660 ","End":"09:45.290","Text":"let\u0027s call it Theta 0 and we\u0027re told that"},{"Start":"09:45.290 ","End":"09:50.375","Text":"our body is rotating at Alpha is equal to Alpha 0,"},{"Start":"09:50.375 ","End":"10:00.455","Text":"so it has constant angular acceleration and its angular velocity at t is equal to 0,"},{"Start":"10:00.455 ","End":"10:04.220","Text":"so its initial angular velocity is equal to 0."},{"Start":"10:04.220 ","End":"10:05.795","Text":"It starts from rest."},{"Start":"10:05.795 ","End":"10:10.620","Text":"We\u0027re trying to find its position as a function of time,"},{"Start":"10:15.640 ","End":"10:17.780","Text":"the equation for that,"},{"Start":"10:17.780 ","End":"10:21.290","Text":"so that means its angle as a function of time."},{"Start":"10:21.290 ","End":"10:23.960","Text":"Just like in linear motion,"},{"Start":"10:23.960 ","End":"10:27.095","Text":"we\u0027re going to use this equation over here for Theta,"},{"Start":"10:27.095 ","End":"10:29.405","Text":"because we\u0027re dealing with circular motion."},{"Start":"10:29.405 ","End":"10:35.165","Text":"That means that our Theta is going to be equal to our Theta 0,"},{"Start":"10:35.165 ","End":"10:42.185","Text":"so our initial position plus our Omega_0 t. What is our Omega_0?"},{"Start":"10:42.185 ","End":"10:44.885","Text":"Its our initial angular velocity."},{"Start":"10:44.885 ","End":"10:47.720","Text":"It was given at t equals 0,"},{"Start":"10:47.720 ","End":"10:49.475","Text":"our Omega is equal to 0,"},{"Start":"10:49.475 ","End":"10:55.580","Text":"so that will equal 0 and then plus 1/2 Alpha,"},{"Start":"10:55.580 ","End":"10:57.380","Text":"which is the acceleration,"},{"Start":"10:57.380 ","End":"11:03.750","Text":"so here we know that it\u0027s Alpha 0, multiplied by t^2."},{"Start":"11:04.810 ","End":"11:07.790","Text":"This is our equation."},{"Start":"11:07.790 ","End":"11:10.970","Text":"That\u0027s it. That\u0027s the answer to that question."},{"Start":"11:10.970 ","End":"11:17.850","Text":"Now. what if they ask me how long it will take to complete 10 circles?"},{"Start":"11:18.460 ","End":"11:20.990","Text":"Let\u0027s see how we do that."},{"Start":"11:20.990 ","End":"11:23.675","Text":"In order to complete 10 circles,"},{"Start":"11:23.675 ","End":"11:25.535","Text":"so let\u0027s start with 1 circle."},{"Start":"11:25.535 ","End":"11:27.560","Text":"In order to complete 1 circle,"},{"Start":"11:27.560 ","End":"11:32.290","Text":"so this over here is Theta 0."},{"Start":"11:32.290 ","End":"11:36.390","Text":"What we would need is that our position,"},{"Start":"11:36.390 ","End":"11:43.190","Text":"so our Theta, minus its initial position will have to equal to Pi."},{"Start":"11:43.190 ","End":"11:47.350","Text":"That would mean that our Theta will be equal over"},{"Start":"11:47.350 ","End":"11:52.195","Text":"here and all of this distance over here would have to be 2 Pi."},{"Start":"11:52.195 ","End":"11:54.265","Text":"Let\u0027s write that down."},{"Start":"11:54.265 ","End":"12:02.165","Text":"That would mean that our position minus our initial position will have to equal 2 Pi."},{"Start":"12:02.165 ","End":"12:06.605","Text":"That means that 1 full circle has been completed."},{"Start":"12:06.605 ","End":"12:13.320","Text":"Now, we\u0027re being asked for 10 circles so then we just have to multiply that by 10."},{"Start":"12:13.690 ","End":"12:16.595","Text":"Now, we have Theta minus Theta 0."},{"Start":"12:16.595 ","End":"12:18.710","Text":"What we can do over in this equation,"},{"Start":"12:18.710 ","End":"12:21.056","Text":"we can write Theta and then minus,"},{"Start":"12:21.056 ","End":"12:23.105","Text":"from both sides, Theta 0."},{"Start":"12:23.105 ","End":"12:28.280","Text":"Then we\u0027ll have Theta minus Theta 0 is going to be equal"},{"Start":"12:28.280 ","End":"12:34.235","Text":"to 1/2 Alpha 0 multiplied by t^2."},{"Start":"12:34.235 ","End":"12:37.700","Text":"Then we know that our Theta minus Theta 0,"},{"Start":"12:37.700 ","End":"12:40.625","Text":"so that denotes 10 circles,"},{"Start":"12:40.625 ","End":"12:51.260","Text":"so then we\u0027ll have the 20 Pi is equal to 1/2 Alpha 0 multiplied by t^2."},{"Start":"12:51.260 ","End":"12:56.495","Text":"Now, all we have to do in order to get the time taken is isolate our t,"},{"Start":"12:56.495 ","End":"13:02.465","Text":"so then we can say that our t is equal to so we\u0027ll have 40"},{"Start":"13:02.465 ","End":"13:11.670","Text":"Pi divided by Alpha 0 and then the square root because we have t^2."},{"Start":"13:13.600 ","End":"13:17.750","Text":"That\u0027s the answer. Exactly like working in"},{"Start":"13:17.750 ","End":"13:23.130","Text":"linear motion or motion in a straight line except with Theta."},{"Start":"13:24.760 ","End":"13:30.245","Text":"The thing to take out of this lesson is 1, this table,"},{"Start":"13:30.245 ","End":"13:36.230","Text":"and even write it in your notes that you can take with you into the exam."},{"Start":"13:36.230 ","End":"13:41.855","Text":"To just really to understand that motion in a straight line and circular motion,"},{"Start":"13:41.855 ","End":"13:44.615","Text":"you solve the questions at exactly the same way,"},{"Start":"13:44.615 ","End":"13:48.860","Text":"just you have different letters symbolizing the different things."},{"Start":"13:48.860 ","End":"13:51.305","Text":"As you can see in simple questions like this,"},{"Start":"13:51.305 ","End":"13:56.945","Text":"it really is exactly the same equation as in motion in a straight line."},{"Start":"13:56.945 ","End":"13:59.850","Text":"That\u0027s the end of this lesson."}],"ID":9421},{"Watched":false,"Name":"Rolling Without Slipping","Duration":"10m 19s","ChapterTopicVideoID":9152,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"Hello. This lesson is going to be a very important lesson."},{"Start":"00:04.080 ","End":"00:05.970","Text":"What we\u0027re going to be speaking about"},{"Start":"00:05.970 ","End":"00:10.260","Text":"here comes back time and time again in exam questions,"},{"Start":"00:10.260 ","End":"00:12.570","Text":"in homework questions, all the time."},{"Start":"00:12.570 ","End":"00:16.930","Text":"We\u0027re dealing with rolling without slipping."},{"Start":"00:17.540 ","End":"00:21.995","Text":"In this lesson we\u0027re going to speak about what that exactly means,"},{"Start":"00:21.995 ","End":"00:24.445","Text":"and what we\u0027re dealing with is some wheel,"},{"Start":"00:24.445 ","End":"00:26.910","Text":"or a cylinder which is going down the road,"},{"Start":"00:26.910 ","End":"00:28.980","Text":"and it\u0027s rolling and not slipping."},{"Start":"00:28.980 ","End":"00:31.660","Text":"Let\u0027s see what this means."},{"Start":"00:31.760 ","End":"00:36.510","Text":"Let\u0027s define what rolling without slipping is."},{"Start":"00:36.510 ","End":"00:41.000","Text":"We\u0027re going to write the definition over here."},{"Start":"00:41.000 ","End":"00:46.230","Text":"At our point of contact between this wheel and the ground,"},{"Start":"00:46.230 ","End":"00:49.320","Text":"let\u0027s call that point A."},{"Start":"00:49.320 ","End":"00:51.925","Text":"At this specific point A,"},{"Start":"00:51.925 ","End":"00:54.455","Text":"the only point which is touching the ground,"},{"Start":"00:54.455 ","End":"01:00.080","Text":"the velocity at that point has to equal 0."},{"Start":"01:00.080 ","End":"01:04.380","Text":"This is the definition for rolling without slipping."},{"Start":"01:04.610 ","End":"01:07.125","Text":"Let\u0027s take a look at this."},{"Start":"01:07.125 ","End":"01:10.110","Text":"Here is the center of the wheel."},{"Start":"01:10.110 ","End":"01:12.705","Text":"This is our center of mass,"},{"Start":"01:12.705 ","End":"01:15.800","Text":"and it has some velocity."},{"Start":"01:15.800 ","End":"01:17.890","Text":"The center of mass has a velocity."},{"Start":"01:17.890 ","End":"01:25.855","Text":"As well, the entire wheel is rotating at an angular velocity of Omega 0,"},{"Start":"01:25.855 ","End":"01:29.299","Text":"and let\u0027s say that this is the positive direction."},{"Start":"01:29.299 ","End":"01:33.845","Text":"Even though usually we say that the anticlockwise direction is the positive direction,"},{"Start":"01:33.845 ","End":"01:37.580","Text":"it doesn\u0027t matter, right now the clockwise is the positive direction."},{"Start":"01:37.580 ","End":"01:43.220","Text":"Then this direction is also the positive x-direction,"},{"Start":"01:43.220 ","End":"01:45.240","Text":"whatever you want to call it."},{"Start":"01:46.610 ","End":"01:51.135","Text":"What we want to show, is that this point is equal to 0."},{"Start":"01:51.135 ","End":"01:57.760","Text":"We want to show how we get to this definition. Let\u0027s see how we do that."},{"Start":"01:57.830 ","End":"02:02.570","Text":"Let\u0027s take a look at the velocity of A."},{"Start":"02:02.570 ","End":"02:05.725","Text":"Let\u0027s see what is equal to."},{"Start":"02:05.725 ","End":"02:11.535","Text":"On the 1 hand, we have our V_cm going in this rightwards direction."},{"Start":"02:11.535 ","End":"02:16.565","Text":"If our center of mass is moving at a velocity of V_cm,"},{"Start":"02:16.565 ","End":"02:19.940","Text":"that means that every single point on this wheel is"},{"Start":"02:19.940 ","End":"02:24.170","Text":"also moving in the rightwards direction with a velocity of V_cm."},{"Start":"02:24.170 ","End":"02:28.705","Text":"The wheel is moving as a whole to this direction."},{"Start":"02:28.705 ","End":"02:30.430","Text":"On the 1 hand,"},{"Start":"02:30.430 ","End":"02:38.200","Text":"we have our V_cm going in this direction, that\u0027s number 1."},{"Start":"02:38.200 ","End":"02:40.880","Text":"However, on the other hand,"},{"Start":"02:40.880 ","End":"02:42.470","Text":"something a little bit tricky."},{"Start":"02:42.470 ","End":"02:46.320","Text":"We also have our Omega 0 moving it."},{"Start":"02:46.320 ","End":"02:50.405","Text":"If we imagine for 1 second that there isn\u0027t this V_cm,"},{"Start":"02:50.405 ","End":"02:53.810","Text":"we can see, imagine our wheel isn\u0027t moving,"},{"Start":"02:53.810 ","End":"02:57.600","Text":"it\u0027s just rotating around without moving."},{"Start":"02:57.680 ","End":"03:01.865","Text":"Our Omega 0 will mean that at this point"},{"Start":"03:01.865 ","End":"03:05.390","Text":"exactly at the contact with the ground, if it\u0027s just spinning,"},{"Start":"03:05.390 ","End":"03:06.800","Text":"not moving as well,"},{"Start":"03:06.800 ","End":"03:10.445","Text":"its velocity in this direction will"},{"Start":"03:10.445 ","End":"03:14.695","Text":"equal to its angular velocity multiplied by its radius."},{"Start":"03:14.695 ","End":"03:21.995","Text":"That will be Omega_0 multiplied by R. Why is that?"},{"Start":"03:21.995 ","End":"03:26.900","Text":"Because this point is just moving along this radius, and in circular motion,"},{"Start":"03:26.900 ","End":"03:30.380","Text":"in regular circular motion,"},{"Start":"03:30.380 ","End":"03:38.410","Text":"the whole circle shape itself isn\u0027t moving and it\u0027s just rotating around itself."},{"Start":"03:39.590 ","End":"03:42.480","Text":"Now what we\u0027re going to do is,"},{"Start":"03:42.480 ","End":"03:45.315","Text":"we\u0027re going to put these two things together,"},{"Start":"03:45.315 ","End":"03:49.430","Text":"because we also have the center of mass moving in this direction,"},{"Start":"03:49.430 ","End":"03:51.500","Text":"which we said was the positive direction,"},{"Start":"03:51.500 ","End":"03:56.000","Text":"and we have our movement from the rotation itself."},{"Start":"03:56.000 ","End":"03:57.665","Text":"If we draw this,"},{"Start":"03:57.665 ","End":"03:59.570","Text":"we have in the right direction,"},{"Start":"03:59.570 ","End":"04:01.025","Text":"in the positive direction,"},{"Start":"04:01.025 ","End":"04:03.020","Text":"we\u0027re going to have our V_cm,"},{"Start":"04:03.020 ","End":"04:05.570","Text":"and then going leftwards in the negative direction,"},{"Start":"04:05.570 ","End":"04:11.915","Text":"we\u0027re going to have Omega_0 R. We can write these both into our equation."},{"Start":"04:11.915 ","End":"04:14.210","Text":"We\u0027ll have V_cm,"},{"Start":"04:14.210 ","End":"04:17.179","Text":"and then negative, because this is in the negative direction,"},{"Start":"04:17.179 ","End":"04:25.965","Text":"Omega_0 R. Now because we\u0027re dealing with our rolling without slipping,"},{"Start":"04:25.965 ","End":"04:32.195","Text":"I have to say that this our V_A=0,"},{"Start":"04:32.195 ","End":"04:37.265","Text":"because this is the condition that we have rolling without slipping."},{"Start":"04:37.265 ","End":"04:39.290","Text":"If there was rolling with slipping,"},{"Start":"04:39.290 ","End":"04:41.015","Text":"then this wouldn\u0027t be the case."},{"Start":"04:41.015 ","End":"04:43.700","Text":"But right now this is rolling without slipping,"},{"Start":"04:43.700 ","End":"04:50.360","Text":"I say that this is equal to 0 using the definition."},{"Start":"04:50.360 ","End":"04:53.540","Text":"That means if I rearrange this, that my V,"},{"Start":"04:53.540 ","End":"04:59.990","Text":"center of mass is going to equal Omega_0 R. That means my V,"},{"Start":"04:59.990 ","End":"05:06.125","Text":"center of mass is equal to my velocity, my rotational velocity."},{"Start":"05:06.125 ","End":"05:12.660","Text":"This similarly is a condition for rolling without slipping."},{"Start":"05:13.940 ","End":"05:19.130","Text":"That means that our velocity of the center of mass is equal"},{"Start":"05:19.130 ","End":"05:23.885","Text":"to our angular velocity multiplied by the radius."},{"Start":"05:23.885 ","End":"05:27.305","Text":"All the time when there is rolling without slipping,"},{"Start":"05:27.305 ","End":"05:30.040","Text":"this is the equation that we use."},{"Start":"05:30.040 ","End":"05:33.200","Text":"Then we can also say that if this happens,"},{"Start":"05:33.200 ","End":"05:35.180","Text":"then there\u0027s rolling without slipping,"},{"Start":"05:35.180 ","End":"05:37.340","Text":"and if there\u0027s rolling without slipping,"},{"Start":"05:37.340 ","End":"05:40.830","Text":"then that means this equation is correct."},{"Start":"05:40.830 ","End":"05:42.875","Text":"It\u0027s if and only if,"},{"Start":"05:42.875 ","End":"05:45.210","Text":"if you know what that means."},{"Start":"05:45.410 ","End":"05:48.605","Text":"In mathematics we symbolize it like this,"},{"Start":"05:48.605 ","End":"05:50.750","Text":"with this 2-headed arrow."},{"Start":"05:50.750 ","End":"05:52.775","Text":"If this is happening,"},{"Start":"05:52.775 ","End":"05:54.785","Text":"then that means there\u0027s rolling without slipping."},{"Start":"05:54.785 ","End":"05:56.225","Text":"If there\u0027s rolling without slipping,"},{"Start":"05:56.225 ","End":"05:58.720","Text":"then that means that this is correct."},{"Start":"05:58.720 ","End":"06:00.960","Text":"Very important to remember this,"},{"Start":"06:00.960 ","End":"06:03.970","Text":"write this down in your notebooks."},{"Start":"06:04.460 ","End":"06:08.490","Text":"Another way of looking at this equation is looking at"},{"Start":"06:08.490 ","End":"06:13.940","Text":"it with the velocity relative to another velocity."},{"Start":"06:13.940 ","End":"06:16.775","Text":"If this is the velocity of the center of mass,"},{"Start":"06:16.775 ","End":"06:20.180","Text":"and this is the velocity relative to the center of mass,"},{"Start":"06:20.180 ","End":"06:21.830","Text":"that\u0027s another way to look at it."},{"Start":"06:21.830 ","End":"06:30.835","Text":"Now this rolling without slipping is a little bit complicated to explain, let\u0027s try it."},{"Start":"06:30.835 ","End":"06:33.710","Text":"Let\u0027s say we have this point over here,"},{"Start":"06:33.710 ","End":"06:36.950","Text":"and we call it B, it\u0027s another point."},{"Start":"06:36.950 ","End":"06:38.854","Text":"If you can imagine,"},{"Start":"06:38.854 ","End":"06:44.510","Text":"in order to try and explain why the velocity at the contact point is equal to 0."},{"Start":"06:44.510 ","End":"06:52.140","Text":"As our V reaches over to here, if you notice,"},{"Start":"06:52.140 ","End":"06:54.030","Text":"we\u0027ll have a point like this,"},{"Start":"06:54.030 ","End":"06:56.625","Text":"and here\u0027s our point that\u0027s about to touch,"},{"Start":"06:56.625 ","End":"07:00.835","Text":"but right now this point here is touching."},{"Start":"07:00.835 ","End":"07:03.075","Text":"As the ground,"},{"Start":"07:03.075 ","End":"07:05.410","Text":"if this is the ground,"},{"Start":"07:06.170 ","End":"07:08.945","Text":"as the ball rolls,"},{"Start":"07:08.945 ","End":"07:13.760","Text":"suddenly this point will be touching, the green."},{"Start":"07:13.760 ","End":"07:15.260","Text":"This is our point B."},{"Start":"07:15.260 ","End":"07:17.330","Text":"Suddenly this point will touch,"},{"Start":"07:17.330 ","End":"07:23.340","Text":"and the rest of the ball will carry on rolling in this direction."},{"Start":"07:26.800 ","End":"07:32.660","Text":"All the time, whilst the ground is tangent to this point,"},{"Start":"07:32.660 ","End":"07:34.625","Text":"this point is stationary."},{"Start":"07:34.625 ","End":"07:39.510","Text":"Then at a certain stage the ball"},{"Start":"07:39.510 ","End":"07:44.595","Text":"will carry on rolling and the point will then lift as well."},{"Start":"07:44.595 ","End":"07:52.170","Text":"But for a certain period of time the ground rolls on this specific point."},{"Start":"07:52.170 ","End":"07:56.175","Text":"That\u0027s in fact what is happening."},{"Start":"07:56.175 ","End":"08:00.170","Text":"If you can imagine the whole ball carries on rolling,"},{"Start":"08:00.170 ","End":"08:03.740","Text":"and at this specific point this point is stationary for"},{"Start":"08:03.740 ","End":"08:07.670","Text":"1 second or for a certain very short period of time,"},{"Start":"08:07.670 ","End":"08:11.900","Text":"as the ground is still tangent to it and then it moves on to the next point."},{"Start":"08:11.900 ","End":"08:17.555","Text":"For very short period of time this specific point right over here is stationary,"},{"Start":"08:17.555 ","End":"08:22.525","Text":"and that\u0027s the exact idea behind rolling without slipping."},{"Start":"08:22.525 ","End":"08:28.520","Text":"Another equation that we can use in order to speak about rolling without slipping"},{"Start":"08:28.520 ","End":"08:34.430","Text":"is to differentiate this equation in order to get a similar condition,"},{"Start":"08:34.430 ","End":"08:36.955","Text":"but when speaking about acceleration."},{"Start":"08:36.955 ","End":"08:44.000","Text":"The differential of V is obviously the acceleration of the center of mass,"},{"Start":"08:44.000 ","End":"08:47.465","Text":"which is equal to the differential of R,"},{"Start":"08:47.465 ","End":"08:53.605","Text":"Omega_0 is going to equal Alpha angular acceleration multiplied by the radius."},{"Start":"08:53.605 ","End":"08:58.500","Text":"This equation we can also use and we\u0027ll find it very useful."},{"Start":"08:58.810 ","End":"09:06.100","Text":"Another thing, because our point of contact in the wheel with the ground,"},{"Start":"09:06.100 ","End":"09:08.500","Text":"because it\u0027s really not moving,"},{"Start":"09:08.500 ","End":"09:10.310","Text":"the whole body is moving around but"},{"Start":"09:10.310 ","End":"09:13.940","Text":"this specific point for that period of time is static,"},{"Start":"09:13.940 ","End":"09:20.564","Text":"it\u0027s stationary, the type of friction that we have is going to be static friction."},{"Start":"09:20.564 ","End":"09:23.135","Text":"There\u0027s static friction,"},{"Start":"09:23.135 ","End":"09:30.709","Text":"and if there\u0027s static friction then that means that there\u0027s conservation of energy."},{"Start":"09:30.709 ","End":"09:34.280","Text":"This is obviously assuming that there\u0027s"},{"Start":"09:34.280 ","End":"09:38.675","Text":"no other force acting which will cancel out the conservation of energy."},{"Start":"09:38.675 ","End":"09:41.690","Text":"But if you\u0027re speaking about this specific point of contact with"},{"Start":"09:41.690 ","End":"09:44.930","Text":"the ground there\u0027s static coefficient of friction,"},{"Start":"09:44.930 ","End":"09:48.990","Text":"which means that there\u0027s at that point conservation of energy."},{"Start":"09:49.930 ","End":"09:55.130","Text":"That\u0027s the end of our lesson for rolling without slipping."},{"Start":"09:55.130 ","End":"09:58.550","Text":"The important thing to remember is that at the point"},{"Start":"09:58.550 ","End":"10:01.985","Text":"of contact between the wheel over this cylinder with the ground,"},{"Start":"10:01.985 ","End":"10:04.970","Text":"the velocity at that point is going to be equal to 0."},{"Start":"10:04.970 ","End":"10:07.925","Text":"Then through that we get these 2 conditions,"},{"Start":"10:07.925 ","End":"10:10.700","Text":"which also mean that there\u0027s rolling without slipping,"},{"Start":"10:10.700 ","End":"10:13.205","Text":"and if those rolling without slipping this happens."},{"Start":"10:13.205 ","End":"10:15.710","Text":"At the point of contact there\u0027s static friction,"},{"Start":"10:15.710 ","End":"10:20.380","Text":"which means conservation of energy. That\u0027s the end of this lesson."}],"ID":9422},{"Watched":false,"Name":"Approaching Questions","Duration":"8m 19s","ChapterTopicVideoID":9153,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:03.570","Text":"Hello. In this lesson,"},{"Start":"00:03.570 ","End":"00:07.320","Text":"we\u0027re going to speak about how to approach different questions,"},{"Start":"00:07.320 ","End":"00:10.080","Text":"different techniques to make it a little bit"},{"Start":"00:10.080 ","End":"00:13.725","Text":"easier to know what to do and where to start."},{"Start":"00:13.725 ","End":"00:17.980","Text":"Then we\u0027re going to follow up with an easy example."},{"Start":"00:18.830 ","End":"00:22.784","Text":"I\u0027m going to split this up into 2 options."},{"Start":"00:22.784 ","End":"00:25.890","Text":"Now, these 2 options will solve around"},{"Start":"00:25.890 ","End":"00:29.880","Text":"about 80 percent or 90 percent of questions that you can be given."},{"Start":"00:29.880 ","End":"00:31.650","Text":"If they don\u0027t solve those questions,"},{"Start":"00:31.650 ","End":"00:33.570","Text":"then you can try and do some other things."},{"Start":"00:33.570 ","End":"00:37.155","Text":"But in general, this is a good start."},{"Start":"00:37.155 ","End":"00:41.680","Text":"The first thing will be what is being conserved."},{"Start":"00:42.050 ","End":"00:46.830","Text":"We\u0027re checking if anything is being conserved."},{"Start":"00:46.830 ","End":"00:53.285","Text":"The first thing that we need to check is if our energy is conserved."},{"Start":"00:53.285 ","End":"00:57.125","Text":"In order to check if there\u0027s conservation of energy,"},{"Start":"00:57.125 ","End":"01:01.830","Text":"we need to check if all of the forces are conservative."},{"Start":"01:01.990 ","End":"01:06.455","Text":"We\u0027re checking if our forces are being conserved."},{"Start":"01:06.455 ","End":"01:09.650","Text":"Now, another way of saying that is to"},{"Start":"01:09.650 ","End":"01:14.880","Text":"check if the work of all the non-conservative forces,"},{"Start":"01:14.880 ","End":"01:18.830","Text":"so the work of all the forces which are not conservative,"},{"Start":"01:18.830 ","End":"01:22.200","Text":"if that is equal to 0."},{"Start":"01:23.450 ","End":"01:26.330","Text":"If the answer to this is yes,"},{"Start":"01:26.330 ","End":"01:28.520","Text":"then our energy is conserved."},{"Start":"01:28.520 ","End":"01:32.975","Text":"The next thing that we need to check is if"},{"Start":"01:32.975 ","End":"01:37.965","Text":"our momentum is being conserved, P for momentum."},{"Start":"01:37.965 ","End":"01:41.735","Text":"How do we know that? If the sum of all of"},{"Start":"01:41.735 ","End":"01:48.935","Text":"our external forces is equal to 0 and this is what we have to check."},{"Start":"01:48.935 ","End":"01:51.005","Text":"If the answer to this is yes,"},{"Start":"01:51.005 ","End":"01:54.095","Text":"then our momentum is being conserved."},{"Start":"01:54.095 ","End":"01:59.045","Text":"Now, notice that here we\u0027re only speaking about our external forces."},{"Start":"01:59.045 ","End":"02:02.825","Text":"Not the internal forces inside the body,"},{"Start":"02:02.825 ","End":"02:05.680","Text":"but external forces acting on the body."},{"Start":"02:05.680 ","End":"02:11.215","Text":"Which is different to in the case of energy where we\u0027re speaking about all forces."},{"Start":"02:11.215 ","End":"02:17.490","Text":"In our momentum, we\u0027re simply speaking about the outside forces."},{"Start":"02:18.370 ","End":"02:25.655","Text":"The third thing that we can check is if our angular momentum is being conserved."},{"Start":"02:25.655 ","End":"02:34.475","Text":"The way we know this is if the sum of the external torques is equal to 0."},{"Start":"02:34.475 ","End":"02:36.380","Text":"This is what we have to check."},{"Start":"02:36.380 ","End":"02:41.000","Text":"What do we have to do is we have to check the moments or the torque of"},{"Start":"02:41.000 ","End":"02:47.555","Text":"our external forces and then we have to see if they equal 0."},{"Start":"02:47.555 ","End":"02:50.278","Text":"Sometimes they will, sometimes they won\u0027t,"},{"Start":"02:50.278 ","End":"02:52.710","Text":"and that\u0027s what we have to check."},{"Start":"02:52.910 ","End":"02:57.605","Text":"Now, bear in mind that if the sum of our external forces equals to 0,"},{"Start":"02:57.605 ","End":"03:03.365","Text":"that doesn\u0027t mean that the sum of our external forces will be equal to 0 and vice versa."},{"Start":"03:03.365 ","End":"03:04.790","Text":"If this is equal to 0,"},{"Start":"03:04.790 ","End":"03:07.920","Text":"it doesn\u0027t mean that this will be equal to 0."},{"Start":"03:08.120 ","End":"03:13.430","Text":"The fourth thing, which isn\u0027t to do with anything being conserved,"},{"Start":"03:13.430 ","End":"03:21.540","Text":"but it\u0027s useful to know is the connection between the velocities or speeds."},{"Start":"03:22.220 ","End":"03:25.520","Text":"Connecting the expression so for instance,"},{"Start":"03:25.520 ","End":"03:28.040","Text":"if we have v is equal to Omega r,"},{"Start":"03:28.040 ","End":"03:33.390","Text":"so connecting our linear motion to our angular motion."},{"Start":"03:35.480 ","End":"03:40.340","Text":"This side is going to solve us most of our questions."},{"Start":"03:40.340 ","End":"03:41.975","Text":"If that doesn\u0027t work,"},{"Start":"03:41.975 ","End":"03:45.365","Text":"then we\u0027re going to move to the other option,"},{"Start":"03:45.365 ","End":"03:49.015","Text":"which is forces and moments."},{"Start":"03:49.015 ","End":"03:56.885","Text":"In this case, the first thing that I would do is to write down our usual equation,"},{"Start":"03:56.885 ","End":"04:04.620","Text":"which is that the sum of all of the forces is going to equal m multiplied by a_cm."},{"Start":"04:04.620 ","End":"04:07.265","Text":"Why do I have that this is equal to"},{"Start":"04:07.265 ","End":"04:11.270","Text":"my mass multiplied by my acceleration of center of mass?"},{"Start":"04:11.270 ","End":"04:13.789","Text":"If you remember with point masses,"},{"Start":"04:13.789 ","End":"04:17.345","Text":"so when you write down the sum of all of the forces,"},{"Start":"04:17.345 ","End":"04:20.390","Text":"all of the forces are external forces."},{"Start":"04:20.390 ","End":"04:24.845","Text":"You get the mass multiplied by the acceleration of the center of mass."},{"Start":"04:24.845 ","End":"04:28.115","Text":"But because our point mass is a single-point,"},{"Start":"04:28.115 ","End":"04:32.525","Text":"our center of mass is the acceleration of the point itself."},{"Start":"04:32.525 ","End":"04:35.895","Text":"Now, over here, when we\u0027re dealing with rigid bodies."},{"Start":"04:35.895 ","End":"04:38.735","Text":"Our body is not a point mass,"},{"Start":"04:38.735 ","End":"04:42.545","Text":"it\u0027s a large body so it\u0027s the same over here."},{"Start":"04:42.545 ","End":"04:46.955","Text":"We\u0027re writing out the sum of all of the external forces and when we do that,"},{"Start":"04:46.955 ","End":"04:52.610","Text":"that means that we get mass multiplied by the acceleration of the center of mass."},{"Start":"04:52.610 ","End":"04:54.485","Text":"This is always correct."},{"Start":"04:54.485 ","End":"05:00.664","Text":"The acceleration of other points in the body is dependent on the internal forces,"},{"Start":"05:00.664 ","End":"05:02.570","Text":"the forces within the body."},{"Start":"05:02.570 ","End":"05:04.625","Text":"In fact, when dealing with a rigid body,"},{"Start":"05:04.625 ","End":"05:07.325","Text":"if we look at all of the other points in the body,"},{"Start":"05:07.325 ","End":"05:09.935","Text":"the points which aren\u0027t the center of mass,"},{"Start":"05:09.935 ","End":"05:13.895","Text":"then we\u0027ll see that every single point has a slightly different acceleration."},{"Start":"05:13.895 ","End":"05:16.580","Text":"Because their accelerations are determined"},{"Start":"05:16.580 ","End":"05:19.870","Text":"not just by the acceleration of the body as a whole,"},{"Start":"05:19.870 ","End":"05:21.215","Text":"of the center of mass,"},{"Start":"05:21.215 ","End":"05:24.995","Text":"but also dependent on the forces within the body,"},{"Start":"05:24.995 ","End":"05:27.630","Text":"between the atoms and molecules."},{"Start":"05:27.630 ","End":"05:33.830","Text":"It\u0027s very important to write that it\u0027s the sum of all of"},{"Start":"05:33.830 ","End":"05:40.225","Text":"the forces is equal to the mass multiplied by the acceleration of the center of mass."},{"Start":"05:40.225 ","End":"05:43.625","Text":"Then the second thing that I would do is write down"},{"Start":"05:43.625 ","End":"05:47.135","Text":"the sum of the moments or the sum of the torques."},{"Start":"05:47.135 ","End":"05:50.165","Text":"As we know, that is equal to I,"},{"Start":"05:50.165 ","End":"05:51.605","Text":"the moment of inertia,"},{"Start":"05:51.605 ","End":"05:55.550","Text":"multiplied by angular acceleration."},{"Start":"05:55.550 ","End":"05:59.020","Text":"Now, when you\u0027re working out the sum of all of the moments,"},{"Start":"05:59.020 ","End":"06:01.810","Text":"it might be around the center of mass and it might not be"},{"Start":"06:01.810 ","End":"06:06.010","Text":"which is not what\u0027s happening over here. It doesn\u0027t really matter."},{"Start":"06:06.010 ","End":"06:09.520","Text":"We\u0027ll see an examples later on."},{"Start":"06:09.520 ","End":"06:12.730","Text":"It might be center of mass, might not."},{"Start":"06:12.730 ","End":"06:16.750","Text":"Then the third thing that I would do,"},{"Start":"06:16.750 ","End":"06:20.005","Text":"so like here, we connected the velocity expressions."},{"Start":"06:20.005 ","End":"06:24.890","Text":"Here, we\u0027ll be connecting the acceleration expressions."},{"Start":"06:24.980 ","End":"06:34.340","Text":"For example, a is equal to Alpha I connecting linear and angular acceleration."},{"Start":"06:34.760 ","End":"06:39.695","Text":"These are the 2 main ways to approach a question."},{"Start":"06:39.695 ","End":"06:43.820","Text":"Obviously, sometimes you\u0027ll get questions that are a little bit different."},{"Start":"06:43.820 ","End":"06:46.670","Text":"Then you\u0027ll have to think of other ways,"},{"Start":"06:46.670 ","End":"06:51.120","Text":"but this will really solve most of the questions that they could ask you."},{"Start":"06:51.310 ","End":"06:56.840","Text":"Also, another tip is before starting to try and understand what your intuition,"},{"Start":"06:56.840 ","End":"07:00.755","Text":"what is going to happen in a given question and then trying to"},{"Start":"07:00.755 ","End":"07:05.675","Text":"find the correct mathematical expression in order to work this out,"},{"Start":"07:05.675 ","End":"07:09.965","Text":"I would suggest doing these things so working out"},{"Start":"07:09.965 ","End":"07:14.485","Text":"what is conserved or working out the forces and moments."},{"Start":"07:14.485 ","End":"07:17.390","Text":"Then once you\u0027ve done that and you get your answer,"},{"Start":"07:17.390 ","End":"07:21.790","Text":"then see if your answer works well with your intuition."},{"Start":"07:21.790 ","End":"07:24.500","Text":"That\u0027s a good way of checking if you\u0027ve made any mistakes,"},{"Start":"07:24.500 ","End":"07:28.620","Text":"especially in your calculations with minuses or whatever it might be."},{"Start":"07:30.340 ","End":"07:34.910","Text":"There will also be some questions that require you to check what\u0027s"},{"Start":"07:34.910 ","End":"07:38.915","Text":"being conserved and write forces and moments equations."},{"Start":"07:38.915 ","End":"07:41.465","Text":"But usually it\u0027s going to be one or the other."},{"Start":"07:41.465 ","End":"07:44.930","Text":"Now, a little trick is that if in"},{"Start":"07:44.930 ","End":"07:49.189","Text":"the question you\u0027re being asked something about velocity."},{"Start":"07:49.189 ","End":"07:51.005","Text":"I would go into,"},{"Start":"07:51.005 ","End":"07:53.345","Text":"if anything is conserved because obviously,"},{"Start":"07:53.345 ","End":"08:00.465","Text":"you have here an equation linking velocity and angular velocity in these equations."},{"Start":"08:00.465 ","End":"08:05.030","Text":"If they\u0027re asking you something about forces or something about accelerations,"},{"Start":"08:05.030 ","End":"08:11.420","Text":"then I would go into your force and moments section,"},{"Start":"08:11.420 ","End":"08:15.770","Text":"because here, you can see that there\u0027s expressions involving also forces,"},{"Start":"08:15.770 ","End":"08:19.680","Text":"and also acceleration, and angular acceleration."}],"ID":9423},{"Watched":false,"Name":"Example- Ball On a Slope","Duration":"23m ","ChapterTopicVideoID":9154,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.255","Text":"Hello. In the last lesson,"},{"Start":"00:03.255 ","End":"00:05.820","Text":"we learned how to approach different types of"},{"Start":"00:05.820 ","End":"00:09.930","Text":"questions and we said that we were going to do an easy example."},{"Start":"00:09.930 ","End":"00:12.390","Text":"Here\u0027s the easy example."},{"Start":"00:12.390 ","End":"00:14.580","Text":"In the question, we\u0027re being asked,"},{"Start":"00:14.580 ","End":"00:17.880","Text":"what is the velocity of the ball at the bottom of the slope?"},{"Start":"00:17.880 ","End":"00:21.420","Text":"We have some slope at an angle of Theta to"},{"Start":"00:21.420 ","End":"00:26.325","Text":"the ground and some ball with radius R and mass m,"},{"Start":"00:26.325 ","End":"00:31.360","Text":"which is located at a height h off the ground."},{"Start":"00:31.490 ","End":"00:39.310","Text":"The ball rolls down the slope until it gets to this point,"},{"Start":"00:39.310 ","End":"00:44.940","Text":"and it has some velocity v final."},{"Start":"00:44.940 ","End":"00:48.570","Text":"They\u0027re asking us to find out what this is."},{"Start":"00:48.570 ","End":"00:51.970","Text":"Let\u0027s see how to answer this question."},{"Start":"00:52.730 ","End":"00:56.730","Text":"Let\u0027s take a look if anything conserved."},{"Start":"00:56.730 ","End":"00:58.695","Text":"We\u0027re going to start off with our energy."},{"Start":"00:58.695 ","End":"01:01.640","Text":"That means that we have to see that all the forces are"},{"Start":"01:01.640 ","End":"01:07.290","Text":"conserved or that the work on the non-conservative forces is equal to 0."},{"Start":"01:07.400 ","End":"01:10.950","Text":"The forces acting on the ball is mg,"},{"Start":"01:10.950 ","End":"01:16.420","Text":"and mg is a conserving force."},{"Start":"01:16.420 ","End":"01:21.604","Text":"Then we also have our normal force acting now on normal force."},{"Start":"01:21.604 ","End":"01:25.890","Text":"Our movement is in this direction and so"},{"Start":"01:25.890 ","End":"01:30.735","Text":"this is our Delta x where our position is changing like this."},{"Start":"01:30.735 ","End":"01:35.465","Text":"Our normal force is perpendicular to it, 90 degrees,"},{"Start":"01:35.465 ","End":"01:45.890","Text":"which means that it doesn\u0027t interest us because the work on the normal is equal to 0."},{"Start":"01:45.890 ","End":"01:50.605","Text":"The work on the non-conservative force is equal to 0."},{"Start":"01:50.605 ","End":"01:53.235","Text":"This is fine, it\u0027s conserving energy."},{"Start":"01:53.235 ","End":"01:55.430","Text":"Then the next force that we have because"},{"Start":"01:55.430 ","End":"01:59.540","Text":"the ball is starting from rest and then it starts rolling."},{"Start":"01:59.540 ","End":"02:03.419","Text":"We know that it\u0027s going to be rolling without slipping."},{"Start":"02:03.430 ","End":"02:08.605","Text":"Which means that we\u0027re going to have at the beginning static friction,"},{"Start":"02:08.605 ","End":"02:11.045","Text":"f_s is static friction."},{"Start":"02:11.045 ","End":"02:18.830","Text":"This is also conserving force because there\u0027s no energy loss in this."},{"Start":"02:18.830 ","End":"02:22.110","Text":"We have conservation of energy."},{"Start":"02:22.550 ","End":"02:28.310","Text":"Now we can look at if there\u0027s a conservation of momentum or angular momentum."},{"Start":"02:28.310 ","End":"02:34.850","Text":"There is no conservation of momentum because we have friction and we have"},{"Start":"02:34.850 ","End":"02:41.420","Text":"the force due to gravity and both of these forces are external forces,"},{"Start":"02:41.420 ","End":"02:46.180","Text":"which means that the sum of the external forces does not equal 0."},{"Start":"02:46.180 ","End":"02:49.340","Text":"The same thing with our angular momentum and also isn\u0027t"},{"Start":"02:49.340 ","End":"02:53.225","Text":"conserved because the sum of the torques if you work it out,"},{"Start":"02:53.225 ","End":"02:55.250","Text":"it also isn\u0027t equal to 0."},{"Start":"02:55.250 ","End":"02:58.350","Text":"We just have conservation of energy."},{"Start":"02:59.420 ","End":"03:01.760","Text":"Now our fourth thing,"},{"Start":"03:01.760 ","End":"03:05.968","Text":"so we have connecting the velocity expressions."},{"Start":"03:05.968 ","End":"03:08.854","Text":"Because we\u0027re dealing with rolling without slipping,"},{"Start":"03:08.854 ","End":"03:14.240","Text":"so that means that our ball is rotating as it goes down the slope,"},{"Start":"03:14.240 ","End":"03:17.680","Text":"and it also has a velocity in this direction."},{"Start":"03:17.680 ","End":"03:21.230","Text":"That means that we can write down another equation"},{"Start":"03:21.230 ","End":"03:24.860","Text":"which links the angular velocity to the linear velocity,"},{"Start":"03:24.860 ","End":"03:29.794","Text":"which is v is equal to Omega multiplied by the radius,"},{"Start":"03:29.794 ","End":"03:33.560","Text":"which here is R. Now what"},{"Start":"03:33.560 ","End":"03:37.760","Text":"we\u0027re going to do is we\u0027re going to start by writing out the equations for the energy."},{"Start":"03:37.760 ","End":"03:39.800","Text":"We have the initial energy,"},{"Start":"03:39.800 ","End":"03:42.130","Text":"which is the energy right at the start."},{"Start":"03:42.130 ","End":"03:45.230","Text":"Because we know that our ball is stationary,"},{"Start":"03:45.230 ","End":"03:49.070","Text":"so it has no kinetic energy because it\u0027s v=0."},{"Start":"03:49.070 ","End":"03:51.770","Text":"However, it is located at a certain height,"},{"Start":"03:51.770 ","End":"03:53.675","Text":"so it\u0027s going to have potential energy."},{"Start":"03:53.675 ","End":"03:58.474","Text":"Potential energy is its mass multiplied by gravity,"},{"Start":"03:58.474 ","End":"04:01.260","Text":"multiplied by its height."},{"Start":"04:02.960 ","End":"04:10.490","Text":"Now I\u0027m going to write the equation for the energy at the final stage over here."},{"Start":"04:10.490 ","End":"04:13.565","Text":"First of all, we know that our ball is moving,"},{"Start":"04:13.565 ","End":"04:16.225","Text":"so it\u0027s going to have kinetic energy."},{"Start":"04:16.225 ","End":"04:20.460","Text":"The usual, so it\u0027s going to be 1/2 multiplied by m,"},{"Start":"04:20.460 ","End":"04:24.750","Text":"multiplied by the v center of mass squared."},{"Start":"04:24.750 ","End":"04:27.545","Text":"Now because we\u0027re rolling without slipping,"},{"Start":"04:27.545 ","End":"04:31.415","Text":"so the ball is also rotating around itself,"},{"Start":"04:31.415 ","End":"04:36.380","Text":"so it\u0027s going to have the added kinetic energy of the rotation,"},{"Start":"04:36.380 ","End":"04:37.970","Text":"which is going to be 1/2."},{"Start":"04:37.970 ","End":"04:40.130","Text":"Then when we\u0027re speaking about angular momentum,"},{"Start":"04:40.130 ","End":"04:46.310","Text":"if you remember the lesson where we compare linear and angular equations."},{"Start":"04:46.310 ","End":"04:52.370","Text":"Instead of m, we\u0027ll have I and this will be around the center of mass."},{"Start":"04:52.370 ","End":"04:56.690","Text":"Here this is the velocity that the center of mass is moving at and"},{"Start":"04:56.690 ","End":"05:00.710","Text":"here the I is the moment of inertia about the center of mass."},{"Start":"05:00.710 ","End":"05:03.369","Text":"The center of mass being the axis of rotation."},{"Start":"05:03.369 ","End":"05:12.325","Text":"Then instead of velocity we\u0027ll have angular velocity squared."},{"Start":"05:12.325 ","End":"05:17.470","Text":"Also this equation we learned when we were speaking about"},{"Start":"05:17.470 ","End":"05:24.445","Text":"the energy when we have an axis of rotation which is also moving, which isn\u0027t fixed."},{"Start":"05:24.445 ","End":"05:29.240","Text":"If you don\u0027t remember this, go back to one of the earlier lessons in this chapter."},{"Start":"05:30.500 ","End":"05:35.380","Text":"Now I need my I_cm of a ball."},{"Start":"05:35.510 ","End":"05:38.605","Text":"They should be written on your equation sheet."},{"Start":"05:38.605 ","End":"05:40.480","Text":"The I_cm of a ball,"},{"Start":"05:40.480 ","End":"05:50.390","Text":"it\u0027s right down over here is going to be equal to 2/5 mR^2."},{"Start":"05:50.880 ","End":"05:55.320","Text":"Now, we also know that here"},{"Start":"05:55.320 ","End":"05:58.230","Text":"our V_cm and it\u0027s linked to"},{"Start":"05:58.230 ","End":"06:01.875","Text":"the Omega because we have two unknowns that\u0027s our V and our Omega."},{"Start":"06:01.875 ","End":"06:07.245","Text":"We can just instead of Omega write in our V_cm."},{"Start":"06:07.245 ","End":"06:10.850","Text":"Then we say that because we have conservation of energy,"},{"Start":"06:10.850 ","End":"06:13.490","Text":"so our E_i is equal to our E_f."},{"Start":"06:13.490 ","End":"06:17.215","Text":"That means that we have mgh,"},{"Start":"06:17.215 ","End":"06:19.650","Text":"E_i is equal to E_f."},{"Start":"06:19.650 ","End":"06:27.720","Text":"It\u0027s equal to 1/2 mV_cm^2 plus 1/2."},{"Start":"06:27.720 ","End":"06:30.940","Text":"Then we can substitute in our I_cm."},{"Start":"06:31.760 ","End":"06:35.415","Text":"2/5mR^2."},{"Start":"06:35.415 ","End":"06:39.480","Text":"That\u0027s our I_cm multiplied by our Omega squared."},{"Start":"06:39.480 ","End":"06:41.630","Text":"If we see what our Omega is,"},{"Start":"06:41.630 ","End":"06:44.105","Text":"it\u0027s going to be V_cm divided by I."},{"Start":"06:44.105 ","End":"06:47.300","Text":"Because it\u0027s squared, it\u0027s going to be V_cm divided"},{"Start":"06:47.300 ","End":"06:52.485","Text":"by I and both of these are going to be squared."},{"Start":"06:52.485 ","End":"06:55.330","Text":"Then I can cancel out this."},{"Start":"06:55.330 ","End":"07:01.968","Text":"Then I can rearrange this in order to isolate out my V_cm."},{"Start":"07:01.968 ","End":"07:05.020","Text":"Then through simple algebra,"},{"Start":"07:05.020 ","End":"07:08.875","Text":"I\u0027ll isolate out my V_cm and I\u0027ll get this root,"},{"Start":"07:08.875 ","End":"07:11.785","Text":"10/7 multiplied by g times"},{"Start":"07:11.785 ","End":"07:18.430","Text":"h. Question Number 2 is asking us what is the ball\u0027s acceleration?"},{"Start":"07:18.430 ","End":"07:21.640","Text":"Now a common mistake is to see the word acceleration and"},{"Start":"07:21.640 ","End":"07:25.765","Text":"think acceleration is the derivative of my velocity,"},{"Start":"07:25.765 ","End":"07:27.970","Text":"so I\u0027m just going to take the derivative,"},{"Start":"07:27.970 ","End":"07:29.635","Text":"I\u0027m going to differentiate this."},{"Start":"07:29.635 ","End":"07:34.450","Text":"That\u0027s not correct because acceleration is a function of time"},{"Start":"07:34.450 ","End":"07:41.245","Text":"and right now our V_cm is just the velocity right at the bottom over here."},{"Start":"07:41.245 ","End":"07:43.570","Text":"I don\u0027t have this equation."},{"Start":"07:43.570 ","End":"07:46.855","Text":"As you can see, there is no time variable in here,"},{"Start":"07:46.855 ","End":"07:50.090","Text":"so I can\u0027t differentiate this."},{"Start":"07:50.850 ","End":"07:57.290","Text":"What I\u0027m going to have to do is we\u0027re going to have to think of another way to do this."},{"Start":"07:57.540 ","End":"08:01.945","Text":"Also before we discuss the way to do it,"},{"Start":"08:01.945 ","End":"08:06.640","Text":"I\u0027m also taking the acceleration of the center of mass because obviously,"},{"Start":"08:06.640 ","End":"08:10.900","Text":"every single point on the ball is going to have a different acceleration."},{"Start":"08:10.900 ","End":"08:14.575","Text":"I\u0027m specifically taking the center of mass,"},{"Start":"08:14.575 ","End":"08:16.850","Text":"just like I took the velocity."},{"Start":"08:16.850 ","End":"08:20.715","Text":"How am I going to do this? I need to find the acceleration."},{"Start":"08:20.715 ","End":"08:24.705","Text":"In questions like this where we\u0027re being asked to find something with acceleration,"},{"Start":"08:24.705 ","End":"08:27.687","Text":"I\u0027m going to go to this arrow forces and moments,"},{"Start":"08:27.687 ","End":"08:29.505","Text":"because already over here,"},{"Start":"08:29.505 ","End":"08:32.440","Text":"I have a very simple equation with"},{"Start":"08:32.440 ","End":"08:36.940","Text":"my unknown that I\u0027m trying to find out right over here."},{"Start":"08:36.940 ","End":"08:43.040","Text":"We\u0027re going to work this out by finding the equation of forces."},{"Start":"08:44.220 ","End":"08:51.445","Text":"We\u0027re going to be using the equation for the sum of forces is equal to ma_cm."},{"Start":"08:51.445 ","End":"08:58.855","Text":"The first thing that we\u0027re going to do is we\u0027re going to draw the diagram of forces."},{"Start":"08:58.855 ","End":"09:00.944","Text":"We\u0027re working, of course,"},{"Start":"09:00.944 ","End":"09:03.535","Text":"from the center of mass and we know"},{"Start":"09:03.535 ","End":"09:06.700","Text":"that from the center of mass acting straight downwards,"},{"Start":"09:06.700 ","End":"09:14.605","Text":"we\u0027re going to have our mg. Then we\u0027re going to have 90 degrees to the surface."},{"Start":"09:14.605 ","End":"09:17.980","Text":"In this direction we\u0027re going to have our normal force."},{"Start":"09:17.980 ","End":"09:24.970","Text":"Then the last force acting on the body is going to be our static friction."},{"Start":"09:24.970 ","End":"09:26.489","Text":"Now the static friction,"},{"Start":"09:26.489 ","End":"09:29.095","Text":"we can see in this case that it\u0027s pointing upwards."},{"Start":"09:29.095 ","End":"09:31.210","Text":"However, it doesn\u0027t matter if you draw it in"},{"Start":"09:31.210 ","End":"09:34.375","Text":"the wrong direction because you\u0027ll see that everything will work out."},{"Start":"09:34.375 ","End":"09:37.285","Text":"Just for the sake of proving my point,"},{"Start":"09:37.285 ","End":"09:40.030","Text":"so I\u0027ll draw it in this direction,"},{"Start":"09:40.030 ","End":"09:41.545","Text":"the wrong direction,"},{"Start":"09:41.545 ","End":"09:46.790","Text":"so f_s and you\u0027ll see that the equation is still going to work out."},{"Start":"09:48.750 ","End":"09:53.754","Text":"Even though we\u0027re dealing with a large ball, a large sphere,"},{"Start":"09:53.754 ","End":"09:55.165","Text":"which is a rigid body,"},{"Start":"09:55.165 ","End":"09:58.930","Text":"it doesn\u0027t really matter because we\u0027re focusing only on the center of mass."},{"Start":"09:58.930 ","End":"10:04.210","Text":"We\u0027re just looking at it as a point mass which is located at the center of mass."},{"Start":"10:04.210 ","End":"10:08.410","Text":"Now what we\u0027re going to do is we\u0027re going to write out the equation for the sum of"},{"Start":"10:08.410 ","End":"10:12.025","Text":"all of the forces and we\u0027re going to do this according to the axes."},{"Start":"10:12.025 ","End":"10:15.265","Text":"First, we\u0027re going to do this according to the x-axis."},{"Start":"10:15.265 ","End":"10:18.730","Text":"The sum of all of the forces on the x-axis is equal to"},{"Start":"10:18.730 ","End":"10:24.110","Text":"its mass multiplied by its acceleration on the x-axis."},{"Start":"10:24.630 ","End":"10:33.895","Text":"If we say that the axes in this direction is the x-axis in the positive direction and"},{"Start":"10:33.895 ","End":"10:43.405","Text":"perpendicular to the slope is our y-axis in the positive direction."},{"Start":"10:43.405 ","End":"10:47.455","Text":"Which forces are working on the x-axis?"},{"Start":"10:47.455 ","End":"10:49.780","Text":"We have our static friction."},{"Start":"10:49.780 ","End":"10:51.805","Text":"We\u0027re going to write down the sum of the forces."},{"Start":"10:51.805 ","End":"10:58.923","Text":"We have our static friction plus our x component for our mg."},{"Start":"10:58.923 ","End":"11:04.390","Text":"When we\u0027re dealing with our force on the x-axis,"},{"Start":"11:04.390 ","End":"11:06.700","Text":"or on the slope, so parallel to it,"},{"Start":"11:06.700 ","End":"11:11.275","Text":"not perpendicular, it\u0027s always going to be sine Theta."},{"Start":"11:11.275 ","End":"11:13.210","Text":"If you want to see why it\u0027s that,"},{"Start":"11:13.210 ","End":"11:14.920","Text":"you can work it out with the angles."},{"Start":"11:14.920 ","End":"11:18.070","Text":"But as a rule, parallel to the slope is"},{"Start":"11:18.070 ","End":"11:22.310","Text":"going to be sine Theta and perpendicular will be cosine of Theta."},{"Start":"11:23.040 ","End":"11:32.990","Text":"This is going to be equal to our mass times our acceleration in our x-direction in cm."},{"Start":"11:33.390 ","End":"11:38.740","Text":"We\u0027ll notice from the diagram that we have"},{"Start":"11:38.740 ","End":"11:43.960","Text":"our normal force acting from our center of mass."},{"Start":"11:43.960 ","End":"11:46.690","Text":"We have our mg acting from the center of mass."},{"Start":"11:46.690 ","End":"11:53.725","Text":"However, our static friction force is acting from the edge of the sphere."},{"Start":"11:53.725 ","End":"11:56.680","Text":"When I wrote out this equation,"},{"Start":"11:56.680 ","End":"12:00.115","Text":"I moved it to the center of mass."},{"Start":"12:00.115 ","End":"12:03.100","Text":"However, in order to balance this out,"},{"Start":"12:03.100 ","End":"12:09.845","Text":"I have to write the equation of the torques and then I have to fix this."},{"Start":"12:09.845 ","End":"12:12.760","Text":"Let\u0027s see how to do this."},{"Start":"12:13.680 ","End":"12:18.535","Text":"Now we\u0027re going to write the equation of our torques."},{"Start":"12:18.535 ","End":"12:21.805","Text":"Our axes of rotation is our center of mass."},{"Start":"12:21.805 ","End":"12:24.140","Text":"It\u0027s our red point over here."},{"Start":"12:24.800 ","End":"12:27.090","Text":"Let\u0027s look at the forces."},{"Start":"12:27.090 ","End":"12:30.810","Text":"We can write that the torque our mg is going to be"},{"Start":"12:30.810 ","End":"12:35.595","Text":"equal to 0 because it\u0027s acting out from the center of mass."},{"Start":"12:35.595 ","End":"12:41.365","Text":"Which means that it\u0027s acting out of the axis of rotation,"},{"Start":"12:41.365 ","End":"12:46.390","Text":"which means that its moment or its torque will be equal to 0."},{"Start":"12:46.390 ","End":"12:50.965","Text":"Next, we can speak about the torque of the normal force."},{"Start":"12:50.965 ","End":"12:55.075","Text":"Technically the normal force is acting from the edge,"},{"Start":"12:55.075 ","End":"12:57.460","Text":"not from the center of mass, however,"},{"Start":"12:57.460 ","End":"13:03.505","Text":"we can see that our radius will be acting in this direction,"},{"Start":"13:03.505 ","End":"13:07.360","Text":"which is 180 degrees from the normal."},{"Start":"13:07.360 ","End":"13:09.982","Text":"It\u0027s still parallel."},{"Start":"13:09.982 ","End":"13:13.090","Text":"When you work this out,"},{"Start":"13:13.090 ","End":"13:18.850","Text":"you\u0027ll see that it will still be equal to 0 because they\u0027re both parallel."},{"Start":"13:18.850 ","End":"13:23.120","Text":"The torque of the normal will also be equal to 0."},{"Start":"13:23.820 ","End":"13:31.360","Text":"Now we\u0027re going to write out our moment of force for our static friction."},{"Start":"13:31.360 ","End":"13:34.585","Text":"That\u0027s going to be equal to"},{"Start":"13:34.585 ","End":"13:40.570","Text":"our static friction multiplied by the radius because where is it acting from?"},{"Start":"13:40.570 ","End":"13:42.310","Text":"It\u0027s acting right from the edge,"},{"Start":"13:42.310 ","End":"13:45.325","Text":"the section which is in contact with the slope,"},{"Start":"13:45.325 ","End":"13:50.050","Text":"which is exactly a distance R from our axis of rotation."},{"Start":"13:50.050 ","End":"13:55.300","Text":"It\u0027s the force multiplied by its distance from the axis of rotation,"},{"Start":"13:55.300 ","End":"14:01.945","Text":"which is R. Then we\u0027re going to multiply it by sine of the angle."},{"Start":"14:01.945 ","End":"14:04.575","Text":"Our angle here is 90 degrees."},{"Start":"14:04.575 ","End":"14:08.375","Text":"As you can see, our radius is perpendicular to our f_s,"},{"Start":"14:08.375 ","End":"14:11.060","Text":"so it\u0027s going to be sine of 90 degrees and this,"},{"Start":"14:11.060 ","End":"14:13.965","Text":"as we know, is equal to 1."},{"Start":"14:13.965 ","End":"14:17.560","Text":"We can just write 1 over here."},{"Start":"14:17.560 ","End":"14:23.130","Text":"Now what we\u0027re going to do is we\u0027re going to speak about the sine."},{"Start":"14:23.820 ","End":"14:30.620","Text":"Because I want my v to equal Omega R and not negative Omega R, it doesn\u0027t really matter."},{"Start":"14:30.620 ","End":"14:32.060","Text":"You can have it either way."},{"Start":"14:32.060 ","End":"14:33.590","Text":"But just for the sake of it,"},{"Start":"14:33.590 ","End":"14:35.105","Text":"for making it easier for me,"},{"Start":"14:35.105 ","End":"14:41.990","Text":"I want my v to equal Omega R. What I\u0027m going to do is I\u0027m going to say that"},{"Start":"14:41.990 ","End":"14:50.354","Text":"my positive direction for angular velocity is in this direction."},{"Start":"14:50.354 ","End":"14:53.070","Text":"Because of that,"},{"Start":"14:53.070 ","End":"14:57.820","Text":"my y-axis is going to switch directions."},{"Start":"14:57.820 ","End":"15:00.100","Text":"Instead of pointing in this direction,"},{"Start":"15:00.100 ","End":"15:02.540","Text":"it\u0027s going to point in this."},{"Start":"15:02.970 ","End":"15:07.520","Text":"Now my y-axis is pointing in this direction."},{"Start":"15:07.680 ","End":"15:10.165","Text":"Again, it doesn\u0027t really matter."},{"Start":"15:10.165 ","End":"15:14.480","Text":"It\u0027s just going to make sure that this turns out how I want it to."},{"Start":"15:15.300 ","End":"15:22.285","Text":"Now in order to see which direction my static friction is pointing in,"},{"Start":"15:22.285 ","End":"15:27.490","Text":"to see what sign it is if here is going to be a positive or a negative."},{"Start":"15:27.490 ","End":"15:29.230","Text":"I can take a look."},{"Start":"15:29.230 ","End":"15:32.110","Text":"If I hold my sphere at its center,"},{"Start":"15:32.110 ","End":"15:34.083","Text":"at its axis of rotation,"},{"Start":"15:34.083 ","End":"15:37.870","Text":"and then from my point of contact where my force,"},{"Start":"15:37.870 ","End":"15:42.100","Text":"my static friction is acting."},{"Start":"15:42.100 ","End":"15:47.275","Text":"Then I pull this section down in this direction."},{"Start":"15:47.275 ","End":"15:51.175","Text":"Then I can see if I\u0027m pulling in this direction,"},{"Start":"15:51.175 ","End":"15:57.010","Text":"my sphere will turn in the anticlockwise direction. Can you see that?"},{"Start":"15:57.010 ","End":"16:01.045","Text":"Can you imagine if I have, let\u0027s enlarge this."},{"Start":"16:01.045 ","End":"16:05.200","Text":"I have my sphere and it\u0027s stuck here with some screw."},{"Start":"16:05.200 ","End":"16:09.310","Text":"I am pulling it over here in this direction."},{"Start":"16:09.310 ","End":"16:14.395","Text":"That means that the whole sphere is going to move around in this direction."},{"Start":"16:14.395 ","End":"16:18.970","Text":"Like that, which is in the exact opposite direction to"},{"Start":"16:18.970 ","End":"16:24.445","Text":"my positive direction for angular velocity."},{"Start":"16:24.445 ","End":"16:28.160","Text":"I\u0027m going to put a minus over here."},{"Start":"16:29.040 ","End":"16:32.440","Text":"Now I have a few equations."},{"Start":"16:32.440 ","End":"16:35.770","Text":"I have this equation over here."},{"Start":"16:35.770 ","End":"16:39.385","Text":"I have this equation over here,"},{"Start":"16:39.385 ","End":"16:41.530","Text":"and now I need my third equation."},{"Start":"16:41.530 ","End":"16:43.510","Text":"What\u0027s my third equation going to be?"},{"Start":"16:43.510 ","End":"16:46.555","Text":"Because I\u0027m dealing with linear motion and also angular motion."},{"Start":"16:46.555 ","End":"16:50.665","Text":"I can use this equation connecting there accelerations."},{"Start":"16:50.665 ","End":"16:55.300","Text":"My third and final equation is going to be my acceleration is equal"},{"Start":"16:55.300 ","End":"17:00.645","Text":"to my circular acceleration multiplied by the radius,"},{"Start":"17:00.645 ","End":"17:05.005","Text":"which is R. Lastly,"},{"Start":"17:05.005 ","End":"17:11.245","Text":"we also know that this is equal to the sum of all my torques is equal to I Alpha."},{"Start":"17:11.245 ","End":"17:16.690","Text":"I know that this is going to equal to because this and this is 0."},{"Start":"17:16.690 ","End":"17:22.900","Text":"This will just be the total sum of my torques is equal to I Alpha."},{"Start":"17:22.900 ","End":"17:27.445","Text":"Now, of course, this is my acceleration on the x-axis."},{"Start":"17:27.445 ","End":"17:32.540","Text":"These are my 3 equations."},{"Start":"17:34.170 ","End":"17:38.005","Text":"Now I have my unknowns."},{"Start":"17:38.005 ","End":"17:40.195","Text":"My F_s is my unknown,"},{"Start":"17:40.195 ","End":"17:43.075","Text":"my axcm is my unknown,"},{"Start":"17:43.075 ","End":"17:44.830","Text":"and also my Alphas are known."},{"Start":"17:44.830 ","End":"17:47.140","Text":"I have 3 unknowns and 3 equations."},{"Start":"17:47.140 ","End":"17:49.195","Text":"That means that I can solve it."},{"Start":"17:49.195 ","End":"17:52.330","Text":"Right now I\u0027m just going to do some algebra in order to solve it,"},{"Start":"17:52.330 ","End":"17:54.760","Text":"to find out the ball\u0027s acceleration."},{"Start":"17:54.760 ","End":"17:59.060","Text":"If you don\u0027t have time, you can skip this step."},{"Start":"18:00.780 ","End":"18:04.360","Text":"Because I want to find out my acceleration,"},{"Start":"18:04.360 ","End":"18:11.785","Text":"I\u0027m just going to write my Alpha in terms of the acceleration."},{"Start":"18:11.785 ","End":"18:14.845","Text":"That means that my Alpha is going to be a_x"},{"Start":"18:14.845 ","End":"18:19.270","Text":"divided by R. That\u0027s what I\u0027m going to substitute in over here."},{"Start":"18:19.270 ","End":"18:22.750","Text":"I\u0027m going to have that my I."},{"Start":"18:22.750 ","End":"18:25.270","Text":"Let\u0027s scroll back up to see what my I is."},{"Start":"18:25.270 ","End":"18:32.245","Text":"My I is 2/5mR^2 multiplied by Alpha,"},{"Start":"18:32.245 ","End":"18:38.335","Text":"multiplied by ax divided by R. That\u0027s my I Alpha."},{"Start":"18:38.335 ","End":"18:46.765","Text":"This is equal to negative f_s multiplied by R. Then on this side,"},{"Start":"18:46.765 ","End":"18:49.210","Text":"I also have this equation."},{"Start":"18:49.210 ","End":"18:52.180","Text":"I\u0027m going to fight, I\u0027m going to isolate out my f_s from"},{"Start":"18:52.180 ","End":"18:55.795","Text":"this equation and then substitute it into here."},{"Start":"18:55.795 ","End":"19:02.950","Text":"My f_s is going to equal to my max."},{"Start":"19:02.950 ","End":"19:04.630","Text":"This is also cm."},{"Start":"19:04.630 ","End":"19:06.400","Text":"I\u0027m just going to skip the cm already,"},{"Start":"19:06.400 ","End":"19:11.125","Text":"but we\u0027re using the acceleration of the center of mass the whole time."},{"Start":"19:11.125 ","End":"19:20.125","Text":"That\u0027s going to equal that minus mg sine of Theta."},{"Start":"19:20.125 ","End":"19:23.965","Text":"Now I\u0027m going to substitute in this f_s into here."},{"Start":"19:23.965 ","End":"19:29.140","Text":"I\u0027m going to get 2/5mR^2."},{"Start":"19:29.140 ","End":"19:33.820","Text":"I can already cross out this R and this R and also"},{"Start":"19:33.820 ","End":"19:40.045","Text":"this R. That means also this R. I\u0027ve gotten rid of all of my R,"},{"Start":"19:40.045 ","End":"19:41.830","Text":"because I had R^2 divided by R,"},{"Start":"19:41.830 ","End":"19:45.460","Text":"so that\u0027s R. Then I crossed out the 2 Rs from both sides."},{"Start":"19:45.460 ","End":"19:49.350","Text":"I have 2/5max,"},{"Start":"19:49.350 ","End":"19:52.215","Text":"which is going to be equal to just my f_s,"},{"Start":"19:52.215 ","End":"20:00.360","Text":"which will equal max minus mg sine of Theta."},{"Start":"20:00.360 ","End":"20:02.640","Text":"Now I can cross out my m\u0027s."},{"Start":"20:02.640 ","End":"20:05.080","Text":"I can cancel them out."},{"Start":"20:05.080 ","End":"20:12.560","Text":"Now I can take my ax to the other side and then isolate all my ax and I have my answer."},{"Start":"20:14.250 ","End":"20:21.895","Text":"I\u0027m going to add g sine of Theta to both sides and minus 2/5a(x) from both sides."},{"Start":"20:21.895 ","End":"20:32.480","Text":"I\u0027ll have g sine of Theta is going to equal ax minus 2/5ax."},{"Start":"20:36.060 ","End":"20:44.110","Text":"Now I can say that my g sine of Theta is going to equal ax."},{"Start":"20:44.110 ","End":"20:47.680","Text":"That\u0027s like 5/5ax minus 2/5ax."},{"Start":"20:47.680 ","End":"20:53.725","Text":"That\u0027s going to be 3/5(ax)=g sine Theta."},{"Start":"20:53.725 ","End":"21:04.040","Text":"That\u0027s going to mean that my ax=5/3(g) sine Theta."},{"Start":"21:06.090 ","End":"21:08.785","Text":"This is the final answer."},{"Start":"21:08.785 ","End":"21:14.410","Text":"Now we can say, what if they asked us to find the position as a function of time?"},{"Start":"21:14.410 ","End":"21:19.585","Text":"That would say our xt is equal to our beginning position,"},{"Start":"21:19.585 ","End":"21:22.660","Text":"which is equal to 0 plus"},{"Start":"21:22.660 ","End":"21:28.255","Text":"our beginning velocity multiplied by t. Our beginning velocity is also equal to 0."},{"Start":"21:28.255 ","End":"21:33.370","Text":"Then plus 1/2at^2."},{"Start":"21:33.370 ","End":"21:35.125","Text":"Then obviously this a,"},{"Start":"21:35.125 ","End":"21:38.360","Text":"I would substitute this in over here."},{"Start":"21:39.030 ","End":"21:44.560","Text":"Then if they asked us to find the angle as a function of time,"},{"Start":"21:44.560 ","End":"21:51.385","Text":"so that\u0027s equal to x divided by R. This simply this"},{"Start":"21:51.385 ","End":"21:55.660","Text":"1/2at^2 divided by R. But let\u0027s say that we didn\u0027t find out"},{"Start":"21:55.660 ","End":"22:00.700","Text":"our x and we\u0027re just going straight into the angle divided by time."},{"Start":"22:00.700 ","End":"22:05.680","Text":"We know that our angular acceleration is equal to"},{"Start":"22:05.680 ","End":"22:12.475","Text":"our linear acceleration divided by R. If we substitute that in over here,"},{"Start":"22:12.475 ","End":"22:16.540","Text":"so we can start with our initial angle,"},{"Start":"22:16.540 ","End":"22:18.505","Text":"which is equal to 0,"},{"Start":"22:18.505 ","End":"22:22.885","Text":"plus our initial angular velocity multiplied by t,"},{"Start":"22:22.885 ","End":"22:31.300","Text":"which is also equal to 0 because we\u0027re starting at rest plus our 1/2 Alpha t^2,"},{"Start":"22:31.300 ","End":"22:34.600","Text":"where Alpha is just going to be equal to"},{"Start":"22:34.600 ","End":"22:42.115","Text":"5/3R g sine of Theta."},{"Start":"22:42.115 ","End":"22:44.810","Text":"It\u0027s the exact same thing."},{"Start":"22:45.420 ","End":"22:50.995","Text":"If they were asking us how many rotations the ball did,"},{"Start":"22:50.995 ","End":"22:58.120","Text":"we can just work this out by Theta over 2Pi and then solving this with our time."},{"Start":"22:58.120 ","End":"23:01.250","Text":"That\u0027s the end of this question."}],"ID":9424},{"Watched":false,"Name":"Choosing an Axis of Rotation at The Surface","Duration":"12m 35s","ChapterTopicVideoID":9155,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:02.670","Text":"Hello. In this lesson,"},{"Start":"00:02.670 ","End":"00:06.970","Text":"we\u0027re going to be speaking about choosing an axis of rotation."},{"Start":"00:07.010 ","End":"00:12.495","Text":"Let\u0027s assume for now that this is our center of mass."},{"Start":"00:12.495 ","End":"00:15.690","Text":"Usually, when we\u0027re dealing with these types of questions,"},{"Start":"00:15.690 ","End":"00:17.655","Text":"we choose our axis of rotation,"},{"Start":"00:17.655 ","End":"00:22.035","Text":"or we\u0027re told that our axis of rotation is at the center of mass."},{"Start":"00:22.035 ","End":"00:26.145","Text":"Or at least that\u0027s what we\u0027ve come across up until now."},{"Start":"00:26.145 ","End":"00:29.660","Text":"Now, what would happen if we choose our axis of"},{"Start":"00:29.660 ","End":"00:34.510","Text":"rotation to be at the point of contact between the wheel and the ground?"},{"Start":"00:34.510 ","End":"00:37.490","Text":"If this is turning without slipping,"},{"Start":"00:37.490 ","End":"00:41.060","Text":"then we know that our velocity at that point,"},{"Start":"00:41.060 ","End":"00:47.550","Text":"so our V at this point, P is going to be equal to 0."},{"Start":"00:49.370 ","End":"00:54.215","Text":"If we imagine that this point P is our axis of rotation,"},{"Start":"00:54.215 ","End":"01:01.090","Text":"so that means that this point has velocity 0."},{"Start":"01:01.090 ","End":"01:05.120","Text":"Also because we\u0027re told that and also because it\u0027s the axis of rotation,"},{"Start":"01:05.120 ","End":"01:08.270","Text":"so the point of the axis of rotation never moves."},{"Start":"01:08.270 ","End":"01:14.380","Text":"This whole wheel is just going to rotate around it so after a while,"},{"Start":"01:14.380 ","End":"01:18.805","Text":"our ball will be here, our wheel sorry."},{"Start":"01:18.805 ","End":"01:23.215","Text":"Sorry, I had to draw them attached to this point though here,"},{"Start":"01:23.215 ","End":"01:24.835","Text":"then it will move to here,"},{"Start":"01:24.835 ","End":"01:31.740","Text":"and then it will move to here until it returns back to its original points."},{"Start":"01:32.240 ","End":"01:38.510","Text":"We know though that because we have the ground over here so"},{"Start":"01:38.510 ","End":"01:45.815","Text":"wheel as a whole is going to move forward in this direction."},{"Start":"01:45.815 ","End":"01:47.990","Text":"How can we think about that?"},{"Start":"01:47.990 ","End":"01:51.185","Text":"We\u0027re always going to think as this point is stationary,"},{"Start":"01:51.185 ","End":"01:53.165","Text":"however, it will move along."},{"Start":"01:53.165 ","End":"01:55.085","Text":"Because as the ball rolls,"},{"Start":"01:55.085 ","End":"02:01.260","Text":"then this point over here will be the one in contact with the ground,"},{"Start":"02:01.260 ","End":"02:05.480","Text":"and then it will be the point P with a velocity of 0,"},{"Start":"02:05.480 ","End":"02:07.925","Text":"and it will be the axis of rotation."},{"Start":"02:07.925 ","End":"02:12.710","Text":"What we\u0027re actually speaking about is a temporary axis."},{"Start":"02:12.710 ","End":"02:16.625","Text":"The axis is always going to be on the same point"},{"Start":"02:16.625 ","End":"02:20.669","Text":"at contact between the wheel and the ground."},{"Start":"02:20.669 ","End":"02:25.295","Text":"However, it moves along the wheel as it rotates."},{"Start":"02:25.295 ","End":"02:28.175","Text":"Then the point of contact will be over here,"},{"Start":"02:28.175 ","End":"02:29.240","Text":"and then over here,"},{"Start":"02:29.240 ","End":"02:30.515","Text":"and then over here,"},{"Start":"02:30.515 ","End":"02:32.755","Text":"and so on, and so on."},{"Start":"02:32.755 ","End":"02:38.310","Text":"The velocity at that point is always going to be equal to 0."},{"Start":"02:38.750 ","End":"02:42.030","Text":"We\u0027re used to having a question,"},{"Start":"02:42.030 ","End":"02:46.700","Text":"where if this is our axis of rotation and this is our rod,"},{"Start":"02:46.700 ","End":"02:49.110","Text":"and its point of contact is here,"},{"Start":"02:49.110 ","End":"02:52.145","Text":"so as the rod rotates about this axis,"},{"Start":"02:52.145 ","End":"02:54.680","Text":"so we know that every section on the rod has"},{"Start":"02:54.680 ","End":"03:00.645","Text":"a different velocity and it\u0027s going around this axis of rotation."},{"Start":"03:00.645 ","End":"03:04.925","Text":"Every piece is going to have a certain velocity."},{"Start":"03:04.925 ","End":"03:10.070","Text":"We\u0027ve also seen questions where we have our axis of rotation and"},{"Start":"03:10.070 ","End":"03:15.795","Text":"then we have a rod which is rotating around the axis."},{"Start":"03:15.795 ","End":"03:19.534","Text":"Again, every section has its velocity,"},{"Start":"03:19.534 ","End":"03:25.370","Text":"but then we also have where the axis itself has some velocity."},{"Start":"03:25.370 ","End":"03:28.285","Text":"We have 2 components of velocity."},{"Start":"03:28.285 ","End":"03:31.940","Text":"In this example over here, with our wheel,"},{"Start":"03:31.940 ","End":"03:39.860","Text":"as it\u0027s rotating, it will be moving in this direction so it will have a VCM."},{"Start":"03:39.860 ","End":"03:43.130","Text":"Its center of mass will be moving so it has a velocity."},{"Start":"03:43.130 ","End":"03:48.650","Text":"However, this point P is always going to be at velocity is equal to 0."},{"Start":"03:48.650 ","End":"03:54.600","Text":"It\u0027s always going to be stationary because we\u0027re speaking about rolling without slipping."},{"Start":"03:56.720 ","End":"04:00.890","Text":"That\u0027s why we can say that this axis isn\u0027t"},{"Start":"04:00.890 ","End":"04:05.875","Text":"moving because the velocity at that point is always equal to 0."},{"Start":"04:05.875 ","End":"04:09.170","Text":"However, if this was our axis of rotation,"},{"Start":"04:09.170 ","End":"04:14.940","Text":"then we would say that our axis is moving and that would affect our equations."},{"Start":"04:15.170 ","End":"04:19.790","Text":"Let\u0027s take a look at what this means when we\u0027re dealing with the equations."},{"Start":"04:19.790 ","End":"04:21.470","Text":"The kinetic energy,"},{"Start":"04:21.470 ","End":"04:29.745","Text":"in red of the center of mass is going to equal 1/2 multiplied by I,"},{"Start":"04:29.745 ","End":"04:34.447","Text":"center of mass multiplied by our angular velocity ^2"},{"Start":"04:34.447 ","End":"04:43.310","Text":"plus 1/2 the mass multiplied by the velocity of the center of mass ^2."},{"Start":"04:43.310 ","End":"04:48.530","Text":"We\u0027re dealing with the angular velocity around our axis of"},{"Start":"04:48.530 ","End":"04:54.210","Text":"rotation and the linear velocity of our axis of rotation."},{"Start":"04:54.210 ","End":"04:56.495","Text":"Conversely, in blue,"},{"Start":"04:56.495 ","End":"05:03.285","Text":"I\u0027m going to be speaking about this point over here p. The kinetic energy of point P,"},{"Start":"05:03.285 ","End":"05:07.718","Text":"because its axis of rotation, is non-moving,"},{"Start":"05:07.718 ","End":"05:13.515","Text":"its velocity is 0, but there is still angular velocity."},{"Start":"05:13.515 ","End":"05:19.760","Text":"The equation for the kinetic energy is going to be 1/2I around that point,"},{"Start":"05:19.760 ","End":"05:25.940","Text":"p. None of the center of mass but of the point P multiplied by Omega ^2,"},{"Start":"05:25.940 ","End":"05:27.725","Text":"the angular velocity ^2."},{"Start":"05:27.725 ","End":"05:30.980","Text":"This section over here will be just plus 0"},{"Start":"05:30.980 ","End":"05:35.520","Text":"because our velocity about point P is equal to 0."},{"Start":"05:35.870 ","End":"05:38.790","Text":"Let\u0027s write out our equations."},{"Start":"05:38.790 ","End":"05:45.115","Text":"My ICM, I\u0027ll write that over here so my I center of mass,"},{"Start":"05:45.115 ","End":"05:47.740","Text":"so it\u0027s off a wheel,"},{"Start":"05:47.740 ","End":"05:51.350","Text":"so it\u0027s going to just be 2/5 MR^2."},{"Start":"05:53.220 ","End":"06:04.060","Text":"Then my I at point P is going to be with Steiner so it will be 2/5MR^2 plus where"},{"Start":"06:04.060 ","End":"06:08.740","Text":"the correction of it being away from the axis of rotation so it\u0027s going to be plus"},{"Start":"06:08.740 ","End":"06:11.800","Text":"MR^2 which is simply"},{"Start":"06:11.800 ","End":"06:20.340","Text":"going to be 7/5MR^2."},{"Start":"06:20.340 ","End":"06:25.745","Text":"Then another thing that we can do is what our Omega is and what our VCM is."},{"Start":"06:25.745 ","End":"06:29.510","Text":"We can say that the connection between our V center of"},{"Start":"06:29.510 ","End":"06:33.613","Text":"mass is obviously going to equal Omega multiplied by r,"},{"Start":"06:33.613 ","End":"06:36.505","Text":"the angular velocity multiplied by the radius."},{"Start":"06:36.505 ","End":"06:38.960","Text":"Now if we substitute this in,"},{"Start":"06:38.960 ","End":"06:43.030","Text":"so our E_k will become,"},{"Start":"06:43.030 ","End":"06:45.030","Text":"it\u0027s 2/5 divided by 1/2,"},{"Start":"06:45.030 ","End":"06:50.220","Text":"so it will be 1/5Icm so"},{"Start":"06:50.220 ","End":"06:53.341","Text":"MR^2 Omega"},{"Start":"06:53.341 ","End":"07:00.570","Text":"squared plus 1/2m Omega squared, R^2."},{"Start":"07:00.570 ","End":"07:03.710","Text":"I just substituted in this into this equation over"},{"Start":"07:03.710 ","End":"07:08.600","Text":"here and then my kinetic energy at point P,"},{"Start":"07:08.600 ","End":"07:14.105","Text":"again, red is speaking at the center of mass being the axis of rotation,"},{"Start":"07:14.105 ","End":"07:18.005","Text":"and my blue is this point P being the axis of rotation."},{"Start":"07:18.005 ","End":"07:22.925","Text":"I\u0027m just going to have 1/2 multiplied by my I at point P,"},{"Start":"07:22.925 ","End":"07:26.760","Text":"which is going to be 7/5MR^2."},{"Start":"07:29.780 ","End":"07:31.980","Text":"You don\u0027t have to write the brackets,"},{"Start":"07:31.980 ","End":"07:33.210","Text":"I don\u0027t know why I wrote that,"},{"Start":"07:33.210 ","End":"07:37.500","Text":"so 7/5 MR^2 multiplied by Omega squared,"},{"Start":"07:37.500 ","End":"07:46.930","Text":"so that\u0027s going to equal 7/10MR^2, Omega squared."},{"Start":"07:47.480 ","End":"07:53.580","Text":"Now, for those of you that can look at what\u0027s going on here."},{"Start":"07:53.580 ","End":"07:57.530","Text":"You\u0027ll see that our kinetic energy at"},{"Start":"07:57.530 ","End":"08:02.580","Text":"our center of mass is still going to equal 7/10MR^2 Omega squared."},{"Start":"08:08.570 ","End":"08:11.325","Text":"Notice if you add all of this up,"},{"Start":"08:11.325 ","End":"08:12.745","Text":"you\u0027re going to get this expression,"},{"Start":"08:12.745 ","End":"08:19.940","Text":"which is the exact same thing as the kinetic energy at our point P. Of course,"},{"Start":"08:19.940 ","End":"08:21.680","Text":"the kinetic energy of this wheel,"},{"Start":"08:21.680 ","End":"08:26.480","Text":"no matter which point you look at it it\u0027s going to have the exact same kinetic energy."},{"Start":"08:26.480 ","End":"08:28.935","Text":"That\u0027s the whole point,"},{"Start":"08:28.935 ","End":"08:32.270","Text":"and that\u0027s also how you can see that this is actually correct."},{"Start":"08:32.270 ","End":"08:36.695","Text":"That this axis the velocity at it is actually 0,"},{"Start":"08:36.695 ","End":"08:39.335","Text":"even though the whole body is moving around it."},{"Start":"08:39.335 ","End":"08:44.015","Text":"Then we can see that this does actually work out."},{"Start":"08:44.015 ","End":"08:49.325","Text":"Even if I was to choose to measure the kinetic energy or calculate it at this point,"},{"Start":"08:49.325 ","End":"08:50.885","Text":"it will be the same as at this point,"},{"Start":"08:50.885 ","End":"08:52.710","Text":"and at this point, and at this point, and at this point."},{"Start":"08:55.070 ","End":"08:58.865","Text":"Now something which could confuse is that"},{"Start":"08:58.865 ","End":"09:03.290","Text":"our angular momentum around our axis of rotation at"},{"Start":"09:03.290 ","End":"09:11.015","Text":"the center of mass will be this red arrow going in this way around that axis of rotation."},{"Start":"09:11.015 ","End":"09:15.875","Text":"However, if we\u0027re looking at our point P, the angular velocity."},{"Start":"09:15.875 ","End":"09:22.275","Text":"Sorry, this is angular velocity so let\u0027s call this Omega."},{"Start":"09:22.275 ","End":"09:24.350","Text":"Then if we\u0027re looking at point P,"},{"Start":"09:24.350 ","End":"09:29.855","Text":"the angular velocity around that will be going around that point, so here."},{"Start":"09:29.855 ","End":"09:33.880","Text":"Then let\u0027s call it Omega with a tilde on top."},{"Start":"09:33.880 ","End":"09:40.185","Text":"This could confuse you saying that these both aren\u0027t the same angular velocities."},{"Start":"09:40.185 ","End":"09:42.420","Text":"They aren\u0027t, 1 is going in this direction,"},{"Start":"09:42.420 ","End":"09:44.640","Text":"1 is going in this direction."},{"Start":"09:44.640 ","End":"09:48.590","Text":"Let\u0027s show us that these actually"},{"Start":"09:48.590 ","End":"09:53.290","Text":"are equal to 1 another and that\u0027s why we get the same equations."},{"Start":"09:53.390 ","End":"09:55.685","Text":"First of all, as a rule,"},{"Start":"09:55.685 ","End":"09:59.320","Text":"you should know that if there\u0027s rolling without slipping,"},{"Start":"09:59.320 ","End":"10:04.715","Text":"then R Omega is always going to be equal to r Omega tilde."},{"Start":"10:04.715 ","End":"10:06.515","Text":"Rolling without slipping."},{"Start":"10:06.515 ","End":"10:08.374","Text":"If we\u0027re rolling with slipping,"},{"Start":"10:08.374 ","End":"10:11.030","Text":"then it\u0027s a really big problem to look at"},{"Start":"10:11.030 ","End":"10:14.660","Text":"these points because these points will sometimes be stationary,"},{"Start":"10:14.660 ","End":"10:15.890","Text":"but also it will be slipping,"},{"Start":"10:15.890 ","End":"10:19.850","Text":"so it will be moving axis of rotation."},{"Start":"10:19.850 ","End":"10:23.065","Text":"Then it becomes a little bit more complicated."},{"Start":"10:23.065 ","End":"10:25.645","Text":"Only when we\u0027re doing rolling without slipping,"},{"Start":"10:25.645 ","End":"10:29.050","Text":"then we can really look at the velocity of this point as being equal to 0,"},{"Start":"10:29.050 ","End":"10:33.340","Text":"and then we can say that R Omega tilde is equal to R Omega."},{"Start":"10:33.340 ","End":"10:37.930","Text":"Another way of looking at this is that if we have,"},{"Start":"10:37.930 ","End":"10:40.180","Text":"here is our wheel,"},{"Start":"10:40.180 ","End":"10:41.890","Text":"and here is our point,"},{"Start":"10:41.890 ","End":"10:46.540","Text":"and I shape is rotating around like this."},{"Start":"10:46.540 ","End":"10:51.490","Text":"I center of mass will have a velocity of VCM."},{"Start":"10:51.490 ","End":"10:59.175","Text":"Now we know that the VCM is connected to our angular velocity so this is"},{"Start":"10:59.175 ","End":"11:03.105","Text":"Omega tilde and this will be equal to"},{"Start":"11:03.105 ","End":"11:08.660","Text":"Omega tilde R. Then if I isolate out my Omega tilde,"},{"Start":"11:08.660 ","End":"11:17.585","Text":"so I get that it equals to my VCM divided by R. I know that my VCM,"},{"Start":"11:17.585 ","End":"11:20.210","Text":"whether I\u0027m rotating around the center,"},{"Start":"11:20.210 ","End":"11:21.360","Text":"the axis of rotation,"},{"Start":"11:21.360 ","End":"11:24.365","Text":"or here my VCM is always going to be the same."},{"Start":"11:24.365 ","End":"11:29.650","Text":"My radius is obviously going to be capital R,"},{"Start":"11:29.650 ","End":"11:31.410","Text":"it also doesn\u0027t change."},{"Start":"11:31.410 ","End":"11:37.335","Text":"I can see that my VCM over R is also equal to my Omega,"},{"Start":"11:37.335 ","End":"11:39.435","Text":"not just my Omega tilde."},{"Start":"11:39.435 ","End":"11:44.540","Text":"Then we can see here that my Omega tilde is equal to my Omega because these"},{"Start":"11:44.540 ","End":"11:50.100","Text":"2 are independent of my axis of rotation."},{"Start":"11:51.500 ","End":"11:53.630","Text":"That\u0027s the end of the lesson."},{"Start":"11:53.630 ","End":"11:58.489","Text":"The important things to take away is that with rolling without slipping,"},{"Start":"11:58.489 ","End":"12:01.790","Text":"your angular velocity around the center of"},{"Start":"12:01.790 ","End":"12:05.465","Text":"mass will be the same as the angular velocity around the different points."},{"Start":"12:05.465 ","End":"12:09.005","Text":"That the velocity at the point of contact between the wheel"},{"Start":"12:09.005 ","End":"12:12.560","Text":"and the ground will always be equal to 0."},{"Start":"12:12.560 ","End":"12:15.560","Text":"That also whichever point you take,"},{"Start":"12:15.560 ","End":"12:17.705","Text":"if you\u0027re working out the kinetic energy,"},{"Start":"12:17.705 ","End":"12:21.330","Text":"so it will work out to be exactly the same."},{"Start":"12:21.330 ","End":"12:25.790","Text":"Just like when we worked out the kinetic energy around the axis of rotation,"},{"Start":"12:25.790 ","End":"12:27.590","Text":"when it\u0027s at the center of mass,"},{"Start":"12:27.590 ","End":"12:32.560","Text":"we got the same exact answer when working it out from this point of contact."},{"Start":"12:32.560 ","End":"12:35.410","Text":"That\u0027s the end of this lesson."}],"ID":9425},{"Watched":false,"Name":"Rope and Pulley","Duration":"3m 19s","ChapterTopicVideoID":9156,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:06.015","Text":"I want to speak about if we have some pulley system."},{"Start":"00:06.015 ","End":"00:09.750","Text":"Over here, what you\u0027re seeing is a pulley system with a string on top,"},{"Start":"00:09.750 ","End":"00:16.875","Text":"with a radius R. If the string is passing over the pulley system and it\u0027s not slipping,"},{"Start":"00:16.875 ","End":"00:20.355","Text":"so that means that as the string is moved along,"},{"Start":"00:20.355 ","End":"00:23.310","Text":"this pulley system is spinning."},{"Start":"00:23.310 ","End":"00:29.895","Text":"Then what I can do is find some connection between"},{"Start":"00:29.895 ","End":"00:32.775","Text":"the velocity of the string and"},{"Start":"00:32.775 ","End":"00:37.560","Text":"the angular velocity or the angular acceleration of the pulley."},{"Start":"00:37.560 ","End":"00:46.160","Text":"The equation for that is that the velocity of the rope is going to be"},{"Start":"00:46.160 ","End":"00:54.440","Text":"equal to the Omega of the pulley multiplied by the radius of the pulley,"},{"Start":"00:54.440 ","End":"00:58.505","Text":"and a similar connection when dealing with acceleration."},{"Start":"00:58.505 ","End":"01:00.860","Text":"Let\u0027s see how this happens."},{"Start":"01:00.860 ","End":"01:05.110","Text":"If I take any point on the rope,"},{"Start":"01:05.240 ","End":"01:08.145","Text":"any point, it doesn\u0027t matter."},{"Start":"01:08.145 ","End":"01:16.475","Text":"The velocity of the rope at each point is going to be the same as all the other points."},{"Start":"01:16.475 ","End":"01:20.900","Text":"Why? Because the distance between each point is a constant."},{"Start":"01:20.900 ","End":"01:25.495","Text":"The velocity over here is the same as the velocity over here."},{"Start":"01:25.495 ","End":"01:31.085","Text":"Now let\u0027s say I take 2 points that are close together."},{"Start":"01:31.085 ","End":"01:33.560","Text":"The blue 1 is the point on the rope,"},{"Start":"01:33.560 ","End":"01:38.485","Text":"and the red 1 is the point right at the tip of the pulley system."},{"Start":"01:38.485 ","End":"01:40.710","Text":"It\u0027s the point of contact,"},{"Start":"01:40.710 ","End":"01:44.120","Text":"the red is the pulley system and the blue is the rope,"},{"Start":"01:44.120 ","End":"01:46.055","Text":"the point of contact."},{"Start":"01:46.055 ","End":"01:48.665","Text":"Let\u0027s say we take these 2 points,"},{"Start":"01:48.665 ","End":"01:52.705","Text":"and then as the rope moves down to here,"},{"Start":"01:52.705 ","End":"01:54.545","Text":"so the pulley system,"},{"Start":"01:54.545 ","End":"01:57.065","Text":"the same point moves with it."},{"Start":"01:57.065 ","End":"02:04.765","Text":"That means that the velocity of the rope going in this direction,"},{"Start":"02:04.765 ","End":"02:11.845","Text":"V_r is going to be equal to the linear velocity,"},{"Start":"02:11.845 ","End":"02:17.505","Text":"also going in this direction of the pulley, V_p."},{"Start":"02:17.505 ","End":"02:21.910","Text":"Then what we\u0027re going to get is this red point,"},{"Start":"02:21.910 ","End":"02:28.485","Text":"which represents the point on the pulley is having an angular motion."},{"Start":"02:28.485 ","End":"02:31.895","Text":"That means that my V of the pulley,"},{"Start":"02:31.895 ","End":"02:33.625","Text":"according to angular motion,"},{"Start":"02:33.625 ","End":"02:39.490","Text":"is going to be equal to its angular velocity multiplied by its radius R. As we said,"},{"Start":"02:39.490 ","End":"02:43.005","Text":"that\u0027s going to be equal to this V of the rope,"},{"Start":"02:43.005 ","End":"02:46.770","Text":"which is exactly what we have up over here."},{"Start":"02:46.770 ","End":"02:49.070","Text":"Then if we differentiate it,"},{"Start":"02:49.070 ","End":"02:52.415","Text":"we\u0027ll get our equations for the acceleration."},{"Start":"02:52.415 ","End":"02:56.930","Text":"The acceleration of the pulley is going to be equal"},{"Start":"02:56.930 ","End":"03:02.825","Text":"to the angular acceleration multiplied by its radius,"},{"Start":"03:02.825 ","End":"03:10.075","Text":"which is going to be equal to the acceleration of the rope for the exact same reason."},{"Start":"03:10.075 ","End":"03:14.375","Text":"These are very useful equations to remember,"},{"Start":"03:14.375 ","End":"03:17.180","Text":"so you can write them in your formula sheet."},{"Start":"03:17.180 ","End":"03:19.950","Text":"That\u0027s the end of this lesson."}],"ID":9426},{"Watched":false,"Name":"Exercise 2","Duration":"14m 29s","ChapterTopicVideoID":9157,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.330 ","End":"00:03.205","Text":"Hello. In this question,"},{"Start":"00:03.205 ","End":"00:06.640","Text":"a ball is nailed to the table so it can\u0027t move and"},{"Start":"00:06.640 ","End":"00:10.990","Text":"it can rotate around the axis perpendicular to the table."},{"Start":"00:10.990 ","End":"00:18.100","Text":"A rope is wound around the center of the ball and it rests on a non-ideal pulley."},{"Start":"00:18.100 ","End":"00:21.545","Text":"The rope goes along and here\u0027s a non-ideal pulley."},{"Start":"00:21.545 ","End":"00:26.170","Text":"A mass m_1 is attached to its end over here."},{"Start":"00:26.170 ","End":"00:31.210","Text":"Then we\u0027re being told that m_2 and R_2 is the mass and radius of"},{"Start":"00:31.210 ","End":"00:36.310","Text":"the pulley and m_3 and R_3 is the mass and radius of the ball."},{"Start":"00:36.310 ","End":"00:38.500","Text":"The system begins at rest."},{"Start":"00:38.500 ","End":"00:43.885","Text":"We\u0027re being told to find each body\u0027s acceleration as well as the tension in the rope."},{"Start":"00:43.885 ","End":"00:48.775","Text":"That means that we\u0027re being told to find"},{"Start":"00:48.775 ","End":"00:52.060","Text":"the acceleration of body 1"},{"Start":"00:52.060 ","End":"00:56.350","Text":"which is linear acceleration because it\u0027s going in the downwards direction."},{"Start":"00:56.350 ","End":"01:00.039","Text":"Then the angular acceleration of the pulley,"},{"Start":"01:00.039 ","End":"01:03.700","Text":"so Alpha for angular acceleration,"},{"Start":"01:03.700 ","End":"01:09.530","Text":"and then the angular acceleration of the ball,"},{"Start":"01:10.490 ","End":"01:15.010","Text":"and also finally the tension in the rope."},{"Start":"01:16.040 ","End":"01:19.485","Text":"All of these we have to find."},{"Start":"01:19.485 ","End":"01:23.860","Text":"Let\u0027s take a look at how we are going to do this."},{"Start":"01:23.860 ","End":"01:30.260","Text":"We can use either moments and forces or we can use energy."},{"Start":"01:30.260 ","End":"01:32.360","Text":"Now, because we have acceleration in"},{"Start":"01:32.360 ","End":"01:36.100","Text":"the question so we\u0027re going to use moments and forces."},{"Start":"01:36.100 ","End":"01:41.600","Text":"Conservation of moments and forces in order to find these unknowns."},{"Start":"01:41.600 ","End":"01:47.306","Text":"What we\u0027re going to do first is to start by drawing a free body diagram,"},{"Start":"01:47.306 ","End":"01:49.040","Text":"so all of our forces."},{"Start":"01:49.040 ","End":"01:56.540","Text":"We have going in this downwards direction we have m_1 multiplied by g,"},{"Start":"01:56.540 ","End":"02:03.215","Text":"and then going in the upwards direction over like this we have tension."},{"Start":"02:03.215 ","End":"02:10.265","Text":"Now, notice that because this is resting on a non-ideal pulley system,"},{"Start":"02:10.265 ","End":"02:14.350","Text":"that means that there\u0027s some kind of friction over here."},{"Start":"02:14.350 ","End":"02:21.425","Text":"That means that here the tension is going to be different to the tension over here,"},{"Start":"02:21.425 ","End":"02:24.175","Text":"so we\u0027re going to call this T_1."},{"Start":"02:24.175 ","End":"02:27.690","Text":"Now let\u0027s look at body Number 2."},{"Start":"02:27.690 ","End":"02:32.430","Text":"Similarly, going in this direction we have T_1 and"},{"Start":"02:32.430 ","End":"02:39.345","Text":"then going in this direction we have a different tension."},{"Start":"02:39.345 ","End":"02:41.695","Text":"We have T_2."},{"Start":"02:41.695 ","End":"02:45.290","Text":"Now, there\u0027s also a force coming from the pulley"},{"Start":"02:45.290 ","End":"02:50.310","Text":"itself but I\u0027m not going to write it and I\u0027ll explain why in a second."},{"Start":"02:50.390 ","End":"02:58.595","Text":"Now in body Number 3 similarly going in this direction we have our T_2."},{"Start":"02:58.595 ","End":"03:03.005","Text":"Again, from the connection over here, the axis,"},{"Start":"03:03.005 ","End":"03:10.640","Text":"there\u0027s a force acting however it doesn\u0027t matter either for the same reason as over here."},{"Start":"03:10.640 ","End":"03:16.500","Text":"Now let\u0027s begin by writing our force equations."},{"Start":"03:16.730 ","End":"03:19.260","Text":"For body Number 1,"},{"Start":"03:19.260 ","End":"03:22.955","Text":"so we\u0027re going to write,"},{"Start":"03:22.955 ","End":"03:30.570","Text":"we can say that this is the y-axis and this is the positive direction going downwards."},{"Start":"03:30.570 ","End":"03:36.170","Text":"We can say that the sum of all of the forces in the y direction is"},{"Start":"03:36.170 ","End":"03:41.990","Text":"going to equal to m_1 g going in the positive direction,"},{"Start":"03:41.990 ","End":"03:47.210","Text":"negative T_1 because T_1 is going in the negative direction,"},{"Start":"03:47.210 ","End":"03:49.460","Text":"downwards is the positive direction."},{"Start":"03:49.460 ","End":"03:54.120","Text":"This is going to be equal to m_1"},{"Start":"03:54.120 ","End":"03:58.815","Text":"multiplied by the acceleration and the acceleration is our a_1."},{"Start":"03:58.815 ","End":"04:01.480","Text":"This is our first unknown."},{"Start":"04:01.580 ","End":"04:04.290","Text":"Now for body Number 2,"},{"Start":"04:04.290 ","End":"04:10.760","Text":"so what I can do is I can write the forces for body Number 2,"},{"Start":"04:10.760 ","End":"04:12.755","Text":"the sum of all of the forces."},{"Start":"04:12.755 ","End":"04:16.145","Text":"Now, the problem with this is that as I said,"},{"Start":"04:16.145 ","End":"04:19.220","Text":"there\u0027s the force in the middle over here and I"},{"Start":"04:19.220 ","End":"04:23.030","Text":"don\u0027t know what it is and I don\u0027t know how to work it out at this stage."},{"Start":"04:23.030 ","End":"04:25.010","Text":"That will mean that I\u0027ll have"},{"Start":"04:25.010 ","End":"04:28.865","Text":"a very complicated equation that I don\u0027t know if I\u0027ll be able to solve."},{"Start":"04:28.865 ","End":"04:34.050","Text":"Instead, what I\u0027m going to write is the sum of all of the moments."},{"Start":"04:34.760 ","End":"04:39.240","Text":"Let\u0027s write the sum of all of the moments."},{"Start":"04:39.240 ","End":"04:42.826","Text":"The sum of all of the moments which is Tau,"},{"Start":"04:42.826 ","End":"04:46.705","Text":"well, the sum of the torques for body Number 2."},{"Start":"04:46.705 ","End":"04:51.720","Text":"It\u0027s going to be equal to my force, my T_1,"},{"Start":"04:51.850 ","End":"04:56.090","Text":"multiplied by its distance from the axis of"},{"Start":"04:56.090 ","End":"05:00.395","Text":"rotation and its distance is R_2, the radius."},{"Start":"05:00.395 ","End":"05:03.265","Text":"This distance over here."},{"Start":"05:03.265 ","End":"05:09.575","Text":"Now we have to decide if it\u0027s a positive or a negative sign."},{"Start":"05:09.575 ","End":"05:14.375","Text":"What I want to do is I want the direction of travel to be the positive direction."},{"Start":"05:14.375 ","End":"05:20.300","Text":"I can say that this clockwise direction of rotation is the positive,"},{"Start":"05:20.300 ","End":"05:22.200","Text":"so going like this."},{"Start":"05:22.200 ","End":"05:26.550","Text":"I can see that my T_1 is turning my pulley in"},{"Start":"05:26.550 ","End":"05:28.890","Text":"the positive direction because the force is going"},{"Start":"05:28.890 ","End":"05:32.505","Text":"down and the pulley is moving in a clockwise direction."},{"Start":"05:32.505 ","End":"05:34.205","Text":"This is a positive."},{"Start":"05:34.205 ","End":"05:37.360","Text":"Then what I have to deal with is my T_2."},{"Start":"05:37.360 ","End":"05:40.820","Text":"Now, my T_2 as we can see when I apply"},{"Start":"05:40.820 ","End":"05:45.410","Text":"the force it\u0027s trying to turn the pulley in the negative direction,"},{"Start":"05:45.410 ","End":"05:47.150","Text":"in the anticlockwise direction."},{"Start":"05:47.150 ","End":"05:50.720","Text":"It\u0027s going to be negative T_2 and"},{"Start":"05:50.720 ","End":"05:55.410","Text":"again its distance from the axis of rotation which is again R_2."},{"Start":"05:55.850 ","End":"06:00.130","Text":"T_2 multiplied by R_2."},{"Start":"06:00.130 ","End":"06:06.540","Text":"Then this is going to be equal to my moment of inertia of body Number 2,"},{"Start":"06:06.540 ","End":"06:13.910","Text":"so I_2 multiplied by the angular acceleration of my second body, so that\u0027s Alpha_2."},{"Start":"06:13.910 ","End":"06:15.650","Text":"This is my other unknown."},{"Start":"06:15.650 ","End":"06:17.690","Text":"Now, what is my I_2?"},{"Start":"06:17.690 ","End":"06:20.840","Text":"My moment of inertia of body Number 2?"},{"Start":"06:20.840 ","End":"06:24.290","Text":"If I look at the pulley system as a disc,"},{"Start":"06:24.290 ","End":"06:30.470","Text":"so I know that the moment of inertia of a disc is equal to half multiplied by the mass of"},{"Start":"06:30.470 ","End":"06:38.710","Text":"the disc which is m_2 multiplied by its radius squared."},{"Start":"06:40.220 ","End":"06:45.335","Text":"This is the moment of inertia of a disc rotating about its center."},{"Start":"06:45.335 ","End":"06:50.490","Text":"If you can\u0027t remember how I did this please go back to the chapter on moment of inertia."},{"Start":"06:51.110 ","End":"06:57.350","Text":"Now what I want to do is I want to rearrange this equation and give some explanation."},{"Start":"06:57.770 ","End":"07:00.795","Text":"If I just do some algebra so I can have"},{"Start":"07:00.795 ","End":"07:05.530","Text":"T_1 minus T_2 because they have a common factor of R_2,"},{"Start":"07:05.530 ","End":"07:09.025","Text":"so multiplied by R_2 is equal to"},{"Start":"07:09.025 ","End":"07:14.050","Text":"the moment of inertia multiplied by the angular acceleration."},{"Start":"07:14.050 ","End":"07:19.060","Text":"Now, what I want to show is why T_1 does not equal T_2."},{"Start":"07:19.060 ","End":"07:23.560","Text":"Remember we said, I mean,"},{"Start":"07:23.560 ","End":"07:25.960","Text":"the tension in this section of the rope will be different to"},{"Start":"07:25.960 ","End":"07:29.865","Text":"this section because it\u0027s a non-ideal pulley system."},{"Start":"07:29.865 ","End":"07:34.715","Text":"Let\u0027s really show that these T\u0027s are different and not equal to each other."},{"Start":"07:34.715 ","End":"07:37.130","Text":"If T_1 was equal to T_2,"},{"Start":"07:37.130 ","End":"07:38.770","Text":"then this would equal to 0."},{"Start":"07:38.770 ","End":"07:42.795","Text":"However, if this side of the equation is equal to 0,"},{"Start":"07:42.795 ","End":"07:45.260","Text":"then this has to be equal to 0,"},{"Start":"07:45.260 ","End":"07:48.230","Text":"which means either the moment of inertia is equal to 0."},{"Start":"07:48.230 ","End":"07:52.790","Text":"Which wouldn\u0027t make sense because we have a mass and radius,"},{"Start":"07:52.790 ","End":"07:54.920","Text":"and they\u0027re not equal to 0."},{"Start":"07:54.920 ","End":"07:58.375","Text":"Or, our angular acceleration is equal to 0."},{"Start":"07:58.375 ","End":"08:00.349","Text":"However, our angular acceleration,"},{"Start":"08:00.349 ","End":"08:01.835","Text":"we know from the question,"},{"Start":"08:01.835 ","End":"08:05.430","Text":"isn\u0027t equal to 0 because we know that it\u0027s rotating."},{"Start":"08:05.470 ","End":"08:09.850","Text":"That means that if one side of the equation can be equal to 0,"},{"Start":"08:09.850 ","End":"08:11.480","Text":"then neither can the other side,"},{"Start":"08:11.480 ","End":"08:21.140","Text":"which means that T_1 does not equal to T_2."},{"Start":"08:21.140 ","End":"08:26.175","Text":"Now, this is only because we\u0027re dealing with a non-ideal pulley system."},{"Start":"08:26.175 ","End":"08:29.995","Text":"If we were dealing with an ideal pulley system in previous questions,"},{"Start":"08:29.995 ","End":"08:33.870","Text":"we\u0027ve been able to say that our T_1 is equal to T_2."},{"Start":"08:33.870 ","End":"08:35.410","Text":"Then how does that work out?"},{"Start":"08:35.410 ","End":"08:36.985","Text":"As we know, it\u0027s rotating,"},{"Start":"08:36.985 ","End":"08:40.070","Text":"so Alpha_2 can\u0027t be equal to 0."},{"Start":"08:40.070 ","End":"08:43.005","Text":"When we\u0027re dealing with an ideal pulley system,"},{"Start":"08:43.005 ","End":"08:46.125","Text":"we\u0027re saying that our mass here,"},{"Start":"08:46.125 ","End":"08:49.085","Text":"m_2, would be equal to 0."},{"Start":"08:49.085 ","End":"08:52.145","Text":"This is an ideal pulley system."},{"Start":"08:52.145 ","End":"08:54.165","Text":"For an ideal pulley,"},{"Start":"08:54.165 ","End":"08:56.060","Text":"the mass would be equal to 0,"},{"Start":"08:56.060 ","End":"08:58.340","Text":"which means that our moment of inertia,"},{"Start":"08:58.340 ","End":"09:01.360","Text":"our I_2 would be equal to 0."},{"Start":"09:02.950 ","End":"09:06.245","Text":"Then this side would be equal to 0,"},{"Start":"09:06.245 ","End":"09:07.855","Text":"and this side would be equal to 0,"},{"Start":"09:07.855 ","End":"09:09.260","Text":"and that would be fine."},{"Start":"09:09.260 ","End":"09:13.860","Text":"However, because we\u0027re not dealing with an ideal pulley, our mass 2,"},{"Start":"09:13.860 ","End":"09:16.085","Text":"does not equal 0."},{"Start":"09:16.085 ","End":"09:18.500","Text":"This isn\u0027t relevant."},{"Start":"09:18.500 ","End":"09:25.420","Text":"We can rub all of this out which means that our moment of inertia isn\u0027t equal to 0."},{"Start":"09:25.420 ","End":"09:30.370","Text":"Then again we\u0027re left with our T_1 does not equal T_2."},{"Start":"09:34.030 ","End":"09:40.900","Text":"Now we\u0027re going to write the equation for our third body, ball."},{"Start":"09:40.900 ","End":"09:43.850","Text":"Now, as discussed with our body number 2,"},{"Start":"09:43.850 ","End":"09:46.435","Text":"because we have a force coming from"},{"Start":"09:46.435 ","End":"09:50.470","Text":"the axis of rotation and we don\u0027t know what this force is,"},{"Start":"09:50.470 ","End":"09:52.885","Text":"and at this stage, we don\u0027t know how to work it out."},{"Start":"09:52.885 ","End":"09:57.480","Text":"Instead of writing an equation for the force is like body number 2,"},{"Start":"09:57.480 ","End":"10:02.855","Text":"we\u0027re going to write an equation for the moments or an equation for the torques."},{"Start":"10:02.855 ","End":"10:06.510","Text":"For body number 3 it\u0027s going to be equal to,"},{"Start":"10:06.510 ","End":"10:08.880","Text":"so we have our T_2."},{"Start":"10:10.660 ","End":"10:17.070","Text":"This is the positive direction of rotation."},{"Start":"10:17.240 ","End":"10:20.875","Text":"We have our T_2 going in the positive direction,"},{"Start":"10:20.875 ","End":"10:23.350","Text":"because we want it in the direction of travel,"},{"Start":"10:23.350 ","End":"10:27.055","Text":"multiplied by its distance from the axis of rotation,"},{"Start":"10:27.055 ","End":"10:29.215","Text":"which is I_3,"},{"Start":"10:29.215 ","End":"10:32.185","Text":"so T_2 multiplied by I_3."},{"Start":"10:32.185 ","End":"10:36.255","Text":"That\u0027s the only force acting here that we know."},{"Start":"10:36.255 ","End":"10:40.345","Text":"That\u0027s going to be equal to our moment of inertia of"},{"Start":"10:40.345 ","End":"10:44.710","Text":"the third body, multiplied by Alpha_3."},{"Start":"10:44.710 ","End":"10:47.780","Text":"Its angular acceleration."},{"Start":"10:47.780 ","End":"10:50.175","Text":"What is our I_3?"},{"Start":"10:50.175 ","End":"10:57.505","Text":"Our I_3 is the moment of inertia of a sphere rotating about its center,"},{"Start":"10:57.505 ","End":"11:02.815","Text":"which is 2/5 multiplied by its mass, m_3,"},{"Start":"11:02.815 ","End":"11:11.140","Text":"multiplied by its radius, R_3^2."},{"Start":"11:11.140 ","End":"11:13.385","Text":"Now we can see that we have three equations."},{"Start":"11:13.385 ","End":"11:15.610","Text":"Obviously the equations for our Is,"},{"Start":"11:15.610 ","End":"11:16.900","Text":"our moment of inertia,"},{"Start":"11:16.900 ","End":"11:18.715","Text":"we just substitute in."},{"Start":"11:18.715 ","End":"11:20.725","Text":"We actually have three equations,"},{"Start":"11:20.725 ","End":"11:23.070","Text":"and let\u0027s see what are our unknowns are."},{"Start":"11:23.070 ","End":"11:24.725","Text":"Our T_1 is unknown,"},{"Start":"11:24.725 ","End":"11:26.845","Text":"our a_1 is unknown."},{"Start":"11:26.845 ","End":"11:29.210","Text":"Our T_2 is unknown,"},{"Start":"11:29.210 ","End":"11:31.480","Text":"our Alpha_2 is unknown,"},{"Start":"11:31.480 ","End":"11:36.680","Text":"and our Alpha_3 is also unknown."},{"Start":"11:36.680 ","End":"11:38.950","Text":"We have three equations, and 1, 2,"},{"Start":"11:38.950 ","End":"11:41.575","Text":"3, 4, 5 unknowns."},{"Start":"11:41.575 ","End":"11:43.788","Text":"That means that we cannot solve this,"},{"Start":"11:43.788 ","End":"11:47.190","Text":"so we have to find some more equations."},{"Start":"11:47.530 ","End":"11:52.170","Text":"The first thing that I can start with is to write"},{"Start":"11:52.170 ","End":"11:57.320","Text":"down the connection between linear and angular acceleration."},{"Start":"11:57.320 ","End":"12:01.710","Text":"Whenever you\u0027re working with accelerations and you need more equations,"},{"Start":"12:01.710 ","End":"12:03.670","Text":"this is a good way to go."},{"Start":"12:03.670 ","End":"12:08.325","Text":"I can write that the acceleration of the rope,"},{"Start":"12:08.325 ","End":"12:11.560","Text":"so a_r, acceleration of the rope,"},{"Start":"12:11.560 ","End":"12:15.215","Text":"if I\u0027m looking at the pulley system,"},{"Start":"12:15.215 ","End":"12:20.040","Text":"is going to be equal to the acceleration of the pulley system,"},{"Start":"12:20.040 ","End":"12:22.364","Text":"multiplied by its radius."},{"Start":"12:22.364 ","End":"12:27.920","Text":"That means the acceleration of the pulley system is obviously with angular acceleration."},{"Start":"12:27.920 ","End":"12:29.785","Text":"It\u0027s my Alpha_2,"},{"Start":"12:29.785 ","End":"12:34.090","Text":"multiplied by its radius, which is R_2."},{"Start":"12:34.090 ","End":"12:36.455","Text":"If you can\u0027t remember this,"},{"Start":"12:36.455 ","End":"12:39.970","Text":"then please go one or two lessons back."},{"Start":"12:40.420 ","End":"12:45.590","Text":"This equation comes from the fact that we\u0027re dealing with a non-ideal pulley system,"},{"Start":"12:45.590 ","End":"12:51.780","Text":"which means that the rope going over the pulley system isn\u0027t slipping."},{"Start":"12:53.360 ","End":"12:59.110","Text":"Then the next equation will go to this over here,"},{"Start":"12:59.110 ","End":"13:01.735","Text":"because our ball is also rotating."},{"Start":"13:01.735 ","End":"13:05.405","Text":"Now, because our rope is one rope,"},{"Start":"13:05.405 ","End":"13:09.710","Text":"so the distance between this point and this point is"},{"Start":"13:09.710 ","End":"13:14.150","Text":"going to remain constant throughout its journey."},{"Start":"13:14.150 ","End":"13:16.705","Text":"It doesn\u0027t matter that the tensions are different."},{"Start":"13:16.705 ","End":"13:20.160","Text":"The rope doesn\u0027t stretch, for instance."},{"Start":"13:20.160 ","End":"13:22.460","Text":"It\u0027s the same uniform rope,"},{"Start":"13:22.460 ","End":"13:27.475","Text":"which means that I can write also that the acceleration of the rope,"},{"Start":"13:27.475 ","End":"13:33.250","Text":"and writes how it\u0027s connected to the angular acceleration of the sphere."},{"Start":"13:33.250 ","End":"13:38.110","Text":"Then I can write that it\u0027s equal to the angular acceleration of the sphere a_3,"},{"Start":"13:38.110 ","End":"13:42.580","Text":"multiplied by the radius of the sphere, R_3."},{"Start":"13:42.580 ","End":"13:45.785","Text":"Obviously these two are the same."},{"Start":"13:45.785 ","End":"13:48.980","Text":"The acceleration of the rope over"},{"Start":"13:48.980 ","End":"13:52.300","Text":"here will be the same as the angular acceleration of the rope over here,"},{"Start":"13:52.300 ","End":"13:53.810","Text":"because it\u0027s the same rope."},{"Start":"13:53.810 ","End":"13:56.830","Text":"These two are equal to the same,"},{"Start":"13:56.830 ","End":"13:59.695","Text":"so then I can say that this is equation number 4,"},{"Start":"13:59.695 ","End":"14:02.380","Text":"and this is equation number 5."},{"Start":"14:02.380 ","End":"14:06.100","Text":"Now, I\u0027m going to skip substituting these all in."},{"Start":"14:06.100 ","End":"14:09.580","Text":"All you have to do is some simple algebra,"},{"Start":"14:09.580 ","End":"14:14.995","Text":"and rearrange all these equations in order to find these unknowns."},{"Start":"14:14.995 ","End":"14:16.910","Text":"You\u0027re more than welcome to do this."},{"Start":"14:16.910 ","End":"14:20.065","Text":"It\u0027s good practice to do this fast on your own,"},{"Start":"14:20.065 ","End":"14:21.690","Text":"because if you get this in the exam,"},{"Start":"14:21.690 ","End":"14:26.220","Text":"it would be nice that you can solve algebra as quickly as possible."},{"Start":"14:26.330 ","End":"14:29.550","Text":"That\u0027s the end of this lesson."}],"ID":9427},{"Watched":false,"Name":"Exercise 3","Duration":"22m 25s","ChapterTopicVideoID":9158,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.745","Text":"Hello. In this question,"},{"Start":"00:02.745 ","End":"00:07.335","Text":"we have a ball of radius R of mass m_2,"},{"Start":"00:07.335 ","End":"00:10.605","Text":"which is attached by a rope that goes over a pulley,"},{"Start":"00:10.605 ","End":"00:13.215","Text":"which we\u0027re being told is an ideal pulley."},{"Start":"00:13.215 ","End":"00:17.505","Text":"Then on the other side of the rope we have a mass m_1 attached."},{"Start":"00:17.505 ","End":"00:23.325","Text":"Now we\u0027ve told the ball is rolling along the slope without slipping,"},{"Start":"00:23.325 ","End":"00:25.995","Text":"which means that we have a static friction."},{"Start":"00:25.995 ","End":"00:28.230","Text":"Whenever we have static friction,"},{"Start":"00:28.230 ","End":"00:30.885","Text":"it means that there\u0027s no slipping."},{"Start":"00:30.885 ","End":"00:34.495","Text":"I wrote that over here for you to remember."},{"Start":"00:34.495 ","End":"00:39.200","Text":"Then we\u0027re being told that the angle of the slope is Theta."},{"Start":"00:39.200 ","End":"00:41.885","Text":"Now in the first question we\u0027re being asked,"},{"Start":"00:41.885 ","End":"00:44.660","Text":"in which direction is the static friction?"},{"Start":"00:44.660 ","End":"00:46.850","Text":"Is it in this direction,"},{"Start":"00:46.850 ","End":"00:49.505","Text":"or in this direction?"},{"Start":"00:49.505 ","End":"00:55.400","Text":"We\u0027re being asked to find the direction and to find the movement of the system,"},{"Start":"00:55.400 ","End":"01:00.500","Text":"so does that mean that the ball is moving up and the mass is moving down or opposite,"},{"Start":"01:00.500 ","End":"01:03.485","Text":"the mass is moving up and the ball is moving down."},{"Start":"01:03.485 ","End":"01:08.600","Text":"The second question, we\u0027re being told to find the accelerations of the bodies."},{"Start":"01:08.600 ","End":"01:11.719","Text":"The acceleration of this and of this,"},{"Start":"01:11.719 ","End":"01:16.854","Text":"and of the pulley and the size of the friction."},{"Start":"01:16.854 ","End":"01:21.925","Text":"Let\u0027s take a look at how I can solve this question."},{"Start":"01:21.925 ","End":"01:24.425","Text":"There\u0027s 2 ways that I can do this,"},{"Start":"01:24.425 ","End":"01:27.780","Text":"either by using conservation of energy,"},{"Start":"01:27.780 ","End":"01:34.355","Text":"or conservation of forces and moments and their equations."},{"Start":"01:34.355 ","End":"01:36.635","Text":"Now, looking at this question,"},{"Start":"01:36.635 ","End":"01:39.785","Text":"because here I have the word accelerations,"},{"Start":"01:39.785 ","End":"01:44.600","Text":"so what that means that I\u0027m going to use the equations for"},{"Start":"01:44.600 ","End":"01:50.140","Text":"sum of the forces and sum of the moments or the sum of the torques,"},{"Start":"01:50.140 ","End":"01:53.960","Text":"it\u0027s the same thing because it just seems like it will be"},{"Start":"01:53.960 ","End":"01:58.765","Text":"easier to flow through and find the accelerations because of this."},{"Start":"01:58.765 ","End":"02:03.455","Text":"What I\u0027m going to do is I\u0027m going to start with question Number 2,"},{"Start":"02:03.455 ","End":"02:07.585","Text":"to find the acceleration of the bodies and the size of the friction."},{"Start":"02:07.585 ","End":"02:10.850","Text":"Then through solving this,"},{"Start":"02:10.850 ","End":"02:13.490","Text":"I\u0027ll also inadvertently solve this."},{"Start":"02:13.490 ","End":"02:16.664","Text":"Why? Because if I just decide,"},{"Start":"02:16.664 ","End":"02:22.935","Text":"I\u0027m going to decide that my positive direction is going to be this direction,"},{"Start":"02:22.935 ","End":"02:28.939","Text":"and then my positive direction for the friction is going to be in this direction."},{"Start":"02:28.939 ","End":"02:30.725","Text":"I\u0027ll do this in red actually."},{"Start":"02:30.725 ","End":"02:36.695","Text":"Then if I work out my accelerations and they have a minus sign next to them,"},{"Start":"02:36.695 ","End":"02:40.240","Text":"then I know that the ball is moving in this direction,"},{"Start":"02:40.240 ","End":"02:42.500","Text":"and the mass is moving in this direction."},{"Start":"02:42.500 ","End":"02:46.655","Text":"If I find that my accelerations have positives next to them,"},{"Start":"02:46.655 ","End":"02:49.085","Text":"then I know that they\u0027re moving in the correct direction,"},{"Start":"02:49.085 ","End":"02:51.080","Text":"which means that the ball is moving in this direction,"},{"Start":"02:51.080 ","End":"02:52.850","Text":"and the mass is in this direction."},{"Start":"02:52.850 ","End":"02:54.635","Text":"Similarly with the friction,"},{"Start":"02:54.635 ","End":"02:59.555","Text":"if I find the size of the friction with the positive then it is in this direction."},{"Start":"02:59.555 ","End":"03:02.420","Text":"If I signed the size of the friction with a negative,"},{"Start":"03:02.420 ","End":"03:07.060","Text":"then it\u0027s in this direction going against this positive direction."},{"Start":"03:07.060 ","End":"03:11.480","Text":"I could have also chosen that my positive blue arrow is in this direction,"},{"Start":"03:11.480 ","End":"03:13.250","Text":"and then my positive red arrow is in"},{"Start":"03:13.250 ","End":"03:17.605","Text":"this direction and any combination of the 2, it doesn\u0027t matter."},{"Start":"03:17.605 ","End":"03:19.855","Text":"This is the easiest way to do it,"},{"Start":"03:19.855 ","End":"03:25.625","Text":"I suggest everyone does it this way and now let\u0027s start solving this."},{"Start":"03:25.625 ","End":"03:28.880","Text":"The first thing that we\u0027re going to do is we\u0027re going"},{"Start":"03:28.880 ","End":"03:32.360","Text":"to write all the forces on the free body diagram."},{"Start":"03:32.360 ","End":"03:36.014","Text":"Over in this direction,"},{"Start":"03:36.014 ","End":"03:38.590","Text":"we have m_1g,"},{"Start":"03:38.590 ","End":"03:40.805","Text":"and then going up here,"},{"Start":"03:40.805 ","End":"03:45.470","Text":"we have our tension T. Now because this is an ideal pulley,"},{"Start":"03:45.470 ","End":"03:50.930","Text":"the tension over here and this tension will be the same as"},{"Start":"03:50.930 ","End":"03:57.015","Text":"well as this tension over here and this tension over here."},{"Start":"03:57.015 ","End":"04:00.380","Text":"The tension on this side of the rope and the tension on this side"},{"Start":"04:00.380 ","End":"04:04.530","Text":"of the rope because it\u0027s an ideal pulley, will be the same."},{"Start":"04:04.600 ","End":"04:07.610","Text":"Then we have also,"},{"Start":"04:07.610 ","End":"04:10.360","Text":"from the center of the ball,"},{"Start":"04:10.360 ","End":"04:15.085","Text":"going downwards, we have m_2g,"},{"Start":"04:15.085 ","End":"04:18.995","Text":"and then of course we have perpendicular."},{"Start":"04:18.995 ","End":"04:20.390","Text":"Let\u0027s draw this here."},{"Start":"04:20.390 ","End":"04:23.990","Text":"Perpendicular to the slope,"},{"Start":"04:23.990 ","End":"04:26.990","Text":"we have our normal force,"},{"Start":"04:26.990 ","End":"04:28.895","Text":"T we\u0027ve already done,"},{"Start":"04:28.895 ","End":"04:36.780","Text":"and in this direction we have our static friction, like we said."},{"Start":"04:36.780 ","End":"04:41.370","Text":"Now let\u0027s write the equations for the forces."},{"Start":"04:41.370 ","End":"04:44.400","Text":"For body Number 1,"},{"Start":"04:44.400 ","End":"04:46.215","Text":"which is our mass,"},{"Start":"04:46.215 ","End":"04:49.909","Text":"so will have that the sum of all of the forces."},{"Start":"04:49.909 ","End":"04:54.530","Text":"Now, this is the y direction and this can be the x-direction."},{"Start":"04:54.530 ","End":"04:56.015","Text":"Let\u0027s write this down,"},{"Start":"04:56.015 ","End":"04:59.390","Text":"this is the x direction and this is the y direction."},{"Start":"04:59.390 ","End":"05:02.660","Text":"Now we\u0027re going to say that the positive direction is in"},{"Start":"05:02.660 ","End":"05:08.725","Text":"this direction because then it coincides with this direction,"},{"Start":"05:08.725 ","End":"05:11.935","Text":"which we also said as the positive direction."},{"Start":"05:11.935 ","End":"05:15.969","Text":"If we would have said that this is the positive direction going downwards,"},{"Start":"05:15.969 ","End":"05:21.550","Text":"then it would have been very hard to combine the equations on this axis,"},{"Start":"05:21.550 ","End":"05:25.085","Text":"on the y axis to the equation on the x axis."},{"Start":"05:25.085 ","End":"05:30.070","Text":"This is the x. We do it in the same direction."},{"Start":"05:30.860 ","End":"05:35.365","Text":"Sum of all of the forces on this mass"},{"Start":"05:35.365 ","End":"05:40.805","Text":"is going to be in the y direction because it\u0027s only working on this axis."},{"Start":"05:40.805 ","End":"05:47.030","Text":"This is going to be equal to the tension which is going in the positive direction,"},{"Start":"05:47.030 ","End":"05:49.470","Text":"because we\u0027re dealing with over here."},{"Start":"05:49.470 ","End":"05:54.350","Text":"We have the tension in the positive direction minus this m_1g,"},{"Start":"05:54.350 ","End":"05:56.690","Text":"which is going in the negative direction,"},{"Start":"05:56.690 ","End":"06:00.158","Text":"so negative m_1g,"},{"Start":"06:00.158 ","End":"06:06.760","Text":"and this is going to be equal to the mass multiplied by the acceleration."},{"Start":"06:06.760 ","End":"06:13.220","Text":"It\u0027s going to be a_1 because we\u0027re going to see even though the tensions are the same,"},{"Start":"06:13.220 ","End":"06:21.625","Text":"that the acceleration of this body and the acceleration of this body will not be equal."},{"Start":"06:21.625 ","End":"06:24.945","Text":"Now for body Number 2,"},{"Start":"06:24.945 ","End":"06:30.125","Text":"so the sum of all of the forces now we\u0027re dealing with on the x axis,"},{"Start":"06:30.125 ","End":"06:36.845","Text":"so the x. I can also write equations for the y axis because we have the normal acting."},{"Start":"06:36.845 ","End":"06:39.140","Text":"However, it\u0027s not relevant because we\u0027re trying to"},{"Start":"06:39.140 ","End":"06:41.900","Text":"find the acceleration which is in the x direction,"},{"Start":"06:41.900 ","End":"06:45.525","Text":"and the friction which is also on the x axis."},{"Start":"06:45.525 ","End":"06:49.250","Text":"Whatever is happening over here doesn\u0027t matter."},{"Start":"06:49.250 ","End":"06:57.140","Text":"That\u0027s going to be, so we can see that our m_2g is pointing in the positive direction."},{"Start":"06:57.140 ","End":"07:01.760","Text":"If we take the component in the x direction on the x-axis."},{"Start":"07:01.760 ","End":"07:04.910","Text":"Then, because we\u0027re trying to find in the direction,"},{"Start":"07:04.910 ","End":"07:11.230","Text":"so the m_2g in the direction parallel to the x-axis."},{"Start":"07:11.230 ","End":"07:13.750","Text":"Whenever we\u0027re dealing with parallel,"},{"Start":"07:13.750 ","End":"07:21.690","Text":"we write m_2g multiplied by sine of Theta,"},{"Start":"07:21.690 ","End":"07:24.100","Text":"Theta being this angle over here."},{"Start":"07:24.100 ","End":"07:27.520","Text":"If we were trying to find the perpendicular component,"},{"Start":"07:27.520 ","End":"07:30.920","Text":"this component so we would use cos."},{"Start":"07:30.920 ","End":"07:34.533","Text":"This parallel is sine of Theta,"},{"Start":"07:34.533 ","End":"07:37.780","Text":"and perpendicular is cosine of Theta."},{"Start":"07:37.780 ","End":"07:43.120","Text":"It\u0027s just something, you can also work it out if you want to take a look at it properly,"},{"Start":"07:43.120 ","End":"07:46.215","Text":"however, it\u0027s simple to remember this."},{"Start":"07:46.215 ","End":"07:48.970","Text":"We have in the positive x-direction,"},{"Start":"07:48.970 ","End":"07:50.575","Text":"m_2g sine Theta,"},{"Start":"07:50.575 ","End":"07:53.565","Text":"and then in the negative x-direction,"},{"Start":"07:53.565 ","End":"07:54.875","Text":"we have over here,"},{"Start":"07:54.875 ","End":"07:59.285","Text":"not forgetting this t. We have a negative t because it\u0027s going"},{"Start":"07:59.285 ","End":"08:04.790","Text":"in the negative x-direction and also here how we defined it,"},{"Start":"08:04.790 ","End":"08:09.280","Text":"negative static frictional force."},{"Start":"08:09.280 ","End":"08:17.850","Text":"This is obviously equal to our m_2 multiplied by its acceleration a_2."},{"Start":"08:21.020 ","End":"08:27.119","Text":"Now we\u0027re going to write the equations for the torques."},{"Start":"08:27.580 ","End":"08:32.580","Text":"First we\u0027re going to write for this body Number 2."},{"Start":"08:34.130 ","End":"08:39.654","Text":"Because I\u0027ve said that this is the positive direction."},{"Start":"08:39.654 ","End":"08:46.550","Text":"What direction will this ball have to move in order for it to move downwards?"},{"Start":"08:46.550 ","End":"08:48.355","Text":"In the positive direction."},{"Start":"08:48.355 ","End":"08:53.520","Text":"That means that it will have to move in this direction."},{"Start":"08:53.530 ","End":"08:56.495","Text":"If the ball is moving in this direction,"},{"Start":"08:56.495 ","End":"08:59.490","Text":"you\u0027ll see it will roll down the slope."},{"Start":"08:59.490 ","End":"09:02.240","Text":"I won\u0027t choose that to be moving in"},{"Start":"09:02.240 ","End":"09:04.490","Text":"this direction because that would mean it will be rolling"},{"Start":"09:04.490 ","End":"09:09.869","Text":"up the slope and the positive direction would have to be this direction."},{"Start":"09:10.120 ","End":"09:13.565","Text":"That\u0027s my positive direction of rotation,"},{"Start":"09:13.565 ","End":"09:18.385","Text":"and I can see over here that my angle is 90 degrees,"},{"Start":"09:18.385 ","End":"09:24.320","Text":"my normal and my fs are perpendicular to each other."},{"Start":"09:24.320 ","End":"09:28.415","Text":"That means that the sum of all of my torques on"},{"Start":"09:28.415 ","End":"09:33.785","Text":"the second body is going to be equal to my force acting,"},{"Start":"09:33.785 ","End":"09:36.720","Text":"which is my static friction,"},{"Start":"09:36.720 ","End":"09:41.415","Text":"so it\u0027s f_s multiplied by the radius,"},{"Start":"09:41.415 ","End":"09:42.720","Text":"which is R,"},{"Start":"09:42.720 ","End":"09:47.725","Text":"and because my direction of rotation is this direction."},{"Start":"09:47.725 ","End":"09:53.310","Text":"That means that this is going to be positive because"},{"Start":"09:53.310 ","End":"09:58.805","Text":"if I I hold my thumb here,"},{"Start":"09:58.805 ","End":"10:03.485","Text":"the axis of rotation over here and I apply my f_s,"},{"Start":"10:03.485 ","End":"10:08.005","Text":"you\u0027ll see that the ball will roll in the positive direction."},{"Start":"10:08.005 ","End":"10:12.400","Text":"If you don\u0027t understand it, just take a second to imagine it."},{"Start":"10:12.400 ","End":"10:17.265","Text":"Now, I have to include my T, my tension."},{"Start":"10:17.265 ","End":"10:20.780","Text":"If again, I put my thumb here in the center at"},{"Start":"10:20.780 ","End":"10:24.635","Text":"the axis of rotation and I apply my force T,"},{"Start":"10:24.635 ","End":"10:32.790","Text":"so you\u0027ll see that it will cause my ball to rotate in the clockwise direction,"},{"Start":"10:32.790 ","End":"10:36.170","Text":"which here as I\u0027ve defined it as the negative direction."},{"Start":"10:36.170 ","End":"10:41.830","Text":"Negative T multiplied by the radius."},{"Start":"10:41.830 ","End":"10:47.839","Text":"Of course, because the angle over here is also 90 degrees,"},{"Start":"10:47.839 ","End":"10:50.675","Text":"because it\u0027s perpendicular one to another"},{"Start":"10:50.675 ","End":"10:54.454","Text":"so I don\u0027t have to multiply it by sine or cosine,"},{"Start":"10:54.454 ","End":"10:56.730","Text":"it\u0027s just by 1."},{"Start":"10:57.530 ","End":"10:59.675","Text":"Now, of course,"},{"Start":"10:59.675 ","End":"11:04.265","Text":"this is going to be equal to the moment of"},{"Start":"11:04.265 ","End":"11:12.265","Text":"inertia of body Number 2 multiplied by its angular acceleration."},{"Start":"11:12.265 ","End":"11:16.425","Text":"What is the I of body Number 2?"},{"Start":"11:16.425 ","End":"11:24.290","Text":"It\u0027s going to be the moment of inertia of a sphere rotating about its center of mass."},{"Start":"11:24.290 ","End":"11:29.705","Text":"That is 2/5 multiplied by its mass,"},{"Start":"11:29.705 ","End":"11:33.890","Text":"multiplied by its radius squared."},{"Start":"11:33.890 ","End":"11:38.000","Text":"Now let\u0027s take a look at our equations,"},{"Start":"11:38.000 ","End":"11:40.080","Text":"and at our unknowns."},{"Start":"11:41.370 ","End":"11:48.160","Text":"What I have here as we can clearly see is I have 1,"},{"Start":"11:48.160 ","End":"11:52.165","Text":"2, and 3 equations."},{"Start":"11:52.165 ","End":"11:54.862","Text":"My I is just substituted into here so 1,"},{"Start":"11:54.862 ","End":"11:56.185","Text":"2, 3 equations."},{"Start":"11:56.185 ","End":"11:57.955","Text":"Let\u0027s see how many unknowns I have."},{"Start":"11:57.955 ","End":"11:59.680","Text":"My tension is unknown,"},{"Start":"11:59.680 ","End":"12:01.660","Text":"my a_1 is unknown,"},{"Start":"12:01.660 ","End":"12:05.020","Text":"my static friction is unknown,"},{"Start":"12:05.020 ","End":"12:09.865","Text":"my a_2 is unknown and my Alpha over here is unknown."},{"Start":"12:09.865 ","End":"12:11.567","Text":"I have 3 equations and 1,"},{"Start":"12:11.567 ","End":"12:12.847","Text":"2, 3, 4,"},{"Start":"12:12.847 ","End":"12:16.600","Text":"5 unknowns which means that I don\u0027t have enough equations."},{"Start":"12:16.600 ","End":"12:19.300","Text":"Let\u0027s see how I solve that."},{"Start":"12:19.300 ","End":"12:25.045","Text":"What I can do is write the equation which connects the linear acceleration of the rope to"},{"Start":"12:25.045 ","End":"12:31.705","Text":"the angular acceleration of my sphere or of my pulley."},{"Start":"12:31.705 ","End":"12:34.375","Text":"Correction, this is another sphere,"},{"Start":"12:34.375 ","End":"12:38.200","Text":"this is of a disk. This is a disk."},{"Start":"12:38.200 ","End":"12:40.060","Text":"It\u0027s as if it\u0027s a yoyo,"},{"Start":"12:40.060 ","End":"12:44.920","Text":"this is a disk and this is the moment of inertia of a disk."},{"Start":"12:44.920 ","End":"12:48.610","Text":"I\u0027ve been referring to it as a sphere but it\u0027s a disk."},{"Start":"12:48.610 ","End":"12:54.655","Text":"My moment of inertia of the disc is substituted into here."},{"Start":"12:54.655 ","End":"12:59.035","Text":"Let\u0027s see what the connection is between them."},{"Start":"12:59.035 ","End":"13:06.190","Text":"I can write that my acceleration of my disk as it moves downwards."},{"Start":"13:06.190 ","End":"13:10.165","Text":"The acceleration of this moving, not including rotating,"},{"Start":"13:10.165 ","End":"13:15.670","Text":"this acceleration of body number 2 in the center of mass of it,"},{"Start":"13:15.670 ","End":"13:25.480","Text":"so this point moving downwards is going to be equal to its angular acceleration."},{"Start":"13:25.480 ","End":"13:28.600","Text":"Sorry, going in this way because it\u0027s rolling."},{"Start":"13:28.600 ","End":"13:34.915","Text":"It\u0027s angular acceleration which is Alpha multiplied by its radius."},{"Start":"13:34.915 ","End":"13:39.895","Text":"Now notice depending on the arrows,"},{"Start":"13:39.895 ","End":"13:45.250","Text":"if I would have put instead of this direction being the positive direction,"},{"Start":"13:45.250 ","End":"13:47.635","Text":"this direction would be in the positive direction,"},{"Start":"13:47.635 ","End":"13:51.850","Text":"then here, I would have had to have added a minus."},{"Start":"13:51.850 ","End":"13:54.340","Text":"Really notice and also with here,"},{"Start":"13:54.340 ","End":"13:56.950","Text":"it depends what you defined as"},{"Start":"13:56.950 ","End":"14:02.290","Text":"a positive direction because it will really change your answer over here."},{"Start":"14:02.290 ","End":"14:07.630","Text":"You have to really make sure and think about the directions that you"},{"Start":"14:07.630 ","End":"14:14.080","Text":"define as positive and label plus or minus correctly."},{"Start":"14:14.080 ","End":"14:18.910","Text":"That means that the acceleration of this disk moving downwards is equal to"},{"Start":"14:18.910 ","End":"14:23.920","Text":"its angular acceleration multiplied by its radius."},{"Start":"14:23.920 ","End":"14:27.070","Text":"This is the connection between accelerations when we\u0027re"},{"Start":"14:27.070 ","End":"14:31.450","Text":"dealing with something rolling without slipping."},{"Start":"14:31.450 ","End":"14:34.840","Text":"Now we have 4 equations."},{"Start":"14:34.840 ","End":"14:38.785","Text":"This is 1, 2, 3,"},{"Start":"14:38.785 ","End":"14:41.560","Text":"and 4 and we have 5 unknowns,"},{"Start":"14:41.560 ","End":"14:44.035","Text":"so we need to find another equation."},{"Start":"14:44.035 ","End":"14:52.345","Text":"Now the equation that might go to your mind is that a_1=a_2."},{"Start":"14:52.345 ","End":"14:55.345","Text":"First of all, I\u0027m going to say this is not correct,"},{"Start":"14:55.345 ","End":"14:57.640","Text":"a_1 does not equal a_2."},{"Start":"14:57.640 ","End":"15:03.295","Text":"Now, this is not a fifth equation and now I\u0027m going to explain why it isn\u0027t."},{"Start":"15:03.295 ","End":"15:07.060","Text":"Usually, if we\u0027re dealing with something moving,"},{"Start":"15:07.060 ","End":"15:09.820","Text":"if the string was just attached to something,"},{"Start":"15:09.820 ","End":"15:15.610","Text":"if this object would go down by 10 centimeters,"},{"Start":"15:15.610 ","End":"15:20.605","Text":"so this object will go up by 10 centimeters because"},{"Start":"15:20.605 ","End":"15:29.170","Text":"the rope gets the same rope and the length of the rope isn\u0027t changing."},{"Start":"15:29.170 ","End":"15:31.360","Text":"That would be in a normal question."},{"Start":"15:31.360 ","End":"15:33.565","Text":"However, in this question,"},{"Start":"15:33.565 ","End":"15:35.500","Text":"we\u0027re dealing with a yoyo."},{"Start":"15:35.500 ","End":"15:39.265","Text":"As it rolls, the string is being wound"},{"Start":"15:39.265 ","End":"15:44.245","Text":"around it which means that the rope available free over here,"},{"Start":"15:44.245 ","End":"15:47.485","Text":"its length that is constantly changing."},{"Start":"15:47.485 ","End":"15:50.530","Text":"That means that as this body moves down,"},{"Start":"15:50.530 ","End":"15:54.490","Text":"it also winds some rope around it which means"},{"Start":"15:54.490 ","End":"15:59.200","Text":"that this body will move up more than this body move down."},{"Start":"15:59.200 ","End":"16:06.145","Text":"This body can move down this much and this body will move up this much, for instance."},{"Start":"16:06.145 ","End":"16:10.240","Text":"Similarly, if we\u0027re moving in the opposite direction,"},{"Start":"16:10.240 ","End":"16:12.070","Text":"if the ball is moving upwards,"},{"Start":"16:12.070 ","End":"16:15.370","Text":"then it will release the rope which is wound around it,"},{"Start":"16:15.370 ","End":"16:18.565","Text":"which means that it will move up"},{"Start":"16:18.565 ","End":"16:23.500","Text":"a certain amount and this body will move down a certain amount,"},{"Start":"16:23.500 ","End":"16:29.545","Text":"and the distances that they each move will not be the same and equal to each other,"},{"Start":"16:29.545 ","End":"16:33.685","Text":"which means that if the distances they move in a given time aren\u0027t equal,"},{"Start":"16:33.685 ","End":"16:36.535","Text":"that means their accelerations aren\u0027t equal."},{"Start":"16:36.535 ","End":"16:39.565","Text":"This is not an equation, I\u0027m going to rub it out."},{"Start":"16:39.565 ","End":"16:42.920","Text":"Let\u0027s think what our 5th equation is."},{"Start":"16:43.700 ","End":"16:51.560","Text":"Now we\u0027re going to have a look at how to work out the distance that each one moves."},{"Start":"16:51.560 ","End":"16:57.160","Text":"The relationship between this body moving and this body moving and"},{"Start":"16:57.160 ","End":"17:02.620","Text":"how much and that will be our 5th equation. Let\u0027s break this up."},{"Start":"17:02.620 ","End":"17:04.660","Text":"We can look at this as,"},{"Start":"17:04.660 ","End":"17:07.525","Text":"let\u0027s say that this disk isn\u0027t rotating,"},{"Start":"17:07.525 ","End":"17:10.150","Text":"it\u0027s just moving downwards."},{"Start":"17:10.150 ","End":"17:15.610","Text":"If our disk moves downwards this distance,"},{"Start":"17:15.610 ","End":"17:18.085","Text":"let\u0027s call it X_2,"},{"Start":"17:18.085 ","End":"17:20.950","Text":"then that means just like we said previously,"},{"Start":"17:20.950 ","End":"17:22.630","Text":"if this isn\u0027t rotating,"},{"Start":"17:22.630 ","End":"17:24.220","Text":"the disk isn\u0027t rotating,"},{"Start":"17:24.220 ","End":"17:25.750","Text":"it\u0027s just moving downwards."},{"Start":"17:25.750 ","End":"17:28.075","Text":"If it\u0027s moving downwards X_2,"},{"Start":"17:28.075 ","End":"17:30.280","Text":"because it\u0027s the same rope,"},{"Start":"17:30.280 ","End":"17:34.360","Text":"this mass will move up also X_2."},{"Start":"17:34.360 ","End":"17:37.255","Text":"But because it\u0027s referring to body 1,"},{"Start":"17:37.255 ","End":"17:40.225","Text":"we\u0027ll call it X_1 and here,"},{"Start":"17:40.225 ","End":"17:46.375","Text":"we\u0027re going to say that our X_1 is equal to our X_2."},{"Start":"17:46.375 ","End":"17:51.835","Text":"Without rotating, if this body moves down a certain amount,"},{"Start":"17:51.835 ","End":"17:57.070","Text":"this body will move up the same amount. That\u0027s that."},{"Start":"17:57.070 ","End":"18:03.385","Text":"Now let\u0027s rub this out and let\u0027s deal with this mass."},{"Start":"18:03.385 ","End":"18:08.020","Text":"This disk isn\u0027t moving downwards but it\u0027s just"},{"Start":"18:08.020 ","End":"18:17.830","Text":"rotating around the same spot and not moving downwards,"},{"Start":"18:17.830 ","End":"18:20.420","Text":"just rotating around the same spot."},{"Start":"18:21.690 ","End":"18:27.940","Text":"Let\u0027s imagine that it rotates a distance of Theta."},{"Start":"18:27.940 ","End":"18:31.480","Text":"From here until here, not Theta."},{"Start":"18:31.480 ","End":"18:34.090","Text":"Because this is Theta, let\u0027s say Phi."},{"Start":"18:34.090 ","End":"18:37.660","Text":"This is Phi, this angle over here."},{"Start":"18:37.660 ","End":"18:41.270","Text":"It\u0027s rotated an angle of Phi."},{"Start":"18:41.760 ","End":"18:52.435","Text":"Now we can say that the rope has moved this length,"},{"Start":"18:52.435 ","End":"18:56.440","Text":"it\u0027s wound this length of rope around it."},{"Start":"18:56.440 ","End":"18:59.035","Text":"What is this length of rope?"},{"Start":"18:59.035 ","End":"19:03.115","Text":"As we know, the arc length is going to be,"},{"Start":"19:03.115 ","End":"19:11.845","Text":"our L is going to be equal to our angle which is Phi multiplied by the radius."},{"Start":"19:11.845 ","End":"19:17.410","Text":"It\u0027s Phi multiplied by R. This is if this is just rotating."},{"Start":"19:17.410 ","End":"19:25.315","Text":"What happens if this body is both moving downwards and rotating simultaneously?"},{"Start":"19:25.315 ","End":"19:28.660","Text":"That will mean, we\u0027re dealing with,"},{"Start":"19:28.660 ","End":"19:33.385","Text":"it\u0027s also moving down in this direction X_2."},{"Start":"19:33.385 ","End":"19:36.910","Text":"That will mean that this body is going"},{"Start":"19:36.910 ","End":"19:40.345","Text":"to move up because we know that our X_1 is equal to X_2."},{"Start":"19:40.345 ","End":"19:45.265","Text":"It\u0027s going to move up X_1 and this length which is wound,"},{"Start":"19:45.265 ","End":"19:46.660","Text":"it\u0027s also going to move up,"},{"Start":"19:46.660 ","End":"19:56.680","Text":"so plus its Phi multiplied by R, so it moves up that,"},{"Start":"19:56.680 ","End":"20:02.360","Text":"which means that this distance X_2,"},{"Start":"20:02.400 ","End":"20:05.905","Text":"let\u0027s write this over here, our X_2,"},{"Start":"20:05.905 ","End":"20:15.160","Text":"this total distance moved down is going to be equal to this total distance moved up,"},{"Start":"20:15.160 ","End":"20:23.395","Text":"which is X_1 plus Phi multiplied by R. However,"},{"Start":"20:23.395 ","End":"20:27.445","Text":"what we want is something with accelerations."},{"Start":"20:27.445 ","End":"20:30.175","Text":"All I have to do if I have my position,"},{"Start":"20:30.175 ","End":"20:33.805","Text":"I just have to differentiate to get the velocity,"},{"Start":"20:33.805 ","End":"20:36.625","Text":"and then I differentiate again to get the acceleration."},{"Start":"20:36.625 ","End":"20:39.590","Text":"I differentiate twice,"},{"Start":"20:39.600 ","End":"20:46.015","Text":"I\u0027ll have my X_2 dot dot, which is acceleration,"},{"Start":"20:46.015 ","End":"20:52.885","Text":"is going to equal to my X_1 dot plus"},{"Start":"20:52.885 ","End":"21:01.210","Text":"our angle Phi differentiated twice which is going to equal angular acceleration,"},{"Start":"21:01.210 ","End":"21:04.750","Text":"multiplied by the radius which is constant."},{"Start":"21:04.750 ","End":"21:11.275","Text":"That is going to equal to a_2=a_1"},{"Start":"21:11.275 ","End":"21:17.995","Text":"plus Alpha R. These are all equal and this,"},{"Start":"21:17.995 ","End":"21:22.850","Text":"of course, is going to be our 5th equation."},{"Start":"21:23.790 ","End":"21:27.040","Text":"Now just like in previous questions,"},{"Start":"21:27.040 ","End":"21:28.795","Text":"I\u0027m not going to solve this algebra,"},{"Start":"21:28.795 ","End":"21:32.410","Text":"you simply have to substitute in all of the equations."},{"Start":"21:32.410 ","End":"21:34.780","Text":"I suggest you practice it so that you can do it as"},{"Start":"21:34.780 ","End":"21:39.190","Text":"fast and correctly as possible in the exam."},{"Start":"21:39.190 ","End":"21:41.710","Text":"Then once you substitute everything in,"},{"Start":"21:41.710 ","End":"21:45.003","Text":"you would have answered finding the accelerations of the body,"},{"Start":"21:45.003 ","End":"21:46.540","Text":"and the size of the friction."},{"Start":"21:46.540 ","End":"21:51.610","Text":"Then what you have to do is you just have to look if they\u0027re positive or negative and"},{"Start":"21:51.610 ","End":"21:57.355","Text":"then you can decide the directions and that will answer question number 1."},{"Start":"21:57.355 ","End":"21:59.560","Text":"Now in this question specifically,"},{"Start":"21:59.560 ","End":"22:03.580","Text":"you\u0027ll find that whether it\u0027s positive or negative"},{"Start":"22:03.580 ","End":"22:08.635","Text":"depends on the relationship of m_2 to m_1."},{"Start":"22:08.635 ","End":"22:11.890","Text":"You can just find that relationship really easily and if"},{"Start":"22:11.890 ","End":"22:15.910","Text":"m_2 is above some value or below some value,"},{"Start":"22:15.910 ","End":"22:19.915","Text":"then the system will move either in this direction or this direction,"},{"Start":"22:19.915 ","End":"22:21.955","Text":"and that\u0027s all you have to do."},{"Start":"22:21.955 ","End":"22:25.310","Text":"That\u0027s the end of this question."}],"ID":9428},{"Watched":false,"Name":"Exercise 4","Duration":"28m 19s","ChapterTopicVideoID":9159,"CourseChapterTopicPlaylistID":5408,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:03.630","Text":"Hello, this question is going to be"},{"Start":"00:03.630 ","End":"00:07.290","Text":"slightly different to questions that we\u0027ve previously seen."},{"Start":"00:07.290 ","End":"00:11.910","Text":"In this question, we have a rod of length L and of mass M,"},{"Start":"00:11.910 ","End":"00:15.315","Text":"which is attached screw to the wall."},{"Start":"00:15.315 ","End":"00:19.215","Text":"It can move like this and back,"},{"Start":"00:19.215 ","End":"00:23.375","Text":"so falling down, this is its axis of rotation."},{"Start":"00:23.375 ","End":"00:25.265","Text":"Now, in the first question,"},{"Start":"00:25.265 ","End":"00:30.350","Text":"we\u0027re being asked to find the angular acceleration of"},{"Start":"00:30.350 ","End":"00:36.625","Text":"the rod and also the acceleration of the center of mass of the rod."},{"Start":"00:36.625 ","End":"00:39.935","Text":"I have 2 options to solve this."},{"Start":"00:39.935 ","End":"00:47.674","Text":"I can either look at energy conservation or by writing out force and moment equations,"},{"Start":"00:47.674 ","End":"00:50.900","Text":"so because I have acceleration over here,"},{"Start":"00:50.900 ","End":"00:54.035","Text":"I\u0027m going to write force and moments equations."},{"Start":"00:54.035 ","End":"00:58.685","Text":"Let\u0027s get started. The first thing that we\u0027re going to do"},{"Start":"00:58.685 ","End":"01:03.880","Text":"is we\u0027re going to write out all of the forces on the free-body diagram."},{"Start":"01:03.880 ","End":"01:06.860","Text":"Because the rod is of uniform mass,"},{"Start":"01:06.860 ","End":"01:08.585","Text":"so I can say that from its center,"},{"Start":"01:08.585 ","End":"01:10.775","Text":"say that it\u0027s center is over here,"},{"Start":"01:10.775 ","End":"01:15.365","Text":"we have Mg going downwards."},{"Start":"01:15.365 ","End":"01:19.250","Text":"Then, of course, there\u0027s a force coming out of"},{"Start":"01:19.250 ","End":"01:23.210","Text":"the axis of rotation where it\u0027s attached to the wall,"},{"Start":"01:23.210 ","End":"01:27.065","Text":"so we don\u0027t know what the force is or which direction it\u0027s in."},{"Start":"01:27.065 ","End":"01:29.660","Text":"There\u0027s 2 ways to define this."},{"Start":"01:29.660 ","End":"01:35.390","Text":"Either we can say that the force is in some general direction, this force,"},{"Start":"01:35.390 ","End":"01:38.164","Text":"and that there\u0027s this angle,"},{"Start":"01:38.164 ","End":"01:46.840","Text":"let\u0027s call it Theta relative to some axis over here."},{"Start":"01:46.840 ","End":"01:52.040","Text":"This is 1 way to describe this force and its direction."},{"Start":"01:52.040 ","End":"01:55.445","Text":"Alternatively, we can do it in the Cartesian way,"},{"Start":"01:55.445 ","End":"01:57.860","Text":"which is every single force,"},{"Start":"01:57.860 ","End":"02:01.500","Text":"we can say that it has some x-component."},{"Start":"02:01.500 ","End":"02:04.160","Text":"We can say that its x-component is this,"},{"Start":"02:04.160 ","End":"02:05.960","Text":"so call this F(x)"},{"Start":"02:05.960 ","End":"02:15.380","Text":"and y component going in this direction, F_y."},{"Start":"02:15.380 ","End":"02:18.605","Text":"This can represent a force in"},{"Start":"02:18.605 ","End":"02:24.065","Text":"any single direction around this axis of rotation around this origin."},{"Start":"02:24.065 ","End":"02:28.640","Text":"This will describe the force in a Cartesian way."},{"Start":"02:29.810 ","End":"02:34.245","Text":"Let\u0027s write out some force equations."},{"Start":"02:34.245 ","End":"02:42.035","Text":"We can write that the sum of all of the forces in the y-direction."},{"Start":"02:42.035 ","End":"02:46.655","Text":"Let\u0027s first see what is our x and what is our y. I\u0027m actually going to choose"},{"Start":"02:46.655 ","End":"02:54.720","Text":"my x-direction to be like this and my y-direction to be downwards."},{"Start":"02:54.720 ","End":"03:04.830","Text":"The sum of all the forces in the y-direction is going to be my Mg minus my F_y."},{"Start":"03:04.870 ","End":"03:08.750","Text":"This is going to be equal to the mass"},{"Start":"03:08.750 ","End":"03:13.040","Text":"multiplied by the acceleration of the center of mass."},{"Start":"03:13.040 ","End":"03:19.610","Text":"Just of the center of mass and not of any other point in the y-direction."},{"Start":"03:19.610 ","End":"03:23.900","Text":"Then similarly, the sum of all of the forces on"},{"Start":"03:23.900 ","End":"03:28.670","Text":"the x-axis is going to be equal to simply F_x,"},{"Start":"03:28.670 ","End":"03:32.800","Text":"but going in this direction, so negative F_x."},{"Start":"03:32.800 ","End":"03:35.420","Text":"Because F_x is pointing in the negative direction,"},{"Start":"03:35.420 ","End":"03:43.250","Text":"which is also going to be the mass multiplied by the acceleration,"},{"Start":"03:43.250 ","End":"03:46.700","Text":"again, of the center of mass not of any other point,"},{"Start":"03:46.700 ","End":"03:50.070","Text":"but this time, on the x-axis."},{"Start":"03:51.410 ","End":"03:55.440","Text":"My unknowns are F_y, F_x,"},{"Start":"03:55.440 ","End":"03:59.915","Text":"a_CM in the y-direction and a_CM in the x-direction."},{"Start":"03:59.915 ","End":"04:04.760","Text":"I\u0027m going to write out another equation and it\u0027s going to be the sum of the torques."},{"Start":"04:04.760 ","End":"04:10.280","Text":"The sum of the torques is going to be equal to our force,"},{"Start":"04:10.280 ","End":"04:17.630","Text":"which is mg, multiplied by its distance from the axis of rotation."},{"Start":"04:17.630 ","End":"04:23.770","Text":"The distance of the entire rod is L and its center of mass is in the center at L/2."},{"Start":"04:23.770 ","End":"04:28.535","Text":"It\u0027s going to be multiplied by L/2,"},{"Start":"04:28.535 ","End":"04:35.375","Text":"its distance from the axis of rotation and multiplied by sine of the angle."},{"Start":"04:35.375 ","End":"04:41.150","Text":"Now, because the angle between Mg and our radius,"},{"Start":"04:41.150 ","End":"04:42.935","Text":"they\u0027re perpendicular, the angle is 90,"},{"Start":"04:42.935 ","End":"04:45.230","Text":"sine of 90 is equal to 1."},{"Start":"04:45.230 ","End":"04:46.985","Text":"We can just leave it at that."},{"Start":"04:46.985 ","End":"04:51.900","Text":"This is going to be equal to I Alpha."},{"Start":"04:52.090 ","End":"04:58.235","Text":"Then just to see the signs because of how we defined the axis,"},{"Start":"04:58.235 ","End":"05:00.950","Text":"we can see that we go from x to y,"},{"Start":"05:00.950 ","End":"05:05.700","Text":"which means that we\u0027re going in this direction, which is anticlockwise."},{"Start":"05:05.700 ","End":"05:12.200","Text":"We can see that our force Mg is also rotating the rod in the anticlockwise direction,"},{"Start":"05:12.200 ","End":"05:14.780","Text":"which means that we\u0027re in the positive direction,"},{"Start":"05:14.780 ","End":"05:17.585","Text":"so this is positive."},{"Start":"05:17.585 ","End":"05:22.445","Text":"Now, of course, we just have to write out the equation for the I."},{"Start":"05:22.445 ","End":"05:27.650","Text":"We\u0027re dealing with a rod whose axis of rotation is at the edge."},{"Start":"05:27.650 ","End":"05:33.995","Text":"The I for a rod whose axis of rotation is at the edge is m,"},{"Start":"05:33.995 ","End":"05:38.120","Text":"the mass of the rod, multiplied by L squared,"},{"Start":"05:38.120 ","End":"05:41.940","Text":"its length squared divided by 3."},{"Start":"05:42.680 ","End":"05:47.130","Text":"Now, because we have 3 equations,"},{"Start":"05:47.130 ","End":"05:50.395","Text":"but we have this F_y and this F_x,"},{"Start":"05:50.395 ","End":"05:55.460","Text":"which means that we can\u0027t solve to find what are a_CMy and a_CMx is."},{"Start":"05:55.460 ","End":"05:59.480","Text":"However, we can see that by substituting in this I over here,"},{"Start":"05:59.480 ","End":"06:01.715","Text":"we can actually find what our Alpha is,"},{"Start":"06:01.715 ","End":"06:04.435","Text":"which is what we\u0027re trying to find over here."},{"Start":"06:04.435 ","End":"06:07.065","Text":"By simply rearranging this,"},{"Start":"06:07.065 ","End":"06:08.825","Text":"because m is given, G is given,"},{"Start":"06:08.825 ","End":"06:12.785","Text":"L is given to the number and I same,"},{"Start":"06:12.785 ","End":"06:16.090","Text":"we can rearrange to find what Alpha is."},{"Start":"06:16.090 ","End":"06:21.190","Text":"We\u0027ll get that Alpha is equal to"},{"Start":"06:22.400 ","End":"06:30.860","Text":"MgL divided by 2I which is going to be equal to substituting in what our I is,"},{"Start":"06:30.860 ","End":"06:31.460","Text":"so we\u0027ll"},{"Start":"06:31.460 ","End":"06:41.255","Text":"have MgL"},{"Start":"06:41.255 ","End":"06:48.730","Text":"divided by 2mL^2 divided by 3,"},{"Start":"06:48.730 ","End":"06:51.574","Text":"which then once we rearrange,"},{"Start":"06:51.574 ","End":"07:00.550","Text":"is going to be 3 over 2 multiplied by g over L. This is our Alpha."},{"Start":"07:01.700 ","End":"07:06.800","Text":"We found our Alpha, and now we want to find what are a_CM is."},{"Start":"07:06.800 ","End":"07:08.375","Text":"How can we do this?"},{"Start":"07:08.375 ","End":"07:12.395","Text":"Now, we\u0027ll remember that there\u0027s the equation where we can find"},{"Start":"07:12.395 ","End":"07:18.725","Text":"the relationship between the linear acceleration to the angular acceleration."},{"Start":"07:18.725 ","End":"07:23.715","Text":"Now, the only problem is we here have this,"},{"Start":"07:23.715 ","End":"07:29.290","Text":"our a_CM, which means that it\u0027s of our center of mass of only 1 point."},{"Start":"07:29.290 ","End":"07:30.925","Text":"But when we\u0027re dealing with a rigid body,"},{"Start":"07:30.925 ","End":"07:32.395","Text":"we have many points."},{"Start":"07:32.395 ","End":"07:36.050","Text":"You might remember this statement."},{"Start":"07:36.200 ","End":"07:42.070","Text":"Here, it says that all the points on a rigid body which rotates about"},{"Start":"07:42.070 ","End":"07:47.860","Text":"an axis undergo circular motion with the exact same angular velocity."},{"Start":"07:47.860 ","End":"07:51.640","Text":"That means that the angular velocity of our center of"},{"Start":"07:51.640 ","End":"07:56.440","Text":"mass is going to be the same angular velocity as points over here,"},{"Start":"07:56.440 ","End":"08:00.415","Text":"here, here, and right on the edge as well."},{"Start":"08:00.415 ","End":"08:05.275","Text":"That means that if we know what our V_cm is,"},{"Start":"08:05.275 ","End":"08:13.360","Text":"so that\u0027s going to be equal to our angular velocity multiplied by our radius,"},{"Start":"08:13.360 ","End":"08:15.580","Text":"which over here is going to be"},{"Start":"08:15.580 ","End":"08:18.430","Text":"our angular velocity Omega multiplied"},{"Start":"08:18.430 ","End":"08:22.015","Text":"by our radius is our distance from the axis of rotation."},{"Start":"08:22.015 ","End":"08:24.040","Text":"Because we\u0027re going from our center of mass,"},{"Start":"08:24.040 ","End":"08:29.538","Text":"so our radius is L divided by 2."},{"Start":"08:29.538 ","End":"08:34.230","Text":"Now, if we differentiate this,"},{"Start":"08:34.230 ","End":"08:44.470","Text":"we\u0027ll get that our a_CM is going to be equal to Alpha multiplied by L over 2."},{"Start":"08:44.540 ","End":"08:48.610","Text":"We have our Alpha, this is our Alpha."},{"Start":"08:49.970 ","End":"08:57.750","Text":"Now, we can find by substituting in our Alpha is 3 over 2,"},{"Start":"08:57.750 ","End":"09:03.490","Text":"g over L multiplied by this, L over 2."},{"Start":"09:03.650 ","End":"09:06.960","Text":"Then we can see our L\u0027s cancel out."},{"Start":"09:06.960 ","End":"09:08.820","Text":"We\u0027ll get that are a_CM,"},{"Start":"09:08.820 ","End":"09:13.680","Text":"so the size of our acceleration of the center of mass,"},{"Start":"09:13.680 ","End":"09:19.020","Text":"is going to be 3g divided by 4."},{"Start":"09:19.020 ","End":"09:21.038","Text":"This is the size,"},{"Start":"09:21.038 ","End":"09:24.420","Text":"3/4 g. Now we just have to find the direction because we"},{"Start":"09:24.420 ","End":"09:28.455","Text":"have a y-component and an x-component."},{"Start":"09:28.455 ","End":"09:30.495","Text":"Let\u0027s see how we do this."},{"Start":"09:30.495 ","End":"09:34.170","Text":"Right now, we can take a look at this equation."},{"Start":"09:34.170 ","End":"09:40.395","Text":"We have that our velocity at the center of mass is equal to our Omega,"},{"Start":"09:40.395 ","End":"09:42.180","Text":"which is the same for all of the points,"},{"Start":"09:42.180 ","End":"09:46.035","Text":"multiplied by its distance from the axis of rotation,"},{"Start":"09:46.035 ","End":"09:48.310","Text":"which here is L over 2."},{"Start":"09:49.580 ","End":"09:52.695","Text":"That means that our velocity,"},{"Start":"09:52.695 ","End":"09:55.590","Text":"because we also know that this is undergoing circular motion,"},{"Start":"09:55.590 ","End":"10:00.210","Text":"so the shape that"},{"Start":"10:00.210 ","End":"10:06.360","Text":"the center of mass is going to be going through is something like this, in a circle."},{"Start":"10:06.360 ","End":"10:08.834","Text":"Then if we\u0027re going to look closely,"},{"Start":"10:08.834 ","End":"10:13.590","Text":"our initial velocity at this point right over here,"},{"Start":"10:13.590 ","End":"10:15.090","Text":"which is caused by the mg,"},{"Start":"10:15.090 ","End":"10:18.280","Text":"which is where our V_cm is at."},{"Start":"10:18.440 ","End":"10:22.500","Text":"Here we have also our V_cm."},{"Start":"10:22.500 ","End":"10:29.655","Text":"We\u0027ll notice that our V_cm over here is at a tangent to the circle,"},{"Start":"10:29.655 ","End":"10:34.965","Text":"which means that this acceleration is also tangential."},{"Start":"10:34.965 ","End":"10:39.885","Text":"If we\u0027re dealing with tangential acceleration,"},{"Start":"10:39.885 ","End":"10:44.010","Text":"that means that we\u0027re dealing with Theta,"},{"Start":"10:44.010 ","End":"10:46.210","Text":"in the Theta direction."},{"Start":"10:46.250 ","End":"10:50.130","Text":"What does this mean if something is in the Theta direction?"},{"Start":"10:50.130 ","End":"10:52.980","Text":"If we look, because our V_cm is going downwards,"},{"Start":"10:52.980 ","End":"10:56.080","Text":"we said that this is in the y-direction."},{"Start":"10:57.050 ","End":"11:01.710","Text":"Then that means that this is going to be over here,"},{"Start":"11:01.710 ","End":"11:05.625","Text":"like this, in the y-direction."},{"Start":"11:05.625 ","End":"11:08.040","Text":"Now, we found our a_y,"},{"Start":"11:08.040 ","End":"11:13.200","Text":"which is the acceleration of the center of mass in the y-direction,"},{"Start":"11:13.200 ","End":"11:15.360","Text":"in the tangential direction."},{"Start":"11:15.360 ","End":"11:17.880","Text":"Now what we have to do is we have to find our a,"},{"Start":"11:17.880 ","End":"11:21.405","Text":"center of mass in the x-direction."},{"Start":"11:21.405 ","End":"11:24.480","Text":"That\u0027s going to be our radial acceleration,"},{"Start":"11:24.480 ","End":"11:26.850","Text":"because we\u0027re moving in circular motion."},{"Start":"11:26.850 ","End":"11:29.280","Text":"We have our tangential acceleration,"},{"Start":"11:29.280 ","End":"11:30.315","Text":"which is Theta,"},{"Start":"11:30.315 ","End":"11:31.785","Text":"which is what we just found here."},{"Start":"11:31.785 ","End":"11:34.365","Text":"Now, we\u0027re going to have our radial,"},{"Start":"11:34.365 ","End":"11:36.390","Text":"which is going to be our x."},{"Start":"11:36.390 ","End":"11:38.820","Text":"Let\u0027s see how we do this."},{"Start":"11:38.820 ","End":"11:42.780","Text":"We can say that our a center of"},{"Start":"11:42.780 ","End":"11:49.960","Text":"mass for the x-direction is going to be a center of mass in the radial direction."},{"Start":"11:50.210 ","End":"11:54.660","Text":"That\u0027s going to be equal to our a_x and this is"},{"Start":"11:54.660 ","End":"11:58.590","Text":"going to be equal to our circular motion that we have going."},{"Start":"11:58.590 ","End":"12:05.685","Text":"It\u0027s going to be Omega^2 r. This is the equation for acceleration."},{"Start":"12:05.685 ","End":"12:09.000","Text":"Now we have to see if our sign over here is going to be a"},{"Start":"12:09.000 ","End":"12:13.110","Text":"positive or a negative. Let\u0027s take a look."},{"Start":"12:13.110 ","End":"12:17.610","Text":"As we know, when we\u0027re dealing with polar coordinates that"},{"Start":"12:17.610 ","End":"12:24.970","Text":"our radial direction is going out from the axis of rotation."},{"Start":"12:25.010 ","End":"12:30.015","Text":"That means that it will be in this direction."},{"Start":"12:30.015 ","End":"12:31.560","Text":"This is our radial direction,"},{"Start":"12:31.560 ","End":"12:33.270","Text":"as we know in polar coordinates."},{"Start":"12:33.270 ","End":"12:36.645","Text":"However, when we\u0027re dealing with circular motion,"},{"Start":"12:36.645 ","End":"12:45.705","Text":"we know that our radial acceleration is actually going in to the axis of rotation."},{"Start":"12:45.705 ","End":"12:47.445","Text":"It\u0027s going in this direction,"},{"Start":"12:47.445 ","End":"12:49.400","Text":"which means that over here,"},{"Start":"12:49.400 ","End":"12:51.300","Text":"we have to put a negative."},{"Start":"12:51.300 ","End":"12:55.804","Text":"Because the radial direction is out in vector format"},{"Start":"12:55.804 ","End":"13:00.665","Text":"and here our acceleration and circular motion is going inside,"},{"Start":"13:00.665 ","End":"13:02.705","Text":"so we have a minus."},{"Start":"13:02.705 ","End":"13:05.555","Text":"Just to explain a little bit better what I mean,"},{"Start":"13:05.555 ","End":"13:11.310","Text":"if we have here our y-axis and here our x-axis, nothing to do with this."},{"Start":"13:11.310 ","End":"13:14.670","Text":"Whenever we draw our vectors, if you remember,"},{"Start":"13:14.670 ","End":"13:18.885","Text":"we always will draw it in this direction and this will be our r,"},{"Start":"13:18.885 ","End":"13:22.080","Text":"and this will be our Theta, in polar coordinates."},{"Start":"13:22.080 ","End":"13:26.880","Text":"Right now, the situation that we have is instead of the arrow going out,"},{"Start":"13:26.880 ","End":"13:31.920","Text":"we have our arrow of radial acceleration going in,"},{"Start":"13:31.920 ","End":"13:36.010","Text":"which means that we\u0027ll have a minus over here."},{"Start":"13:36.920 ","End":"13:39.675","Text":"Now, the next thing that we have to do,"},{"Start":"13:39.675 ","End":"13:42.614","Text":"we have an Omega here and in our question,"},{"Start":"13:42.614 ","End":"13:45.090","Text":"we\u0027re not given what the value of Omega is,"},{"Start":"13:45.090 ","End":"13:47.460","Text":"so we have to figure out what Omega is."},{"Start":"13:47.460 ","End":"13:49.710","Text":"Let\u0027s write that here."},{"Start":"13:49.710 ","End":"13:54.930","Text":"In our question, if you remember right at the beginning,"},{"Start":"13:54.930 ","End":"13:58.935","Text":"we were told that the rod was released from rest."},{"Start":"13:58.935 ","End":"14:03.929","Text":"That means that our angular velocity"},{"Start":"14:03.929 ","End":"14:08.925","Text":"is going to be equal to 0 because the rod was released from rest."},{"Start":"14:08.925 ","End":"14:14.820","Text":"That means that our a_x is also going to be equal to 0."},{"Start":"14:14.820 ","End":"14:22.360","Text":"We have no angular acceleration of the center of mass in the x-direction."},{"Start":"14:22.970 ","End":"14:29.355","Text":"Now, we have our value for Alpha and we also have our values for a_CM."},{"Start":"14:29.355 ","End":"14:32.340","Text":"We have our a_y and our a_x."},{"Start":"14:32.340 ","End":"14:34.515","Text":"We\u0027ve answered question number 1."},{"Start":"14:34.515 ","End":"14:37.510","Text":"Now let\u0027s move on to question number 2."},{"Start":"14:37.970 ","End":"14:40.395","Text":"In question number 2,"},{"Start":"14:40.395 ","End":"14:44.400","Text":"we\u0027re being asked to find what F_x and F_y is."},{"Start":"14:44.400 ","End":"14:50.025","Text":"In other words, we\u0027re trying to find what force our axis of rotation is applying,"},{"Start":"14:50.025 ","End":"14:53.440","Text":"our F_x and F_y."},{"Start":"14:54.020 ","End":"14:56.805","Text":"If we look back in question number 1,"},{"Start":"14:56.805 ","End":"14:59.565","Text":"we can see that we have the sum of all of the forces,"},{"Start":"14:59.565 ","End":"15:03.315","Text":"F_x is equal to negative F_x and we have that it\u0027s"},{"Start":"15:03.315 ","End":"15:07.695","Text":"mass multiplied by our acceleration in the x-direction,"},{"Start":"15:07.695 ","End":"15:09.240","Text":"which here is 0."},{"Start":"15:09.240 ","End":"15:11.879","Text":"Here our mg, we know,"},{"Start":"15:11.879 ","End":"15:18.285","Text":"minus our F_y is equal to our mass multiplied by acceleration in the y-direction,"},{"Start":"15:18.285 ","End":"15:20.355","Text":"which is what we have over here."},{"Start":"15:20.355 ","End":"15:25.570","Text":"We can simply just substitute this in and solve."},{"Start":"15:26.270 ","End":"15:32.490","Text":"I simply substituted it in and this is the answer for F_x and F_y."},{"Start":"15:32.490 ","End":"15:34.875","Text":"If you would like to do this on a piece of paper,"},{"Start":"15:34.875 ","End":"15:36.840","Text":"you\u0027re more than welcome to."},{"Start":"15:36.840 ","End":"15:39.015","Text":"Now, in question number 3,"},{"Start":"15:39.015 ","End":"15:43.020","Text":"we\u0027re being told that the rod falls and faces vertically downwards."},{"Start":"15:43.020 ","End":"15:45.885","Text":"What is the angular velocity in this case?"},{"Start":"15:45.885 ","End":"15:53.460","Text":"Here, we\u0027re being asked to find what our Omega is when the rod is facing downwards."},{"Start":"15:53.460 ","End":"15:59.260","Text":"Let\u0027s go back to the diagram so I can show you what they mean."},{"Start":"16:00.110 ","End":"16:05.085","Text":"All I\u0027ve done is move the rod so that it\u0027s facing vertically downwards."},{"Start":"16:05.085 ","End":"16:10.365","Text":"The center of mass is moved from this position to this position over here."},{"Start":"16:10.365 ","End":"16:15.100","Text":"Now, we\u0027re being asked what is the angular velocity?"},{"Start":"16:15.410 ","End":"16:19.200","Text":"The Omega that we\u0027re now being asked to find,"},{"Start":"16:19.200 ","End":"16:21.570","Text":"is right after it\u0027s fallen,"},{"Start":"16:21.570 ","End":"16:24.760","Text":"what is this Omega?"},{"Start":"16:25.160 ","End":"16:28.484","Text":"The way that we can solve this very easily,"},{"Start":"16:28.484 ","End":"16:33.105","Text":"whenever you deal with an object which has fallen from some height,"},{"Start":"16:33.105 ","End":"16:36.630","Text":"a simple way is to use conservation of energy."},{"Start":"16:36.630 ","End":"16:39.465","Text":"We can see that our axis of rotation,"},{"Start":"16:39.465 ","End":"16:41.565","Text":"which is applying our forces."},{"Start":"16:41.565 ","End":"16:44.370","Text":"It hasn\u0027t moved, it\u0027s remained stationary,"},{"Start":"16:44.370 ","End":"16:47.730","Text":"which means that it hasn\u0027t done any work."},{"Start":"16:47.730 ","End":"16:51.840","Text":"This is a conservational force and also the other force that we have"},{"Start":"16:51.840 ","End":"16:56.655","Text":"acting is our Mg and our Mg is always a conservational force."},{"Start":"16:56.655 ","End":"17:01.985","Text":"We\u0027re going to use the idea of conservation of energy in order to find what our Omega is."},{"Start":"17:01.985 ","End":"17:07.820","Text":"Again, we\u0027re going to use the relationship between our Omega and our velocity."},{"Start":"17:08.390 ","End":"17:10.705","Text":"Let\u0027s see how we do this."},{"Start":"17:10.705 ","End":"17:15.440","Text":"We\u0027re going to start by writing out the initial energy of the system."},{"Start":"17:15.440 ","End":"17:19.090","Text":"Our E_i is going to be equal to,"},{"Start":"17:19.090 ","End":"17:22.270","Text":"so we have a 1/2 and then,"},{"Start":"17:22.270 ","End":"17:23.560","Text":"because we\u0027re dealing with omega,"},{"Start":"17:23.560 ","End":"17:28.000","Text":"so it\u0027s going to be 1/2 I Omega^2."},{"Start":"17:28.000 ","End":"17:30.235","Text":"Now as we know, initially,"},{"Start":"17:30.235 ","End":"17:32.365","Text":"our angular velocity, our Omega,"},{"Start":"17:32.365 ","End":"17:34.025","Text":"is equal to 0."},{"Start":"17:34.025 ","End":"17:37.330","Text":"That means that this is going to be equal to 0,"},{"Start":"17:37.330 ","End":"17:42.300","Text":"and then we\u0027re going to have plus our mgh."},{"Start":"17:42.300 ","End":"17:45.170","Text":"Now, if we say that our h is equal to 0,"},{"Start":"17:45.170 ","End":"17:50.200","Text":"because our height, let\u0027s go back to the diagram."},{"Start":"17:50.200 ","End":"17:53.590","Text":"This is our axis of rotation."},{"Start":"17:53.590 ","End":"18:01.670","Text":"Our rod in its original position is over here, at height 0."},{"Start":"18:02.010 ","End":"18:08.800","Text":"That means that we can say that this height is equal to 0,"},{"Start":"18:08.800 ","End":"18:12.890","Text":"which means that this is also equal to 0."},{"Start":"18:14.280 ","End":"18:19.360","Text":"My total initial energy is equal to 0."},{"Start":"18:19.360 ","End":"18:26.770","Text":"Now, notice in this equation that when dealing with my kinetic energy,"},{"Start":"18:26.770 ","End":"18:30.220","Text":"the expression for kinetic energy only has the expression"},{"Start":"18:30.220 ","End":"18:34.135","Text":"involving the Omega and nothing involving the velocity."},{"Start":"18:34.135 ","End":"18:38.365","Text":"That\u0027s because my axis of rotation is stationary, it\u0027s not moving."},{"Start":"18:38.365 ","End":"18:40.375","Text":"If my axis of rotation was moving,"},{"Start":"18:40.375 ","End":"18:44.840","Text":"then I would have to include another expression over here."},{"Start":"18:46.020 ","End":"18:50.110","Text":"Now, we\u0027re going to work out our final energy,"},{"Start":"18:50.110 ","End":"18:52.840","Text":"so our E_f of the system."},{"Start":"18:52.840 ","End":"18:58.165","Text":"Again, we\u0027re going to have 1/2(I) Omega^2."},{"Start":"18:58.165 ","End":"19:00.940","Text":"This is the Omega that we\u0027re trying to find."},{"Start":"19:00.940 ","End":"19:05.890","Text":"I\u0027ll put over here in red a tilde to show that we\u0027re trying to find this."},{"Start":"19:05.890 ","End":"19:07.555","Text":"This is our unknown."},{"Start":"19:07.555 ","End":"19:14.005","Text":"Our I of course, is what we worked out before for a rod rotating along its edge,"},{"Start":"19:14.005 ","End":"19:16.070","Text":"which is this over here."},{"Start":"19:17.430 ","End":"19:24.415","Text":"Then we\u0027re going to add on plus mgh."},{"Start":"19:24.415 ","End":"19:27.310","Text":"Again, I don\u0027t have another expression when"},{"Start":"19:27.310 ","End":"19:29.710","Text":"dealing with the kinetic energy section because,"},{"Start":"19:29.710 ","End":"19:34.765","Text":"again, my axis of rotation is stationary. I have this."},{"Start":"19:34.765 ","End":"19:37.030","Text":"Now, when I\u0027m dealing with my potential energy,"},{"Start":"19:37.030 ","End":"19:38.365","Text":"I have my mass,"},{"Start":"19:38.365 ","End":"19:40.120","Text":"the mass of the system,"},{"Start":"19:40.120 ","End":"19:45.865","Text":"multiplied by g and then my h. Now my h here is going to be different."},{"Start":"19:45.865 ","End":"19:47.230","Text":"How do I solve this?"},{"Start":"19:47.230 ","End":"19:50.680","Text":"I\u0027m going to write it according to my h center of mass."},{"Start":"19:50.680 ","End":"19:53.380","Text":"I\u0027m going to take it from my center of mass."},{"Start":"19:53.380 ","End":"19:59.935","Text":"Now, we can say my h center of mass is going to be equal to,"},{"Start":"19:59.935 ","End":"20:04.165","Text":"so my center of mass is at L/2."},{"Start":"20:04.165 ","End":"20:05.440","Text":"Now, this section,"},{"Start":"20:05.440 ","End":"20:07.150","Text":"let\u0027s see what our sign is."},{"Start":"20:07.150 ","End":"20:09.595","Text":"Is it going to be a positive or a negative?"},{"Start":"20:09.595 ","End":"20:13.045","Text":"It\u0027s going to be a negative and I\u0027ll tell you why."},{"Start":"20:13.045 ","End":"20:15.355","Text":"Let\u0027s go back over here."},{"Start":"20:15.355 ","End":"20:17.907","Text":"My center of mass has moved,"},{"Start":"20:17.907 ","End":"20:22.510","Text":"it\u0027s located at L/2 in the rod."},{"Start":"20:22.510 ","End":"20:27.070","Text":"Even though we said that my positive y-direction is facing downwards,"},{"Start":"20:27.070 ","End":"20:30.865","Text":"I don\u0027t put a positive over here in front of my a_CM."},{"Start":"20:30.865 ","End":"20:36.670","Text":"This is because it doesn\u0027t matter which direction I\u0027ve said that my axes are going in."},{"Start":"20:36.670 ","End":"20:40.615","Text":"We\u0027re working on Earth and gravity."},{"Start":"20:40.615 ","End":"20:44.410","Text":"As we can see that my height began over here,"},{"Start":"20:44.410 ","End":"20:46.300","Text":"and it moves downwards."},{"Start":"20:46.300 ","End":"20:49.495","Text":"My height was reduced from this movement."},{"Start":"20:49.495 ","End":"20:53.680","Text":"The center of mass has moved now closer to my Earth."},{"Start":"20:53.680 ","End":"20:57.980","Text":"The height has reduced, which means that it\u0027s negative."},{"Start":"20:59.760 ","End":"21:02.515","Text":"When dealing with where your height is,"},{"Start":"21:02.515 ","End":"21:04.611","Text":"it doesn\u0027t matter which direction you said is"},{"Start":"21:04.611 ","End":"21:09.685","Text":"positive direction on your x or on your y-axis specifically."},{"Start":"21:09.685 ","End":"21:15.440","Text":"What matters is if you\u0027ve gone closer to Earth."},{"Start":"21:16.440 ","End":"21:21.070","Text":"Now, this is of course also going to be equal to"},{"Start":"21:21.070 ","End":"21:25.330","Text":"0 because we\u0027re using the idea of conservation of energy."},{"Start":"21:25.330 ","End":"21:35.125","Text":"Now, all I have to do is isolate out my Omega and then I will get my answer. Let\u0027s see."},{"Start":"21:35.125 ","End":"21:36.970","Text":"We can substitute everything in."},{"Start":"21:36.970 ","End":"21:40.765","Text":"I\u0027ll have my 1/2 multiplied by my I,"},{"Start":"21:40.765 ","End":"21:43.075","Text":"we can substitute in our value from over there."},{"Start":"21:43.075 ","End":"21:51.550","Text":"Omega^2 is going to be equal to negative Mg multiplied by negative L/2,"},{"Start":"21:51.550 ","End":"21:52.795","Text":"so it\u0027s going to be a positive."},{"Start":"21:52.795 ","End":"21:56.800","Text":"It will be Mg L/2."},{"Start":"21:56.800 ","End":"22:00.384","Text":"The M I wrote here as a capital because in the question,"},{"Start":"22:00.384 ","End":"22:07.510","Text":"the mass is denoted by M. Then I can multiply both sides by 2 and get rid of that."},{"Start":"22:07.510 ","End":"22:09.715","Text":"Then I can just isolate out my Omega."},{"Start":"22:09.715 ","End":"22:14.950","Text":"My Omega is going to be equal to the square root of"},{"Start":"22:14.950 ","End":"22:23.090","Text":"my MgL divided by my I."},{"Start":"22:25.560 ","End":"22:28.420","Text":"This is going to be equal to,"},{"Start":"22:28.420 ","End":"22:30.505","Text":"once you substitute in your I,"},{"Start":"22:30.505 ","End":"22:34.150","Text":"is going to be equal to the square root of 3g"},{"Start":"22:34.150 ","End":"22:40.225","Text":"divided by L. That\u0027s the answer to question 3."},{"Start":"22:40.225 ","End":"22:43.970","Text":"Let\u0027s go on to question 4."},{"Start":"22:44.430 ","End":"22:50.910","Text":"In question number 4, I\u0027m again asked to find what I was asked in question 1 and 2."},{"Start":"22:50.910 ","End":"22:52.575","Text":"Here to find my Alpha,"},{"Start":"22:52.575 ","End":"22:54.765","Text":"my a_CM, my F_x, and F_y."},{"Start":"22:54.765 ","End":"22:58.875","Text":"However, this time when the rod is facing downwards."},{"Start":"22:58.875 ","End":"23:01.940","Text":"In the situation that we had over here."},{"Start":"23:01.940 ","End":"23:05.560","Text":"Just to draw it for you."},{"Start":"23:05.560 ","End":"23:07.975","Text":"If I draw the rod in pink,"},{"Start":"23:07.975 ","End":"23:11.725","Text":"I\u0027ll have here is my rod."},{"Start":"23:11.725 ","End":"23:15.055","Text":"There\u0027s a little draft. Here\u0027s my axis of rotation."},{"Start":"23:15.055 ","End":"23:18.020","Text":"Here\u0027s my center of mass."},{"Start":"23:18.290 ","End":"23:24.530","Text":"Let\u0027s begin by what I did in my first section."},{"Start":"23:24.530 ","End":"23:30.760","Text":"What I did is I wrote my equations for sum of all of the forces and moments,"},{"Start":"23:30.760 ","End":"23:32.215","Text":"and it worked very well for me."},{"Start":"23:32.215 ","End":"23:35.960","Text":"I\u0027m going to start in the exact same way."},{"Start":"23:36.240 ","End":"23:41.890","Text":"Now, also we know that we have our Omega from the previous question."},{"Start":"23:41.890 ","End":"23:43.780","Text":"Let\u0027s see what forces I have."},{"Start":"23:43.780 ","End":"23:46.225","Text":"I have it going in this downwards direction."},{"Start":"23:46.225 ","End":"23:51.460","Text":"I have my Mg. Then again my other forces are"},{"Start":"23:51.460 ","End":"23:58.210","Text":"my F_x and F_y."},{"Start":"23:58.210 ","End":"24:01.940","Text":"But there of course going to be different F_x and F_y\u0027s."},{"Start":"24:03.630 ","End":"24:07.645","Text":"The first thing we\u0027re going to try and find is our Alpha."},{"Start":"24:07.645 ","End":"24:11.125","Text":"We\u0027re going to start by working out the sum of"},{"Start":"24:11.125 ","End":"24:14.950","Text":"our moments or the sum of our torques. Let\u0027s see."},{"Start":"24:14.950 ","End":"24:20.755","Text":"Our F_x and F_y don\u0027t have a moment and our Mg also doesn\u0027t have a moment."},{"Start":"24:20.755 ","End":"24:26.030","Text":"That means that the sum of all of our moments is going to be equal to 0."},{"Start":"24:26.790 ","End":"24:33.460","Text":"That means that my Alpha is going to equal to 0."},{"Start":"24:33.460 ","End":"24:36.235","Text":"Next, my a_Theta,"},{"Start":"24:36.235 ","End":"24:41.080","Text":"so my tangential acceleration is also going to be equal"},{"Start":"24:41.080 ","End":"24:46.510","Text":"to 0 because it equals to Alpha R. I\u0027m actually going to write this out."},{"Start":"24:46.510 ","End":"24:52.375","Text":"It equals to Alpha multiplied by the radius of rotation, which is L/2."},{"Start":"24:52.375 ","End":"24:54.700","Text":"However, my Alpha is equal to 0."},{"Start":"24:54.700 ","End":"24:58.580","Text":"My a_Theta is also going to be equal to 0."},{"Start":"24:58.620 ","End":"25:01.390","Text":"Then if we look over here,"},{"Start":"25:01.390 ","End":"25:06.685","Text":"our a_Theta is going to be in this direction."},{"Start":"25:06.685 ","End":"25:09.385","Text":"This is our a_Theta."},{"Start":"25:09.385 ","End":"25:13.580","Text":"That means that it also equals to our a_x."},{"Start":"25:14.580 ","End":"25:26.030","Text":"That means that 0 is also going to be equal to the a center of mass in the x-direction."},{"Start":"25:26.460 ","End":"25:30.950","Text":"Now, we can see that our a_y,"},{"Start":"25:30.990 ","End":"25:34.975","Text":"our a center of mass in the y-direction,"},{"Start":"25:34.975 ","End":"25:39.850","Text":"is going to be equal to our a in the radial direction."},{"Start":"25:39.850 ","End":"25:46.690","Text":"Because this is now going to be the direction of our a_r which is on the y-axis."},{"Start":"25:46.690 ","End":"25:51.025","Text":"You just have to follow through and see if it\u0027s on the x-axis, on the y-axis."},{"Start":"25:51.025 ","End":"25:56.730","Text":"This is equal to negative Omega^2 multiplied by r,"},{"Start":"25:56.730 ","End":"26:01.400","Text":"and r over here is L divided by 2."},{"Start":"26:02.160 ","End":"26:05.725","Text":"We have our Omega over here."},{"Start":"26:05.725 ","End":"26:08.350","Text":"We can just multiply this."},{"Start":"26:08.350 ","End":"26:10.719","Text":"We\u0027ll have negative Omega^2,"},{"Start":"26:10.719 ","End":"26:20.690","Text":"which will be negative 3g/L multiplied by L divided by 2."},{"Start":"26:21.000 ","End":"26:27.070","Text":"Then we\u0027ll get that our a center of mass in the y-direction"},{"Start":"26:27.070 ","End":"26:33.710","Text":"is simply going to be negative 3g divided by 2."},{"Start":"26:34.470 ","End":"26:38.830","Text":"Now, we\u0027ve found our Alpha and our a_CM."},{"Start":"26:38.830 ","End":"26:41.935","Text":"Now, what\u0027s left to find is our F_x and F_y."},{"Start":"26:41.935 ","End":"26:50.020","Text":"If we go back to this section over here where we wrote our force equations."},{"Start":"26:50.020 ","End":"26:53.245","Text":"We\u0027re going to use the exact same force equations."},{"Start":"26:53.245 ","End":"26:57.175","Text":"Although of course our F_x and F_y are going to"},{"Start":"26:57.175 ","End":"27:01.975","Text":"equal different things as our a_CMs are different."},{"Start":"27:01.975 ","End":"27:06.620","Text":"Now, we can just rewrite that out."},{"Start":"27:06.720 ","End":"27:11.215","Text":"These are the exact same force equations from question number 1."},{"Start":"27:11.215 ","End":"27:16.510","Text":"Now, we can simply substitute this in to find out what our F_y is."},{"Start":"27:16.510 ","End":"27:24.220","Text":"To find out F_y, we can get our M\u0027s and then we\u0027re going to have"},{"Start":"27:24.220 ","End":"27:33.100","Text":"a g minus our a center of mass in the y-direction, which is this."},{"Start":"27:33.100 ","End":"27:34.780","Text":"Over here what we have."},{"Start":"27:34.780 ","End":"27:39.535","Text":"It\u0027s going to be minus 3/2g."},{"Start":"27:39.535 ","End":"27:42.670","Text":"We\u0027ll cancel that out in a second."},{"Start":"27:42.670 ","End":"27:52.555","Text":"That\u0027s going to be M. It\u0027s going to equal to negative 1/2 Mg,"},{"Start":"27:52.555 ","End":"27:56.990","Text":"which is equal to our F_y."},{"Start":"27:57.030 ","End":"28:01.405","Text":"Then regarding our F_x,"},{"Start":"28:01.405 ","End":"28:06.320","Text":"we know that our a_CM in the x-direction is 0."},{"Start":"28:06.630 ","End":"28:09.745","Text":"We have negative F_x is equal to 0."},{"Start":"28:09.745 ","End":"28:14.660","Text":"That means that our F_x is equal to 0."},{"Start":"28:15.510 ","End":"28:20.450","Text":"That is the end of the question."}],"ID":9429}],"Thumbnail":null,"ID":5408},{"Name":"5. Rolling With Slipping","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Rolling With Slipping","Duration":"10m 50s","ChapterTopicVideoID":9160,"CourseChapterTopicPlaylistID":5409,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this lesson,"},{"Start":"00:02.235 ","End":"00:05.715","Text":"we\u0027re going to be speaking about rolling without slipping."},{"Start":"00:05.715 ","End":"00:08.100","Text":"Now, I\u0027m just going to remind you what we spoke"},{"Start":"00:08.100 ","End":"00:11.505","Text":"about when we were dealing with rolling with slipping."},{"Start":"00:11.505 ","End":"00:16.785","Text":"We said that if we have this wheel moving,"},{"Start":"00:16.785 ","End":"00:19.560","Text":"its center of mass, which is located over here,"},{"Start":"00:19.560 ","End":"00:23.775","Text":"we\u0027ll have some velocity going in this direction,"},{"Start":"00:23.775 ","End":"00:27.930","Text":"which we said was called the V_cm velocity of the center of mass."},{"Start":"00:27.930 ","End":"00:30.765","Text":"Because the wheel is rolling along,"},{"Start":"00:30.765 ","End":"00:36.900","Text":"so it\u0027s also going to have some angular velocity Omega."},{"Start":"00:36.900 ","End":"00:39.830","Text":"Then we said that given this,"},{"Start":"00:39.830 ","End":"00:45.125","Text":"so if we have some point over here and we call this point, let\u0027s say A."},{"Start":"00:45.125 ","End":"00:50.720","Text":"We said that the velocity at point A is going to be"},{"Start":"00:50.720 ","End":"00:56.075","Text":"equal to a v center of mass multiplied by Omega,"},{"Start":"00:56.075 ","End":"00:58.920","Text":"multiplied by the radius."},{"Start":"01:00.020 ","End":"01:02.195","Text":"This is of course,"},{"Start":"01:02.195 ","End":"01:07.715","Text":"given that this clockwise direction is the positive direction"},{"Start":"01:07.715 ","End":"01:14.430","Text":"and this direction in the x-axis is the positive direction."},{"Start":"01:14.660 ","End":"01:20.815","Text":"If you remember, we said that rolling without slipping,"},{"Start":"01:20.815 ","End":"01:26.675","Text":"the condition was that this would equal to 0 without slipping."},{"Start":"01:26.675 ","End":"01:28.400","Text":"Now when we\u0027re dealing with rolling,"},{"Start":"01:28.400 ","End":"01:32.165","Text":"with slipping, it\u0027s exactly the opposite."},{"Start":"01:32.165 ","End":"01:37.320","Text":"The condition is that this does not equal 0,"},{"Start":"01:37.320 ","End":"01:40.450","Text":"anything else but not 0."},{"Start":"01:40.700 ","End":"01:45.240","Text":"I\u0027ve written up over here the conditions."},{"Start":"01:45.240 ","End":"01:47.780","Text":"When we\u0027re speaking about no slipping,"},{"Start":"01:47.780 ","End":"01:49.475","Text":"then our V_A,"},{"Start":"01:49.475 ","End":"01:53.330","Text":"the velocity of this point over here will equal to 0."},{"Start":"01:53.330 ","End":"01:55.025","Text":"When there is slipping,"},{"Start":"01:55.025 ","End":"01:59.670","Text":"the velocity at this point over here will be different to 0."},{"Start":"01:59.800 ","End":"02:07.075","Text":"This is useful to write in your books and that also means that there\u0027s only one case."},{"Start":"02:07.075 ","End":"02:10.425","Text":"When our V_cm is equal to our Omega R,"},{"Start":"02:10.425 ","End":"02:13.005","Text":"and that\u0027s when there\u0027s no slipping."},{"Start":"02:13.005 ","End":"02:19.565","Text":"That only happens in one circumstance when V_cm is equal to Omega R. However,"},{"Start":"02:19.565 ","End":"02:21.425","Text":"in the case of width slipping,"},{"Start":"02:21.425 ","End":"02:22.900","Text":"because it doesn\u0027t equal to 0,"},{"Start":"02:22.900 ","End":"02:29.070","Text":"we have an infinite amount of cases that will fit this expression."},{"Start":"02:29.930 ","End":"02:32.105","Text":"When we\u0027re dealing with the case,"},{"Start":"02:32.105 ","End":"02:36.515","Text":"rolling with slipping, when our V_A does not equal to 0."},{"Start":"02:36.515 ","End":"02:41.690","Text":"That means that our point over here has some velocity,"},{"Start":"02:41.690 ","End":"02:50.080","Text":"which means that we\u0027re going to have kinetic friction."},{"Start":"02:54.080 ","End":"02:59.735","Text":"Our frictional force is going to be kinetic, not static."},{"Start":"02:59.735 ","End":"03:09.030","Text":"Because our point has some velocity relative to the ground or relative to the surface."},{"Start":"03:09.740 ","End":"03:15.890","Text":"In order to explain to you what this actually means that we\u0027re rolling without slipping,"},{"Start":"03:15.890 ","End":"03:19.640","Text":"there are a few examples that fit this case."},{"Start":"03:19.640 ","End":"03:25.260","Text":"That either means that our wheel is not rotating."},{"Start":"03:25.260 ","End":"03:26.775","Text":"We can rub this out,"},{"Start":"03:26.775 ","End":"03:28.275","Text":"it has no omega."},{"Start":"03:28.275 ","End":"03:32.135","Text":"Just the center of mass is moving so you\u0027re just pushing it along."},{"Start":"03:32.135 ","End":"03:34.175","Text":"Which means that if it\u0027s not rotating,"},{"Start":"03:34.175 ","End":"03:37.610","Text":"this point is moving from this to here to here to here,"},{"Start":"03:37.610 ","End":"03:41.190","Text":"as the whole wheel moves along."},{"Start":"03:41.190 ","End":"03:45.410","Text":"Obviously, the velocity of this point is not going to be equal to 0."},{"Start":"03:45.410 ","End":"03:50.770","Text":"V_A will not equal 0 because it itself is moving along the surface."},{"Start":"03:50.770 ","End":"03:54.065","Text":"The next example is exactly the opposite."},{"Start":"03:54.065 ","End":"03:57.215","Text":"If we rub out V_cm,"},{"Start":"03:57.215 ","End":"04:01.655","Text":"that means that I wheel isn\u0027t moving along the surface."},{"Start":"04:01.655 ","End":"04:04.630","Text":"But if we have some Omega,"},{"Start":"04:04.630 ","End":"04:08.145","Text":"so it\u0027s just rotating in the same spot."},{"Start":"04:08.145 ","End":"04:10.125","Text":"Obviously, our point here,"},{"Start":"04:10.125 ","End":"04:13.550","Text":"point A is going to have some velocity,"},{"Start":"04:13.550 ","End":"04:16.520","Text":"and we\u0027re going to have kinetic friction because"},{"Start":"04:16.520 ","End":"04:21.740","Text":"this point is rubbing against the surface every time it goes past."},{"Start":"04:22.910 ","End":"04:27.675","Text":"Those were the 2 simplest ways to"},{"Start":"04:27.675 ","End":"04:32.830","Text":"explain what our V_A is when we\u0027re talking about width slipping."},{"Start":"04:32.830 ","End":"04:37.280","Text":"However, there\u0027s obviously much more complicated movements where we still"},{"Start":"04:37.280 ","End":"04:42.005","Text":"have both V_cm and Omega."},{"Start":"04:42.005 ","End":"04:45.440","Text":"Both of these are numbers that are different to 0."},{"Start":"04:45.440 ","End":"04:53.330","Text":"However, our V_cm is so much larger than Omega R or vice versa,"},{"Start":"04:53.330 ","End":"04:55.160","Text":"or omega is much larger."},{"Start":"04:55.160 ","End":"04:59.385","Text":"Then we\u0027ll have that this is a negative expression."},{"Start":"04:59.385 ","End":"05:05.795","Text":"That will just mean that we\u0027re rotating at a high angular velocity for instance."},{"Start":"05:05.795 ","End":"05:11.050","Text":"However, we\u0027re moving along the surface at a relatively low velocity,"},{"Start":"05:11.050 ","End":"05:13.340","Text":"so you can imagine if you have a car,"},{"Start":"05:13.340 ","End":"05:16.670","Text":"for instance, the wheels of the car,"},{"Start":"05:16.670 ","End":"05:19.710","Text":"excuse my drawing skills."},{"Start":"05:20.540 ","End":"05:24.020","Text":"Here\u0027s your car, and if for instance,"},{"Start":"05:24.020 ","End":"05:26.170","Text":"you\u0027re stuck in sand,"},{"Start":"05:26.170 ","End":"05:28.950","Text":"this is sand, these dots."},{"Start":"05:28.950 ","End":"05:31.610","Text":"If you can imagine how your tires"},{"Start":"05:31.610 ","End":"05:34.160","Text":"would look when they\u0027re spinning, when they\u0027re stuck in sand."},{"Start":"05:34.160 ","End":"05:36.395","Text":"They will be spinning very, very fast."},{"Start":"05:36.395 ","End":"05:41.180","Text":"But the distance that your car will move forwards is going to"},{"Start":"05:41.180 ","End":"05:47.730","Text":"be much smaller in comparison to this angular velocity over here."},{"Start":"05:50.810 ","End":"05:54.110","Text":"Whenever we\u0027re dealing with rolling, with slipping,"},{"Start":"05:54.110 ","End":"05:59.225","Text":"then we\u0027re going to be dealing with kinetic friction."},{"Start":"05:59.225 ","End":"06:02.240","Text":"Generally what we\u0027re going to want to know is in"},{"Start":"06:02.240 ","End":"06:05.555","Text":"which direction is the kinetic friction acting."},{"Start":"06:05.555 ","End":"06:11.405","Text":"This kinetic friction comes from the fact that this point is slipping,"},{"Start":"06:11.405 ","End":"06:14.120","Text":"which means that it has friction with"},{"Start":"06:14.120 ","End":"06:18.865","Text":"the surface that are wheel or the object is rolling on."},{"Start":"06:18.865 ","End":"06:23.500","Text":"Now when we\u0027re dealing with the directions."},{"Start":"06:24.260 ","End":"06:29.110","Text":"The direction of kinetic friction"},{"Start":"06:29.540 ","End":"06:36.600","Text":"has to be in the opposite direction to the velocity."},{"Start":"06:36.830 ","End":"06:40.005","Text":"Now of which velocity are we speaking of?"},{"Start":"06:40.005 ","End":"06:45.750","Text":"We are speaking of the velocity of A,"},{"Start":"06:45.750 ","End":"06:49.170","Text":"in the opposite direction of V_ A."},{"Start":"06:50.150 ","End":"06:53.120","Text":"Our kinetic friction is meant to be in"},{"Start":"06:53.120 ","End":"06:57.570","Text":"the opposite direction to the direction of our V_A,"},{"Start":"06:57.610 ","End":"07:00.620","Text":"maybe in the same direction of our V_cm,"},{"Start":"07:00.620 ","End":"07:06.435","Text":"but always in the opposite direction of point A in the direction that it\u0027s going in."},{"Start":"07:06.435 ","End":"07:10.070","Text":"Now we need to know the direction of our V_A."},{"Start":"07:10.070 ","End":"07:13.070","Text":"How would we know the direction of our V_A?"},{"Start":"07:13.070 ","End":"07:17.970","Text":"what we have to do is we have to look at this expression over here."},{"Start":"07:18.200 ","End":"07:23.240","Text":"Let\u0027s take a look. Now I\u0027m writing this in red because it\u0027s important"},{"Start":"07:23.240 ","End":"07:28.290","Text":"and I want you to write this in your sheets, in your books."},{"Start":"07:28.360 ","End":"07:35.565","Text":"If we look at our V_cm minus Omega R,"},{"Start":"07:35.565 ","End":"07:37.289","Text":"and if this expression,"},{"Start":"07:37.289 ","End":"07:38.830","Text":"expression for V_A,"},{"Start":"07:38.830 ","End":"07:43.630","Text":"if this is bigger than 0, so it\u0027s positive,"},{"Start":"07:43.630 ","End":"07:54.285","Text":"then we know that our kinetic friction is going to be equal to negative Mu_k."},{"Start":"07:54.285 ","End":"08:00.410","Text":"The coefficient of kinetic friction multiplied by the normal force."},{"Start":"08:00.410 ","End":"08:10.080","Text":"Here the key thing is the negative if this is bigger than 0."},{"Start":"08:10.780 ","End":"08:16.060","Text":"In this case, what will be having is that our force,"},{"Start":"08:16.060 ","End":"08:20.205","Text":"if I velocity is positive, V_A is positive."},{"Start":"08:20.205 ","End":"08:23.730","Text":"Then our frictional force is going to be negative."},{"Start":"08:23.730 ","End":"08:27.990","Text":"It\u0027s going to be in this direction."},{"Start":"08:27.990 ","End":"08:33.840","Text":"That means it\u0027s also in the opposite direction to our V_cm and that\u0027s fine."},{"Start":"08:35.900 ","End":"08:39.545","Text":"Now, when dealing with the other case,"},{"Start":"08:39.545 ","End":"08:47.795","Text":"so if we have that V_cm minus our Omega R is smaller than 0."},{"Start":"08:47.795 ","End":"08:53.570","Text":"Then kinetic friction is going to be equal to positive,"},{"Start":"08:53.570 ","End":"08:59.405","Text":"the coefficient of friction Mu_k multiplied by the normal force."},{"Start":"08:59.405 ","End":"09:02.870","Text":"Here, if it\u0027s bigger than here,"},{"Start":"09:02.870 ","End":"09:04.850","Text":"we have a positive."},{"Start":"09:04.850 ","End":"09:08.840","Text":"Let\u0027s explain this. What happens here?"},{"Start":"09:08.840 ","End":"09:13.615","Text":"Our Omega is much larger than our V_cm."},{"Start":"09:13.615 ","End":"09:15.710","Text":"This section is much larger,"},{"Start":"09:15.710 ","End":"09:18.590","Text":"meaning that we\u0027ll get something smaller to 0."},{"Start":"09:18.590 ","End":"09:20.345","Text":"When it\u0027s much larger,"},{"Start":"09:20.345 ","End":"09:23.690","Text":"our wheel or whatever the object is,"},{"Start":"09:23.690 ","End":"09:27.080","Text":"it\u0027s going to rotate much more than it\u0027s going to move"},{"Start":"09:27.080 ","End":"09:34.625","Text":"forwards Then if you can imagine that it\u0027s rotating much faster than it\u0027s moving forward,"},{"Start":"09:34.625 ","End":"09:41.640","Text":"then you can see that our point A its movement is going to be in the left direction."},{"Start":"09:41.640 ","End":"09:44.040","Text":"It\u0027s going to move like this,"},{"Start":"09:44.040 ","End":"09:49.025","Text":"our point A which is going leftwards in the negative direction."},{"Start":"09:49.025 ","End":"09:55.070","Text":"That means that our force has to be in the opposite direction to our V_A,"},{"Start":"09:55.070 ","End":"09:59.180","Text":"which means that our f_k our kinetic friction is going to"},{"Start":"09:59.180 ","End":"10:05.300","Text":"be in this direction because our shape is rotating much faster,"},{"Start":"10:05.300 ","End":"10:11.460","Text":"which means that before this point has enough time to move forwards,"},{"Start":"10:11.460 ","End":"10:12.780","Text":"like with the V_cm,"},{"Start":"10:12.780 ","End":"10:17.080","Text":"so it\u0027s being kicked back much faster."},{"Start":"10:17.750 ","End":"10:24.230","Text":"Now, the last thing to remember is that if we\u0027re dealing with the kinetic friction,"},{"Start":"10:24.230 ","End":"10:34.365","Text":"so our f_k, that means that there\u0027s no conservation of energy."},{"Start":"10:34.365 ","End":"10:39.060","Text":"No conservation of energy if we have kinetic friction."},{"Start":"10:40.100 ","End":"10:45.185","Text":"These are the most important points for this section."},{"Start":"10:45.185 ","End":"10:50.970","Text":"Just try and remember all the things that I wrote in red, especially."}],"ID":9430},{"Watched":false,"Name":"Exercise 1","Duration":"20m 37s","ChapterTopicVideoID":9161,"CourseChapterTopicPlaylistID":5409,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:03.165","Text":"Hello. In this question,"},{"Start":"00:03.165 ","End":"00:05.280","Text":"a homogeneous ball of"},{"Start":"00:05.280 ","End":"00:11.710","Text":"mass m and initial velocity v_0"},{"Start":"00:15.380 ","End":"00:18.630","Text":"is moving without rotating."},{"Start":"00:18.630 ","End":"00:22.380","Text":"There\u0027s no angular velocity, so no Omega."},{"Start":"00:22.380 ","End":"00:26.740","Text":"The whole body is just moving as a whole."},{"Start":"00:26.890 ","End":"00:30.770","Text":"We\u0027re being told to find its final velocity if it is"},{"Start":"00:30.770 ","End":"00:33.901","Text":"known that the coefficient of friction is kinetic,"},{"Start":"00:33.901 ","End":"00:36.530","Text":"and what it is."},{"Start":"00:36.530 ","End":"00:40.660","Text":"Our Mu k is given to us."},{"Start":"00:40.660 ","End":"00:43.835","Text":"Now we\u0027re trying to find out its final velocity,"},{"Start":"00:43.835 ","End":"00:46.265","Text":"given that we know this value."},{"Start":"00:46.265 ","End":"00:49.140","Text":"Let\u0027s see how we do this."},{"Start":"00:49.370 ","End":"00:54.395","Text":"What\u0027s happening here is that because there\u0027s no rotation,"},{"Start":"00:54.395 ","End":"00:57.605","Text":"and we\u0027re just moving forwards with a velocity of v_0,"},{"Start":"00:57.605 ","End":"01:02.765","Text":"every single points in this ball is going to have a velocity v_0."},{"Start":"01:02.765 ","End":"01:06.985","Text":"That means that our point over here,"},{"Start":"01:06.985 ","End":"01:10.065","Text":"which is in contact with the surface,"},{"Start":"01:10.065 ","End":"01:12.840","Text":"our point A, we\u0027ll call it."},{"Start":"01:12.840 ","End":"01:15.510","Text":"Because it\u0027s also a point on the ball,"},{"Start":"01:15.510 ","End":"01:18.655","Text":"so it has an initial velocity of v_0."},{"Start":"01:18.655 ","End":"01:21.450","Text":"We can write this down."},{"Start":"01:21.450 ","End":"01:27.090","Text":"Our V_A initially is going to be equal to our v_0."},{"Start":"01:27.890 ","End":"01:31.200","Text":"If our V_A is equal to our v_0,"},{"Start":"01:31.200 ","End":"01:33.795","Text":"and our v_0 is going in this direction,"},{"Start":"01:33.795 ","End":"01:40.130","Text":"so we know from the previous lesson that a frictional force,"},{"Start":"01:40.130 ","End":"01:41.825","Text":"when dealing with kinetic friction,"},{"Start":"01:41.825 ","End":"01:45.575","Text":"is in the opposite direction to our V_A."},{"Start":"01:45.575 ","End":"01:49.980","Text":"If our V_A is going in this direction because it\u0027s equal to v_0,"},{"Start":"01:49.980 ","End":"01:55.260","Text":"then we know that our kinetic friction,"},{"Start":"01:55.260 ","End":"01:59.770","Text":"our f_k, is going to be in this direction."},{"Start":"02:00.470 ","End":"02:06.800","Text":"Now what I can do is I can start writing my equations for the sum of all of the forces."},{"Start":"02:06.800 ","End":"02:09.290","Text":"First, I\u0027m going to decide my axis."},{"Start":"02:09.290 ","End":"02:15.110","Text":"I\u0027m going to say that this direction of rotation is positive,"},{"Start":"02:15.110 ","End":"02:21.055","Text":"and that this direction is the positive x-direction."},{"Start":"02:21.055 ","End":"02:23.385","Text":"Now I can write down."},{"Start":"02:23.385 ","End":"02:28.795","Text":"We can start off with the sum of all of my forces on the x-axis,"},{"Start":"02:28.795 ","End":"02:31.925","Text":"so the only force that\u0027s acting on my system over here,"},{"Start":"02:31.925 ","End":"02:34.925","Text":"I can see is my kinetic friction."},{"Start":"02:34.925 ","End":"02:43.880","Text":"It\u0027s going in the negative x-direction, so negative f_k."},{"Start":"02:43.880 ","End":"02:46.130","Text":"That\u0027s equal to, as we know,"},{"Start":"02:46.130 ","End":"02:47.700","Text":"so it\u0027s negative because of here,"},{"Start":"02:47.700 ","End":"02:52.280","Text":"and then the coefficient of friction multiplied by the normal force."},{"Start":"02:52.280 ","End":"02:54.185","Text":"Then, as we know,"},{"Start":"02:54.185 ","End":"02:58.520","Text":"this is going to be equal to our mass multiplied"},{"Start":"02:58.520 ","End":"03:03.965","Text":"by our acceleration of the center of mass in the x-direction."},{"Start":"03:03.965 ","End":"03:08.120","Text":"Now, something important to notice over here is"},{"Start":"03:08.120 ","End":"03:12.185","Text":"that we\u0027re always dealing with our acceleration of the center of mass."},{"Start":"03:12.185 ","End":"03:14.285","Text":"Specifically of the center of mass,"},{"Start":"03:14.285 ","End":"03:16.530","Text":"not of the individual points,"},{"Start":"03:16.530 ","End":"03:19.115","Text":"and that\u0027s because when I\u0027m looking at my forces,"},{"Start":"03:19.115 ","End":"03:21.530","Text":"my kinetic friction, for instance,"},{"Start":"03:21.530 ","End":"03:23.675","Text":"this is an external force."},{"Start":"03:23.675 ","End":"03:26.240","Text":"I\u0027m not calculating the forces within"},{"Start":"03:26.240 ","End":"03:29.240","Text":"the ball that\u0027s happening between the atoms and whatever."},{"Start":"03:29.240 ","End":"03:31.250","Text":"I\u0027m looking at, the ball as a whole,"},{"Start":"03:31.250 ","End":"03:32.630","Text":"as some kind of body,"},{"Start":"03:32.630 ","End":"03:35.465","Text":"and the external force is acting on it."},{"Start":"03:35.465 ","End":"03:37.730","Text":"If I\u0027m dealing with that, then I have to deal with"},{"Start":"03:37.730 ","End":"03:39.980","Text":"the acceleration of the ball as a whole,"},{"Start":"03:39.980 ","End":"03:44.305","Text":"and not taking into account the acceleration of each individual atom."},{"Start":"03:44.305 ","End":"03:48.305","Text":"As we know, whenever we\u0027re dealing with something as a whole,"},{"Start":"03:48.305 ","End":"03:51.140","Text":"so we\u0027re going to look at it as a center of mass."},{"Start":"03:51.140 ","End":"03:53.569","Text":"When we\u0027re dealing with rigid bodies,"},{"Start":"03:53.569 ","End":"03:56.610","Text":"so we deal with the center of mass."},{"Start":"03:57.950 ","End":"04:01.830","Text":"Now looking at the y-axis,"},{"Start":"04:01.830 ","End":"04:04.220","Text":"we can say that the sum of all the forces in"},{"Start":"04:04.220 ","End":"04:08.675","Text":"the y-axis is going to be equal to. Let\u0027s take a look."},{"Start":"04:08.675 ","End":"04:13.670","Text":"We have mg going in the downwards direction."},{"Start":"04:13.670 ","End":"04:24.120","Text":"We can say that y-axis is going in this direction."},{"Start":"04:24.120 ","End":"04:27.495","Text":"Then we have the normal force. Let\u0027s draw this."},{"Start":"04:27.495 ","End":"04:31.255","Text":"We have a normal force going like this,"},{"Start":"04:31.255 ","End":"04:35.655","Text":"and our mg going downwards."},{"Start":"04:35.655 ","End":"04:41.095","Text":"Then what we can just write here is N minus mg."},{"Start":"04:41.095 ","End":"04:43.630","Text":"This is going to be equal to our mass times"},{"Start":"04:43.630 ","End":"04:47.530","Text":"our acceleration of the center of mass in the y-direction."},{"Start":"04:47.530 ","End":"04:50.240","Text":"Now because our ball isn\u0027t moving up or down,"},{"Start":"04:50.240 ","End":"04:55.975","Text":"the acceleration of the center of mass in the y-direction will equal to 0,"},{"Start":"04:55.975 ","End":"05:02.300","Text":"which means that this is going to be equal to 0."},{"Start":"05:02.780 ","End":"05:10.335","Text":"Then we can see that our N is equal to our mg. We can substitute that in over here."},{"Start":"05:10.335 ","End":"05:19.050","Text":"Then we can get that our kinetic friction is equal to Mu k Mg,"},{"Start":"05:19.050 ","End":"05:27.480","Text":"which is equal to M multiplied by our a center of mass in the x-direction."},{"Start":"05:27.480 ","End":"05:32.920","Text":"Now what we can do is we can isolate out our a center"},{"Start":"05:32.920 ","End":"05:37.780","Text":"of mass in the x-direction to get what its value is."},{"Start":"05:37.780 ","End":"05:40.630","Text":"We have our a center of mass in the x-direction,"},{"Start":"05:40.630 ","End":"05:46.470","Text":"can divide both sides by M. Take this away."},{"Start":"05:46.470 ","End":"05:51.990","Text":"Then we will just get that our value for"},{"Start":"05:51.990 ","End":"05:57.170","Text":"this is Mu k multiplied by g. Of course,"},{"Start":"05:57.170 ","End":"06:00.325","Text":"because I forgot my negative over here,"},{"Start":"06:00.325 ","End":"06:03.135","Text":"so also over here we have a negative."},{"Start":"06:03.135 ","End":"06:08.690","Text":"Now we can see that our a center of mass in the x-direction is negative,"},{"Start":"06:08.690 ","End":"06:10.820","Text":"and then Mu k is a constant,"},{"Start":"06:10.820 ","End":"06:12.320","Text":"and our g is a constant."},{"Start":"06:12.320 ","End":"06:16.830","Text":"We can tell that we\u0027re dealing with a constant deceleration."},{"Start":"06:16.830 ","End":"06:21.965","Text":"When we\u0027re dealing with a constant deceleration or a constant acceleration,"},{"Start":"06:21.965 ","End":"06:26.490","Text":"the equation for the velocity, again,"},{"Start":"06:26.490 ","End":"06:34.295","Text":"of the center of mass as a function of time is going to be equal to our initial velocity."},{"Start":"06:34.295 ","End":"06:41.285","Text":"Here, it\u0027s v_0. Plus our acceleration multiplied by time."},{"Start":"06:41.285 ","End":"06:47.195","Text":"Here, it\u0027s going to be v_0 plus our at,"},{"Start":"06:47.195 ","End":"06:48.440","Text":"which is a negative,"},{"Start":"06:48.440 ","End":"06:53.150","Text":"so negative Mu_k g multiplied"},{"Start":"06:53.150 ","End":"06:58.880","Text":"by t. Up until now,"},{"Start":"06:58.880 ","End":"07:01.025","Text":"we\u0027ve been considering our ball as"},{"Start":"07:01.025 ","End":"07:04.645","Text":"a point mass located at the center of mass of the ball."},{"Start":"07:04.645 ","End":"07:11.015","Text":"I haven\u0027t taken into account the different sections on the ball and the size of it."},{"Start":"07:11.015 ","End":"07:14.600","Text":"Now what I\u0027m going to start working on is looking at"},{"Start":"07:14.600 ","End":"07:19.144","Text":"the angular or the circular motion of the ball,"},{"Start":"07:19.144 ","End":"07:23.270","Text":"which means that I\u0027m now going to start looking at"},{"Start":"07:23.270 ","End":"07:28.385","Text":"it as an object made up of lots of different pieces."},{"Start":"07:28.385 ","End":"07:32.645","Text":"I\u0027m going to start looking at all of the forces acting on the ball,"},{"Start":"07:32.645 ","End":"07:34.850","Text":"and of all of the moments."},{"Start":"07:34.850 ","End":"07:41.130","Text":"We\u0027re going to write down the equation for the moments."},{"Start":"07:41.130 ","End":"07:46.100","Text":"What I can look at is that the only force that I have acting,"},{"Start":"07:46.100 ","End":"07:49.730","Text":"which I can write an equation for the sum of all of the moments,"},{"Start":"07:49.730 ","End":"07:52.130","Text":"is my frictional force over here."},{"Start":"07:52.130 ","End":"07:55.595","Text":"Because the only other forces that I have is my mg,"},{"Start":"07:55.595 ","End":"07:58.740","Text":"which is acting from the center of mass,"},{"Start":"07:58.740 ","End":"08:02.205","Text":"so it doesn\u0027t come into the calculation,"},{"Start":"08:02.205 ","End":"08:06.140","Text":"and my normal force which is acting into the center of mass."},{"Start":"08:06.140 ","End":"08:11.385","Text":"Again, it doesn\u0027t count into my calculations. Let\u0027s write this out."},{"Start":"08:11.385 ","End":"08:14.630","Text":"I\u0027m going to have the sum of all of my moments or the sum"},{"Start":"08:14.630 ","End":"08:18.720","Text":"of all my torques is going to be equal to."},{"Start":"08:20.630 ","End":"08:26.380","Text":"We\u0027re going to choose our center of mass to be our axis of rotation."},{"Start":"08:26.380 ","End":"08:32.665","Text":"Now, notice that when we\u0027re dealing with rolling without slipping,"},{"Start":"08:32.665 ","End":"08:36.595","Text":"then I can also choose this point a to be my axis of rotation."},{"Start":"08:36.595 ","End":"08:41.020","Text":"However, because we\u0027re dealing with rolling with slipping right now,"},{"Start":"08:41.020 ","End":"08:43.510","Text":"that means that my point a is moving."},{"Start":"08:43.510 ","End":"08:48.890","Text":"It\u0027s not stationary, which means that I can\u0027t choose it to be my axis of rotation."},{"Start":"08:49.160 ","End":"08:52.329","Text":"When dealing with rolling with slipping,"},{"Start":"08:52.329 ","End":"08:56.785","Text":"the axis of rotation is the center of mass, not this point."},{"Start":"08:56.785 ","End":"09:02.725","Text":"As we can see, the force that I have acting here is my frictional force."},{"Start":"09:02.725 ","End":"09:05.660","Text":"Now, in order to find the size of the force,"},{"Start":"09:05.660 ","End":"09:13.715","Text":"I\u0027m going to take the size of my kinetic friction multiplied by the size,"},{"Start":"09:13.715 ","End":"09:16.580","Text":"the distance it is from the axis of rotation,"},{"Start":"09:16.580 ","End":"09:18.840","Text":"which is obviously R."},{"Start":"09:19.290 ","End":"09:24.960","Text":"Then I\u0027m going to multiply it by sine of the angle between the two."},{"Start":"09:24.960 ","End":"09:27.345","Text":"As we can see very clearly,"},{"Start":"09:27.345 ","End":"09:29.985","Text":"the angle between my radius."},{"Start":"09:29.985 ","End":"09:34.954","Text":"This is the radius and my frictional force is 90 degrees."},{"Start":"09:34.954 ","End":"09:37.180","Text":"I multiply this by sine of 90,"},{"Start":"09:37.180 ","End":"09:38.770","Text":"which as we know is equal to 1,"},{"Start":"09:38.770 ","End":"09:41.620","Text":"and this is going to be equal to my moment of"},{"Start":"09:41.620 ","End":"09:45.130","Text":"inertia multiplied by my angular acceleration."},{"Start":"09:45.130 ","End":"09:50.750","Text":"Now, the next thing I have to do is to find out my sine over here."},{"Start":"09:51.570 ","End":"09:57.100","Text":"What I can do is if I see that this is my radius going"},{"Start":"09:57.100 ","End":"10:02.470","Text":"in this direction and my frictional force is going in the leftwards direction."},{"Start":"10:02.470 ","End":"10:05.619","Text":"If I take a look over here,"},{"Start":"10:05.619 ","End":"10:09.445","Text":"and I redraw this with some origins."},{"Start":"10:09.445 ","End":"10:11.800","Text":"My origin is at the point where they meet."},{"Start":"10:11.800 ","End":"10:19.405","Text":"Going down here, I have my radius and going in this direction,"},{"Start":"10:19.405 ","End":"10:22.675","Text":"I have my frictional force."},{"Start":"10:22.675 ","End":"10:30.115","Text":"Now when I do my vector multiplication."},{"Start":"10:30.115 ","End":"10:35.680","Text":"We can see that I multiply k from r until f,"},{"Start":"10:35.680 ","End":"10:38.304","Text":"which means that we\u0027re going in the clockwise direction,"},{"Start":"10:38.304 ","End":"10:40.355","Text":"which as we saw at the beginning,"},{"Start":"10:40.355 ","End":"10:44.540","Text":"I denoted the clockwise direction as the positive direction."},{"Start":"10:44.540 ","End":"10:48.310","Text":"That means that I can put a positive over here."},{"Start":"10:48.310 ","End":"10:51.160","Text":"This expression is going to be positive."},{"Start":"10:51.160 ","End":"10:54.610","Text":"Now when dealing with the moment of inertia,"},{"Start":"10:54.610 ","End":"10:56.200","Text":"this is something you should know."},{"Start":"10:56.200 ","End":"10:58.630","Text":"The moment of inertia for a ball,"},{"Start":"10:58.630 ","End":"11:01.360","Text":"or I should say rather a homogenous ball."},{"Start":"11:01.360 ","End":"11:05.155","Text":"Let\u0027s just go back to the question, a homogenous ball."},{"Start":"11:05.155 ","End":"11:10.945","Text":"This, you should just be writing out in your equation sheets is"},{"Start":"11:10.945 ","End":"11:16.720","Text":"simply 2/5 multiplied by the mass,"},{"Start":"11:16.720 ","End":"11:20.830","Text":"multiplied by its radius squared."},{"Start":"11:20.830 ","End":"11:23.215","Text":"This is the moment of inertia,"},{"Start":"11:23.215 ","End":"11:25.090","Text":"and then we have our Alpha."},{"Start":"11:25.090 ","End":"11:27.160","Text":"Now to find what our Alpha is,"},{"Start":"11:27.160 ","End":"11:29.455","Text":"which is our angular acceleration."},{"Start":"11:29.455 ","End":"11:35.215","Text":"We just have to rearrange this equation in order to isolate out our Alpha."},{"Start":"11:35.215 ","End":"11:41.300","Text":"It\u0027s going to be equal to our Mu_k multiplied by mgR"},{"Start":"11:41.850 ","End":"11:51.295","Text":"divided by our moment of inertia, which is 2/5mR^2."},{"Start":"11:51.295 ","End":"11:54.355","Text":"Then once we cross everything out, this R this,"},{"Start":"11:54.355 ","End":"11:57.535","Text":"our m and our m, move the 5 up."},{"Start":"11:57.535 ","End":"12:05.875","Text":"What we\u0027re going to have is 5Mu_kg divided by 2R."},{"Start":"12:05.875 ","End":"12:10.345","Text":"This is our angular acceleration."},{"Start":"12:10.345 ","End":"12:15.430","Text":"A few lessons ago, you\u0027ll remember that we had a table where we compared"},{"Start":"12:15.430 ","End":"12:21.400","Text":"our linear velocities and accelerations to our angular velocities and accelerations."},{"Start":"12:21.400 ","End":"12:27.220","Text":"Now what I\u0027m going to show you is just like we did over here,"},{"Start":"12:27.220 ","End":"12:33.310","Text":"where we found our acceleration and then we worked out our velocity through that."},{"Start":"12:33.310 ","End":"12:41.995","Text":"We can apply the exact same conditions and the exact same almost equation, 2 over here."},{"Start":"12:41.995 ","End":"12:47.178","Text":"Our equation, because we can see that our angular acceleration is positive,"},{"Start":"12:47.178 ","End":"12:48.925","Text":"and it\u0027s also constant."},{"Start":"12:48.925 ","End":"12:50.530","Text":"It\u0027s never changing."},{"Start":"12:50.530 ","End":"12:53.815","Text":"We can say, we can use this equation."},{"Start":"12:53.815 ","End":"12:57.220","Text":"Our Omega or angular velocity is a function of time,"},{"Start":"12:57.220 ","End":"13:05.750","Text":"is going to be equal to our original angular velocity plus at."},{"Start":"13:06.150 ","End":"13:08.995","Text":"Our initial angular velocity,"},{"Start":"13:08.995 ","End":"13:11.780","Text":"again, let\u0027s go back to the question."},{"Start":"13:12.600 ","End":"13:18.400","Text":"We can see, without rotating."},{"Start":"13:18.400 ","End":"13:21.020","Text":"There\u0027s no angular velocity."},{"Start":"13:21.150 ","End":"13:24.415","Text":"Our initial angular velocity is 0,"},{"Start":"13:24.415 ","End":"13:27.505","Text":"and then we\u0027re just adding in our at."},{"Start":"13:27.505 ","End":"13:35.740","Text":"Our a is 5Mu_kg divided by 2R and then multiplied by"},{"Start":"13:35.740 ","End":"13:45.820","Text":"t. What I\u0027ve done up until now is to find my linear movements."},{"Start":"13:45.820 ","End":"13:52.645","Text":"My movement in the linear direction and my movement in the angular or circular motion,"},{"Start":"13:52.645 ","End":"13:54.910","Text":"and both of them I found separately."},{"Start":"13:54.910 ","End":"13:59.330","Text":"Now what I want do is make a connection between the two."},{"Start":"13:59.570 ","End":"14:03.990","Text":"As we know, we have our equation that our V as"},{"Start":"14:03.990 ","End":"14:07.620","Text":"a function of time is equal to our Omega as a function of"},{"Start":"14:07.620 ","End":"14:15.595","Text":"time multiplied by R. Now what I\u0027m in fact trying to do is to find out the exact time,"},{"Start":"14:15.595 ","End":"14:18.370","Text":"not a roundabout, but the exact time that"},{"Start":"14:18.370 ","End":"14:22.540","Text":"my equation for my V as a function of time is going to be"},{"Start":"14:22.540 ","End":"14:29.680","Text":"equal to what my Omega equals at that time multiplied by R. At this time,"},{"Start":"14:29.680 ","End":"14:32.005","Text":"if it\u0027s 5 seconds, 10 seconds, whatever."},{"Start":"14:32.005 ","End":"14:36.010","Text":"My V at 5 seconds is equal to my Omega at"},{"Start":"14:36.010 ","End":"14:41.289","Text":"5 seconds multiplied by R. Now, when this happens,"},{"Start":"14:41.289 ","End":"14:44.590","Text":"when my V as a function of t is equal to my Omega"},{"Start":"14:44.590 ","End":"14:49.930","Text":"as a function of t multiplied by R. That will"},{"Start":"14:49.930 ","End":"14:54.790","Text":"mean that my rolling with slipping is going"},{"Start":"14:54.790 ","End":"15:00.070","Text":"to stop and rolling without slipping is going to commence."},{"Start":"15:00.070 ","End":"15:03.760","Text":"I just want to take you back for a second."},{"Start":"15:03.760 ","End":"15:09.100","Text":"If you remember the equation for your velocity at point A,"},{"Start":"15:09.100 ","End":"15:13.490","Text":"I\u0027m reminding you, point A is this point over here."},{"Start":"15:15.630 ","End":"15:22.120","Text":"This was equal to our V of the center of mass minus"},{"Start":"15:22.120 ","End":"15:29.530","Text":"Omega R. Then we said that when we\u0027re dealing with out slipping,"},{"Start":"15:29.530 ","End":"15:33.439","Text":"over here no slipping."},{"Start":"15:34.650 ","End":"15:37.750","Text":"That means that this is equal to 0."},{"Start":"15:37.750 ","End":"15:39.970","Text":"Then we said that when this happens,"},{"Start":"15:39.970 ","End":"15:43.645","Text":"we can say that our V center of mass is equal to"},{"Start":"15:43.645 ","End":"15:49.915","Text":"our Omega R. This is exactly the equation that we have over here."},{"Start":"15:49.915 ","End":"15:52.465","Text":"When our V,"},{"Start":"15:52.465 ","End":"15:56.800","Text":"and notice that in this equation it\u0027s really a V,"},{"Start":"15:56.800 ","End":"16:00.880","Text":"center of mass as a function of time. It\u0027s correct."},{"Start":"16:00.880 ","End":"16:03.745","Text":"This is cm. When our V,"},{"Start":"16:03.745 ","End":"16:06.190","Text":"center of mass as a function of time is equal to"},{"Start":"16:06.190 ","End":"16:09.685","Text":"our Omega R. We\u0027re in this exact situation,"},{"Start":"16:09.685 ","End":"16:11.785","Text":"which means that when this situation,"},{"Start":"16:11.785 ","End":"16:14.125","Text":"our V_A is equal to 0,"},{"Start":"16:14.125 ","End":"16:18.350","Text":"which means that we\u0027re dealing with no slipping."},{"Start":"16:19.230 ","End":"16:22.960","Text":"Now when we\u0027re dealing with no slipping,"},{"Start":"16:22.960 ","End":"16:29.905","Text":"rolling without slipping, then that means that we no longer have kinetic friction."},{"Start":"16:29.905 ","End":"16:34.750","Text":"Potentially, we might have static friction or no friction at all."},{"Start":"16:34.750 ","End":"16:38.875","Text":"In this case specifically will have no friction at all because there\u0027s no other forces"},{"Start":"16:38.875 ","End":"16:43.450","Text":"on the x-axis with which the friction can oppose."},{"Start":"16:43.450 ","End":"16:47.860","Text":"Static friction only occurs when there are other forces to be opposed by it."},{"Start":"16:47.860 ","End":"16:49.240","Text":"Here there\u0027s no other forces,"},{"Start":"16:49.240 ","End":"16:52.460","Text":"so our friction will be equal to 0."},{"Start":"16:52.680 ","End":"16:59.485","Text":"The way that we can look at this system is as if we just placed a ball onto a surface."},{"Start":"16:59.485 ","End":"17:01.525","Text":"It\u0027s not moving, it\u0027s stationary."},{"Start":"17:01.525 ","End":"17:03.970","Text":"There\u0027s no frictional forces acting."},{"Start":"17:03.970 ","End":"17:08.230","Text":"Which means that from this moment and onwards,"},{"Start":"17:08.230 ","End":"17:16.630","Text":"we can say that the sum of all of our forces in the x-direction is equal to 0."},{"Start":"17:16.630 ","End":"17:18.565","Text":"The sum of the forces equal to 0,"},{"Start":"17:18.565 ","End":"17:21.835","Text":"because there\u0027s no frictional force over here after this moment,"},{"Start":"17:21.835 ","End":"17:27.970","Text":"when V is equal to Omega R. If the sum of all of the forces in the x axis is equal to 0,"},{"Start":"17:27.970 ","End":"17:35.230","Text":"that means that our acceleration in the x-axis is also equal to 0."},{"Start":"17:35.230 ","End":"17:38.875","Text":"Now the trick to see here is that we have"},{"Start":"17:38.875 ","End":"17:43.120","Text":"all these equations for our linear acceleration and angular acceleration."},{"Start":"17:43.120 ","End":"17:45.040","Text":"However, at this exact moment,"},{"Start":"17:45.040 ","End":"17:47.905","Text":"when a V is equal to our Omega R,"},{"Start":"17:47.905 ","End":"17:51.175","Text":"we\u0027ll have no forces acting,"},{"Start":"17:51.175 ","End":"17:54.190","Text":"which means that there\u0027ll be no acceleration acting."},{"Start":"17:54.190 ","End":"17:58.900","Text":"Which means that up until this point our velocity"},{"Start":"17:58.900 ","End":"18:00.700","Text":"is changing as a function of time and"},{"Start":"18:00.700 ","End":"18:04.060","Text":"our angular velocity is changing as a function of time."},{"Start":"18:04.060 ","End":"18:06.940","Text":"However, when we get to this exact point,"},{"Start":"18:06.940 ","End":"18:11.605","Text":"the velocity suddenly becomes a constant and it\u0027s unchanging."},{"Start":"18:11.605 ","End":"18:15.070","Text":"That means for infinity time in another 30 years,"},{"Start":"18:15.070 ","End":"18:18.250","Text":"the ball will still be moving at the same velocity,"},{"Start":"18:18.250 ","End":"18:22.795","Text":"until some of external force will come and change that."},{"Start":"18:22.795 ","End":"18:27.730","Text":"Now, all that\u0027s left for me to do is to just substitute all of this in."},{"Start":"18:27.730 ","End":"18:35.300","Text":"My equation for my V_cm is equal to V_0 minus Mu_kgt."},{"Start":"18:36.150 ","End":"18:39.370","Text":"Then for my Omega,"},{"Start":"18:39.370 ","End":"18:46.870","Text":"this is equal to Omega R. My Omega I wrote over here."},{"Start":"18:46.870 ","End":"18:57.075","Text":"This is 5Mu_kg divided by 2R multiplied by t. Then because it\u0027s Omega R,"},{"Start":"18:57.075 ","End":"18:59.180","Text":"I multiply by an R over here."},{"Start":"18:59.180 ","End":"19:03.649","Text":"Now all I have to do is I have to substitute"},{"Start":"19:03.649 ","End":"19:09.139","Text":"out my t. I have to find out at which time this happens."},{"Start":"19:09.139 ","End":"19:11.030","Text":"Then once I find the time,"},{"Start":"19:11.030 ","End":"19:15.388","Text":"then I can substitute that time into my equation for velocity,"},{"Start":"19:15.388 ","End":"19:18.480","Text":"and then find the final velocity."},{"Start":"19:18.790 ","End":"19:21.875","Text":"Let\u0027s see how we do this."},{"Start":"19:21.875 ","End":"19:26.120","Text":"The first thing we can notice is that this R and this R can cancel out."},{"Start":"19:26.120 ","End":"19:29.630","Text":"Now I\u0027m just going to isolate out my t, you can do this."},{"Start":"19:29.630 ","End":"19:32.480","Text":"You\u0027ll get that your t is equal to"},{"Start":"19:32.480 ","End":"19:40.370","Text":"2V_0 divided by 7Mu_kg."},{"Start":"19:40.370 ","End":"19:47.600","Text":"Then that means when we substitute this into my equation for velocity over here."},{"Start":"19:47.600 ","End":"19:50.975","Text":"Let me just make this a little bit clearer."},{"Start":"19:50.975 ","End":"19:54.095","Text":"I substitute this t into over here."},{"Start":"19:54.095 ","End":"19:58.670","Text":"Then what you\u0027ll get is your equation for your final V. That"},{"Start":"19:58.670 ","End":"20:03.185","Text":"means that my V final is going to be"},{"Start":"20:03.185 ","End":"20:13.970","Text":"equal to my V at a time of 2V_0 divided by 7Mu_kg,"},{"Start":"20:14.970 ","End":"20:17.633","Text":"my V at this time."},{"Start":"20:17.633 ","End":"20:19.970","Text":"Then I simply substitute this in,"},{"Start":"20:19.970 ","End":"20:21.229","Text":"and you can do this."},{"Start":"20:21.229 ","End":"20:28.580","Text":"You will get V_0 minus 2 over 7 V_0,"},{"Start":"20:28.580 ","End":"20:35.400","Text":"which is just going to be equal to 5 over 7 V_0."},{"Start":"20:35.500 ","End":"20:38.700","Text":"That\u0027s the end of this lesson."}],"ID":9431},{"Watched":false,"Name":"Exercise 2","Duration":"10m 46s","ChapterTopicVideoID":9162,"CourseChapterTopicPlaylistID":5409,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.475","Text":"Hello. This question is very similar to the previous question."},{"Start":"00:05.475 ","End":"00:09.090","Text":"Here, we have a homogeneous ball of mass M,"},{"Start":"00:09.090 ","End":"00:13.020","Text":"which is held in the air where it rotates about its center of mass,"},{"Start":"00:13.020 ","End":"00:15.570","Text":"with an angular velocity of Omega 0."},{"Start":"00:15.570 ","End":"00:19.650","Text":"We\u0027re being told that the ball is lower to the ground whilst rotating."},{"Start":"00:19.650 ","End":"00:25.425","Text":"What is the ball\u0027s final velocity if the coefficient of friction is Mu k?"},{"Start":"00:25.425 ","End":"00:31.750","Text":"Which means, if we have Mu k that we\u0027re dealing with kinetic friction."},{"Start":"00:31.760 ","End":"00:34.965","Text":"Let\u0027s begin to see what will happen."},{"Start":"00:34.965 ","End":"00:40.600","Text":"Let\u0027s imagine that we\u0027re lowering the ball onto the floor."},{"Start":"00:41.690 ","End":"00:46.800","Text":"Now the bowl has been placed onto the surface,"},{"Start":"00:46.800 ","End":"00:53.460","Text":"and we can see that it has this Omega 0, this angular velocity."},{"Start":"00:53.460 ","End":"00:56.510","Text":"Currently, the center of mass isn\u0027t moving,"},{"Start":"00:56.510 ","End":"01:00.320","Text":"it\u0027s stationary, and the ball is rotating around this."},{"Start":"01:00.320 ","End":"01:07.620","Text":"That we can write that V center of mass is equal to 0 over here."},{"Start":"01:07.910 ","End":"01:10.370","Text":"On the other hand,"},{"Start":"01:10.370 ","End":"01:16.880","Text":"here we have our point of contact and we\u0027re going to call this point A."},{"Start":"01:16.880 ","End":"01:21.665","Text":"Now because we can see that our ball is rotating around itself,"},{"Start":"01:21.665 ","End":"01:25.210","Text":"so we know that this point a is moving,"},{"Start":"01:25.210 ","End":"01:30.735","Text":"and it\u0027s moving in this direction with the angular velocity."},{"Start":"01:30.735 ","End":"01:34.444","Text":"That means that it\u0027s moving to the left currently,"},{"Start":"01:34.444 ","End":"01:38.120","Text":"which means that we can label over here,"},{"Start":"01:38.120 ","End":"01:41.795","Text":"as we know, we\u0027re dealing with rolling, with slipping."},{"Start":"01:41.795 ","End":"01:47.480","Text":"That means that our kinetic friction is working in"},{"Start":"01:47.480 ","End":"01:53.075","Text":"this direction because it has to be working in the opposite direction to our VA,"},{"Start":"01:53.075 ","End":"01:57.270","Text":"and we can write that our V_A"},{"Start":"01:57.350 ","End":"02:04.570","Text":"is equal to negative Omega 0R."},{"Start":"02:06.410 ","End":"02:10.280","Text":"Now the first thing that we\u0027re going to start doing is"},{"Start":"02:10.280 ","End":"02:14.555","Text":"writing our equations for the sum of all of my forces."},{"Start":"02:14.555 ","End":"02:20.575","Text":"I can say that the sum of all of the forces in the x-axis is going to be equal to,"},{"Start":"02:20.575 ","End":"02:27.110","Text":"so let\u0027s say that this direction is the positive x-direction,"},{"Start":"02:27.110 ","End":"02:34.015","Text":"and that this is the positive angular velocity direction."},{"Start":"02:34.015 ","End":"02:40.410","Text":"That means that we will have our"},{"Start":"02:40.410 ","End":"02:47.045","Text":"positive because our frictional or kinetic friction is going in the positive x-direction."},{"Start":"02:47.045 ","End":"02:50.524","Text":"We\u0027ll have our positive kinetic friction,"},{"Start":"02:50.524 ","End":"02:51.980","Text":"which is equal to,"},{"Start":"02:51.980 ","End":"02:57.815","Text":"as we know, Mu k multiplied by N. I can substitute in."},{"Start":"02:57.815 ","End":"03:00.800","Text":"I know that my N is equal to Mg."},{"Start":"03:00.800 ","End":"03:07.080","Text":"It\u0027s Mu k multiplied by Mg."},{"Start":"03:07.080 ","End":"03:13.725","Text":"That we know is equal to our mass multiplied by acceleration."},{"Start":"03:13.725 ","End":"03:18.035","Text":"The acceleration of the center of mass in the x-direction."},{"Start":"03:18.035 ","End":"03:24.730","Text":"Now we can isolate out our a center of mass in the x-direction."},{"Start":"03:24.860 ","End":"03:28.320","Text":"We can see what it\u0027s equal to."},{"Start":"03:28.320 ","End":"03:33.760","Text":"This will be Mu k multiplied by g. Very similar to our last question,"},{"Start":"03:33.760 ","End":"03:38.380","Text":"but this time we get that our acceleration is a positive acceleration,"},{"Start":"03:38.380 ","End":"03:39.760","Text":"whereas last time we had a negative,"},{"Start":"03:39.760 ","End":"03:42.155","Text":"so it was a deceleration."},{"Start":"03:42.155 ","End":"03:46.614","Text":"We can see that Mu k and also our g are constants,"},{"Start":"03:46.614 ","End":"03:48.565","Text":"which means that we have constant acceleration,"},{"Start":"03:48.565 ","End":"03:54.880","Text":"which means that we can write in that our V center of mass as a function of time,"},{"Start":"03:54.880 ","End":"03:58.105","Text":"is equal to our initial velocity,"},{"Start":"03:58.105 ","End":"04:03.830","Text":"which is 0, plus our acceleration multiplied by time."},{"Start":"04:03.830 ","End":"04:09.385","Text":"Plus Mu kg multiplied by"},{"Start":"04:09.385 ","End":"04:16.415","Text":"t. Now I\u0027m going to do the exact same thing,"},{"Start":"04:16.415 ","End":"04:21.540","Text":"but now referring to the angular acceleration."},{"Start":"04:21.650 ","End":"04:24.725","Text":"If we take a look at the diagram,"},{"Start":"04:24.725 ","End":"04:30.275","Text":"we can see that our radius is going down in this direction."},{"Start":"04:30.275 ","End":"04:32.675","Text":"Then, we can see that our fk,"},{"Start":"04:32.675 ","End":"04:36.840","Text":"our frictional force, is going in the right direction."},{"Start":"04:36.840 ","End":"04:39.870","Text":"Now if we draw this out onto,"},{"Start":"04:39.870 ","End":"04:42.634","Text":"around, or about the same axis,"},{"Start":"04:42.634 ","End":"04:44.000","Text":"so the same origin,"},{"Start":"04:44.000 ","End":"04:49.415","Text":"so we\u0027ll see that we have our radius going down in this direction,"},{"Start":"04:49.415 ","End":"04:53.240","Text":"and then we have a force going in this direction."},{"Start":"04:53.240 ","End":"04:56.900","Text":"Which means when we do cross multiplication,"},{"Start":"04:56.900 ","End":"05:00.260","Text":"that we take our radius in the direction of the force,"},{"Start":"05:00.260 ","End":"05:02.420","Text":"which means that it\u0027s going anticlockwise."},{"Start":"05:02.420 ","End":"05:03.980","Text":"Now, we overhear said that"},{"Start":"05:03.980 ","End":"05:06.845","Text":"the clockwise direction was going to be the positive direction,"},{"Start":"05:06.845 ","End":"05:09.275","Text":"which means that we have a minus."},{"Start":"05:09.275 ","End":"05:11.540","Text":"I\u0027ll just label this."},{"Start":"05:11.540 ","End":"05:15.240","Text":"There."},{"Start":"05:15.240 ","End":"05:19.270","Text":"Now, we can write out the sum of all of the torques."},{"Start":"05:19.270 ","End":"05:24.135","Text":"This is going to be equal to minus because of their situation over here,"},{"Start":"05:24.135 ","End":"05:28.385","Text":"multiplied by the size of kinetic friction,"},{"Start":"05:28.385 ","End":"05:31.190","Text":"multiplied by the size of our radius,"},{"Start":"05:31.190 ","End":"05:35.185","Text":"multiplied by sine of the angle between the 2."},{"Start":"05:35.185 ","End":"05:37.430","Text":"As we can see over here,"},{"Start":"05:37.430 ","End":"05:41.360","Text":"the angle between the 2 is at 90 degrees."},{"Start":"05:41.360 ","End":"05:45.020","Text":"Here we\u0027ll have sine of 90, which as we know,"},{"Start":"05:45.020 ","End":"05:49.885","Text":"is equal to 1, and this is going to be equal to I Alpha."},{"Start":"05:49.885 ","End":"05:51.750","Text":"Where our I for"},{"Start":"05:51.750 ","End":"06:00.315","Text":"a homogeneous ball is 2/5 MR^2."},{"Start":"06:00.315 ","End":"06:05.495","Text":"Now what we can do is we can substitute everything in and then isolate out our Alpha."},{"Start":"06:05.495 ","End":"06:13.020","Text":"We\u0027ll get that we have negative Mu kMgR."},{"Start":"06:13.020 ","End":"06:19.050","Text":"This is for negative our M_k Mg is our f_k over here, and then our I,"},{"Start":"06:19.050 ","End":"06:29.320","Text":"sine of 90 is equal to 1 which is equal to I 2/5MR^2 multiplied by Alpha."},{"Start":"06:29.320 ","End":"06:32.420","Text":"Now, if we isolate out our Alpha,"},{"Start":"06:32.420 ","End":"06:35.405","Text":"so we\u0027ll see that this M and this M cancel out."},{"Start":"06:35.405 ","End":"06:38.180","Text":"This squared and this I cancel out."},{"Start":"06:38.180 ","End":"06:47.070","Text":"Then, what we will get is our Alpha will equal to 5 Mu kg/2."},{"Start":"06:47.570 ","End":"06:50.435","Text":"Now, just like what we did over here,"},{"Start":"06:50.435 ","End":"06:55.700","Text":"where we found our linear acceleration and then we wrote out our velocity equation."},{"Start":"06:55.700 ","End":"07:00.730","Text":"Similarly here, so we can say that our Omega as a function of time,"},{"Start":"07:00.730 ","End":"07:03.604","Text":"is equal to our initial Omega,"},{"Start":"07:03.604 ","End":"07:07.940","Text":"which is equal to our Omega 0 given in the question,"},{"Start":"07:07.940 ","End":"07:14.095","Text":"plus Alpha our angular acceleration, which is this,"},{"Start":"07:14.095 ","End":"07:18.945","Text":"5 Mu kg divided by"},{"Start":"07:18.945 ","End":"07:26.625","Text":"2 multiplied by t. I made a mistake."},{"Start":"07:26.625 ","End":"07:30.560","Text":"Here, I Alpha is negative because we had the negative over here."},{"Start":"07:30.560 ","End":"07:33.230","Text":"That means that over here we have a negative,"},{"Start":"07:33.230 ","End":"07:40.560","Text":"and also I forgot to divide by R. Just add in the I\u0027s and the denominator over there."},{"Start":"07:41.000 ","End":"07:44.255","Text":"Just like in the previous question,"},{"Start":"07:44.255 ","End":"07:49.850","Text":"now what we have to do is we have to find the relationship between our VCM and our Omega."},{"Start":"07:49.850 ","End":"07:54.605","Text":"As we know, our VCM as a function of time,"},{"Start":"07:54.605 ","End":"08:01.310","Text":"is going to be equal to our Omega as a function of time multiplied by the radius."},{"Start":"08:01.310 ","End":"08:05.485","Text":"Now we know that at this time exactly,"},{"Start":"08:05.485 ","End":"08:09.470","Text":"we\u0027re going to lose our kinetic friction and get"},{"Start":"08:09.470 ","End":"08:13.830","Text":"some static friction or no friction law sometimes,"},{"Start":"08:13.830 ","End":"08:17.120","Text":"and that means that our velocities,"},{"Start":"08:17.120 ","End":"08:24.060","Text":"where here we can see that our velocity and our angular velocity are changing with time."},{"Start":"08:24.060 ","End":"08:26.875","Text":"When this condition is met,"},{"Start":"08:26.875 ","End":"08:33.260","Text":"then our velocity and our angular velocity will remain constant until infinity."},{"Start":"08:33.260 ","End":"08:36.420","Text":"They will be never changing."},{"Start":"08:37.190 ","End":"08:40.490","Text":"Now we can just substitute in our values."},{"Start":"08:40.490 ","End":"08:43.910","Text":"Our VCM as a function of time is this."},{"Start":"08:43.910 ","End":"08:51.350","Text":"We have Mu kgt is equal to our Omega as a function of time,"},{"Start":"08:51.350 ","End":"09:00.400","Text":"which is Omega 0 minus 5 Mu kgt divided by 2R,"},{"Start":"09:00.400 ","End":"09:03.555","Text":"and then all of this multiplied by"},{"Start":"09:03.555 ","End":"09:13.180","Text":"R. Now what we have to do is we have to isolate out our t. Really quickly,"},{"Start":"09:13.180 ","End":"09:18.194","Text":"we\u0027ll get that our t is equal to 2 Omega 0"},{"Start":"09:18.194 ","End":"09:25.210","Text":"divided by 7 Mu kg."},{"Start":"09:25.310 ","End":"09:28.950","Text":"I\u0027m missing an R from here."},{"Start":"09:28.950 ","End":"09:35.040","Text":"Then that means therefore that our V at this time,"},{"Start":"09:35.040 ","End":"09:39.165","Text":"so at 2 Omega 0R divided by"},{"Start":"09:39.165 ","End":"09:45.305","Text":"7 Mu kg is going to be"},{"Start":"09:45.305 ","End":"09:52.550","Text":"equal to over 7 Omega 0R."},{"Start":"09:52.610 ","End":"09:55.605","Text":"This answers our question."},{"Start":"09:55.605 ","End":"10:00.605","Text":"The question was, what is the ball\u0027s final velocity if the coefficient of friction"},{"Start":"10:00.605 ","End":"10:06.455","Text":"is Mu k. We found our final velocity."},{"Start":"10:06.455 ","End":"10:10.625","Text":"This question is exactly the opposite of our last question."},{"Start":"10:10.625 ","End":"10:15.110","Text":"What I\u0027m given here is that I don\u0027t have an initial velocity."},{"Start":"10:15.110 ","End":"10:20.535","Text":"I just have an initial angular velocity over here."},{"Start":"10:20.535 ","End":"10:22.925","Text":"Then as a result of this,"},{"Start":"10:22.925 ","End":"10:25.355","Text":"I have my kinetic friction,"},{"Start":"10:25.355 ","End":"10:27.900","Text":"which is going in the right direction."},{"Start":"10:27.900 ","End":"10:33.215","Text":"This friction is going to push the center of mass on 1 hand,"},{"Start":"10:33.215 ","End":"10:38.585","Text":"but on the other hand is also going to slow down this angular velocity until"},{"Start":"10:38.585 ","End":"10:45.310","Text":"the 2 values for my VCM and my Omega will be equal."}],"ID":9432}],"Thumbnail":null,"ID":5409},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Falling Pencil","Duration":"32m 19s","ChapterTopicVideoID":9163,"CourseChapterTopicPlaylistID":5410,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:08.069","Text":"Hello. This is a type of question that you can get in the exam to do with a rigid body."},{"Start":"00:08.069 ","End":"00:10.739","Text":"Let\u0027s take a look at this question, and how to solve it."},{"Start":"00:10.739 ","End":"00:16.304","Text":"We\u0027re being told that a pencil is standing perpendicular to the table, 90 degrees."},{"Start":"00:16.304 ","End":"00:20.654","Text":"Then we\u0027re told that the pencil begins to fall to the right."},{"Start":"00:20.654 ","End":"00:23.700","Text":"When the angle between the pencil and the axis"},{"Start":"00:23.700 ","End":"00:27.180","Text":"perpendicular to the table reaches Theta 1,"},{"Start":"00:27.180 ","End":"00:30.135","Text":"the pencil begins to slip."},{"Start":"00:30.135 ","End":"00:32.460","Text":"Let\u0027s take a look at this,"},{"Start":"00:32.460 ","End":"00:36.880","Text":"now I\u0027m going to draw a rough diagram."},{"Start":"00:37.250 ","End":"00:42.764","Text":"If this is our pencil,"},{"Start":"00:42.764 ","End":"00:47.344","Text":"and it\u0027s standing at the beginning,"},{"Start":"00:47.344 ","End":"00:49.360","Text":"90 degrees to the table."},{"Start":"00:49.360 ","End":"00:52.099","Text":"Then we\u0027re told that the pencil"},{"Start":"00:52.099 ","End":"00:54.859","Text":"begins to fall to the right when the angle between the pencil,"},{"Start":"00:54.859 ","End":"00:58.975","Text":"and the axis perpendicular to the table reaches Theta 1."},{"Start":"00:58.975 ","End":"01:02.794","Text":"The pencil begins to fall like this,"},{"Start":"01:02.794 ","End":"01:04.700","Text":"relative to the table,"},{"Start":"01:04.700 ","End":"01:08.299","Text":"and when this angle over here,"},{"Start":"01:08.299 ","End":"01:10.955","Text":"so this is the axis perpendicular."},{"Start":"01:10.955 ","End":"01:15.270","Text":"This angle over here, Theta 1."},{"Start":"01:15.270 ","End":"01:17.160","Text":"Up until this angle,"},{"Start":"01:17.160 ","End":"01:24.364","Text":"our point of contact between the pencil and the table is static."},{"Start":"01:24.364 ","End":"01:26.615","Text":"If this is our A,"},{"Start":"01:26.615 ","End":"01:32.905","Text":"so our V_A is equal to 0 up until the angle of Theta 1."},{"Start":"01:32.905 ","End":"01:35.335","Text":"After the angle Theta 1,"},{"Start":"01:35.335 ","End":"01:41.900","Text":"what we\u0027re going to have is that if here I draw this,"},{"Start":"01:41.900 ","End":"01:47.644","Text":"so the pencil from originally being located here at its point of contact."},{"Start":"01:47.644 ","End":"01:50.105","Text":"The point of contact will move here,"},{"Start":"01:50.105 ","End":"01:52.489","Text":"and then will carry on falling."},{"Start":"01:52.489 ","End":"01:58.324","Text":"Which means that between here and here, there\u0027s been movement,"},{"Start":"01:58.324 ","End":"02:08.509","Text":"which means that our V_A over here will not equal 0 because we\u0027re slipping."},{"Start":"02:09.170 ","End":"02:11.355","Text":"Question Number 1,"},{"Start":"02:11.355 ","End":"02:13.020","Text":"for all angles Theta,"},{"Start":"02:13.020 ","End":"02:15.839","Text":"such that Theta is smaller than Theta 1,"},{"Start":"02:15.839 ","End":"02:19.600","Text":"so for all the angles until this point is reached,"},{"Start":"02:19.600 ","End":"02:22.629","Text":"where the pencil is still not slipping,"},{"Start":"02:22.629 ","End":"02:24.675","Text":"where our V_A is equal to 0."},{"Start":"02:24.675 ","End":"02:29.125","Text":"We\u0027re being asked, what is the angular velocity of the pencil?"},{"Start":"02:29.125 ","End":"02:32.569","Text":"Let\u0027s take a look at how we\u0027re going to do this."},{"Start":"02:33.430 ","End":"02:37.959","Text":"The first thing that we\u0027re going to look at is what is being conserved."},{"Start":"02:37.959 ","End":"02:41.125","Text":"Now because we\u0027re being asked about velocity,"},{"Start":"02:41.125 ","End":"02:43.645","Text":"even though it\u0027s angular but its velocity."},{"Start":"02:43.645 ","End":"02:48.195","Text":"We\u0027re going to think that potentially there\u0027s going to be energy conservation,"},{"Start":"02:48.195 ","End":"02:51.320","Text":"and another thing that gives us a clue that potentially this is"},{"Start":"02:51.320 ","End":"02:54.934","Text":"energy conservation is because we\u0027re being told that the pencil is falling,"},{"Start":"02:54.934 ","End":"02:56.899","Text":"so the height is changing."},{"Start":"02:56.899 ","End":"03:01.084","Text":"Anytime that we have a change in height and we\u0027re being asked about velocity,"},{"Start":"03:01.084 ","End":"03:05.789","Text":"most probably, there\u0027s conservation of energy."},{"Start":"03:07.030 ","End":"03:10.850","Text":"In angles of Theta which are smaller than Theta 1,"},{"Start":"03:10.850 ","End":"03:14.824","Text":"we\u0027re being told that the pencil isn\u0027t slipping."},{"Start":"03:14.824 ","End":"03:17.224","Text":"Which means if the pencil isn\u0027t slipping,"},{"Start":"03:17.224 ","End":"03:21.455","Text":"that means that we have static friction over here at the point of contact,"},{"Start":"03:21.455 ","End":"03:23.044","Text":"and if we have static friction,"},{"Start":"03:23.044 ","End":"03:26.760","Text":"that means that we have energy conservation."},{"Start":"03:26.810 ","End":"03:30.485","Text":"Of course, there are 2 other forces which are acting here,"},{"Start":"03:30.485 ","End":"03:33.695","Text":"which is our mg downwards and our normal force upwards,"},{"Start":"03:33.695 ","End":"03:37.620","Text":"and as we know, these forces are always conservational forces."},{"Start":"03:37.620 ","End":"03:40.339","Text":"I\u0027ve written over here a little flow."},{"Start":"03:40.339 ","End":"03:42.229","Text":"If we have no slipping,"},{"Start":"03:42.229 ","End":"03:44.105","Text":"then that means that we have static friction,"},{"Start":"03:44.105 ","End":"03:45.319","Text":"and if there is static friction,"},{"Start":"03:45.319 ","End":"03:46.760","Text":"then it means that at that point,"},{"Start":"03:46.760 ","End":"03:49.340","Text":"specifically there\u0027s energy conservation,"},{"Start":"03:49.340 ","End":"03:53.314","Text":"and then we have to look at the other forces and our normal force"},{"Start":"03:53.314 ","End":"03:59.469","Text":"and our mg, our conservational forces."},{"Start":"04:00.140 ","End":"04:10.010","Text":"Now let\u0027s begin by answering Question Number 1a."},{"Start":"04:10.010 ","End":"04:11.419","Text":"What we\u0027re going to do is first,"},{"Start":"04:11.419 ","End":"04:16.530","Text":"we\u0027re going to write the initial energy of the system, so Ei."},{"Start":"04:16.530 ","End":"04:21.350","Text":"Now, something that was forgotten to be given in the question is that the length of"},{"Start":"04:21.350 ","End":"04:28.364","Text":"the pencil is L. We can write over here,"},{"Start":"04:28.364 ","End":"04:32.070","Text":"this length is L. Let\u0027s see."},{"Start":"04:32.070 ","End":"04:35.374","Text":"First, let\u0027s take a look at our kinetic energy at the beginning."},{"Start":"04:35.374 ","End":"04:37.294","Text":"Right at the start,"},{"Start":"04:37.294 ","End":"04:39.367","Text":"the pencil begins to fall to the right,"},{"Start":"04:39.367 ","End":"04:41.107","Text":"which means that right at the beginning,"},{"Start":"04:41.107 ","End":"04:42.409","Text":"before it begins to fall,"},{"Start":"04:42.409 ","End":"04:43.939","Text":"it has 0 movement,"},{"Start":"04:43.939 ","End":"04:46.179","Text":"which means 0 kinetic energy."},{"Start":"04:46.179 ","End":"04:48.679","Text":"Now we have to look at our potential energy."},{"Start":"04:48.679 ","End":"04:51.110","Text":"We\u0027ll add the mass of the pencil,"},{"Start":"04:51.110 ","End":"05:00.574","Text":"which is also m multiplied by g multiplied by h. Now, what is our h?"},{"Start":"05:00.574 ","End":"05:02.900","Text":"Because we\u0027re dealing with a rigid body,"},{"Start":"05:02.900 ","End":"05:07.910","Text":"it means that we\u0027re always going to take our h to be at the center of mass."},{"Start":"05:07.910 ","End":"05:11.119","Text":"We can say that here is our center of mass,"},{"Start":"05:11.119 ","End":"05:16.476","Text":"and that is going to be our h. Our h is going to be at L over 2,"},{"Start":"05:16.476 ","End":"05:25.800","Text":"which means that our initial energy is going to be mg multiplied by L divided by 2."},{"Start":"05:26.450 ","End":"05:32.479","Text":"Now, what we want to do is we want to find an equation that is as a function of"},{"Start":"05:32.479 ","End":"05:35.570","Text":"Theta so that we can know what the energy of"},{"Start":"05:35.570 ","End":"05:39.715","Text":"the system is at every single angle Theta that we will put it in."},{"Start":"05:39.715 ","End":"05:42.915","Text":"We can call this Ef,"},{"Start":"05:42.915 ","End":"05:46.084","Text":"our final and this is going to be as a function of Theta."},{"Start":"05:46.084 ","End":"05:54.359","Text":"As we know, we\u0027re going to have to add in our kinetic energy plus our potential energy."},{"Start":"05:54.680 ","End":"05:59.779","Text":"Let\u0027s see what our kinetic energy is going to be."},{"Start":"05:59.779 ","End":"06:07.405","Text":"In general, we know that our kinetic energy is going to be equal to half MV^2."},{"Start":"06:07.405 ","End":"06:12.680","Text":"However, because we\u0027re dealing with a rigid body which has a circular motion,"},{"Start":"06:12.680 ","End":"06:20.510","Text":"so the equation for kinetic energy is equal to 1/2 I Omega^2."},{"Start":"06:20.810 ","End":"06:25.015","Text":"I\u0027ve just written here on the side to remind you that"},{"Start":"06:25.015 ","End":"06:29.620","Text":"this equation for kinetic energy is correct when dealing with a rigid body,"},{"Start":"06:29.620 ","End":"06:36.025","Text":"which has circular motion and whose axis of rotation is stationary."},{"Start":"06:36.025 ","End":"06:38.840","Text":"Stationary axis of rotation."},{"Start":"06:39.510 ","End":"06:43.185","Text":"Currently, up until Theta 1,"},{"Start":"06:43.185 ","End":"06:45.360","Text":"the axis of rotation is stationary,"},{"Start":"06:45.360 ","End":"06:48.440","Text":"because we have no slipping and only at Theta 1,"},{"Start":"06:48.440 ","End":"06:51.020","Text":"and afterwards, then we have slipping,"},{"Start":"06:51.020 ","End":"06:53.914","Text":"so the axis of rotation is not stationary."},{"Start":"06:53.914 ","End":"06:58.590","Text":"Remember when this equation applies."},{"Start":"06:58.730 ","End":"07:01.689","Text":"Now we have to find what our I is."},{"Start":"07:01.689 ","End":"07:03.950","Text":"Our moments of inertia,"},{"Start":"07:03.950 ","End":"07:05.345","Text":"let\u0027s write it over here,"},{"Start":"07:05.345 ","End":"07:08.562","Text":"our moment of inertia of the pencil,"},{"Start":"07:08.562 ","End":"07:16.004","Text":"so we can regard the pencil as some rod of length L rotating along its end."},{"Start":"07:16.004 ","End":"07:19.940","Text":"The moment of inertia of a rod rotating around its end is"},{"Start":"07:19.940 ","End":"07:25.765","Text":"1/3 m multiplied by its length squared."},{"Start":"07:25.765 ","End":"07:29.509","Text":"Now, this is something that you should know off by heart."},{"Start":"07:29.509 ","End":"07:31.159","Text":"If you want to know how to work it out,"},{"Start":"07:31.159 ","End":"07:35.669","Text":"then please go back to the chapter which discusses this."},{"Start":"07:36.410 ","End":"07:40.714","Text":"Now we can just substitute in our I over here,"},{"Start":"07:40.714 ","End":"07:43.924","Text":"and we get our expression for kinetic energy."},{"Start":"07:43.924 ","End":"07:47.260","Text":"Now we have to find out our expression for our U."},{"Start":"07:47.260 ","End":"07:53.644","Text":"What is our U? Our U is equal to mgh, our potential energy."},{"Start":"07:53.644 ","End":"07:57.980","Text":"Now what we have to see is what is our h over here?"},{"Start":"07:57.980 ","End":"08:01.134","Text":"The most general h that we can find."},{"Start":"08:01.134 ","End":"08:03.100","Text":"Our h is, of course,"},{"Start":"08:03.100 ","End":"08:05.209","Text":"going to be the height of our center of mass."},{"Start":"08:05.209 ","End":"08:08.659","Text":"As we said over here when dealing with a rigid body,"},{"Start":"08:08.659 ","End":"08:10.824","Text":"we deal with the center of mass."},{"Start":"08:10.824 ","End":"08:14.175","Text":"Let\u0027s take a look at what this means."},{"Start":"08:14.175 ","End":"08:17.909","Text":"If we go to this diagram over here,"},{"Start":"08:17.980 ","End":"08:22.773","Text":"we can say that our center of mass is over here,"},{"Start":"08:22.773 ","End":"08:32.450","Text":"which means that this over here is our height."},{"Start":"08:32.450 ","End":"08:36.720","Text":"Our hcm, is this over here."},{"Start":"08:38.060 ","End":"08:40.900","Text":"If this is our hcm,"},{"Start":"08:40.900 ","End":"08:43.655","Text":"then we can know that"},{"Start":"08:43.655 ","End":"08:53.330","Text":"this length from the tip until the center of mass is going to be L over 2,"},{"Start":"08:53.330 ","End":"08:56.250","Text":"because it\u0027s halfway up the pencil,"},{"Start":"08:56.250 ","End":"08:59.435","Text":"so that black line is L over 2."},{"Start":"08:59.435 ","End":"09:01.505","Text":"That means if this is Theta,"},{"Start":"09:01.505 ","End":"09:05.370","Text":"we can see that this angle here is also Theta."},{"Start":"09:06.110 ","End":"09:10.879","Text":"I drew this diagram a little bit bigger over here."},{"Start":"09:10.879 ","End":"09:16.160","Text":"We see that the angle between the pen and the axis perpendicular to the table is Theta."},{"Start":"09:16.160 ","End":"09:17.562","Text":"Here\u0027s our center of mass,"},{"Start":"09:17.562 ","End":"09:21.619","Text":"this purple arrow represents the height of our center of mass."},{"Start":"09:21.619 ","End":"09:28.460","Text":"From the tip of the pencil until the center of mass is length L over 2."},{"Start":"09:28.460 ","End":"09:33.049","Text":"Now we can see through alternate angles that this angle over"},{"Start":"09:33.049 ","End":"09:39.040","Text":"here is also Theta, this angle."},{"Start":"09:39.590 ","End":"09:44.244","Text":"If this angle over here is Theta,"},{"Start":"09:44.244 ","End":"09:47.775","Text":"so then we can go back to what we were writing,"},{"Start":"09:47.775 ","End":"09:53.560","Text":"and we can say that our hcm is simply going to be when we work with cosine and sine,"},{"Start":"09:53.560 ","End":"10:00.170","Text":"L over 2 multiplied by cosine of Theta."},{"Start":"10:01.480 ","End":"10:04.810","Text":"Now we can substitute everything in."},{"Start":"10:04.810 ","End":"10:13.329","Text":"Our energy as a function of Theta is going to be equal to our kinetic energy,"},{"Start":"10:13.329 ","End":"10:18.015","Text":"which is 1/2 I,"},{"Start":"10:18.015 ","End":"10:20.021","Text":"so it\u0027s 1/3,"},{"Start":"10:20.021 ","End":"10:29.269","Text":"so it\u0027s 1 over 6 mL^2 multiplied by Omega^2 plus our potential energy,"},{"Start":"10:29.269 ","End":"10:30.830","Text":"which is mg,"},{"Start":"10:30.830 ","End":"10:37.979","Text":"and then substituting in the h multiplied by L over 2 cosine of Theta."},{"Start":"10:37.979 ","End":"10:39.740","Text":"This, because of conservation of energy,"},{"Start":"10:39.740 ","End":"10:41.720","Text":"is equal to our initial energy,"},{"Start":"10:41.720 ","End":"10:46.275","Text":"which is equal to mg L over 2."},{"Start":"10:46.275 ","End":"10:54.979","Text":"Now, what we can do is we can go ahead and cross out all of our like terms."},{"Start":"10:54.979 ","End":"10:57.229","Text":"We can cross out the m\u0027s from everywhere,"},{"Start":"10:57.229 ","End":"11:06.214","Text":"divide both sides by m. We can divide both sides by L. We can multiply both sides by 2,"},{"Start":"11:06.214 ","End":"11:10.760","Text":"so then we\u0027ll get over here 3."},{"Start":"11:10.760 ","End":"11:14.120","Text":"Then we can see that everything is a constant."},{"Start":"11:14.120 ","End":"11:18.205","Text":"Now, all we have to do is we have to isolate out our Omega."},{"Start":"11:18.205 ","End":"11:20.989","Text":"You\u0027re more than welcome to do this,"},{"Start":"11:20.989 ","End":"11:23.059","Text":"but what you will get at the end is that"},{"Start":"11:23.059 ","End":"11:26.779","Text":"your Omega is going to be equal to the square root of"},{"Start":"11:26.779 ","End":"11:37.360","Text":"3 g over L multiplied by 1 minus cosine of Theta."},{"Start":"11:38.300 ","End":"11:41.595","Text":"Now you can see that we have an expression for"},{"Start":"11:41.595 ","End":"11:47.620","Text":"our angular velocity of the pencil as a function of Theta."},{"Start":"11:47.900 ","End":"11:53.640","Text":"That\u0027s question 1a so now we\u0027re going to go over to 1b."},{"Start":"11:53.640 ","End":"11:57.569","Text":"Over here we\u0027re being asked what is the angular acceleration of the pencil?"},{"Start":"11:57.569 ","End":"12:00.990","Text":"Now, we\u0027ve already used the idea of conservation of"},{"Start":"12:00.990 ","End":"12:05.220","Text":"energy and also to find velocity and also when we\u0027re dealing with acceleration,"},{"Start":"12:05.220 ","End":"12:08.715","Text":"usually it\u0027s going to be some conservation,"},{"Start":"12:08.715 ","End":"12:13.215","Text":"forces and torques and writing out those equations."},{"Start":"12:13.215 ","End":"12:15.809","Text":"That is what we\u0027re going to do right now."},{"Start":"12:15.809 ","End":"12:20.500","Text":"I\u0027m going to rub out this diagram so that we can draw it more clearly."},{"Start":"12:21.440 ","End":"12:26.550","Text":"The first thing that we want to do is we want to choose our axis."},{"Start":"12:26.550 ","End":"12:30.179","Text":"I\u0027m going to say that my x-axis is in this direction,"},{"Start":"12:30.179 ","End":"12:34.184","Text":"and that my y-axis is in this direction."},{"Start":"12:34.184 ","End":"12:37.620","Text":"Now I\u0027m going to choose my positive direction of"},{"Start":"12:37.620 ","End":"12:42.540","Text":"rotation to be in the clockwise direction."},{"Start":"12:42.540 ","End":"12:45.990","Text":"Here it\u0027s different, usually we go from x to y as being positive,"},{"Start":"12:45.990 ","End":"12:48.210","Text":"but here I\u0027m going to go from y to x."},{"Start":"12:48.210 ","End":"12:50.129","Text":"If this is a little bit confusing for you,"},{"Start":"12:50.129 ","End":"12:53.070","Text":"then you can draw your x-axis in the opposite direction,"},{"Start":"12:53.070 ","End":"12:54.570","Text":"going in this direction."},{"Start":"12:54.570 ","End":"12:57.359","Text":"The reason I\u0027m doing this is because we\u0027ve"},{"Start":"12:57.359 ","End":"13:00.750","Text":"defined our Theta angle coming from our y and opening up,"},{"Start":"13:00.750 ","End":"13:04.499","Text":"gradually becoming bigger as we approach the x-axis."},{"Start":"13:04.499 ","End":"13:08.399","Text":"Which means that if I would choose my arrow to be counterclockwise,"},{"Start":"13:08.399 ","End":"13:10.598","Text":"then we\u0027ll get negative angles,"},{"Start":"13:10.598 ","End":"13:14.890","Text":"and that can sometimes lead to confusion and silly mistakes."},{"Start":"13:16.010 ","End":"13:23.445","Text":"Now we\u0027re going to write an equation for the conservation of momentum. Let\u0027s see."},{"Start":"13:23.445 ","End":"13:25.154","Text":"We can say, as we know,"},{"Start":"13:25.154 ","End":"13:27.690","Text":"our axis of rotation is going to be over here"},{"Start":"13:27.690 ","End":"13:30.734","Text":"because this is actually the point of contact where it\u0027s rotating."},{"Start":"13:30.734 ","End":"13:33.285","Text":"Remember again, our Theta is smaller than Theta 1,"},{"Start":"13:33.285 ","End":"13:38.639","Text":"which means that we\u0027re not slipping so the axis of rotation,"},{"Start":"13:38.639 ","End":"13:42.040","Text":"is stationary, still in this area."},{"Start":"13:42.860 ","End":"13:49.335","Text":"Now we can take a look and we can see that from our center of mass going downwards,"},{"Start":"13:49.335 ","End":"13:52.740","Text":"we have our force mg,"},{"Start":"13:52.740 ","End":"13:56.054","Text":"over here perpendicular we have"},{"Start":"13:56.054 ","End":"14:00.345","Text":"our normal force and of course we have our static friction."},{"Start":"14:00.345 ","End":"14:03.539","Text":"We don\u0027t know which direction it\u0027s in, it doesn\u0027t really matter."},{"Start":"14:03.539 ","End":"14:07.290","Text":"We can draw it in this direction. It doesn\u0027t matter."},{"Start":"14:07.290 ","End":"14:11.325","Text":"Because our axis of rotation is over here when we\u0027re going to be"},{"Start":"14:11.325 ","End":"14:15.495","Text":"racing out our equation for our torque."},{"Start":"14:15.495 ","End":"14:20.024","Text":"Our normal force and our static friction,"},{"Start":"14:20.024 ","End":"14:22.215","Text":"they cancel out,"},{"Start":"14:22.215 ","End":"14:25.170","Text":"because they\u0027re coming out of the axis of rotation."},{"Start":"14:25.170 ","End":"14:30.570","Text":"Right now we\u0027re answering question 1b. Let\u0027s see."},{"Start":"14:30.570 ","End":"14:37.150","Text":"We\u0027re writing our torques for force mg. That\u0027s the only force that has a moment."},{"Start":"14:37.640 ","End":"14:42.585","Text":"The equation is going to be equal to our force"},{"Start":"14:42.585 ","End":"14:47.175","Text":"mg multiplied by its distance from the axis of rotation."},{"Start":"14:47.175 ","End":"14:49.560","Text":"Because it\u0027s located at the center of mass,"},{"Start":"14:49.560 ","End":"14:54.420","Text":"its distance from the axis of rotation is from the center of the pencil,"},{"Start":"14:54.420 ","End":"14:57.450","Text":"it\u0027s L/2, multiply by L/2,"},{"Start":"14:57.450 ","End":"15:04.065","Text":"and then we\u0027re going to multiply it by sine of the angle between the 2 forces."},{"Start":"15:04.065 ","End":"15:09.284","Text":"As we can see, this is Theta and this is also Theta, alternate angles."},{"Start":"15:09.284 ","End":"15:14.080","Text":"It\u0027s going to be multiplied by sine of Theta."},{"Start":"15:14.870 ","End":"15:21.390","Text":"This is in fact our equation for the sum of all of the torques."},{"Start":"15:21.390 ","End":"15:27.765","Text":"As we know, this expression is going to be equal to our I Alpha."},{"Start":"15:27.765 ","End":"15:34.215","Text":"We know that our I is the same I that we wrote over here."},{"Start":"15:34.215 ","End":"15:39.810","Text":"It\u0027s the I for a rod rotating around its end."},{"Start":"15:39.810 ","End":"15:42.269","Text":"Now if we substitute all of this in,"},{"Start":"15:42.269 ","End":"15:46.950","Text":"so what we\u0027ll get is that we can isolate out our Alpha."},{"Start":"15:46.950 ","End":"15:49.259","Text":"Then we\u0027ve answered our question."},{"Start":"15:49.259 ","End":"15:53.159","Text":"Because our question b is,"},{"Start":"15:53.159 ","End":"15:55.755","Text":"what is the angular acceleration of the pencils?"},{"Start":"15:55.755 ","End":"15:57.705","Text":"We\u0027re trying to find what are Alpha is."},{"Start":"15:57.705 ","End":"16:01.214","Text":"We\u0027re going to substitute in our I and isolate out our Alpha,"},{"Start":"16:01.214 ","End":"16:06.439","Text":"and we\u0027re just going to get that our Alpha is equal to 3g divided"},{"Start":"16:06.439 ","End":"16:12.630","Text":"by 2L multiplied by sine of Theta."},{"Start":"16:12.630 ","End":"16:15.210","Text":"That\u0027s the end of this question."},{"Start":"16:15.210 ","End":"16:17.589","Text":"Let\u0027s move on to the next 1."},{"Start":"16:17.780 ","End":"16:22.590","Text":"Before we move on, 1 small thing sine over here is going to be"},{"Start":"16:22.590 ","End":"16:27.029","Text":"positive because we can see that our torque is in the positive direction."},{"Start":"16:27.029 ","End":"16:28.410","Text":"If we go to our diagram,"},{"Start":"16:28.410 ","End":"16:31.664","Text":"we can see that our mg force"},{"Start":"16:31.664 ","End":"16:36.180","Text":"is pulling the rod downwards With this being the axis of rotation."},{"Start":"16:36.180 ","End":"16:39.884","Text":"We can see that the rod is going to be rotating in a clockwise direction,"},{"Start":"16:39.884 ","End":"16:45.627","Text":"which is the positive direction that we said before."},{"Start":"16:45.627 ","End":"16:51.480","Text":"1c is to find the acceleration vector of the pencils center of mass."},{"Start":"16:51.480 ","End":"16:53.355","Text":"If we look at this diagram,"},{"Start":"16:53.355 ","End":"16:59.339","Text":"we can see that the center of mass is always at a constant distance away from the origin,"},{"Start":"16:59.339 ","End":"17:02.264","Text":"the radius of L/2 from the origin."},{"Start":"17:02.264 ","End":"17:03.989","Text":"Then it just moves down,"},{"Start":"17:03.989 ","End":"17:07.035","Text":"and we can see that it\u0027s going in circular motion."},{"Start":"17:07.035 ","End":"17:09.405","Text":"When we\u0027re dealing with circular motion,"},{"Start":"17:09.405 ","End":"17:14.969","Text":"we can write down our equation for the acceleration in polar coordinates,"},{"Start":"17:14.969 ","End":"17:17.340","Text":"our acceleration in our radial direction and"},{"Start":"17:17.340 ","End":"17:21.435","Text":"our acceleration in our tangential or our Theta direction."},{"Start":"17:21.435 ","End":"17:25.529","Text":"Our acceleration in the radial direction is going to be equal"},{"Start":"17:25.529 ","End":"17:29.730","Text":"to Omega squared multiplied by r and"},{"Start":"17:29.730 ","End":"17:34.499","Text":"our acceleration in the tangential or in the Theta direction is going to be equal"},{"Start":"17:34.499 ","End":"17:39.449","Text":"to Alpha multiplied by r. Now,"},{"Start":"17:39.449 ","End":"17:44.500","Text":"these are good equations to remember and to write in your notes."},{"Start":"17:44.780 ","End":"17:50.800","Text":"Of course, this is only when dealing with circular motion, nothing else."},{"Start":"17:52.340 ","End":"17:54.705","Text":"We know what our r is."},{"Start":"17:54.705 ","End":"17:58.350","Text":"Our r is going to be equal to our L/2,"},{"Start":"17:58.350 ","End":"18:00.660","Text":"because our center of mass is at a distance"},{"Start":"18:00.660 ","End":"18:04.275","Text":"of L/2 from the axis of rotation or from the origin."},{"Start":"18:04.275 ","End":"18:10.649","Text":"Now we can just write an acceleration vector consisting of these 2."},{"Start":"18:10.649 ","End":"18:14.445","Text":"This is going to be equal to our ar in"},{"Start":"18:14.445 ","End":"18:21.000","Text":"the r direction plus our a Theta in the Theta direction."},{"Start":"18:21.000 ","End":"18:22.905","Text":"What is this equal to?"},{"Start":"18:22.905 ","End":"18:29.955","Text":"As we know, the direction of this vector is going in this direction,"},{"Start":"18:29.955 ","End":"18:34.964","Text":"our r-hat is always going like this outwards."},{"Start":"18:34.964 ","End":"18:36.209","Text":"However, as we know,"},{"Start":"18:36.209 ","End":"18:38.490","Text":"when we\u0027re dealing with circular motion,"},{"Start":"18:38.490 ","End":"18:43.410","Text":"our acceleration in the radial direction is always going inwards,"},{"Start":"18:43.410 ","End":"18:47.100","Text":"in this direction, which is not minus."},{"Start":"18:47.100 ","End":"18:49.935","Text":"In order to make this more correct,"},{"Start":"18:49.935 ","End":"18:52.650","Text":"we\u0027re going to add a minus over here."},{"Start":"18:52.650 ","End":"18:56.654","Text":"Now all we have to do is we have to substitute in."},{"Start":"18:56.654 ","End":"19:04.695","Text":"We have a negative and our ar is Omega squared multiplied by r. Our Omega is this,"},{"Start":"19:04.695 ","End":"19:07.200","Text":"squared is just going to be without the square root."},{"Start":"19:07.200 ","End":"19:15.690","Text":"We\u0027re going to have 3g divided by L 1 minus cosine of Theta."},{"Start":"19:15.690 ","End":"19:17.730","Text":"Then this is going to be multiplied by r,"},{"Start":"19:17.730 ","End":"19:22.088","Text":"which is L/2 so our L and our L will cancel out,"},{"Start":"19:22.088 ","End":"19:25.874","Text":"and then over 2 so we can add in our 2 over here."},{"Start":"19:25.874 ","End":"19:29.910","Text":"This is going to be in the radial direction."},{"Start":"19:29.910 ","End":"19:33.750","Text":"Now we can add in our Theta direction."},{"Start":"19:33.750 ","End":"19:36.870","Text":"Our a Theta is going to be our Alpha, which is this."},{"Start":"19:36.870 ","End":"19:41.490","Text":"We have 3g divided by 2L sine"},{"Start":"19:41.490 ","End":"19:47.190","Text":"of Theta multiplied by our radius, which is L/2."},{"Start":"19:47.190 ","End":"19:51.345","Text":"Our L and our L will cancel out and then we have"},{"Start":"19:51.345 ","End":"19:56.400","Text":"here divided by 2 and here another 2 so it will be divided by 4."},{"Start":"19:56.400 ","End":"20:00.240","Text":"This is going to be in the Theta direction."},{"Start":"20:00.240 ","End":"20:03.810","Text":"That\u0027s the answer to this section."},{"Start":"20:03.810 ","End":"20:05.445","Text":"Now we\u0027re on to 1d,"},{"Start":"20:05.445 ","End":"20:09.255","Text":"to find the size and direction of the frictional force."},{"Start":"20:09.255 ","End":"20:11.715","Text":"This is going to be nice and easy."},{"Start":"20:11.715 ","End":"20:15.945","Text":"All we have to do is write out an equation for the sum of all of our forces."},{"Start":"20:15.945 ","End":"20:18.479","Text":"Our frictional force is, as we can see,"},{"Start":"20:18.479 ","End":"20:21.285","Text":"just acting on the x-axis."},{"Start":"20:21.285 ","End":"20:27.675","Text":"We only have to write an equation for the sum of all of the forces on the x axis."},{"Start":"20:27.675 ","End":"20:31.470","Text":"we know that that\u0027s going to be equal to our f_s."},{"Start":"20:31.470 ","End":"20:33.330","Text":"We don\u0027t really know which direction it\u0027s in right now,"},{"Start":"20:33.330 ","End":"20:35.055","Text":"but we\u0027ll work it out in a second."},{"Start":"20:35.055 ","End":"20:42.190","Text":"As we know, this is going to be equal to our mass times acceleration in the x-axis."},{"Start":"20:42.370 ","End":"20:48.229","Text":"Our mass given, and now all we have to do is we have to work out what our a_x is."},{"Start":"20:48.229 ","End":"20:52.010","Text":"Now, we have this vector for acceleration,"},{"Start":"20:52.010 ","End":"20:53.510","Text":"but it\u0027s in polar coordinates."},{"Start":"20:53.510 ","End":"20:59.285","Text":"Now we just have to find out how it works on the x-axis."},{"Start":"20:59.285 ","End":"21:01.370","Text":"In order to demonstrate this,"},{"Start":"21:01.370 ","End":"21:07.020","Text":"I\u0027m going to draw out this again in a slightly more clear way."},{"Start":"21:07.390 ","End":"21:10.985","Text":"We have our pencil over here."},{"Start":"21:10.985 ","End":"21:16.589","Text":"Then we have our ground over"},{"Start":"21:16.589 ","End":"21:24.430","Text":"here and then we have our axis over here and this angle over here is Theta."},{"Start":"21:24.770 ","End":"21:34.025","Text":"Now what we can do is we can draw out our vector for acceleration, very easily."},{"Start":"21:34.025 ","End":"21:37.850","Text":"We know that our tangential acceleration,"},{"Start":"21:37.850 ","End":"21:43.575","Text":"our a Theta is going to be in this direction and our a_r,"},{"Start":"21:43.575 ","End":"21:47.370","Text":"as we know, is always going in to the origin,"},{"Start":"21:47.370 ","End":"21:51.180","Text":"this is going to be our radial acceleration."},{"Start":"21:51.180 ","End":"21:54.030","Text":"Now if we take our x-axis,"},{"Start":"21:54.030 ","End":"21:55.920","Text":"remember when we\u0027re dealing with vectors,"},{"Start":"21:55.920 ","End":"21:57.510","Text":"it doesn\u0027t really matter,"},{"Start":"21:57.510 ","End":"22:01.125","Text":"we can move them up or down as long as they keep their size and direction."},{"Start":"22:01.125 ","End":"22:03.810","Text":"We can move our x-axis onto here."},{"Start":"22:03.810 ","End":"22:08.025","Text":"This is the x-axis and this is the x-axis as well."},{"Start":"22:08.025 ","End":"22:17.250","Text":"Now all we have to do is we have to find our projection onto the x-axis."},{"Start":"22:17.250 ","End":"22:26.940","Text":"What we can see is that this angle over here is going to be equal to 90 minus Theta."},{"Start":"22:26.940 ","End":"22:30.270","Text":"In order to find what our projection is going to be,"},{"Start":"22:30.270 ","End":"22:33.400","Text":"it\u0027s going to look something like this."},{"Start":"22:33.730 ","End":"22:37.655","Text":"Our projection is going to be,"},{"Start":"22:37.655 ","End":"22:42.715","Text":"or rather its size is going to be on the x-axis."},{"Start":"22:42.715 ","End":"22:45.765","Text":"Our a_x is going to equal 2."},{"Start":"22:45.765 ","End":"22:52.065","Text":"When dealing with our a_r we\u0027ll have a_r multiplied by cosine."},{"Start":"22:52.065 ","End":"22:54.839","Text":"Because it\u0027s the adjacent angle,"},{"Start":"22:54.839 ","End":"23:02.740","Text":"cosine of and then this angle is equal to 90 minus Theta."},{"Start":"23:05.700 ","End":"23:10.420","Text":"Notice that because it\u0027s going in the negative x-direction,"},{"Start":"23:10.420 ","End":"23:13.435","Text":"in this direction, so we have a negative over here."},{"Start":"23:13.435 ","End":"23:17.155","Text":"Then we can add in our value for a Theta."},{"Start":"23:17.155 ","End":"23:21.160","Text":"Again, if this is 90 minus Theta,"},{"Start":"23:21.160 ","End":"23:23.905","Text":"over here this angle is Theta."},{"Start":"23:23.905 ","End":"23:27.430","Text":"This angle over here will be 90 minus Theta,"},{"Start":"23:27.430 ","End":"23:31.660","Text":"which means this angle will also be Theta."},{"Start":"23:31.660 ","End":"23:36.130","Text":"This one is going to be going in the positive x-direction."},{"Start":"23:36.130 ","End":"23:38.934","Text":"What we\u0027re going to have is we\u0027re going to have plus"},{"Start":"23:38.934 ","End":"23:44.965","Text":"a Theta multiplied by cosine of Theta."},{"Start":"23:44.965 ","End":"23:49.390","Text":"Now we can simplify this a little bit to negative a_r,"},{"Start":"23:49.390 ","End":"23:52.705","Text":"and then instead of cosine of 90 minus Theta,"},{"Start":"23:52.705 ","End":"23:55.570","Text":"that\u0027s the same as sine of Theta,"},{"Start":"23:55.570 ","End":"24:02.420","Text":"and then plus a Theta cosine of Theta."},{"Start":"24:03.660 ","End":"24:07.149","Text":"Now in order to find out what this is,"},{"Start":"24:07.149 ","End":"24:10.900","Text":"we can substitute in what our a_r and a Theta is,"},{"Start":"24:10.900 ","End":"24:13.765","Text":"which is what we found up here."},{"Start":"24:13.765 ","End":"24:19.720","Text":"Then we substitute that back into this equation in order to find our f_s."},{"Start":"24:19.720 ","End":"24:21.100","Text":"Then we\u0027ll find its size,"},{"Start":"24:21.100 ","End":"24:25.315","Text":"and you\u0027ll notice that its sign can be either positive or negative,"},{"Start":"24:25.315 ","End":"24:28.299","Text":"because as the pencil is falling,"},{"Start":"24:28.299 ","End":"24:29.710","Text":"its direction changes,"},{"Start":"24:29.710 ","End":"24:32.199","Text":"so one time it will be facing this way and the other time it will"},{"Start":"24:32.199 ","End":"24:35.720","Text":"be facing in the opposite direction."},{"Start":"24:36.300 ","End":"24:39.355","Text":"Depending on what our angle Theta is,"},{"Start":"24:39.355 ","End":"24:42.819","Text":"our expression for f_s might be negative or positive,"},{"Start":"24:42.819 ","End":"24:45.220","Text":"and then you\u0027ll know in which direction it is,"},{"Start":"24:45.220 ","End":"24:48.460","Text":"which was part of the question over here,"},{"Start":"24:48.460 ","End":"24:50.950","Text":"find the size and direction of the frictional force."},{"Start":"24:50.950 ","End":"24:56.240","Text":"That\u0027s the size and the direction is dependent on the angle."},{"Start":"24:56.940 ","End":"25:00.504","Text":"You\u0027re f_s will work out something like this."},{"Start":"25:00.504 ","End":"25:08.890","Text":"You could potentially cancel this down or simplify this by writing in trig identities."},{"Start":"25:08.890 ","End":"25:12.145","Text":"This is your f_s, and you can see that it will change"},{"Start":"25:12.145 ","End":"25:16.090","Text":"signs depending on the size of the angle."},{"Start":"25:16.090 ","End":"25:21.385","Text":"Now, let\u0027s take a look at the next section."},{"Start":"25:21.385 ","End":"25:26.290","Text":"We have e, find the normal force."},{"Start":"25:26.290 ","End":"25:28.675","Text":"It\u0027s going to be very similar to our d,"},{"Start":"25:28.675 ","End":"25:30.744","Text":"the size of direction of the frictional force."},{"Start":"25:30.744 ","End":"25:34.764","Text":"However, this time we\u0027re writing out the sum of all the forces,"},{"Start":"25:34.764 ","End":"25:39.040","Text":"except not on the x-axis but rather on the y-axis."},{"Start":"25:39.040 ","End":"25:41.810","Text":"Let\u0027s see how we do this."},{"Start":"25:42.630 ","End":"25:45.220","Text":"Here is my diagram,"},{"Start":"25:45.220 ","End":"25:48.894","Text":"so we\u0027re going to write the sum of all of the forces on the y-axis."},{"Start":"25:48.894 ","End":"25:53.349","Text":"We have the sum of all the forces on the y-axis is going to be equal to."},{"Start":"25:53.349 ","End":"25:54.790","Text":"In our positive y-direction,"},{"Start":"25:54.790 ","End":"26:00.310","Text":"we have our N and in the negative y-direction we have our m_g, so negative m_g."},{"Start":"26:00.310 ","End":"26:07.285","Text":"This is, of course, equal to our mass multiplied by our acceleration in the y-direction."},{"Start":"26:07.285 ","End":"26:12.340","Text":"Just like what we did with our acceleration in the x-direction,"},{"Start":"26:12.340 ","End":"26:15.910","Text":"we use this and we worked out what it was in the y-direction,"},{"Start":"26:15.910 ","End":"26:19.790","Text":"so we\u0027re going to do the exact same thing."},{"Start":"26:20.580 ","End":"26:25.930","Text":"If I draw my new y-axis because it\u0027s vectors,"},{"Start":"26:25.930 ","End":"26:29.365","Text":"as long as I keep the size and direction then it\u0027s fine."},{"Start":"26:29.365 ","End":"26:31.465","Text":"This is my new y-axis."},{"Start":"26:31.465 ","End":"26:34.540","Text":"Now let\u0027s take a look at our acceleration vector."},{"Start":"26:34.540 ","End":"26:43.047","Text":"Again, I have my a_r in this direction and I have my a Theta."},{"Start":"26:43.047 ","End":"26:46.825","Text":"I can\u0027t remember the perpendicular in this direction."},{"Start":"26:46.825 ","End":"26:52.720","Text":"Now what I have to do is I have to find their projection onto my y-axis."},{"Start":"26:52.720 ","End":"26:55.675","Text":"This I can simply do,"},{"Start":"26:55.675 ","End":"27:00.385","Text":"we can see that they\u0027re both going in this direction."},{"Start":"27:00.385 ","End":"27:04.270","Text":"Now, the equation for them, let\u0027s write this,"},{"Start":"27:04.270 ","End":"27:09.940","Text":"our a_y is going to be equal to a_r and this,"},{"Start":"27:09.940 ","End":"27:12.205","Text":"let\u0027s just write this over here."},{"Start":"27:12.205 ","End":"27:16.610","Text":"This angle over here remember is Theta."},{"Start":"27:16.890 ","End":"27:19.960","Text":"The gray arrow here is the adjacent,"},{"Start":"27:19.960 ","End":"27:24.175","Text":"so it\u0027s going to be a_r cosine of Theta,"},{"Start":"27:24.175 ","End":"27:27.729","Text":"and then we\u0027re also going to have a Theta."},{"Start":"27:27.729 ","End":"27:32.719","Text":"Again, this angle over here"},{"Start":"27:32.840 ","End":"27:40.650","Text":"is going to be 90 minus Theta, this angle."},{"Start":"27:40.650 ","End":"27:45.405","Text":"We\u0027re going to have a Theta multiplied by cosine,"},{"Start":"27:45.405 ","End":"27:48.135","Text":"because it\u0027s again the adjacent,"},{"Start":"27:48.135 ","End":"27:55.975","Text":"so cosine of the angle there which is 90 minus Theta."},{"Start":"27:55.975 ","End":"27:57.879","Text":"Now to look at the signs,"},{"Start":"27:57.879 ","End":"28:00.069","Text":"we can see that they\u0027re both facing downwards,"},{"Start":"28:00.069 ","End":"28:01.930","Text":"which is the negative y-direction,"},{"Start":"28:01.930 ","End":"28:03.880","Text":"so we\u0027re going to have negatives over here."},{"Start":"28:03.880 ","End":"28:11.049","Text":"Then, of course, we can simplify this into negative a_r cosine Theta,"},{"Start":"28:11.049 ","End":"28:19.060","Text":"negative a Theta and cosine of 90 minus Theta is simply sine of Theta."},{"Start":"28:19.060 ","End":"28:22.270","Text":"Then just like over here what we did,"},{"Start":"28:22.270 ","End":"28:26.710","Text":"we just have to substitute in what our a_r is and what our a Theta is,"},{"Start":"28:26.710 ","End":"28:31.390","Text":"then substitute all of that into this expression over"},{"Start":"28:31.390 ","End":"28:36.950","Text":"here into our a_y and then isolate out our normal."},{"Start":"28:37.680 ","End":"28:41.650","Text":"This is what you\u0027re meant to get at the end,"},{"Start":"28:41.650 ","End":"28:47.830","Text":"so bear in mind that this is the bracket for this."},{"Start":"28:47.830 ","End":"28:52.420","Text":"Now let\u0027s go on to question B."},{"Start":"28:52.420 ","End":"28:56.710","Text":"To find the static coefficient of friction Mu_s."},{"Start":"28:56.710 ","End":"29:02.870","Text":"What we\u0027re doing now is we\u0027re trying to find our value for Mu_s."},{"Start":"29:04.320 ","End":"29:09.879","Text":"We know that our equation is f_s max,"},{"Start":"29:09.879 ","End":"29:15.310","Text":"so the maximum value for the static friction is equal to Mu_s."},{"Start":"29:15.310 ","End":"29:16.810","Text":"This is what we\u0027re trying to find,"},{"Start":"29:16.810 ","End":"29:19.495","Text":"multiplied by our normal force."},{"Start":"29:19.495 ","End":"29:23.425","Text":"Now, we have our equation for our f_s over here,"},{"Start":"29:23.425 ","End":"29:27.100","Text":"and we have our equation for our N. However,"},{"Start":"29:27.100 ","End":"29:34.150","Text":"we can\u0027t simply divide this by N because we need to know when our f_s is maximal."},{"Start":"29:34.150 ","End":"29:38.740","Text":"How are we going to know when this value is its largest?"},{"Start":"29:38.740 ","End":"29:44.230","Text":"If we go back to the question, all this way,"},{"Start":"29:44.230 ","End":"29:48.115","Text":"we\u0027ll see that at our angle of Theta_1,"},{"Start":"29:48.115 ","End":"29:50.125","Text":"our pencil starts to slip,"},{"Start":"29:50.125 ","End":"29:52.704","Text":"which means that at that exact angle,"},{"Start":"29:52.704 ","End":"29:57.160","Text":"at Theta_1, our static friction is at its maximum."},{"Start":"29:57.160 ","End":"30:01.130","Text":"That is what we have to substitute in."},{"Start":"30:02.400 ","End":"30:04.480","Text":"Let\u0027s add over here,"},{"Start":"30:04.480 ","End":"30:08.747","Text":"so we know that our f_s is as a function of Theta,"},{"Start":"30:08.747 ","End":"30:10.464","Text":"and so is our normal force."},{"Start":"30:10.464 ","End":"30:12.685","Text":"It\u0027s also as a function of Theta."},{"Start":"30:12.685 ","End":"30:19.120","Text":"Now all we have to do is we have to substitute N. We saw that when our Theta_ 1,"},{"Start":"30:19.120 ","End":"30:21.130","Text":"when our pencil reaches that angle,"},{"Start":"30:21.130 ","End":"30:23.905","Text":"then that\u0027s when our f_s is maximum."},{"Start":"30:23.905 ","End":"30:27.955","Text":"We\u0027re just going to add over here a 1 because that\u0027s when it\u0027s maximum,"},{"Start":"30:27.955 ","End":"30:30.895","Text":"and then we just have to substitute N over here,"},{"Start":"30:30.895 ","End":"30:33.220","Text":"and then divide both sides by our N,"},{"Start":"30:33.220 ","End":"30:37.460","Text":"and then we\u0027re going to get our Mu_s."},{"Start":"30:38.220 ","End":"30:41.079","Text":"Once you\u0027ve done all of this algebra,"},{"Start":"30:41.079 ","End":"30:45.910","Text":"so you\u0027ll get that the Mu_k is equal to this long and complicated expression."},{"Start":"30:45.910 ","End":"30:47.380","Text":"You can cancel it down,"},{"Start":"30:47.380 ","End":"30:48.655","Text":"it doesn\u0027t really matter,"},{"Start":"30:48.655 ","End":"30:51.505","Text":"or leave it like this."},{"Start":"30:51.505 ","End":"30:55.370","Text":"That is the end of this question."},{"Start":"30:55.440 ","End":"31:02.005","Text":"This was a pretty large question with a lot of sections,"},{"Start":"31:02.005 ","End":"31:04.419","Text":"but really once you see how to solve it,"},{"Start":"31:04.419 ","End":"31:06.400","Text":"it\u0027s actually pretty easy."},{"Start":"31:06.400 ","End":"31:10.225","Text":"You just have to remember what this means,"},{"Start":"31:10.225 ","End":"31:12.760","Text":"when something is slipping and when it isn\u0027t slipping,"},{"Start":"31:12.760 ","End":"31:14.200","Text":"that something is stationary,"},{"Start":"31:14.200 ","End":"31:16.794","Text":"and then you have conservation of energy."},{"Start":"31:16.794 ","End":"31:21.295","Text":"You also have to remember that your N and your mg are conservation forces."},{"Start":"31:21.295 ","End":"31:24.100","Text":"Then whenever you have a question involving velocity,"},{"Start":"31:24.100 ","End":"31:26.409","Text":"then you should use conservation of energy."},{"Start":"31:26.409 ","End":"31:27.850","Text":"Whenever you have acceleration,"},{"Start":"31:27.850 ","End":"31:31.329","Text":"then you should use a torque or accelerations for torque."},{"Start":"31:31.329 ","End":"31:33.834","Text":"When finding your vectors,"},{"Start":"31:33.834 ","End":"31:37.270","Text":"you can also just simply by drawing it on the diagram,"},{"Start":"31:37.270 ","End":"31:42.865","Text":"it makes it a lot easier to work out and also to remember these equations,"},{"Start":"31:42.865 ","End":"31:46.645","Text":"that your a_r in circular motion is Omega^2 r and"},{"Start":"31:46.645 ","End":"31:51.820","Text":"a Theta in circular motion is Alpha r. Then when you write this out,"},{"Start":"31:51.820 ","End":"31:56.994","Text":"then it\u0027s very easy and all of the other questions come as a gift."},{"Start":"31:56.994 ","End":"32:00.445","Text":"Then when you\u0027re told to find the size,"},{"Start":"32:00.445 ","End":"32:03.310","Text":"the direction just depends on the sign,"},{"Start":"32:03.310 ","End":"32:06.895","Text":"but specifically the size of the frictional force or the normal force,"},{"Start":"32:06.895 ","End":"32:10.795","Text":"then you use your acceleration that you have,"},{"Start":"32:10.795 ","End":"32:12.849","Text":"you work out your force equations,"},{"Start":"32:12.849 ","End":"32:19.579","Text":"and you just substitute that in by using the projection onto the various axis."}],"ID":9437}],"Thumbnail":null,"ID":5410}]

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