Basic Explanation
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Motion Due To External Forces
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Exercises
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Calculating The Center Of Mass Using Integrals
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- Review on Coordinates (Same Lecture as In The Math Intro)
- Calculating Center of Mass of Large Body
- Exercise- Center Of Mass Of Rod
- Exercise- Triangles Center of Mass
- Exercise- Gate Center of Mass
- Exercise - Finding the Area of a Sector
- Exercise- Sector Center of Mass
- Exercise- Center of Mass Half Sphere Full
- Exercise- Center Of Mass Solid Cone
- Exercise- Half Hoop With Two Masses

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[{"Name":"Basic Explanation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro","Duration":"7m 5s","ChapterTopicVideoID":9084,"CourseChapterTopicPlaylistID":5338,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9084.jpeg","UploadDate":"2017-03-22T10:20:49.9100000","DurationForVideoObject":"PT7M5S","Description":null,"MetaTitle":"Intro: Video + Workbook | Proprep","MetaDescription":"Center of Mass - Basic Explanation. 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/center-of-mass/basic-explanation/vid9357","VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:03.450","Text":"Hello. In this lesson,"},{"Start":"00:03.450 ","End":"00:09.705","Text":"we\u0027re going to learn about the introduction to the center of mass."},{"Start":"00:09.705 ","End":"00:14.655","Text":"Let\u0027s start off very easily with the equation for the center of mass."},{"Start":"00:14.655 ","End":"00:23.490","Text":"The position of the center of mass is equal to the sum of all of"},{"Start":"00:23.490 ","End":"00:28.320","Text":"the masses multiplied by the position of each mass"},{"Start":"00:28.320 ","End":"00:34.350","Text":"divided by the sum of all of the masses."},{"Start":"00:34.350 ","End":"00:38.630","Text":"What will happen in the end that the sum of all of"},{"Start":"00:38.630 ","End":"00:44.255","Text":"the masses is just equal to the total mass of the system."},{"Start":"00:44.255 ","End":"00:47.600","Text":"Here you do the mass multiplied by"},{"Start":"00:47.600 ","End":"00:49.550","Text":"the distance plus the mass of the second body"},{"Start":"00:49.550 ","End":"00:52.070","Text":"multiplied by the distance of the second body."},{"Start":"00:52.070 ","End":"00:55.740","Text":"Let\u0027s see a simple example."},{"Start":"00:55.880 ","End":"01:01.005","Text":"In our example we\u0027ll have 3 bodies."},{"Start":"01:01.005 ","End":"01:04.290","Text":"Now the first thing we have to do is we have to find"},{"Start":"01:04.290 ","End":"01:09.320","Text":"the axis that from there we\u0027re trying to find the center of mass."},{"Start":"01:09.320 ","End":"01:11.120","Text":"We just choose 1 of the bodies."},{"Start":"01:11.120 ","End":"01:14.075","Text":"Let\u0027s say, it doesn\u0027t really matter which 1 you choose."},{"Start":"01:14.075 ","End":"01:21.600","Text":"Let\u0027s say that we\u0027re saying that this mass is our starting point."},{"Start":"01:21.600 ","End":"01:24.275","Text":"We draw the axis on top of that."},{"Start":"01:24.275 ","End":"01:26.800","Text":"Let\u0027s give the masses names."},{"Start":"01:26.800 ","End":"01:29.110","Text":"This will be mass 1,"},{"Start":"01:29.110 ","End":"01:31.705","Text":"and its position will, of course, be 0,0."},{"Start":"01:31.705 ","End":"01:34.275","Text":"Then this is mass 2,"},{"Start":"01:34.275 ","End":"01:38.355","Text":"and its position, let\u0027s say is 3, 2."},{"Start":"01:38.355 ","End":"01:40.560","Text":"This will be mass number 3,"},{"Start":"01:40.560 ","End":"01:45.120","Text":"and its position will be 4, negative 1."},{"Start":"01:45.120 ","End":"01:48.100","Text":"Now to find our X center of mass,"},{"Start":"01:48.100 ","End":"01:50.620","Text":"and of course, there\u0027s also Y center of mass."},{"Start":"01:50.620 ","End":"01:52.120","Text":"Let\u0027s draw it over here."},{"Start":"01:52.120 ","End":"01:56.410","Text":"Our Y center of mass is going to be the exact same thing,"},{"Start":"01:56.410 ","End":"02:00.145","Text":"so the sum of the masses multiplied by"},{"Start":"02:00.145 ","End":"02:06.880","Text":"the y position divided by the sum of all of the masses."},{"Start":"02:07.430 ","End":"02:17.570","Text":"Then we can say that our X center of mass is going to be equal to our mass number 1,"},{"Start":"02:17.570 ","End":"02:21.710","Text":"so mass 1 multiplied by its x component,"},{"Start":"02:21.710 ","End":"02:23.150","Text":"which is multiplied by 0,"},{"Start":"02:23.150 ","End":"02:24.545","Text":"so this equals 0,"},{"Start":"02:24.545 ","End":"02:30.660","Text":"plus our mass number 2 multiplied by its x component, so 3,"},{"Start":"02:30.660 ","End":"02:35.710","Text":"plus our mass number 3 multiplied by its x component which is 4,"},{"Start":"02:35.710 ","End":"02:39.890","Text":"divided by the total mass of all the systems,"},{"Start":"02:39.890 ","End":"02:44.970","Text":"so it\u0027s m1 plus m2 plus m3."},{"Start":"02:44.970 ","End":"02:49.670","Text":"Of course, it\u0027s the same thing for our Y_cm."},{"Start":"02:49.670 ","End":"02:56.150","Text":"The sum of all of the masses will be our m_1 multiplied by its y component,"},{"Start":"02:56.150 ","End":"02:58.100","Text":"which again is 0, so this goes,"},{"Start":"02:58.100 ","End":"03:01.729","Text":"plus our m_2 multiplied by its y component,"},{"Start":"03:01.729 ","End":"03:08.435","Text":"so multiplied by 2 plus our m3 multiplied by its y component which is negative 1,"},{"Start":"03:08.435 ","End":"03:11.480","Text":"divided by the total mass of the system,"},{"Start":"03:11.480 ","End":"03:16.385","Text":"which again is m1 plus m2 plus m3."},{"Start":"03:16.385 ","End":"03:22.930","Text":"This will give us that the center of mass is somewhere around about here."},{"Start":"03:22.930 ","End":"03:26.820","Text":"This will be the center of mass."},{"Start":"03:26.820 ","End":"03:33.470","Text":"Now to give some intuition for what the center of mass means,"},{"Start":"03:33.470 ","End":"03:37.825","Text":"what it represents, it means that it\u0027s the point of equilibrium."},{"Start":"03:37.825 ","End":"03:41.460","Text":"If I have some uneven body,"},{"Start":"03:41.460 ","End":"03:42.885","Text":"let\u0027s say it\u0027s like that,"},{"Start":"03:42.885 ","End":"03:46.025","Text":"and if I say that this is its center of mass,"},{"Start":"03:46.025 ","End":"03:49.670","Text":"then that means that that is the point of equilibrium."},{"Start":"03:49.670 ","End":"03:52.919","Text":"If I were to hold it with my finger,"},{"Start":"03:53.180 ","End":"03:56.120","Text":"this is what my finger looks like,"},{"Start":"03:56.120 ","End":"04:02.820","Text":"then the shape would be balanced because that is its point of equilibrium."},{"Start":"04:03.650 ","End":"04:07.880","Text":"Now let\u0027s take a look at an example."},{"Start":"04:07.880 ","End":"04:11.570","Text":"Let\u0027s say we have some half sphere or half disk,"},{"Start":"04:11.570 ","End":"04:12.980","Text":"it doesn\u0027t really matter,"},{"Start":"04:12.980 ","End":"04:18.440","Text":"and we\u0027re being asked what the center of mass is of this."},{"Start":"04:18.440 ","End":"04:20.690","Text":"We\u0027re not dealing with a bunch of point masses."},{"Start":"04:20.690 ","End":"04:23.920","Text":"We\u0027re dealing with some weird shape."},{"Start":"04:23.920 ","End":"04:29.420","Text":"What we do is we take a small section of the shape and"},{"Start":"04:29.420 ","End":"04:35.105","Text":"then we use the same equation over here,"},{"Start":"04:35.105 ","End":"04:37.580","Text":"except instead of doing the sum of"},{"Start":"04:37.580 ","End":"04:41.075","Text":"the mass times by the distance divided by the total mass,"},{"Start":"04:41.075 ","End":"04:44.805","Text":"we use the integral, we integrate."},{"Start":"04:44.805 ","End":"04:46.430","Text":"This is a bit complicated,"},{"Start":"04:46.430 ","End":"04:51.485","Text":"and there\u0027s a separate lesson for this which we\u0027re going to look at a bit later."},{"Start":"04:51.485 ","End":"04:55.805","Text":"But imagine that we knew how to do it and we would find that the center of masses,"},{"Start":"04:55.805 ","End":"04:57.650","Text":"something around about here."},{"Start":"04:57.650 ","End":"05:01.370","Text":"But we\u0027ll have that lesson later on in"},{"Start":"05:01.370 ","End":"05:07.335","Text":"this section of the course because it\u0027s too long for over here."},{"Start":"05:07.335 ","End":"05:10.870","Text":"But another question that we could be asked is,"},{"Start":"05:10.870 ","End":"05:15.015","Text":"for instance, if we have some disk over here,"},{"Start":"05:15.015 ","End":"05:22.480","Text":"and we\u0027re told that this is the center and it has a radius of R and it has a mass M,"},{"Start":"05:22.480 ","End":"05:29.260","Text":"and there\u0027s a smaller disk on top of it of a different radius and mass small"},{"Start":"05:29.260 ","End":"05:38.035","Text":"m. Now the way that I find the center of mass of these 2 shapes is,"},{"Start":"05:38.035 ","End":"05:41.785","Text":"first of all, I find the center of mass of each shape."},{"Start":"05:41.785 ","End":"05:48.525","Text":"Obviously, the center of mass of the big shape will be here, center mass 1."},{"Start":"05:48.525 ","End":"05:55.640","Text":"Of course, the center mass of the smaller shape will be over there."},{"Start":"05:55.640 ","End":"06:01.610","Text":"Then let\u0027s say that the distance between this center of mass and this center of mass is,"},{"Start":"06:01.610 ","End":"06:03.595","Text":"let\u0027s say, a."},{"Start":"06:03.595 ","End":"06:10.355","Text":"Now in order to find the center of mass of the 2 shapes in total,"},{"Start":"06:10.355 ","End":"06:15.870","Text":"I find the center of mass between each 1 center of mass."},{"Start":"06:15.870 ","End":"06:24.320","Text":"Then my X_cm will just equal my large mass and it\u0027s at the origin."},{"Start":"06:24.320 ","End":"06:31.410","Text":"Imagine we drew the axis over here multiplied by 0,"},{"Start":"06:31.410 ","End":"06:33.080","Text":"so this is obviously 0,"},{"Start":"06:33.080 ","End":"06:38.180","Text":"plus my small mass multiplied by its distance from the origin, which is a,"},{"Start":"06:38.180 ","End":"06:46.010","Text":"divided by the total mass which is m plus m. The way to find the center of mass of"},{"Start":"06:46.010 ","End":"06:51.020","Text":"2 large bodies is to find the center of mass of each body and then find"},{"Start":"06:51.020 ","End":"06:56.370","Text":"the center of mass between the bodies centers of mass."},{"Start":"06:56.370 ","End":"07:01.665","Text":"Bit confusing, maybe listen to this section again."},{"Start":"07:01.665 ","End":"07:05.670","Text":"That is the end of this introduction."}],"ID":9357},{"Watched":false,"Name":"Example- Disk With Hole","Duration":"5m 12s","ChapterTopicVideoID":10280,"CourseChapterTopicPlaylistID":5338,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.820","Text":"Hello. In this question,"},{"Start":"00:02.820 ","End":"00:05.520","Text":"we\u0027re being told that a disk of"},{"Start":"00:05.520 ","End":"00:14.805","Text":"radius R and a mass of M has had a hole drilled in it,"},{"Start":"00:14.805 ","End":"00:21.090","Text":"a distance of 1/2R away from the center and that"},{"Start":"00:21.090 ","End":"00:28.620","Text":"the radius of the hole that was drilled is 1/4R."},{"Start":"00:28.620 ","End":"00:36.080","Text":"We\u0027re being told to find what the center of mass is of this disk with a hole in it."},{"Start":"00:36.080 ","End":"00:39.535","Text":"Let\u0027s see how we do this."},{"Start":"00:39.535 ","End":"00:46.430","Text":"The first thing that we want to do is find out what the mass of the hole is."},{"Start":"00:46.430 ","End":"00:55.355","Text":"How we can refer to the hole is as if we\u0027ve stuck on a smaller disk onto the larger disk."},{"Start":"00:55.355 ","End":"00:59.365","Text":"But then we just say that the mass is a negative number."},{"Start":"00:59.365 ","End":"01:01.350","Text":"That\u0027s how we can refer to the hole."},{"Start":"01:01.350 ","End":"01:03.820","Text":"We\u0027re going to say that the mass of the hole,"},{"Start":"01:03.820 ","End":"01:10.580","Text":"or the mass of the small disk is going to be called m. Then we just have to"},{"Start":"01:10.580 ","End":"01:19.350","Text":"find what its mass is relative to the larger disk."},{"Start":"01:19.350 ","End":"01:27.364","Text":"As we know, the mass is the area multiplied by the density."},{"Start":"01:27.364 ","End":"01:30.710","Text":"But we\u0027re just going to use the area currently."},{"Start":"01:30.710 ","End":"01:36.110","Text":"We know that the area of the small disk is pi R^2."},{"Start":"01:36.110 ","End":"01:42.745","Text":"It\u0027s pi times 1/4R^2."},{"Start":"01:42.745 ","End":"01:48.050","Text":"The area of the large disk is pi R^2."},{"Start":"01:48.050 ","End":"01:50.930","Text":"All of this is multiplied by M,"},{"Start":"01:50.930 ","End":"01:56.225","Text":"the mass of the entire desk when there was no hole."},{"Start":"01:56.225 ","End":"02:02.450","Text":"Then we\u0027ll see that our pi and our pi cross out and our R^2 and our R^2 cross out."},{"Start":"02:02.450 ","End":"02:05.105","Text":"Then we just have (1/4)^2 M,"},{"Start":"02:05.105 ","End":"02:10.860","Text":"which is 1/16 M. The mass of"},{"Start":"02:10.860 ","End":"02:17.905","Text":"the hole is 1/16 of the mass of the entire disk."},{"Start":"02:17.905 ","End":"02:21.590","Text":"Now we\u0027re going to decide the axis that we\u0027re working on,"},{"Start":"02:21.590 ","End":"02:25.085","Text":"because this is very important depending on our axis,"},{"Start":"02:25.085 ","End":"02:29.405","Text":"our center of mass will be in a different area."},{"Start":"02:29.405 ","End":"02:32.430","Text":"We\u0027re going to say that our axis is going to be"},{"Start":"02:32.430 ","End":"02:40.500","Text":"this z-axis from the screen into your eye."},{"Start":"02:40.500 ","End":"02:45.435","Text":"The z axis is going from the screen into your eye,"},{"Start":"02:45.435 ","End":"02:47.955","Text":"imagine in 3 dimensions."},{"Start":"02:47.955 ","End":"02:52.710","Text":"We have the person standing here at the origin."},{"Start":"02:53.420 ","End":"02:57.215","Text":"We\u0027re going to say that our z center of mass,"},{"Start":"02:57.215 ","End":"03:00.820","Text":"exact same formula as our x center of mass."},{"Start":"03:00.820 ","End":"03:06.005","Text":"What do we do is we\u0027re going to say the mass of the large disk,"},{"Start":"03:06.005 ","End":"03:10.265","Text":"and then it\u0027s going to be multiplied by its distance away from the origin."},{"Start":"03:10.265 ","End":"03:12.950","Text":"Now, the center of mass of the large disk"},{"Start":"03:12.950 ","End":"03:16.715","Text":"is obviously going to be at the center of the circle,"},{"Start":"03:16.715 ","End":"03:20.615","Text":"which is where we said that our origin is going to be,"},{"Start":"03:20.615 ","End":"03:22.885","Text":"so it\u0027s multiplied by 0."},{"Start":"03:22.885 ","End":"03:28.965","Text":"Then we\u0027re going to add then the mass of the small disk,"},{"Start":"03:28.965 ","End":"03:34.140","Text":"and I\u0027m reminding you that we\u0027re thinking"},{"Start":"03:34.140 ","End":"03:40.970","Text":"about this hole over here as a disk with negative mass,"},{"Start":"03:40.970 ","End":"03:48.230","Text":"so negative 1/16 M multiplied by its distance from the origin,"},{"Start":"03:48.230 ","End":"03:51.540","Text":"which we were told is 1/2R."},{"Start":"03:52.760 ","End":"04:01.505","Text":"Then we are just dividing this entire thing by the total mass, which is M,"},{"Start":"04:01.505 ","End":"04:04.940","Text":"plus the mass of the small disk,"},{"Start":"04:04.940 ","End":"04:08.570","Text":"which again is a negative mass,"},{"Start":"04:08.570 ","End":"04:16.800","Text":"1/16 M. Then obviously this is equal to 0."},{"Start":"04:16.800 ","End":"04:25.364","Text":"Then we will get negative MR divided by 32,"},{"Start":"04:25.364 ","End":"04:35.330","Text":"divided by 15/16 M. Then if we just rearrange this,"},{"Start":"04:35.330 ","End":"04:43.370","Text":"we will see that we will get negative 1/30R is where the center of mass is."},{"Start":"04:43.370 ","End":"04:48.755","Text":"This is the answer to the question."},{"Start":"04:48.755 ","End":"04:51.245","Text":"That is where the center of mass is."},{"Start":"04:51.245 ","End":"04:53.645","Text":"That means that our center of mass,"},{"Start":"04:53.645 ","End":"04:56.180","Text":"if this is the origin over here,"},{"Start":"04:56.180 ","End":"05:01.075","Text":"will be negative 1/30thR."},{"Start":"05:01.075 ","End":"05:05.610","Text":"The center of mass will be somewhere around here."},{"Start":"05:05.610 ","End":"05:09.670","Text":"This will be Z_cm."},{"Start":"05:09.670 ","End":"05:13.130","Text":"That is the end of our lesson."}],"ID":29468}],"Thumbnail":null,"ID":5338},{"Name":"Motion Due To External Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"External Forces","Duration":"4m 29s","ChapterTopicVideoID":9085,"CourseChapterTopicPlaylistID":5339,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:05.010","Text":"Hello. In this lesson we\u0027re going to be speaking about"},{"Start":"00:05.010 ","End":"00:09.240","Text":"how we work with the center of mass and a few more details about it."},{"Start":"00:09.240 ","End":"00:11.220","Text":"The first question is,"},{"Start":"00:11.220 ","End":"00:14.715","Text":"how do we work using the center of mass, the notion?"},{"Start":"00:14.715 ","End":"00:21.540","Text":"The center of mass of a system is only moved by external forces."},{"Start":"00:21.540 ","End":"00:28.335","Text":"The only forces which act on a system or the center of mass are external forces."},{"Start":"00:28.335 ","End":"00:33.600","Text":"For instance, if here we have the Earth and here we have all sorts of"},{"Start":"00:33.600 ","End":"00:39.245","Text":"bodies and we are told that this is our center of mass."},{"Start":"00:39.245 ","End":"00:42.350","Text":"Now, these bodies can be bouncing off each other with"},{"Start":"00:42.350 ","End":"00:45.905","Text":"different forces acting all over the place."},{"Start":"00:45.905 ","End":"00:48.890","Text":"Also over here it doesn\u0027t matter."},{"Start":"00:48.890 ","End":"00:53.495","Text":"The only force that we can say is moving"},{"Start":"00:53.495 ","End":"00:59.070","Text":"the system is the external force which is due to gravity."},{"Start":"00:59.870 ","End":"01:03.740","Text":"This is the only force which matters which"},{"Start":"01:03.740 ","End":"01:06.935","Text":"will move the system and this is due to gravity,"},{"Start":"01:06.935 ","End":"01:13.795","Text":"the forces within the system acting on one another aren\u0027t important."},{"Start":"01:13.795 ","End":"01:19.430","Text":"Another example as to why working with a center of mass can also make"},{"Start":"01:19.430 ","End":"01:25.145","Text":"answering certain questions significantly easier is if we take over here,"},{"Start":"01:25.145 ","End":"01:34.040","Text":"we have a girl and inside her body she has all sorts of vessels,"},{"Start":"01:34.040 ","End":"01:40.370","Text":"molecules, cells, whatever it will be and all of these things"},{"Start":"01:40.370 ","End":"01:43.400","Text":"inside the body as you know have"},{"Start":"01:43.400 ","End":"01:47.285","Text":"forces and are affecting what\u0027s going on within the body."},{"Start":"01:47.285 ","End":"01:52.760","Text":"However, this doesn\u0027t matter to us because if we just look at the center of mass of"},{"Start":"01:52.760 ","End":"01:58.430","Text":"the girl which is around about just above the stomach,"},{"Start":"01:58.430 ","End":"02:01.865","Text":"then it makes dealing with a question"},{"Start":"02:01.865 ","End":"02:06.955","Text":"of someone applying a force to the girl and pushing her."},{"Start":"02:06.955 ","End":"02:11.450","Text":"Instead of doing the whole calculation and every single molecule and what\u0027s happening,"},{"Start":"02:11.450 ","End":"02:17.285","Text":"we can look at some body just as a point mass."},{"Start":"02:17.285 ","End":"02:21.805","Text":"Then we say if a force is applied then the girl will move forwards."},{"Start":"02:21.805 ","End":"02:26.820","Text":"It can really simplify answering certain questions."},{"Start":"02:27.040 ","End":"02:30.635","Text":"Another example that we can give,"},{"Start":"02:30.635 ","End":"02:37.810","Text":"let\u0027s split this over here is if we look for instance at a rocket launcher."},{"Start":"02:37.810 ","End":"02:46.280","Text":"The launcher launches the rocket and it flies forward at some velocity V_0."},{"Start":"02:46.280 ","End":"02:51.380","Text":"Now, as we know, the rocket will shoot in"},{"Start":"02:51.380 ","End":"02:57.170","Text":"a ballistic fashion and land on the ground like so."},{"Start":"02:57.170 ","End":"03:03.695","Text":"However, if we say that for instance the rocket splits over here,"},{"Start":"03:03.695 ","End":"03:06.890","Text":"so this is Section 1 of the rocket and this is Section 2."},{"Start":"03:06.890 ","End":"03:10.430","Text":"Then from the split there\u0027s some kind of force which"},{"Start":"03:10.430 ","End":"03:14.325","Text":"means that Section 2 will fall over here."},{"Start":"03:14.325 ","End":"03:18.440","Text":"This is Section 2 and Section 1 will"},{"Start":"03:18.440 ","End":"03:23.370","Text":"carry on in this direction and say it will fall over here."},{"Start":"03:23.900 ","End":"03:29.030","Text":"What we can say is if the full rocket,"},{"Start":"03:29.030 ","End":"03:30.500","Text":"if it wouldn\u0027t split,"},{"Start":"03:30.500 ","End":"03:32.750","Text":"it would land over here."},{"Start":"03:32.750 ","End":"03:38.210","Text":"After the split the center of mass will still be here."},{"Start":"03:38.210 ","End":"03:43.010","Text":"The center of mass after the rocket is split into 2 pieces"},{"Start":"03:43.010 ","End":"03:48.560","Text":"and each piece has gone its separate way depending on the force is acting,"},{"Start":"03:48.560 ","End":"03:52.670","Text":"the center of mass will still remain and fall in"},{"Start":"03:52.670 ","End":"03:57.940","Text":"the exact same place that the whole full rocket head would have fallen."},{"Start":"03:57.940 ","End":"04:04.110","Text":"This is another good trick that you can apply to certain questions."},{"Start":"04:05.930 ","End":"04:10.640","Text":"This can show that the internal forces which are acting inside"},{"Start":"04:10.640 ","End":"04:15.170","Text":"the rocket head during the split don\u0027t affect again,"},{"Start":"04:15.170 ","End":"04:20.570","Text":"like what we said, don\u0027t affect where the center of mass will fall."},{"Start":"04:20.570 ","End":"04:26.090","Text":"It will fall in the exact same place as if the rocket hadn\u0027t split."},{"Start":"04:26.090 ","End":"04:30.150","Text":"That is the end of this lesson."}],"ID":9358}],"Thumbnail":null,"ID":5339},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Man On A Boat","Duration":"12m 31s","ChapterTopicVideoID":9086,"CourseChapterTopicPlaylistID":5340,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:03.280","Text":"Hello, in this question,"},{"Start":"00:03.280 ","End":"00:06.670","Text":"we\u0027re told that a man is standing at one end of a boat,"},{"Start":"00:06.670 ","End":"00:09.520","Text":"and that the boat is of length 3 meters."},{"Start":"00:09.520 ","End":"00:14.530","Text":"We\u0027re told that the man weighs 70 kilograms and that the boat weighs 100 kilograms."},{"Start":"00:14.530 ","End":"00:17.470","Text":"The man moves then 2 meters up the boat."},{"Start":"00:17.470 ","End":"00:22.490","Text":"He\u0027s starting at one end and he walks until here."},{"Start":"00:22.490 ","End":"00:25.635","Text":"How much does the boat move?"},{"Start":"00:25.635 ","End":"00:30.215","Text":"Then we\u0027re being told that we can ignore the friction between the boat and the water."},{"Start":"00:30.215 ","End":"00:35.455","Text":"Let\u0027s just speak about why the boat moves when the man is walking."},{"Start":"00:35.455 ","End":"00:37.405","Text":"In order for the man to move,"},{"Start":"00:37.405 ","End":"00:42.100","Text":"there must be some friction between his feet and the boat,"},{"Start":"00:42.100 ","End":"00:46.785","Text":"which is allowing him to push himself forward."},{"Start":"00:46.785 ","End":"00:51.480","Text":"The friction that is acting over here is static friction."},{"Start":"00:53.210 ","End":"00:57.830","Text":"His foot lands on the floor of the boat."},{"Start":"00:57.830 ","End":"01:01.160","Text":"This is his foot on the floor of the boat and as he lifts"},{"Start":"01:01.160 ","End":"01:07.730","Text":"his foot he\u0027s pushing"},{"Start":"01:07.730 ","End":"01:11.912","Text":"the floor of the boat backwards slightly."},{"Start":"01:11.912 ","End":"01:14.330","Text":"That\u0027s him pushing forward."},{"Start":"01:14.330 ","End":"01:18.890","Text":"As we know, the floor is moving back slightly because of Newton\u0027s third law"},{"Start":"01:18.890 ","End":"01:23.735","Text":"saying that there\u0027s an equal and opposite force to every force."},{"Start":"01:23.735 ","End":"01:25.415","Text":"That means that here,"},{"Start":"01:25.415 ","End":"01:29.010","Text":"there\u0027s also static friction."},{"Start":"01:29.870 ","End":"01:35.765","Text":"Now, if we look at these 2 forces and then we try and work out this question,"},{"Start":"01:35.765 ","End":"01:39.865","Text":"it\u0027s going to be very tricky and very difficult."},{"Start":"01:39.865 ","End":"01:43.415","Text":"An easier way to do this is by using,"},{"Start":"01:43.415 ","End":"01:46.835","Text":"our now new found trick,"},{"Start":"01:46.835 ","End":"01:49.585","Text":"which is by using the center of mass."},{"Start":"01:49.585 ","End":"01:51.915","Text":"Watch how easy this is."},{"Start":"01:51.915 ","End":"01:54.440","Text":"Let\u0027s see, first of all,"},{"Start":"01:54.440 ","End":"01:57.130","Text":"we can rub this out because we don\u0027t need this."},{"Start":"01:57.130 ","End":"02:02.765","Text":"We already know that the man weighs 70 kilograms and the boat weighs 100 kilograms."},{"Start":"02:02.765 ","End":"02:06.050","Text":"Then we can say that the center of mass is around about,"},{"Start":"02:06.050 ","End":"02:07.930","Text":"let\u0027s say it\u0027s over here."},{"Start":"02:07.930 ","End":"02:10.760","Text":"Soon we\u0027ll work it out formally,"},{"Start":"02:10.760 ","End":"02:13.200","Text":"but let\u0027s say it\u0027s around about here."},{"Start":"02:13.360 ","End":"02:18.784","Text":"Now we can say that all the friction that\u0027s happening"},{"Start":"02:18.784 ","End":"02:24.925","Text":"between the man walking on the boat is now an internal force."},{"Start":"02:24.925 ","End":"02:27.725","Text":"As we know in dealing with center of mass,"},{"Start":"02:27.725 ","End":"02:31.055","Text":"it\u0027s only the external forces which matter."},{"Start":"02:31.055 ","End":"02:34.520","Text":"The friction between the man in the boat is internal,"},{"Start":"02:34.520 ","End":"02:37.460","Text":"so we can just ignore it and because we\u0027re"},{"Start":"02:37.460 ","End":"02:40.595","Text":"being told that we can ignore the friction between the boat and the water."},{"Start":"02:40.595 ","End":"02:43.945","Text":"That means that we have no external forces."},{"Start":"02:43.945 ","End":"02:49.000","Text":"Now, what does this mean if there are no external forces?"},{"Start":"02:49.160 ","End":"02:52.460","Text":"Also the friction between the man and the boat is internal,"},{"Start":"02:52.460 ","End":"02:55.220","Text":"so not counted. I wrote down these points."},{"Start":"02:55.220 ","End":"03:02.570","Text":"That means that we can use this idea that because there\u0027s no external forces,"},{"Start":"03:02.570 ","End":"03:07.040","Text":"it means that the center of mass must remain constant."},{"Start":"03:07.040 ","End":"03:12.185","Text":"It means that the center of mass is not moving and therefore,"},{"Start":"03:12.185 ","End":"03:14.525","Text":"when the man is standing at one edge of the boat,"},{"Start":"03:14.525 ","End":"03:18.575","Text":"and when the man is standing 2 meters in the boat,"},{"Start":"03:18.575 ","End":"03:22.235","Text":"the center of mass, must have stayed in the same place."},{"Start":"03:22.235 ","End":"03:24.830","Text":"What I\u0027m going to do is I\u0027m going to work out"},{"Start":"03:24.830 ","End":"03:27.245","Text":"the center of mass of the man at the beginning,"},{"Start":"03:27.245 ","End":"03:29.900","Text":"the center of mass of the man at the end and then"},{"Start":"03:29.900 ","End":"03:33.385","Text":"I\u0027m going to equate them one to the other."},{"Start":"03:33.385 ","End":"03:37.755","Text":"It means that the center of mass is conserved."},{"Start":"03:37.755 ","End":"03:41.690","Text":"Then we can find how much the boat has moved through that."},{"Start":"03:41.690 ","End":"03:44.340","Text":"Let\u0027s see this in action."},{"Start":"03:44.930 ","End":"03:49.955","Text":"Now let\u0027s start by finding the center of mass at the first moment."},{"Start":"03:49.955 ","End":"03:52.070","Text":"First I have to find where I\u0027m going to put"},{"Start":"03:52.070 ","End":"03:56.240","Text":"my axes and from there I work out the center of mass."},{"Start":"03:56.240 ","End":"03:59.720","Text":"I\u0027m going to say that my axes is right over"},{"Start":"03:59.720 ","End":"04:06.575","Text":"here and we\u0027re going to work this out according to our x center of mass."},{"Start":"04:06.575 ","End":"04:09.020","Text":"Now of course, we can also do our y center of mass,"},{"Start":"04:09.020 ","End":"04:14.100","Text":"but here it doesn\u0027t matter because we can see that the boat is only moving in the x axis."},{"Start":"04:14.720 ","End":"04:23.480","Text":"As we know, we have the sum of all of the masses multiplied by their possessions,"},{"Start":"04:23.480 ","End":"04:31.493","Text":"divided by the sum of all of the masses."},{"Start":"04:31.493 ","End":"04:36.020","Text":"We\u0027ll put in 1 over here."},{"Start":"04:36.020 ","End":"04:37.970","Text":"At the beginning, the man,"},{"Start":"04:37.970 ","End":"04:41.240","Text":"which is 70 kilograms,"},{"Start":"04:41.240 ","End":"04:43.635","Text":"is located 3 meters away,"},{"Start":"04:43.635 ","End":"04:51.130","Text":"so multiplied by 3 meters plus the boat which weighs 100 kilograms."},{"Start":"04:51.130 ","End":"04:54.610","Text":"Then we\u0027re going to multiply it by,"},{"Start":"04:54.610 ","End":"04:57.050","Text":"what are we multiplying it by?"},{"Start":"04:57.050 ","End":"05:00.845","Text":"Because we can see clearly that the boat isn\u0027t a point-mass."},{"Start":"05:00.845 ","End":"05:03.785","Text":"It\u0027s some long body."},{"Start":"05:03.785 ","End":"05:08.320","Text":"What we have to do is we have to find its center of mass."},{"Start":"05:08.320 ","End":"05:12.500","Text":"Now in the question, we don\u0027t know if the boat is homogeneous,"},{"Start":"05:12.500 ","End":"05:14.885","Text":"we don\u0027t know if one end is heavier than the other,"},{"Start":"05:14.885 ","End":"05:16.475","Text":"so it doesn\u0027t matter."},{"Start":"05:16.475 ","End":"05:18.840","Text":"You\u0027ll see in just a moment."},{"Start":"05:18.840 ","End":"05:26.075","Text":"Let\u0027s say that its center of mass is over here at a distance d away from the origin."},{"Start":"05:26.075 ","End":"05:32.410","Text":"I\u0027ll just multiply the center of mass by the mass of the boat."},{"Start":"05:32.410 ","End":"05:35.900","Text":"Then we divide it by the total mass,"},{"Start":"05:35.900 ","End":"05:41.285","Text":"which as we can see, is 170 kilograms."},{"Start":"05:41.285 ","End":"05:48.260","Text":"This is the center of mass right at the start."},{"Start":"05:48.260 ","End":"05:56.795","Text":"Now in front of us is the position of the boat after the man has walked these 2 meters."},{"Start":"05:56.795 ","End":"05:59.855","Text":"Now notice the boat has moved."},{"Start":"05:59.855 ","End":"06:05.075","Text":"Our axes that we started off with is in the exact same position."},{"Start":"06:05.075 ","End":"06:09.530","Text":"Exactly the same place, and the boat has just moved away from it."},{"Start":"06:09.530 ","End":"06:15.830","Text":"Let\u0027s say that we\u0027re going to call this distance between the axis and the boat,"},{"Start":"06:15.830 ","End":"06:19.830","Text":"let\u0027s say that this is a distance x."},{"Start":"06:20.690 ","End":"06:25.655","Text":"Now, our d which was"},{"Start":"06:25.655 ","End":"06:31.655","Text":"the distance from the axis until the center of mass is still going to be the same."},{"Start":"06:31.655 ","End":"06:34.890","Text":"We can draw it over here."},{"Start":"06:35.660 ","End":"06:38.940","Text":"This is our d it\u0027s exactly the same."},{"Start":"06:38.940 ","End":"06:44.850","Text":"Our boat\u0027s length is still exactly the same 3 meters,"},{"Start":"06:44.850 ","End":"06:51.270","Text":"and our guy now is standing at 2 meters into the boat."},{"Start":"06:51.270 ","End":"07:00.190","Text":"Now let\u0027s work out what the new center of mass is. We have xcm_2."},{"Start":"07:01.160 ","End":"07:04.970","Text":"Again, that\u0027s the sum of the masses multiplied by"},{"Start":"07:04.970 ","End":"07:10.435","Text":"their positions divided by the total mass of the system."},{"Start":"07:10.435 ","End":"07:14.741","Text":"Now we\u0027re going to look at the position of the man."},{"Start":"07:14.741 ","End":"07:22.715","Text":"The man is at distance x away from the axes plus 1 meter, because he\u0027s standing."},{"Start":"07:22.715 ","End":"07:31.650","Text":"This distance is 1 meter and then this distance is 2 meters,"},{"Start":"07:31.650 ","End":"07:34.545","Text":"because the boat is 3 meters long."},{"Start":"07:34.545 ","End":"07:37.080","Text":"We have an x plus 1 meter,"},{"Start":"07:37.080 ","End":"07:44.540","Text":"so x plus 1 meter multiplied by his mass,"},{"Start":"07:44.540 ","End":"07:47.240","Text":"which is 70 kilograms."},{"Start":"07:47.240 ","End":"07:53.345","Text":"Now notice also that the center of mass is in the exact same position."},{"Start":"07:53.345 ","End":"07:57.260","Text":"It also hasn\u0027t moved because that is the basis of this calculation."},{"Start":"07:57.260 ","End":"07:59.440","Text":"We\u0027re saying that the center of mass has not moved."},{"Start":"07:59.440 ","End":"08:05.790","Text":"You can see it in the picture here and here it\u0027s at the exact same position."},{"Start":"08:05.950 ","End":"08:10.170","Text":"Let me just clear this."},{"Start":"08:11.720 ","End":"08:16.495","Text":"Now we\u0027re going to do this for the boat as well."},{"Start":"08:16.495 ","End":"08:19.195","Text":"Now, as we know with the boat,"},{"Start":"08:19.195 ","End":"08:23.425","Text":"we have to find the center of mass of the boat just like before."},{"Start":"08:23.425 ","End":"08:25.165","Text":"The center of mass, again,"},{"Start":"08:25.165 ","End":"08:33.840","Text":"is a distance d from one end of the boat until the center of mass plus the x."},{"Start":"08:33.840 ","End":"08:36.715","Text":"The center of mass of the boat, which is here,"},{"Start":"08:36.715 ","End":"08:43.530","Text":"is now a distance of d plus x away from the origin of the axes."},{"Start":"08:43.530 ","End":"08:50.815","Text":"Plus, and then we\u0027re going to have x plus d multiplied by the mass of the boat,"},{"Start":"08:50.815 ","End":"08:57.200","Text":"which is 100 kilograms and then divided by the total mass of the system,"},{"Start":"08:57.200 ","End":"09:00.815","Text":"which is 170 kilograms."},{"Start":"09:00.815 ","End":"09:08.250","Text":"Now what I have over here in xcm_2 has to equal what I have in xcm_1,"},{"Start":"09:08.250 ","End":"09:13.185","Text":"because that means that the center of mass has not moved."},{"Start":"09:13.185 ","End":"09:15.780","Text":"Let\u0027s equate both sides."},{"Start":"09:15.780 ","End":"09:19.310","Text":"We can see immediately that the denominator,"},{"Start":"09:19.310 ","End":"09:22.070","Text":"we don\u0027t have to write it because we can just multiply"},{"Start":"09:22.070 ","End":"09:25.700","Text":"both sides by 170 kilograms and get rid of it."},{"Start":"09:25.700 ","End":"09:29.015","Text":"Then we\u0027ll see that we have 70 times 3."},{"Start":"09:29.015 ","End":"09:31.940","Text":"We can not look at the units in the meantime."},{"Start":"09:31.940 ","End":"09:35.640","Text":"We\u0027ll have 210 plus"},{"Start":"09:35.640 ","End":"09:44.895","Text":"100d is going to be equal to 70x I\u0027m opening up the brackets here,"},{"Start":"09:44.895 ","End":"09:52.245","Text":"plus 70 plus 100x plus 100d."},{"Start":"09:52.245 ","End":"09:58.455","Text":"Then we see that we have a 100d on both sides so we can subtract from both sides 100d."},{"Start":"09:58.455 ","End":"10:01.925","Text":"That\u0027s why at the beginning when I said what this d is,"},{"Start":"10:01.925 ","End":"10:05.075","Text":"it doesn\u0027t really matter because at the end it\u0027s going to cancel out."},{"Start":"10:05.075 ","End":"10:07.430","Text":"Which is a really good thing because if you\u0027re trying to"},{"Start":"10:07.430 ","End":"10:10.520","Text":"find in real life the center of mass of a boat,"},{"Start":"10:10.520 ","End":"10:13.165","Text":"it could be quite tricky."},{"Start":"10:13.165 ","End":"10:15.045","Text":"Here it cancels out."},{"Start":"10:15.045 ","End":"10:18.030","Text":"Then we can subtract 70 from each side,"},{"Start":"10:18.030 ","End":"10:23.970","Text":"so we\u0027ll have a 140 is equal to 170x."},{"Start":"10:24.040 ","End":"10:30.125","Text":"Then in order to isolate out the x and the x is in fact what we were looking for,"},{"Start":"10:30.125 ","End":"10:32.555","Text":"it\u0027s how much the boats moved."},{"Start":"10:32.555 ","End":"10:40.530","Text":"It\u0027s just going to be equal to 14 divided by 17 meters."},{"Start":"10:40.660 ","End":"10:43.190","Text":"Meters, are the units,"},{"Start":"10:43.190 ","End":"10:46.200","Text":"I\u0027ll just symbolize it like that."},{"Start":"10:46.200 ","End":"10:49.790","Text":"That means that when the man is standing on one end of"},{"Start":"10:49.790 ","End":"10:53.313","Text":"the boat and moves 2 meters down the boat,"},{"Start":"10:53.313 ","End":"10:55.790","Text":"the boat moves almost 1 meter,"},{"Start":"10:55.790 ","End":"10:58.845","Text":"not 1 meter a bit less this distance."},{"Start":"10:58.845 ","End":"11:06.570","Text":"That\u0027s how to solve a relatively complicated question using this idea of center of mass."},{"Start":"11:06.830 ","End":"11:10.864","Text":"Also, when answering this type of question,"},{"Start":"11:10.864 ","End":"11:15.560","Text":"mostly you\u0027ll be told that the boat moves and you"},{"Start":"11:15.560 ","End":"11:20.090","Text":"just have to know that the center of mass remains in the exact same position."},{"Start":"11:20.090 ","End":"11:22.190","Text":"That\u0027s how we answer this type of question,"},{"Start":"11:22.190 ","End":"11:27.810","Text":"but we have to make sure that the center of mass starts at rest."},{"Start":"11:28.010 ","End":"11:34.280","Text":"Before there\u0027s any movement the boat is stationary and the center of mass is stationary."},{"Start":"11:34.280 ","End":"11:38.005","Text":"If the boat is moving at the beginning,"},{"Start":"11:38.005 ","End":"11:40.265","Text":"then we can\u0027t solve it like this."},{"Start":"11:40.265 ","End":"11:45.440","Text":"We\u0027re going to have to use a combination of kinematics and the center of mass."},{"Start":"11:45.440 ","End":"11:51.285","Text":"But that\u0027s less common in these types of exam questions."},{"Start":"11:51.285 ","End":"11:54.950","Text":"We have to make sure that the center of mass starts at rest,"},{"Start":"11:54.950 ","End":"11:58.550","Text":"and very important that there are no external forces acting."},{"Start":"11:58.550 ","End":"12:00.305","Text":"If they are external force is acting,"},{"Start":"12:00.305 ","End":"12:02.555","Text":"then it\u0027s a different type of question."},{"Start":"12:02.555 ","End":"12:04.370","Text":"Now, also note that here,"},{"Start":"12:04.370 ","End":"12:08.765","Text":"we were asking the position where the boat will be."},{"Start":"12:08.765 ","End":"12:10.670","Text":"How much it will move, where the boat will be."},{"Start":"12:10.670 ","End":"12:13.325","Text":"We\u0027re being asked about the position."},{"Start":"12:13.325 ","End":"12:15.950","Text":"If however, we were asked about something else,"},{"Start":"12:15.950 ","End":"12:17.060","Text":"the motion of the boat,"},{"Start":"12:17.060 ","End":"12:19.865","Text":"the velocity with which the boat would be moving,"},{"Start":"12:19.865 ","End":"12:24.275","Text":"then we would be looking at a different way of solving this type of question."},{"Start":"12:24.275 ","End":"12:26.630","Text":"Not necessarily using the center of mass,"},{"Start":"12:26.630 ","End":"12:32.850","Text":"but rather using maybe conservation of energy or momentum and things like that."}],"ID":9359},{"Watched":false,"Name":"Ball On Ramp","Duration":"21m 55s","ChapterTopicVideoID":9087,"CourseChapterTopicPlaylistID":5340,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.880","Text":"Hello. In this question,"},{"Start":"00:02.880 ","End":"00:07.140","Text":"we\u0027re being told that a ball is resting on a sloped ramp,"},{"Start":"00:07.140 ","End":"00:08.895","Text":"which is also at rest."},{"Start":"00:08.895 ","End":"00:11.310","Text":"The height of the ball is 1 meter,"},{"Start":"00:11.310 ","End":"00:15.900","Text":"so we\u0027re saying that this distance over here"},{"Start":"00:15.900 ","End":"00:20.880","Text":"is 1 meter and that it\u0027s 5 meters away from the edge of the ramp."},{"Start":"00:20.880 ","End":"00:24.735","Text":"We can see 5 meters the edge of the ramp."},{"Start":"00:24.735 ","End":"00:28.950","Text":"The mass of the ramp m_1 is equal to 10 kilograms and the mass of the ball,"},{"Start":"00:28.950 ","End":"00:31.245","Text":"m_2 is equal to 2 kilograms."},{"Start":"00:31.245 ","End":"00:33.705","Text":"Our first question is to find the displacement"},{"Start":"00:33.705 ","End":"00:36.435","Text":"of the ramp when the ball reaches its edge."},{"Start":"00:36.435 ","End":"00:42.220","Text":"Again, we\u0027re going to look at this using center of mass."},{"Start":"00:42.530 ","End":"00:47.330","Text":"That way answering this question will be a lot easier."},{"Start":"00:47.330 ","End":"00:51.620","Text":"Let\u0027s look at this with regards to the different axes."},{"Start":"00:51.620 ","End":"00:56.330","Text":"Now, we can say that in the x-direction there is no friction because we can see that"},{"Start":"00:56.330 ","End":"00:58.400","Text":"there are these wheels over here and we\u0027re not"},{"Start":"00:58.400 ","End":"01:01.615","Text":"told about any coefficient of friction in the question."},{"Start":"01:01.615 ","End":"01:04.325","Text":"We know that in the y-direction,"},{"Start":"01:04.325 ","End":"01:08.390","Text":"we have the normal pointing upwards and we also"},{"Start":"01:08.390 ","End":"01:15.075","Text":"have Mg pointing downwards."},{"Start":"01:15.075 ","End":"01:18.050","Text":"There are external forces in the y-direction,"},{"Start":"01:18.050 ","End":"01:22.465","Text":"but there aren\u0027t external forces in the x-direction."},{"Start":"01:22.465 ","End":"01:29.345","Text":"What we\u0027re assuming here is just like in the question that we looked at just before,"},{"Start":"01:29.345 ","End":"01:32.900","Text":"we can see that this slope, this ramp,"},{"Start":"01:32.900 ","End":"01:38.490","Text":"it\u0027s moving because it has wheels and we\u0027re being told that it\u0027s starting at rest."},{"Start":"01:40.070 ","End":"01:42.920","Text":"Because it\u0027s starting at rest and there\u0027s"},{"Start":"01:42.920 ","End":"01:47.705","Text":"no external forces on the x-axis we know that its center of mass,"},{"Start":"01:47.705 ","End":"01:50.285","Text":"which we can say is,"},{"Start":"01:50.285 ","End":"01:56.745","Text":"let\u0027s say around about over here and let\u0027s say that this is our axis."},{"Start":"01:56.745 ","End":"01:59.345","Text":"From here, we\u0027re going to work this out."},{"Start":"01:59.345 ","End":"02:01.980","Text":"Sorry, not like this."},{"Start":"02:03.620 ","End":"02:09.485","Text":"We can say that our center of mass is around about over here and so now we can find out"},{"Start":"02:09.485 ","End":"02:14.720","Text":"what our initial center of mass is and we\u0027ll see that as the ball rolls,"},{"Start":"02:14.720 ","End":"02:19.070","Text":"the center of mass and the axis stay in the exact same location."},{"Start":"02:19.070 ","End":"02:21.950","Text":"Because of the conservation of the center of mass,"},{"Start":"02:21.950 ","End":"02:26.090","Text":"which is something that we looked at in the previous question."},{"Start":"02:26.090 ","End":"02:33.575","Text":"Let\u0027s do this, so we have Xcm initially is equal to"},{"Start":"02:33.575 ","End":"02:43.060","Text":"our m_1 multiplied by its position plus m_2 multiplied by its position,"},{"Start":"02:43.060 ","End":"02:51.975","Text":"divided by the total mass, which is m_1+m_2."},{"Start":"02:51.975 ","End":"02:53.145","Text":"What is X_2?"},{"Start":"02:53.145 ","End":"02:57.080","Text":"X_2 represents the center of mass of this ramp,"},{"Start":"02:57.080 ","End":"02:59.215","Text":"because the ramp isn\u0027t a point-mass."},{"Start":"02:59.215 ","End":"03:05.575","Text":"We can say that X_2 the center of mass of this is probably also a roundabout over here."},{"Start":"03:05.575 ","End":"03:11.080","Text":"This distance over here in the x-direction is going to be X_2."},{"Start":"03:11.080 ","End":"03:14.990","Text":"We don\u0027t know this and we also currently don\u0027t know what our X_1 is,"},{"Start":"03:14.990 ","End":"03:17.435","Text":"but we\u0027re going to leave it like this in the meantime because"},{"Start":"03:17.435 ","End":"03:21.055","Text":"we\u0027ll see that in the end it will all work out."},{"Start":"03:21.055 ","End":"03:25.985","Text":"First we\u0027re going to say that this is our X_1 initial and this is our X_2 initial."},{"Start":"03:25.985 ","End":"03:29.720","Text":"Now let\u0027s find out what our Xcm is once"},{"Start":"03:29.720 ","End":"03:33.865","Text":"the ball has reached this position, the final position."},{"Start":"03:33.865 ","End":"03:41.480","Text":"We can say that our Xcm final is going to be equal to our m_1 multiplied by"},{"Start":"03:41.480 ","End":"03:50.535","Text":"our X_1 final plus our m_2 multiplied by our center of mass of the ramp, final."},{"Start":"03:50.535 ","End":"03:56.160","Text":"Then of course again divided by the total mass, which is m_1+m_2."},{"Start":"03:56.800 ","End":"04:00.320","Text":"Because this question is very similar to the question that"},{"Start":"04:00.320 ","End":"04:03.230","Text":"we just looked at a moment earlier,"},{"Start":"04:03.230 ","End":"04:08.510","Text":"then we know that because center of mass of"},{"Start":"04:08.510 ","End":"04:10.640","Text":"the entire system remains in"},{"Start":"04:10.640 ","End":"04:15.575","Text":"the same points even after the ball has reached the end of the cart."},{"Start":"04:15.575 ","End":"04:21.635","Text":"We can say that our Xcm initial is equal to our Xcm final."},{"Start":"04:21.635 ","End":"04:24.050","Text":"Now let\u0027s just equate them."},{"Start":"04:24.050 ","End":"04:26.686","Text":"With simple algebra, we\u0027ll have that m_1,"},{"Start":"04:26.686 ","End":"04:31.109","Text":"X_1 initial plus m_2,"},{"Start":"04:31.109 ","End":"04:36.253","Text":"X_2 initial is equal to m_1,"},{"Start":"04:36.253 ","End":"04:42.210","Text":"X_1 final plus m_2, X_2 final."},{"Start":"04:42.210 ","End":"04:48.230","Text":"I didn\u0027t write in my denominators because they just cancel out immediately,"},{"Start":"04:48.230 ","End":"04:50.760","Text":"so it\u0027s silly to write them in."},{"Start":"04:51.260 ","End":"04:57.035","Text":"Now because I\u0027m being asked to find the displacement."},{"Start":"04:57.035 ","End":"04:58.955","Text":"We know that the displacement,"},{"Start":"04:58.955 ","End":"05:02.955","Text":"which is aka Delta X."},{"Start":"05:02.955 ","End":"05:05.895","Text":"Let\u0027s look at Delta X number 1 first."},{"Start":"05:05.895 ","End":"05:12.225","Text":"The distance traveled from point A to point B is the displacement,"},{"Start":"05:12.225 ","End":"05:14.860","Text":"and how we work it out is B-A."},{"Start":"05:15.950 ","End":"05:25.380","Text":"We\u0027re just going to have X_1 final minus X-1 initial."},{"Start":"05:25.380 ","End":"05:29.960","Text":"Then the same for the displacement of Delta X_2,"},{"Start":"05:29.960 ","End":"05:35.995","Text":"which will be X_2 final minus X_2 initial."},{"Start":"05:35.995 ","End":"05:43.830","Text":"Now I\u0027m just going to substitute that into my equation."},{"Start":"05:50.360 ","End":"05:55.040","Text":"I\u0027m going to put all of the expressions on 1 side."},{"Start":"05:55.040 ","End":"05:59.225","Text":"I\u0027ll have that 0 is simply equal to"},{"Start":"05:59.225 ","End":"06:07.440","Text":"m_1 Delta X_1 plus m_2 Delta X_2."},{"Start":"06:07.440 ","End":"06:11.090","Text":"I just substituted these in into this equation and"},{"Start":"06:11.090 ","End":"06:15.575","Text":"moved all of my expressions to just the right side of the equation,"},{"Start":"06:15.575 ","End":"06:17.570","Text":"leaving me with a 0."},{"Start":"06:17.570 ","End":"06:21.170","Text":"Therefore, if I separate this out,"},{"Start":"06:21.170 ","End":"06:26.280","Text":"so I move my m_1 Delta X_1 to the other side so I\u0027ll get"},{"Start":"06:26.280 ","End":"06:35.490","Text":"that negative m_1 Delta X_1=m_2 Delta X_2."},{"Start":"06:35.490 ","End":"06:41.875","Text":"Now let\u0027s take a look at what that means on our drawing over here."},{"Start":"06:41.875 ","End":"06:49.010","Text":"What happens is our ball starts sliding down until it reaches the end."},{"Start":"06:49.010 ","End":"06:52.600","Text":"Now, the confusing bit is over here,"},{"Start":"06:52.600 ","End":"06:59.540","Text":"according to this equation it means that according to the ratio of the masses,"},{"Start":"06:59.540 ","End":"07:03.035","Text":"so obviously the weight of the ramp,"},{"Start":"07:03.035 ","End":"07:05.540","Text":"the mass of the ramp is going to be greater than the mass of"},{"Start":"07:05.540 ","End":"07:09.440","Text":"our 2 kilogram ball, we\u0027re told that."},{"Start":"07:09.440 ","End":"07:11.884","Text":"But according to that ratio,"},{"Start":"07:11.884 ","End":"07:15.304","Text":"as the ball moves forwards,"},{"Start":"07:15.304 ","End":"07:19.680","Text":"it\u0027s simultaneously pushing the ramp back."},{"Start":"07:20.500 ","End":"07:26.045","Text":"The ball is moving to the end of the ramp but the ramp is moving."},{"Start":"07:26.045 ","End":"07:34.475","Text":"The ball isn\u0027t in fact going to roll this entire 5 meters when it gets here."},{"Start":"07:34.475 ","End":"07:41.620","Text":"Because the ramp is going to look now something like this."},{"Start":"07:42.050 ","End":"07:51.420","Text":"It would have moved and now the end of the ramp will be here and the ball will be here."},{"Start":"07:51.420 ","End":"07:54.840","Text":"We see that our 5 meters,"},{"Start":"07:54.840 ","End":"07:58.095","Text":"our ruler over here, hasn\u0027t moved."},{"Start":"07:58.095 ","End":"08:02.100","Text":"The ball hasn\u0027t moved exactly 5 meters."},{"Start":"08:02.100 ","End":"08:04.665","Text":"Let\u0027s see how we deal with this."},{"Start":"08:04.665 ","End":"08:10.095","Text":"Now let\u0027s just draw on our new good diagram what are X_1,"},{"Start":"08:10.095 ","End":"08:13.230","Text":"X_2 and everything was okay."},{"Start":"08:13.230 ","End":"08:18.840","Text":"Our X_2 we drew right at the beginning."},{"Start":"08:18.840 ","End":"08:22.520","Text":"That was where we said, for instance,"},{"Start":"08:22.520 ","End":"08:27.410","Text":"that the ball was starting to roll from."},{"Start":"08:27.410 ","End":"08:33.435","Text":"Now we can say that it was the X_2 initial,"},{"Start":"08:33.435 ","End":"08:39.725","Text":"and that the actual displacement of the ball is going to be from here until here."},{"Start":"08:39.725 ","End":"08:47.250","Text":"This was Delta X_2."},{"Start":"08:47.250 ","End":"08:52.565","Text":"Our X_1, as we can see,"},{"Start":"08:52.565 ","End":"08:59.655","Text":"it started off over here and it moved backwards, this distance."},{"Start":"08:59.655 ","End":"09:03.810","Text":"This will be our X_1 final."},{"Start":"09:03.810 ","End":"09:07.485","Text":"Then we can say that"},{"Start":"09:07.485 ","End":"09:16.050","Text":"our total Delta X_1 will be all of this."},{"Start":"09:17.440 ","End":"09:20.495","Text":"Why up until here? Because remember,"},{"Start":"09:20.495 ","End":"09:23.375","Text":"this was our center of mass."},{"Start":"09:23.375 ","End":"09:28.875","Text":"This was our X_1 initial and our center of mass,"},{"Start":"09:28.875 ","End":"09:35.680","Text":"so we move back our center of mass, this Delta X_1."},{"Start":"09:36.110 ","End":"09:42.725","Text":"Now our problem is that we don\u0027t know what our Delta X_2 is."},{"Start":"09:42.725 ","End":"09:45.980","Text":"We need that in order to find the displacement of"},{"Start":"09:45.980 ","End":"09:49.310","Text":"the ramp when the ball reaches its edge."},{"Start":"09:49.310 ","End":"09:52.055","Text":"How we going to solve this problem,"},{"Start":"09:52.055 ","End":"09:58.080","Text":"we\u0027re going to use relative motion."},{"Start":"09:58.080 ","End":"10:01.525","Text":"I know that relative to the ramp,"},{"Start":"10:01.525 ","End":"10:03.580","Text":"the ball moves 5 meters,"},{"Start":"10:03.580 ","End":"10:07.030","Text":"because it begins right at the head of the ramp and"},{"Start":"10:07.030 ","End":"10:11.350","Text":"goes all the way to the end and we\u0027ve been told in the question that that\u0027s 5 meters."},{"Start":"10:11.350 ","End":"10:17.560","Text":"If there\u0027s someone here standing on the ramp relative to the person on the ramp,"},{"Start":"10:17.560 ","End":"10:20.230","Text":"the ball moves 5 meters."},{"Start":"10:20.230 ","End":"10:26.605","Text":"What I\u0027m going to say is that my Delta x tag of the ball,"},{"Start":"10:26.605 ","End":"10:34.120","Text":"so 2 and our equation we know is going to be equal"},{"Start":"10:34.120 ","End":"10:42.835","Text":"to the displacement of the ball minus the displacement of the ramp,"},{"Start":"10:42.835 ","End":"10:48.265","Text":"and we know that all of this is going to be equal to 5."},{"Start":"10:48.265 ","End":"10:52.375","Text":"Because we know that the ball\u0027s movement,"},{"Start":"10:52.375 ","End":"10:57.920","Text":"the ball\u0027s displacement relative to the ramp is 5 from the question."},{"Start":"10:58.290 ","End":"11:04.645","Text":"Now you\u0027ll see that what I\u0027ve gotten here is another equation that I can use."},{"Start":"11:04.645 ","End":"11:11.665","Text":"What we\u0027re trying to find is the displacement of the ramp when the ball reaches its edge."},{"Start":"11:11.665 ","End":"11:12.910","Text":"Or in other words,"},{"Start":"11:12.910 ","End":"11:18.700","Text":"we\u0027re trying to find out what our Delta x_1 is equal to."},{"Start":"11:18.700 ","End":"11:20.650","Text":"This is our question."},{"Start":"11:20.650 ","End":"11:23.815","Text":"So because I want to know what my Delta x_1 is,"},{"Start":"11:23.815 ","End":"11:27.220","Text":"I can just isolate out my Delta x_2."},{"Start":"11:27.220 ","End":"11:35.125","Text":"I can say that my Delta x_2 is equal to 5 plus Delta x_1."},{"Start":"11:35.125 ","End":"11:41.440","Text":"Then I simply substitute this into our equation that we worked out here,"},{"Start":"11:41.440 ","End":"11:51.130","Text":"so then I will get that negative m_1 Delta x_1 is equal to m_2."},{"Start":"11:51.130 ","End":"11:52.990","Text":"What is my m_2?"},{"Start":"11:52.990 ","End":"11:57.370","Text":"It\u0027s 2 kilograms multiplied by my Delta x_2."},{"Start":"11:57.370 ","End":"12:03.970","Text":"My Delta x_2 is 5 plus Delta x_1."},{"Start":"12:03.970 ","End":"12:07.810","Text":"Then of course, I can say instead of m_1,"},{"Start":"12:07.810 ","End":"12:09.760","Text":"I can write down its value."},{"Start":"12:09.760 ","End":"12:12.490","Text":"M_1 is 10 kilograms."},{"Start":"12:12.490 ","End":"12:16.135","Text":"Then if I open up the brackets,"},{"Start":"12:16.135 ","End":"12:25.925","Text":"I\u0027ll get negative 10 Delta x_1 is equal to 10 plus 2 Delta x_1."},{"Start":"12:25.925 ","End":"12:31.345","Text":"Then I can just say that"},{"Start":"12:31.345 ","End":"12:38.380","Text":"negative 12 Delta x_1 is equal to 10,"},{"Start":"12:38.380 ","End":"12:47.395","Text":"and therefore Delta x_1 is equal to negative 10 divided by 12 meters."},{"Start":"12:47.395 ","End":"12:50.575","Text":"This is our answer to question Number 1."},{"Start":"12:50.575 ","End":"12:53.270","Text":"This is the displacement."},{"Start":"12:53.550 ","End":"12:59.425","Text":"Now let\u0027s just take a look at why our answer to question Number 1 make sense,"},{"Start":"12:59.425 ","End":"13:01.570","Text":"because we get a negative number."},{"Start":"13:01.570 ","End":"13:05.455","Text":"As we can see, if the ball is moving forwards,"},{"Start":"13:05.455 ","End":"13:09.070","Text":"the ramp, even in our picture, was moving backwards."},{"Start":"13:09.070 ","End":"13:10.840","Text":"If it\u0027s moving backwards,"},{"Start":"13:10.840 ","End":"13:15.790","Text":"it will have a negative sign. It makes sense."},{"Start":"13:15.790 ","End":"13:20.720","Text":"Also we can see that Delta x_1."},{"Start":"13:20.730 ","End":"13:24.790","Text":"In fact, I can rub this little thing out."},{"Start":"13:24.790 ","End":"13:27.670","Text":"It starts from right over here,"},{"Start":"13:27.670 ","End":"13:30.355","Text":"and it\u0027s a backwards facing vector,"},{"Start":"13:30.355 ","End":"13:33.655","Text":"which again means that there\u0027ll be a negative sign over here."},{"Start":"13:33.655 ","End":"13:37.360","Text":"It\u0027s a backward facing vector."},{"Start":"13:37.360 ","End":"13:39.040","Text":"That is question Number 1."},{"Start":"13:39.040 ","End":"13:42.080","Text":"Now let\u0027s move on to question Number 2,"},{"Start":"13:42.870 ","End":"13:50.170","Text":"so I\u0027ve rubbed out all of my diagrams from question Number 1 because question Number 1,"},{"Start":"13:50.170 ","End":"13:53.680","Text":"what I wrote there isn\u0027t relevant to question Number 2."},{"Start":"13:53.680 ","End":"13:57.640","Text":"Here we\u0027re being asked to find the velocities of the bodies."},{"Start":"13:57.640 ","End":"14:00.400","Text":"If it is given that the velocity of the ball at"},{"Start":"14:00.400 ","End":"14:04.120","Text":"the edge of the ramp is only in the x-direction."},{"Start":"14:04.120 ","End":"14:09.230","Text":"That means our v is just in this direction."},{"Start":"14:09.930 ","End":"14:12.505","Text":"Let\u0027s see how we do this."},{"Start":"14:12.505 ","End":"14:15.295","Text":"Now, the second that I\u0027m being told about velocity,"},{"Start":"14:15.295 ","End":"14:17.605","Text":"I\u0027m going to think about momentum and energy."},{"Start":"14:17.605 ","End":"14:22.675","Text":"Why? Because I have in there the velocity variable."},{"Start":"14:22.675 ","End":"14:25.070","Text":"I know that I can use that."},{"Start":"14:25.080 ","End":"14:28.870","Text":"Now what I\u0027m going to ask myself is if there\u0027s"},{"Start":"14:28.870 ","End":"14:33.010","Text":"conservation of momentum and conservation of energy."},{"Start":"14:33.010 ","End":"14:35.350","Text":"Now in question Number 1,"},{"Start":"14:35.350 ","End":"14:37.720","Text":"I use the idea of the center of mass"},{"Start":"14:37.720 ","End":"14:41.200","Text":"because I knew that the sum of all of the external forces was"},{"Start":"14:41.200 ","End":"14:48.160","Text":"equal to 0 and that therefore there was conservation of the center of mass."},{"Start":"14:48.160 ","End":"14:53.800","Text":"Here I\u0027m asking if there\u0027s conservation of momentum or of energy."},{"Start":"14:53.800 ","End":"15:00.220","Text":"Now because we saw in question Number 1 that there is conservation of the center of mass."},{"Start":"15:00.220 ","End":"15:03.940","Text":"I can now know that there\u0027s conservation of momentum."},{"Start":"15:03.940 ","End":"15:07.300","Text":"Therefore, if there\u0027s conservation of center of mass,"},{"Start":"15:07.300 ","End":"15:11.095","Text":"then this conservation of momentum."},{"Start":"15:11.095 ","End":"15:15.670","Text":"Let\u0027s begin by writing our equations for the momentum."},{"Start":"15:15.670 ","End":"15:24.830","Text":"Because we know that the sum of all of the forces on the x-axis is equal to 0."},{"Start":"15:24.840 ","End":"15:34.570","Text":"Then we know that the momentum P on the x-axis is going to be like this."},{"Start":"15:34.570 ","End":"15:39.610","Text":"At the beginning, the momentum is equal to 0 because both bodies are at rest."},{"Start":"15:39.610 ","End":"15:42.325","Text":"It will be 0,"},{"Start":"15:42.325 ","End":"15:46.540","Text":"and then our momentum at the end is going to be equal"},{"Start":"15:46.540 ","End":"15:53.635","Text":"to m_1 u_1 plus m_2, u_2."},{"Start":"15:53.635 ","End":"15:56.800","Text":"Because the ball is moving in the right direction"},{"Start":"15:56.800 ","End":"16:00.790","Text":"and the ramp is moving in the left direction,"},{"Start":"16:00.790 ","End":"16:02.810","Text":"so it has a u_2."},{"Start":"16:03.300 ","End":"16:08.545","Text":"Now let\u0027s write out the equations for the energy."},{"Start":"16:08.545 ","End":"16:12.385","Text":"The initial energy, the energy at the beginning,"},{"Start":"16:12.385 ","End":"16:15.250","Text":"so we know because both bodies are at rest,"},{"Start":"16:15.250 ","End":"16:17.785","Text":"so our velocity is going to be equal to 0."},{"Start":"16:17.785 ","End":"16:20.935","Text":"We know that there\u0027s no kinetic energy."},{"Start":"16:20.935 ","End":"16:27.655","Text":"There\u0027s only potential energy which has just going to be m_2 multiplied by g,"},{"Start":"16:27.655 ","End":"16:31.375","Text":"multiplied by H,"},{"Start":"16:31.375 ","End":"16:34.195","Text":"which we know is 1 meter."},{"Start":"16:34.195 ","End":"16:39.550","Text":"This is going to be equal to through conservation of energy,"},{"Start":"16:39.550 ","End":"16:41.815","Text":"the energy at the end,"},{"Start":"16:41.815 ","End":"16:46.630","Text":"the final energy, which is solely going to be kinetic energy."},{"Start":"16:46.630 ","End":"16:52.467","Text":"We\u0027re going to have 1/2 m_1 u_1^2"},{"Start":"16:52.467 ","End":"16:59.410","Text":"plus 1/2 m_2 u_2^2."},{"Start":"16:59.410 ","End":"17:04.610","Text":"This is equal to energy at the end E final."},{"Start":"17:04.800 ","End":"17:11.379","Text":"The only trick to answering question Number 2 now is just to substitute"},{"Start":"17:11.379 ","End":"17:17.440","Text":"in the numbers because we have 2 equations with 2 unknowns,"},{"Start":"17:17.440 ","End":"17:19.135","Text":"so we can deal with this."},{"Start":"17:19.135 ","End":"17:22.405","Text":"The only trick was to figure out that"},{"Start":"17:22.405 ","End":"17:27.685","Text":"both the momentum and the energy were conserved throughout and therefore,"},{"Start":"17:27.685 ","End":"17:30.340","Text":"we can apply these equations."},{"Start":"17:30.340 ","End":"17:32.860","Text":"Now how do we know that the momentum is conserved?"},{"Start":"17:32.860 ","End":"17:36.805","Text":"Because the sum of all of the forces was equal to 0."},{"Start":"17:36.805 ","End":"17:39.355","Text":"How do we know that the energy was conserved?"},{"Start":"17:39.355 ","End":"17:42.940","Text":"Because normal isn\u0027t doing work,"},{"Start":"17:42.940 ","End":"17:47.440","Text":"and all of our other forces are being conserved."},{"Start":"17:47.440 ","End":"17:51.115","Text":"Whoever doesn\u0027t want to look at the algebra now,"},{"Start":"17:51.115 ","End":"17:54.295","Text":"because they feel like they know how to do it can turn away."},{"Start":"17:54.295 ","End":"17:57.760","Text":"Otherwise, let\u0027s go over how we solve this."},{"Start":"17:57.760 ","End":"18:03.745","Text":"Now we can substitute in numbers over here in the equation for the momentum."},{"Start":"18:03.745 ","End":"18:07.120","Text":"We can say we know that m_1 is 10,"},{"Start":"18:07.120 ","End":"18:12.040","Text":"so we\u0027ll have 10 u_1 plus 2,"},{"Start":"18:12.040 ","End":"18:14.785","Text":"u_2 is equal to 0."},{"Start":"18:14.785 ","End":"18:16.945","Text":"Then if we isolate out,"},{"Start":"18:16.945 ","End":"18:22.195","Text":"Let\u0027s say we isolate out u_2,"},{"Start":"18:22.195 ","End":"18:30.625","Text":"so we will get that r U2 is equal 2 negative 10 over 2,"},{"Start":"18:30.625 ","End":"18:35.965","Text":"u_1, which is equal to negative 5, u_1."},{"Start":"18:35.965 ","End":"18:42.190","Text":"Now we can substitute this into our equation. Let\u0027s put this in."},{"Start":"18:42.190 ","End":"18:46.270","Text":"We know that our m_2 is equal to 2 kilograms or g,"},{"Start":"18:46.270 ","End":"18:48.340","Text":"Let\u0027s say it\u0027s equal to 10."},{"Start":"18:48.340 ","End":"18:51.530","Text":"Our H is equal to 1."},{"Start":"18:53.370 ","End":"18:57.700","Text":"Let\u0027s multiply both sides by 2 to get rid of these halves."},{"Start":"18:57.700 ","End":"18:59.365","Text":"I\u0027ll have 2 times this,"},{"Start":"18:59.365 ","End":"19:01.210","Text":"which will equal to m_1,"},{"Start":"19:01.210 ","End":"19:09.840","Text":"which is 10 times u_1^2 plus 2 times u 2^2."},{"Start":"19:09.840 ","End":"19:14.325","Text":"Now we can substitute n what are u_2 is equal 2."},{"Start":"19:14.325 ","End":"19:17.925","Text":"On this side we\u0027ll have 40,"},{"Start":"19:17.925 ","End":"19:28.750","Text":"which will equal to 10 u_1^2 plus 2,"},{"Start":"19:28.750 ","End":"19:33.730","Text":"and then our u_2^2 is negative 5 u_1."},{"Start":"19:33.730 ","End":"19:38.860","Text":"That squared will be 25 u_1^2."},{"Start":"19:38.860 ","End":"19:43.105","Text":"Then 10 u_1^2 plus"},{"Start":"19:43.105 ","End":"19:50.410","Text":"2 times 25 u_n^2 is going to be 10 plus 50,"},{"Start":"19:50.410 ","End":"19:56.065","Text":"so it\u0027s going to be 60 u_1^2."},{"Start":"19:56.065 ","End":"20:01.565","Text":"Then therefore we have 40 divided by 60."},{"Start":"20:01.565 ","End":"20:06.985","Text":"Then, which is just going to be 4 divided by 6,"},{"Start":"20:06.985 ","End":"20:09.475","Text":"is going to be equal to u_1^2."},{"Start":"20:09.475 ","End":"20:14.030","Text":"Therefore, the square root of this is equal to u_1."},{"Start":"20:14.030 ","End":"20:22.910","Text":"Therefore, u_1 is equal to 2 over root 3 meters per second."},{"Start":"20:22.920 ","End":"20:34.020","Text":"Then therefore u_2 is going to be equal negative 5 of u_1."},{"Start":"20:34.410 ","End":"20:43.780","Text":"That is going to be equal to negative 10 over root 3 meters per second."},{"Start":"20:43.780 ","End":"20:51.170","Text":"Now, one thing we need to notice is that we know that our u_1 is the ramp."},{"Start":"20:51.170 ","End":"20:57.150","Text":"We\u0027ve already said from question Number 1 that our ramp is going backwards."},{"Start":"20:57.150 ","End":"21:02.870","Text":"This is our u_1 and in the forwards direction we have a u_2,"},{"Start":"21:02.870 ","End":"21:06.995","Text":"because the ball is sliding in a positive direction."},{"Start":"21:06.995 ","End":"21:10.355","Text":"Therefore, right at the end, even though,"},{"Start":"21:10.355 ","End":"21:13.220","Text":"because of how we isolated out this u_2,"},{"Start":"21:13.220 ","End":"21:18.060","Text":"we got the u_2 will be the negative 5 of u_1."},{"Start":"21:18.060 ","End":"21:24.700","Text":"We should actually change this so that we have negative u_2 is equal to 5 u_1."},{"Start":"21:24.700 ","End":"21:30.200","Text":"Then here we should have the negative and here we should have the positive."},{"Start":"21:30.200 ","End":"21:34.505","Text":"Because these velocities then make more sense"},{"Start":"21:34.505 ","End":"21:39.940","Text":"regarding the direction that we know each body is traveling in."},{"Start":"21:39.940 ","End":"21:44.855","Text":"Now we have u_2 as a positive velocity. Which is correct."},{"Start":"21:44.855 ","End":"21:48.860","Text":"Our ball is moving in the positive direction and our u_1 is a negative velocity,"},{"Start":"21:48.860 ","End":"21:53.055","Text":"which is also correct because it\u0027s moving in the negative direction."},{"Start":"21:53.055 ","End":"21:56.640","Text":"That is it. That\u0027s the end of the lesson."}],"ID":9360}],"Thumbnail":null,"ID":5340},{"Name":"Calculating The Center Of Mass Using Integrals","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Review on Coordinates (Same Lecture as In The Math Intro)","Duration":"16m 33s","ChapterTopicVideoID":10281,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.555","Text":"Hello. In this video, we\u0027re going to talk about coordinates."},{"Start":"00:03.555 ","End":"00:07.785","Text":"In order to describe the position of an object in space,"},{"Start":"00:07.785 ","End":"00:09.705","Text":"we need 3 data points."},{"Start":"00:09.705 ","End":"00:11.580","Text":"There are multiple coordinate systems,"},{"Start":"00:11.580 ","End":"00:16.425","Text":"but in physics we tend to use 1 of 3 systems for any given problem."},{"Start":"00:16.425 ","End":"00:21.630","Text":"The 3 most common systems are Cartesian coordinates, x, y,"},{"Start":"00:21.630 ","End":"00:27.430","Text":"and z, polar coordinates, and spherical coordinates."},{"Start":"00:27.430 ","End":"00:30.980","Text":"What I\u0027d like to do is look at these 3 types,"},{"Start":"00:30.980 ","End":"00:34.160","Text":"see some of the elements of them,"},{"Start":"00:34.160 ","End":"00:36.460","Text":"and talk about how we can transition from one to the other"},{"Start":"00:36.460 ","End":"00:39.925","Text":"mathematically when looking to describe a given point."},{"Start":"00:39.925 ","End":"00:43.700","Text":"The first coordinates we\u0027ll talk about are Cartesian coordinates."},{"Start":"00:43.700 ","End":"00:47.180","Text":"You might be familiar with them from prior class you\u0027ve taken particularly in math."},{"Start":"00:47.180 ","End":"00:50.390","Text":"We describe a point with values of x, y,"},{"Start":"00:50.390 ","End":"00:54.325","Text":"or z, and use these variables to locate our points in space."},{"Start":"00:54.325 ","End":"00:57.210","Text":"If we want to describe this red point, we just circled,"},{"Start":"00:57.210 ","End":"00:59.900","Text":"what we can do is find the values in x, y,"},{"Start":"00:59.900 ","End":"01:02.190","Text":"and z using orthogonal lines,"},{"Start":"01:02.190 ","End":"01:04.429","Text":"that is lines that have right angles."},{"Start":"01:04.429 ","End":"01:06.140","Text":"We draw from the red point,"},{"Start":"01:06.140 ","End":"01:09.560","Text":"we can draw an orthogonal line to the z-axis,"},{"Start":"01:09.560 ","End":"01:14.420","Text":"and that will give us our value in z in terms of the height by measuring that down,"},{"Start":"01:14.420 ","End":"01:17.160","Text":"and then we draw a line to the x,"},{"Start":"01:17.160 ","End":"01:19.900","Text":"y plane, and from the point where it comes out in x, y plane,"},{"Start":"01:19.900 ","End":"01:23.440","Text":"we draw 1 orthogonal line to the x-axis and another one to the y-axis,"},{"Start":"01:23.440 ","End":"01:27.020","Text":"and we can use these 3 lines to measure the values of x,"},{"Start":"01:27.020 ","End":"01:29.155","Text":"y, and z and locate our point."},{"Start":"01:29.155 ","End":"01:34.625","Text":"Now, let\u0027s describe the same point using cylindrical or polar coordinates."},{"Start":"01:34.625 ","End":"01:36.860","Text":"We\u0027re going to look instead of for x, y,"},{"Start":"01:36.860 ","End":"01:39.205","Text":"and z, for r, Theta, and z."},{"Start":"01:39.205 ","End":"01:40.820","Text":"We can start with z, it\u0027s rather simple."},{"Start":"01:40.820 ","End":"01:41.975","Text":"It\u0027s the same z from before,"},{"Start":"01:41.975 ","End":"01:44.300","Text":"it\u0027s the height rising from the plane,"},{"Start":"01:44.300 ","End":"01:46.630","Text":"we can find it in the same way."},{"Start":"01:46.630 ","End":"01:50.765","Text":"If we want to find r, what we\u0027re going to do is draw that same"},{"Start":"01:50.765 ","End":"01:54.995","Text":"orthogonal or perpendicular line from the red point to the z-axis,"},{"Start":"01:54.995 ","End":"01:56.660","Text":"and that distance is r,"},{"Start":"01:56.660 ","End":"02:01.010","Text":"r is the distance that the point is from the z-axis."},{"Start":"02:01.010 ","End":"02:03.605","Text":"Now, to find the Theta angle,"},{"Start":"02:03.605 ","End":"02:05.210","Text":"what we\u0027re going to do is from the red point,"},{"Start":"02:05.210 ","End":"02:08.080","Text":"we\u0027re going to drop down a line like we did before to the plane,"},{"Start":"02:08.080 ","End":"02:12.570","Text":"and the angle between the r line along this plane,"},{"Start":"02:12.570 ","End":"02:15.435","Text":"the same r you drew at the z-axis before,"},{"Start":"02:15.435 ","End":"02:20.570","Text":"the angle between that and the x-axis from before is your Theta angle,"},{"Start":"02:20.570 ","End":"02:22.400","Text":"and when we find the angle of Theta,"},{"Start":"02:22.400 ","End":"02:30.815","Text":"we can describe the angle of this object in relationship to the x-axis."},{"Start":"02:30.815 ","End":"02:34.790","Text":"If you\u0027re given a point described in Cartesian coordinates,"},{"Start":"02:34.790 ","End":"02:37.145","Text":"you can easily transition into cylindrical,"},{"Start":"02:37.145 ","End":"02:40.340","Text":"and if you\u0027re given a cylindrical coordinate object,"},{"Start":"02:40.340 ","End":"02:44.420","Text":"you can easily transpose it into Cartesian coordinate form."},{"Start":"02:44.420 ","End":"02:46.940","Text":"Basically what you need to do is if you\u0027re given,"},{"Start":"02:46.940 ","End":"02:49.220","Text":"let\u0027s say something described in r,"},{"Start":"02:49.220 ","End":"02:52.715","Text":"Theta, and z, you need to transpose that into x, y, and z."},{"Start":"02:52.715 ","End":"02:54.980","Text":"For example, if we\u0027re do that here,"},{"Start":"02:54.980 ","End":"02:57.775","Text":"the first thing we knew is z is the same z,"},{"Start":"02:57.775 ","End":"02:59.205","Text":"so we can stick with that,"},{"Start":"02:59.205 ","End":"03:06.020","Text":"and to find x, what we do is take our r value and multiply it by cosine of Theta."},{"Start":"03:06.020 ","End":"03:10.250","Text":"The way we derive this is if we transpose the point down to the x,"},{"Start":"03:10.250 ","End":"03:14.495","Text":"y plane, we can draw a line from that point to the x-axis,"},{"Start":"03:14.495 ","End":"03:15.740","Text":"which is going to be at a right angle,"},{"Start":"03:15.740 ","End":"03:19.040","Text":"and that way we know that x over r equals cosine Theta,"},{"Start":"03:19.040 ","End":"03:20.450","Text":"so if we multiply that out,"},{"Start":"03:20.450 ","End":"03:23.510","Text":"we end up with x equaling r times cosine Theta."},{"Start":"03:23.510 ","End":"03:26.120","Text":"To find the y-value, we do a very similar thing."},{"Start":"03:26.120 ","End":"03:27.940","Text":"Y is equal to r sine Theta,"},{"Start":"03:27.940 ","End":"03:31.340","Text":"and the reason for that is because the angle Theta as I just drew there,"},{"Start":"03:31.340 ","End":"03:35.360","Text":"is between r and the right angle towards the y-axis,"},{"Start":"03:35.360 ","End":"03:38.300","Text":"so y over r equals sine Theta,"},{"Start":"03:38.300 ","End":"03:41.260","Text":"and therefore y equals r sine Theta."},{"Start":"03:41.260 ","End":"03:43.855","Text":"Now, if you want to do things the other way around,"},{"Start":"03:43.855 ","End":"03:45.985","Text":"that\u0027s also rather simple."},{"Start":"03:45.985 ","End":"03:51.025","Text":"The easiest way to find r is to think of it in terms of Pythagorean theorem."},{"Start":"03:51.025 ","End":"03:55.240","Text":"R is equal to the square root of x^2 plus y^2"},{"Start":"03:55.240 ","End":"03:59.260","Text":"because if you take that y-axis line and move it forward a little bit,"},{"Start":"03:59.260 ","End":"04:02.185","Text":"you can see it makes a perfect right-angle triangle."},{"Start":"04:02.185 ","End":"04:04.540","Text":"If we\u0027re trying to find the angle of Theta,"},{"Start":"04:04.540 ","End":"04:07.555","Text":"we use tangent of Theta equals y over x, because again,"},{"Start":"04:07.555 ","End":"04:08.815","Text":"with that same triangle,"},{"Start":"04:08.815 ","End":"04:14.515","Text":"you can use the geometrical formulas to find that that is the tangent angle."},{"Start":"04:14.515 ","End":"04:17.320","Text":"You could do it with the sine using r,"},{"Start":"04:17.320 ","End":"04:23.875","Text":"but if you see x as the side next to the angle and y is the side across from it,"},{"Start":"04:23.875 ","End":"04:26.090","Text":"you can use the tangent it\u0027s rather simple."},{"Start":"04:26.090 ","End":"04:29.250","Text":"Now the last important thing is just to realize what values r"},{"Start":"04:29.250 ","End":"04:32.780","Text":"and Theta can take on in a cylindrical coordinate graph."},{"Start":"04:32.780 ","End":"04:36.965","Text":"What we see here is that r can only be positive,"},{"Start":"04:36.965 ","End":"04:38.615","Text":"r can be from 0 to infinity."},{"Start":"04:38.615 ","End":"04:41.630","Text":"The reason for that is we\u0027re going to use the Theta angle to"},{"Start":"04:41.630 ","End":"04:45.440","Text":"position it around the z-axis or around the origin,"},{"Start":"04:45.440 ","End":"04:47.960","Text":"and we don\u0027t need it to take on negative values because"},{"Start":"04:47.960 ","End":"04:50.570","Text":"its data can go from anywhere from 0-2 Pi,"},{"Start":"04:50.570 ","End":"04:53.930","Text":"meaning we can move it entirely around the axis in a full circle."},{"Start":"04:53.930 ","End":"04:56.705","Text":"So that line that I drew there in blue,"},{"Start":"04:56.705 ","End":"04:58.490","Text":"instead of describing with a negative r,"},{"Start":"04:58.490 ","End":"05:03.620","Text":"you can describe with a positive r and the Theta would be something like maybe 1.6 Pi,"},{"Start":"05:03.620 ","End":"05:07.370","Text":"1.7 Pi, you could make it go"},{"Start":"05:07.370 ","End":"05:11.420","Text":"290 maybe that\u0027s 300 degrees around the circle o describe that,"},{"Start":"05:11.420 ","End":"05:13.820","Text":"instead of using a negative r. In general,"},{"Start":"05:13.820 ","End":"05:16.850","Text":"we could describe it as a negative angle of negative 60,"},{"Start":"05:16.850 ","End":"05:20.180","Text":"negative 80, but we don\u0027t need to deal with that right now."},{"Start":"05:20.180 ","End":"05:24.575","Text":"The last step is describing things in terms of a spherical coordinate system."},{"Start":"05:24.575 ","End":"05:26.705","Text":"We\u0027re going to describe things in terms of r,"},{"Start":"05:26.705 ","End":"05:28.945","Text":"Theta, and Phi."},{"Start":"05:28.945 ","End":"05:31.520","Text":"Now, it\u0027s important to know that the difference between"},{"Start":"05:31.520 ","End":"05:37.020","Text":"cylindrical and spherical graphing is that the r is not the same thing."},{"Start":"05:37.020 ","End":"05:40.460","Text":"In spherical graphing, we tend to keep it as the symbol r,"},{"Start":"05:40.460 ","End":"05:44.870","Text":"people tend to describe the r in cylindrical graphing as Rho."},{"Start":"05:44.870 ","End":"05:47.480","Text":"The reason that we\u0027re using the letter r,"},{"Start":"05:47.480 ","End":"05:50.480","Text":"the symbol r for our class is that oftentimes"},{"Start":"05:50.480 ","End":"05:53.750","Text":"in physics we use the symbol Rho to describe density."},{"Start":"05:53.750 ","End":"05:56.180","Text":"You just need to know that we\u0027re not talking about the same thing,"},{"Start":"05:56.180 ","End":"05:57.980","Text":"you\u0027ll see why in a second."},{"Start":"05:57.980 ","End":"06:02.795","Text":"With that said, let\u0027s turn our attention for a minute to spherical coordinates."},{"Start":"06:02.795 ","End":"06:07.190","Text":"The r is different because instead of being measured from the z-axis,"},{"Start":"06:07.190 ","End":"06:08.780","Text":"the r is measured from the origin."},{"Start":"06:08.780 ","End":"06:12.950","Text":"Again, the r is measuring a distance from the origin and not from the z-axis,"},{"Start":"06:12.950 ","End":"06:14.480","Text":"this is the main difference here."},{"Start":"06:14.480 ","End":"06:19.430","Text":"Now, we use that r value to find our Phi value."},{"Start":"06:19.430 ","End":"06:25.380","Text":"Phi is the angle from the r line that is there to the z-axis."},{"Start":"06:25.380 ","End":"06:27.675","Text":"Remember we\u0027re describing the same point,"},{"Start":"06:27.675 ","End":"06:31.940","Text":"and if we drop that point down to the x-y plane like we did before,"},{"Start":"06:31.940 ","End":"06:34.730","Text":"transpose it there, we can find the same Theta angles before,"},{"Start":"06:34.730 ","End":"06:37.990","Text":"Theta is the same as it was before."},{"Start":"06:37.990 ","End":"06:46.445","Text":"One last note is that the line between the origin and the transposed object point,"},{"Start":"06:46.445 ","End":"06:48.470","Text":"we\u0027re going to call that r, x,"},{"Start":"06:48.470 ","End":"06:50.960","Text":"y for our drop down to the x, y plane."},{"Start":"06:50.960 ","End":"06:52.400","Text":"It\u0027s not always called that,"},{"Start":"06:52.400 ","End":"06:55.850","Text":"but it\u0027s really useful for some of the functions as we\u0027ll see below, and of course,"},{"Start":"06:55.850 ","End":"06:57.695","Text":"because this whole thing is a rectangle,"},{"Start":"06:57.695 ","End":"07:03.041","Text":"the same line that\u0027s parallel to it above is also r, x, y."},{"Start":"07:03.041 ","End":"07:05.810","Text":"Again, measuring that length from the z-axis that we would"},{"Start":"07:05.810 ","End":"07:08.870","Text":"normally in cylindrical graphing measure is r,"},{"Start":"07:08.870 ","End":"07:10.130","Text":"we\u0027re going to call here r, x,"},{"Start":"07:10.130 ","End":"07:12.665","Text":"y, and it\u0027s going to be important for us."},{"Start":"07:12.665 ","End":"07:15.710","Text":"In order to find the relationships between"},{"Start":"07:15.710 ","End":"07:21.215","Text":"these spherical and the Cartesian and cylindrical coordinates,"},{"Start":"07:21.215 ","End":"07:24.035","Text":"we\u0027re going to use those things that are equal between some of them."},{"Start":"07:24.035 ","End":"07:27.080","Text":"So between cylindrical and spherical coordinates,"},{"Start":"07:27.080 ","End":"07:29.240","Text":"we know that Theta is the same, we know that r, x,"},{"Start":"07:29.240 ","End":"07:32.645","Text":"y from spherical coordinates is the same as r from cylindrical coordinates."},{"Start":"07:32.645 ","End":"07:34.580","Text":"When we want to find similar things,"},{"Start":"07:34.580 ","End":"07:36.605","Text":"we\u0027re going to use those terms a lot."},{"Start":"07:36.605 ","End":"07:39.410","Text":"For example, if we\u0027re going to Cartesian coordinates,"},{"Start":"07:39.410 ","End":"07:42.155","Text":"if we want to find the value of z,"},{"Start":"07:42.155 ","End":"07:45.170","Text":"we know that it\u0027s on a right angle with the line r, x,"},{"Start":"07:45.170 ","End":"07:48.715","Text":"y, and we can use the angle Phi to find that."},{"Start":"07:48.715 ","End":"07:51.875","Text":"Of course, we know Phi it\u0027s given in our spherical coordinates."},{"Start":"07:51.875 ","End":"07:56.975","Text":"We know that our z over r is going to equal the cosine of Phi."},{"Start":"07:56.975 ","End":"08:03.785","Text":"Because again, you can see the angle Phi is adjacent to r and adjacent to z,"},{"Start":"08:03.785 ","End":"08:05.300","Text":"so now if we multiply that out,"},{"Start":"08:05.300 ","End":"08:08.590","Text":"we get z equals r cosine Phi."},{"Start":"08:08.590 ","End":"08:10.120","Text":"Now that we have z,"},{"Start":"08:10.120 ","End":"08:12.460","Text":"we should go and find x and y, of course."},{"Start":"08:12.460 ","End":"08:15.475","Text":"Where we\u0027re going to do that is find the value first of r_xy,"},{"Start":"08:15.475 ","End":"08:19.495","Text":"and we\u0027re going to use the exact same way we used r in our cylindrical format earlier."},{"Start":"08:19.495 ","End":"08:23.890","Text":"To find r_xy, we\u0027re going to do a very similar operation that we just did with z."},{"Start":"08:23.890 ","End":"08:28.900","Text":"We know that r_xy is across from the angle Phi,"},{"Start":"08:28.900 ","End":"08:36.205","Text":"so we know that r_xy over r equals sine of Phi."},{"Start":"08:36.205 ","End":"08:40.180","Text":"If we multiply that out, we get r_xy equals r sine of Phi."},{"Start":"08:40.180 ","End":"08:44.815","Text":"What we can do is transpose that r_xy down below onto the xy-plane,"},{"Start":"08:44.815 ","End":"08:46.615","Text":"which is where the name comes from again."},{"Start":"08:46.615 ","End":"08:51.580","Text":"We can do the exact same functions that we did before in our cylindrical coordinates."},{"Start":"08:51.580 ","End":"08:56.020","Text":"We know that x as it was before equal to r cosine of Theta,"},{"Start":"08:56.020 ","End":"08:59.350","Text":"now we know that it\u0027s equal to r_xy cosine of Theta."},{"Start":"08:59.350 ","End":"09:02.830","Text":"If we take that and extrapolate it to the next step,"},{"Start":"09:02.830 ","End":"09:09.715","Text":"we know that r_xy is actually equal to r sine of Phi,"},{"Start":"09:09.715 ","End":"09:13.855","Text":"so r sine of Phi times cosine of Theta equals x."},{"Start":"09:13.855 ","End":"09:17.740","Text":"Similarly for the y, we do the same thing that we did before,"},{"Start":"09:17.740 ","End":"09:20.410","Text":"r sine of Theta equaled y,"},{"Start":"09:20.410 ","End":"09:25.300","Text":"so now r_xy sine of Theta equals y. Extrapolating that to the next step,"},{"Start":"09:25.300 ","End":"09:29.485","Text":"r sine of Phi times r sine of Theta equals y,"},{"Start":"09:29.485 ","End":"09:33.865","Text":"giving us both x and y in terms of our polar coordinates."},{"Start":"09:33.865 ","End":"09:36.655","Text":"Now that you can take spherical coordinates"},{"Start":"09:36.655 ","End":"09:38.680","Text":"and transpose them into Cartesian coordinates,"},{"Start":"09:38.680 ","End":"09:41.440","Text":"it\u0027s useful to know how to do things the other way around so you really can"},{"Start":"09:41.440 ","End":"09:45.400","Text":"describe things any way you want in any way you need in terms of a problem you\u0027re given."},{"Start":"09:45.400 ","End":"09:48.130","Text":"You need to know how to take something given to you in x, y,"},{"Start":"09:48.130 ","End":"09:52.345","Text":"and z terms and describe it in terms of r, Theta, and Phi."},{"Start":"09:52.345 ","End":"09:55.810","Text":"The first thing we can start with is r. We know that r is"},{"Start":"09:55.810 ","End":"09:59.530","Text":"the hypotenuse of the triangle z, r, r _xy."},{"Start":"09:59.530 ","End":"10:03.040","Text":"R_xy again is our value r from the cylindrical format."},{"Start":"10:03.040 ","End":"10:10.105","Text":"The way that we find the value of r in spherical terms is we do a Pythagorean theorem."},{"Start":"10:10.105 ","End":"10:17.380","Text":"We take r as equal to the square root of r_xy^2 and z^2."},{"Start":"10:17.380 ","End":"10:20.560","Text":"Now if you recall from the cylindrical model,"},{"Start":"10:20.560 ","End":"10:22.315","Text":"the cylindrical coordinates,"},{"Start":"10:22.315 ","End":"10:27.835","Text":"r_xy is actually the hypotenuse of the triangle x, y, r_xy."},{"Start":"10:27.835 ","End":"10:33.340","Text":"We can describe that instead of as r_xy^2, as x^2+y^2."},{"Start":"10:33.340 ","End":"10:36.355","Text":"We can extrapolate that out as r,"},{"Start":"10:36.355 ","End":"10:38.095","Text":"in our spherical format,"},{"Start":"10:38.095 ","End":"10:41.675","Text":"equaling the square root of x^2+y^2,"},{"Start":"10:41.675 ","End":"10:47.250","Text":"which is the same as r_xy^2+z^2."},{"Start":"10:47.250 ","End":"10:53.975","Text":"In the end, what we come out to simply put is r equals the square root of x^2+y^2+z^2."},{"Start":"10:53.975 ","End":"10:55.990","Text":"Now to further prove to you that the r in"},{"Start":"10:55.990 ","End":"10:59.050","Text":"spherical format and the r in cylindrical format are not the same,"},{"Start":"10:59.050 ","End":"11:03.115","Text":"you can see that they\u0027re equal to different things in terms of Cartesian coordinates."},{"Start":"11:03.115 ","End":"11:08.170","Text":"The r in cylindrical format is equal to the square root of x^2+y^2,"},{"Start":"11:08.170 ","End":"11:13.690","Text":"and the r in spherical form is equal to square root of x^2+y^2+z^2."},{"Start":"11:13.690 ","End":"11:17.230","Text":"Now, the Theta stays the same, so it\u0027s easy to find."},{"Start":"11:17.230 ","End":"11:20.650","Text":"Tangent of Theta is still equal to y/x, that hasn\u0027t changed."},{"Start":"11:20.650 ","End":"11:24.400","Text":"But to find Phi, we\u0027re going to actually use the r that we just found."},{"Start":"11:24.400 ","End":"11:29.635","Text":"We know that we can use the cosine of Phi to find it because"},{"Start":"11:29.635 ","End":"11:36.220","Text":"r and z are both adjacent to the Phi angle."},{"Start":"11:36.220 ","End":"11:39.850","Text":"We know that the cosine of Phi equals z/r,"},{"Start":"11:39.850 ","End":"11:42.490","Text":"or more accurately, z/r equals cosine Phi."},{"Start":"11:42.490 ","End":"11:45.415","Text":"We can simplify that or really extrapolate that out by"},{"Start":"11:45.415 ","End":"11:48.505","Text":"making r in terms of Cartesian coordinates that we found above."},{"Start":"11:48.505 ","End":"11:50.155","Text":"Instead this being z over r,"},{"Start":"11:50.155 ","End":"11:56.530","Text":"we\u0027re going to say cosine of Phi is equal to z over the square root of x^2+y^2+z^2."},{"Start":"11:56.530 ","End":"12:01.240","Text":"Now we can describe all of our spherical coordinates in terms of Cartesian values."},{"Start":"12:01.240 ","End":"12:05.410","Text":"The next thing to do is describe how these values work."},{"Start":"12:05.410 ","End":"12:07.480","Text":"R is the same as the r from"},{"Start":"12:07.480 ","End":"12:11.440","Text":"the cylindrical coordinates in the sense that it can only be positive."},{"Start":"12:11.440 ","End":"12:13.060","Text":"It goes from 0 to infinity,"},{"Start":"12:13.060 ","End":"12:16.510","Text":"but it must be positive for a similar reason as before."},{"Start":"12:16.510 ","End":"12:18.760","Text":"Theta again is going to stay as it was."},{"Start":"12:18.760 ","End":"12:20.050","Text":"It goes from 0 to 2Pi."},{"Start":"12:20.050 ","End":"12:24.790","Text":"It really functions identically to the way it functions in a cylindrical coordinates."},{"Start":"12:24.790 ","End":"12:27.610","Text":"Phi only from 0 to Pi."},{"Start":"12:27.610 ","End":"12:29.350","Text":"Now I\u0027m sure that you\u0027re asking yourself,"},{"Start":"12:29.350 ","End":"12:33.730","Text":"why does Theta go from 0 to 2P and Phi only goes from 0 to Pi."},{"Start":"12:33.730 ","End":"12:35.320","Text":"I\u0027ll describe that in a second,"},{"Start":"12:35.320 ","End":"12:37.930","Text":"that\u0027s our last subject for this lecture."},{"Start":"12:37.930 ","End":"12:41.860","Text":"We\u0027re faced with an issue here where we can describe things"},{"Start":"12:41.860 ","End":"12:45.025","Text":"in terms of Phi being from 0 to 2Pi."},{"Start":"12:45.025 ","End":"12:47.035","Text":"We could do that if we wanted,"},{"Start":"12:47.035 ","End":"12:51.880","Text":"but it would cause a problem for us in terms of a double representation."},{"Start":"12:51.880 ","End":"12:54.460","Text":"What I mean by that is the following."},{"Start":"12:54.460 ","End":"12:59.270","Text":"If we have Phi going all the way around to any point,"},{"Start":"13:00.440 ","End":"13:04.200","Text":"0 to 2Pi, we can describe the same point in"},{"Start":"13:04.200 ","End":"13:07.545","Text":"terms of Phi being a value between 1Pi and 2Pi,"},{"Start":"13:07.545 ","End":"13:10.245","Text":"or in terms of Theta being a value between 1Pi and 2Pi."},{"Start":"13:10.245 ","End":"13:12.600","Text":"I\u0027ll illustrate what I mean here with an example."},{"Start":"13:12.600 ","End":"13:16.650","Text":"Let\u0027s say we want to describe an object at this red point,"},{"Start":"13:16.650 ","End":"13:20.370","Text":"which can be described 1 way as taking"},{"Start":"13:20.370 ","End":"13:24.415","Text":"Phi and giving it a value of something like 1.7 Pi."},{"Start":"13:24.415 ","End":"13:28.240","Text":"Let\u0027s say it\u0027s 30 degrees from the z-axis."},{"Start":"13:28.240 ","End":"13:31.705","Text":"That would be 330 degrees around."},{"Start":"13:31.705 ","End":"13:34.720","Text":"We could describe it that way and then have our r be positive"},{"Start":"13:34.720 ","End":"13:38.575","Text":"and have a minimal Theta value and be fine."},{"Start":"13:38.575 ","End":"13:41.830","Text":"Now the problem here is that we can also describe"},{"Start":"13:41.830 ","End":"13:49.360","Text":"the same point with keeping our Phi at 30 degrees instead of 330 degrees."},{"Start":"13:49.360 ","End":"13:53.560","Text":"If we just keep our Phi at 30 degrees,"},{"Start":"13:53.560 ","End":"13:58.060","Text":"we keep our r in the same place,"},{"Start":"13:58.060 ","End":"14:00.805","Text":"we can move like that."},{"Start":"14:00.805 ","End":"14:09.610","Text":"When we give our Theta some value of around 330 degrees,"},{"Start":"14:09.610 ","End":"14:12.970","Text":"we bring our point to that same place there,"},{"Start":"14:12.970 ","End":"14:14.410","Text":"I guess it\u0027s 270 degrees,"},{"Start":"14:14.410 ","End":"14:19.945","Text":"to that same place there without having to use Phi value that\u0027s greater than 1Pi."},{"Start":"14:19.945 ","End":"14:22.270","Text":"Now, this seems like it would be an advantage,"},{"Start":"14:22.270 ","End":"14:23.560","Text":"we can describe things in multiple ways,"},{"Start":"14:23.560 ","End":"14:28.405","Text":"but actually, it gives us a problem when we\u0027re trying to skim out integrals."},{"Start":"14:28.405 ","End":"14:33.940","Text":"An integral is essentially the sum of different locations, as you\u0027ll see later."},{"Start":"14:33.940 ","End":"14:37.840","Text":"When we do this, we\u0027re going to be counting each point twice,"},{"Start":"14:37.840 ","End":"14:40.795","Text":"and this can give us a lot of problems going forward."},{"Start":"14:40.795 ","End":"14:42.805","Text":"Just to illustrate this further,"},{"Start":"14:42.805 ","End":"14:49.420","Text":"you can see here how if our Phi is limited to being up to 1Pi and greater,"},{"Start":"14:49.420 ","End":"14:51.160","Text":"we keep it at an angle of 30."},{"Start":"14:51.160 ","End":"14:53.680","Text":"We can have our Theta going round up to 2Pi,"},{"Start":"14:53.680 ","End":"14:55.270","Text":"so it can make a full circle,"},{"Start":"14:55.270 ","End":"14:59.365","Text":"that we were only counting that point once."},{"Start":"14:59.365 ","End":"15:03.760","Text":"Same is true here, if we have a greater angle for Phi."},{"Start":"15:03.760 ","End":"15:07.960","Text":"The same goes true as you get lower and lower and lower on our graph here."},{"Start":"15:07.960 ","End":"15:13.165","Text":"But if we could have Phi given a value greater than 1Pi,"},{"Start":"15:13.165 ","End":"15:20.095","Text":"say that value there where it\u0027s going more than halfway around the circle,"},{"Start":"15:20.095 ","End":"15:22.210","Text":"we\u0027re going to count these same points twice,"},{"Start":"15:22.210 ","End":"15:23.290","Text":"and this really is problematic,"},{"Start":"15:23.290 ","End":"15:27.950","Text":"we don\u0027t want to be over-counting our points when we\u0027re making an integral."},{"Start":"15:28.320 ","End":"15:34.525","Text":"Let it suffice to say for now that we\u0027re just going to have Phi go from 0 to 1Pi,"},{"Start":"15:34.525 ","End":"15:37.240","Text":"and Theta go from 0 up to 2Pi,"},{"Start":"15:37.240 ","End":"15:42.535","Text":"meaning that Phi will be responsible for smaller angles up to 180 degrees,"},{"Start":"15:42.535 ","End":"15:47.800","Text":"whereas Theta can go up to 360 degrees and go all the way around our origin,"},{"Start":"15:47.800 ","End":"15:50.120","Text":"all the way around the z-axis."},{"Start":"15:50.340 ","End":"15:52.600","Text":"We\u0027ll leave it at that for now."},{"Start":"15:52.600 ","End":"15:53.740","Text":"If you remember that,"},{"Start":"15:53.740 ","End":"15:55.960","Text":"things will be simple and we can perform any function we"},{"Start":"15:55.960 ","End":"15:59.305","Text":"want within spherical coordinates."},{"Start":"15:59.305 ","End":"16:04.180","Text":"1 last point to mention is that other lectures, other textbooks,"},{"Start":"16:04.180 ","End":"16:08.680","Text":"other people you come across may change Theta and Phi from time to time,"},{"Start":"16:08.680 ","End":"16:10.930","Text":"meaning that everything that we just wrote,"},{"Start":"16:10.930 ","End":"16:13.990","Text":"we would switch our Theta and our Phi in terms of their places."},{"Start":"16:13.990 ","End":"16:16.930","Text":"Phi would describe the angle on the xy-plane,"},{"Start":"16:16.930 ","End":"16:21.560","Text":"and Theta would describe the angle between r and the z-axis."},{"Start":"16:21.560 ","End":"16:23.150","Text":"Of course, as a result,"},{"Start":"16:23.150 ","End":"16:26.885","Text":"your Theta would only go from 0 to 1Pi and Phi would go from 0 to 2Pi,"},{"Start":"16:26.885 ","End":"16:28.400","Text":"and all the angles below,"},{"Start":"16:28.400 ","End":"16:33.300","Text":"and our formulas here would be switched. That ends our lecture."}],"ID":10625},{"Watched":false,"Name":"Calculating Center of Mass of Large Body","Duration":"8m 47s","ChapterTopicVideoID":10421,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.685","Text":"Hello. In previous lessons,"},{"Start":"00:02.685 ","End":"00:09.270","Text":"we worked out how to find the center of mass of a system of many point masses."},{"Start":"00:09.270 ","End":"00:13.950","Text":"What we\u0027re going to learn in this lesson is how to find the center of mass of"},{"Start":"00:13.950 ","End":"00:18.600","Text":"a large body where the mass is spread out throughout this large body,"},{"Start":"00:18.600 ","End":"00:22.530","Text":"or the center of mass of a rigid body, in other words."},{"Start":"00:22.530 ","End":"00:26.025","Text":"Let\u0027s remind ourselves of how to find the center of mass in"},{"Start":"00:26.025 ","End":"00:29.400","Text":"the x direction of a system of point masses."},{"Start":"00:29.400 ","End":"00:33.347","Text":"We have that the x center of mass was equal to,"},{"Start":"00:33.347 ","End":"00:39.545","Text":"and then we did the sum on i of the mass"},{"Start":"00:39.545 ","End":"00:46.400","Text":"of the individual point mass multiplied by its position in the x direction,"},{"Start":"00:46.400 ","End":"00:53.835","Text":"divided by the sum of i of all of the point masses."},{"Start":"00:53.835 ","End":"01:04.570","Text":"This section in the denominator is just the total mass of the system."},{"Start":"01:07.940 ","End":"01:12.305","Text":"What do we do if we have a large body?"},{"Start":"01:12.305 ","End":"01:17.035","Text":"Let\u0027s say that we have half of a disc."},{"Start":"01:17.035 ","End":"01:20.970","Text":"Ignore my poor drawing skills."},{"Start":"01:20.970 ","End":"01:25.215","Text":"Here\u0027s a large body,"},{"Start":"01:25.215 ","End":"01:30.110","Text":"half disc, and then we want to find out where the center of mass is."},{"Start":"01:30.110 ","End":"01:33.845","Text":"The first thing that we have to deal with in these type of questions"},{"Start":"01:33.845 ","End":"01:38.945","Text":"is to choose which point we are taking as our origin,"},{"Start":"01:38.945 ","End":"01:42.920","Text":"and then we\u0027re going to find the center of mass relative to that origin."},{"Start":"01:42.920 ","End":"01:47.300","Text":"Let\u0027s say that we\u0027re going to take our center of mass from this point over here,"},{"Start":"01:47.300 ","End":"01:51.370","Text":"the center of our disk."},{"Start":"01:51.680 ","End":"01:56.315","Text":"Now what we\u0027re going to do is we\u0027re going to split up our disk"},{"Start":"01:56.315 ","End":"02:02.145","Text":"into many little small pieces,"},{"Start":"02:02.145 ","End":"02:06.355","Text":"and then we\u0027ll see that the distance over here,"},{"Start":"02:06.355 ","End":"02:11.175","Text":"this distance here is our xi,"},{"Start":"02:11.175 ","End":"02:14.370","Text":"its value along the x-axis."},{"Start":"02:14.370 ","End":"02:18.880","Text":"This piece has mass dm."},{"Start":"02:19.180 ","End":"02:25.370","Text":"It\u0027s a small part of this large mass."},{"Start":"02:25.370 ","End":"02:29.270","Text":"Then we can see that our equation over here becomes in"},{"Start":"02:29.270 ","End":"02:33.770","Text":"the numerator and integral along its x coordinate,"},{"Start":"02:33.770 ","End":"02:39.305","Text":"so xi multiplied by that element of mass,"},{"Start":"02:39.305 ","End":"02:42.826","Text":"so multiplied by dm."},{"Start":"02:42.826 ","End":"02:49.300","Text":"Then it\u0027s going to be divided by an integral on dm,"},{"Start":"02:49.300 ","End":"02:53.210","Text":"which of course is still the total mass of the system,"},{"Start":"02:53.210 ","End":"02:56.360","Text":"we\u0027re just adding up all the masses of these dm\u0027s,"},{"Start":"02:56.360 ","End":"03:00.460","Text":"which will equal the total mass of our half disk."},{"Start":"03:00.460 ","End":"03:04.035","Text":"Now of course, this i over here is arbitrary,"},{"Start":"03:04.035 ","End":"03:10.940","Text":"we only wrote over here i because we were doing summation on all of our i\u0027s."},{"Start":"03:10.940 ","End":"03:12.605","Text":"But when we\u0027re integrating,"},{"Start":"03:12.605 ","End":"03:16.070","Text":"this doesn\u0027t need to be here anymore."},{"Start":"03:16.070 ","End":"03:20.010","Text":"We\u0027re integrating along xdm."},{"Start":"03:21.140 ","End":"03:25.250","Text":"This is the equation which is important to remember."},{"Start":"03:25.250 ","End":"03:31.660","Text":"Now the next thing that we have to do is we have to figure out what our dm is equal to."},{"Start":"03:31.660 ","End":"03:34.005","Text":"Let\u0027s write it over here."},{"Start":"03:34.005 ","End":"03:41.980","Text":"Our dm or element of mass is always going to be equal to 1 of 3 things."},{"Start":"03:41.980 ","End":"03:46.500","Text":"If we\u0027re dealing with 1d, a line,"},{"Start":"03:46.500 ","End":"03:54.840","Text":"so we\u0027re always going to have Lambda dl,"},{"Start":"03:54.840 ","End":"04:01.860","Text":"where Lambda is the mass density per unit length multiplied by the total length."},{"Start":"04:01.860 ","End":"04:03.926","Text":"If we\u0027re dealing with 2d,"},{"Start":"04:03.926 ","End":"04:08.365","Text":"we\u0027re going to have Sigma ds,"},{"Start":"04:08.365 ","End":"04:15.270","Text":"where Sigma is the density per unit area multiplied by the area."},{"Start":"04:15.270 ","End":"04:16.965","Text":"If we\u0027re dealing with 3d,"},{"Start":"04:16.965 ","End":"04:21.075","Text":"so then we\u0027re going to have Rho dv,"},{"Start":"04:21.075 ","End":"04:30.330","Text":"where Rho is the mass per unit volume multiplied by the volume."},{"Start":"04:31.640 ","End":"04:37.515","Text":"Now let\u0027s see how we can work out our Lambda Sigma and Rho are equal to."},{"Start":"04:37.515 ","End":"04:41.370","Text":"Sometimes in the question will be given the values,"},{"Start":"04:41.370 ","End":"04:44.245","Text":"but if we\u0027re not, then we can work it out."},{"Start":"04:44.245 ","End":"04:50.540","Text":"If we\u0027re dealing with a mass which is evenly distributed,"},{"Start":"04:50.540 ","End":"04:59.780","Text":"so then our Lambda is going to be equal to the total mass divided by the total length."},{"Start":"04:59.780 ","End":"05:01.895","Text":"Our Sigma, again,"},{"Start":"05:01.895 ","End":"05:05.135","Text":"only if our mass is evenly distributed,"},{"Start":"05:05.135 ","End":"05:10.355","Text":"is going to be the total mass divided by the total area,"},{"Start":"05:10.355 ","End":"05:13.210","Text":"and our Rho again,"},{"Start":"05:13.210 ","End":"05:16.925","Text":"if the mass is evenly distributed across the whole body,"},{"Start":"05:16.925 ","End":"05:22.340","Text":"is going to be the total mass divided by the total volume."},{"Start":"05:24.170 ","End":"05:28.380","Text":"Now let\u0027s talk about how we can work out what the dl,"},{"Start":"05:28.380 ","End":"05:30.060","Text":"ds or dv is."},{"Start":"05:30.060 ","End":"05:32.780","Text":"This is obviously a unit length,"},{"Start":"05:32.780 ","End":"05:36.250","Text":"unit area, and unit volume."},{"Start":"05:36.250 ","End":"05:42.280","Text":"Let\u0027s see how we work out what our dl is equal to."},{"Start":"05:42.280 ","End":"05:46.580","Text":"Our unit length is equal to if we\u0027re working in Cartesian,"},{"Start":"05:46.580 ","End":"05:50.165","Text":"so it can let say be an example is dx,"},{"Start":"05:50.165 ","End":"05:55.410","Text":"it can also be dy if we\u0027re working in Cartesian coordinates."},{"Start":"05:55.410 ","End":"05:58.854","Text":"If however we\u0027re working in polar coordinates,"},{"Start":"05:58.854 ","End":"06:02.210","Text":"it can either be rd Theta,"},{"Start":"06:02.210 ","End":"06:07.425","Text":"or alternatively, it can just be di,"},{"Start":"06:07.425 ","End":"06:11.050","Text":"so this is in polar coordinates."},{"Start":"06:11.720 ","End":"06:14.615","Text":"Next we have ds,"},{"Start":"06:14.615 ","End":"06:17.863","Text":"which is our unit of area."},{"Start":"06:17.863 ","End":"06:20.850","Text":"Again, if we\u0027re using Cartesian,"},{"Start":"06:20.850 ","End":"06:22.970","Text":"this can be dx,"},{"Start":"06:22.970 ","End":"06:29.865","Text":"dy unit of area or if we\u0027re dealing with polar coordinates."},{"Start":"06:29.865 ","End":"06:39.035","Text":"This can be rdrd Theta and if we\u0027re using spherical coordinates,"},{"Start":"06:39.035 ","End":"06:44.750","Text":"this will be equal to r^2 sin of"},{"Start":"06:44.750 ","End":"06:51.840","Text":"Phi d Theta d Phi."},{"Start":"06:51.840 ","End":"06:54.615","Text":"Lastly, we have our dv,"},{"Start":"06:54.615 ","End":"06:57.615","Text":"so our dv is our unit of volume,"},{"Start":"06:57.615 ","End":"07:00.260","Text":"so if we\u0027re dealing with Cartesian coordinates,"},{"Start":"07:00.260 ","End":"07:06.440","Text":"our dv will be dx, dy, dz."},{"Start":"07:06.440 ","End":"07:09.635","Text":"If we\u0027re dealing with cylindrical coordinates,"},{"Start":"07:09.635 ","End":"07:16.660","Text":"this will be equal to rdrd Theta dz."},{"Start":"07:16.660 ","End":"07:19.714","Text":"If we\u0027re working with spherical coordinates,"},{"Start":"07:19.714 ","End":"07:24.350","Text":"this will be equal to r^2 sin"},{"Start":"07:24.350 ","End":"07:31.230","Text":"Phi drd Theta d Phi."},{"Start":"07:32.810 ","End":"07:37.005","Text":"This is the basis of the coordinates."},{"Start":"07:37.005 ","End":"07:40.390","Text":"If you want to go over this in a little bit more detail,"},{"Start":"07:40.390 ","End":"07:43.639","Text":"there\u0027s in the mathematical section,"},{"Start":"07:43.639 ","End":"07:44.960","Text":"we go into detail of how"},{"Start":"07:44.960 ","End":"07:51.725","Text":"to convert between Cartesian and polar and spherical coordinates."},{"Start":"07:51.725 ","End":"07:55.130","Text":"Now, all we have to do in order to find the center of mass of"},{"Start":"07:55.130 ","End":"07:59.270","Text":"the large body is we plug in let\u0027s say here we\u0027re dealing"},{"Start":"07:59.270 ","End":"08:06.545","Text":"with this x coordinate and then dm here because we\u0027re working with area,"},{"Start":"08:06.545 ","End":"08:12.245","Text":"it will be the total mass of the system divided by the total area."},{"Start":"08:12.245 ","End":"08:13.940","Text":"The total area of half a disc,"},{"Start":"08:13.940 ","End":"08:16.198","Text":"you know how to work out,"},{"Start":"08:16.198 ","End":"08:18.020","Text":"and then you multiply it by dm,"},{"Start":"08:18.020 ","End":"08:22.270","Text":"where dm here will be Sigma ds."},{"Start":"08:22.270 ","End":"08:26.015","Text":"The Sigma is this multiplied by ds,"},{"Start":"08:26.015 ","End":"08:32.900","Text":"which here I would suggest using the rdrd Theta, the polar coordinates."},{"Start":"08:32.900 ","End":"08:36.110","Text":"Then we can see that because there\u0027s a drd Theta,"},{"Start":"08:36.110 ","End":"08:37.355","Text":"so it\u0027s a double integrals."},{"Start":"08:37.355 ","End":"08:44.840","Text":"You would add another integral into here and then the dm would simply be the total mass."},{"Start":"08:44.840 ","End":"08:47.940","Text":"That\u0027s the end of this lesson."}],"ID":10777},{"Watched":false,"Name":"Exercise- Center Of Mass Of Rod","Duration":"4m 3s","ChapterTopicVideoID":9088,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.330 ","End":"00:02.995","Text":"Hello. In this question,"},{"Start":"00:02.995 ","End":"00:08.320","Text":"we\u0027re being told to find the center of mass of a non-homogenous rod of"},{"Start":"00:08.320 ","End":"00:12.280","Text":"length L. What is given to us in the question is"},{"Start":"00:12.280 ","End":"00:17.680","Text":"length L and we\u0027re being told that the density changes as a function of x,"},{"Start":"00:17.680 ","End":"00:21.040","Text":"by the equation of Lambda 0,"},{"Start":"00:21.040 ","End":"00:24.720","Text":"x divided by L, which is our length."},{"Start":"00:24.720 ","End":"00:28.210","Text":"Let\u0027s see how to solve this question."},{"Start":"00:28.880 ","End":"00:34.975","Text":"Now what we\u0027re going to do is we\u0027re going to take some arbitrary length,"},{"Start":"00:34.975 ","End":"00:40.615","Text":"which we\u0027re going to say is our x and this is what is varying."},{"Start":"00:40.615 ","End":"00:46.505","Text":"Then we can say that we\u0027re summing along"},{"Start":"00:46.505 ","End":"00:55.240","Text":"small segments of the length and we\u0027re going to say that these segments are of length dx,"},{"Start":"00:55.240 ","End":"00:57.465","Text":"this is our x."},{"Start":"00:57.465 ","End":"01:03.745","Text":"Now let\u0027s remind ourselves of our equation for our center of mass when using integration."},{"Start":"01:03.745 ","End":"01:08.600","Text":"We\u0027re going to say that our x center of mass is equal to the integral"},{"Start":"01:08.600 ","End":"01:14.960","Text":"of x_dm divided by the integral of dm."},{"Start":"01:14.960 ","End":"01:17.780","Text":"If we remember our dm\u0027s,"},{"Start":"01:17.780 ","End":"01:23.350","Text":"that we have the options of using was Lambda dl,"},{"Start":"01:23.350 ","End":"01:28.665","Text":"Sigma ds, and rho dv."},{"Start":"01:28.665 ","End":"01:35.620","Text":"We are now going to use our Lambda dl because we\u0027re using the unit length."},{"Start":"01:35.620 ","End":"01:42.455","Text":"Therefore, our dm are changing mass is going to be equal to Lambda,"},{"Start":"01:42.455 ","End":"01:44.360","Text":"and then our dl,"},{"Start":"01:44.360 ","End":"01:45.620","Text":"as we said before,"},{"Start":"01:45.620 ","End":"01:53.290","Text":"because we\u0027re working on the x-axis so we\u0027re going to say that it is equal to Lambda dx,"},{"Start":"01:53.290 ","End":"01:59.920","Text":"so density per unit length multiplied by our length."},{"Start":"02:00.350 ","End":"02:04.850","Text":"Now the next thing that we can do here is substitute in our Lambda."},{"Start":"02:04.850 ","End":"02:06.755","Text":"Our Lambda is given in the question,"},{"Start":"02:06.755 ","End":"02:14.890","Text":"equals Lambda 0 multiplied by x divided by L and then dx."},{"Start":"02:14.890 ","End":"02:19.460","Text":"That means that our equation for our xcm is going to be"},{"Start":"02:19.460 ","End":"02:23.785","Text":"equal to the integral of x multiplied by our dm,"},{"Start":"02:23.785 ","End":"02:26.200","Text":"which is Lambda 0,"},{"Start":"02:26.200 ","End":"02:33.860","Text":"x divided by Ldx divided by our integral of just our dm,"},{"Start":"02:33.860 ","End":"02:37.715","Text":"which is just Lambda 0 x divided by L dx."},{"Start":"02:37.715 ","End":"02:46.560","Text":"Our dm is just the small masses."},{"Start":"02:46.970 ","End":"02:56.220","Text":"Every unit length of the rod has a mass so we\u0027re summing up these small masses."},{"Start":"02:56.220 ","End":"03:01.340","Text":"Now let\u0027s see what the bounds of the integral is."},{"Start":"03:01.340 ","End":"03:04.970","Text":"We\u0027re summing up our integral between the bounds of"},{"Start":"03:04.970 ","End":"03:09.380","Text":"0 until our total length of L so it\u0027s from"},{"Start":"03:09.380 ","End":"03:13.340","Text":"0 to L and the same in the denominator from 0 to"},{"Start":"03:13.340 ","End":"03:19.595","Text":"L. Now we\u0027ll notice that our Lambda 0 crosses out in both sides."},{"Start":"03:19.595 ","End":"03:23.975","Text":"Then it\u0027s just a simple integration from here."},{"Start":"03:23.975 ","End":"03:26.030","Text":"So over here,"},{"Start":"03:26.030 ","End":"03:30.465","Text":"we will have L^3 divided by 3L,"},{"Start":"03:30.465 ","End":"03:32.630","Text":"and in the denominator,"},{"Start":"03:32.630 ","End":"03:36.770","Text":"we\u0027ll have L^2 divided by 2L."},{"Start":"03:36.770 ","End":"03:39.710","Text":"Then when we just rearrange this,"},{"Start":"03:39.710 ","End":"03:48.280","Text":"we\u0027ll have (2/3)L and this is the final answer."},{"Start":"03:49.100 ","End":"03:55.550","Text":"What that means is 2/3 of the way through the rod will have our center of mass."},{"Start":"03:55.550 ","End":"03:59.915","Text":"So around about here will be our center of mass."},{"Start":"03:59.915 ","End":"04:03.749","Text":"That\u0027s the end of this question."}],"ID":9361},{"Watched":false,"Name":"Exercise- Triangles Center of Mass","Duration":"12m 33s","ChapterTopicVideoID":9089,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:03.000","Text":"Hello. In this lesson,"},{"Start":"00:03.000 ","End":"00:06.915","Text":"we\u0027re going to find the triangle\u0027s center of mass."},{"Start":"00:06.915 ","End":"00:09.030","Text":"How are we going to do this?"},{"Start":"00:09.030 ","End":"00:14.205","Text":"Here we have that the side over here is of length a."},{"Start":"00:14.205 ","End":"00:17.100","Text":"We have that the side at the bottom is of length b and"},{"Start":"00:17.100 ","End":"00:21.580","Text":"obviously the angle between them is 90 degrees."},{"Start":"00:21.980 ","End":"00:24.930","Text":"Our equation for the center of mass,"},{"Start":"00:24.930 ","End":"00:27.225","Text":"if you remember, is x_cm,"},{"Start":"00:27.225 ","End":"00:31.365","Text":"which is equal to 1 over the total mass,"},{"Start":"00:31.365 ","End":"00:37.825","Text":"m_total and then the integral of x dm."},{"Start":"00:37.825 ","End":"00:45.320","Text":"Because here we\u0027re dealing with area so the integral is going to be a double integral."},{"Start":"00:45.320 ","End":"00:52.290","Text":"Let\u0027s see how we apply this equation into our question over here."},{"Start":"00:52.290 ","End":"00:56.060","Text":"What is going to be our dm?"},{"Start":"00:56.060 ","End":"00:58.229","Text":"Because we\u0027re dealing with area,"},{"Start":"00:58.229 ","End":"01:03.815","Text":"the easiest way to do where the right-angled triangle is by using Cartesian coordinates."},{"Start":"01:03.815 ","End":"01:06.799","Text":"When we\u0027re using Cartesian coordinates,"},{"Start":"01:06.799 ","End":"01:16.475","Text":"dm in Cartesian becomes dxdy and of course we have to multiply it by Sigma,"},{"Start":"01:16.475 ","End":"01:21.088","Text":"which is our density,"},{"Start":"01:21.088 ","End":"01:23.820","Text":"so our mass per area."},{"Start":"01:23.820 ","End":"01:27.230","Text":"How we\u0027re going to do this, because our Sigma is a constant,"},{"Start":"01:27.230 ","End":"01:29.645","Text":"we can move it out of the integration."},{"Start":"01:29.645 ","End":"01:37.400","Text":"I\u0027ll have Sigma divided by our total mass of the system and then double integral and"},{"Start":"01:37.400 ","End":"01:40.400","Text":"we\u0027re going to set the bounds in a second and then we\u0027re going to"},{"Start":"01:40.400 ","End":"01:46.200","Text":"multiply this by x and then dxdy."},{"Start":"01:47.870 ","End":"01:53.480","Text":"The complicated bit here is to put in the correct bounds"},{"Start":"01:53.480 ","End":"01:59.645","Text":"so what do we have to decide is if we\u0027re going to go along the x-axis,"},{"Start":"01:59.645 ","End":"02:02.705","Text":"let\u0027s say from 0 until b."},{"Start":"02:02.705 ","End":"02:06.320","Text":"Then when we\u0027re going to go along the y-axis,"},{"Start":"02:06.320 ","End":"02:08.660","Text":"we\u0027re not integrating on full square,"},{"Start":"02:08.660 ","End":"02:10.945","Text":"we\u0027re integrating on the triangle."},{"Start":"02:10.945 ","End":"02:16.220","Text":"We have to go up to an upper limit which is constantly"},{"Start":"02:16.220 ","End":"02:22.330","Text":"changing depending on where we are on the x-axis. Can you see that?"},{"Start":"02:22.330 ","End":"02:25.880","Text":"How are we going to do this? On the x we\u0027ll say that"},{"Start":"02:25.880 ","End":"02:29.165","Text":"it\u0027s from 0 until b. That\u0027s pretty simple."},{"Start":"02:29.165 ","End":"02:34.010","Text":"On the y-axis, imagine"},{"Start":"02:34.010 ","End":"02:40.845","Text":"that our triangle is on x-y axis."},{"Start":"02:40.845 ","End":"02:44.340","Text":"This is x and this is y."},{"Start":"02:44.340 ","End":"02:51.130","Text":"What we want to write down is the equation of this line over here."},{"Start":"02:51.500 ","End":"02:56.870","Text":"How are we going to do that? Let\u0027s say that our line,"},{"Start":"02:56.870 ","End":"03:00.200","Text":"we\u0027ll call it y equals."},{"Start":"03:00.200 ","End":"03:03.365","Text":"We can see that the slope is"},{"Start":"03:03.365 ","End":"03:06.880","Text":"going downwards in a negative direction so we\u0027ll write negative."},{"Start":"03:06.880 ","End":"03:10.469","Text":"Just like you write a normal line equation,"},{"Start":"03:10.469 ","End":"03:13.205","Text":"let\u0027s see what our gradient is going to be."},{"Start":"03:13.205 ","End":"03:17.540","Text":"As we know, our gradient is our y divided by our x values."},{"Start":"03:17.540 ","End":"03:23.390","Text":"We know that our y value is going to be a divided by our x value,"},{"Start":"03:23.390 ","End":"03:32.855","Text":"which is our b multiplied by x and then we see plus where it crosses the y-axis,"},{"Start":"03:32.855 ","End":"03:35.220","Text":"which is at a."},{"Start":"03:36.310 ","End":"03:39.350","Text":"If you don\u0027t know how to do this,"},{"Start":"03:39.350 ","End":"03:45.360","Text":"please go over how you write the equation of a line in Cartesian coordinates."},{"Start":"03:46.190 ","End":"03:52.475","Text":"We have, this is going to be our bound when we\u0027re integrating according to y."},{"Start":"03:52.475 ","End":"03:54.605","Text":"What is the problem over here?"},{"Start":"03:54.605 ","End":"03:57.200","Text":"When we\u0027re integrating according to y,"},{"Start":"03:57.200 ","End":"04:00.634","Text":"we\u0027re going to be left with an x in our expression,"},{"Start":"04:00.634 ","End":"04:02.405","Text":"which we don\u0027t want."},{"Start":"04:02.405 ","End":"04:03.650","Text":"Because when we\u0027re integrating,"},{"Start":"04:03.650 ","End":"04:06.505","Text":"we want to get to a final answer."},{"Start":"04:06.505 ","End":"04:09.020","Text":"What we\u0027re going to do is we\u0027re going to switch"},{"Start":"04:09.020 ","End":"04:11.480","Text":"the order with which we integrate and first we\u0027re going to"},{"Start":"04:11.480 ","End":"04:16.855","Text":"integrate according to y and then we\u0027ll integrate according to x."},{"Start":"04:16.855 ","End":"04:22.845","Text":"Here we\u0027ll put in from 0 until our y,"},{"Start":"04:22.845 ","End":"04:25.290","Text":"which is this over here."},{"Start":"04:25.290 ","End":"04:28.935","Text":"Let\u0027s rewrite this in a proper way."},{"Start":"04:28.935 ","End":"04:33.875","Text":"We\u0027ll have Sigma divided by total mass."},{"Start":"04:33.875 ","End":"04:39.575","Text":"We\u0027re integrating according to x last, so 0 to b."},{"Start":"04:39.575 ","End":"04:47.915","Text":"Our next bound is going to be 0 to negative a over bx"},{"Start":"04:47.915 ","End":"04:57.240","Text":"plus a and then we\u0027re doing x,"},{"Start":"04:57.240 ","End":"05:01.935","Text":"then dy by dx."},{"Start":"05:01.935 ","End":"05:05.115","Text":"We\u0027ve just changed the order."},{"Start":"05:05.115 ","End":"05:09.695","Text":"Now that we have our expression for our integration,"},{"Start":"05:09.695 ","End":"05:15.030","Text":"from now on it\u0027s pretty plain sailing and we just have to integrate."},{"Start":"05:15.380 ","End":"05:18.705","Text":"Let\u0027s do this integration."},{"Start":"05:18.705 ","End":"05:22.475","Text":"If we\u0027re integrating x by y,"},{"Start":"05:22.475 ","End":"05:24.115","Text":"so we\u0027ll get,"},{"Start":"05:24.115 ","End":"05:28.130","Text":"so we still have our Sigma mass total on the outside,"},{"Start":"05:28.130 ","End":"05:35.540","Text":"x from 0 to b and then we\u0027ll have xy and"},{"Start":"05:35.540 ","End":"05:43.220","Text":"this will be bound by 0 and negative a divided by bx plus a."},{"Start":"05:43.220 ","End":"05:48.143","Text":"Because the 0 here will just fall,"},{"Start":"05:48.143 ","End":"05:50.345","Text":"we can just write in,"},{"Start":"05:50.345 ","End":"05:53.185","Text":"substitute in instead of y,"},{"Start":"05:53.185 ","End":"05:56.390","Text":"our expression for the upper bound."},{"Start":"05:56.390 ","End":"06:06.105","Text":"We\u0027re just going to get Sigma over m_tot 0 to b of multiplying"},{"Start":"06:06.105 ","End":"06:16.350","Text":"by everything so we\u0027re going to get negative a over bx^2 plus ax dx."},{"Start":"06:16.350 ","End":"06:22.225","Text":"All I did here was I integrated with the bounds and because the lower bound was 0,"},{"Start":"06:22.225 ","End":"06:30.150","Text":"x multiplied by 0 is 0 and then x multiplied by substituting into the upper bound into"},{"Start":"06:30.150 ","End":"06:37.925","Text":"our y so we get this expression over here and now integrating according to x."},{"Start":"06:37.925 ","End":"06:44.135","Text":"We have our Sigma m divided by total mass,"},{"Start":"06:44.135 ","End":"06:51.320","Text":"x^2 when integrated is going to be x^3 divided by 3."},{"Start":"06:51.320 ","End":"06:57.950","Text":"We\u0027re going to then have negative a over b x^3 divided by"},{"Start":"06:57.950 ","End":"07:06.110","Text":"3 plus x when integrated so it\u0027s going to be x^2 divided by 2."},{"Start":"07:06.110 ","End":"07:09.947","Text":"We\u0027re bounded between 0 and b."},{"Start":"07:09.947 ","End":"07:13.550","Text":"If I substitute in my 0,"},{"Start":"07:13.550 ","End":"07:15.540","Text":"my whole expression becomes 0."},{"Start":"07:15.540 ","End":"07:17.645","Text":"You don\u0027t even need to bother writing that"},{"Start":"07:17.645 ","End":"07:21.260","Text":"down and then I substitute in my upper bound my b into"},{"Start":"07:21.260 ","End":"07:28.774","Text":"everywhere where we see x. Sigma over m total multiplied by"},{"Start":"07:28.774 ","End":"07:34.500","Text":"negative a over b multiplied by b^3 over"},{"Start":"07:34.500 ","End":"07:41.110","Text":"3 plus a times b^2 over 2."},{"Start":"07:42.350 ","End":"07:46.280","Text":"I can substitute away,"},{"Start":"07:46.280 ","End":"07:48.770","Text":"so I can cross that out."},{"Start":"07:48.770 ","End":"07:53.885","Text":"Notice that when I\u0027m trying to find my center of mass,"},{"Start":"07:53.885 ","End":"07:57.380","Text":"now notice this is just the center of mass on the x-axis,"},{"Start":"07:57.380 ","End":"08:00.755","Text":"not taking into account the center of mass on the y-axis."},{"Start":"08:00.755 ","End":"08:09.550","Text":"I need my units to be in meters or some units in the distance."},{"Start":"08:09.620 ","End":"08:12.570","Text":"Let\u0027s see what is my Sigma."},{"Start":"08:12.570 ","End":"08:17.445","Text":"The definition for what Sigma is,"},{"Start":"08:17.445 ","End":"08:21.150","Text":"is mass divided by area."},{"Start":"08:21.150 ","End":"08:23.445","Text":"My area is going to be a,"},{"Start":"08:23.445 ","End":"08:26.235","Text":"so let\u0027s see what that means."},{"Start":"08:26.235 ","End":"08:30.080","Text":"That\u0027s going to be mass divided by what is my area?"},{"Start":"08:30.080 ","End":"08:34.590","Text":"My area is 1/2 a times b,"},{"Start":"08:35.030 ","End":"08:41.450","Text":"which is just equal to ab over 2 so instead of writing the 2 here,"},{"Start":"08:41.450 ","End":"08:43.750","Text":"I can put it in the numerator."},{"Start":"08:43.750 ","End":"08:45.940","Text":"I have 2m over ab."},{"Start":"08:45.940 ","End":"08:52.270","Text":"If I scroll down a bit and substitute in my Sigma into this,"},{"Start":"08:52.270 ","End":"08:53.950","Text":"let\u0027s see what we get."},{"Start":"08:53.950 ","End":"09:03.300","Text":"This will equal 2m over ab multiplied by 1 over m"},{"Start":"09:03.300 ","End":"09:08.890","Text":"total multiplied by negative ab^2"},{"Start":"09:08.890 ","End":"09:17.120","Text":"over 3 plus ab^2 over 2."},{"Start":"09:17.300 ","End":"09:19.505","Text":"Here I have ab,"},{"Start":"09:19.505 ","End":"09:24.755","Text":"so I can cross out all my a\u0027s and then my b\u0027s,"},{"Start":"09:24.755 ","End":"09:28.680","Text":"so I can cross them out as well."},{"Start":"09:28.970 ","End":"09:36.050","Text":"Notice this m is actually equal to total mass,"},{"Start":"09:36.050 ","End":"09:40.565","Text":"because when we were speaking about our Sigma,"},{"Start":"09:40.565 ","End":"09:45.215","Text":"we\u0027re speaking about the total mass of the system divided by the total area."},{"Start":"09:45.215 ","End":"09:52.700","Text":"We can cross out these as well because they are equal and then what we will be left"},{"Start":"09:52.700 ","End":"10:00.755","Text":"with is going to be 2 and then we can take our b out, it\u0027s a common value."},{"Start":"10:00.755 ","End":"10:07.055","Text":"We\u0027re going to be left with 1/5 minus 1/3,"},{"Start":"10:07.055 ","End":"10:11.680","Text":"which as we know, is equal to 1/6."},{"Start":"10:11.680 ","End":"10:16.235","Text":"When we multiply everything out,"},{"Start":"10:16.235 ","End":"10:23.880","Text":"we\u0027re going to have 1/3b as our final answer."},{"Start":"10:24.920 ","End":"10:29.560","Text":"That means that if we scroll back up,"},{"Start":"10:31.420 ","End":"10:41.370","Text":"that our center of mass is going to be somewhere on the line where our b is a third."},{"Start":"10:43.540 ","End":"10:46.940","Text":"If we do the same process,"},{"Start":"10:46.940 ","End":"10:47.960","Text":"which I won\u0027t do right now,"},{"Start":"10:47.960 ","End":"10:51.413","Text":"but it\u0027s the exact same formula in the exact same process,"},{"Start":"10:51.413 ","End":"10:55.205","Text":"to find the center of mass in the y-direction,"},{"Start":"10:55.205 ","End":"11:01.605","Text":"we\u0027ll see that our center of mass in the y-direction is 1/3a."},{"Start":"11:01.605 ","End":"11:05.990","Text":"We can join up these 2 lines and then we can see that"},{"Start":"11:05.990 ","End":"11:11.160","Text":"our center of mass is right over there in the middle."},{"Start":"11:11.890 ","End":"11:17.950","Text":"Giving our center of mass in coordinates is just going to be 1/3b,"},{"Start":"11:17.950 ","End":"11:22.545","Text":"where the x-component goes and 1/3a with the y component goes."},{"Start":"11:22.545 ","End":"11:25.770","Text":"Let\u0027s just quickly recap what we did."},{"Start":"11:25.770 ","End":"11:29.285","Text":"We wrote down the equation for the center of mass."},{"Start":"11:29.285 ","End":"11:31.685","Text":"We saw that because in this case we\u0027re dealing with"},{"Start":"11:31.685 ","End":"11:36.004","Text":"a triangle so we\u0027re going to use Cartesian coordinates,"},{"Start":"11:36.004 ","End":"11:40.685","Text":"which means that our dm turns into Sigma dxdy."},{"Start":"11:40.685 ","End":"11:43.535","Text":"Because we\u0027re using area,"},{"Start":"11:43.535 ","End":"11:46.415","Text":"so it\u0027s going to be the double integral."},{"Start":"11:46.415 ","End":"11:49.340","Text":"What we did is we worked out our bounds,"},{"Start":"11:49.340 ","End":"11:52.760","Text":"taking into account that our bound on"},{"Start":"11:52.760 ","End":"11:58.345","Text":"the y-axis is changing dependent on what\u0027s happening in our x-axis."},{"Start":"11:58.345 ","End":"12:02.300","Text":"We just integrated until we"},{"Start":"12:02.300 ","End":"12:07.520","Text":"found some equation and we saw that because the units didn\u0027t work out,"},{"Start":"12:07.520 ","End":"12:09.695","Text":"that we had to define what our Sigma was."},{"Start":"12:09.695 ","End":"12:12.230","Text":"Our Sigma as we know in this case,"},{"Start":"12:12.230 ","End":"12:13.415","Text":"because we\u0027re dealing with area,"},{"Start":"12:13.415 ","End":"12:17.105","Text":"it\u0027s total mass divided by the total area."},{"Start":"12:17.105 ","End":"12:21.455","Text":"When we wrote that down and substituted that into our equation,"},{"Start":"12:21.455 ","End":"12:24.080","Text":"we saw that we got an answer for 1/3b."},{"Start":"12:24.080 ","End":"12:27.830","Text":"What you can do as an exercise is repeat the exact same thing,"},{"Start":"12:27.830 ","End":"12:31.610","Text":"but for the y-axis and that is how we got the coordinates."},{"Start":"12:31.610 ","End":"12:34.260","Text":"That\u0027s the end of this lesson."}],"ID":9362},{"Watched":false,"Name":"Exercise- Gate Center of Mass","Duration":"9m 17s","ChapterTopicVideoID":9090,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"Hello. In this question,"},{"Start":"00:02.160 ","End":"00:06.810","Text":"we\u0027re being told that we have in electric gates of mass m and length l,"},{"Start":"00:06.810 ","End":"00:09.120","Text":"which is resting on an axis,"},{"Start":"00:09.120 ","End":"00:10.635","Text":"this is the axis,"},{"Start":"00:10.635 ","End":"00:14.790","Text":"which is located a distance d from its edge."},{"Start":"00:14.790 ","End":"00:18.570","Text":"This distance over here is d,"},{"Start":"00:18.570 ","End":"00:20.835","Text":"and then we\u0027re being told to explain why"},{"Start":"00:20.835 ","End":"00:23.835","Text":"a heavy mass is attached to one end of the gates."},{"Start":"00:23.835 ","End":"00:26.760","Text":"That\u0027s this square block,"},{"Start":"00:26.760 ","End":"00:32.810","Text":"and to find its mass if its length is L. We have to find what our mass is,"},{"Start":"00:32.810 ","End":"00:36.815","Text":"and we\u0027re being told that the length of the mass is equal to"},{"Start":"00:36.815 ","End":"00:41.810","Text":"L. Let\u0027s start with the first section."},{"Start":"00:41.810 ","End":"00:45.335","Text":"Explain why a heavy mass is attached to one end of the gate."},{"Start":"00:45.335 ","End":"00:46.970","Text":"Now, if we don\u0027t have the mass,"},{"Start":"00:46.970 ","End":"00:49.235","Text":"let\u0027s look at this diagram over here."},{"Start":"00:49.235 ","End":"00:54.395","Text":"Then, we know that our center of mass will be somewhere in the center of the rod."},{"Start":"00:54.395 ","End":"00:57.590","Text":"Now, if we have our axis over here,"},{"Start":"00:57.590 ","End":"01:00.545","Text":"then in order to lift this rod,"},{"Start":"01:00.545 ","End":"01:02.615","Text":"to open the gate,"},{"Start":"01:02.615 ","End":"01:06.590","Text":"we\u0027re going to have to apply a really big force because we\u0027re going"},{"Start":"01:06.590 ","End":"01:10.550","Text":"to be lifting it through its center of mass."},{"Start":"01:10.550 ","End":"01:12.545","Text":"Then aside from that,"},{"Start":"01:12.545 ","End":"01:14.285","Text":"once it is up in the air,"},{"Start":"01:14.285 ","End":"01:18.430","Text":"the center of mass is going to be applying a downwards force,"},{"Start":"01:18.430 ","End":"01:24.350","Text":"so it\u0027s also going to be very difficult to hold our rod in place because of that."},{"Start":"01:24.350 ","End":"01:28.790","Text":"What we do is we apply a mass over here,"},{"Start":"01:28.790 ","End":"01:32.315","Text":"which means that after this is applied,"},{"Start":"01:32.315 ","End":"01:37.775","Text":"our center of mass will move to a different point, and here,"},{"Start":"01:37.775 ","End":"01:45.860","Text":"a different point will be if we add a certain mass that the engineers will come up with,"},{"Start":"01:45.860 ","End":"01:49.880","Text":"which we will also come up with in the next part of the question."},{"Start":"01:49.880 ","End":"01:53.450","Text":"Then we will move our center of mass from the center of"},{"Start":"01:53.450 ","End":"01:57.275","Text":"the gate onto exactly where it\u0027s axis lies."},{"Start":"01:57.275 ","End":"02:00.455","Text":"Then when we left the gate, first of all,"},{"Start":"02:00.455 ","End":"02:03.950","Text":"it will be a lot easier because we won\u0027t be lifting it from its center of"},{"Start":"02:03.950 ","End":"02:08.915","Text":"mass so the forces required to lift it will be much less."},{"Start":"02:08.915 ","End":"02:12.770","Text":"Also because our center of mass will be on the axes,"},{"Start":"02:12.770 ","End":"02:14.450","Text":"when we lift the gates,"},{"Start":"02:14.450 ","End":"02:16.550","Text":"we won\u0027t be lifting the center of mass,"},{"Start":"02:16.550 ","End":"02:19.910","Text":"we\u0027ll simply be rotating the center of mass."},{"Start":"02:19.910 ","End":"02:24.605","Text":"The center of mass will be in the exact same place, just slightly rotated,"},{"Start":"02:24.605 ","End":"02:26.425","Text":"which means that 1,"},{"Start":"02:26.425 ","End":"02:30.155","Text":"to open the gate we\u0027ll require a smaller force, and 2,"},{"Start":"02:30.155 ","End":"02:31.895","Text":"once the gate is open,"},{"Start":"02:31.895 ","End":"02:36.065","Text":"the center of mass won\u0027t be applying a downwards force,"},{"Start":"02:36.065 ","End":"02:40.580","Text":"so the gate will just stay in place."},{"Start":"02:40.580 ","End":"02:46.495","Text":"We won\u0027t have to exert extra forces in order to keep the gate open."},{"Start":"02:46.495 ","End":"02:51.320","Text":"That\u0027s the idea of why a mass is attached to one end of the gate."},{"Start":"02:51.320 ","End":"03:00.505","Text":"Now let\u0027s find its mass if its length is L. A mass that we\u0027re going to be adding"},{"Start":"03:00.505 ","End":"03:04.850","Text":"needs to be a mass such that the center of mass of the gate"},{"Start":"03:04.850 ","End":"03:09.600","Text":"will be exactly where this triangle is,"},{"Start":"03:09.600 ","End":"03:10.825","Text":"where it\u0027s axis is,"},{"Start":"03:10.825 ","End":"03:15.605","Text":"so we\u0027re going to see that the points where the axis is, is our 0."},{"Start":"03:15.605 ","End":"03:17.665","Text":"That\u0027s our origin."},{"Start":"03:17.665 ","End":"03:22.035","Text":"Now if the length of the mass is L,"},{"Start":"03:22.035 ","End":"03:24.960","Text":"then we know that half of it."},{"Start":"03:24.960 ","End":"03:30.390","Text":"This section is going to be L divided by 2."},{"Start":"03:30.430 ","End":"03:35.690","Text":"Now, the center of mass of the gates is usually going to be"},{"Start":"03:35.690 ","End":"03:41.315","Text":"at L divided by 2 if we were just dealing with the gate."},{"Start":"03:41.315 ","End":"03:44.600","Text":"However, because we know that it\u0027s attached to an axis."},{"Start":"03:44.600 ","End":"03:47.360","Text":"It\u0027s not going to be L divided by 2."},{"Start":"03:47.360 ","End":"03:50.510","Text":"It\u0027s going to be L divided by 2 minus"},{"Start":"03:50.510 ","End":"03:58.640","Text":"d. This is because the center of mass of the gate is at L over 2,"},{"Start":"03:58.640 ","End":"04:00.200","Text":"right in the center of the gauge,"},{"Start":"04:00.200 ","End":"04:02.255","Text":"and because it\u0027s resting on an axis,"},{"Start":"04:02.255 ","End":"04:07.460","Text":"we minus the distance between the edge of the gates and the axis,"},{"Start":"04:07.460 ","End":"04:12.380","Text":"which in this case is d. This center of mass is going to be here,"},{"Start":"04:12.380 ","End":"04:16.460","Text":"and then, because we have the center of mass,"},{"Start":"04:16.460 ","End":"04:21.570","Text":"we can consider this point over here as a point-mass."},{"Start":"04:21.570 ","End":"04:26.600","Text":"We can pretend as if the gate isn\u0027t there and we just have a point mass,"},{"Start":"04:26.600 ","End":"04:32.690","Text":"a distance of L over 2 minus d away from the origin."},{"Start":"04:32.690 ","End":"04:36.110","Text":"Now what we\u0027re going to do is we\u0027re going to"},{"Start":"04:36.110 ","End":"04:39.350","Text":"find the center of mass of the entire system."},{"Start":"04:39.350 ","End":"04:40.760","Text":"What does that mean?"},{"Start":"04:40.760 ","End":"04:46.860","Text":"It means the center of mass between point mass at L over 2 minus d,"},{"Start":"04:46.860 ","End":"04:49.590","Text":"which we know is a mass of m because it\u0027s given in"},{"Start":"04:49.590 ","End":"04:55.400","Text":"the question and our large mass over here,"},{"Start":"04:55.400 ","End":"04:57.475","Text":"which is in the box."},{"Start":"04:57.475 ","End":"05:00.540","Text":"Which we know is an unknown,"},{"Start":"05:00.540 ","End":"05:06.425","Text":"so it\u0027s going to be over here and it\u0027s an unknown and we know it\u0027s a capital M and it\u0027s"},{"Start":"05:06.425 ","End":"05:12.920","Text":"a distance of L over d plus d away from the center of mass."},{"Start":"05:12.920 ","End":"05:15.335","Text":"Sorry, I wrote a mistake here."},{"Start":"05:15.335 ","End":"05:21.930","Text":"This isn\u0027t L over d. This is L over 2."},{"Start":"05:21.930 ","End":"05:26.945","Text":"We\u0027re taking that the center of mass of the box is going to be in the center of the box,"},{"Start":"05:26.945 ","End":"05:29.375","Text":"which if we know that the length of the box is L,"},{"Start":"05:29.375 ","End":"05:32.479","Text":"then its center is going to be L over 2."},{"Start":"05:32.479 ","End":"05:36.590","Text":"Now we\u0027re considering this box of mass,"},{"Start":"05:36.590 ","End":"05:39.190","Text":"capital M as another point mass,"},{"Start":"05:39.190 ","End":"05:42.110","Text":"and we want to find the center of mass of"},{"Start":"05:42.110 ","End":"05:47.150","Text":"this entire system such that its center of mass is over here at the origin,"},{"Start":"05:47.150 ","End":"05:50.640","Text":"right where the axis is."},{"Start":"05:50.690 ","End":"05:56.690","Text":"Now we\u0027re going to find the center of mass of these 2 point masses."},{"Start":"05:56.690 ","End":"05:59.640","Text":"Let\u0027s see how we do that."},{"Start":"05:59.900 ","End":"06:03.860","Text":"As we know a formula for the center of mass, now,"},{"Start":"06:03.860 ","End":"06:06.320","Text":"we\u0027re only going to be looking on the x-axis because"},{"Start":"06:06.320 ","End":"06:11.440","Text":"we\u0027re just concerned with the length not the height."},{"Start":"06:11.440 ","End":"06:16.265","Text":"Our x center of mass is going to be the sum of all of"},{"Start":"06:16.265 ","End":"06:21.830","Text":"the masses multiplied by their distances away from the origin,"},{"Start":"06:21.830 ","End":"06:25.100","Text":"divided by the sum of all of the masses,"},{"Start":"06:25.100 ","End":"06:27.680","Text":"divided by the total mass of the system."},{"Start":"06:27.680 ","End":"06:32.490","Text":"Let\u0027s see how we apply this to our question."},{"Start":"06:33.380 ","End":"06:37.815","Text":"Let\u0027s take our first mass to be small m,"},{"Start":"06:37.815 ","End":"06:40.895","Text":"so we\u0027re going to have its displacement."},{"Start":"06:40.895 ","End":"06:43.070","Text":"Now it\u0027s displacement is from the origin."},{"Start":"06:43.070 ","End":"06:48.650","Text":"We\u0027re going to say that it\u0027s L divided by 2 minus d. That\u0027s"},{"Start":"06:48.650 ","End":"06:54.545","Text":"this distance multiplied by its mass plus our second point mass."},{"Start":"06:54.545 ","End":"06:57.860","Text":"Now, notice that it\u0027s on the other side of the origin,"},{"Start":"06:57.860 ","End":"07:03.814","Text":"so it\u0027s in the negative section of our imaginary x-axis."},{"Start":"07:03.814 ","End":"07:06.955","Text":"We\u0027re going to have plus negative,"},{"Start":"07:06.955 ","End":"07:11.555","Text":"and then its distance from the origin is of course L divided by"},{"Start":"07:11.555 ","End":"07:16.685","Text":"2 plus d multiplied by its mass, which is,"},{"Start":"07:16.685 ","End":"07:19.580","Text":"of course our unknown, which is capital M. Then we"},{"Start":"07:19.580 ","End":"07:23.300","Text":"have all of this is divided by the total mass of the system,"},{"Start":"07:23.300 ","End":"07:25.340","Text":"which is going to be m plus"},{"Start":"07:25.340 ","End":"07:31.260","Text":"M. Then we want our x center of mass to be right in the origin,"},{"Start":"07:31.260 ","End":"07:33.710","Text":"so we\u0027re going to say that this is equal to"},{"Start":"07:33.710 ","End":"07:40.565","Text":"0 because we want our center of mass to be at the origin at point 0."},{"Start":"07:40.565 ","End":"07:47.000","Text":"Now all we have to do is we have to isolate out this m. How do we do that?"},{"Start":"07:47.000 ","End":"07:49.880","Text":"We can multiply both sides by m plus M,"},{"Start":"07:49.880 ","End":"07:56.340","Text":"which of course will just give us L divided by 2 minus d m"},{"Start":"07:56.340 ","End":"08:04.634","Text":"minus L over 2 plus d M will be equal to 0."},{"Start":"08:04.634 ","End":"08:15.765","Text":"Then we\u0027ll have L divided by 2 m minus dm will be equal to L"},{"Start":"08:15.765 ","End":"08:21.470","Text":"over 2 plus d M. Then we can"},{"Start":"08:21.470 ","End":"08:27.530","Text":"just divide both sides by capital L over 2 plus d,"},{"Start":"08:27.530 ","End":"08:30.540","Text":"and then we have our M isolated out,"},{"Start":"08:30.540 ","End":"08:34.590","Text":"so we\u0027ll have l divided by 2m minus dm."},{"Start":"08:34.590 ","End":"08:37.160","Text":"I didn\u0027t have to open these brackets, but never mind."},{"Start":"08:37.160 ","End":"08:42.310","Text":"Divided by L over 2 plus d is equal to"},{"Start":"08:42.310 ","End":"08:51.510","Text":"M. Then this is our final answer."},{"Start":"08:52.060 ","End":"08:59.075","Text":"Our mass for our large mass at the edge of the gates,"},{"Start":"08:59.075 ","End":"09:02.000","Text":"which will therefore make our center of mass of"},{"Start":"09:02.000 ","End":"09:06.845","Text":"the entire system be raised at the pivotal point at the axis."},{"Start":"09:06.845 ","End":"09:12.920","Text":"The origin is going to have to be this expression over here."},{"Start":"09:12.920 ","End":"09:17.310","Text":"This is the end of our example."}],"ID":9363},{"Watched":false,"Name":"Exercise - Finding the Area of a Sector","Duration":"3m 4s","ChapterTopicVideoID":13101,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"Hello. In this lesson,"},{"Start":"00:02.205 ","End":"00:06.585","Text":"we\u0027re going to see how to find the area of a sector."},{"Start":"00:06.585 ","End":"00:11.355","Text":"A sector is this type of shape that you will have in a circle,"},{"Start":"00:11.355 ","End":"00:16.515","Text":"where we have some angle and the edge is the curve of the circle."},{"Start":"00:16.515 ","End":"00:18.345","Text":"It\u0027s not a straight edge."},{"Start":"00:18.345 ","End":"00:21.720","Text":"Here we have a circle of radius R,"},{"Start":"00:21.720 ","End":"00:27.550","Text":"and the angle of the sector is equal to Theta."},{"Start":"00:27.650 ","End":"00:32.855","Text":"The big idea when solving this type of question is by using the idea"},{"Start":"00:32.855 ","End":"00:38.300","Text":"that the sector is some kind of fraction of the entire circle."},{"Start":"00:38.300 ","End":"00:47.105","Text":"Using that, we know that the angle of the sector is equal to Theta,"},{"Start":"00:47.105 ","End":"00:49.565","Text":"and so we\u0027re using this as a ratio,"},{"Start":"00:49.565 ","End":"00:51.935","Text":"as a fraction of the entire circle,"},{"Start":"00:51.935 ","End":"01:01.660","Text":"where of course the angle of the entire circle is equal to 2 Pi or 360 degrees."},{"Start":"01:02.750 ","End":"01:07.760","Text":"This ratio has to correspond to the total area of the sector."},{"Start":"01:07.760 ","End":"01:10.725","Text":"Let\u0027s call the area of the sector S."},{"Start":"01:10.725 ","End":"01:16.580","Text":"The total area of the sector is also a fraction of the total area of the circle."},{"Start":"01:16.580 ","End":"01:20.690","Text":"We have S divided by the area of a circle,"},{"Start":"01:20.690 ","End":"01:27.065","Text":"which is of course equal to Pi r^2."},{"Start":"01:27.065 ","End":"01:30.195","Text":"We\u0027re using r for general,"},{"Start":"01:30.195 ","End":"01:33.830","Text":"and here of course our radius is R. Therefore,"},{"Start":"01:33.830 ","End":"01:38.875","Text":"if we want to find the area of this segment S,"},{"Start":"01:38.875 ","End":"01:41.495","Text":"we can see that the Pi\u0027s will cancel out,"},{"Start":"01:41.495 ","End":"01:43.490","Text":"and we\u0027ll have that S is equal to,"},{"Start":"01:43.490 ","End":"01:46.140","Text":"in our case, the radius squared,"},{"Start":"01:46.140 ","End":"01:53.625","Text":"so R^2 multiplied by Theta and divided by 2."},{"Start":"01:53.625 ","End":"01:57.755","Text":"This is of course the area of this sector,"},{"Start":"01:57.755 ","End":"02:00.155","Text":"and this was one way of solving it."},{"Start":"02:00.155 ","End":"02:02.630","Text":"Another way is just by integration."},{"Start":"02:02.630 ","End":"02:08.150","Text":"What do we can do is we can say that the area of the sector is equal to"},{"Start":"02:08.150 ","End":"02:13.895","Text":"the integral on a unit area inside this sector."},{"Start":"02:13.895 ","End":"02:15.350","Text":"We\u0027re dealing with a circle,"},{"Start":"02:15.350 ","End":"02:16.970","Text":"so we have 2 dimensions,"},{"Start":"02:16.970 ","End":"02:27.265","Text":"so we\u0027re integrating along rd Theta dr. Now we can plug in our bounds."},{"Start":"02:27.265 ","End":"02:29.370","Text":"Follow the bounds from the radius,"},{"Start":"02:29.370 ","End":"02:36.065","Text":"so we\u0027re integrating from 0 up until a maximum radius of R and our Theta,"},{"Start":"02:36.065 ","End":"02:39.490","Text":"we\u0027re integrating from an angle of 0 over here,"},{"Start":"02:39.490 ","End":"02:44.015","Text":"and all the way to the angle Theta,"},{"Start":"02:44.015 ","End":"02:46.740","Text":"the angle of the sector."},{"Start":"02:46.740 ","End":"02:49.475","Text":"This is a very easy integral to do,"},{"Start":"02:49.475 ","End":"02:56.210","Text":"and of course the answer that we\u0027ll get is R^2 Theta divided by 2."},{"Start":"02:56.210 ","End":"03:02.075","Text":"Just like we got in this way using fractions or ratios."},{"Start":"03:02.075 ","End":"03:05.400","Text":"That is the end of this lesson."}],"ID":13580},{"Watched":false,"Name":"Exercise- Sector Center of Mass","Duration":"7m 48s","ChapterTopicVideoID":9091,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:07.739","Text":"Hello, in this section we\u0027re going to try to find the center of mass of a segment."},{"Start":"00:07.739 ","End":"00:11.505","Text":"If we put in our axis,"},{"Start":"00:11.505 ","End":"00:17.670","Text":"we\u0027re going to say that our origin is right at this spike of the segment."},{"Start":"00:17.670 ","End":"00:21.645","Text":"Now we\u0027re going to try and find its center of mass."},{"Start":"00:21.645 ","End":"00:24.435","Text":"As we remember, the equation is X,"},{"Start":"00:24.435 ","End":"00:30.130","Text":"center of mass is equal to the integral of xdm,"},{"Start":"00:30.130 ","End":"00:33.990","Text":"divided by the integral on dm."},{"Start":"00:33.990 ","End":"00:41.460","Text":"Then this integral on dm is just equal to the total mass of the system."},{"Start":"00:41.460 ","End":"00:45.945","Text":"Now, let\u0027s first see what our dm is going to be."},{"Start":"00:45.945 ","End":"00:49.085","Text":"When we\u0027re dealing with a change in mass,"},{"Start":"00:49.085 ","End":"00:55.670","Text":"we can say that it\u0027s either Lambda dl when dealing with a line."},{"Start":"00:55.670 ","End":"00:59.870","Text":"In one dimension, Sigma ds,"},{"Start":"00:59.870 ","End":"01:03.590","Text":"when dealing with an area which is what we\u0027re dealing with now"},{"Start":"01:03.590 ","End":"01:07.955","Text":"and of course we have Rho dv when dealing with a volume."},{"Start":"01:07.955 ","End":"01:09.755","Text":"Now I\u0027ve already said we\u0027re dealing with area,"},{"Start":"01:09.755 ","End":"01:13.025","Text":"so we\u0027re going to use our segment ds."},{"Start":"01:13.025 ","End":"01:19.995","Text":"Therefore, we\u0027re going to say that dm is equal to Sigma ds."},{"Start":"01:19.995 ","End":"01:24.465","Text":"Now we have to see what Sigma is equal to."},{"Start":"01:24.465 ","End":"01:29.105","Text":"Our Sigma is equal to the total mass of"},{"Start":"01:29.105 ","End":"01:34.235","Text":"the shapes so here on the segment divided by the total area."},{"Start":"01:34.235 ","End":"01:37.490","Text":"Here I\u0027m saying that the area is"},{"Start":"01:37.490 ","End":"01:44.310","Text":"s. That is going to be equal to the total mass divided by now,"},{"Start":"01:44.310 ","End":"01:45.870","Text":"what is the area."},{"Start":"01:45.870 ","End":"01:52.100","Text":"We have our angle which is Theta 0 so it\u0027s going to be Theta 0 multiplied"},{"Start":"01:52.100 ","End":"02:00.100","Text":"by our radius squared and then divided by 2."},{"Start":"02:00.740 ","End":"02:11.910","Text":"This equation just comes from the equation for finding a sector of a circle."},{"Start":"02:12.920 ","End":"02:16.470","Text":"Now before we start doing all of our integration,"},{"Start":"02:16.470 ","End":"02:23.480","Text":"we have to decide which units we\u0027re going to be integrating via."},{"Start":"02:23.480 ","End":"02:26.615","Text":"Now, because we\u0027re using a circular shape,"},{"Start":"02:26.615 ","End":"02:31.855","Text":"we can see that our small sections of dm,"},{"Start":"02:31.855 ","End":"02:34.395","Text":"will also be circular."},{"Start":"02:34.395 ","End":"02:37.265","Text":"Using Cartesian coordinates isn\u0027t going to be"},{"Start":"02:37.265 ","End":"02:41.060","Text":"the best idea what we should use are polar coordinates."},{"Start":"02:41.060 ","End":"02:43.360","Text":"Now, in polar coordinates,"},{"Start":"02:43.360 ","End":"02:46.410","Text":"our ds, is going to be measured in rd,"},{"Start":"02:46.410 ","End":"02:51.265","Text":"Theta multiplied by dr,"},{"Start":"02:51.265 ","End":"02:59.270","Text":"where r, is the radius from the origin until our section of mass."},{"Start":"02:59.270 ","End":"03:04.535","Text":"Theta is the small angles that we\u0027re moving relative to a small section of"},{"Start":"03:04.535 ","End":"03:10.620","Text":"mass and then dr integrating along our radius."},{"Start":"03:10.670 ","End":"03:13.900","Text":"Let us begin."},{"Start":"03:14.450 ","End":"03:17.865","Text":"Let\u0027s write down our equation."},{"Start":"03:17.865 ","End":"03:22.250","Text":"We have that our x center of mass is going to be equal"},{"Start":"03:22.250 ","End":"03:28.495","Text":"to our integral on xdm."},{"Start":"03:28.495 ","End":"03:31.970","Text":"We\u0027re going to start divided by the total mass."},{"Start":"03:31.970 ","End":"03:37.785","Text":"We can already write on the outside 1/m,"},{"Start":"03:37.785 ","End":"03:41.660","Text":"and then the integral because we\u0027re going on d Theta dr,"},{"Start":"03:41.660 ","End":"03:48.745","Text":"it\u0027s going to be a double integral and then on the inside we have Sigma,"},{"Start":"03:48.745 ","End":"03:52.035","Text":"because it\u0027s Sigma ds."},{"Start":"03:52.035 ","End":"03:55.515","Text":"Our Sigma was m over"},{"Start":"03:55.515 ","End":"04:03.030","Text":"Theta 0 r^2/2 multiplied by dr,"},{"Start":"04:03.030 ","End":"04:08.835","Text":"d Theta dr. Of course,"},{"Start":"04:08.835 ","End":"04:14.045","Text":"we have to add in our x over here because it\u0027s in the equation."},{"Start":"04:14.045 ","End":"04:22.140","Text":"Now this M and this M can cross out and then we know that our R^2 2 and R Theta 0 are"},{"Start":"04:22.140 ","End":"04:31.440","Text":"constants so we can rewrite this as 1 over Theta 0 R^2 squared."},{"Start":"04:31.440 ","End":"04:38.070","Text":"We can write the 2 here instead of the 2 at the bottom and then double integral on xr,"},{"Start":"04:38.070 ","End":"04:43.150","Text":"dr, d Theta."},{"Start":"04:43.280 ","End":"04:49.095","Text":"Now our x obviously is a function of"},{"Start":"04:49.095 ","End":"04:55.980","Text":"R. Because as our R moves along,"},{"Start":"04:55.980 ","End":"05:00.855","Text":"we can see that depending on the angle,"},{"Start":"05:00.855 ","End":"05:04.935","Text":"then our x also goes further away."},{"Start":"05:04.935 ","End":"05:09.020","Text":"If we say that the angle between the x-axis"},{"Start":"05:09.020 ","End":"05:13.320","Text":"and wherever our radius is is going to be equal to Theta,"},{"Start":"05:13.320 ","End":"05:14.540","Text":"not Theta naught,"},{"Start":"05:14.540 ","End":"05:19.700","Text":"but Theta, then we can see that our x value is"},{"Start":"05:19.700 ","End":"05:25.470","Text":"going to equal to R cosine of Theta."},{"Start":"05:25.470 ","End":"05:28.815","Text":"Again, not Theta naught of Theta."},{"Start":"05:28.815 ","End":"05:33.800","Text":"Then we can substitute that into our equation so we can have 2"},{"Start":"05:33.800 ","End":"05:38.820","Text":"divided by Theta naught R^2, double integral,"},{"Start":"05:38.820 ","End":"05:44.315","Text":"we\u0027ll soon add in the bounds of R cosine of Theta multiplied by R. It\u0027s going to be"},{"Start":"05:44.315 ","End":"05:53.715","Text":"R^2 cosine of Theta and then let\u0027s integrate by Theta first so d Theta dr. Now,"},{"Start":"05:53.715 ","End":"05:57.135","Text":"what are our bounds going to be?"},{"Start":"05:57.135 ","End":"06:01.820","Text":"Our bounds on our r is going to be from our 0 until our maximum radius,"},{"Start":"06:01.820 ","End":"06:06.650","Text":"which is R. From 0 until R."},{"Start":"06:06.650 ","End":"06:12.725","Text":"Then our bounds for Theta is going to be,"},{"Start":"06:12.725 ","End":"06:16.580","Text":"if we look, we have theta naught."},{"Start":"06:16.580 ","End":"06:17.675","Text":"I want to remind you,"},{"Start":"06:17.675 ","End":"06:20.195","Text":"is all of this over here."},{"Start":"06:20.195 ","End":"06:23.490","Text":"This entire angle is Theta naught."},{"Start":"06:24.380 ","End":"06:28.515","Text":"That means that until the x-axis,"},{"Start":"06:28.515 ","End":"06:37.650","Text":"this will be Theta naught divided by 2 and this will also be Theta naught divided by 2."},{"Start":"06:37.650 ","End":"06:44.120","Text":"Now notice that this Theta naught over here is below the x-axis so we\u0027re going to go from"},{"Start":"06:44.120 ","End":"06:51.585","Text":"negative Theta naught over 2 until positive Theta naught over 2."},{"Start":"06:51.585 ","End":"06:56.345","Text":"Also notice from symmetry that our Y center of mass,"},{"Start":"06:56.345 ","End":"07:00.290","Text":"because we\u0027ve split our sector into 2 so we"},{"Start":"07:00.290 ","End":"07:04.685","Text":"know that our Y center of mass is going to be somewhere on the line of Y equals 0."},{"Start":"07:04.685 ","End":"07:07.610","Text":"It\u0027s going to be somewhere on the x-axis and now we\u0027re just"},{"Start":"07:07.610 ","End":"07:10.250","Text":"trying to find exactly where on"},{"Start":"07:10.250 ","End":"07:16.860","Text":"the x-axis by finding the center of mass on the x-axis."},{"Start":"07:17.300 ","End":"07:21.980","Text":"Now all you have to do is to integrate"},{"Start":"07:21.980 ","End":"07:26.810","Text":"this expression and then I won\u0027t waste time now, you can do this,"},{"Start":"07:26.810 ","End":"07:34.980","Text":"but your final answer should be 4 of sine Theta naught divided by 2,"},{"Start":"07:34.980 ","End":"07:38.805","Text":"divided by 3 Theta naught."},{"Start":"07:38.805 ","End":"07:41.270","Text":"Once you do the integration,"},{"Start":"07:41.270 ","End":"07:46.040","Text":"you will see that this is the answer that you will get."},{"Start":"07:46.040 ","End":"07:48.990","Text":"That\u0027s the end of this lesson."}],"ID":9364},{"Watched":false,"Name":"Exercise- Center of Mass Half Sphere Full","Duration":"13m 15s","ChapterTopicVideoID":9093,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.700","Text":"Hello, in this question,"},{"Start":"00:02.700 ","End":"00:07.275","Text":"we\u0027re being told to find the center of mass of a filled half sphere."},{"Start":"00:07.275 ","End":"00:16.770","Text":"That means that we have half a sphere over here, and it\u0027s solid."},{"Start":"00:16.770 ","End":"00:21.330","Text":"Let\u0027s say that it has some radius R,"},{"Start":"00:21.330 ","End":"00:25.245","Text":"and now we\u0027re going to find the center of mass."},{"Start":"00:25.245 ","End":"00:27.900","Text":"First I\u0027m going to put down my axis,"},{"Start":"00:27.900 ","End":"00:31.140","Text":"so I\u0027m going to say that my origin is right in"},{"Start":"00:31.140 ","End":"00:35.160","Text":"the center of the base of the half sphere, so right over here."},{"Start":"00:35.160 ","End":"00:38.370","Text":"Then I\u0027m going to say that going like this,"},{"Start":"00:38.370 ","End":"00:42.500","Text":"I have my z-axis in this direction,"},{"Start":"00:42.500 ","End":"00:44.300","Text":"I have my x-axis,"},{"Start":"00:44.300 ","End":"00:48.095","Text":"and in this direction I have my y-axis."},{"Start":"00:48.095 ","End":"00:55.530","Text":"As we can see, our center of masses for xcm and ycm it\u0027s going to be at the origin,"},{"Start":"00:55.530 ","End":"00:58.845","Text":"at x=0 and y=0."},{"Start":"00:58.845 ","End":"01:00.710","Text":"It\u0027s symmetrical, it doesn\u0027t matter,"},{"Start":"01:00.710 ","End":"01:02.510","Text":"but in the z direction,"},{"Start":"01:02.510 ","End":"01:05.735","Text":"that\u0027s where we have to find where our center of mass is,"},{"Start":"01:05.735 ","End":"01:09.815","Text":"so it\u0027s going to be somewhere on the z-axis."},{"Start":"01:09.815 ","End":"01:12.260","Text":"Let\u0027s put in our equation."},{"Start":"01:12.260 ","End":"01:16.745","Text":"We know that our zcm is going to be equal to the integral"},{"Start":"01:16.745 ","End":"01:22.585","Text":"on zdm divided by the integral on dm."},{"Start":"01:22.585 ","End":"01:28.145","Text":"Now we know that our integral on dm is just equal to our total mass,"},{"Start":"01:28.145 ","End":"01:31.680","Text":"so we\u0027re going to say that that\u0027s equal to capital M."},{"Start":"01:32.150 ","End":"01:38.585","Text":"Now let\u0027s start doing the integration and finding out what our different variables are."},{"Start":"01:38.585 ","End":"01:41.000","Text":"We see that we have dm over here."},{"Start":"01:41.000 ","End":"01:43.335","Text":"What exactly is dm?"},{"Start":"01:43.335 ","End":"01:47.045","Text":"Because we\u0027re dealing with a filled half sphere,"},{"Start":"01:47.045 ","End":"01:51.200","Text":"then we know that we\u0027re dealing with volume because it\u0027s not an area."},{"Start":"01:51.200 ","End":"01:56.540","Text":"That means that our m whenever we\u0027re dealing with volume,"},{"Start":"01:56.540 ","End":"01:59.480","Text":"is Rho dv beautiful."},{"Start":"01:59.480 ","End":"02:01.535","Text":"Now, what is our Rho?"},{"Start":"02:01.535 ","End":"02:03.750","Text":"I hear you asking."},{"Start":"02:04.520 ","End":"02:07.370","Text":"Our Rho is our density,"},{"Start":"02:07.370 ","End":"02:12.425","Text":"which just means our mass divided by our total volume."},{"Start":"02:12.425 ","End":"02:18.995","Text":"Now, the volume of a sphere is 4 over 3 Pi R cubed."},{"Start":"02:18.995 ","End":"02:25.145","Text":"But now because we have half a sphere is going to be divided by 2."},{"Start":"02:25.145 ","End":"02:32.780","Text":"Then we can cross out this over here and then we can rearrange this to"},{"Start":"02:32.780 ","End":"02:40.740","Text":"say that it\u0027s 3m divided by 2 Pi R cubed."},{"Start":"02:40.760 ","End":"02:44.260","Text":"This is our Rho."},{"Start":"02:44.420 ","End":"02:46.920","Text":"Now we see that that\u0027s a Rho."},{"Start":"02:46.920 ","End":"02:48.970","Text":"Now what is our dv?"},{"Start":"02:49.610 ","End":"02:52.205","Text":"Because we\u0027re dealing with the half sphere,"},{"Start":"02:52.205 ","End":"02:55.730","Text":"it\u0027s going to be easiest to deal with spherical coordinates."},{"Start":"02:55.730 ","End":"02:59.540","Text":"As you remember, we have the Jacobian in the spherical coordinates,"},{"Start":"02:59.540 ","End":"03:05.945","Text":"which when we\u0027re using this spherical coordinates as r squared sine of"},{"Start":"03:05.945 ","End":"03:13.380","Text":"Phi multiplied by dr d Theta d Phi."},{"Start":"03:14.360 ","End":"03:19.760","Text":"Now let\u0027s put all our variables into our integration."},{"Start":"03:19.760 ","End":"03:28.660","Text":"We\u0027re going to have that our zcm is equal to 1 over m from the original equation,"},{"Start":"03:28.700 ","End":"03:31.520","Text":"and then we\u0027ll do the integration,"},{"Start":"03:31.520 ","End":"03:33.680","Text":"which because there\u0027s dr d Theta, d Phi,"},{"Start":"03:33.680 ","End":"03:37.745","Text":"so we have the 3 integral signs"},{"Start":"03:37.745 ","End":"03:43.550","Text":"multiplied by z from the original in duration multiplied by dm,"},{"Start":"03:43.550 ","End":"03:44.870","Text":"which is Rho dv,"},{"Start":"03:44.870 ","End":"03:51.215","Text":"so our Rho is 3m over 2 Pi R^3,"},{"Start":"03:51.215 ","End":"04:01.620","Text":"and then rdv is multiplied by r^2 sine of Phi dr d Theta d Phi."},{"Start":"04:01.620 ","End":"04:05.685","Text":"Now we can cross out this 1 over m with this m,"},{"Start":"04:05.685 ","End":"04:10.730","Text":"and notice that we cannot cross out this R cubed with"},{"Start":"04:10.730 ","End":"04:16.595","Text":"this r^2 because obviously they\u0027re not the same r,"},{"Start":"04:16.595 ","End":"04:19.475","Text":"so we cannot cross that out."},{"Start":"04:19.475 ","End":"04:24.275","Text":"Now what we have to do before we add in our bounds"},{"Start":"04:24.275 ","End":"04:29.510","Text":"is that we have to see what our z is because we\u0027re integrating with regards to z,"},{"Start":"04:29.510 ","End":"04:30.860","Text":"but we only have dr,"},{"Start":"04:30.860 ","End":"04:32.450","Text":"d Theta and d Phi."},{"Start":"04:32.450 ","End":"04:41.640","Text":"We have to find some way to express z through r Theta and Phi. Let\u0027s see how we do that."},{"Start":"04:42.170 ","End":"04:45.965","Text":"If we look at our diagram over here,"},{"Start":"04:45.965 ","End":"04:50.965","Text":"so we can say that our r,"},{"Start":"04:50.965 ","End":"04:54.120","Text":"that we\u0027re integrating by is up until here,"},{"Start":"04:54.120 ","End":"04:58.020","Text":"and our z value is up until here,"},{"Start":"04:58.020 ","End":"05:06.655","Text":"so then this forms a right angle triangle and the angle over here is going to be Phi."},{"Start":"05:06.655 ","End":"05:10.530","Text":"Let\u0027s draw that in big."},{"Start":"05:10.530 ","End":"05:12.320","Text":"From here to here,"},{"Start":"05:12.320 ","End":"05:14.285","Text":"we\u0027re going to have our height z,"},{"Start":"05:14.285 ","End":"05:19.360","Text":"and then here we\u0027re going to have our radius,"},{"Start":"05:19.360 ","End":"05:22.200","Text":"and then this is a right angle triangle,"},{"Start":"05:22.200 ","End":"05:25.830","Text":"and the angle over here is going to be Phi."},{"Start":"05:26.980 ","End":"05:32.420","Text":"Now as we can remember via our SOHCAHTOA that we\u0027re going to be"},{"Start":"05:32.420 ","End":"05:39.665","Text":"using Cos Phi is equal to adjacent over hypotenuse,"},{"Start":"05:39.665 ","End":"05:43.325","Text":"because adjacent to the Phi angle, e have z,"},{"Start":"05:43.325 ","End":"05:50.900","Text":"and the hypotenuse of the triangle is our r. That equals to z divided by r,"},{"Start":"05:50.900 ","End":"05:59.130","Text":"and then rearranging this will get that z is equal to r cosine of Phi."},{"Start":"05:59.130 ","End":"06:04.115","Text":"Now we can substitute that back into our integration,"},{"Start":"06:04.115 ","End":"06:07.205","Text":"so we\u0027ll have a triple integration,"},{"Start":"06:07.205 ","End":"06:12.555","Text":"and then we\u0027ll have r cosine Phi multiplied by our r^2 over here,"},{"Start":"06:12.555 ","End":"06:19.935","Text":"so we\u0027ll have r cubed cosine of Phi sine of Phi,"},{"Start":"06:19.935 ","End":"06:28.780","Text":"and then we\u0027ll have on the outside 3/2 Pi r^3."},{"Start":"06:28.780 ","End":"06:33.035","Text":"Then of course we have to multiply by dr,"},{"Start":"06:33.035 ","End":"06:36.815","Text":"d Theta and d Phi."},{"Start":"06:36.815 ","End":"06:41.375","Text":"Now all we have to do is add in our bounds."},{"Start":"06:41.375 ","End":"06:44.255","Text":"When we\u0027re integrating on our r,"},{"Start":"06:44.255 ","End":"06:47.510","Text":"so obviously it\u0027s going from 0 until our maximum radius,"},{"Start":"06:47.510 ","End":"06:51.770","Text":"which is R, then d Theta."},{"Start":"06:51.770 ","End":"06:56.320","Text":"Just 1 moment."},{"Start":"06:56.990 ","End":"07:03.870","Text":"Our d Theta is going all around the circle like this."},{"Start":"07:07.820 ","End":"07:11.360","Text":"Therefore, when we\u0027re adding in our bounds,"},{"Start":"07:11.360 ","End":"07:14.530","Text":"we\u0027ll say 0-2 Pi because it\u0027s a full circle."},{"Start":"07:14.530 ","End":"07:16.835","Text":"D Phi on the other hand, now,"},{"Start":"07:16.835 ","End":"07:22.460","Text":"usually when we\u0027re integrating on the sphere we\u0027ll have from 0 to Pi."},{"Start":"07:22.460 ","End":"07:25.965","Text":"Because usually, let\u0027s do that again,"},{"Start":"07:25.965 ","End":"07:31.935","Text":"so we\u0027ll be going all the way from here and all the way down half a sphere,"},{"Start":"07:31.935 ","End":"07:36.680","Text":"and then in order to get to the other side over here, like that,"},{"Start":"07:36.680 ","End":"07:40.535","Text":"then we simply just rotate by Theta,"},{"Start":"07:40.535 ","End":"07:42.245","Text":"and then we\u0027ll get over there."},{"Start":"07:42.245 ","End":"07:44.285","Text":"Now, here on the other hand,"},{"Start":"07:44.285 ","End":"07:45.950","Text":"we\u0027re only doing half a sphere,"},{"Start":"07:45.950 ","End":"07:49.470","Text":"so we don\u0027t have to go Pi all the way down,"},{"Start":"07:49.470 ","End":"07:53.595","Text":"we have to go just Pi divided by 2."},{"Start":"07:53.595 ","End":"07:56.880","Text":"Then when we\u0027re rotating by Theta,"},{"Start":"07:56.880 ","End":"08:01.770","Text":"so we\u0027ll do Pi over 2 all around this half sphere."},{"Start":"08:01.770 ","End":"08:12.400","Text":"That means that our bounds for Phi is going to be 0 to Pi divided by 2."},{"Start":"08:12.560 ","End":"08:17.065","Text":"Now, to make this integration slightly easier,"},{"Start":"08:17.065 ","End":"08:22.300","Text":"notice that we have cosine Phi multiplied by sine Phi,"},{"Start":"08:22.300 ","End":"08:33.380","Text":"so we can change this using a trig identity and say that it\u0027s half sine of 2 Phi."},{"Start":"08:34.340 ","End":"08:38.595","Text":"Now let\u0027s do the integration."},{"Start":"08:38.595 ","End":"08:45.235","Text":"First of all, we can have on the outside 3 divided by 2 Pi R^3,"},{"Start":"08:45.235 ","End":"08:46.779","Text":"which is a constant."},{"Start":"08:46.779 ","End":"08:53.075","Text":"Then we\u0027re going to be multiplying by what\u0027s happening inside the integration."},{"Start":"08:53.075 ","End":"08:55.325","Text":"When we\u0027re integrating by r,"},{"Start":"08:55.325 ","End":"09:04.245","Text":"so we\u0027re going to have r^4 divided by 4,"},{"Start":"09:04.245 ","End":"09:06.850","Text":"and then if we substitute our bounds into this,"},{"Start":"09:06.850 ","End":"09:09.290","Text":"so we\u0027ll have 0 to R,"},{"Start":"09:09.290 ","End":"09:12.830","Text":"so when it\u0027s 0, it will obviously just cross out,"},{"Start":"09:12.830 ","End":"09:13.895","Text":"and when it\u0027s R,"},{"Start":"09:13.895 ","End":"09:16.985","Text":"we simply substitute in our capital"},{"Start":"09:16.985 ","End":"09:24.005","Text":"R. Then we\u0027re going to be multiplying by the integration according to Theta."},{"Start":"09:24.005 ","End":"09:25.265","Text":"Now, as we can see,"},{"Start":"09:25.265 ","End":"09:27.160","Text":"we don\u0027t have a variable Theta,"},{"Start":"09:27.160 ","End":"09:33.370","Text":"so all we have to do is multiply this by 2Pi."},{"Start":"09:35.060 ","End":"09:37.430","Text":"Again, when we\u0027re integrating,"},{"Start":"09:37.430 ","End":"09:38.840","Text":"we just add a Theta,"},{"Start":"09:38.840 ","End":"09:40.790","Text":"multiply the whole expression by Theta."},{"Start":"09:40.790 ","End":"09:42.725","Text":"Then we add in the balance of 0-2 Pi."},{"Start":"09:42.725 ","End":"09:45.035","Text":"Obviously when it\u0027s 0, everything will cross out,"},{"Start":"09:45.035 ","End":"09:47.120","Text":"and when it\u0027s 2Pi,"},{"Start":"09:47.120 ","End":"09:49.415","Text":"so we\u0027re just multiplying by 2Pi."},{"Start":"09:49.415 ","End":"09:53.680","Text":"Now we\u0027re going to be integrating by Phi."},{"Start":"09:53.680 ","End":"09:56.145","Text":"Let\u0027s see how we do this."},{"Start":"09:56.145 ","End":"10:00.785","Text":"We\u0027re integrating from 0 to Pi over 2"},{"Start":"10:00.785 ","End":"10:07.675","Text":"of a half sine of 2 Phi d Phi."},{"Start":"10:07.675 ","End":"10:11.305","Text":"Obviously the half can go outside,"},{"Start":"10:11.305 ","End":"10:17.340","Text":"so I can just cross out this half and cross out this 2 and it balances out."},{"Start":"10:19.010 ","End":"10:22.860","Text":"Let\u0027s use substitution."},{"Start":"10:22.860 ","End":"10:29.045","Text":"We can say that u is equal to 2 Phi,"},{"Start":"10:29.045 ","End":"10:33.530","Text":"and that means that du is equal to 2d Phi,"},{"Start":"10:33.530 ","End":"10:39.800","Text":"and that means that half du is equal to d Phi."},{"Start":"10:39.800 ","End":"10:48.290","Text":"Now we\u0027re just going to integrate according to u."},{"Start":"10:48.290 ","End":"10:53.225","Text":"We\u0027re going to have 0 to Pi over 2 of"},{"Start":"10:53.225 ","End":"11:00.345","Text":"sine u multiplied by a half du."},{"Start":"11:00.345 ","End":"11:05.650","Text":"As we know, sine u integrates into negative cosine,"},{"Start":"11:05.650 ","End":"11:11.360","Text":"so we\u0027re going to have negative a half cosine of u."},{"Start":"11:11.360 ","End":"11:13.340","Text":"We\u0027re substituting back in our u,"},{"Start":"11:13.340 ","End":"11:15.485","Text":"which is our 2 Phi."},{"Start":"11:15.485 ","End":"11:17.990","Text":"Then we\u0027re going to substitute in our bounds,"},{"Start":"11:17.990 ","End":"11:21.820","Text":"which is from Pi over 2to 0."},{"Start":"11:21.820 ","End":"11:28.505","Text":"Now let\u0027s draw the graph of cosine because it will be easier to solve this way."},{"Start":"11:28.505 ","End":"11:31.985","Text":"Cosine of 0 is 1,"},{"Start":"11:31.985 ","End":"11:35.205","Text":"and this is Pi over 2,"},{"Start":"11:35.205 ","End":"11:37.785","Text":"and this is at Pi."},{"Start":"11:37.785 ","End":"11:41.470","Text":"When we substitute for Phi Pi over 2,"},{"Start":"11:41.470 ","End":"11:46.300","Text":"notice that we\u0027re multiplying by 2 because it\u0027s 2 Phi, not just Phi."},{"Start":"11:46.300 ","End":"11:49.175","Text":"We\u0027re actually working out cosine of Pi."},{"Start":"11:49.175 ","End":"11:53.159","Text":"Now, cosine of Pi is over here and equals to negative 1."},{"Start":"11:53.159 ","End":"12:00.860","Text":"We\u0027re going to have that this is equal to negative a half of negative 1."},{"Start":"12:00.860 ","End":"12:04.460","Text":"It\u0027s just going to be a half minus,"},{"Start":"12:04.460 ","End":"12:08.539","Text":"and then cosine of 0,"},{"Start":"12:08.539 ","End":"12:13.830","Text":"cosine of 2 times 0 is going to be 1 multiplied by a half,"},{"Start":"12:13.830 ","End":"12:18.630","Text":"so then it\u0027s plus another half, which equals 1."},{"Start":"12:19.190 ","End":"12:24.770","Text":"Now we can substitute all of that back in to our integral over here."},{"Start":"12:24.770 ","End":"12:32.820","Text":"We\u0027re going to have that all of this is equal to 1."},{"Start":"12:33.170 ","End":"12:37.155","Text":"Now we can cross out this Pi with this Pi,"},{"Start":"12:37.155 ","End":"12:40.410","Text":"R_4 with this R cubed,"},{"Start":"12:40.410 ","End":"12:51.660","Text":"and then we\u0027re going to be left with 3R divided by 8."},{"Start":"12:51.660 ","End":"12:57.060","Text":"This is the final answer."},{"Start":"12:57.060 ","End":"13:00.930","Text":"This equals zcm,"},{"Start":"13:00.930 ","End":"13:04.365","Text":"so at 3/8 of R,"},{"Start":"13:04.365 ","End":"13:12.600","Text":"so around about over here is going to be our zcm."},{"Start":"13:12.600 ","End":"13:15.820","Text":"That\u0027s the end of this lesson."}],"ID":9366},{"Watched":false,"Name":"Exercise- Center Of Mass Solid Cone","Duration":"10m 21s","ChapterTopicVideoID":9094,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:02.970","Text":"Hello. In this question,"},{"Start":"00:02.970 ","End":"00:07.200","Text":"we\u0027re going to be working out the center of mass of a solid cone."},{"Start":"00:07.200 ","End":"00:11.310","Text":"A cone will look something like this, again,"},{"Start":"00:11.310 ","End":"00:16.380","Text":"it\u0027s 3-dimensional, and it has radius capital R,"},{"Start":"00:16.380 ","End":"00:25.590","Text":"and its height is going to be H. Because of symmetry in the cone,"},{"Start":"00:25.590 ","End":"00:29.580","Text":"we can say that our axis start at the origin right in the center,"},{"Start":"00:29.580 ","End":"00:32.490","Text":"and then we have z-axis going up,"},{"Start":"00:32.490 ","End":"00:34.315","Text":"and because of symmetry,"},{"Start":"00:34.315 ","End":"00:40.430","Text":"the center of mass for x and y is going to be at the origin, at 0."},{"Start":"00:40.430 ","End":"00:44.950","Text":"Then all we have to work out is Z_cm,"},{"Start":"00:44.950 ","End":"00:46.620","Text":"and let\u0027s write the formula,"},{"Start":"00:46.620 ","End":"00:53.390","Text":"so it\u0027s the integral of Z_dm divided by the integral on dm,"},{"Start":"00:53.390 ","End":"00:54.635","Text":"which as we know,"},{"Start":"00:54.635 ","End":"00:57.815","Text":"is the total mass of the system."},{"Start":"00:57.815 ","End":"01:02.315","Text":"We can just leave it as a capital M because we\u0027ll see that soon it will cross"},{"Start":"01:02.315 ","End":"01:07.125","Text":"out and it\u0027s irrelevant in the calculations."},{"Start":"01:07.125 ","End":"01:11.845","Text":"Now we have to figure out what our dm is."},{"Start":"01:11.845 ","End":"01:17.105","Text":"Our dm is going to be because we\u0027re dealing with a solid cone,"},{"Start":"01:17.105 ","End":"01:21.950","Text":"so it\u0027s a 3-dimensional shape and we\u0027re dealing with volume,"},{"Start":"01:21.950 ","End":"01:25.285","Text":"so we\u0027re going to use Rho dv."},{"Start":"01:25.285 ","End":"01:28.085","Text":"Now we have to figure out what our Rho is."},{"Start":"01:28.085 ","End":"01:31.325","Text":"Our Rho is our mass,"},{"Start":"01:31.325 ","End":"01:36.530","Text":"the total mass of the system divided by the total volume."},{"Start":"01:36.530 ","End":"01:39.370","Text":"Soon we\u0027re going to see how to work out the volume of a cone."},{"Start":"01:39.370 ","End":"01:44.310","Text":"In the meantime, let\u0027s just leave it as volume c, volume of a cone."},{"Start":"01:44.330 ","End":"01:48.615","Text":"Then let\u0027s work out what our dv is,"},{"Start":"01:48.615 ","End":"01:52.550","Text":"so that means what we are going to be integrating via what coordinate,"},{"Start":"01:52.550 ","End":"01:55.090","Text":"so we\u0027re going to be using."},{"Start":"01:55.090 ","End":"02:00.450","Text":"Because we\u0027re integrating on a cone,"},{"Start":"02:00.450 ","End":"02:04.005","Text":"so it\u0027s easiest to use cylindrical coordinates,"},{"Start":"02:04.005 ","End":"02:06.685","Text":"not spherical, but cylindrical."},{"Start":"02:06.685 ","End":"02:14.490","Text":"What does that mean? It means r dr d Theta dz,"},{"Start":"02:15.980 ","End":"02:19.725","Text":"cylindrical coordinates in 3-dimensions."},{"Start":"02:19.725 ","End":"02:21.380","Text":"There\u0027s also polar coordinates,"},{"Start":"02:21.380 ","End":"02:22.820","Text":"which is r dr d Theta,"},{"Start":"02:22.820 ","End":"02:26.090","Text":"but because we\u0027re dealing with 3-dimensions,"},{"Start":"02:26.090 ","End":"02:30.450","Text":"so we also multiply here by rdz."},{"Start":"02:30.530 ","End":"02:34.260","Text":"Let\u0027s plug everything into our equation."},{"Start":"02:34.260 ","End":"02:36.670","Text":"So Z_cm is going to be equal to,"},{"Start":"02:36.670 ","End":"02:37.800","Text":"from our equation,"},{"Start":"02:37.800 ","End":"02:46.455","Text":"1 divided by m multiplied by 3 integral signs multiplied by z, from our equation,"},{"Start":"02:46.455 ","End":"02:49.700","Text":"and dm, which is our Rho dv,"},{"Start":"02:49.700 ","End":"02:58.200","Text":"which is M over volume of cone multiplied by r dr d Theta dz."},{"Start":"02:58.700 ","End":"03:00.725","Text":"As we can see,"},{"Start":"03:00.725 ","End":"03:04.490","Text":"as I promised, my masses cross out."},{"Start":"03:04.490 ","End":"03:10.430","Text":"Beautiful. Before we start integrating,"},{"Start":"03:10.430 ","End":"03:13.045","Text":"we have to add in our bounds."},{"Start":"03:13.045 ","End":"03:16.845","Text":"What our bounds for the integration?"},{"Start":"03:16.845 ","End":"03:20.790","Text":"We have 2 options."},{"Start":"03:20.790 ","End":"03:27.780","Text":"So one of the options is if we\u0027re going to integrate first by R,"},{"Start":"03:33.650 ","End":"03:39.480","Text":"so then our z will be from 0 until H and then we"},{"Start":"03:39.480 ","End":"03:45.225","Text":"have to figure out how our R changes as we go up the z axis."},{"Start":"03:45.225 ","End":"03:47.280","Text":"Let\u0027s just draw that in."},{"Start":"03:47.280 ","End":"03:51.360","Text":"So largest R is going to be capital R,"},{"Start":"03:51.360 ","End":"03:52.680","Text":"that\u0027s our largest radius,"},{"Start":"03:52.680 ","End":"03:56.615","Text":"but notice that as we go up the z axis,"},{"Start":"03:56.615 ","End":"03:59.915","Text":"our radiuses gets smaller and smaller and smaller."},{"Start":"03:59.915 ","End":"04:07.440","Text":"That means that we would have to find how"},{"Start":"04:07.440 ","End":"04:15.140","Text":"R changes according to z and substitute that in into the bounds for R. Alternatively,"},{"Start":"04:15.140 ","End":"04:21.380","Text":"we can say that we\u0027re integrating first according to our radius and then we\u0027ll"},{"Start":"04:21.380 ","End":"04:23.990","Text":"have to do the exact same thing but find the equation of"},{"Start":"04:23.990 ","End":"04:28.490","Text":"how z changes according to the radius."},{"Start":"04:28.490 ","End":"04:33.170","Text":"Let\u0027s do, in my opinion, the simpler option."},{"Start":"04:33.170 ","End":"04:38.975","Text":"Let\u0027s say that we\u0027re going to integrate by z from the maximum height,"},{"Start":"04:38.975 ","End":"04:42.235","Text":"which is from 0 until H,"},{"Start":"04:42.235 ","End":"04:45.130","Text":"because H is over here."},{"Start":"04:45.470 ","End":"04:51.060","Text":"Then we\u0027ll figure out how our R changes according to our z,"},{"Start":"04:51.060 ","End":"04:52.500","Text":"which we\u0027ll do in a second,"},{"Start":"04:52.500 ","End":"04:55.100","Text":"and then our Theta because it has nothing"},{"Start":"04:55.100 ","End":"04:57.800","Text":"to do with our z or our R. We can just straight away say,"},{"Start":"04:57.800 ","End":"05:00.650","Text":"because we\u0027re going full circle around the cone,"},{"Start":"05:00.650 ","End":"05:03.380","Text":"so it will be 0 to 2 Pi."},{"Start":"05:03.380 ","End":"05:06.755","Text":"Now our R how it changes according to z."},{"Start":"05:06.755 ","End":"05:10.295","Text":"Obviously, our R is going to start from 0 and now"},{"Start":"05:10.295 ","End":"05:14.770","Text":"we just have to find what our upper bound is."},{"Start":"05:14.770 ","End":"05:22.550","Text":"How do we do that? We just find the line equation because notice we\u0027re integrating,"},{"Start":"05:22.550 ","End":"05:23.705","Text":"this is our bound,"},{"Start":"05:23.705 ","End":"05:25.715","Text":"this line over here."},{"Start":"05:25.715 ","End":"05:29.870","Text":"Let\u0027s draw it in red. This is our bound and"},{"Start":"05:29.870 ","End":"05:33.710","Text":"then it just goes around according to Theta in the circle,"},{"Start":"05:33.710 ","End":"05:41.820","Text":"but the equation for the line will always stay the same with the variables r and z."},{"Start":"05:43.970 ","End":"05:48.000","Text":"Let\u0027s see what this is. So we start by writing,"},{"Start":"05:48.000 ","End":"05:53.500","Text":"we have the equation y equals mx plus c,"},{"Start":"05:53.500 ","End":"05:54.800","Text":"so the exact same thing."},{"Start":"05:54.800 ","End":"05:56.670","Text":"But instead of y,"},{"Start":"05:56.670 ","End":"05:59.605","Text":"we\u0027re substituting in z,"},{"Start":"05:59.605 ","End":"06:02.045","Text":"and instead of x,"},{"Start":"06:02.045 ","End":"06:06.710","Text":"we\u0027re substituting in r. We\u0027re going to say that z is,"},{"Start":"06:06.710 ","End":"06:12.645","Text":"and now we have to find the gradient or the slope."},{"Start":"06:12.645 ","End":"06:21.165","Text":"We know that we have a rise of H divided by R steps to the side."},{"Start":"06:21.165 ","End":"06:27.960","Text":"So we\u0027ll have H divided by R. Remember rise over run, H over R,"},{"Start":"06:27.960 ","End":"06:31.430","Text":"and then because we can see that it\u0027s going in the negative direction,"},{"Start":"06:31.430 ","End":"06:33.830","Text":"it\u0027s not going in this direction,"},{"Start":"06:33.830 ","End":"06:35.420","Text":"which is the positive direction,"},{"Start":"06:35.420 ","End":"06:37.255","Text":"it\u0027s going downwards,"},{"Start":"06:37.255 ","End":"06:39.965","Text":"so we add a negative sign over here."},{"Start":"06:39.965 ","End":"06:41.750","Text":"Then instead of x,"},{"Start":"06:41.750 ","End":"06:43.820","Text":"we have our r over here."},{"Start":"06:43.820 ","End":"06:49.200","Text":"Then our c represents where the line cuts the axis,"},{"Start":"06:49.200 ","End":"06:51.705","Text":"which as we can see here, it cuts at H,"},{"Start":"06:51.705 ","End":"06:56.195","Text":"so plus H. But now,"},{"Start":"06:56.195 ","End":"06:58.715","Text":"because we\u0027re integrating according to r,"},{"Start":"06:58.715 ","End":"07:01.565","Text":"when we substitute this as the upper bound,"},{"Start":"07:01.565 ","End":"07:09.210","Text":"it means that we have to isolate out r because we want to see that as we change our z,"},{"Start":"07:09.210 ","End":"07:12.915","Text":"what our r is going to be equal to."},{"Start":"07:12.915 ","End":"07:17.705","Text":"Then all we have to do is simply isolate our r. So we\u0027ll have that r is"},{"Start":"07:17.705 ","End":"07:23.040","Text":"equal to negative R divided by H,"},{"Start":"07:23.040 ","End":"07:26.145","Text":"z minus H,"},{"Start":"07:26.145 ","End":"07:28.200","Text":"simple algebra, you rearrange it."},{"Start":"07:28.200 ","End":"07:34.510","Text":"Then we\u0027ll see that as z changes what our r is going to be equal to."},{"Start":"07:35.720 ","End":"07:41.960","Text":"Now let\u0027s substitute this back into Z_cm equation,"},{"Start":"07:41.960 ","End":"07:46.310","Text":"so we\u0027re going to have 0 to H,"},{"Start":"07:46.310 ","End":"07:49.440","Text":"0 to 2Pi,"},{"Start":"07:50.810 ","End":"07:58.180","Text":"0 until negative R divided by H,"},{"Start":"08:03.170 ","End":"08:07.425","Text":"z minus H,"},{"Start":"08:07.425 ","End":"08:13.400","Text":"and then we\u0027ll have z divided by volume of a cone,"},{"Start":"08:13.400 ","End":"08:18.120","Text":"r dr d Theta, and then dz."},{"Start":"08:19.730 ","End":"08:23.705","Text":"Now we have to see what our volume of a cone is."},{"Start":"08:23.705 ","End":"08:25.490","Text":"So we can just substitute it in,"},{"Start":"08:25.490 ","End":"08:31.515","Text":"you can find it anywhere on the Internet and it should also be in your equation sheets."},{"Start":"08:31.515 ","End":"08:40.545","Text":"So volume of a cone is equal to Pi R^2 H divided by 3."},{"Start":"08:40.545 ","End":"08:42.765","Text":"Then we can substitute that in."},{"Start":"08:42.765 ","End":"08:48.000","Text":"Because Pi capital R^2 H and 3"},{"Start":"08:48.000 ","End":"08:54.450","Text":"are all independent of small r,"},{"Start":"08:54.450 ","End":"08:58.865","Text":"Theta, and z, we can put it outside of the equation."},{"Start":"08:58.865 ","End":"09:05.330","Text":"We\u0027ll have that Z_cm is equal to Pi R^2 H divided by 3,"},{"Start":"09:05.330 ","End":"09:09.410","Text":"0 to H,"},{"Start":"09:09.410 ","End":"09:12.180","Text":"0 to 2Pi,"},{"Start":"09:12.180 ","End":"09:16.620","Text":"0 to negative R divided by H,"},{"Start":"09:16.620 ","End":"09:19.770","Text":"z minus H,"},{"Start":"09:19.770 ","End":"09:24.955","Text":"of Z r dr d Theta dz."},{"Start":"09:24.955 ","End":"09:27.095","Text":"In order to save time,"},{"Start":"09:27.095 ","End":"09:29.540","Text":"I\u0027m not going to do the integration,"},{"Start":"09:29.540 ","End":"09:33.156","Text":"but please do it because it\u0027s good practice to do this."},{"Start":"09:33.156 ","End":"09:37.490","Text":"Also remember that once you integrate according to R and you substitute in the bounds,"},{"Start":"09:37.490 ","End":"09:40.120","Text":"you\u0027ll have the variable z as well."},{"Start":"09:40.120 ","End":"09:46.895","Text":"So don\u0027t forget to incorporate that into your equation when you\u0027re integrating via z."},{"Start":"09:46.895 ","End":"09:51.590","Text":"Also notice that because there\u0027s no Theta in the equation, you can just,"},{"Start":"09:51.590 ","End":"09:55.220","Text":"instead of writing d Theta and then the bounds 0 to 2Pi,"},{"Start":"09:55.220 ","End":"10:00.855","Text":"you can just multiply the entire expression just by the constant 2Pi."},{"Start":"10:00.855 ","End":"10:07.610","Text":"Just in this case specifically because there\u0027s no variable Theta in this equation."},{"Start":"10:07.610 ","End":"10:14.550","Text":"Then the final answer that you should get is going to be H divided by 4."},{"Start":"10:14.550 ","End":"10:19.595","Text":"So this is the final answer that you should get."},{"Start":"10:19.595 ","End":"10:22.470","Text":"That\u0027s the end of this lesson."}],"ID":9367},{"Watched":false,"Name":"Exercise- Half Hoop With Two Masses","Duration":"17m 5s","ChapterTopicVideoID":9096,"CourseChapterTopicPlaylistID":5341,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.680 ","End":"00:03.060","Text":"Hello. In this question,"},{"Start":"00:03.060 ","End":"00:09.690","Text":"we\u0027re being told to find the center of mass of the half hoop of mass M and radius R,"},{"Start":"00:09.690 ","End":"00:12.630","Text":"when we\u0027re also told that at each end of"},{"Start":"00:12.630 ","End":"00:18.719","Text":"the half, hoop there is also a ball of mass, m attached."},{"Start":"00:18.719 ","End":"00:22.590","Text":"How we\u0027re going to do this is first we\u0027re going to find the center of"},{"Start":"00:22.590 ","End":"00:27.195","Text":"mass of the half hoop,"},{"Start":"00:27.195 ","End":"00:28.950","Text":"say it\u0027s somewhere around here."},{"Start":"00:28.950 ","End":"00:31.890","Text":"Then we already know that because"},{"Start":"00:31.890 ","End":"00:35.460","Text":"these balls at each end can be considered already point masses."},{"Start":"00:35.460 ","End":"00:39.050","Text":"Then we\u0027ll find the center of mass of this,"},{"Start":"00:39.050 ","End":"00:41.715","Text":"this, and this."},{"Start":"00:41.715 ","End":"00:44.090","Text":"We\u0027re doing this in 2 stages."},{"Start":"00:44.090 ","End":"00:47.210","Text":"We\u0027re finding the center of mass of the hoop,"},{"Start":"00:47.210 ","End":"00:51.090","Text":"and then using that to find the center of mass of the system."},{"Start":"00:51.530 ","End":"00:54.590","Text":"What we\u0027re going to do now is we\u0027re going to start by"},{"Start":"00:54.590 ","End":"00:57.060","Text":"finding the center of mass of just the hoop,"},{"Start":"00:57.060 ","End":"01:02.475","Text":"without these masses, just the hoop."},{"Start":"01:02.475 ","End":"01:03.650","Text":"Let\u0027s see how we do that."},{"Start":"01:03.650 ","End":"01:05.299","Text":"Now, because of symmetry,"},{"Start":"01:05.299 ","End":"01:08.935","Text":"we\u0027ll set up our axes right in the center."},{"Start":"01:08.935 ","End":"01:12.410","Text":"We\u0027re just going to be working with x-y axes."},{"Start":"01:12.410 ","End":"01:17.190","Text":"So we\u0027re going to say that this is the origin."},{"Start":"01:17.740 ","End":"01:21.980","Text":"Now what we\u0027re going to do is write down the equation for y_cm,"},{"Start":"01:21.980 ","End":"01:25.370","Text":"because we want to find our center of mass is going to be,"},{"Start":"01:25.370 ","End":"01:27.005","Text":"this is the y-axis,"},{"Start":"01:27.005 ","End":"01:28.310","Text":"this is the x-axis,"},{"Start":"01:28.310 ","End":"01:32.074","Text":"our y center of mass is going to be somewhere around here on the y-axis."},{"Start":"01:32.074 ","End":"01:38.145","Text":"Because of symmetry, it\u0027s just going to be at the origin on the x-axis."},{"Start":"01:38.145 ","End":"01:39.965","Text":"y_cm, as you know,"},{"Start":"01:39.965 ","End":"01:49.670","Text":"is the integral of y_dm divided by the integral of dm."},{"Start":"01:49.670 ","End":"01:54.050","Text":"As we know, this is equal to the mass of the system,"},{"Start":"01:54.050 ","End":"01:58.860","Text":"which here is M. That means that we are going"},{"Start":"01:58.860 ","End":"02:04.090","Text":"to do 1 divided by m multiplied by the integral of y_dm."},{"Start":"02:04.090 ","End":"02:07.805","Text":"The first thing we have to see is what is our dm?"},{"Start":"02:07.805 ","End":"02:09.170","Text":"Let\u0027s write it on this side."},{"Start":"02:09.170 ","End":"02:10.730","Text":"We have dm."},{"Start":"02:10.730 ","End":"02:15.900","Text":"Now because we\u0027re not dealing with volume and we\u0027re not dealing with area,"},{"Start":"02:15.900 ","End":"02:20.455","Text":"we\u0027re dealing with arc length, so length."},{"Start":"02:20.455 ","End":"02:24.615","Text":"We\u0027re going to be saying that it\u0027s Lambda dl."},{"Start":"02:24.615 ","End":"02:28.450","Text":"Then over here, our dl,"},{"Start":"02:28.450 ","End":"02:33.432","Text":"because we\u0027re going in a radial direction,"},{"Start":"02:33.432 ","End":"02:41.430","Text":"we\u0027re going to say that this is going to be equal to Lambda Rd Theta."},{"Start":"02:41.870 ","End":"02:45.510","Text":"Remember, Lambda is our density,"},{"Start":"02:45.510 ","End":"02:47.880","Text":"R is our Jacobian,"},{"Start":"02:47.880 ","End":"02:55.480","Text":"and our Theta is our variable because we\u0027re moving along the radial direction."},{"Start":"02:56.450 ","End":"02:58.760","Text":"If it\u0027s in the radial direction,"},{"Start":"02:58.760 ","End":"03:00.080","Text":"then we\u0027re going around in a circle,"},{"Start":"03:00.080 ","End":"03:04.200","Text":"so the angle is changing, so that\u0027s Theta."},{"Start":"03:05.270 ","End":"03:07.370","Text":"Now that we have that,"},{"Start":"03:07.370 ","End":"03:11.600","Text":"we can say also what our Lambda is equal to."},{"Start":"03:11.600 ","End":"03:16.070","Text":"Our Lambda is equal to the mass of the 1/2 hoop,"},{"Start":"03:16.070 ","End":"03:20.365","Text":"just the 1/2 hoop, divided by its length."},{"Start":"03:20.365 ","End":"03:21.800","Text":"Because this is a 1/2 hoop,"},{"Start":"03:21.800 ","End":"03:27.590","Text":"it\u0027s like a half circle and the circumference is 2PiR."},{"Start":"03:27.590 ","End":"03:30.774","Text":"Now because we\u0027re dealing with 1/2 the circumference of a circle,"},{"Start":"03:30.774 ","End":"03:32.135","Text":"instead of 2PiR,"},{"Start":"03:32.135 ","End":"03:40.720","Text":"it\u0027s going to be just PiR multiplied by R multiplied by d Theta."},{"Start":"03:41.360 ","End":"03:44.805","Text":"That deals with what our dm is,"},{"Start":"03:44.805 ","End":"03:49.860","Text":"and reminding you that dm is the mass."},{"Start":"03:49.860 ","End":"03:58.040","Text":"The mass that we have in infinitely small sections along the circumference,"},{"Start":"03:58.040 ","End":"04:00.740","Text":"along the length of this 1/2 hoop."},{"Start":"04:00.740 ","End":"04:04.835","Text":"Now we see that we\u0027re integrating according to Theta."},{"Start":"04:04.835 ","End":"04:07.550","Text":"But as we can see in our integral,"},{"Start":"04:07.550 ","End":"04:10.760","Text":"we have the variable y. y,"},{"Start":"04:10.760 ","End":"04:13.070","Text":"as we can see, is constantly changing."},{"Start":"04:13.070 ","End":"04:15.770","Text":"Because if we\u0027re over here,"},{"Start":"04:15.770 ","End":"04:19.580","Text":"then our y is all the way down here. Draw this."},{"Start":"04:19.580 ","End":"04:23.015","Text":"Then as we move along,"},{"Start":"04:23.015 ","End":"04:25.985","Text":"our y suddenly is way up here."},{"Start":"04:25.985 ","End":"04:29.540","Text":"It\u0027s constantly changing height and constantly changing value."},{"Start":"04:29.540 ","End":"04:36.960","Text":"What we have to do is we have to find a way to express this according to Theta."},{"Start":"04:36.960 ","End":"04:43.393","Text":"Luckily for us, we know that there\u0027s the cosine and sine functions and the unit circle,"},{"Start":"04:43.393 ","End":"04:44.765","Text":"so this should be easy,"},{"Start":"04:44.765 ","End":"04:46.750","Text":"so let\u0027s take a look."},{"Start":"04:46.750 ","End":"04:50.550","Text":"We\u0027re going to be dealing with the angle Theta,"},{"Start":"04:50.550 ","End":"04:56.040","Text":"and what do we want to do is we want to find the height of where our y is."},{"Start":"04:56.080 ","End":"05:02.090","Text":"Let me just stop right here and let\u0027s quickly say what our bounds will be for the Theta,"},{"Start":"05:02.090 ","End":"05:05.450","Text":"because depending on what we say our bounds will be,"},{"Start":"05:05.450 ","End":"05:08.830","Text":"that will make the difference between if we say that Theta"},{"Start":"05:08.830 ","End":"05:12.880","Text":"is here or if our Theta is here,"},{"Start":"05:12.880 ","End":"05:16.530","Text":"it makes a whole world of difference."},{"Start":"05:16.630 ","End":"05:19.685","Text":"Now our bounds, as we can see,"},{"Start":"05:19.685 ","End":"05:21.950","Text":"we know that for a circle,"},{"Start":"05:21.950 ","End":"05:25.190","Text":"then we would integrate from 0 until 2Pi."},{"Start":"05:25.190 ","End":"05:29.835","Text":"That\u0027s a full circle."},{"Start":"05:29.835 ","End":"05:32.190","Text":"However, we\u0027re dealing with the 1/2."},{"Start":"05:32.190 ","End":"05:37.700","Text":"What we can do for half a circle is we can either say that it\u0027s from 0 until Pi"},{"Start":"05:37.700 ","End":"05:47.570","Text":"or we can say that it\u0027s from negative Pi over 4 until positive Pi over 4."},{"Start":"05:47.570 ","End":"05:50.570","Text":"That means that we\u0027re starting from this point over here,"},{"Start":"05:50.570 ","End":"05:55.670","Text":"and then we either go down here or in this direction."},{"Start":"05:55.670 ","End":"06:00.170","Text":"That\u0027s from negative Pi over 4 to Pi over 4."},{"Start":"06:00.170 ","End":"06:03.575","Text":"If instead we want do it from 0 to Pi,"},{"Start":"06:03.575 ","End":"06:07.680","Text":"so it would be just going like this."},{"Start":"06:09.110 ","End":"06:15.830","Text":"For this question, we\u0027re going to say that it\u0027s from negative Pi over 4 to Pi over 4,"},{"Start":"06:15.830 ","End":"06:20.165","Text":"because that will make our integration a little bit easier."},{"Start":"06:20.165 ","End":"06:22.820","Text":"But it doesn\u0027t really matter, you can do it either way,"},{"Start":"06:22.820 ","End":"06:25.220","Text":"you\u0027ll get the same answer in the end,"},{"Start":"06:25.220 ","End":"06:30.320","Text":"but we\u0027re specifically going to do from negative Pi over 4 to Pi over 4."},{"Start":"06:30.320 ","End":"06:35.780","Text":"In that case, if this is our R,"},{"Start":"06:35.780 ","End":"06:45.860","Text":"because we are now"},{"Start":"06:45.860 ","End":"06:52.890","Text":"going in this direction from negative Pi over 4 to Pi over 4,"},{"Start":"06:52.890 ","End":"06:58.110","Text":"our Theta is in that case going to be over here."},{"Start":"06:58.110 ","End":"07:00.315","Text":"This is our Theta."},{"Start":"07:00.315 ","End":"07:04.455","Text":"That means that what is our y value going to be?"},{"Start":"07:04.455 ","End":"07:08.468","Text":"We have y and R and Theta."},{"Start":"07:08.468 ","End":"07:17.970","Text":"As we know from our SOHCAHTOA that if we\u0027re going to use our cos,"},{"Start":"07:17.970 ","End":"07:23.525","Text":"then we\u0027ll have cos of Theta is equal to our adjacent over our hypotenuse,"},{"Start":"07:23.525 ","End":"07:29.795","Text":"which here our adjacent angle is our y and our hypotenuse is our radius."},{"Start":"07:29.795 ","End":"07:33.275","Text":"Then if we just rearrange this out to get what y is,"},{"Start":"07:33.275 ","End":"07:39.455","Text":"then we\u0027ll get the R cosine of Theta is equal to y."},{"Start":"07:39.455 ","End":"07:45.511","Text":"Then we can just substitute that into our equation."},{"Start":"07:45.511 ","End":"07:48.535","Text":"Let\u0027s do this now."},{"Start":"07:48.535 ","End":"07:50.080","Text":"We have ycm,"},{"Start":"07:50.080 ","End":"07:53.545","Text":"we\u0027re going to substitute everything into the equation."},{"Start":"07:53.545 ","End":"07:59.905","Text":"We have 1 over m and then the integral of ydm."},{"Start":"07:59.905 ","End":"08:03.070","Text":"This, as we know, is going to be 1 over m,"},{"Start":"08:03.070 ","End":"08:10.255","Text":"and then the integral from negative Pi over 4 until Pi over 4."},{"Start":"08:10.255 ","End":"08:12.910","Text":"Then we\u0027re going to substitute in our y,"},{"Start":"08:12.910 ","End":"08:20.920","Text":"which is R cosine of theta multiplied by our dm and our dm,"},{"Start":"08:20.920 ","End":"08:29.575","Text":"as we saw, with m divided by Pi R multiplied by Rd Theta."},{"Start":"08:29.575 ","End":"08:32.995","Text":"This and this, can just cross off."},{"Start":"08:32.995 ","End":"08:37.480","Text":"This, and this 1 over m can cross off."},{"Start":"08:37.480 ","End":"08:40.855","Text":"Then we\u0027re just left with the integral,"},{"Start":"08:40.855 ","End":"08:42.460","Text":"we can move out all the constants,"},{"Start":"08:42.460 ","End":"08:44.319","Text":"so R and Pi are constants."},{"Start":"08:44.319 ","End":"08:50.200","Text":"R divided by Pi and the integral between negative Pi over 4 to"},{"Start":"08:50.200 ","End":"08:56.680","Text":"Pi over 4 of cosine Theta d Theta."},{"Start":"08:56.680 ","End":"09:01.735","Text":"Now because we chose the bounds of negative Pi over 4 and Pi over 4,"},{"Start":"09:01.735 ","End":"09:05.210","Text":"you\u0027re going to see why this was a shortcut in this case."},{"Start":"09:05.210 ","End":"09:08.315","Text":"Cosine of Theta."},{"Start":"09:08.315 ","End":"09:11.080","Text":"Let\u0027s take a look at this."},{"Start":"09:11.080 ","End":"09:14.800","Text":"When we integrated is going to be sine Theta,"},{"Start":"09:14.800 ","End":"09:18.880","Text":"because the derivative of sine Theta is cosine Theta."},{"Start":"09:18.880 ","End":"09:22.930","Text":"Then we\u0027re saying that the integral of cosine Theta,"},{"Start":"09:22.930 ","End":"09:29.845","Text":"sine Theta between the bounds of negative Pi over 4 to Pi over 4."},{"Start":"09:29.845 ","End":"09:32.740","Text":"Now notice that sine Theta,"},{"Start":"09:32.740 ","End":"09:34.825","Text":"because it\u0027s piracy is even,"},{"Start":"09:34.825 ","End":"09:37.450","Text":"it means that instead of substituting"},{"Start":"09:37.450 ","End":"09:46.405","Text":"sine of Pi over 4 minus sine of negative Pi over 4,"},{"Start":"09:46.405 ","End":"09:52.840","Text":"we can just say that this is equal to 2 sine Theta,"},{"Start":"09:52.840 ","End":"09:55.390","Text":"and then we can just cross out this negative Pi over 4,"},{"Start":"09:55.390 ","End":"09:59.290","Text":"and just substitute in Pi over 4."},{"Start":"09:59.290 ","End":"10:00.775","Text":"The bound is just going to be,"},{"Start":"10:00.775 ","End":"10:06.265","Text":"we just substitute in Pi over 4 is equal to Theta."},{"Start":"10:06.265 ","End":"10:13.600","Text":"Then sine of Pi over 4 is just sine of 45 degrees."},{"Start":"10:13.600 ","End":"10:16.300","Text":"Let\u0027s draw this out."},{"Start":"10:16.300 ","End":"10:21.010","Text":"We can see that over here is Pi over 2,"},{"Start":"10:21.010 ","End":"10:23.785","Text":"so somewhere along here is Pi over 4."},{"Start":"10:23.785 ","End":"10:26.740","Text":"If you remember from the unit circle,"},{"Start":"10:26.740 ","End":"10:33.710","Text":"sine of Pi over 4 is simply root 2 over 2."},{"Start":"10:33.990 ","End":"10:37.240","Text":"We just have 2 times root 2 over 2,"},{"Start":"10:37.240 ","End":"10:40.885","Text":"which is simply equal to root 2,"},{"Start":"10:40.885 ","End":"10:43.615","Text":"so that\u0027s where this integration,"},{"Start":"10:43.615 ","End":"10:48.085","Text":"so you can just draw a line."},{"Start":"10:48.085 ","End":"10:52.570","Text":"Therefore we can say that y center of mass is"},{"Start":"10:52.570 ","End":"10:57.145","Text":"equal to R divided by Pi multiplied by hands over here,"},{"Start":"10:57.145 ","End":"11:03.475","Text":"which is just root 2R divided by Pi,"},{"Start":"11:03.475 ","End":"11:13.375","Text":"and so this is our answer for center of mass of hoop."},{"Start":"11:13.375 ","End":"11:18.790","Text":"But now you remember that we have these 2 further masses attached to the hoop."},{"Start":"11:18.790 ","End":"11:20.725","Text":"Now in order to find,"},{"Start":"11:20.725 ","End":"11:24.820","Text":"because our original question is to find the center of mass of"},{"Start":"11:24.820 ","End":"11:30.775","Text":"the 1/2 hoop and these 2 masses of the entire system,"},{"Start":"11:30.775 ","End":"11:32.575","Text":"so let\u0027s see how we do that."},{"Start":"11:32.575 ","End":"11:36.520","Text":"We can say that root 2 over Pi R,"},{"Start":"11:36.520 ","End":"11:39.205","Text":"is where the center of mass of this is."},{"Start":"11:39.205 ","End":"11:42.490","Text":"I have no idea, something like this."},{"Start":"11:42.490 ","End":"11:45.520","Text":"This is center of mass H,"},{"Start":"11:45.520 ","End":"11:47.815","Text":"which has center of mass hoop."},{"Start":"11:47.815 ","End":"11:50.560","Text":"Now I\u0027m going to cross out all the working out and we\u0027re going to find"},{"Start":"11:50.560 ","End":"11:53.960","Text":"the center of mass of the entire system."},{"Start":"11:54.180 ","End":"11:58.375","Text":"Right now, we\u0027re dealing with 3 points,"},{"Start":"11:58.375 ","End":"12:02.560","Text":"which means that we have a finite amount of points and it\u0027s a small Number 3."},{"Start":"12:02.560 ","End":"12:11.185","Text":"We can use the equation to find the center of mass via using the summation equation,"},{"Start":"12:11.185 ","End":"12:16.630","Text":"so now we\u0027re going to do ycm system,"},{"Start":"12:16.630 ","End":"12:21.745","Text":"which is going to be equal to the sum of yi,"},{"Start":"12:21.745 ","End":"12:32.810","Text":"so its position and the y-axis multiplied by its mass divided by the sum of masses."},{"Start":"12:33.470 ","End":"12:37.665","Text":"Then in this case, again,"},{"Start":"12:37.665 ","End":"12:46.255","Text":"we\u0027re setting up that the origin is right in the center of the half sphere."},{"Start":"12:46.255 ","End":"12:48.249","Text":"Which means that because of symmetry,"},{"Start":"12:48.249 ","End":"12:51.580","Text":"that the x center of mass is just going to be at the origin."},{"Start":"12:51.580 ","End":"12:55.270","Text":"We\u0027re just dealing with on the y-axis."},{"Start":"12:55.270 ","End":"12:57.625","Text":"We\u0027re being told that this mass is m,"},{"Start":"12:57.625 ","End":"12:58.720","Text":"this mass is also m,"},{"Start":"12:58.720 ","End":"13:08.079","Text":"so we can say that that equals 2m multiplied by its distance in the y-axis,"},{"Start":"13:08.079 ","End":"13:10.405","Text":"which is multiplied by 0."},{"Start":"13:10.405 ","End":"13:13.480","Text":"This whole thing crosses off to 0."},{"Start":"13:13.480 ","End":"13:18.670","Text":"Plus capital M, the mass of the hoop,"},{"Start":"13:18.670 ","End":"13:21.505","Text":"multiplied by its distance,"},{"Start":"13:21.505 ","End":"13:27.020","Text":"so it\u0027s distance on the y-axis from the origin."},{"Start":"13:27.870 ","End":"13:30.250","Text":"We just sorted it out,"},{"Start":"13:30.250 ","End":"13:38.890","Text":"we found that its center of mass is located root 2 over Pi R on the y-axis."},{"Start":"13:38.890 ","End":"13:40.255","Text":"Sorry, this is y,"},{"Start":"13:40.255 ","End":"13:45.970","Text":"so it\u0027s center of mass is that distance away,"},{"Start":"13:45.970 ","End":"13:47.860","Text":"so that is the equation,"},{"Start":"13:47.860 ","End":"13:50.410","Text":"and then divide it by the sum of all of the masses."},{"Start":"13:50.410 ","End":"13:54.145","Text":"We have divided by M plus 2m."},{"Start":"13:54.145 ","End":"14:00.170","Text":"Then that just equals root 2MR."},{"Start":"14:00.720 ","End":"14:04.090","Text":"Then we can move the Pi down,"},{"Start":"14:04.090 ","End":"14:09.200","Text":"so divided by Pi multiplied by M plus 2m."},{"Start":"14:10.050 ","End":"14:18.835","Text":"This is our final answer for the center of mass of the entire system."},{"Start":"14:18.835 ","End":"14:25.210","Text":"Now what I can do is I can check myself to see if the units add up."},{"Start":"14:25.210 ","End":"14:27.595","Text":"The y center of mass of the hoop,"},{"Start":"14:27.595 ","End":"14:34.315","Text":"we need it to be in units of length because it\u0027s not an area, it\u0027s not a volume."},{"Start":"14:34.315 ","End":"14:37.360","Text":"We can see R, is in meters, it\u0027s in length."},{"Start":"14:37.360 ","End":"14:40.480","Text":"Perfect. Now the center of mass of the system."},{"Start":"14:40.480 ","End":"14:42.685","Text":"Let\u0027s see how we check this."},{"Start":"14:42.685 ","End":"14:47.650","Text":"With regards to the units of the y center of mass of the entire system."},{"Start":"14:47.650 ","End":"14:55.315","Text":"We again want it to be in length and we can see that we have mass, kilograms."},{"Start":"14:55.315 ","End":"14:57.985","Text":"This M plus 2m is also kilograms,"},{"Start":"14:57.985 ","End":"14:59.305","Text":"so they cross out."},{"Start":"14:59.305 ","End":"15:02.770","Text":"Then again we have which hasn\u0027t meters and of course root"},{"Start":"15:02.770 ","End":"15:06.880","Text":"2 and Pi in both cases are unitless."},{"Start":"15:06.880 ","End":"15:11.905","Text":"Now we also want to see if our answer is logical with regards to"},{"Start":"15:11.905 ","End":"15:19.495","Text":"the masses themselves the mass of each mass."},{"Start":"15:19.495 ","End":"15:23.650","Text":"If we see if the 2 small masses,"},{"Start":"15:23.650 ","End":"15:26.470","Text":"this one and this one,"},{"Start":"15:26.470 ","End":"15:32.530","Text":"so if m is tending to 0,"},{"Start":"15:32.530 ","End":"15:36.820","Text":"then we\u0027ll see that we get root 2 over"},{"Start":"15:36.820 ","End":"15:42.830","Text":"Pi multiplied by R. Because this m and this one will cross off,"},{"Start":"15:42.830 ","End":"15:45.360","Text":"this 2m will go to 0."},{"Start":"15:45.360 ","End":"15:48.920","Text":"When the small masses are tending to 0,"},{"Start":"15:48.920 ","End":"15:50.720","Text":"so as if they don\u0027t exist,"},{"Start":"15:50.720 ","End":"16:00.205","Text":"then we get that the ycm of the system is just equal to the ycm of the hoop,"},{"Start":"16:00.205 ","End":"16:07.260","Text":"which makes sense because then we\u0027re taking the case that these masses aren\u0027t existent,"},{"Start":"16:07.260 ","End":"16:09.800","Text":"so then it\u0027s like in the first case what we did."},{"Start":"16:09.800 ","End":"16:14.240","Text":"Then, if we\u0027re saying that the small masses are tending to infinity,"},{"Start":"16:14.240 ","End":"16:16.445","Text":"so they\u0027re infinitely heavy,"},{"Start":"16:16.445 ","End":"16:20.825","Text":"so then 2m will go to infinity,"},{"Start":"16:20.825 ","End":"16:24.185","Text":"which means that the whole denominator will go to infinity,"},{"Start":"16:24.185 ","End":"16:27.260","Text":"which means that ycm of the system will go to"},{"Start":"16:27.260 ","End":"16:35.190","Text":"0 or rather like this."},{"Start":"16:35.190 ","End":"16:38.300","Text":"If the ycm of the system is going to 0,"},{"Start":"16:38.300 ","End":"16:41.015","Text":"then that means that the center of mass will be at the origin."},{"Start":"16:41.015 ","End":"16:42.750","Text":"Which makes sense,"},{"Start":"16:42.750 ","End":"16:46.909","Text":"because that means that all of the mass will be pulled between"},{"Start":"16:46.909 ","End":"16:52.920","Text":"these 2 small masses because each one is infinitely heavy."},{"Start":"16:52.920 ","End":"16:58.060","Text":"Then the mass of the hoop is irrelevant in comparison."},{"Start":"16:58.060 ","End":"17:03.335","Text":"That\u0027s a way to check that what we got is actually logical."},{"Start":"17:03.335 ","End":"17:06.450","Text":"That\u0027s the end of this lesson."}],"ID":9369}],"Thumbnail":null,"ID":5341}]

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