[{"Name":"Introduction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Definition - Rigid Body","Duration":"1m 33s","ChapterTopicVideoID":10427,"CourseChapterTopicPlaylistID":139576,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.535","Text":"Hello. In this part of the course,"},{"Start":"00:02.535 ","End":"00:06.974","Text":"we\u0027re going to be speaking about rigid bodies and axis of rotation."},{"Start":"00:06.974 ","End":"00:10.709","Text":"The definition of a rigid body is that the distance"},{"Start":"00:10.709 ","End":"00:15.780","Text":"between any 2 points on the body remains constant."},{"Start":"00:15.780 ","End":"00:22.236","Text":"In other words, we have some body which isn\u0027t a point-mass,"},{"Start":"00:22.236 ","End":"00:24.090","Text":"it has a few points on it."},{"Start":"00:24.090 ","End":"00:27.900","Text":"It\u0027s a larger body and it\u0027s a solid."},{"Start":"00:27.900 ","End":"00:34.170","Text":"That means that the distance between any 2 points on this body will remain constant."},{"Start":"00:34.730 ","End":"00:39.305","Text":"The main things to know that this is a large body,"},{"Start":"00:39.305 ","End":"00:40.655","Text":"which is a solid."},{"Start":"00:40.655 ","End":"00:43.025","Text":"That means not a liquid or a gas,"},{"Start":"00:43.025 ","End":"00:47.270","Text":"and that the shape of the body therefore is not changing."},{"Start":"00:47.270 ","End":"00:51.920","Text":"So the difference between this unit and previous units that we\u0027ve looked"},{"Start":"00:51.920 ","End":"00:57.110","Text":"at is that before we\u0027ve either been considering a point-mass,"},{"Start":"00:57.110 ","End":"01:00.020","Text":"when we did our calculations,"},{"Start":"01:00.020 ","End":"01:04.085","Text":"or we\u0027ve worked out the center of mass and then"},{"Start":"01:04.085 ","End":"01:08.885","Text":"calculated the answers to our question based on the movement of the center of mass."},{"Start":"01:08.885 ","End":"01:14.350","Text":"We\u0027ve been looking at our bodies from an outward perspective."},{"Start":"01:14.350 ","End":"01:16.355","Text":"In this unit, however,"},{"Start":"01:16.355 ","End":"01:21.890","Text":"we\u0027re going to be looking at the inner forces that are acting within the body,"},{"Start":"01:21.890 ","End":"01:25.760","Text":"that\u0027s why we\u0027re looking at a rigid body and we\u0027re now not considering"},{"Start":"01:25.760 ","End":"01:29.300","Text":"it as a point-mass or looking at the center of mass,"},{"Start":"01:29.300 ","End":"01:34.410","Text":"but we\u0027re rather looking at what is happening within the body itself."}],"ID":10783},{"Watched":false,"Name":"Definition - Axis of Rotation","Duration":"7m 35s","ChapterTopicVideoID":10428,"CourseChapterTopicPlaylistID":139576,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"Now what we\u0027re going to look at is the rigid body,"},{"Start":"00:04.830 ","End":"00:10.440","Text":"some body rotating around some axis."},{"Start":"00:10.440 ","End":"00:13.065","Text":"It will be the axis of rotation."},{"Start":"00:13.065 ","End":"00:14.850","Text":"This is the axis of rotation,"},{"Start":"00:14.850 ","End":"00:18.615","Text":"and now I\u0027m drawing some random shaped body,"},{"Start":"00:18.615 ","End":"00:21.150","Text":"can be anything random."},{"Start":"00:21.150 ","End":"00:27.870","Text":"We can see that the axis of rotation is passing through this rigid body."},{"Start":"00:27.870 ","End":"00:30.110","Text":"Now, it\u0027s going to be very,"},{"Start":"00:30.110 ","End":"00:35.870","Text":"very useful looking at the axis of rotation of the rigid body because this axis"},{"Start":"00:35.870 ","End":"00:42.620","Text":"of rotation is going to help us to describe the inner motions inside the rigid body."},{"Start":"00:42.620 ","End":"00:49.700","Text":"Now, this axis of rotation can be at any point within the rigid body,"},{"Start":"00:49.700 ","End":"00:51.860","Text":"at some random arbitrary point."},{"Start":"00:51.860 ","End":"00:55.190","Text":"It can also go through the center of mass of the rigid body."},{"Start":"00:55.190 ","End":"00:59.810","Text":"Or the axis of rotation might even be located outside of the body,"},{"Start":"00:59.810 ","End":"01:01.925","Text":"say, somewhere over here."},{"Start":"01:01.925 ","End":"01:10.190","Text":"Now what\u0027s important to know about the axis of rotation is that every single point on"},{"Start":"01:10.190 ","End":"01:19.370","Text":"the rigid body has to be doing the exact same circular motion about the axis of rotation."},{"Start":"01:19.370 ","End":"01:23.240","Text":"The exact same circular motion in the same direction and with"},{"Start":"01:23.240 ","End":"01:27.650","Text":"the exact same angular velocity about the axis of rotation."},{"Start":"01:27.650 ","End":"01:30.210","Text":"That\u0027s what\u0027s important to know."},{"Start":"01:30.830 ","End":"01:33.455","Text":"I\u0027ve written out what I just said,"},{"Start":"01:33.455 ","End":"01:34.970","Text":"that the axis of rotation,"},{"Start":"01:34.970 ","End":"01:36.575","Text":"this is the definition."},{"Start":"01:36.575 ","End":"01:39.740","Text":"All the points on the rigid body rotate about"},{"Start":"01:39.740 ","End":"01:43.565","Text":"the axis of rotation with the same angular velocity."},{"Start":"01:43.565 ","End":"01:46.430","Text":"It\u0027s dependent of whether the axis of rotation is"},{"Start":"01:46.430 ","End":"01:49.715","Text":"located at some arbitrary point on the body,"},{"Start":"01:49.715 ","End":"01:51.485","Text":"at the center of mass of the body,"},{"Start":"01:51.485 ","End":"01:53.435","Text":"or outside of the body."},{"Start":"01:53.435 ","End":"01:56.495","Text":"Now I\u0027ve put this definition in"},{"Start":"01:56.495 ","End":"02:00.710","Text":"red because I would recommend writing this even in your equation sheets."},{"Start":"02:00.710 ","End":"02:03.350","Text":"This is very, very useful, very important."},{"Start":"02:03.350 ","End":"02:06.935","Text":"You\u0027re also going to see later on in this chapter"},{"Start":"02:06.935 ","End":"02:12.780","Text":"how this is really going to help us solve a lot of the questions."},{"Start":"02:12.850 ","End":"02:18.095","Text":"Now let\u0027s take a look at what this actually means."},{"Start":"02:18.095 ","End":"02:21.805","Text":"If we take some point over here."},{"Start":"02:21.805 ","End":"02:27.950","Text":"We can see from this sentence over here that this point"},{"Start":"02:27.950 ","End":"02:34.115","Text":"over here is going to be rotating about the axis of rotation,"},{"Start":"02:34.115 ","End":"02:38.090","Text":"where its radius of rotation,"},{"Start":"02:38.090 ","End":"02:39.860","Text":"about the axis of rotation,"},{"Start":"02:39.860 ","End":"02:45.080","Text":"so this dotted circle is a circle around the axis of rotation,"},{"Start":"02:45.080 ","End":"02:49.440","Text":"and it\u0027s located perpendicular to the axis of rotation."},{"Start":"02:50.390 ","End":"02:55.685","Text":"It\u0027s perpendicular, and we can see that if this point is rotating around,"},{"Start":"02:55.685 ","End":"03:00.275","Text":"it\u0027s going to have some angular velocity of Omega."},{"Start":"03:00.275 ","End":"03:04.265","Text":"Similarly, if we take a look at any other point,"},{"Start":"03:04.265 ","End":"03:07.800","Text":"so let\u0027s say this point over here,"},{"Start":"03:07.960 ","End":"03:16.755","Text":"it\u0027s also going to be rotating about this axis of rotation over here."},{"Start":"03:16.755 ","End":"03:22.205","Text":"It\u0027s going to have a different radius away from the axis of rotation."},{"Start":"03:22.205 ","End":"03:29.840","Text":"It\u0027s also going to have the exact same angular velocity Omega rotating around."},{"Start":"03:29.840 ","End":"03:31.130","Text":"Again, this circle,"},{"Start":"03:31.130 ","End":"03:32.330","Text":"this blue dotted line,"},{"Start":"03:32.330 ","End":"03:35.610","Text":"is perpendicular to the axis of rotation."},{"Start":"03:36.130 ","End":"03:39.815","Text":"What we can see is that both of these points are"},{"Start":"03:39.815 ","End":"03:44.120","Text":"traveling in circular motion about the axis of rotation,"},{"Start":"03:44.120 ","End":"03:47.000","Text":"and they\u0027re rotating in the same direction."},{"Start":"03:47.000 ","End":"03:50.945","Text":"They have the exact same angular velocity."},{"Start":"03:50.945 ","End":"03:52.625","Text":"Now, what we can see though,"},{"Start":"03:52.625 ","End":"03:56.600","Text":"is that their position from the axis of rotation is different."},{"Start":"03:56.600 ","End":"04:04.515","Text":"Let\u0027s say that this point is a distance of r_1 away from the axis of rotation."},{"Start":"04:04.515 ","End":"04:11.890","Text":"This point over here has a distance or radius of r_2 away from the axis of rotation."},{"Start":"04:11.890 ","End":"04:17.630","Text":"Then we can see that although their angular velocities are going to be the same,"},{"Start":"04:17.630 ","End":"04:23.930","Text":"their linear velocity is V_1 is going to be different to V_2."},{"Start":"04:23.930 ","End":"04:31.325","Text":"V_1 is going to be equal to their angular velocity multiplied by r_1, and V_2,"},{"Start":"04:31.325 ","End":"04:33.740","Text":"the linear velocity of this point over here,"},{"Start":"04:33.740 ","End":"04:36.605","Text":"is also going to be equal to the same Omega,"},{"Start":"04:36.605 ","End":"04:41.010","Text":"however, multiplied by r_2."},{"Start":"04:42.260 ","End":"04:46.480","Text":"This equation is an equation that you should know"},{"Start":"04:46.480 ","End":"04:50.380","Text":"for connecting linear velocity to angular velocity."},{"Start":"04:50.380 ","End":"04:55.630","Text":"Linear velocity is equal to angular velocity multiplied by the radius."},{"Start":"04:55.630 ","End":"05:00.710","Text":"The distance that point is from the center of rotation."},{"Start":"05:00.710 ","End":"05:03.700","Text":"This is important, and that means that in"},{"Start":"05:03.700 ","End":"05:07.555","Text":"questions dealing with the rigid body and the axis of rotation,"},{"Start":"05:07.555 ","End":"05:11.110","Text":"we have to remember that every point on the rigid body has"},{"Start":"05:11.110 ","End":"05:15.055","Text":"the exact same angular velocity always."},{"Start":"05:15.055 ","End":"05:21.670","Text":"But the linear velocity won\u0027t be the same necessarily unless"},{"Start":"05:21.670 ","End":"05:25.550","Text":"the distance between the point and"},{"Start":"05:25.550 ","End":"05:30.003","Text":"the axis of rotation for both the points is equidistant,"},{"Start":"05:30.003 ","End":"05:31.750","Text":"which can sometimes happen."},{"Start":"05:31.750 ","End":"05:34.010","Text":"But usually, as a general rule,"},{"Start":"05:34.010 ","End":"05:37.175","Text":"the linear velocity is going to be different."},{"Start":"05:37.175 ","End":"05:43.400","Text":"Now, what happens if we\u0027re looking at a point which is on the rigid body and it\u0027s"},{"Start":"05:43.400 ","End":"05:50.905","Text":"located on the axis of rotation."},{"Start":"05:50.905 ","End":"05:54.075","Text":"That\u0027s this point over here in the green."},{"Start":"05:54.075 ","End":"05:58.115","Text":"This point is going to be rotating about itself."},{"Start":"05:58.115 ","End":"06:00.530","Text":"How do we do this calculation?"},{"Start":"06:00.530 ","End":"06:02.705","Text":"We can see that it\u0027s going to,"},{"Start":"06:02.705 ","End":"06:06.477","Text":"because it\u0027s rotating about itself can also mean that it\u0027s not moving"},{"Start":"06:06.477 ","End":"06:10.705","Text":"because it\u0027s on the axis of rotation."},{"Start":"06:10.705 ","End":"06:13.130","Text":"That also means that its radius,"},{"Start":"06:13.130 ","End":"06:16.220","Text":"it\u0027s a distance away from the axis of rotation is 0,"},{"Start":"06:16.220 ","End":"06:18.785","Text":"because it\u0027s located on the axis of rotation,"},{"Start":"06:18.785 ","End":"06:24.570","Text":"which means that its linear velocity is going to be equal to 0."},{"Start":"06:25.670 ","End":"06:30.640","Text":"The main points to take away from this lesson is to remember"},{"Start":"06:30.640 ","End":"06:36.690","Text":"this box in red over here and write this in your equation sheet."},{"Start":"06:36.690 ","End":"06:40.360","Text":"It\u0027s also to remember that every single point on a rigid body,"},{"Start":"06:40.360 ","End":"06:43.300","Text":"if it\u0027s rotating about an axis of rotation,"},{"Start":"06:43.300 ","End":"06:47.305","Text":"it doesn\u0027t matter where the axis of rotation is in relation to the rigid body."},{"Start":"06:47.305 ","End":"06:51.550","Text":"Every single point on that rigid body is going to"},{"Start":"06:51.550 ","End":"06:55.809","Text":"have the exact same angular velocity about the axis of rotation."},{"Start":"06:55.809 ","End":"06:59.560","Text":"However, because the points on the rigid body are"},{"Start":"06:59.560 ","End":"07:04.165","Text":"located at different radius is away from the axis of rotation,"},{"Start":"07:04.165 ","End":"07:07.765","Text":"that means that although their angular velocity will be identical,"},{"Start":"07:07.765 ","End":"07:11.510","Text":"their linear velocity will not."},{"Start":"07:11.510 ","End":"07:16.460","Text":"The next important thing to remember is that the point"},{"Start":"07:16.460 ","End":"07:21.830","Text":"that is located on the rigid body and on the axis of rotation."},{"Start":"07:21.830 ","End":"07:27.215","Text":"That point, its linear velocity is going to be equal to 0,"},{"Start":"07:27.215 ","End":"07:32.760","Text":"which means that it\u0027s not moving relative to the axis of rotation."},{"Start":"07:32.760 ","End":"07:35.680","Text":"That\u0027s the end of this lesson."}],"ID":10784}],"Thumbnail":null,"ID":139576},{"Name":"Introduction to Moment of Inertia","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation - Moment of Inertia","Duration":"7m 15s","ChapterTopicVideoID":10426,"CourseChapterTopicPlaylistID":139577,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this lesson,"},{"Start":"00:01.845 ","End":"00:05.235","Text":"we\u0027re going to be learning about moment of inertia."},{"Start":"00:05.235 ","End":"00:12.135","Text":"The moment of inertia is denoted by the letter or symbol capital"},{"Start":"00:12.135 ","End":"00:15.330","Text":"I and now we\u0027re going to start"},{"Start":"00:15.330 ","End":"00:20.085","Text":"the explanation by going over the equation for the moment of inertia."},{"Start":"00:20.085 ","End":"00:28.485","Text":"The equation for the moment of inertia is equal to the sum on I of all of"},{"Start":"00:28.485 ","End":"00:38.235","Text":"the masses multiplied by their distances from the axis of rotation squared."},{"Start":"00:38.235 ","End":"00:44.015","Text":"What does that mean? Let\u0027s say that we have over here axis of rotation"},{"Start":"00:44.015 ","End":"00:50.700","Text":"it\u0027s going upwards in the positive z direction and then we have some rod,"},{"Start":"00:50.700 ","End":"00:57.015","Text":"like so, and it\u0027s rotating in this direction about the axis of rotation."},{"Start":"00:57.015 ","End":"01:03.230","Text":"What we\u0027re going to do is we\u0027re going to split up our rod into tiny little segments,"},{"Start":"01:03.230 ","End":"01:08.150","Text":"where each segment is of mass mi and we\u0027re going to"},{"Start":"01:08.150 ","End":"01:13.690","Text":"measure the distance or so of each segment from the axis of rotation."},{"Start":"01:13.690 ","End":"01:17.467","Text":"This distance over here is going to be ri."},{"Start":"01:17.467 ","End":"01:23.375","Text":"Then we take our ri^2 and multiply it by mi,"},{"Start":"01:23.375 ","End":"01:27.580","Text":"and then do the same thing for every single bits along this rod,"},{"Start":"01:27.580 ","End":"01:31.940","Text":"and that is how we find out the moment of inertia."},{"Start":"01:32.570 ","End":"01:36.625","Text":"That\u0027s a little bit of a mathematical explanation,"},{"Start":"01:36.625 ","End":"01:41.545","Text":"intuitively we\u0027ll understand it and explain it a little bit more later,"},{"Start":"01:41.545 ","End":"01:43.300","Text":"I don\u0027t want to go into it now."},{"Start":"01:43.300 ","End":"01:49.715","Text":"But a way of thinking of the moment of inertia is thinking of it as mass."},{"Start":"01:49.715 ","End":"01:53.050","Text":"Now I\u0027ll explain that a little bit later why it\u0027s a good idea and"},{"Start":"01:53.050 ","End":"01:56.080","Text":"why it makes it a lot easier to consider"},{"Start":"01:56.080 ","End":"02:03.620","Text":"the moment of inertia as similar to a mass and let\u0027s carry on with explanation."},{"Start":"02:03.620 ","End":"02:07.310","Text":"Now what we\u0027re going to do is we\u0027re going to do"},{"Start":"02:07.310 ","End":"02:11.970","Text":"a worked example of a much easier example."},{"Start":"02:11.970 ","End":"02:18.330","Text":"Here again, we have our axis of rotation,"},{"Start":"02:18.330 ","End":"02:23.085","Text":"and here we have a point mass of mass m,"},{"Start":"02:23.085 ","End":"02:30.135","Text":"and it\u0027s located a distance R away from the axis of rotation,"},{"Start":"02:30.135 ","End":"02:35.400","Text":"and this point mass is rotating around the axis of rotation."},{"Start":"02:35.400 ","End":"02:38.440","Text":"Let\u0027s work out the moment of inertia,"},{"Start":"02:38.440 ","End":"02:41.210","Text":"so I is simply going to be equal to,"},{"Start":"02:41.210 ","End":"02:43.468","Text":"so we only have one point mass,"},{"Start":"02:43.468 ","End":"02:46.395","Text":"so it\u0027s going to be the mass of the point mass,"},{"Start":"02:46.395 ","End":"02:50.715","Text":"that\u0027s m multiplied by its radius squared."},{"Start":"02:50.715 ","End":"02:58.350","Text":"Its radius as its distance from the axis of rotation squared, that\u0027s R^2."},{"Start":"02:59.270 ","End":"03:03.950","Text":"This I advice you writing in your equation pages,"},{"Start":"03:03.950 ","End":"03:06.590","Text":"this is the moment of inertia of a point"},{"Start":"03:06.590 ","End":"03:10.610","Text":"mass and we\u0027re going to be using this in examples,"},{"Start":"03:10.610 ","End":"03:13.195","Text":"so it\u0027s useful to remember this."},{"Start":"03:13.195 ","End":"03:17.780","Text":"Now what happens if I add in another point mass of"},{"Start":"03:17.780 ","End":"03:22.070","Text":"equal mass m and the same distance over here,"},{"Start":"03:22.070 ","End":"03:24.970","Text":"R away from the axis of rotation."},{"Start":"03:24.970 ","End":"03:29.300","Text":"Now, what will my moment of inertia be for this system of"},{"Start":"03:29.300 ","End":"03:33.920","Text":"2 masses rotating around as we can see from the equation,"},{"Start":"03:33.920 ","End":"03:38.350","Text":"it\u0027s the sum of mr^2,"},{"Start":"03:38.350 ","End":"03:44.820","Text":"we\u0027re just going to have mR^2 plus mR^2 from this side."},{"Start":"03:44.820 ","End":"03:50.830","Text":"That\u0027s simply going to be 2mR^2."},{"Start":"03:51.200 ","End":"03:57.110","Text":"This is the moment of inertia of two point masses rotating around an axis of"},{"Start":"03:57.110 ","End":"04:04.620","Text":"rotation and also don\u0027t forget to copy out this equation over here,"},{"Start":"04:04.620 ","End":"04:06.810","Text":"this is very important."},{"Start":"04:06.810 ","End":"04:10.335","Text":"Now let\u0027s take another example,"},{"Start":"04:10.335 ","End":"04:14.160","Text":"here we have the axis of rotation,"},{"Start":"04:14.160 ","End":"04:16.474","Text":"and about the axis of rotation,"},{"Start":"04:16.474 ","End":"04:19.610","Text":"we have a disk,"},{"Start":"04:19.610 ","End":"04:27.750","Text":"and the disk is of radius R and of total mass"},{"Start":"04:27.750 ","End":"04:31.550","Text":"M. Now what we want to do is we want to find the moment of"},{"Start":"04:31.550 ","End":"04:35.855","Text":"inertia of this disk which is rotating around here."},{"Start":"04:35.855 ","End":"04:43.725","Text":"What are we going to do is we\u0027re going to take small portions of this disk of mass,"},{"Start":"04:43.725 ","End":"04:45.315","Text":"each 1 dm,"},{"Start":"04:45.315 ","End":"04:52.085","Text":"and then we\u0027re going to sum up and all these portions and plug it into this equation."},{"Start":"04:52.085 ","End":"04:55.790","Text":"Let\u0027s see how we do that."},{"Start":"04:55.790 ","End":"04:59.945","Text":"So when we deal with a shape like this,"},{"Start":"04:59.945 ","End":"05:03.665","Text":"where we have lots of masses dm,"},{"Start":"05:03.665 ","End":"05:05.960","Text":"the way that we\u0027re going to use"},{"Start":"05:05.960 ","End":"05:08.825","Text":"this equation is we\u0027re going to use the exact same equation,"},{"Start":"05:08.825 ","End":"05:10.760","Text":"but with an integral,"},{"Start":"05:10.760 ","End":"05:14.015","Text":"and we\u0027re going to integrate along the entire disk."},{"Start":"05:14.015 ","End":"05:20.675","Text":"Now, we\u0027re going to do worked examples in future questions in this chapter,"},{"Start":"05:20.675 ","End":"05:23.780","Text":"I advice to go over that afterwards,"},{"Start":"05:23.780 ","End":"05:25.565","Text":"but we\u0027re simply going to integrate,"},{"Start":"05:25.565 ","End":"05:29.450","Text":"right now I don\u0027t want to do that I just wanted to carry on with explanation."},{"Start":"05:29.450 ","End":"05:33.365","Text":"If you want, you can pause the video now and do a practice,"},{"Start":"05:33.365 ","End":"05:37.550","Text":"and I\u0027m going to tell you now what the moment of inertia for"},{"Start":"05:37.550 ","End":"05:42.305","Text":"a disk rotating about its center of mass is equal to."},{"Start":"05:42.305 ","End":"05:51.115","Text":"That is simply equal to 1/2 the mass of the disk multiplied by its radius squared."},{"Start":"05:51.115 ","End":"05:56.360","Text":"This is of course, only when the mass is uniformly distributed and"},{"Start":"05:56.360 ","End":"06:01.475","Text":"where the axis of rotation is going through the center of mass of the disk."},{"Start":"06:01.475 ","End":"06:04.610","Text":"Now, whenever we have the moment of inertia,"},{"Start":"06:04.610 ","End":"06:09.650","Text":"where the axis of rotation is at the center of mass,"},{"Start":"06:09.650 ","End":"06:14.385","Text":"so we call this ICM, and of course,"},{"Start":"06:14.385 ","End":"06:16.995","Text":"this is of a disk,"},{"Start":"06:16.995 ","End":"06:22.490","Text":"and this you should also write in your equation sheets."},{"Start":"06:22.490 ","End":"06:24.890","Text":"Now, a lot of the time,"},{"Start":"06:24.890 ","End":"06:31.145","Text":"if we\u0027re talking about a disk with its axis of rotation going through its center,"},{"Start":"06:31.145 ","End":"06:37.620","Text":"it will be noted by drawing the disk from a bird\u0027s eye view, for instance,"},{"Start":"06:37.620 ","End":"06:41.375","Text":"and then denoting the axis of rotation as this dot,"},{"Start":"06:41.375 ","End":"06:45.505","Text":"meaning that the axis of rotation is going into the page."},{"Start":"06:45.505 ","End":"06:49.655","Text":"If this dot was replaced by an x,"},{"Start":"06:49.655 ","End":"06:54.750","Text":"that would mean that the axis of rotation was going out of the page."},{"Start":"06:56.000 ","End":"07:01.880","Text":"What we\u0027ve gone over now is how to solve questions when you\u0027re being asked to"},{"Start":"07:01.880 ","End":"07:07.384","Text":"find the moment of inertia using the simple equation."},{"Start":"07:07.384 ","End":"07:09.200","Text":"Now aside from this equation,"},{"Start":"07:09.200 ","End":"07:13.470","Text":"there are a few tricks that you can use in order to find the moment of inertia,"},{"Start":"07:13.470 ","End":"07:16.020","Text":"so let\u0027s take a look at them"}],"ID":10785}],"Thumbnail":null,"ID":139577},{"Name":"Steiners Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Steiners Parallel Axis Theorem","Duration":"6m ","ChapterTopicVideoID":9097,"CourseChapterTopicPlaylistID":5384,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.160 ","End":"00:04.860","Text":"In this lesson, we\u0027re going to be learning a trick of"},{"Start":"00:04.860 ","End":"00:09.870","Text":"a very easy way to work out the moment of inertia of different bodies."},{"Start":"00:09.870 ","End":"00:12.360","Text":"This is called Steiner\u0027s theorem."},{"Start":"00:12.360 ","End":"00:15.975","Text":"It\u0027s also known in different textbooks and"},{"Start":"00:15.975 ","End":"00:21.165","Text":"different teachers or professors as the Huygens Steiner theorem."},{"Start":"00:21.165 ","End":"00:24.630","Text":"I\u0027m probably butchering the name Huygens, it\u0027s a Dutch name."},{"Start":"00:24.630 ","End":"00:26.790","Text":"I\u0027m not pronouncing it correctly."},{"Start":"00:26.790 ","End":"00:30.720","Text":"Or also known as the parallel axis theorem."},{"Start":"00:30.720 ","End":"00:33.495","Text":"If you want to search on the internet online,"},{"Start":"00:33.495 ","End":"00:40.120","Text":"you can search all of these 3 names and you will find information about it."},{"Start":"00:40.610 ","End":"00:43.140","Text":"What is the Steiner\u0027s theorem?"},{"Start":"00:43.140 ","End":"00:46.260","Text":"In this course, I\u0027m going to be calling it the Steiner\u0027s theorem."},{"Start":"00:46.260 ","End":"00:56.675","Text":"It states that the moment of inertia of a body tag is equal to"},{"Start":"00:56.675 ","End":"01:02.375","Text":"the moment of inertia of the center of mass of the body plus"},{"Start":"01:02.375 ","End":"01:12.115","Text":"the total mass of the body multiplied by its distance squared from the center of mass."},{"Start":"01:12.115 ","End":"01:14.160","Text":"What does this mean?"},{"Start":"01:14.160 ","End":"01:16.755","Text":"Let us speak about the disk again."},{"Start":"01:16.755 ","End":"01:19.950","Text":"Here I have my circular disk."},{"Start":"01:19.950 ","End":"01:22.985","Text":"Then if I\u0027m trying to work out the moment of inertia,"},{"Start":"01:22.985 ","End":"01:26.075","Text":"the I from the center of mass."},{"Start":"01:26.075 ","End":"01:32.525","Text":"That means that the disk is rotating about this point in the center."},{"Start":"01:32.525 ","End":"01:37.610","Text":"Let\u0027s say that the mass of the disk is m and its radius is"},{"Start":"01:37.610 ","End":"01:43.025","Text":"r. That means that the disk is spinning around this central point."},{"Start":"01:43.025 ","End":"01:47.375","Text":"However, what if I wanted to find the moment of inertia at some point over here,"},{"Start":"01:47.375 ","End":"01:49.040","Text":"right at the edge of the disk."},{"Start":"01:49.040 ","End":"01:54.585","Text":"Then instead, the disk will be rotating about this point."},{"Start":"01:54.585 ","End":"01:57.800","Text":"It will be going around and around like this,"},{"Start":"01:57.800 ","End":"02:03.420","Text":"which is different to the moment of inertia around the central point."},{"Start":"02:03.420 ","End":"02:08.540","Text":"How would we find out the moment of inertia around this point over here on the side."},{"Start":"02:08.540 ","End":"02:11.615","Text":"This would be our I tag,"},{"Start":"02:11.615 ","End":"02:14.080","Text":"the moment of inertia on this point,"},{"Start":"02:14.080 ","End":"02:15.890","Text":"not in the center of mass,"},{"Start":"02:15.890 ","End":"02:19.460","Text":"which will equal to the I center of mass of the disk,"},{"Start":"02:19.460 ","End":"02:25.610","Text":"which as we saw in the previous lesson, is half mr^2."},{"Start":"02:25.610 ","End":"02:30.020","Text":"Then we\u0027re going to be adding the total mass,"},{"Start":"02:30.020 ","End":"02:36.155","Text":"so plus m multiplied by the distance squared from the center of mass."},{"Start":"02:36.155 ","End":"02:38.720","Text":"Now because the disc is of radius r,"},{"Start":"02:38.720 ","End":"02:44.945","Text":"this point over here is a distance r away from the center of mass."},{"Start":"02:44.945 ","End":"02:50.340","Text":"It\u0027s going to be multiplied by r^2."},{"Start":"02:50.340 ","End":"02:53.780","Text":"This was just an example of how to use Steiner\u0027s theorem."},{"Start":"02:53.780 ","End":"02:57.710","Text":"But now let\u0027s speak about a very important point which is also"},{"Start":"02:57.710 ","End":"03:01.370","Text":"hidden in 1 of the as known as names,"},{"Start":"03:01.370 ","End":"03:06.240","Text":"which is it\u0027s also called the parallel axis theorem."},{"Start":"03:06.860 ","End":"03:09.200","Text":"The clue is in the name."},{"Start":"03:09.200 ","End":"03:13.340","Text":"It\u0027s very important that the axis of rotations,"},{"Start":"03:13.340 ","End":"03:16.310","Text":"when we\u0027re using our I tag equation,"},{"Start":"03:16.310 ","End":"03:22.005","Text":"the axis of rotation must both be parallel to each other."},{"Start":"03:22.005 ","End":"03:24.745","Text":"Very important. What does this mean?"},{"Start":"03:24.745 ","End":"03:26.770","Text":"This means that if here when we were speaking"},{"Start":"03:26.770 ","End":"03:29.725","Text":"about the axis going through the center of the disk."},{"Start":"03:29.725 ","End":"03:34.240","Text":"As we said, the axis is going into the page. Let\u0027s see."},{"Start":"03:34.240 ","End":"03:39.955","Text":"The axis is perpendicular to the disk and it\u0027s going into the page."},{"Start":"03:39.955 ","End":"03:42.670","Text":"Now, when we want to find out the moment of"},{"Start":"03:42.670 ","End":"03:45.850","Text":"inertia of this point at the edge of the disk,"},{"Start":"03:45.850 ","End":"03:49.150","Text":"its axis of rotation must also be"},{"Start":"03:49.150 ","End":"03:55.480","Text":"perpendicular to the disk and also we going into the page."},{"Start":"03:56.780 ","End":"04:00.950","Text":"For instance, I\u0027ll show you"},{"Start":"04:00.950 ","End":"04:05.120","Text":"from a side view if this is my axis of rotation, let\u0027s call it z,"},{"Start":"04:05.120 ","End":"04:15.455","Text":"and I\u0027m working out the moment of inertia of this stick from this point."},{"Start":"04:15.455 ","End":"04:19.985","Text":"If I then want to find out the moment of inertia from this point,"},{"Start":"04:19.985 ","End":"04:21.980","Text":"I can have my axis of rotation,"},{"Start":"04:21.980 ","End":"04:24.064","Text":"let\u0027s say going in this direction."},{"Start":"04:24.064 ","End":"04:25.925","Text":"No, not good."},{"Start":"04:25.925 ","End":"04:30.830","Text":"What I have to do is I have to say that axis must be parallel."},{"Start":"04:30.830 ","End":"04:34.100","Text":"My new axis of rotation will be going in"},{"Start":"04:34.100 ","End":"04:38.565","Text":"the exact same distance as this axis that I\u0027ve labeled z,"},{"Start":"04:38.565 ","End":"04:41.820","Text":"but it\u0027s moved slightly along."},{"Start":"04:41.820 ","End":"04:44.625","Text":"I can call it z tag for instance."},{"Start":"04:44.625 ","End":"04:47.090","Text":"The axis must be parallel."},{"Start":"04:47.090 ","End":"04:52.010","Text":"They must be pointing in the exact same distance and the angle"},{"Start":"04:52.010 ","End":"04:57.860","Text":"relative to the object must be the same as that when it\u0027s in the center of mass."},{"Start":"04:57.860 ","End":"05:01.300","Text":"It\u0027s very important, I\u0027ll write a note."},{"Start":"05:01.300 ","End":"05:06.005","Text":"I wrote it here. Important, the axis of rotation must be parallel."},{"Start":"05:06.005 ","End":"05:13.235","Text":"The second important point is that we must also know the I,"},{"Start":"05:13.235 ","End":"05:16.235","Text":"the moment of inertia for the center of mass."},{"Start":"05:16.235 ","End":"05:19.255","Text":"As you can see in our equation,"},{"Start":"05:19.255 ","End":"05:23.845","Text":"for I tag, we have I center of mass over here."},{"Start":"05:23.845 ","End":"05:28.730","Text":"In order to use Steiner\u0027s theorem or the parallel axis theorem,"},{"Start":"05:28.730 ","End":"05:33.275","Text":"we have to know the moment of inertia for the center of mass."},{"Start":"05:33.275 ","End":"05:37.205","Text":"We have to know how to work it out or have"},{"Start":"05:37.205 ","End":"05:41.810","Text":"the equation written in our notes when we go into the exam."},{"Start":"05:41.810 ","End":"05:45.350","Text":"You cannot use Steiner\u0027s theorem without first working"},{"Start":"05:45.350 ","End":"05:50.060","Text":"out the moment of inertia for the center of mass."},{"Start":"05:50.060 ","End":"05:56.415","Text":"Very important. That\u0027s the end of this lesson."},{"Start":"05:56.415 ","End":"06:00.600","Text":"Now we\u0027re going to go on to some more questions."}],"ID":9370}],"Thumbnail":null,"ID":5384},{"Name":"Additivity","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Additivity of I","Duration":"7m 51s","ChapterTopicVideoID":9098,"CourseChapterTopicPlaylistID":5385,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.560 ","End":"00:05.720","Text":"Hello. In this lesson we\u0027re going to be learning about the additivity of I."},{"Start":"00:05.720 ","End":"00:09.045","Text":"The additivity of the moment of inertia."},{"Start":"00:09.045 ","End":"00:11.490","Text":"In the last lesson, we were learning about"},{"Start":"00:11.490 ","End":"00:15.300","Text":"Steiner\u0027s theorem or the parallel axis theorem,"},{"Start":"00:15.300 ","End":"00:21.840","Text":"which was a useful trick for working out the moment of inertia of certain shapes."},{"Start":"00:21.840 ","End":"00:24.630","Text":"Now we\u0027re going to be speaking about another trip"},{"Start":"00:24.630 ","End":"00:27.075","Text":"which is additivity. Now, what does that mean?"},{"Start":"00:27.075 ","End":"00:29.625","Text":"It means that if we have 2 shapes,"},{"Start":"00:29.625 ","End":"00:32.820","Text":"then the moment of inertia of the 2 shapes together,"},{"Start":"00:32.820 ","End":"00:36.015","Text":"I_tot, the total moment of inertia,"},{"Start":"00:36.015 ","End":"00:41.085","Text":"is equal to I_1 plus I_2."},{"Start":"00:41.085 ","End":"00:46.345","Text":"The sum of the moments of inertia of the shapes."},{"Start":"00:46.345 ","End":"00:48.365","Text":"Let\u0027s give an example."},{"Start":"00:48.365 ","End":"00:50.900","Text":"If we have a disc,"},{"Start":"00:50.900 ","End":"00:53.135","Text":"imagine that this is round,"},{"Start":"00:53.135 ","End":"00:57.860","Text":"of radius R_1 and of mass M_1."},{"Start":"00:57.860 ","End":"01:01.125","Text":"Then we put in another disk,"},{"Start":"01:01.125 ","End":"01:08.610","Text":"a smaller disc right on top of radius R_2 and mass M_2."},{"Start":"01:08.610 ","End":"01:11.825","Text":"What would be the moment of inertia of this?"},{"Start":"01:11.825 ","End":"01:14.645","Text":"We would have I_tot is equal to,"},{"Start":"01:14.645 ","End":"01:18.005","Text":"the equation for the moment of inertia of a disc"},{"Start":"01:18.005 ","End":"01:22.265","Text":"is 1/2 the mass multiplied by the radius^2."},{"Start":"01:22.265 ","End":"01:29.150","Text":"Half M_1 R_1^2 plus that of the smaller desk,"},{"Start":"01:29.150 ","End":"01:36.030","Text":"which is plus half M_2 R_2^2."},{"Start":"01:36.080 ","End":"01:40.295","Text":"Now another question that can be asked is,"},{"Start":"01:40.295 ","End":"01:42.005","Text":"what if we made a hole?"},{"Start":"01:42.005 ","End":"01:43.910","Text":"Instead of 2 discs,"},{"Start":"01:43.910 ","End":"01:45.560","Text":"one on top of the other."},{"Start":"01:45.560 ","End":"01:49.080","Text":"We have a disc with a hole in the middle."},{"Start":"01:49.080 ","End":"01:53.505","Text":"Some ring shape."},{"Start":"01:53.505 ","End":"01:56.160","Text":"Again, we\u0027re going to say that"},{"Start":"01:56.160 ","End":"02:01.415","Text":"the radius from the center of the hole until the end is R_1."},{"Start":"02:01.415 ","End":"02:03.875","Text":"From the center to the end of the hole,"},{"Start":"02:03.875 ","End":"02:06.140","Text":"this is going to be R_2."},{"Start":"02:06.140 ","End":"02:14.255","Text":"Then the mass here is M_1 and the mass of the hole we\u0027ll say is M_2. How do we do that?"},{"Start":"02:14.255 ","End":"02:18.680","Text":"We do the exact same thing as with the 2 discs,"},{"Start":"02:18.680 ","End":"02:19.910","Text":"one on top of the other."},{"Start":"02:19.910 ","End":"02:23.480","Text":"However, this time we consider that the mass of"},{"Start":"02:23.480 ","End":"02:28.845","Text":"the central disc is a negative and that will represent a hole."},{"Start":"02:28.845 ","End":"02:30.585","Text":"It\u0027s a negative mass."},{"Start":"02:30.585 ","End":"02:39.575","Text":"Then our I_tot of this ring will be the same as 1/2 M_1 R_1^2,"},{"Start":"02:39.575 ","End":"02:44.675","Text":"minus, because we have a negative mass because it\u0027s a hole,"},{"Start":"02:44.675 ","End":"02:50.310","Text":"1/2 of M_2 R_2^2."},{"Start":"02:50.310 ","End":"02:53.660","Text":"This is a really, really easy way of figuring"},{"Start":"02:53.660 ","End":"02:57.350","Text":"out the moment of inertia of more complex shapes,"},{"Start":"02:57.350 ","End":"02:59.420","Text":"such as 2 discs,"},{"Start":"02:59.420 ","End":"03:00.755","Text":"one on top of the other,"},{"Start":"03:00.755 ","End":"03:04.970","Text":"a ring of certain diameter."},{"Start":"03:04.970 ","End":"03:07.475","Text":"It\u0027s very useful."},{"Start":"03:07.475 ","End":"03:13.460","Text":"Now, an important note is that all the moments of inertia,"},{"Start":"03:13.460 ","End":"03:17.360","Text":"that means I_1 and I_2 and if there are more I_3, I_4,"},{"Start":"03:17.360 ","End":"03:21.725","Text":"whatever it might be, must share an axis of rotation."},{"Start":"03:21.725 ","End":"03:26.630","Text":"That means that if the axis of rotation is, for instance,"},{"Start":"03:26.630 ","End":"03:32.165","Text":"over here, so this is the axis of rotation of the small desk and of the large disc."},{"Start":"03:32.165 ","End":"03:36.290","Text":"Same over here, if the axis of rotation is here,"},{"Start":"03:36.290 ","End":"03:42.780","Text":"then both this hole, entire ring,"},{"Start":"03:42.780 ","End":"03:47.135","Text":"which means the disc and the negative mass representing the hole,"},{"Start":"03:47.135 ","End":"03:51.215","Text":"must both spin around this point."},{"Start":"03:51.215 ","End":"03:55.070","Text":"Now let\u0027s give an example to see what"},{"Start":"03:55.070 ","End":"04:00.710","Text":"happens if the moments don\u0027t share an axis of rotation."},{"Start":"04:00.710 ","End":"04:03.740","Text":"Let\u0027s see how we can still use this rule of additivity,"},{"Start":"04:03.740 ","End":"04:05.720","Text":"but in a different way."},{"Start":"04:05.720 ","End":"04:08.155","Text":"Let\u0027s see an example."},{"Start":"04:08.155 ","End":"04:11.840","Text":"Now the question that we\u0027re being asked is what happens"},{"Start":"04:11.840 ","End":"04:16.085","Text":"if the 2 shapes don\u0027t share the same axis of rotation."},{"Start":"04:16.085 ","End":"04:18.410","Text":"For instance, these 2,"},{"Start":"04:18.410 ","End":"04:21.410","Text":"let\u0027s say that there are clocks and they\u0027re hanging from the wall"},{"Start":"04:21.410 ","End":"04:27.100","Text":"from a hook or screw, whatever."},{"Start":"04:27.100 ","End":"04:32.765","Text":"Now, as we can see these clocks aren\u0027t one on top of the other,"},{"Start":"04:32.765 ","End":"04:36.145","Text":"meaning that they don\u0027t share an axis of rotation."},{"Start":"04:36.145 ","End":"04:38.000","Text":"From the previous lesson,"},{"Start":"04:38.000 ","End":"04:40.370","Text":"we learned about Steiner\u0027s theorem."},{"Start":"04:40.370 ","End":"04:43.285","Text":"Let\u0027s put that theorem to the test."},{"Start":"04:43.285 ","End":"04:48.600","Text":"The upper clock is called 1 and the lower clock is called 2."},{"Start":"04:48.600 ","End":"04:53.345","Text":"Let\u0027s write the moment of inertia for clock number 1."},{"Start":"04:53.345 ","End":"04:56.660","Text":"That\u0027s I_1. As we know from Steiner\u0027s theorem,"},{"Start":"04:56.660 ","End":"05:01.445","Text":"we write down the moment of inertia for the shape."},{"Start":"05:01.445 ","End":"05:03.935","Text":"The shape here is that of a disc."},{"Start":"05:03.935 ","End":"05:11.860","Text":"It\u0027s 1/2 MR^2 plus its distance from the axis of rotation."},{"Start":"05:11.860 ","End":"05:13.690","Text":"What\u0027s our axis of rotation?"},{"Start":"05:13.690 ","End":"05:16.760","Text":"This drawing isn\u0027t exactly accurate."},{"Start":"05:16.760 ","End":"05:20.870","Text":"The axis of rotation is right over here. Right on the edge."},{"Start":"05:20.870 ","End":"05:24.605","Text":"We can see that it\u0027s a distance, R, away."},{"Start":"05:24.605 ","End":"05:27.320","Text":"The center of the circle, which right now,"},{"Start":"05:27.320 ","End":"05:34.040","Text":"1/2 MR^2 is the moment of inertia for a disc spinning around its center."},{"Start":"05:34.040 ","End":"05:36.139","Text":"But we want it right on the edge,"},{"Start":"05:36.139 ","End":"05:41.975","Text":"so we add its mass multiplied by its distance."},{"Start":"05:41.975 ","End":"05:46.355","Text":"The center of mass is distance from its point of rotation."},{"Start":"05:46.355 ","End":"05:52.970","Text":"This 1/2 MR^2 is the moment of inertia for a disc rotating around its center."},{"Start":"05:52.970 ","End":"05:55.445","Text":"But because here we\u0027ve said that"},{"Start":"05:55.445 ","End":"05:58.940","Text":"the axis of rotation is this point right at the top over here,"},{"Start":"05:58.940 ","End":"06:00.620","Text":"the screw holding the clock up,"},{"Start":"06:00.620 ","End":"06:03.730","Text":"so we have to use Steiner\u0027s theorem."},{"Start":"06:03.730 ","End":"06:06.415","Text":"Now we have I_2,"},{"Start":"06:06.415 ","End":"06:09.515","Text":"the moment of inertia for this second shape."},{"Start":"06:09.515 ","End":"06:13.290","Text":"Again, we can see it\u0027s a disc with the same radius and the same mass."},{"Start":"06:13.290 ","End":"06:17.150","Text":"We\u0027re going to say right now that it\u0027s rotating about its center."},{"Start":"06:17.150 ","End":"06:19.310","Text":"The equation is again,"},{"Start":"06:19.310 ","End":"06:25.165","Text":"1/2 MR^2, but of course it\u0027s not rotating."},{"Start":"06:25.165 ","End":"06:28.625","Text":"Our actual point in reality it isn\u0027t rotating around the center,"},{"Start":"06:28.625 ","End":"06:31.835","Text":"but it\u0027s rotating around this point over here."},{"Start":"06:31.835 ","End":"06:33.860","Text":"Let\u0027s see the distance from this point."},{"Start":"06:33.860 ","End":"06:34.910","Text":"It\u0027s 1 radius,"},{"Start":"06:34.910 ","End":"06:37.325","Text":"2 radius, 3 radiuses."},{"Start":"06:37.325 ","End":"06:42.630","Text":"It\u0027s going to be plus M multiplied by the distance^2."},{"Start":"06:42.630 ","End":"06:47.580","Text":"It\u0027s going to be multiplied by the distance which is 3R^2."},{"Start":"06:47.720 ","End":"06:53.430","Text":"I_1 is going to equal to 3 over 2 MR^2,"},{"Start":"06:53.430 ","End":"07:01.360","Text":"and then I_2 is going to be equal to 19 divided by 2MR^2."},{"Start":"07:01.360 ","End":"07:05.645","Text":"Now, to find the total moment of inertia of the whole system,"},{"Start":"07:05.645 ","End":"07:08.345","Text":"of the 2 discs, of the 2 clocks."},{"Start":"07:08.345 ","End":"07:12.020","Text":"Now we can use what we learned just earlier in this lesson,"},{"Start":"07:12.020 ","End":"07:14.135","Text":"which was the additivity of I."},{"Start":"07:14.135 ","End":"07:17.510","Text":"That means that I_tot is equal to,"},{"Start":"07:17.510 ","End":"07:21.710","Text":"so now we can add these both up because our frame of reference,"},{"Start":"07:21.710 ","End":"07:26.585","Text":"our point, our axis of rotation is the same axis of rotation, this point over here."},{"Start":"07:26.585 ","End":"07:32.270","Text":"Now, we can use additivity because these 2 share the same axis of rotation."},{"Start":"07:32.270 ","End":"07:36.820","Text":"I_tot is equal to I_1 plus I_2,"},{"Start":"07:36.820 ","End":"07:38.630","Text":"which as we can see,"},{"Start":"07:38.630 ","End":"07:46.240","Text":"is going to be equal to 11 MR^2."},{"Start":"07:46.240 ","End":"07:48.515","Text":"That is the end of the lesson."},{"Start":"07:48.515 ","End":"07:49.610","Text":"If you don\u0027t understand it,"},{"Start":"07:49.610 ","End":"07:52.020","Text":"please see it again."}],"ID":10781},{"Watched":false,"Name":"Example - Additivity","Duration":"4m 47s","ChapterTopicVideoID":10425,"CourseChapterTopicPlaylistID":5385,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:05.925","Text":"we\u0027re going to be doing a worked example of"},{"Start":"00:05.925 ","End":"00:10.965","Text":"solving the moment of inertia using the idea of additivity."},{"Start":"00:10.965 ","End":"00:15.495","Text":"Here we have 2 discs which are identical."},{"Start":"00:15.495 ","End":"00:19.410","Text":"Both of them have a radius of r and a total mass"},{"Start":"00:19.410 ","End":"00:23.475","Text":"of m. The 2 desks are joined at some point."},{"Start":"00:23.475 ","End":"00:29.370","Text":"Then the top desk is joined via some nail which"},{"Start":"00:29.370 ","End":"00:36.040","Text":"is in the wall and they can rotate about this nail over here."},{"Start":"00:36.040 ","End":"00:39.680","Text":"What we want to do is we want to find the moment of"},{"Start":"00:39.680 ","End":"00:45.325","Text":"inertia of this entire body about this point."},{"Start":"00:45.325 ","End":"00:48.555","Text":"What are we going to do? First,"},{"Start":"00:48.555 ","End":"00:54.770","Text":"we\u0027re going to find the moment of inertia of this first disk about this point,"},{"Start":"00:54.770 ","End":"01:01.040","Text":"then we\u0027re going to find the moment of inertia of this bottom desk about this point,"},{"Start":"01:01.040 ","End":"01:02.210","Text":"the axes of rotation,"},{"Start":"01:02.210 ","End":"01:06.630","Text":"and then we\u0027re just going to add them both together."},{"Start":"01:07.100 ","End":"01:10.510","Text":"Let\u0027s call this desk Number 1."},{"Start":"01:10.510 ","End":"01:18.150","Text":"Let\u0027s work out the moment of inertia of desk Number 1. i1 is going to be equal to."},{"Start":"01:18.150 ","End":"01:24.260","Text":"If we work out the moment of inertia about the center of mass of the disk,"},{"Start":"01:24.260 ","End":"01:27.020","Text":"as we know, that is equal to 1/2mr^."},{"Start":"01:27.020 ","End":"01:36.460","Text":"Great. But now we want to know what the moment of inertia is over here about this point."},{"Start":"01:36.460 ","End":"01:39.870","Text":"We\u0027re going to use Steno\u0027s theorem."},{"Start":"01:39.870 ","End":"01:41.960","Text":"What we\u0027re going to do is we\u0027re going to"},{"Start":"01:41.960 ","End":"01:48.020","Text":"add mr^ and then"},{"Start":"01:48.020 ","End":"01:53.375","Text":"that moves the axis of rotation from the center of mass to this point over here."},{"Start":"01:53.375 ","End":"01:56.090","Text":"Now if we add all of that together,"},{"Start":"01:56.090 ","End":"02:04.180","Text":"so we\u0027ll get that the moment of inertia for desk Number 1 is 3 over 2mr^."},{"Start":"02:04.520 ","End":"02:11.360","Text":"Now let\u0027s work out the moment of inertia for desk Number 2. i2 is equal"},{"Start":"02:11.360 ","End":"02:17.810","Text":"to the moment of inertia of this disk when the axis of rotation is at the center of mass."},{"Start":"02:17.810 ","End":"02:24.630","Text":"That\u0027s again going to be 1/2mr^. Now plus."},{"Start":"02:24.630 ","End":"02:28.090","Text":"Now we know that the axis of rotation is over here,"},{"Start":"02:28.090 ","End":"02:30.835","Text":"not at the center of mass of this disk."},{"Start":"02:30.835 ","End":"02:35.670","Text":"What we want to do is move our axis of rotation from here all the way to here."},{"Start":"02:35.670 ","End":"02:37.735","Text":"Let\u0027s see what this distance is."},{"Start":"02:37.735 ","End":"02:41.905","Text":"The distance from here to here is r,"},{"Start":"02:41.905 ","End":"02:47.820","Text":"and then from here to halfway through this disk is another r. Then from"},{"Start":"02:47.820 ","End":"02:53.769","Text":"this halfway point until where our axis of rotation is actually located,"},{"Start":"02:53.769 ","End":"02:58.080","Text":"is another distance r. It\u0027s 3 radiuses."},{"Start":"02:58.080 ","End":"03:03.505","Text":"That\u0027s a total of 3 r. Using Steno\u0027s theorem,"},{"Start":"03:03.505 ","End":"03:08.600","Text":"we\u0027re going to add on the mass of the disk multiplied"},{"Start":"03:08.600 ","End":"03:14.320","Text":"by the distance that we have to move our axis of rotation squared."},{"Start":"03:14.320 ","End":"03:18.275","Text":"We have to move our axis of rotation a distance of 3r,"},{"Start":"03:18.275 ","End":"03:23.030","Text":"so m multiply by 3r^2."},{"Start":"03:23.030 ","End":"03:25.280","Text":"Then once we work this out,"},{"Start":"03:25.280 ","End":"03:26.885","Text":"this is simply going to be equal to"},{"Start":"03:26.885 ","End":"03:35.880","Text":"19 divided by 2mr^."},{"Start":"03:35.880 ","End":"03:39.965","Text":"Now we have the moment of inertia of disc Number 1 and of disc Number 2."},{"Start":"03:39.965 ","End":"03:42.740","Text":"What do we want to do is we want to find the total moment of"},{"Start":"03:42.740 ","End":"03:47.795","Text":"inertia of the 2 disks together rotating about this point."},{"Start":"03:47.795 ","End":"03:51.560","Text":"Our total moment of inertia i total,"},{"Start":"03:51.560 ","End":"03:55.505","Text":"is going to be equal to using our additivity rule,"},{"Start":"03:55.505 ","End":"04:01.150","Text":"is going to be equal to i1 plus i2."},{"Start":"04:01.150 ","End":"04:03.515","Text":"Now we can add those up."},{"Start":"04:03.515 ","End":"04:09.390","Text":"We have 19 over 2mr^ plus 3 over 2mr^,"},{"Start":"04:09.390 ","End":"04:13.305","Text":"which is going to be 22 over 2mr^,"},{"Start":"04:13.305 ","End":"04:21.795","Text":"which simplifying the fraction is 11mr^."},{"Start":"04:21.795 ","End":"04:23.940","Text":"This is the final answer."},{"Start":"04:23.940 ","End":"04:27.230","Text":"What we can see here is that if we have a number of"},{"Start":"04:27.230 ","End":"04:31.279","Text":"bodies rotating about the same axis of rotation,"},{"Start":"04:31.279 ","End":"04:36.770","Text":"we can find individually the moment of inertia of each body and"},{"Start":"04:36.770 ","End":"04:40.100","Text":"then we can use additivity to simply"},{"Start":"04:40.100 ","End":"04:44.450","Text":"add them up and find the total moment of inertia of the whole system."},{"Start":"04:44.450 ","End":"04:48.120","Text":"Let\u0027s move on to the next trick."}],"ID":10782}],"Thumbnail":null,"ID":5385},{"Name":"Iz = Ix plus Iy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Shape on x-y plane","Duration":"5m 49s","ChapterTopicVideoID":9099,"CourseChapterTopicPlaylistID":5386,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:06.660","Text":"Now a third trick for solving complicated moment of inertia questions is by using"},{"Start":"00:06.660 ","End":"00:13.630","Text":"the equation that goes like this I_z=I_x plus I_y."},{"Start":"00:13.760 ","End":"00:21.460","Text":"What does that mean? It means that if I have a flat surface, some plane,"},{"Start":"00:21.460 ","End":"00:29.390","Text":"which is just 2-dimensional just on the x and y-plane,"},{"Start":"00:29.390 ","End":"00:33.025","Text":"then that means I can use this formula."},{"Start":"00:33.025 ","End":"00:35.535","Text":"Let\u0027s see how it goes."},{"Start":"00:35.535 ","End":"00:42.400","Text":"If I have over here my z-axis and my z-axis is going to be my axis of rotation."},{"Start":"00:42.400 ","End":"00:50.825","Text":"Then I\u0027m going to have my y-axis over here and my x-axis over here."},{"Start":"00:50.825 ","End":"00:57.500","Text":"If I have my 2-dimensional flat, plain sheet, this,"},{"Start":"00:57.500 ","End":"01:01.550","Text":"so it\u0027s obviously very easy for me to work out"},{"Start":"01:01.550 ","End":"01:08.405","Text":"my I_z because my shape is rotating about the z-axis."},{"Start":"01:08.405 ","End":"01:11.820","Text":"If this is going to be a disk."},{"Start":"01:12.650 ","End":"01:18.305","Text":"This is a desk. Now, let\u0027s say for a moment that it\u0027s rotating about the center."},{"Start":"01:18.305 ","End":"01:20.975","Text":"Then my I, in this case z,"},{"Start":"01:20.975 ","End":"01:24.710","Text":"because the z-axis is my center of rotation,"},{"Start":"01:24.710 ","End":"01:26.455","Text":"my axis of rotation,"},{"Start":"01:26.455 ","End":"01:36.420","Text":"this is going to be equal to 1/2(mR)^2."},{"Start":"01:36.420 ","End":"01:44.105","Text":"Now, if I would turn the x-axis or the y-axis into the axis of rotation."},{"Start":"01:44.105 ","End":"01:50.470","Text":"Instead of the coin spinning around like this."},{"Start":"01:51.770 ","End":"01:54.720","Text":"If this was the coin,"},{"Start":"01:54.720 ","End":"01:58.115","Text":"the z-axis is coming out of the page,"},{"Start":"01:58.115 ","End":"02:00.290","Text":"out of the screen and into your I."},{"Start":"02:00.290 ","End":"02:05.165","Text":"If this is going to be the y-axis like in the picture,"},{"Start":"02:05.165 ","End":"02:09.245","Text":"so that the coin would spin around the y-axis."},{"Start":"02:09.245 ","End":"02:11.330","Text":"Or alternatively, if I would have"},{"Start":"02:11.330 ","End":"02:16.840","Text":"the coin spinning around if this is the x-axis like in the photo."},{"Start":"02:16.840 ","End":"02:20.660","Text":"Spinning around the x-axis."},{"Start":"02:20.670 ","End":"02:24.530","Text":"I can use this equation to find what my moment of"},{"Start":"02:24.530 ","End":"02:32.400","Text":"inertia around the x-axis and my moment of inertia around the y-axis is equal to."},{"Start":"02:33.530 ","End":"02:36.020","Text":"Let\u0027s see how we do this."},{"Start":"02:36.020 ","End":"02:39.460","Text":"Now because of symmetry in the coin,"},{"Start":"02:39.460 ","End":"02:42.170","Text":"we can see that the moment of inertia,"},{"Start":"02:42.170 ","End":"02:47.725","Text":"if it\u0027s spinning around the x or the y-axis will be exactly the same."},{"Start":"02:47.725 ","End":"02:52.415","Text":"Just a further explanation of what do I mean because of symmetry."},{"Start":"02:52.415 ","End":"02:55.129","Text":"If you imagine a coin spinning,"},{"Start":"02:55.129 ","End":"03:02.345","Text":"if you put it on its side and you spin it on the table like a dreidel,"},{"Start":"03:02.345 ","End":"03:06.725","Text":"that would be like spinning around this x-axis."},{"Start":"03:06.725 ","End":"03:13.320","Text":"If you rotate it so that it\u0027s spinning on the imaginary y-axis."},{"Start":"03:13.320 ","End":"03:15.090","Text":"You just turn it around."},{"Start":"03:15.090 ","End":"03:18.155","Text":"You\u0027ll see that because it\u0027s the same shape,"},{"Start":"03:18.155 ","End":"03:20.540","Text":"it\u0027s always going to be a circular coin spinning."},{"Start":"03:20.540 ","End":"03:25.325","Text":"It doesn\u0027t matter which axis it\u0027s spinning around the x or the y."},{"Start":"03:25.325 ","End":"03:28.400","Text":"That means because the shape is symmetrical,"},{"Start":"03:28.400 ","End":"03:30.770","Text":"it\u0027s the same shape on every axis that we can"},{"Start":"03:30.770 ","End":"03:35.190","Text":"use this equation for I that equals I_x plus I_y."},{"Start":"03:35.260 ","End":"03:46.020","Text":"As we can see, so I_z=I_x plus I_y."},{"Start":"03:46.360 ","End":"03:57.175","Text":"Then I can say that my 1/2mR^2=I_x plus I_y."},{"Start":"03:57.175 ","End":"03:59.585","Text":"Now because we said that because of symmetry,"},{"Start":"03:59.585 ","End":"04:02.440","Text":"I_x and I_y are the same."},{"Start":"04:02.440 ","End":"04:05.795","Text":"We can say that I_x=I_y."},{"Start":"04:05.795 ","End":"04:15.410","Text":"Then we can say that 1/2mR^2=2I_x,"},{"Start":"04:15.410 ","End":"04:17.760","Text":"which also equals to 2I_y."},{"Start":"04:17.990 ","End":"04:21.810","Text":"You could do it either way. Therefore,"},{"Start":"04:21.810 ","End":"04:26.120","Text":"we can say that I_x."},{"Start":"04:26.120 ","End":"04:29.580","Text":"If we have I_z=2I_x."},{"Start":"04:30.010 ","End":"04:40.210","Text":"Then we can say that our I_x=1/2RI_z."},{"Start":"04:40.210 ","End":"04:46.000","Text":"In this case, it\u0027s going to be 1/2 multiplied by 1/2mR^1,"},{"Start":"04:46.100 ","End":"04:51.480","Text":"which is equal to 1/4mR^2."},{"Start":"04:51.480 ","End":"04:56.650","Text":"This is also equal to I_y because I_x=I_y."},{"Start":"04:58.400 ","End":"05:01.555","Text":"This is something important to remember."},{"Start":"05:01.555 ","End":"05:05.555","Text":"This equation, I_z=I_x plus I_y."},{"Start":"05:05.555 ","End":"05:12.455","Text":"You can only use this when dealing with a 2-dimensional shape on the xy-plane."},{"Start":"05:12.455 ","End":"05:20.130","Text":"If the shape is 3-dimensional or has any values on the z-axis,"},{"Start":"05:20.130 ","End":"05:23.125","Text":"then you cannot use this equation."},{"Start":"05:23.125 ","End":"05:27.440","Text":"Only when dealing with a 2-dimensional shape on the xy-plane."},{"Start":"05:27.440 ","End":"05:32.840","Text":"Obviously, if it\u0027s on the xz-plane or yz-plane, it\u0027s the same equation."},{"Start":"05:32.840 ","End":"05:37.985","Text":"You just change these letters around, the axis around."},{"Start":"05:37.985 ","End":"05:45.170","Text":"But this can only be a 2-dimensional shape without any depth."},{"Start":"05:45.170 ","End":"05:49.530","Text":"This is also another good trick to use."}],"ID":9372}],"Thumbnail":null,"ID":5386},{"Name":"Z Symmetry","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Symmetry on Z Axis","Duration":"5m 1s","ChapterTopicVideoID":9100,"CourseChapterTopicPlaylistID":5387,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:04.125","Text":"Hello. This lesson is going to speak of"},{"Start":"00:04.125 ","End":"00:08.700","Text":"yet another trick for working out the moment of inertia in a very easy way."},{"Start":"00:08.700 ","End":"00:13.365","Text":"Now, this one especially is going to seem like a cheat, but it isn\u0027t."},{"Start":"00:13.365 ","End":"00:15.840","Text":"It\u0027s actually correct. Let\u0027s see what it means."},{"Start":"00:15.840 ","End":"00:20.295","Text":"It\u0027s called symmetry on the z-axis."},{"Start":"00:20.295 ","End":"00:22.710","Text":"What does this actually mean?"},{"Start":"00:22.710 ","End":"00:28.035","Text":"If we have a disc which has center of rotation right in the middle,"},{"Start":"00:28.035 ","End":"00:31.620","Text":"it has a radius of R,"},{"Start":"00:31.620 ","End":"00:35.250","Text":"and it\u0027s rotating about the z-axis."},{"Start":"00:35.250 ","End":"00:36.825","Text":"What is our I going to be?"},{"Start":"00:36.825 ","End":"00:40.660","Text":"We know it\u0027s the I of a desk, which is,"},{"Start":"00:40.660 ","End":"00:42.830","Text":"let\u0027s say that it\u0027s of mass m,"},{"Start":"00:42.830 ","End":"00:46.405","Text":"so it\u0027s going to be 1/2mR^2."},{"Start":"00:46.405 ","End":"00:49.020","Text":"Till here, we\u0027ve got this."},{"Start":"00:49.020 ","End":"00:57.965","Text":"Now, what would happen if we copied this disc all along the z-axis?"},{"Start":"00:57.965 ","End":"01:00.990","Text":"Let\u0027s see what it would look like."},{"Start":"01:01.070 ","End":"01:10.820","Text":"As we can see, this shape has been copied along the z-axis."},{"Start":"01:10.820 ","End":"01:13.995","Text":"Then we can see that this looks just like a cylinder."},{"Start":"01:13.995 ","End":"01:17.370","Text":"If we collect everything,"},{"Start":"01:17.370 ","End":"01:19.820","Text":"and this is obviously a solid shape,"},{"Start":"01:19.820 ","End":"01:22.640","Text":"so we can color this in."},{"Start":"01:22.690 ","End":"01:25.025","Text":"I hope that this is clear."},{"Start":"01:25.025 ","End":"01:27.739","Text":"This now is a cylinder."},{"Start":"01:27.739 ","End":"01:32.760","Text":"What is the cheating part that I was speaking about?"},{"Start":"01:33.220 ","End":"01:37.705","Text":"This was the I of disc."},{"Start":"01:37.705 ","End":"01:40.170","Text":"Now we can say,"},{"Start":"01:40.170 ","End":"01:42.645","Text":"and this is 100 percent correct,"},{"Start":"01:42.645 ","End":"01:52.420","Text":"that the I of a cylinder is equal to the I of the disc, so it\u0027s also equal to 1/2."},{"Start":"01:53.540 ","End":"01:59.870","Text":"But here the mass is going to be different because the mass of 1 disc was m. We\u0027ll"},{"Start":"01:59.870 ","End":"02:05.630","Text":"say that the mass of all the single discs together making the cylinders,"},{"Start":"02:05.630 ","End":"02:08.850","Text":"just call it M. It\u0027s of the same radius,"},{"Start":"02:08.850 ","End":"02:10.699","Text":"so it\u0027s going to be I^2."},{"Start":"02:10.699 ","End":"02:13.725","Text":"Sounds weird, but it\u0027s 100 percent true."},{"Start":"02:13.725 ","End":"02:17.890","Text":"This is a great cheat to use if you have some axis of"},{"Start":"02:17.890 ","End":"02:24.710","Text":"rotation and a simple 2-dimensional shape rotating around it."},{"Start":"02:25.820 ","End":"02:30.410","Text":"Let\u0027s say if you have a cylinder rotating around it and they ask you,"},{"Start":"02:30.410 ","End":"02:31.865","Text":"oh, what\u0027s the moment of inertia?"},{"Start":"02:31.865 ","End":"02:36.210","Text":"The shape. Simple cheats like this."},{"Start":"02:36.740 ","End":"02:39.660","Text":"Here\u0027s another example."},{"Start":"02:39.660 ","End":"02:45.350","Text":"Say you have this rod rotating around the z-axis."},{"Start":"02:45.350 ","End":"02:55.280","Text":"As we know, the equation for the moment of inertia of this rod is equal to m,"},{"Start":"02:55.280 ","End":"02:59.435","Text":"the mass of the rod, multiplied by L^2,"},{"Start":"02:59.435 ","End":"03:03.500","Text":"L being the length of the rod, divided by 3."},{"Start":"03:03.500 ","End":"03:09.380","Text":"Then again, if I stretch this out, let\u0027s see."},{"Start":"03:09.380 ","End":"03:11.420","Text":"If I stretch this out,"},{"Start":"03:11.420 ","End":"03:17.840","Text":"so we can see that it\u0027s the same trick as with the disc and the cylinder."},{"Start":"03:17.840 ","End":"03:23.940","Text":"This will also be equal to the I of the board."},{"Start":"03:25.010 ","End":"03:27.240","Text":"It\u0027s a great trick."},{"Start":"03:27.240 ","End":"03:28.600","Text":"If they show you, let\u0027s say,"},{"Start":"03:28.600 ","End":"03:34.175","Text":"a board and they say it\u0027s of length L and of height a,"},{"Start":"03:34.175 ","End":"03:36.965","Text":"and it\u0027s rotating around the z-axis,"},{"Start":"03:36.965 ","End":"03:40.130","Text":"you can say that it looks like a rod rotating around"},{"Start":"03:40.130 ","End":"03:44.910","Text":"the z-axis and apply this trick of symmetry on the z-axis."},{"Start":"03:45.200 ","End":"03:49.880","Text":"Finally, if we\u0027re going to be speaking about let\u0027s say this,"},{"Start":"03:49.880 ","End":"03:51.860","Text":"which is the shape of a ring,"},{"Start":"03:51.860 ","End":"03:55.840","Text":"it\u0027s like the disc but it\u0027s empty."},{"Start":"03:55.840 ","End":"03:59.540","Text":"It\u0027s just the circumference of the disc."},{"Start":"03:59.540 ","End":"04:01.910","Text":"Let\u0027s call it a ring."},{"Start":"04:01.910 ","End":"04:05.030","Text":"Again, we can do the same trick as with the disc."},{"Start":"04:05.030 ","End":"04:07.740","Text":"If we have the z-axis like this,"},{"Start":"04:07.740 ","End":"04:12.530","Text":"we can just draw the disc and then we have instead of in"},{"Start":"04:12.530 ","End":"04:17.735","Text":"the example of the disc where we\u0027ve got a solid fill cylinder,"},{"Start":"04:17.735 ","End":"04:19.505","Text":"so when we do this with a ring,"},{"Start":"04:19.505 ","End":"04:22.755","Text":"we get a cylinder which is empty."},{"Start":"04:22.755 ","End":"04:26.190","Text":"Just the outer casing of the cylinder."},{"Start":"04:26.190 ","End":"04:30.590","Text":"Again, we can say that the moment of inertia of the ring,"},{"Start":"04:30.590 ","End":"04:38.175","Text":"which is equal to mR^2 of the disc."},{"Start":"04:38.175 ","End":"04:42.364","Text":"It\u0027s going to be 1/2mR^2 and of the ring it\u0027s going to be mR^2,"},{"Start":"04:42.364 ","End":"04:50.465","Text":"which is also going to be the same as the I of the cylinder casing."},{"Start":"04:50.465 ","End":"04:56.620","Text":"This was the solid cylinder."},{"Start":"04:56.620 ","End":"05:01.380","Text":"Now we\u0027re also done with this trick."}],"ID":9373}],"Thumbnail":null,"ID":5387},{"Name":"Calculating the Moment of Inertia of a Disk About the Z and X Axis","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Disk Rotating Around Z Axis","Duration":"3m 53s","ChapterTopicVideoID":9101,"CourseChapterTopicPlaylistID":5388,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.585","Text":"Hello. In this lesson we\u0027re going to be speaking about how to work out mathematically,"},{"Start":"00:06.585 ","End":"00:10.650","Text":"no cheats, the disks moments of inertia."},{"Start":"00:10.650 ","End":"00:13.710","Text":"As we can see, our disk is rotating about"},{"Start":"00:13.710 ","End":"00:17.775","Text":"its center and its axes of rotation is the z-axis."},{"Start":"00:17.775 ","End":"00:20.870","Text":"Now the desk has a radius of capital R and its mass"},{"Start":"00:20.870 ","End":"00:25.635","Text":"is capital M. If we remember from right at the beginning,"},{"Start":"00:25.635 ","End":"00:30.825","Text":"the equation for a moment of inertia I is equal to"},{"Start":"00:30.825 ","End":"00:37.570","Text":"the sum of M_i multiplied by r_i^2."},{"Start":"00:38.540 ","End":"00:41.660","Text":"But now what we\u0027re doing over here,"},{"Start":"00:41.660 ","End":"00:46.175","Text":"because it\u0027s a circular shape and we don\u0027t have a random point masses that we\u0027re"},{"Start":"00:46.175 ","End":"00:51.649","Text":"summing their mass multiplied by their distance from the center of rotation squared."},{"Start":"00:51.649 ","End":"00:54.385","Text":"We have to do an integral."},{"Start":"00:54.385 ","End":"00:57.555","Text":"We\u0027re integrating on r squared,"},{"Start":"00:57.555 ","End":"00:59.325","Text":"instead of r_i^2,"},{"Start":"00:59.325 ","End":"01:02.205","Text":"so it\u0027s r^2 dm."},{"Start":"01:02.205 ","End":"01:03.525","Text":"Instead of M_i,"},{"Start":"01:03.525 ","End":"01:06.040","Text":"it\u0027s going to be dm."},{"Start":"01:06.050 ","End":"01:10.455","Text":"Now let\u0027s see what our dm is."},{"Start":"01:10.455 ","End":"01:14.610","Text":"Our dm is equal to Sigma,"},{"Start":"01:14.610 ","End":"01:23.220","Text":"which is mass per unit of area multiplied by unit of area, which is ds."},{"Start":"01:23.220 ","End":"01:30.180","Text":"We\u0027re summing on all these little shapes on the disk, going around."},{"Start":"01:30.680 ","End":"01:34.790","Text":"Now let\u0027s see what our Sigma is equal to."},{"Start":"01:34.790 ","End":"01:39.230","Text":"Our Sigma is equal to our mass per area."},{"Start":"01:39.230 ","End":"01:42.605","Text":"Our total mass divided by the total area which is"},{"Start":"01:42.605 ","End":"01:46.790","Text":"S. That is equal because we\u0027re speaking about the desk,"},{"Start":"01:46.790 ","End":"01:52.805","Text":"so its mass divided by the area of the disk is Pi R^2."},{"Start":"01:52.805 ","End":"01:57.610","Text":"Here it\u0027s capital R because in the question we were given the radius."},{"Start":"01:57.610 ","End":"02:00.240","Text":"Then let\u0027s see what ds is."},{"Start":"02:00.240 ","End":"02:04.770","Text":"Our ds is a small section of area."},{"Start":"02:04.770 ","End":"02:07.610","Text":"Because we\u0027re dealing with a circular shape,"},{"Start":"02:07.610 ","End":"02:10.805","Text":"it\u0027s easiest for us to use polar coordinates,"},{"Start":"02:10.805 ","End":"02:14.905","Text":"which means our dr and d Thetas."},{"Start":"02:14.905 ","End":"02:17.930","Text":"We can write dr d Theta."},{"Start":"02:17.930 ","End":"02:20.390","Text":"As we know, whenever we have a d Theta written,"},{"Start":"02:20.390 ","End":"02:22.520","Text":"we have to add in our Jacobian,"},{"Start":"02:22.520 ","End":"02:27.930","Text":"which is r. Our ds is going to be r dr d Theta."},{"Start":"02:29.300 ","End":"02:39.440","Text":"Now we can say that our I is therefore equal to the integral of r^2 multiplied by dm,"},{"Start":"02:39.440 ","End":"02:42.830","Text":"which is going to be multiplied by dm,"},{"Start":"02:42.830 ","End":"02:43.965","Text":"which is Sigma ds."},{"Start":"02:43.965 ","End":"02:50.839","Text":"Our sigma is mass times divided by Pi squared multiplied by our ds."},{"Start":"02:50.839 ","End":"02:56.370","Text":"Our ds is r dr d Theta."},{"Start":"02:57.080 ","End":"02:59.400","Text":"Now a small note,"},{"Start":"02:59.400 ","End":"03:01.680","Text":"this r over here,"},{"Start":"03:01.680 ","End":"03:07.520","Text":"and our r that is our Jacobian won\u0027t always be the same r. In this example they are,"},{"Start":"03:07.520 ","End":"03:12.385","Text":"but we\u0027ll see later on in our examples that sometimes we\u0027ll get something else."},{"Start":"03:12.385 ","End":"03:15.380","Text":"That\u0027s fine. Now, of course,"},{"Start":"03:15.380 ","End":"03:19.310","Text":"because we\u0027re integrating along dr and d Theta,"},{"Start":"03:19.310 ","End":"03:22.835","Text":"along r and Theta, so this is a double integral."},{"Start":"03:22.835 ","End":"03:29.365","Text":"When we\u0027re integrating on r we\u0027re integrating between 0 and the final radius."},{"Start":"03:29.365 ","End":"03:31.460","Text":"I and d Theta,"},{"Start":"03:31.460 ","End":"03:32.750","Text":"because it\u0027s a circle,"},{"Start":"03:32.750 ","End":"03:36.215","Text":"it\u0027s a disk, so it\u0027s going to be from 0 until 2 Pi."},{"Start":"03:36.215 ","End":"03:37.820","Text":"Then in order to save time,"},{"Start":"03:37.820 ","End":"03:39.935","Text":"I won\u0027t do this integration right now."},{"Start":"03:39.935 ","End":"03:42.575","Text":"But if you were doing it, it\u0027s relatively simple."},{"Start":"03:42.575 ","End":"03:47.405","Text":"You\u0027ll get that the final answer as1/2 MR^2."},{"Start":"03:47.405 ","End":"03:50.810","Text":"You\u0027re welcome to check this on your own."},{"Start":"03:50.810 ","End":"03:54.450","Text":"That\u0027s the end of this lesson."}],"ID":9374},{"Watched":false,"Name":"Disk Rotating Around X Axis","Duration":"8m 40s","ChapterTopicVideoID":9102,"CourseChapterTopicPlaylistID":5388,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:05.955","Text":"Hello. In this lesson we\u0027re going to be speaking about the disk\u0027s moment of inertia,"},{"Start":"00:05.955 ","End":"00:07.965","Text":"this time around the x-axis."},{"Start":"00:07.965 ","End":"00:10.680","Text":"Now, if you go back to the previous video of"},{"Start":"00:10.680 ","End":"00:13.875","Text":"the disk\u0027s moment of inertia around the z-axis,"},{"Start":"00:13.875 ","End":"00:20.680","Text":"I would\u0027ve said there that the different rs that we get in the integral,"},{"Start":"00:20.680 ","End":"00:26.525","Text":"the r that we\u0027re integrating by and the r which is the Jacobian,"},{"Start":"00:26.525 ","End":"00:28.505","Text":"will be different rs."},{"Start":"00:28.505 ","End":"00:31.250","Text":"I said that that can happen, not in that video,"},{"Start":"00:31.250 ","End":"00:33.360","Text":"but it will happen in the future,"},{"Start":"00:33.360 ","End":"00:35.325","Text":"so here we are."},{"Start":"00:35.325 ","End":"00:39.045","Text":"This is the future where it\u0027s happening."},{"Start":"00:39.045 ","End":"00:41.300","Text":"Let\u0027s get back to our problem over here."},{"Start":"00:41.300 ","End":"00:44.870","Text":"The disk\u0027s moment of inertia around the x-axis."},{"Start":"00:44.870 ","End":"00:47.690","Text":"We have a disk of radius capital R,"},{"Start":"00:47.690 ","End":"00:49.775","Text":"mass capital M,"},{"Start":"00:49.775 ","End":"00:52.111","Text":"and it\u0027s rotating about the x-axis,"},{"Start":"00:52.111 ","End":"00:55.975","Text":"so something like this."},{"Start":"00:55.975 ","End":"00:57.965","Text":"Let\u0027s see how we do this."},{"Start":"00:57.965 ","End":"01:03.935","Text":"Now, we remember that the equation for the moment of inertia is going to be the sum"},{"Start":"01:03.935 ","End":"01:10.775","Text":"of the masses multiplied by their corresponding radiuses squared."},{"Start":"01:10.775 ","End":"01:14.705","Text":"I\u0027m going to put a little Tilde on top of this r over here,"},{"Start":"01:14.705 ","End":"01:21.530","Text":"because we\u0027re speaking about the distance now from the axis of rotation."},{"Start":"01:21.530 ","End":"01:25.120","Text":"Now, let\u0027s draw this in a diagram."},{"Start":"01:25.120 ","End":"01:27.823","Text":"We\u0027re rotating about the x-axis,"},{"Start":"01:27.823 ","End":"01:32.720","Text":"and we\u0027re going to split the disk off into little sections of mass."},{"Start":"01:32.720 ","End":"01:36.080","Text":"Of course we\u0027re summing along these sections."},{"Start":"01:36.080 ","End":"01:39.535","Text":"Let\u0027s speak about this that we\u0027re summing on."},{"Start":"01:39.535 ","End":"01:46.065","Text":"This little section is going to have mass dm and its radius,"},{"Start":"01:46.065 ","End":"01:51.860","Text":"it\u0027s distance, its displacement from the x-axis is going to be this,"},{"Start":"01:51.860 ","End":"01:54.055","Text":"so this is r Tilde."},{"Start":"01:54.055 ","End":"01:56.775","Text":"Its distance from the x-axis,"},{"Start":"01:56.775 ","End":"01:59.855","Text":"not from its center,"},{"Start":"01:59.855 ","End":"02:02.090","Text":"but rather from the x-axis."},{"Start":"02:02.090 ","End":"02:11.720","Text":"Our integral, we\u0027re going to be integrating along this distance from the x-axis squared."},{"Start":"02:11.720 ","End":"02:16.445","Text":"It\u0027s going to be r Tilde squared and the corresponding masses,"},{"Start":"02:16.445 ","End":"02:18.930","Text":"the small masses, which is dm."},{"Start":"02:18.930 ","End":"02:22.280","Text":"Let\u0027s just write that down."},{"Start":"02:22.280 ","End":"02:27.200","Text":"I made a little note this r Tilde is the distance from the axis of rotation,"},{"Start":"02:27.200 ","End":"02:31.715","Text":"not the distance from the center of the disk."},{"Start":"02:31.715 ","End":"02:33.980","Text":"Now let\u0027s see again."},{"Start":"02:33.980 ","End":"02:37.010","Text":"Now we see that we\u0027re dealing with rdm."},{"Start":"02:37.010 ","End":"02:39.080","Text":"What is rdm?"},{"Start":"02:39.080 ","End":"02:43.745","Text":"Because this is a flat shape,"},{"Start":"02:43.745 ","End":"02:48.500","Text":"so we\u0027re going to be using the dm for a 2-dimensional shape,"},{"Start":"02:48.500 ","End":"02:52.400","Text":"which is going to be Sigma multiplied by ds."},{"Start":"02:52.400 ","End":"02:55.790","Text":"What is Sigma? It\u0027s the density,"},{"Start":"02:55.790 ","End":"02:58.580","Text":"so the mass per unit of area,"},{"Start":"02:58.580 ","End":"03:03.060","Text":"and ds is the unit of area."},{"Start":"03:03.980 ","End":"03:07.925","Text":"Now let\u0027s see, therefore what our sigma is."},{"Start":"03:07.925 ","End":"03:10.820","Text":"Our Sigma, because here our Sigma is constant,"},{"Start":"03:10.820 ","End":"03:16.448","Text":"this is a homogenous uniform shape, so our density,"},{"Start":"03:16.448 ","End":"03:19.490","Text":"our Sigma is going to be equal to the total mass,"},{"Start":"03:19.490 ","End":"03:22.805","Text":"which is M, divided by the total area."},{"Start":"03:22.805 ","End":"03:27.495","Text":"Now as we know, the area of a disk is PiR^2."},{"Start":"03:27.495 ","End":"03:32.190","Text":"Then we\u0027re going to say that our ds is equal to,"},{"Start":"03:32.190 ","End":"03:35.900","Text":"now because we\u0027re dealing with a circular shape, we\u0027re dealing with a disk,"},{"Start":"03:35.900 ","End":"03:40.665","Text":"it\u0027s going to be rdrd Theta."},{"Start":"03:40.665 ","End":"03:43.080","Text":"What is r over here?"},{"Start":"03:43.080 ","End":"03:47.490","Text":"This is r Jacobian and it comes with rd Theta."},{"Start":"03:47.490 ","End":"03:52.310","Text":"Every time we\u0027re integrating and there\u0027s the term d Theta inside,"},{"Start":"03:52.310 ","End":"03:58.715","Text":"we have to remember to multiply it by r. We have rdrd Theta."},{"Start":"03:58.715 ","End":"04:02.198","Text":"Now, lo and behold what I was speaking about."},{"Start":"04:02.198 ","End":"04:08.475","Text":"This r Tilde and this r,"},{"Start":"04:08.475 ","End":"04:14.830","Text":"so r Tilde does not equal this r Jacobian."},{"Start":"04:16.430 ","End":"04:19.875","Text":"Now let\u0027s talk about this for a moment."},{"Start":"04:19.875 ","End":"04:21.960","Text":"What are these different rs?"},{"Start":"04:21.960 ","End":"04:25.790","Text":"As we saw, our r Tilde is the distance from the axis of"},{"Start":"04:25.790 ","End":"04:31.655","Text":"rotation and our r over here r Jacobian, what is that?"},{"Start":"04:31.655 ","End":"04:37.235","Text":"It\u0027s the distance from the center in this diagonal way,"},{"Start":"04:37.235 ","End":"04:39.470","Text":"in the direction of the radius."},{"Start":"04:39.470 ","End":"04:43.670","Text":"This is r. Then,"},{"Start":"04:43.670 ","End":"04:45.650","Text":"because we\u0027re integrating through Theta,"},{"Start":"04:45.650 ","End":"04:49.060","Text":"so this angle over here will be Theta."},{"Start":"04:49.060 ","End":"04:51.830","Text":"Now as we know from trigonometry,"},{"Start":"04:51.830 ","End":"04:58.100","Text":"we have the hypotenuse and we have the opposite angle. What uses that?"},{"Start":"04:58.100 ","End":"05:04.795","Text":"We have the sine of the angle is equal to the opposite over the hypotenuse."},{"Start":"05:04.795 ","End":"05:12.080","Text":"Therefore, we can just simply rearrange that and say that the relationship between r and"},{"Start":"05:12.080 ","End":"05:20.370","Text":"r Tilde is that r Tilde is equal to r multiplied by sine of the angle,"},{"Start":"05:20.380 ","End":"05:22.635","Text":"from this diagram here."},{"Start":"05:22.635 ","End":"05:27.765","Text":"I\u0027m reminding you of the r in the Jacobian is equal to"},{"Start":"05:27.765 ","End":"05:35.005","Text":"the radius from the center of the disk until the rdm."},{"Start":"05:35.005 ","End":"05:38.240","Text":"Until a little square of mass,"},{"Start":"05:38.240 ","End":"05:41.880","Text":"which we are integrating by."},{"Start":"05:42.370 ","End":"05:46.460","Text":"Then through trigonometry, we get the relationship that"},{"Start":"05:46.460 ","End":"05:50.580","Text":"r Tilde is equal to r sine of Theta."},{"Start":"05:51.400 ","End":"05:56.725","Text":"Now we can substitute all of this in to our equation."},{"Start":"05:56.725 ","End":"06:01.400","Text":"We\u0027ll have that I is equal to the double integral because we\u0027re"},{"Start":"06:01.400 ","End":"06:06.275","Text":"integrating on r and Theta of r Tilde squared,"},{"Start":"06:06.275 ","End":"06:10.620","Text":"which is just, r Tilde is equal to r sine Theta,"},{"Start":"06:10.620 ","End":"06:17.255","Text":"so it\u0027s going to be r squared sine squared of Theta multiplied by rdm."},{"Start":"06:17.255 ","End":"06:20.675","Text":"rdm is made up of r Sigma,"},{"Start":"06:20.675 ","End":"06:24.830","Text":"which is mass divided by PiR^2."},{"Start":"06:24.830 ","End":"06:27.275","Text":"Remember it\u0027s a capital R because it\u0027s the area of"},{"Start":"06:27.275 ","End":"06:32.615","Text":"the entire disk multiplied by then rds,"},{"Start":"06:32.615 ","End":"06:38.400","Text":"which is equal to rdrd Theta."},{"Start":"06:38.400 ","End":"06:43.190","Text":"We are integrating our r from 0 until our maximum radius,"},{"Start":"06:43.190 ","End":"06:50.010","Text":"which is R and r Theta because it\u0027s a full circle from 0 until 2Pi."},{"Start":"06:51.860 ","End":"06:57.530","Text":"Now we have this equation and what we will get is that"},{"Start":"06:57.530 ","End":"07:03.430","Text":"the I is equal to 1/4MR^2."},{"Start":"07:03.430 ","End":"07:07.295","Text":"Now, if you remember back to my cheats,"},{"Start":"07:07.295 ","End":"07:10.010","Text":"the lesson of the third cheats,"},{"Start":"07:10.010 ","End":"07:18.180","Text":"I believe it was, which was the equation for I_z equals I_x plus I_y."},{"Start":"07:18.180 ","End":"07:22.205","Text":"If you take yourselves back to that lesson,"},{"Start":"07:22.205 ","End":"07:28.395","Text":"you\u0027ll remember we said that our I_x was equal to our I_y."},{"Start":"07:28.395 ","End":"07:34.140","Text":"Then we said that our I_z is equal to either to 2I_x or 2I_y,"},{"Start":"07:34.140 ","End":"07:35.745","Text":"it makes no difference."},{"Start":"07:35.745 ","End":"07:39.770","Text":"Then remember that just now in our lesson,"},{"Start":"07:39.770 ","End":"07:44.030","Text":"we saw that if we have a disk rotating about the z axis,"},{"Start":"07:44.030 ","End":"07:54.405","Text":"it is equal to 1/2MR^2 and then our cheat lesson we saw that to find the I"},{"Start":"07:54.405 ","End":"07:58.700","Text":"around the x-axis and the I around the y-axis would be equal to"},{"Start":"07:58.700 ","End":"08:06.060","Text":"1/4MR^2 or 1/2 times 1/2MR^2, if you remember."},{"Start":"08:06.060 ","End":"08:12.505","Text":"Here we go. Here\u0027s the proof showing that it was correct."},{"Start":"08:12.505 ","End":"08:15.285","Text":"Now, just a little note,"},{"Start":"08:15.285 ","End":"08:21.200","Text":"a little tip on how to integrate when integrating sine squared Theta."},{"Start":"08:21.200 ","End":"08:28.785","Text":"A handy trick is to say that sine squared of Theta is equal to"},{"Start":"08:28.785 ","End":"08:38.460","Text":"1 minus cosine of 2 Theta divided by 2 and then the integration is very much simpler."},{"Start":"08:38.460 ","End":"08:41.380","Text":"That\u0027s the end of this lesson."}],"ID":9375},{"Watched":false,"Name":"Disk Rotating Around X Axis Using The Sum of Axis","Duration":"1m 48s","ChapterTopicVideoID":9103,"CourseChapterTopicPlaylistID":5388,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.400 ","End":"00:04.410","Text":"This will be working out the Disk\u0027s Moment of"},{"Start":"00:04.410 ","End":"00:07.560","Text":"Inertia around the x-axis as the sum of the axis."},{"Start":"00:07.560 ","End":"00:11.730","Text":"This is just using our equation that we learned in"},{"Start":"00:11.730 ","End":"00:18.790","Text":"our tricks which says that I_z is equal to I_ x plus I_ y."},{"Start":"00:19.180 ","End":"00:27.273","Text":"If we have our z axis coming through the center of the disk and out of your screen,"},{"Start":"00:27.273 ","End":"00:34.060","Text":"we know that our I_z is going to be equal to 1/2 MR^2."},{"Start":"00:34.060 ","End":"00:36.600","Text":"We\u0027ve seen this, we\u0027ve worked this out."},{"Start":"00:36.600 ","End":"00:40.730","Text":"Then we can say that because the shape that\u0027s"},{"Start":"00:40.730 ","End":"00:44.720","Text":"spinning around the x-axis in this direction,"},{"Start":"00:44.720 ","End":"00:49.340","Text":"and the shape that is spinning around the y-axis in this direction,"},{"Start":"00:49.340 ","End":"00:53.755","Text":"is the same shape, which means that it\u0027s going to have the same moment of inertia."},{"Start":"00:53.755 ","End":"00:58.380","Text":"Then, we can say that our I_ x is equal to our I_y."},{"Start":"00:58.380 ","End":"01:02.280","Text":"Then we can then therefore say that I_z is"},{"Start":"01:02.280 ","End":"01:11.820","Text":"equal to 2I_x or 2I_ y."},{"Start":"01:11.820 ","End":"01:14.430","Text":"Let\u0027s take the I_x for example."},{"Start":"01:14.430 ","End":"01:16.830","Text":"We can say therefore that"},{"Start":"01:16.830 ","End":"01:25.305","Text":"1/2 I_z is equal to I_ x. I\u0027ve just rearranged this equation."},{"Start":"01:25.305 ","End":"01:31.605","Text":"Therefore, we can say that our I_x is equal to 1/2 times our I_z,"},{"Start":"01:31.605 ","End":"01:35.670","Text":"which is 1/2 times 1/2 MR^2,"},{"Start":"01:35.670 ","End":"01:40.005","Text":"which is equal to a 1/4 MR^2 squared."},{"Start":"01:40.005 ","End":"01:45.425","Text":"This will obviously be the same as our I in our y-direction."},{"Start":"01:45.425 ","End":"01:48.270","Text":"That\u0027s the end of this lesson."}],"ID":9376}],"Thumbnail":null,"ID":5388},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"I of Non Uniform Rod","Duration":"8m 35s","ChapterTopicVideoID":9104,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9104.jpeg","UploadDate":"2017-03-22T10:37:00.5330000","DurationForVideoObject":"PT8M35S","Description":null,"MetaTitle":"I of Non Uniform Rod: Video + Workbook | Proprep","MetaDescription":"Moment of Inertia - Exercises. 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/moment-of-inertia/exercises/vid9377","VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.880","Text":"Hello. In this question,"},{"Start":"00:02.880 ","End":"00:05.550","Text":"we\u0027re being told to calculate the moment of inertia of"},{"Start":"00:05.550 ","End":"00:09.735","Text":"a rod of density Lambda as a function of x is equal to"},{"Start":"00:09.735 ","End":"00:17.940","Text":"Lambda 0 multiplied by x divided by L. The rod is rotating about the edge of the rod."},{"Start":"00:17.940 ","End":"00:22.800","Text":"Its axis of rotation is right at the edge of the rod and not in the center."},{"Start":"00:22.800 ","End":"00:24.870","Text":"Now, we\u0027re told that L,"},{"Start":"00:24.870 ","End":"00:28.545","Text":"the L over here is the length of the rod, the total length."},{"Start":"00:28.545 ","End":"00:31.245","Text":"Then x is the distance from the edge,"},{"Start":"00:31.245 ","End":"00:37.220","Text":"x is as we\u0027re moving from 0 to L. Now,"},{"Start":"00:37.220 ","End":"00:41.360","Text":"notice when x is equal to 0 the density of the rod is 0."},{"Start":"00:41.360 ","End":"00:45.995","Text":"Notice that right at the other end of the rod at distance L,"},{"Start":"00:45.995 ","End":"00:49.880","Text":"our density is Lambda 0 which is our maximum density."},{"Start":"00:49.880 ","End":"00:51.455","Text":"As we can see,"},{"Start":"00:51.455 ","End":"00:55.535","Text":"our density isn\u0027t uniform and it\u0027s changing with the distance."},{"Start":"00:55.535 ","End":"01:04.119","Text":"As we get further away from the axis of rotation so our density is increasing."},{"Start":"01:04.460 ","End":"01:08.235","Text":"Let\u0027s see how we solve this."},{"Start":"01:08.235 ","End":"01:14.570","Text":"Let\u0027s first find out what our moment of inertia is right over here at the origin."},{"Start":"01:14.570 ","End":"01:18.075","Text":"We\u0027re going at 0, so we\u0027re going to call this I_0."},{"Start":"01:18.075 ","End":"01:25.660","Text":"According to the equation we know that it\u0027s the integral of r squared multiplied by dm."},{"Start":"01:25.660 ","End":"01:30.210","Text":"Let\u0027s speak about what our dm is equal to."},{"Start":"01:30.210 ","End":"01:34.260","Text":"Our dm, I know that here I\u0027ve drawn it in"},{"Start":"01:34.260 ","End":"01:38.970","Text":"a 2 dimensional way but this is in fact just a line."},{"Start":"01:38.970 ","End":"01:41.410","Text":"Let me redraw this."},{"Start":"01:41.840 ","End":"01:44.430","Text":"This is our rod."},{"Start":"01:44.430 ","End":"01:47.790","Text":"So because it\u0027s 1 dimensional, it\u0027s just a line."},{"Start":"01:47.790 ","End":"01:54.510","Text":"I know that I can use for my dm Lambda dx because Lambda"},{"Start":"01:54.510 ","End":"02:00.980","Text":"is my density per unit length and I\u0027m integrating along the x-axis so I can call it dx,"},{"Start":"02:00.980 ","End":"02:02.420","Text":"I could have also called it dL,"},{"Start":"02:02.420 ","End":"02:04.800","Text":"it doesn\u0027t make a difference."},{"Start":"02:05.300 ","End":"02:09.585","Text":"Now, what is my Lambda equal to?"},{"Start":"02:09.585 ","End":"02:12.240","Text":"This is given to me in my question."},{"Start":"02:12.240 ","End":"02:15.945","Text":"My Lambda is equal to Lambda 0x"},{"Start":"02:15.945 ","End":"02:20.450","Text":"divided by l. Here comes in that it\u0027s good that I called it dx"},{"Start":"02:20.450 ","End":"02:24.100","Text":"because we\u0027re integrating along my distance"},{"Start":"02:24.100 ","End":"02:30.420","Text":"from the origin and my distance is changing according to x."},{"Start":"02:31.090 ","End":"02:36.965","Text":"Then my dx remains dx."},{"Start":"02:36.965 ","End":"02:44.040","Text":"Let\u0027s rewrite that so I have my integral of my r^2 is going to actually be,"},{"Start":"02:44.040 ","End":"02:48.980","Text":"so this is going to be equal to x^2 because I\u0027m integrating not"},{"Start":"02:48.980 ","End":"02:54.754","Text":"according to a radius but according to a distance along the x-axis."},{"Start":"02:54.754 ","End":"03:00.050","Text":"It\u0027s going to be equal to x^2 multiplied by my dm which is equal"},{"Start":"03:00.050 ","End":"03:08.205","Text":"to Lambda 0x divided by l multiplied by dx."},{"Start":"03:08.205 ","End":"03:14.440","Text":"My Lambda 0 and my L are constants so I can take them out of the integral."},{"Start":"03:14.440 ","End":"03:16.530","Text":"I\u0027ll have Lambda 0 divided by l,"},{"Start":"03:16.530 ","End":"03:21.740","Text":"the integral between 0 until L because that\u0027s"},{"Start":"03:21.740 ","End":"03:27.870","Text":"the distance that I\u0027m integrating through of x^3dx."},{"Start":"03:27.980 ","End":"03:32.805","Text":"Obviously, this is going to equal to"},{"Start":"03:32.805 ","End":"03:42.100","Text":"L^3 multiplied by Lambda 0 divided by 4."},{"Start":"03:42.380 ","End":"03:46.080","Text":"Now, the answer that we have here of"},{"Start":"03:46.080 ","End":"03:51.240","Text":"L^3 multiplied by Lambda 0 divided by 4 is a great answer,"},{"Start":"03:51.240 ","End":"03:53.675","Text":"if you write it in the test you\u0027ll get 100 percent."},{"Start":"03:53.675 ","End":"03:58.655","Text":"However, just to make a slightly better,"},{"Start":"03:58.655 ","End":"04:00.030","Text":"more correct answer,"},{"Start":"04:00.030 ","End":"04:07.100","Text":"and also just in case you have further questions that relate to your answer over here."},{"Start":"04:07.100 ","End":"04:09.680","Text":"It\u0027s easier instead of working with density,"},{"Start":"04:09.680 ","End":"04:13.050","Text":"our Lambda 0 is our density."},{"Start":"04:13.050 ","End":"04:17.030","Text":"Instead of using density it\u0027s easier to work with mass."},{"Start":"04:17.030 ","End":"04:18.650","Text":"Also with the units,"},{"Start":"04:18.650 ","End":"04:20.450","Text":"also with a lot of equations,"},{"Start":"04:20.450 ","End":"04:22.910","Text":"and a lot of further questions that they can ask you."},{"Start":"04:22.910 ","End":"04:27.050","Text":"On top of this, so to make life easier and to make"},{"Start":"04:27.050 ","End":"04:31.825","Text":"yourself more correct and more fancy let\u0027s see how we do this."},{"Start":"04:31.825 ","End":"04:37.199","Text":"What we want to do is we want to work with mass instead of with density."},{"Start":"04:37.360 ","End":"04:41.125","Text":"Let\u0027s see. Now, what is mass?"},{"Start":"04:41.125 ","End":"04:43.760","Text":"Now, how we figure out the mass of"},{"Start":"04:43.760 ","End":"04:48.820","Text":"this non-uniform shape because obviously here our mass is going to be 0"},{"Start":"04:48.820 ","End":"04:58.945","Text":".0 and L our mass is going to be Lambda 0 divided by the total length."},{"Start":"04:58.945 ","End":"05:01.140","Text":"Let\u0027s see how we do this."},{"Start":"05:01.140 ","End":"05:04.650","Text":"Now, the definition of mass is the integral."},{"Start":"05:04.650 ","End":"05:06.690","Text":"We\u0027re summing along the entire shape,"},{"Start":"05:06.690 ","End":"05:07.890","Text":"so here it\u0027s a line."},{"Start":"05:07.890 ","End":"05:13.085","Text":"It\u0027s from 0 to L multiplied by the density,"},{"Start":"05:13.085 ","End":"05:17.400","Text":"multiplied by the length, so dl."},{"Start":"05:17.400 ","End":"05:25.650","Text":"We\u0027ve already said that our dl is equal to dx and that our density is"},{"Start":"05:25.650 ","End":"05:34.875","Text":"equal to Lambda 0x divided by L. This is the definition of mass,"},{"Start":"05:34.875 ","End":"05:39.275","Text":"it\u0027s the density per unit of"},{"Start":"05:39.275 ","End":"05:45.560","Text":"the object of the shape multiplied by the length or the area or the volume."},{"Start":"05:45.560 ","End":"05:47.435","Text":"Here we\u0027re doing by length,"},{"Start":"05:47.435 ","End":"05:51.115","Text":"and then we put in our bounds."},{"Start":"05:51.115 ","End":"06:00.810","Text":"Let\u0027s put this in so we\u0027re doing from 0 to L of Lambda 0x divided by L dx."},{"Start":"06:00.810 ","End":"06:06.120","Text":"Again, our Lambda 0 and our L are constants so we can take them out of their integral."},{"Start":"06:06.120 ","End":"06:11.955","Text":"We\u0027ll have Lambda 0 divided by L integral from 0 of L of xdx."},{"Start":"06:11.955 ","End":"06:22.015","Text":"As we know, this will equal to Lambda 0 multiplied by L divided by 2,"},{"Start":"06:22.015 ","End":"06:24.245","Text":"I just did the integration."},{"Start":"06:24.245 ","End":"06:30.155","Text":"Now, let\u0027s see that this actually makes sense because we have a rod of non-uniform mass."},{"Start":"06:30.155 ","End":"06:37.475","Text":"We can take, because here it will be 0 and here it will be Lambda 0 divided by L,"},{"Start":"06:37.475 ","End":"06:40.160","Text":"Lambda 0 L, I\u0027m sorry."},{"Start":"06:40.160 ","End":"06:43.255","Text":"We can just take what the mass would be"},{"Start":"06:43.255 ","End":"06:46.995","Text":"let\u0027s say in the middle or what our mean mass will be."},{"Start":"06:46.995 ","End":"06:50.420","Text":"This really does make sense because it\u0027s Lambda 0 divided by L,"},{"Start":"06:50.420 ","End":"06:56.570","Text":"so it\u0027s like the total mass if the entire rod had a uniform mass divided by 2,"},{"Start":"06:56.570 ","End":"06:59.620","Text":"so it\u0027s the mean mass of the rod."},{"Start":"06:59.620 ","End":"07:02.450","Text":"We got this and we see that this makes sense."},{"Start":"07:02.450 ","End":"07:03.620","Text":"Now, how do we use this?"},{"Start":"07:03.620 ","End":"07:09.180","Text":"How do we put this into our answer that we got in our pink over here above?"},{"Start":"07:09.410 ","End":"07:15.575","Text":"I want to make my life a little bit simpler and also avoid making calculation errors"},{"Start":"07:15.575 ","End":"07:18.890","Text":"because if you\u0027re already going this extra mile to really"},{"Start":"07:18.890 ","End":"07:22.040","Text":"get that bonus in that your professor will say,"},{"Start":"07:22.040 ","End":"07:25.455","Text":"wow, this guy, this girl is genius."},{"Start":"07:25.455 ","End":"07:29.915","Text":"We want to avoid that calculation error that will then in the end take off marks."},{"Start":"07:29.915 ","End":"07:34.385","Text":"What I\u0027m going to do is I\u0027m going to rearrange what I got here in my answer in"},{"Start":"07:34.385 ","End":"07:39.650","Text":"pink such that I can just substitute in what my M is."},{"Start":"07:39.650 ","End":"07:44.480","Text":"Instead of Lambda 0 multiplied by L^3 divided by 4 I can just"},{"Start":"07:44.480 ","End":"07:49.815","Text":"write Lambda 0 multiplied by L divided by 2,"},{"Start":"07:49.815 ","End":"07:53.400","Text":"multiplied by L^2 divided by 2."},{"Start":"07:53.400 ","End":"08:00.660","Text":"Notice, if we join these up together we\u0027ll again get Lambda 0 L^3 divided by 4."},{"Start":"08:00.660 ","End":"08:06.960","Text":"Then all I have to do is this and this are equal,"},{"Start":"08:06.960 ","End":"08:12.895","Text":"this is equal to m. Then instead of writing Lambda 0 L divided by 2,"},{"Start":"08:12.895 ","End":"08:18.165","Text":"I just write M multiplied by L^2 divided by 2."},{"Start":"08:18.165 ","End":"08:26.150","Text":"This is equal to the moment of inertia of a non-uniform rod rotating from its edge."},{"Start":"08:26.150 ","End":"08:29.555","Text":"Remember it\u0027s not rotating from the center of the rod,"},{"Start":"08:29.555 ","End":"08:32.840","Text":"it\u0027s rotating from 1 end of the rod."},{"Start":"08:32.840 ","End":"08:35.760","Text":"That\u0027s the end of the lesson."}],"ID":9377},{"Watched":false,"Name":"I of Non Uniform Rod Rotating At Center","Duration":"6m 55s","ChapterTopicVideoID":9105,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"Hello. In this lesson,"},{"Start":"00:02.460 ","End":"00:05.670","Text":"we\u0027re going to be calculating the moment of inertia of a rod of"},{"Start":"00:05.670 ","End":"00:10.890","Text":"density Lambda as a function of x=A multiplied by x^2,"},{"Start":"00:10.890 ","End":"00:14.055","Text":"which is rotating about the center of the rod."},{"Start":"00:14.055 ","End":"00:15.630","Text":"The rod is rotating,"},{"Start":"00:15.630 ","End":"00:17.145","Text":"this is the center."},{"Start":"00:17.145 ","End":"00:19.290","Text":"L is the length of the rod,"},{"Start":"00:19.290 ","End":"00:22.665","Text":"and x is the distance from the center of the rod."},{"Start":"00:22.665 ","End":"00:26.170","Text":"From here until here,"},{"Start":"00:28.040 ","End":"00:32.260","Text":"this is the length of the rod."},{"Start":"00:32.350 ","End":"00:38.330","Text":"We can see that as x increases or as it decreases,"},{"Start":"00:38.330 ","End":"00:41.165","Text":"on either side of the z-axis,"},{"Start":"00:41.165 ","End":"00:45.120","Text":"our density is increasing squared."},{"Start":"00:47.450 ","End":"00:51.870","Text":"Let\u0027s see how we do this calculation."},{"Start":"00:52.640 ","End":"00:58.670","Text":"As we know, our equation for the moment of inertia is equal"},{"Start":"00:58.670 ","End":"01:04.970","Text":"to the integral of r^2 dm."},{"Start":"01:04.970 ","End":"01:09.285","Text":"Let\u0027s see, our r^2 over here because we\u0027re dealing with a one-dimensional rod,"},{"Start":"01:09.285 ","End":"01:14.305","Text":"so it\u0027s just a line along the x-axis."},{"Start":"01:14.305 ","End":"01:17.555","Text":"We can just call it x^2."},{"Start":"01:17.555 ","End":"01:21.810","Text":"Then what is our dm going to be equal to?"},{"Start":"01:21.810 ","End":"01:27.290","Text":"Our dm, because we\u0027re dealing again with a one-dimensional line along the x-axis,"},{"Start":"01:27.290 ","End":"01:31.560","Text":"so we\u0027re going to say that it\u0027s Lambda multiplied by dx."},{"Start":"01:31.560 ","End":"01:35.085","Text":"Let\u0027s just draw this if we take some section."},{"Start":"01:35.085 ","End":"01:43.670","Text":"This will be dx of small length and the distance 2 dx is distance x."},{"Start":"01:43.670 ","End":"01:46.670","Text":"This is what we have."},{"Start":"01:46.670 ","End":"01:48.647","Text":"Now, our Lambda is equal to"},{"Start":"01:48.647 ","End":"01:58.335","Text":"Ax^2 as given in the question, and dx=dx."},{"Start":"01:58.335 ","End":"02:01.955","Text":"Now we can rewrite this in with our bounds."},{"Start":"02:01.955 ","End":"02:03.155","Text":"Now, what are our bounds?"},{"Start":"02:03.155 ","End":"02:11.550","Text":"We\u0027re going to be going from negative L over 2 up until positive L over 2."},{"Start":"02:11.550 ","End":"02:13.655","Text":"Along the entire rod,"},{"Start":"02:13.655 ","End":"02:16.745","Text":"notice we can\u0027t go from 0 until L,"},{"Start":"02:16.745 ","End":"02:23.800","Text":"we can have that as our bounds because then our integration will come out strange."},{"Start":"02:23.800 ","End":"02:28.520","Text":"Because we have to take our integration from our point of symmetry."},{"Start":"02:28.520 ","End":"02:30.620","Text":"Mirror, for instance,"},{"Start":"02:30.620 ","End":"02:36.900","Text":"is at the z-axis until L over 2 and until negative L over 2."},{"Start":"02:36.910 ","End":"02:39.335","Text":"Then we\u0027ll have our bounds,"},{"Start":"02:39.335 ","End":"02:48.720","Text":"and then we\u0027ll have x^2 multiplied by Ax^2 dx."},{"Start":"02:48.720 ","End":"02:50.480","Text":"Now, we can have a look."},{"Start":"02:50.480 ","End":"02:53.255","Text":"We can see that our A is constant, so we\u0027ll have A,"},{"Start":"02:53.255 ","End":"03:01.190","Text":"the integral from negative L over 2 until L over 2 multiplied by x^4 dx."},{"Start":"03:01.190 ","End":"03:04.330","Text":"Then, we can simply do the integration,"},{"Start":"03:04.330 ","End":"03:06.950","Text":"cancel out like terms."},{"Start":"03:06.950 ","End":"03:11.780","Text":"You can do this already at home and you will see that you will"},{"Start":"03:11.780 ","End":"03:18.600","Text":"get A multiplied by L^5 divided by 80."},{"Start":"03:18.600 ","End":"03:21.765","Text":"Now we have this answer but again,"},{"Start":"03:21.765 ","End":"03:26.780","Text":"we have this A in our equation and it\u0027s just"},{"Start":"03:26.780 ","End":"03:29.180","Text":"going to be a lot more complicated if they ask you"},{"Start":"03:29.180 ","End":"03:32.330","Text":"follow-up questions based on this answer,"},{"Start":"03:32.330 ","End":"03:35.495","Text":"and with the units, it\u0027s also a little bit complicated and it\u0027s just"},{"Start":"03:35.495 ","End":"03:39.214","Text":"a lot easier to present"},{"Start":"03:39.214 ","End":"03:41.810","Text":"the moment of inertia when"},{"Start":"03:41.810 ","End":"03:48.390","Text":"there\u0027s units for mass in the final answer, it\u0027s the better way."},{"Start":"03:48.950 ","End":"03:52.485","Text":"What we want to do is we want to get rid of this A."},{"Start":"03:52.485 ","End":"04:00.015","Text":"Now what we have to do is we have to find out what m is as a function of A."},{"Start":"04:00.015 ","End":"04:05.710","Text":"Now it isn\u0027t written here but the question I forgot to add that we\u0027re told that"},{"Start":"04:05.710 ","End":"04:11.370","Text":"the mass of the rod is m. Let\u0027s see what we can do."},{"Start":"04:11.370 ","End":"04:13.095","Text":"We can say that our m,"},{"Start":"04:13.095 ","End":"04:15.300","Text":"which we are given, it\u0027s known,"},{"Start":"04:15.300 ","End":"04:20.670","Text":"is the integral on dm of the entire rod."},{"Start":"04:20.670 ","End":"04:27.505","Text":"We\u0027re going to have an integral again with the same bounds from negative L divided by 2"},{"Start":"04:27.505 ","End":"04:34.880","Text":"until L divided by 2 equal directions around the axis of rotation,"},{"Start":"04:34.880 ","End":"04:39.185","Text":"which is the z-axis by dm."},{"Start":"04:39.185 ","End":"04:42.665","Text":"Now, let\u0027s take a look at what our dm is."},{"Start":"04:42.665 ","End":"04:49.150","Text":"Now, our dm is once again our Lambda dx."},{"Start":"04:49.150 ","End":"04:54.405","Text":"As we know, our Lambda is Ax^2."},{"Start":"04:54.405 ","End":"05:00.065","Text":"What we\u0027re going to do is we\u0027re going to write Ax^2 dx,"},{"Start":"05:00.065 ","End":"05:02.945","Text":"which then because our a is A constant,"},{"Start":"05:02.945 ","End":"05:12.050","Text":"we can take it out of the integral and then negative L over 2 to L over 2 x^2 dx."},{"Start":"05:13.120 ","End":"05:18.410","Text":"Now, once we do the integral and substitute in our bounds"},{"Start":"05:18.410 ","End":"05:24.270","Text":"we\u0027ll get that our m is equal to 1/12 AL^3."},{"Start":"05:25.520 ","End":"05:28.455","Text":"This is what our m equals."},{"Start":"05:28.455 ","End":"05:32.955","Text":"Now, we can see what our A is equal to."},{"Start":"05:32.955 ","End":"05:39.405","Text":"If we say m is equal to 1/12 AL^3,"},{"Start":"05:39.405 ","End":"05:42.800","Text":"then if we isolate out the A,"},{"Start":"05:42.800 ","End":"05:50.020","Text":"then we\u0027ll have that A is equal to 12m divided by L^3."},{"Start":"05:50.020 ","End":"05:54.605","Text":"Now if we substitute that into our I,"},{"Start":"05:54.605 ","End":"05:59.045","Text":"I\u0027m just going to change colors to make this easier on the I,"},{"Start":"05:59.045 ","End":"06:08.280","Text":"then we have that I=AL^5 divided by 80,"},{"Start":"06:08.280 ","End":"06:12.320","Text":"and then we\u0027re going to substitute in for A, what we found."},{"Start":"06:12.320 ","End":"06:21.510","Text":"We\u0027ll have 12m divided by L^3 multiplied by L^5 divided by 80."},{"Start":"06:21.510 ","End":"06:27.790","Text":"Now, we can just cross out like factors,"},{"Start":"06:30.440 ","End":"06:41.910","Text":"and then 12 over 80 will be 3 over 20 multiplied by mL^2."},{"Start":"06:41.910 ","End":"06:46.965","Text":"Then, this is our final answer,"},{"Start":"06:46.965 ","End":"06:53.335","Text":"and this is a far easier answer to use and to manipulate in further questions."},{"Start":"06:53.335 ","End":"06:56.200","Text":"That\u0027s the end of this lesson."}],"ID":9378},{"Watched":false,"Name":"Moment of Inertia L","Duration":"17m 33s","ChapterTopicVideoID":9107,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:02.970","Text":"Hello. In this question,"},{"Start":"00:02.970 ","End":"00:06.465","Text":"we\u0027re being told that there are 2 ways to calculate the value for I,"},{"Start":"00:06.465 ","End":"00:09.390","Text":"so the moment of inertia for the body in"},{"Start":"00:09.390 ","End":"00:13.365","Text":"the sketch and centered around its center of mass."},{"Start":"00:13.365 ","End":"00:17.730","Text":"They\u0027re asking us to show both ways of calculating the moment of inertia,"},{"Start":"00:17.730 ","End":"00:23.700","Text":"and we\u0027re being told that each rod is of length l and mass m. This is of length l,"},{"Start":"00:23.700 ","End":"00:24.840","Text":"this is of length l,"},{"Start":"00:24.840 ","End":"00:26.010","Text":"this is of length m,"},{"Start":"00:26.010 ","End":"00:32.590","Text":"and this is of length m. I\u0027m just going to rub this out in the meantime."},{"Start":"00:33.290 ","End":"00:38.400","Text":"Now, the first way we\u0027ll do is probably the slightly longer way."},{"Start":"00:38.400 ","End":"00:42.560","Text":"If we look, both rods are exactly symmetrical,"},{"Start":"00:42.560 ","End":"00:45.025","Text":"so they\u0027re both the same length."},{"Start":"00:45.025 ","End":"00:49.400","Text":"Then I can say that the center of mass of"},{"Start":"00:49.400 ","End":"00:54.065","Text":"this rod is approximately here and the center of mass of this rod is here,"},{"Start":"00:54.065 ","End":"00:56.250","Text":"right in the middle."},{"Start":"00:56.620 ","End":"00:59.870","Text":"Then the center of mass, therefore,"},{"Start":"00:59.870 ","End":"01:04.760","Text":"between these now 2 masses will be right in the center."},{"Start":"01:04.760 ","End":"01:08.450","Text":"We can see that by either looking at"},{"Start":"01:08.450 ","End":"01:12.770","Text":"the symmetry of the picture and seeing that this point is in the middle of"},{"Start":"01:12.770 ","End":"01:16.580","Text":"this rod and this point is in the middle of this rod and so therefore"},{"Start":"01:16.580 ","End":"01:21.020","Text":"the center between them is just going to be in the middle of the line joining the 2."},{"Start":"01:21.020 ","End":"01:26.855","Text":"Or alternatively, we can look at where this force"},{"Start":"01:26.855 ","End":"01:33.020","Text":"and where this point falls and see that if we look at this as being the x axis,"},{"Start":"01:33.020 ","End":"01:36.710","Text":"for instance, and this has been the y axis in the middle of,"},{"Start":"01:36.710 ","End":"01:39.840","Text":"between the origin to our point,"},{"Start":"01:39.840 ","End":"01:44.440","Text":"and then the middle on the y-axis between the origin and our point,"},{"Start":"01:44.440 ","End":"01:45.680","Text":"and we join them up."},{"Start":"01:45.680 ","End":"01:50.960","Text":"We\u0027ll see that we\u0027ll get to our center of mass over here."},{"Start":"01:50.960 ","End":"02:00.780","Text":"Now we\u0027ll also notice that this point over here will be l divided by 4,"},{"Start":"02:00.780 ","End":"02:07.130","Text":"and this point over here will also be l divided by 4."},{"Start":"02:07.130 ","End":"02:10.190","Text":"Now what we\u0027re going to do is we\u0027re going to use"},{"Start":"02:10.190 ","End":"02:14.580","Text":"this information in order to find the moment of inertia."},{"Start":"02:14.980 ","End":"02:18.590","Text":"As I said, we\u0027re doing right now the long version,"},{"Start":"02:18.590 ","End":"02:25.055","Text":"and just pay attention because it gets a tiny little bit tricky in a second."},{"Start":"02:25.055 ","End":"02:29.240","Text":"We write down that I\u0027m moment of inertia is equal to,"},{"Start":"02:29.240 ","End":"02:30.710","Text":"as we know our equation,"},{"Start":"02:30.710 ","End":"02:33.360","Text":"the integral of r^2dm,"},{"Start":"02:33.520 ","End":"02:37.140","Text":"till here, perfect."},{"Start":"02:37.240 ","End":"02:44.210","Text":"Now what we have to do is we have to take a look at what is r^2, what is dm?"},{"Start":"02:44.210 ","End":"02:46.380","Text":"It\u0027s a little bit complicated."},{"Start":"02:46.380 ","End":"02:52.450","Text":"First of all our dm is going to be easier and let\u0027s take a look at why."},{"Start":"02:52.490 ","End":"02:56.090","Text":"We have this type of l-shape."},{"Start":"02:56.090 ","End":"03:00.260","Text":"Now, a way of looking at it because of the symmetry in the problem,"},{"Start":"03:00.260 ","End":"03:02.240","Text":"because we\u0027re being told that both"},{"Start":"03:02.240 ","End":"03:04.760","Text":"the rods are of the same length and have the same mass,"},{"Start":"03:04.760 ","End":"03:08.630","Text":"we can say that the problem will be symmetrical."},{"Start":"03:08.630 ","End":"03:12.890","Text":"Our problem is rotating about the same axis of rotation,"},{"Start":"03:12.890 ","End":"03:15.630","Text":"which is this midpoint over here,"},{"Start":"03:16.880 ","End":"03:22.445","Text":"and so what we can do is we can just look at one of the rods."},{"Start":"03:22.445 ","End":"03:27.515","Text":"Then we can just multiply it by 2 because we have 2 rods."},{"Start":"03:27.515 ","End":"03:30.815","Text":"This is the trick that we\u0027re going to be using,"},{"Start":"03:30.815 ","End":"03:34.730","Text":"so we\u0027re noticing the symmetry in the problem and then seeing that"},{"Start":"03:34.730 ","End":"03:39.450","Text":"will solve it for 1 rod and then we\u0027ll multiply it by 2 because there are 2 rods."},{"Start":"03:39.920 ","End":"03:43.055","Text":"First of all, we\u0027re going to be multiplying by 2,"},{"Start":"03:43.055 ","End":"03:45.170","Text":"so I\u0027ll write that down over here."},{"Start":"03:45.170 ","End":"03:48.275","Text":"It\u0027s going to be 2 times and then our dm,"},{"Start":"03:48.275 ","End":"03:49.880","Text":"Now we\u0027re looking at the rod,"},{"Start":"03:49.880 ","End":"03:54.420","Text":"which means that it\u0027s a line,it\u0027s a 1 dimensional shape."},{"Start":"03:54.420 ","End":"03:56.960","Text":"As we know, when we\u0027re speaking about 1 dimension,"},{"Start":"03:56.960 ","End":"04:00.680","Text":"we\u0027re speaking about Lambda dx,"},{"Start":"04:00.680 ","End":"04:05.090","Text":"because we\u0027re going to be speaking about the rod on the x axis, but it doesn\u0027t matter."},{"Start":"04:05.090 ","End":"04:07.570","Text":"It could have been dy, never mind."},{"Start":"04:07.570 ","End":"04:11.069","Text":"This is going to be Lambda dx."},{"Start":"04:11.069 ","End":"04:13.380","Text":"Now what is the I^2?"},{"Start":"04:13.380 ","End":"04:15.240","Text":"Have a look over here."},{"Start":"04:15.240 ","End":"04:17.815","Text":"We want to find the distance"},{"Start":"04:17.815 ","End":"04:23.900","Text":"from this center point until whichever point we\u0027re looking at."},{"Start":"04:23.900 ","End":"04:26.030","Text":"This is going to be our I."},{"Start":"04:26.030 ","End":"04:30.950","Text":"This is also I. I is always changing,"},{"Start":"04:30.950 ","End":"04:38.630","Text":"and we wanted to sum all of this over here until here,"},{"Start":"04:38.630 ","End":"04:42.180","Text":"so all of this is still I."},{"Start":"04:42.190 ","End":"04:44.960","Text":"How are we going to do this?"},{"Start":"04:44.960 ","End":"04:47.850","Text":"Let\u0027s just rub everything out."},{"Start":"04:48.050 ","End":"04:55.340","Text":"We know that this height over here is going to be l over 4."},{"Start":"04:55.340 ","End":"04:57.815","Text":"This is going to be l over 4,"},{"Start":"04:57.815 ","End":"05:04.595","Text":"and then we know that this distance over here is going to be changing with our x."},{"Start":"05:04.595 ","End":"05:09.320","Text":"But because we\u0027re starting from over here on the x-axis from l over 4 as well,"},{"Start":"05:09.320 ","End":"05:14.430","Text":"so we\u0027ll always minus the l over 4,"},{"Start":"05:14.430 ","End":"05:16.650","Text":"so what will this look like?"},{"Start":"05:16.650 ","End":"05:17.990","Text":"Because it\u0027s I^2,"},{"Start":"05:17.990 ","End":"05:20.390","Text":"we don\u0027t even have to take the square root of it."},{"Start":"05:20.390 ","End":"05:22.805","Text":"Because we\u0027re just going to use Pythagoras,"},{"Start":"05:22.805 ","End":"05:29.145","Text":"which says that our x^2 plus our y^2 is equal to I^2."},{"Start":"05:29.145 ","End":"05:31.785","Text":"We\u0027re just going to have x^2 plus y^2."},{"Start":"05:31.785 ","End":"05:40.150","Text":"Let\u0027s see what that is, so our x^2 is simply going to be x minus l over 4^2."},{"Start":"05:42.860 ","End":"05:50.535","Text":"Then our y^2 is going to be plus l over 4^2,"},{"Start":"05:50.535 ","End":"05:52.995","Text":"so this is our I^2."},{"Start":"05:52.995 ","End":"05:58.280","Text":"Again, our x^2 is going to be let\u0027s say we\u0027re speaking about this point."},{"Start":"05:58.280 ","End":"06:01.820","Text":"It\u0027s going to be this distance,"},{"Start":"06:01.820 ","End":"06:07.115","Text":"which means the entire distance up until that point,"},{"Start":"06:07.115 ","End":"06:09.675","Text":"which is, this is x,"},{"Start":"06:09.675 ","End":"06:13.580","Text":"and then the distance up until this point is l over 4."},{"Start":"06:13.580 ","End":"06:17.480","Text":"In order to get this distance over here,"},{"Start":"06:17.480 ","End":"06:23.045","Text":"we take the full distance of x minus our l over 4,"},{"Start":"06:23.045 ","End":"06:26.045","Text":"and then our y is obvious."},{"Start":"06:26.045 ","End":"06:32.580","Text":"Now what we can do is we can put this into our equation,"},{"Start":"06:33.890 ","End":"06:40.265","Text":"so what we\u0027ll have is the integral of"},{"Start":"06:40.265 ","End":"06:46.550","Text":"2 Lambda multiplied by"},{"Start":"06:46.550 ","End":"06:53.880","Text":"x minus l over 4^2 plus l over 4^2,"},{"Start":"06:55.300 ","End":"07:00.630","Text":"and then this dx."},{"Start":"07:00.920 ","End":"07:05.885","Text":"Now, because I want to be using the units for my moment of inertia,"},{"Start":"07:05.885 ","End":"07:08.720","Text":"then I can\u0027t use my Lambda,"},{"Start":"07:08.720 ","End":"07:11.090","Text":"I can\u0027t use it as density,"},{"Start":"07:11.090 ","End":"07:13.925","Text":"so I need to convert it into mass."},{"Start":"07:13.925 ","End":"07:22.695","Text":"As we know, Lambda is equal to the total mass divided by length, so that\u0027s it."},{"Start":"07:22.695 ","End":"07:24.960","Text":"Now we can substitute that in,"},{"Start":"07:24.960 ","End":"07:30.075","Text":"so therefore I\u0027ll have that our I is equal to,"},{"Start":"07:30.075 ","End":"07:33.260","Text":"and now the bounds are going to be from 0 until l,"},{"Start":"07:33.260 ","End":"07:35.810","Text":"from the beginning until the final length."},{"Start":"07:35.810 ","End":"07:43.395","Text":"I\u0027ll have the integral from 0 to l and then we have the constants,"},{"Start":"07:43.395 ","End":"07:47.790","Text":"so we have 2 Lambda integral from 0 to"},{"Start":"07:47.790 ","End":"07:53.720","Text":"l. Then let\u0027s simplify the terms inside the square bracket,"},{"Start":"07:53.720 ","End":"08:02.465","Text":"so what we will get is we will have in the end 2l^2 divided by"},{"Start":"08:02.465 ","End":"08:11.905","Text":"16 plus x^2 minus xl divided by 2."},{"Start":"08:11.905 ","End":"08:18.840","Text":"This 2 divided by 16 can cross out to 1 over 8,"},{"Start":"08:18.840 ","End":"08:22.035","Text":"and then this is dx."},{"Start":"08:22.035 ","End":"08:23.795","Text":"All I did was simplify,"},{"Start":"08:23.795 ","End":"08:25.820","Text":"was opened up these brackets,"},{"Start":"08:25.820 ","End":"08:32.070","Text":"and the square and simplify the expression within the square brackets. That\u0027s all I did."},{"Start":"08:32.840 ","End":"08:36.390","Text":"Now we can do this integration,"},{"Start":"08:36.390 ","End":"08:43.150","Text":"so again we\u0027ll have 2 Lambda and then let\u0027s integrate."},{"Start":"08:43.150 ","End":"08:49.895","Text":"l^2 divided by 8 will become l^2 divided by 8 multiplied by x."},{"Start":"08:49.895 ","End":"08:53.525","Text":"When we substitute into that l and 0,"},{"Start":"08:53.525 ","End":"08:59.683","Text":"we\u0027ll get that it will be l^3 divided by 8."},{"Start":"08:59.683 ","End":"09:05.580","Text":"Sorry, and our gamma or lambda,"},{"Start":"09:05.580 ","End":"09:14.040","Text":"sorry, I forgot to put in that it equals m divided by L. There we go."},{"Start":"09:14.040 ","End":"09:19.555","Text":"I just substituted in the lambda with our mL because I forgot to do that."},{"Start":"09:19.555 ","End":"09:21.730","Text":"Then we\u0027ll integrate our x^2,"},{"Start":"09:21.730 ","End":"09:25.000","Text":"which will become x^3 divided by 3."},{"Start":"09:25.000 ","End":"09:27.190","Text":"When we substitute in our L,"},{"Start":"09:27.190 ","End":"09:31.915","Text":"so we\u0027ll have plus L^3 divided by 3,"},{"Start":"09:31.915 ","End":"09:38.965","Text":"and then we\u0027ll have negative x^2 over 2 multiplied by L over 2."},{"Start":"09:38.965 ","End":"09:41.080","Text":"Again, when we substitute in our L,"},{"Start":"09:41.080 ","End":"09:44.620","Text":"will have negative L^3,"},{"Start":"09:44.620 ","End":"09:47.720","Text":"and then divide it by 4."},{"Start":"09:48.240 ","End":"09:58.465","Text":"This is that, and now we can cross off with all of our like terms, which is that,"},{"Start":"09:58.465 ","End":"10:00.920","Text":"and then we\u0027ll be left with"},{"Start":"10:02.280 ","End":"10:11.530","Text":"2mL^2 multiplied by 1 over 8 plus 1 over 3 minus a 1/4."},{"Start":"10:11.530 ","End":"10:15.325","Text":"Now, when you find the common denominator,"},{"Start":"10:15.325 ","End":"10:19.104","Text":"and you deal with simple arithmetic,"},{"Start":"10:19.104 ","End":"10:26.710","Text":"then you will find that the final answer will be 5 over 12mL^2."},{"Start":"10:28.950 ","End":"10:32.935","Text":"This will be the final answer,"},{"Start":"10:32.935 ","End":"10:36.560","Text":"we skipped the algebra."},{"Start":"10:37.350 ","End":"10:42.020","Text":"Now, let\u0027s see the next way of doing this."},{"Start":"10:42.240 ","End":"10:46.045","Text":"Now, I left the answer up over here so that"},{"Start":"10:46.045 ","End":"10:49.855","Text":"we can take a look at it afterwards and compare."},{"Start":"10:49.855 ","End":"10:55.405","Text":"Now the second way of solving this is by using our Steiner theorem."},{"Start":"10:55.405 ","End":"10:56.995","Text":"Now as we remember,"},{"Start":"10:56.995 ","End":"11:00.595","Text":"we have our I_mu,"},{"Start":"11:00.595 ","End":"11:06.535","Text":"which is equal to the I of the center of mass,"},{"Start":"11:06.535 ","End":"11:11.120","Text":"plus mass times distance squared."},{"Start":"11:11.120 ","End":"11:13.770","Text":"We know this equation, this is Steiner."},{"Start":"11:13.770 ","End":"11:16.010","Text":"Now over here,"},{"Start":"11:16.010 ","End":"11:23.945","Text":"we can see that it will be a lot easier for us to figure out what our I_mu is,"},{"Start":"11:23.945 ","End":"11:31.750","Text":"and then do our M multiplied by our distance squared to find our I center of mass,"},{"Start":"11:31.750 ","End":"11:34.585","Text":"which will be over here."},{"Start":"11:34.585 ","End":"11:36.310","Text":"It\u0027s easier to find that,"},{"Start":"11:36.310 ","End":"11:38.695","Text":"to do the equation in the opposite way,"},{"Start":"11:38.695 ","End":"11:40.180","Text":"to find the I_mu,"},{"Start":"11:40.180 ","End":"11:44.440","Text":"and then move it to where the center of mass is."},{"Start":"11:44.440 ","End":"11:48.070","Text":"All we\u0027re going to do is we\u0027re going to rearrange this equation,"},{"Start":"11:48.070 ","End":"11:52.105","Text":"and we will say that I_mu minus"},{"Start":"11:52.105 ","End":"11:57.550","Text":"our mass times distance squared is going to be our I center of mass,"},{"Start":"11:57.550 ","End":"12:00.595","Text":"and this is what we are looking for."},{"Start":"12:00.595 ","End":"12:05.870","Text":"It\u0027s a slightly different way of looking at the same equation."},{"Start":"12:05.880 ","End":"12:09.730","Text":"Here specifically, because we\u0027re dealing with 2 rods,"},{"Start":"12:09.730 ","End":"12:12.790","Text":"finding out what I_mu is,"},{"Start":"12:12.790 ","End":"12:14.485","Text":"is going to be very easy."},{"Start":"12:14.485 ","End":"12:21.655","Text":"We know that the moment of inertia of a rod spinning around,"},{"Start":"12:21.655 ","End":"12:25.405","Text":"where its center of rotation is the edge over here."},{"Start":"12:25.405 ","End":"12:32.080","Text":"We know that it\u0027s 1/3mL^2."},{"Start":"12:32.080 ","End":"12:40.790","Text":"This is for rod rotating around edge."},{"Start":"12:43.380 ","End":"12:48.940","Text":"But now, we have to multiply this expression by 2 because again,"},{"Start":"12:48.940 ","End":"12:50.155","Text":"due to symmetry,"},{"Start":"12:50.155 ","End":"12:55.120","Text":"we can say that 1/3mL^2 refers to both rods,"},{"Start":"12:55.120 ","End":"12:56.230","Text":"and there\u0027s 2 rods,"},{"Start":"12:56.230 ","End":"12:58.480","Text":"so we multiply it by 2."},{"Start":"12:58.480 ","End":"13:01.360","Text":"That\u0027s for that bit."},{"Start":"13:01.360 ","End":"13:07.270","Text":"Now what I have to do is I have to minus,"},{"Start":"13:07.270 ","End":"13:11.050","Text":"I have to subtract md^2."},{"Start":"13:11.050 ","End":"13:13.480","Text":"Now, in this equation,"},{"Start":"13:13.480 ","End":"13:15.070","Text":"the general equation for Steiner,"},{"Start":"13:15.070 ","End":"13:18.340","Text":"it says m which just represents the mass of the system."},{"Start":"13:18.340 ","End":"13:22.405","Text":"Here the mass of the system is the mass of both of the rods together,"},{"Start":"13:22.405 ","End":"13:26.470","Text":"and then the question, we\u0027re told that each rod is of mass m,"},{"Start":"13:26.470 ","End":"13:30.790","Text":"and there\u0027s 2 of them, so we have to minus 2m,"},{"Start":"13:30.790 ","End":"13:34.990","Text":"and then this is multiplied by d^2."},{"Start":"13:34.990 ","End":"13:41.560","Text":"What is d^2? It\u0027s the distance of our current point of reference,"},{"Start":"13:41.560 ","End":"13:45.115","Text":"currently where we set our rotation is"},{"Start":"13:45.115 ","End":"13:50.695","Text":"moved to where we actually want our rotation to be,"},{"Start":"13:50.695 ","End":"13:54.340","Text":"which is right over here in this point."},{"Start":"13:54.340 ","End":"13:56.965","Text":"How do we figure out this?"},{"Start":"13:56.965 ","End":"13:58.945","Text":"It\u0027s easy, it\u0027s just Pythagoras."},{"Start":"13:58.945 ","End":"14:02.920","Text":"As before, this is L divided by 4,"},{"Start":"14:02.920 ","End":"14:09.130","Text":"and this is also L divided by 4 because it\u0027s the midpoint of both the x,"},{"Start":"14:09.130 ","End":"14:13.840","Text":"and y-axis, and this is where our center of mass is meeting."},{"Start":"14:13.840 ","End":"14:18.130","Text":"Then we know that this length over here,"},{"Start":"14:18.130 ","End":"14:22.840","Text":"our hypotenuse or radius is just going to be equal to,"},{"Start":"14:22.840 ","End":"14:26.455","Text":"r^2 is equal to d^2,"},{"Start":"14:26.455 ","End":"14:32.020","Text":"which is equal to x^2 plus y^2."},{"Start":"14:32.020 ","End":"14:34.735","Text":"Our d^2 is simply going to be our r^2,"},{"Start":"14:34.735 ","End":"14:36.535","Text":"which is our x^2 plus y^2."},{"Start":"14:36.535 ","End":"14:42.770","Text":"We\u0027re going to be multiplying by L divided by 4^2,"},{"Start":"14:42.870 ","End":"14:48.670","Text":"and then this multiplied by 2 because it\u0027s just"},{"Start":"14:48.670 ","End":"14:55.105","Text":"going to be 2 times L over 4^2."},{"Start":"14:55.105 ","End":"14:57.895","Text":"Now, if we simplify this out,"},{"Start":"14:57.895 ","End":"15:03.715","Text":"we\u0027ll have 2/3mL^2 minus,"},{"Start":"15:03.715 ","End":"15:06.070","Text":"so we\u0027ll have 2 times 2,"},{"Start":"15:06.070 ","End":"15:08.260","Text":"which is 4m,"},{"Start":"15:08.260 ","End":"15:14.980","Text":"and then we\u0027ll open up this square multiplied by L^2 divided by 16."},{"Start":"15:14.980 ","End":"15:17.410","Text":"Then if we take out a common factor,"},{"Start":"15:17.410 ","End":"15:19.030","Text":"so we have in both mL^2,"},{"Start":"15:19.030 ","End":"15:23.950","Text":"and then also a 2,"},{"Start":"15:23.950 ","End":"15:29.110","Text":"and then we have 1/3 minus, so here,"},{"Start":"15:29.110 ","End":"15:31.360","Text":"we\u0027ll have 2 over 8,"},{"Start":"15:31.360 ","End":"15:35.240","Text":"so this will be 1/4."},{"Start":"15:36.720 ","End":"15:40.210","Text":"Then if we multiply this out,"},{"Start":"15:40.210 ","End":"15:43.525","Text":"the common denominator will have to equal 12,"},{"Start":"15:43.525 ","End":"15:46.495","Text":"and then if we simplify this,"},{"Start":"15:46.495 ","End":"15:49.645","Text":"sorry, I made a calculation error."},{"Start":"15:49.645 ","End":"15:54.340","Text":"This over here is meant to be 1/8, not 1/4."},{"Start":"15:54.340 ","End":"15:57.895","Text":"Then we can see that the common denominator is not 12,"},{"Start":"15:57.895 ","End":"16:01.450","Text":"the common denominator will be 24."},{"Start":"16:01.450 ","End":"16:09.070","Text":"Then what we will see is that we have 2mL^2, and then here,"},{"Start":"16:09.070 ","End":"16:12.993","Text":"we\u0027ll have 8 minus 3 divided by 24,"},{"Start":"16:12.993 ","End":"16:15.620","Text":"8 minus 3 is equal to 5,"},{"Start":"16:15.620 ","End":"16:18.388","Text":"5 times 2 is 10,"},{"Start":"16:18.388 ","End":"16:23.620","Text":"10 divided by 24 is 5 over 12,"},{"Start":"16:23.620 ","End":"16:27.865","Text":"and we\u0027ll get 5 over 12 mL^2."},{"Start":"16:27.865 ","End":"16:34.700","Text":"Look at that, it\u0027s the exact same answer that we got in our first method."},{"Start":"16:35.190 ","End":"16:39.475","Text":"What we\u0027ve learned from this lesson is that,"},{"Start":"16:39.475 ","End":"16:45.175","Text":"to find the moment of inertia of a random body in a weird shape,"},{"Start":"16:45.175 ","End":"16:46.630","Text":"that there are 2 ways of doing it."},{"Start":"16:46.630 ","End":"16:48.850","Text":"You can either do it the classic way through"},{"Start":"16:48.850 ","End":"16:51.610","Text":"the definition of finding the moment of inertia,"},{"Start":"16:51.610 ","End":"16:54.625","Text":"which is integrating by"},{"Start":"16:54.625 ","End":"17:00.145","Text":"r^2dm and then opening it up and working it out for this specific case."},{"Start":"17:00.145 ","End":"17:02.380","Text":"There\u0027s the easier and much shorter way,"},{"Start":"17:02.380 ","End":"17:04.750","Text":"which is by using the Steiner theorem,"},{"Start":"17:04.750 ","End":"17:07.300","Text":"both will get you to the exact same answer."},{"Start":"17:07.300 ","End":"17:12.115","Text":"Then another important thing to remember with the Steiner theorem that,"},{"Start":"17:12.115 ","End":"17:18.820","Text":"although the equation in itself is written like this over here,"},{"Start":"17:18.820 ","End":"17:21.745","Text":"that just like in every algebraic equation,"},{"Start":"17:21.745 ","End":"17:23.455","Text":"we can rearrange it,"},{"Start":"17:23.455 ","End":"17:26.830","Text":"and this will sometimes make it much easier for us to"},{"Start":"17:26.830 ","End":"17:30.505","Text":"solve questions such as how we just did this now."},{"Start":"17:30.505 ","End":"17:33.530","Text":"That\u0027s the end of this lesson."}],"ID":9380},{"Watched":false,"Name":"I of Half Hoop Two Masses","Duration":"7m 48s","ChapterTopicVideoID":9108,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:02.850","Text":"Hello. In this question,"},{"Start":"00:02.850 ","End":"00:06.600","Text":"we\u0027re being told to calculate the moment of inertia of a 1/2 hoop of"},{"Start":"00:06.600 ","End":"00:11.445","Text":"radius R and of mass M. We\u0027re being told that at each edge,"},{"Start":"00:11.445 ","End":"00:14.310","Text":"a mass of small m is attached,"},{"Start":"00:14.310 ","End":"00:18.585","Text":"and that the hoop rotates about a screw at its center."},{"Start":"00:18.585 ","End":"00:23.970","Text":"That\u0027s this over here. What do we know?"},{"Start":"00:23.970 ","End":"00:27.840","Text":"We know that to find the moment of inertia of"},{"Start":"00:27.840 ","End":"00:33.330","Text":"some shape where its center is over here is easy."},{"Start":"00:33.330 ","End":"00:38.625","Text":"Then we know to get from this center,"},{"Start":"00:38.625 ","End":"00:42.550","Text":"if we consider this as a circle,"},{"Start":"00:42.550 ","End":"00:49.415","Text":"from the center up until the center of mass of the 1/2 hoop with the 2 masses."},{"Start":"00:49.415 ","End":"00:52.730","Text":"Then the center of mass of the 1/2 hoop and"},{"Start":"00:52.730 ","End":"00:57.480","Text":"the 2 masses will be something around about here."},{"Start":"00:57.480 ","End":"01:02.690","Text":"According to Steiner, we can use the center of mass of this circle,"},{"Start":"01:02.690 ","End":"01:06.859","Text":"move it up to the center of mass of the system,"},{"Start":"01:06.859 ","End":"01:10.865","Text":"and then use Steiner one more time,"},{"Start":"01:10.865 ","End":"01:13.310","Text":"and then we can move it up to here,"},{"Start":"01:13.310 ","End":"01:19.500","Text":"which is our actual axis of rotation that we want to work this out for."},{"Start":"01:20.590 ","End":"01:23.525","Text":"The first time that I use Steiner,"},{"Start":"01:23.525 ","End":"01:27.845","Text":"I will need to find out what this length is over here,"},{"Start":"01:27.845 ","End":"01:29.210","Text":"and let\u0027s call it C,"},{"Start":"01:29.210 ","End":"01:31.385","Text":"and then the second time I use Steiner,"},{"Start":"01:31.385 ","End":"01:37.200","Text":"I have to find what this length is over here and I\u0027ll call it B."},{"Start":"01:37.200 ","End":"01:39.240","Text":"Now, in the unit,"},{"Start":"01:39.240 ","End":"01:41.955","Text":"talking about the center of mass,"},{"Start":"01:41.955 ","End":"01:43.670","Text":"you can go back to it."},{"Start":"01:43.670 ","End":"01:47.150","Text":"There\u0027s a question exactly like this one,"},{"Start":"01:47.150 ","End":"01:48.755","Text":"a hoop with 2 masses,"},{"Start":"01:48.755 ","End":"01:51.725","Text":"and we\u0027re being asked to find what the center of mass is."},{"Start":"01:51.725 ","End":"01:53.180","Text":"I won\u0027t go into it now."},{"Start":"01:53.180 ","End":"01:54.680","Text":"If you don\u0027t remember how to do it,"},{"Start":"01:54.680 ","End":"01:56.240","Text":"please go back to that video."},{"Start":"01:56.240 ","End":"01:58.430","Text":"It\u0027s the exact question,"},{"Start":"01:58.430 ","End":"02:01.115","Text":"hoop with 2 masses, finding the center of mass."},{"Start":"02:01.115 ","End":"02:03.470","Text":"We\u0027ll know that our y,"},{"Start":"02:03.470 ","End":"02:04.745","Text":"center of mass,"},{"Start":"02:04.745 ","End":"02:15.010","Text":"is equal to 2RM divided by Pi M plus 2m."},{"Start":"02:15.010 ","End":"02:17.670","Text":"This is the answer. Of course,"},{"Start":"02:17.670 ","End":"02:25.275","Text":"we know that our ycm will equal to c. Now we know what our c is."},{"Start":"02:25.275 ","End":"02:31.135","Text":"Then, because we know that our radius is R,"},{"Start":"02:31.135 ","End":"02:40.375","Text":"so therefore, we know that R minus c will be equal to b."},{"Start":"02:40.375 ","End":"02:43.280","Text":"Now we know what these lengths are."},{"Start":"02:43.280 ","End":"02:45.860","Text":"We know what c and b is equal to, are equal to."},{"Start":"02:45.860 ","End":"02:52.830","Text":"Now we will just do a Steiner twice and we will get our answers."},{"Start":"02:53.270 ","End":"02:57.000","Text":"Let\u0027s write down I Steiner,"},{"Start":"02:57.000 ","End":"03:00.335","Text":"but first we\u0027ll actually write down,"},{"Start":"03:00.335 ","End":"03:03.905","Text":"we\u0027re going to say that m tag is equal to"},{"Start":"03:03.905 ","End":"03:11.145","Text":"the mass of the hoop plus the 2 masses attached to it."},{"Start":"03:11.145 ","End":"03:14.340","Text":"Now let\u0027s write down what our Steiner is."},{"Start":"03:14.340 ","End":"03:21.660","Text":"Our I is going to be equal to our I center of mass plus"},{"Start":"03:21.660 ","End":"03:29.534","Text":"our m tag multiplied by our distance c^2."},{"Start":"03:29.534 ","End":"03:31.965","Text":"Then our I final,"},{"Start":"03:31.965 ","End":"03:33.930","Text":"where we want to get to,"},{"Start":"03:33.930 ","End":"03:43.853","Text":"our I final is going to be again our I center of mass plus our m tag multiplied by,"},{"Start":"03:43.853 ","End":"03:46.215","Text":"but this time we\u0027re doing the next step."},{"Start":"03:46.215 ","End":"03:49.960","Text":"It\u0027s going to be b^2."},{"Start":"03:50.300 ","End":"03:54.930","Text":"Let\u0027s find out what our I is equal to."},{"Start":"03:54.930 ","End":"03:59.460","Text":"Our I is going to be equal to,"},{"Start":"03:59.460 ","End":"04:02.850","Text":"we\u0027re going to use additivity."},{"Start":"04:02.850 ","End":"04:07.795","Text":"We\u0027re going to find the moment of inertia of this point mass,"},{"Start":"04:07.795 ","End":"04:09.220","Text":"which as we know,"},{"Start":"04:09.220 ","End":"04:12.210","Text":"is just going to be mR^2."},{"Start":"04:12.210 ","End":"04:13.810","Text":"Our appointed mass is mR^2."},{"Start":"04:13.810 ","End":"04:18.085","Text":"We have 2 of them, so we\u0027re going to have 2mR^2,"},{"Start":"04:18.085 ","End":"04:24.500","Text":"and then we\u0027re going to add on the moment of inertia of a 1/2 hoop."},{"Start":"04:24.500 ","End":"04:27.910","Text":"We know that the moment of inertia of a 1/2 hoop is going to"},{"Start":"04:27.910 ","End":"04:31.255","Text":"be the moment of inertia of a full hoop."},{"Start":"04:31.255 ","End":"04:36.065","Text":"A handy trick here to know is that if you have a full hoop,"},{"Start":"04:36.065 ","End":"04:37.555","Text":"a 1/2 hoop, a quarter hoop,"},{"Start":"04:37.555 ","End":"04:39.155","Text":"an eighth hoop,"},{"Start":"04:39.155 ","End":"04:41.420","Text":"a 1 trillionth of a hoop,"},{"Start":"04:41.420 ","End":"04:46.125","Text":"the moment of inertia is always going to be exactly the same, MI^2."},{"Start":"04:46.125 ","End":"04:50.385","Text":"That is the same for every single other shapes,"},{"Start":"04:50.385 ","End":"04:53.055","Text":"squares, lines, you name it."},{"Start":"04:53.055 ","End":"04:55.350","Text":"You can have it, you can double it,"},{"Start":"04:55.350 ","End":"05:02.270","Text":"the size of them, and your moment of inertia will remain the same."},{"Start":"05:02.270 ","End":"05:04.910","Text":"Now we have what our I is."},{"Start":"05:04.910 ","End":"05:09.290","Text":"Now all we have to do is substitute this"},{"Start":"05:09.290 ","End":"05:15.330","Text":"into 2 equations and then we will get the final answer."},{"Start":"05:15.910 ","End":"05:18.650","Text":"Let\u0027s quickly do this."},{"Start":"05:18.650 ","End":"05:23.295","Text":"Now we can just do this equation."},{"Start":"05:23.295 ","End":"05:27.875","Text":"We\u0027ll say that 2mR^2 plus"},{"Start":"05:27.875 ","End":"05:37.080","Text":"mR^2 is equal to I center of mass plus M tag c^2."},{"Start":"05:37.080 ","End":"05:41.040","Text":"Now we want to isolate out I center of mass."},{"Start":"05:41.040 ","End":"05:49.140","Text":"We\u0027ll say that we have 2mR^2 plus mR^2 minus m tag,"},{"Start":"05:49.140 ","End":"05:54.015","Text":"which is M plus 2m multiplied by c^2."},{"Start":"05:54.015 ","End":"05:55.635","Text":"Now what is our c^2?"},{"Start":"05:55.635 ","End":"05:57.375","Text":"We\u0027re going to have"},{"Start":"05:57.375 ","End":"06:03.500","Text":"4R^2 M^2 divided by"},{"Start":"06:03.500 ","End":"06:10.080","Text":"Pi^2 M plus 2m^2."},{"Start":"06:10.080 ","End":"06:12.140","Text":"Then this can cross out,"},{"Start":"06:12.140 ","End":"06:13.390","Text":"and this can cross out."},{"Start":"06:13.390 ","End":"06:18.340","Text":"This is going to be equal to I center of mass."},{"Start":"06:18.340 ","End":"06:23.655","Text":"Now we can just substitute this."},{"Start":"06:23.655 ","End":"06:29.090","Text":"I\u0027m not going to simplify it too much because we don\u0027t really need to right now."},{"Start":"06:29.090 ","End":"06:35.640","Text":"Now we\u0027re going to substitute that into our equation for our I final,"},{"Start":"06:35.640 ","End":"06:37.530","Text":"which is going to be our I center of mass,"},{"Start":"06:37.530 ","End":"06:39.030","Text":"which is going to be this."},{"Start":"06:39.030 ","End":"06:42.185","Text":"We have I^2 as a common factor,"},{"Start":"06:42.185 ","End":"06:46.980","Text":"and then we\u0027ll have 2m plus M minus"},{"Start":"06:46.980 ","End":"06:55.530","Text":"4M^2 divided by Pi^2 M plus 2m."},{"Start":"06:56.540 ","End":"07:02.085","Text":"We have our I center of mass plus our M tag,"},{"Start":"07:02.085 ","End":"07:07.530","Text":"which is our M plus 2m multiplied by our b^2,"},{"Start":"07:07.530 ","End":"07:15.545","Text":"and our b^2 is r minus c. Then we\u0027re going to have this multiplied by R minus,"},{"Start":"07:15.545 ","End":"07:23.610","Text":"I see which is 2RM divided by Pi M plus 2m,"},{"Start":"07:23.610 ","End":"07:26.565","Text":"and all of this squared."},{"Start":"07:26.565 ","End":"07:30.430","Text":"This is going to be equal to RI final."},{"Start":"07:30.890 ","End":"07:33.920","Text":"I\u0027m not going to waste time now simplifying this."},{"Start":"07:33.920 ","End":"07:35.840","Text":"You can do this later."},{"Start":"07:35.840 ","End":"07:40.745","Text":"This is now our answer for"},{"Start":"07:40.745 ","End":"07:46.430","Text":"the moment of inertia of this 1/2 hoop and 2 masses attached to it."},{"Start":"07:46.430 ","End":"07:49.650","Text":"That\u0027s the end of this lesson."}],"ID":9381},{"Watched":false,"Name":"Moments of Inertia of Squares","Duration":"14m 43s","ChapterTopicVideoID":9109,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:02.745","Text":"Hello. In this lesson,"},{"Start":"00:02.745 ","End":"00:07.050","Text":"we\u0027re going to see how to calculate the moments of inertia of square boards with"},{"Start":"00:07.050 ","End":"00:12.915","Text":"mass M and sides of length a in the following ways."},{"Start":"00:12.915 ","End":"00:16.725","Text":"When the axis of rotation is at the edge of the board,"},{"Start":"00:16.725 ","End":"00:18.945","Text":"so this is number 1."},{"Start":"00:18.945 ","End":"00:22.050","Text":"When the axis of rotation is parallel to the edges and"},{"Start":"00:22.050 ","End":"00:24.870","Text":"goes through the center of the board, so that\u0027s this."},{"Start":"00:24.870 ","End":"00:26.670","Text":"This is number 2."},{"Start":"00:26.670 ","End":"00:30.150","Text":"In the third case, when the axis of rotation is perpendicular"},{"Start":"00:30.150 ","End":"00:34.000","Text":"to the board and goes through its center."},{"Start":"00:34.250 ","End":"00:39.580","Text":"Let\u0027s see how we do at number 1 first."},{"Start":"00:39.830 ","End":"00:43.115","Text":"What we can do is we can start by writing down"},{"Start":"00:43.115 ","End":"00:47.150","Text":"the equation that we know for the moment of inertia."},{"Start":"00:47.150 ","End":"00:51.660","Text":"It\u0027s the integral an r^2 dm."},{"Start":"00:51.660 ","End":"00:59.320","Text":"Over here, our r^2 because our axis of rotation is at the edge,"},{"Start":"00:59.320 ","End":"01:01.892","Text":"our y\u0027s don\u0027t matter,"},{"Start":"01:01.892 ","End":"01:06.950","Text":"our values for y and we just have to take into account our x values,"},{"Start":"01:06.950 ","End":"01:12.060","Text":"so our r^2 is just going to become x^2,"},{"Start":"01:12.060 ","End":"01:22.440","Text":"and then we\u0027re going to have the integral from 0 until a because each side length is a."},{"Start":"01:22.440 ","End":"01:24.690","Text":"Then what is our dm?"},{"Start":"01:24.690 ","End":"01:30.305","Text":"Now because we\u0027re going to use Cartesian coordinates and because we\u0027re dealing with area,"},{"Start":"01:30.305 ","End":"01:34.950","Text":"so it\u0027s going to be sigma dxdy,"},{"Start":"01:34.950 ","End":"01:44.985","Text":"so now our y\u0027s will also be from 0 until a because also this is of length a."},{"Start":"01:44.985 ","End":"01:47.250","Text":"Now, what is our sigma going to be?"},{"Start":"01:47.250 ","End":"01:49.850","Text":"Our sigma is the density."},{"Start":"01:49.850 ","End":"01:55.060","Text":"The density is the total mass of the system divided by its area."},{"Start":"01:55.060 ","End":"01:59.300","Text":"That\u0027s going to be mass divided by a squared,"},{"Start":"01:59.300 ","End":"02:00.680","Text":"and because this is a constant,"},{"Start":"02:00.680 ","End":"02:03.470","Text":"we can take it from outside of the integrals."},{"Start":"02:03.470 ","End":"02:06.530","Text":"We can write m divided by a squared,"},{"Start":"02:06.530 ","End":"02:10.605","Text":"and then we have the double integral because it\u0027s dx by dy,"},{"Start":"02:10.605 ","End":"02:12.525","Text":"we\u0027re summing all along,"},{"Start":"02:12.525 ","End":"02:20.065","Text":"and then we can write our bounds which is from 0 to a and from 0 to a,"},{"Start":"02:20.065 ","End":"02:25.480","Text":"and then it\u0027s going to be x^2 dxdy."},{"Start":"02:25.970 ","End":"02:31.960","Text":"Now if we do this integral, it\u0027s very simple."},{"Start":"02:31.960 ","End":"02:33.940","Text":"I\u0027m just going to do it really quickly."},{"Start":"02:33.940 ","End":"02:41.510","Text":"We\u0027re going to get ma^3 divided by a^2 times 3,"},{"Start":"02:41.510 ","End":"02:43.670","Text":"and then another a over here,"},{"Start":"02:43.670 ","End":"02:50.760","Text":"and then this can cross out to ma^2 divided by 3."},{"Start":"02:51.020 ","End":"02:53.380","Text":"If you don\u0027t see how I did that,"},{"Start":"02:53.380 ","End":"02:57.680","Text":"please do that on a piece of paper and pause the video in the meantime."},{"Start":"02:58.130 ","End":"03:01.885","Text":"Now we\u0027re going to look at case number 2,"},{"Start":"03:01.885 ","End":"03:03.350","Text":"so let\u0027s have a look."},{"Start":"03:03.350 ","End":"03:06.440","Text":"This time it\u0027s a very similar case to case number 1."},{"Start":"03:06.440 ","End":"03:10.340","Text":"However, instead of the axis of rotation being on the edge of the board,"},{"Start":"03:10.340 ","End":"03:16.180","Text":"we have it going through the middle and then the board is just spinning around like so."},{"Start":"03:16.180 ","End":"03:19.350","Text":"Now, just like in the case of number 1,"},{"Start":"03:19.350 ","End":"03:24.425","Text":"we don\u0027t have to take our y\u0027s into consideration because"},{"Start":"03:24.425 ","End":"03:31.980","Text":"every dm is going to be a distance just on the x-axis."},{"Start":"03:33.500 ","End":"03:41.415","Text":"We\u0027re just going to be summing along the x-axis because the y will always be constant."},{"Start":"03:41.415 ","End":"03:45.665","Text":"Each shape is the same distance away on the y-axis,"},{"Start":"03:45.665 ","End":"03:47.660","Text":"if you have a look at that."},{"Start":"03:47.660 ","End":"03:50.450","Text":"What we\u0027ll do is again we\u0027ll write,"},{"Start":"03:50.450 ","End":"03:55.530","Text":"I is equal to the integral of r^2dm,"},{"Start":"03:56.950 ","End":"03:59.690","Text":"so what is our r^2 going to be?"},{"Start":"03:59.690 ","End":"04:03.290","Text":"It\u0027s going to be again our x^2, just because, again,"},{"Start":"04:03.290 ","End":"04:08.029","Text":"the only distances that are changing is our x distances,"},{"Start":"04:08.029 ","End":"04:13.710","Text":"our y is constant because it\u0027s along the axis."},{"Start":"04:13.710 ","End":"04:16.280","Text":"We\u0027ll write in our x^2,"},{"Start":"04:16.280 ","End":"04:19.385","Text":"and now we have to write in our bounds."},{"Start":"04:19.385 ","End":"04:20.825","Text":"What will our bounds be?"},{"Start":"04:20.825 ","End":"04:24.745","Text":"Again, for our y-axis,"},{"Start":"04:24.745 ","End":"04:29.225","Text":"we can see that we\u0027re doing from 0 until a,"},{"Start":"04:29.225 ","End":"04:31.245","Text":"so 0 until a."},{"Start":"04:31.245 ","End":"04:33.525","Text":"But what\u0027s going on our x-axis?"},{"Start":"04:33.525 ","End":"04:36.520","Text":"Because our axis is located right in the middle of the board,"},{"Start":"04:36.520 ","End":"04:40.315","Text":"so we\u0027re going to have the positive axes in this direction,"},{"Start":"04:40.315 ","End":"04:43.790","Text":"and this side is the negative side,"},{"Start":"04:43.790 ","End":"04:45.870","Text":"so we\u0027ll split it into half,"},{"Start":"04:45.870 ","End":"04:51.740","Text":"so we\u0027ll be going from negative 8 divided by 2 until a divided by 2."},{"Start":"04:51.740 ","End":"04:55.200","Text":"Because we\u0027re summing along this half,"},{"Start":"04:55.200 ","End":"04:58.090","Text":"and we\u0027re summing along this half because we"},{"Start":"04:58.090 ","End":"05:02.215","Text":"have the integration on either side of the axis of rotation."},{"Start":"05:02.215 ","End":"05:06.030","Text":"In case 1, it was just on this side,"},{"Start":"05:06.030 ","End":"05:09.150","Text":"we didn\u0027t have anything over here."},{"Start":"05:11.000 ","End":"05:16.115","Text":"Here we\u0027re summing on positive a over 2 and negative a over 2."},{"Start":"05:16.115 ","End":"05:19.400","Text":"Then our dm again because we\u0027re dealing with"},{"Start":"05:19.400 ","End":"05:23.075","Text":"area and again because we\u0027re dealing with Cartesian coordinates,"},{"Start":"05:23.075 ","End":"05:26.625","Text":"so we\u0027re going to say that sigma dxdy,"},{"Start":"05:26.625 ","End":"05:28.125","Text":"and what is our sigma?"},{"Start":"05:28.125 ","End":"05:29.700","Text":"It\u0027s the same as over here."},{"Start":"05:29.700 ","End":"05:32.430","Text":"It\u0027s m divided by a^2."},{"Start":"05:32.430 ","End":"05:34.010","Text":"Because it\u0027s a constant again,"},{"Start":"05:34.010 ","End":"05:36.500","Text":"we can take it out of the integral,"},{"Start":"05:36.500 ","End":"05:39.020","Text":"so we\u0027ll have m divided by a^2,"},{"Start":"05:39.020 ","End":"05:41.585","Text":"the integral from 0 to a,"},{"Start":"05:41.585 ","End":"05:49.780","Text":"from negative a over 2 until a over 2, x^2dxdy."},{"Start":"05:49.820 ","End":"05:57.455","Text":"Then again, you can do this integral at home on your piece of paper and pause the video."},{"Start":"05:57.455 ","End":"06:05.250","Text":"But what you will get in the end is ma^2 divided by 12."},{"Start":"06:05.690 ","End":"06:08.195","Text":"That is for the second case."},{"Start":"06:08.195 ","End":"06:11.310","Text":"Let\u0027s move on now to the third case."},{"Start":"06:12.340 ","End":"06:15.320","Text":"Now, in our third case,"},{"Start":"06:15.320 ","End":"06:21.845","Text":"we can call this the z-axis right now,"},{"Start":"06:21.845 ","End":"06:24.950","Text":"before we could look at these axis as the y-axis,"},{"Start":"06:24.950 ","End":"06:26.735","Text":"but now we can call it the z-axis,"},{"Start":"06:26.735 ","End":"06:31.760","Text":"which means that we have our y-axis going like this,"},{"Start":"06:31.760 ","End":"06:36.005","Text":"and we also have an x-axis going like this,"},{"Start":"06:36.005 ","End":"06:41.810","Text":"so that means that if I\u0027m taking some dm,"},{"Start":"06:41.810 ","End":"06:48.094","Text":"so its distance from the origin won\u0027t be just summing on the x-axis, or on the y-axis."},{"Start":"06:48.094 ","End":"06:50.300","Text":"It\u0027s going to be like the radius,"},{"Start":"06:50.300 ","End":"06:55.400","Text":"so we have to do Pythagoras to find out what this length is."},{"Start":"06:55.400 ","End":"06:58.745","Text":"As you remember, r^2,"},{"Start":"06:58.745 ","End":"07:02.885","Text":"the radius is equal to x^2 plus y^2,"},{"Start":"07:02.885 ","End":"07:05.690","Text":"and if you notice in our equation,"},{"Start":"07:05.690 ","End":"07:09.430","Text":"so our equation is the integral on r^2dm,"},{"Start":"07:09.430 ","End":"07:11.750","Text":"so we don\u0027t have to square root this."},{"Start":"07:11.750 ","End":"07:16.476","Text":"We can just use this as x^2 plus y^2."},{"Start":"07:16.476 ","End":"07:21.520","Text":"Let\u0027s substitute this all in."},{"Start":"07:21.520 ","End":"07:23.650","Text":"We\u0027ll have our r-squared,"},{"Start":"07:23.650 ","End":"07:28.255","Text":"which we know already we\u0027re going to be integrating dxdy again."},{"Start":"07:28.255 ","End":"07:31.375","Text":"It\u0027s going to be a double integral as usual."},{"Start":"07:31.375 ","End":"07:36.580","Text":"Then our r squared is just x squared plus y squared."},{"Start":"07:36.580 ","End":"07:43.105","Text":"Because we have to take the Pythagoras because its distance is a radial distance."},{"Start":"07:43.105 ","End":"07:45.040","Text":"Then again, we\u0027re going to be using,"},{"Start":"07:45.040 ","End":"07:47.395","Text":"but it\u0027s just easier because it\u0027s a square."},{"Start":"07:47.395 ","End":"07:50.905","Text":"We\u0027re going to be using Cartesian coordinates again."},{"Start":"07:50.905 ","End":"07:54.625","Text":"If it was a circle, we would use polar coordinates."},{"Start":"07:54.625 ","End":"07:59.680","Text":"Again, we\u0027re going to be using sigma dxdy."},{"Start":"07:59.680 ","End":"08:02.560","Text":"Then what is our Sigma?"},{"Start":"08:02.560 ","End":"08:04.540","Text":"Again because it\u0027s area."},{"Start":"08:04.540 ","End":"08:09.640","Text":"It\u0027s again going to be the total mass of the system divided by the total area."},{"Start":"08:09.640 ","End":"08:13.570","Text":"It\u0027s again a constant so we can move it out of the integral."},{"Start":"08:13.570 ","End":"08:17.410","Text":"Now, let\u0027s take a look at what our bounds are going to be."},{"Start":"08:17.410 ","End":"08:19.975","Text":"Our bounds, as we can see,"},{"Start":"08:19.975 ","End":"08:23.260","Text":"this is the positive y-direction."},{"Start":"08:23.260 ","End":"08:26.335","Text":"And then, we also have over here,"},{"Start":"08:26.335 ","End":"08:28.240","Text":"this is the negative y direction,"},{"Start":"08:28.240 ","End":"08:30.790","Text":"and this is the positive x-direction."},{"Start":"08:30.790 ","End":"08:32.215","Text":"We can see going back here,"},{"Start":"08:32.215 ","End":"08:34.100","Text":"it\u0027s the negative x-direction."},{"Start":"08:34.100 ","End":"08:35.975","Text":"It\u0027s a similar case,"},{"Start":"08:35.975 ","End":"08:38.635","Text":"like what we saw in case Number 2."},{"Start":"08:38.635 ","End":"08:45.580","Text":"Our bounds are going to be on the y from negative a divided by 2 until a divided by 2."},{"Start":"08:45.580 ","End":"08:50.920","Text":"The same on our x from negative a divided by 2 until a divided by 2."},{"Start":"08:50.920 ","End":"08:55.735","Text":"Because we\u0027re integrating on either side of this z axes."},{"Start":"08:55.735 ","End":"08:58.465","Text":"Now, if we integrate,"},{"Start":"08:58.465 ","End":"09:01.210","Text":"I\u0027m going to skip the steps."},{"Start":"09:01.210 ","End":"09:06.850","Text":"But what we\u0027ll get is we\u0027ll get M 1/12th a"},{"Start":"09:06.850 ","End":"09:14.005","Text":"squared plus another 1/12th a squared."},{"Start":"09:14.005 ","End":"09:24.235","Text":"Which then we can just take out like terms and rewrite this as ma squared divided by 6."},{"Start":"09:24.235 ","End":"09:27.080","Text":"This is the final answer."},{"Start":"09:28.170 ","End":"09:31.420","Text":"Now, what we can do is we can take a look"},{"Start":"09:31.420 ","End":"09:34.330","Text":"at the logic behind the answers in order to make"},{"Start":"09:34.330 ","End":"09:39.625","Text":"it slightly easier to understand the answers and how to remember them,"},{"Start":"09:39.625 ","End":"09:42.415","Text":"and also how to apply them to different shapes."},{"Start":"09:42.415 ","End":"09:45.475","Text":"In the exam that you can save some time."},{"Start":"09:45.475 ","End":"09:47.320","Text":"Let\u0027s look at case number 1."},{"Start":"09:47.320 ","End":"09:53.244","Text":"Now, I mentioned in a previous video and I\u0027ll speak about it now that we can consider"},{"Start":"09:53.244 ","End":"10:00.825","Text":"this entire block of a and a as a single rod of length a."},{"Start":"10:00.825 ","End":"10:05.515","Text":"It\u0027s just a line. And its length is a,"},{"Start":"10:05.515 ","End":"10:09.640","Text":"and so its moment of inertia when it\u0027s spinning with the axes of"},{"Start":"10:09.640 ","End":"10:15.380","Text":"rotation at its edge is always going to be ma squared divided by 3."},{"Start":"10:15.480 ","End":"10:21.385","Text":"Then if I draw many rods across this,"},{"Start":"10:21.385 ","End":"10:24.610","Text":"so the mass will grow as"},{"Start":"10:24.610 ","End":"10:28.240","Text":"long as their axis of rotation is always at the edge of each rod."},{"Start":"10:28.240 ","End":"10:30.459","Text":"Even though the mass will be different,"},{"Start":"10:30.459 ","End":"10:34.885","Text":"the equation m a squared divided by 3 will always be correct."},{"Start":"10:34.885 ","End":"10:37.810","Text":"That\u0027s why, if you have 1 rod,"},{"Start":"10:37.810 ","End":"10:39.820","Text":"10 rods, 3 rods,"},{"Start":"10:39.820 ","End":"10:44.350","Text":"or an infinite amount of rods which make up this square,"},{"Start":"10:44.350 ","End":"10:47.290","Text":"and they\u0027re all rotating at the edge,"},{"Start":"10:47.290 ","End":"10:53.575","Text":"the equation for the moment of inertia will always be ma squared divided by 3."},{"Start":"10:53.575 ","End":"10:57.340","Text":"That\u0027s something very nice trick."},{"Start":"10:57.340 ","End":"11:02.620","Text":"If they ask, you have 1 rod rotating around now you have 3 calculate,"},{"Start":"11:02.620 ","End":"11:03.790","Text":"you don\u0027t even have to calculate,"},{"Start":"11:03.790 ","End":"11:05.665","Text":"you can just state that."},{"Start":"11:05.665 ","End":"11:10.030","Text":"Then let\u0027s take a look at case Number 2."},{"Start":"11:10.030 ","End":"11:14.920","Text":"So we can see that it\u0027s ma squared divided by 12."},{"Start":"11:14.920 ","End":"11:18.985","Text":"Again, we can apply the same trick as what we did in question Number 1."},{"Start":"11:18.985 ","End":"11:21.010","Text":"When we said that it\u0027s a rod."},{"Start":"11:21.010 ","End":"11:25.045","Text":"We can consider that the rod is spinning,"},{"Start":"11:25.045 ","End":"11:26.380","Text":"but its axis of rotation,"},{"Start":"11:26.380 ","End":"11:27.430","Text":"instead of being at the edge,"},{"Start":"11:27.430 ","End":"11:28.615","Text":"like in case Number 1,"},{"Start":"11:28.615 ","End":"11:30.955","Text":"is going to be at the center of the rod."},{"Start":"11:30.955 ","End":"11:34.510","Text":"The equation for the moment of inertia of a rod"},{"Start":"11:34.510 ","End":"11:40.165","Text":"spinning like this is going to be ma squared divided by 12."},{"Start":"11:40.165 ","End":"11:43.855","Text":"Then again, if you add more rods,"},{"Start":"11:43.855 ","End":"11:51.700","Text":"an infinite amount of rods in order to build this square shape,"},{"Start":"11:51.700 ","End":"11:55.555","Text":"so the only thing that would have changed is the mass,"},{"Start":"11:55.555 ","End":"12:00.265","Text":"but the equation for the moment of inertia will still be the same."},{"Start":"12:00.265 ","End":"12:03.625","Text":"Again, we\u0027ll get ma squared divided by 12,"},{"Start":"12:03.625 ","End":"12:08.140","Text":"where the only difference will be that the mass will be different."},{"Start":"12:08.140 ","End":"12:12.760","Text":"Again, if you had 1 rod or 3 rods are"},{"Start":"12:12.760 ","End":"12:14.980","Text":"an infinite amount of rods spinning"},{"Start":"12:14.980 ","End":"12:17.980","Text":"around where the axis of rotation is in the center of the rod."},{"Start":"12:17.980 ","End":"12:23.710","Text":"The equation for the moment of inertia is ma squared divided by 12."},{"Start":"12:23.710 ","End":"12:26.935","Text":"Now let\u0027s take a look at case Number 3."},{"Start":"12:26.935 ","End":"12:30.505","Text":"Now, case Number 3, you\u0027ll see that before I wrote the final answer,"},{"Start":"12:30.505 ","End":"12:34.000","Text":"I wrote the step before the final answer before I"},{"Start":"12:34.000 ","End":"12:39.130","Text":"opened up the brackets and joined up all the common factors."},{"Start":"12:39.130 ","End":"12:41.560","Text":"How come I wrote it like this?"},{"Start":"12:41.560 ","End":"12:48.865","Text":"Because when we\u0027re dealing with this square plane rotating around the z axis,"},{"Start":"12:48.865 ","End":"12:56.185","Text":"so we can see that it\u0027s like I have 1 rod here and I have 1 rod here."},{"Start":"12:56.185 ","End":"13:01.075","Text":"Do you see that? Now,"},{"Start":"13:01.075 ","End":"13:03.025","Text":"if we look back at Case Number 2,"},{"Start":"13:03.025 ","End":"13:07.870","Text":"we saw that each rod is mass times a squared,"},{"Start":"13:07.870 ","End":"13:14.630","Text":"where a squared here is the length of the rod a divided by 12."},{"Start":"13:14.850 ","End":"13:19.810","Text":"Now, I have 2 of them because I have 1 rod and another rod here,"},{"Start":"13:19.810 ","End":"13:22.510","Text":"and then of course I can add infinite amounts of rods"},{"Start":"13:22.510 ","End":"13:25.450","Text":"like this and infinite amounts of rods in this direction."},{"Start":"13:25.450 ","End":"13:26.980","Text":"But as we\u0027ve just said,"},{"Start":"13:26.980 ","End":"13:29.410","Text":"it doesn\u0027t matter how many rods I\u0027ve added."},{"Start":"13:29.410 ","End":"13:31.945","Text":"The equation will still remain the same."},{"Start":"13:31.945 ","End":"13:34.435","Text":"The only thing that will change is the mass,"},{"Start":"13:34.435 ","End":"13:37.375","Text":"because the mass will increase because there\u0027s more rods,"},{"Start":"13:37.375 ","End":"13:39.610","Text":"but it means that I can calculate for"},{"Start":"13:39.610 ","End":"13:44.545","Text":"this 1 rod and for this 1 rod and then just add them together."},{"Start":"13:44.545 ","End":"13:49.525","Text":"We saw that I\u0027ll have ma squared for 1 rod rotating about its center,"},{"Start":"13:49.525 ","End":"13:51.910","Text":"because the axes of rotation,"},{"Start":"13:51.910 ","End":"13:53.650","Text":"if we look is here,"},{"Start":"13:53.650 ","End":"13:57.890","Text":"which is around about in the center and also in the center here."},{"Start":"13:58.500 ","End":"14:03.610","Text":"As we can see, and so we can just say that it\u0027s ma"},{"Start":"14:03.610 ","End":"14:09.470","Text":"squared divided by 12 plus ma squared divided by 12 because there\u0027s 2 rods."},{"Start":"14:09.570 ","End":"14:11.980","Text":"It\u0027s 2 times that."},{"Start":"14:11.980 ","End":"14:14.260","Text":"So we can see here we have 1ma squared"},{"Start":"14:14.260 ","End":"14:17.050","Text":"divided by 12 plus another ma squared divided by 12,"},{"Start":"14:17.050 ","End":"14:19.870","Text":"which will equal to ma squared divided by 12,"},{"Start":"14:19.870 ","End":"14:23.845","Text":"which cancels out to ma squared divided by 6."},{"Start":"14:23.845 ","End":"14:28.150","Text":"That is the logic behind this."},{"Start":"14:28.150 ","End":"14:30.220","Text":"If you\u0027re asked to do this an exam,"},{"Start":"14:30.220 ","End":"14:33.680","Text":"first of all, you should have these equations written down in your notes."},{"Start":"14:33.680 ","End":"14:37.610","Text":"But just in case they give you a more complex shape where you can apply some of"},{"Start":"14:37.610 ","End":"14:42.315","Text":"these little tricks that we spoke about in this lesson, it\u0027s very useful."},{"Start":"14:42.315 ","End":"14:44.870","Text":"That\u0027s the end of the lesson."}],"ID":9382},{"Watched":false,"Name":"Moment of Inertia Triangle","Duration":"14m 41s","ChapterTopicVideoID":9110,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:02.460","Text":"Hello. In this lesson,"},{"Start":"00:02.460 ","End":"00:04.845","Text":"we\u0027re going to be calculating the triangle,"},{"Start":"00:04.845 ","End":"00:08.520","Text":"this triangle\u0027s moment of inertia when"},{"Start":"00:08.520 ","End":"00:13.575","Text":"its axis of rotation is located at the right angled corner."},{"Start":"00:13.575 ","End":"00:15.930","Text":"That means, right over here,"},{"Start":"00:15.930 ","End":"00:19.155","Text":"where I\u0027ve labeled 0 for the origin."},{"Start":"00:19.155 ","End":"00:21.840","Text":"This is a right angle."},{"Start":"00:21.840 ","End":"00:29.400","Text":"Now, we don\u0027t know how to work out this straightaway but what we do know is to"},{"Start":"00:29.400 ","End":"00:32.765","Text":"work out the moment of inertia when the axis of rotation is"},{"Start":"00:32.765 ","End":"00:37.735","Text":"over here in the center of the hypotenuse."},{"Start":"00:37.735 ","End":"00:41.250","Text":"Why do we know how to work this out?"},{"Start":"00:41.250 ","End":"00:45.275","Text":"If we look, we can make this triangle into"},{"Start":"00:45.275 ","End":"00:51.240","Text":"some rectangle with a side lengths of a and of b."},{"Start":"00:51.890 ","End":"00:56.405","Text":"If this is a rectangle like we just learned in the previous lesson,"},{"Start":"00:56.405 ","End":"00:59.495","Text":"and as has been mentioned in a few other lessons,"},{"Start":"00:59.495 ","End":"01:05.180","Text":"if we have a square or a circle or a rod or whatever it is,"},{"Start":"01:05.180 ","End":"01:08.637","Text":"and we fold it to make it smaller."},{"Start":"01:08.637 ","End":"01:14.825","Text":"So if we have a square and we fold it to one side,"},{"Start":"01:14.825 ","End":"01:21.095","Text":"then it becomes this or we fold it in the other way like this."},{"Start":"01:21.095 ","End":"01:25.380","Text":"It becomes just a longer piece."},{"Start":"01:26.270 ","End":"01:29.870","Text":"The moment of inertia will be exactly the same"},{"Start":"01:29.870 ","End":"01:33.665","Text":"for all of these shapes as it was for the original square."},{"Start":"01:33.665 ","End":"01:36.080","Text":"Over here, it\u0027s the same thing;"},{"Start":"01:36.080 ","End":"01:38.980","Text":"we\u0027re just folding it along the diagonal."},{"Start":"01:38.980 ","End":"01:46.640","Text":"We know that the moment of inertia for a square and therefore,"},{"Start":"01:46.640 ","End":"01:49.480","Text":"any shape that can be folded from it"},{"Start":"01:49.480 ","End":"01:52.835","Text":"when we\u0027re folding in a symmetrical way along its diagonal,"},{"Start":"01:52.835 ","End":"01:56.690","Text":"along here or along here,"},{"Start":"01:56.690 ","End":"02:04.325","Text":"will equal to 112 the total mass of the shape,"},{"Start":"02:04.325 ","End":"02:11.970","Text":"multiplied by the length of each edge squared plus the other length."},{"Start":"02:11.970 ","End":"02:17.855","Text":"We\u0027ll have a^2 plus b^2 so each edge squared,"},{"Start":"02:17.855 ","End":"02:22.800","Text":"we sum them, and we multiply this by m divided by 12."},{"Start":"02:23.180 ","End":"02:25.760","Text":"Just imagine in the question we were told that"},{"Start":"02:25.760 ","End":"02:28.920","Text":"the triangles\u0027 mass was M. It doesn\u0027t matter."},{"Start":"02:32.330 ","End":"02:37.045","Text":"Now what we\u0027re going to do is we\u0027re going to remember Steiner\u0027s equation,"},{"Start":"02:37.045 ","End":"02:40.210","Text":"which says that the ICM,"},{"Start":"02:40.210 ","End":"02:45.850","Text":"the moment of inertia for the center of mass plus the mass of"},{"Start":"02:45.850 ","End":"02:52.510","Text":"the system multiplied by the distance squared from our center of mass."},{"Start":"02:52.510 ","End":"02:57.610","Text":"For my axes, that our center of mass is from our axis of rotation sorry,"},{"Start":"02:57.610 ","End":"03:03.530","Text":"is going to be equal to the moment of inertia of the system."},{"Start":"03:04.760 ","End":"03:09.070","Text":"Now we have to consider that when we\u0027re"},{"Start":"03:09.070 ","End":"03:14.610","Text":"taking this as the axis of rotation and this axis,"},{"Start":"03:14.610 ","End":"03:18.475","Text":"a new axis of rotation that we want to get to"},{"Start":"03:18.475 ","End":"03:23.798","Text":"so neither 1 of these points is the center of mass."},{"Start":"03:23.798 ","End":"03:28.595","Text":"Which means that we\u0027re going to have to use the Steiner equation twice,"},{"Start":"03:28.595 ","End":"03:34.360","Text":"because our center of mass is going to be somewhere over here."},{"Start":"03:34.360 ","End":"03:37.850","Text":"This is going to be the center of mass somewhere along here."},{"Start":"03:37.850 ","End":"03:40.520","Text":"What do we have to do is we\u0027re going to have to use"},{"Start":"03:40.520 ","End":"03:43.870","Text":"the Steiner equation to get to this point."},{"Start":"03:43.870 ","End":"03:45.830","Text":"This will be Steinem Number 1."},{"Start":"03:45.830 ","End":"03:49.640","Text":"Then from this center of mass until our new axis of"},{"Start":"03:49.640 ","End":"03:54.820","Text":"rotation and so we\u0027re going to have to do the Steiner equation twice."},{"Start":"03:54.820 ","End":"03:58.670","Text":"The reason that we can\u0027t just do this in 1 step,"},{"Start":"03:58.670 ","End":"04:02.000","Text":"going from this point over here,"},{"Start":"04:02.000 ","End":"04:05.330","Text":"straight on to this point over here,"},{"Start":"04:05.330 ","End":"04:07.055","Text":"we cannot do this,"},{"Start":"04:07.055 ","End":"04:12.425","Text":"and that is because we\u0027re looking in the equation at m distance squared."},{"Start":"04:12.425 ","End":"04:14.135","Text":"Now the distance squared,"},{"Start":"04:14.135 ","End":"04:15.710","Text":"we can see that if,"},{"Start":"04:15.710 ","End":"04:20.120","Text":"let\u0027s say, we have 2 distances."},{"Start":"04:20.120 ","End":"04:24.275","Text":"We have 10 and 3 are the distances"},{"Start":"04:24.275 ","End":"04:31.050","Text":"so 10^2 plus 3^2 does not equal the same as 13^2."},{"Start":"04:31.050 ","End":"04:35.640","Text":"This is why we have to deal with Steiner twice."},{"Start":"04:36.310 ","End":"04:39.970","Text":"Because we have to do this Steiner twice,"},{"Start":"04:39.970 ","End":"04:45.565","Text":"so we just have to take the usual Steiner equation and also,"},{"Start":"04:45.565 ","End":"04:49.070","Text":"we can rearrange it for our first step."},{"Start":"04:49.070 ","End":"04:51.355","Text":"Then our first step will be,"},{"Start":"04:51.355 ","End":"04:52.480","Text":"we\u0027re just going to isolate"},{"Start":"04:52.480 ","End":"04:55.030","Text":"this I center of mass because that\u0027s what we\u0027re trying to find."},{"Start":"04:55.030 ","End":"05:02.535","Text":"We\u0027ll have our I minus our md^2 will equal to our I center of mass."},{"Start":"05:02.535 ","End":"05:08.530","Text":"We\u0027re going to go from our I of the square,"},{"Start":"05:09.230 ","End":"05:15.960","Text":"minus our md^2 and then that will give us the I of the center of mass over here."},{"Start":"05:15.960 ","End":"05:22.354","Text":"Then we\u0027ll plug that into this equation and then we\u0027ll have the I center of mass."},{"Start":"05:22.354 ","End":"05:24.000","Text":"Then plus our md^2,"},{"Start":"05:24.000 ","End":"05:26.550","Text":"which will give us our new I,"},{"Start":"05:26.550 ","End":"05:28.515","Text":"which is what we\u0027re after."},{"Start":"05:28.515 ","End":"05:31.690","Text":"This is what we\u0027re looking for, the answer."},{"Start":"05:32.150 ","End":"05:34.745","Text":"Let\u0027s do this right now."},{"Start":"05:34.745 ","End":"05:41.385","Text":"Now, what I\u0027m going to do is I\u0027m going to draw these distances slightly bigger."},{"Start":"05:41.385 ","End":"05:43.205","Text":"From our origin now,"},{"Start":"05:43.205 ","End":"05:46.100","Text":"if you go back to the unit speaking about center of mass,"},{"Start":"05:46.100 ","End":"05:50.620","Text":"there\u0027s a lesson there on how to find the center of mass of a triangle."},{"Start":"05:50.620 ","End":"05:53.345","Text":"If you don\u0027t remember how to work that out,"},{"Start":"05:53.345 ","End":"05:56.675","Text":"I\u0027m not going to go over it here go back to that lesson."},{"Start":"05:56.675 ","End":"06:01.610","Text":"We know that the center of mass is located a third in,"},{"Start":"06:01.610 ","End":"06:06.065","Text":"so that means that till this dot over here,"},{"Start":"06:06.065 ","End":"06:10.025","Text":"this distance is going to be b/3."},{"Start":"06:10.025 ","End":"06:17.510","Text":"Then the center of mass of this length b is obviously going to be at b/2,"},{"Start":"06:17.510 ","End":"06:26.550","Text":"so b/2, then this is going to be exactly the same on this rod over here, which is a."},{"Start":"06:26.550 ","End":"06:32.720","Text":"The distance, this distance until the center of mass is going to be a divided by"},{"Start":"06:32.720 ","End":"06:39.550","Text":"3 and then at the center of mass of the rod a is going to be at a/2."},{"Start":"06:39.550 ","End":"06:45.005","Text":"Now I\u0027m going to draw this in a slightly more clear way."},{"Start":"06:45.005 ","End":"06:47.765","Text":"We have our origin point over here,"},{"Start":"06:47.765 ","End":"06:52.445","Text":"and then we have a distance of b"},{"Start":"06:52.445 ","End":"06:57.605","Text":"divided by 3 until we get to where the center of mass is."},{"Start":"06:57.605 ","End":"07:00.650","Text":"Then the center of mass is somewhere over here."},{"Start":"07:00.650 ","End":"07:02.690","Text":"This is center of mass."},{"Start":"07:02.690 ","End":"07:08.345","Text":"Then from the center of mass until the center of the rod,"},{"Start":"07:08.345 ","End":"07:14.640","Text":"which is where the center of mass of the square would be,"},{"Start":"07:14.640 ","End":"07:19.100","Text":"we have this distance over here."},{"Start":"07:19.100 ","End":"07:23.045","Text":"Which is going to be just b/2 minus b/3,"},{"Start":"07:23.045 ","End":"07:26.525","Text":"which is just going to be b divided by 6."},{"Start":"07:26.525 ","End":"07:28.700","Text":"If you do that algebra."},{"Start":"07:28.700 ","End":"07:36.270","Text":"Then over here, we\u0027re going to have our I of the square."},{"Start":"07:36.270 ","End":"07:42.880","Text":"Then obviously, with the same pattern of what we just did with the b\u0027s."},{"Start":"07:42.880 ","End":"07:47.620","Text":"This length over here is going to be a divided by 6 and"},{"Start":"07:47.620 ","End":"07:54.445","Text":"this length over here is going to be divided by 3."},{"Start":"07:54.445 ","End":"07:57.280","Text":"For the exact same way."},{"Start":"07:57.280 ","End":"08:04.675","Text":"Now, what we want to do is we want to find the distance."},{"Start":"08:04.675 ","End":"08:06.400","Text":"Because we have our I^2."},{"Start":"08:06.400 ","End":"08:08.080","Text":"We worked it out."},{"Start":"08:08.080 ","End":"08:11.880","Text":"We want to find this distance over"},{"Start":"08:11.880 ","End":"08:16.385","Text":"here from I^2 until our center of mass. How do we do that?"},{"Start":"08:16.385 ","End":"08:18.850","Text":"Using our best friend Pythagoras."},{"Start":"08:18.850 ","End":"08:20.965","Text":"What does he say that this length,"},{"Start":"08:20.965 ","End":"08:23.095","Text":"let\u0027s call it r over here."},{"Start":"08:23.095 ","End":"08:29.860","Text":"Our r is going to be equal to a/6^2."},{"Start":"08:29.860 ","End":"08:38.270","Text":"Sorry, r^2 is going to be a/6^2 plus b/6^2."},{"Start":"08:38.430 ","End":"08:44.620","Text":"For the sake of using this in our equation,"},{"Start":"08:44.620 ","End":"08:46.540","Text":"because we\u0027re using the variable d,"},{"Start":"08:46.540 ","End":"08:50.380","Text":"it\u0027s the same thing, we can just call this d^2."},{"Start":"08:50.380 ","End":"08:53.020","Text":"There\u0027s no need to square root it because in our equation"},{"Start":"08:53.020 ","End":"08:56.225","Text":"itself they\u0027re asking for the distance d^2."},{"Start":"08:56.225 ","End":"09:02.785","Text":"Then we can just plug in all of our values into the equation."},{"Start":"09:02.785 ","End":"09:04.285","Text":"Let\u0027s see how we do that."},{"Start":"09:04.285 ","End":"09:08.875","Text":"We\u0027re going to have the I of the square,"},{"Start":"09:08.875 ","End":"09:17.770","Text":"which is m/12a^2 plus b^2 minus md^2."},{"Start":"09:17.770 ","End":"09:28.970","Text":"Minus m then our d^2 was a/6^2 plus b/6^2."},{"Start":"09:29.850 ","End":"09:35.639","Text":"This is going to equal our I center of mass."},{"Start":"09:35.639 ","End":"09:37.540","Text":"Now we\u0027ve gone into"},{"Start":"09:37.540 ","End":"09:42.680","Text":"our pink point over here that I\u0027ve labeled center of mass, we are now here."},{"Start":"09:44.130 ","End":"09:48.160","Text":"I labeled that this was Steiner Number 1."},{"Start":"09:48.160 ","End":"09:51.715","Text":"This over here was Number 1 and the equation that we used"},{"Start":"09:51.715 ","End":"09:55.760","Text":"was this rearranged equation for a Steiner."},{"Start":"09:55.760 ","End":"09:57.600","Text":"Then now for our second step,"},{"Start":"09:57.600 ","End":"10:00.510","Text":"we\u0027re going to use the second of a second equation,"},{"Start":"10:00.510 ","End":"10:03.540","Text":"which is going to be from the center of"},{"Start":"10:03.540 ","End":"10:07.695","Text":"mass until our actual axes of rotation that we\u0027re trying to work out."},{"Start":"10:07.695 ","End":"10:10.235","Text":"This is stage Number 2."},{"Start":"10:10.235 ","End":"10:12.205","Text":"Again, if you look in the triangle,"},{"Start":"10:12.205 ","End":"10:14.290","Text":"it represents that stage."},{"Start":"10:14.290 ","End":"10:17.095","Text":"Let\u0027s see how we do this."},{"Start":"10:17.095 ","End":"10:18.865","Text":"This is stage Number 2."},{"Start":"10:18.865 ","End":"10:24.340","Text":"Stage Number 2 says that we have to substitute in our equation for our I center of mass,"},{"Start":"10:24.340 ","End":"10:26.335","Text":"which is what we did in stage number 1,"},{"Start":"10:26.335 ","End":"10:30.145","Text":"plus our mass times our distance squared."},{"Start":"10:30.145 ","End":"10:33.824","Text":"Let\u0027s see what our distance squared here is."},{"Start":"10:33.824 ","End":"10:37.225","Text":"This, I\u0027m going to call d1 and this is d2."},{"Start":"10:37.225 ","End":"10:42.550","Text":"Again, we\u0027re going to use Pythagoras."},{"Start":"10:42.550 ","End":"10:48.940","Text":"We\u0027re going to see that our d_2^2 is equal to,"},{"Start":"10:48.940 ","End":"10:55.915","Text":"then it\u0027s just going to be a/3^2 plus b/3^2."},{"Start":"10:55.915 ","End":"10:58.360","Text":"We\u0027re just doing Pythagoras for this."},{"Start":"10:58.360 ","End":"11:03.370","Text":"This is d_2 and this was d_1."},{"Start":"11:03.370 ","End":"11:09.985","Text":"Now we\u0027ll just substitute in our ICM."},{"Start":"11:09.985 ","End":"11:16.525","Text":"I wrote out ICM and then in our equation we have plus the mass."},{"Start":"11:16.525 ","End":"11:22.840","Text":"Plus the mass of the system multiplied by d_2^2."},{"Start":"11:22.840 ","End":"11:28.975","Text":"Our d_2^2, we saw was a/3^2 plus b/3^2."},{"Start":"11:28.975 ","End":"11:37.135","Text":"We\u0027ll just multiply this by a/3^2 plus b/3^2."},{"Start":"11:37.135 ","End":"11:41.515","Text":"This will equal to I that we\u0027re searching for."},{"Start":"11:41.515 ","End":"11:44.335","Text":"Now what we\u0027re going to do is we\u0027re going to"},{"Start":"11:44.335 ","End":"11:48.865","Text":"rearrange this equation and take out like terms,"},{"Start":"11:48.865 ","End":"11:53.410","Text":"in order to get a more simplified version of this I."},{"Start":"11:53.410 ","End":"11:55.840","Text":"Let\u0027s see how we do this."},{"Start":"11:55.840 ","End":"11:59.215","Text":"Now the easiest way is if we open up the brackets."},{"Start":"11:59.215 ","End":"12:07.695","Text":"We\u0027ll have that ma^2 divided by 12 plus mb^2 divided by 12."},{"Start":"12:07.695 ","End":"12:16.270","Text":"Negative ma^2/36, negative mb^2/36."},{"Start":"12:16.270 ","End":"12:18.100","Text":"Now, this comes from this negative over here,"},{"Start":"12:18.100 ","End":"12:20.020","Text":"multiplying all these brackets,"},{"Start":"12:20.020 ","End":"12:27.940","Text":"because we have a/6 ^2 and b/6^2 that\u0027s a^2 divided by 6^2, which is 36."},{"Start":"12:27.940 ","End":"12:32.905","Text":"Then we\u0027ll have plus ma^2/9."},{"Start":"12:32.905 ","End":"12:39.430","Text":"Because it\u0027s 3^2 mb^2 divided by 9."},{"Start":"12:39.430 ","End":"12:44.290","Text":"Now we can see that the common denominator is 36."},{"Start":"12:44.290 ","End":"12:47.545","Text":"Then we can say,"},{"Start":"12:47.545 ","End":"12:49.795","Text":"we can rewrite this."},{"Start":"12:49.795 ","End":"12:52.960","Text":"If we have the common denominators 36,"},{"Start":"12:52.960 ","End":"12:55.420","Text":"what do we have to multiply by"},{"Start":"12:55.420 ","End":"13:02.215","Text":"this numerator in order to get that the denominator is equal to 36?"},{"Start":"13:02.215 ","End":"13:04.420","Text":"We can just multiply it by 3,"},{"Start":"13:04.420 ","End":"13:06.130","Text":"12 times 3 is 36,"},{"Start":"13:06.130 ","End":"13:08.789","Text":"which means that we also have to multiply."},{"Start":"13:08.789 ","End":"13:12.410","Text":"If we have 3 times here and 3 times here,"},{"Start":"13:12.780 ","End":"13:16.075","Text":"then we\u0027ll get 36 and we\u0027re keeping,"},{"Start":"13:16.075 ","End":"13:19.135","Text":"the ratios correct. I have 3."},{"Start":"13:19.135 ","End":"13:22.270","Text":"Then on the 36, we have 1."},{"Start":"13:22.270 ","End":"13:27.860","Text":"We\u0027re going to have 3 minus 1."},{"Start":"13:28.470 ","End":"13:35.620","Text":"Then we\u0027re going to have the 9 has to be multiplied by 4."},{"Start":"13:35.620 ","End":"13:38.200","Text":"If we have 4 times 9,"},{"Start":"13:38.200 ","End":"13:42.490","Text":"we\u0027ll get 36 so 4 times 9,"},{"Start":"13:42.490 ","End":"13:44.605","Text":"it\u0027s going to be at 36."},{"Start":"13:44.605 ","End":"13:47.510","Text":"Then we have plus 4."},{"Start":"13:48.240 ","End":"13:51.520","Text":"Then if we simplify this over here,"},{"Start":"13:51.520 ","End":"13:54.430","Text":"we\u0027re going to have 3 minus 1 is 2 plus 4 is 6."},{"Start":"13:54.430 ","End":"13:56.650","Text":"I have 6/36,"},{"Start":"13:56.650 ","End":"13:59.140","Text":"which simplifies to 1/6."},{"Start":"13:59.140 ","End":"14:02.275","Text":"Then we can put this back into our equation."},{"Start":"14:02.275 ","End":"14:06.160","Text":"Then we\u0027ll have 1/6 multiplied by our m,"},{"Start":"14:06.160 ","End":"14:08.515","Text":"which is our common factor."},{"Start":"14:08.515 ","End":"14:14.680","Text":"Then everywhere we can also see that we also have a^2 and b^2,"},{"Start":"14:14.680 ","End":"14:19.120","Text":"a^2 and b^2, a^2 and b^2."},{"Start":"14:19.120 ","End":"14:24.100","Text":"Then we can put in brackets a^2 plus b^2."},{"Start":"14:24.100 ","End":"14:31.300","Text":"Now this is our final answer for the moment of inertia of"},{"Start":"14:31.300 ","End":"14:39.520","Text":"a triangle which is rotating about its corner where the right angle is."},{"Start":"14:39.520 ","End":"14:42.650","Text":"That\u0027s the end of the lesson."}],"ID":9383},{"Watched":false,"Name":"I of Disk With Hole","Duration":"6m 44s","ChapterTopicVideoID":9111,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.910","Text":"Hello. In this question,"},{"Start":"00:02.910 ","End":"00:06.990","Text":"we\u0027re being told that a disk of mass M and radius R"},{"Start":"00:06.990 ","End":"00:12.210","Text":"has had a hole drilled in at its end at a distance of R over 2 from the center."},{"Start":"00:12.210 ","End":"00:16.425","Text":"Then we\u0027re told that the hole has a radius of R divided by 4."},{"Start":"00:16.425 ","End":"00:18.270","Text":"Now in Question number 1,"},{"Start":"00:18.270 ","End":"00:20.700","Text":"we\u0027re being told that the disk is rotating about"},{"Start":"00:20.700 ","End":"00:25.185","Text":"its center and we\u0027re being asked to calculate its moment of inertia."},{"Start":"00:25.185 ","End":"00:28.380","Text":"The first thing that we\u0027re going to do is we\u0027re going to calculate"},{"Start":"00:28.380 ","End":"00:32.130","Text":"what the mass of the hole is,"},{"Start":"00:32.130 ","End":"00:39.105","Text":"we\u0027re calling it m. The mass of the hole is going to be the radius of the hole,"},{"Start":"00:39.105 ","End":"00:43.430","Text":"which is R divided by 4 divided by the radius of the entire disk,"},{"Start":"00:43.430 ","End":"00:47.430","Text":"which is R^2 multiplied by"},{"Start":"00:47.430 ","End":"00:54.315","Text":"M. Then we\u0027ll see that the R\u0027s cross off and then (1/4)^2 is 1 over 16."},{"Start":"00:54.315 ","End":"00:57.420","Text":"So it\u0027s going to be 1 over 16M."},{"Start":"00:57.420 ","End":"01:00.660","Text":"That\u0027s what the mass of the hole is."},{"Start":"01:00.660 ","End":"01:05.350","Text":"Now let\u0027s see what we have to do."},{"Start":"01:05.810 ","End":"01:12.320","Text":"We\u0027ve learned Steiner\u0027s theorem and we\u0027ve also learned the other trick about additivity."},{"Start":"01:12.320 ","End":"01:14.165","Text":"The moment of inertia,"},{"Start":"01:14.165 ","End":"01:17.060","Text":"you can add on moments of inertia."},{"Start":"01:17.060 ","End":"01:19.310","Text":"What are we going to do is we\u0027re going to"},{"Start":"01:19.310 ","End":"01:26.385","Text":"first use additivity and put in weave into their Steiner\u0027s theorem."},{"Start":"01:26.385 ","End":"01:29.480","Text":"What we want to do, and the first question is to calculate"},{"Start":"01:29.480 ","End":"01:32.390","Text":"the moment of inertia when the disk is rotating about its center."},{"Start":"01:32.390 ","End":"01:35.330","Text":"I\u0027m going to call the moment of inertia"},{"Start":"01:35.330 ","End":"01:38.815","Text":"that we\u0027re wanting I_0 because it\u0027s around its center."},{"Start":"01:38.815 ","End":"01:41.885","Text":"Now into this, we\u0027re going to substitute in"},{"Start":"01:41.885 ","End":"01:46.685","Text":"the equation for the moment of inertia of a disc rotating about its center,"},{"Start":"01:46.685 ","End":"01:51.335","Text":"we\u0027re going to substitute in the moment of inertia and just a regular disk,"},{"Start":"01:51.335 ","End":"01:53.135","Text":"which as we know,"},{"Start":"01:53.135 ","End":"01:59.725","Text":"is just MR^2 divided by 2."},{"Start":"01:59.725 ","End":"02:04.840","Text":"This is the moment of inertia of the disk."},{"Start":"02:06.460 ","End":"02:14.160","Text":"Now we\u0027re going to use our additivity trick and say that"},{"Start":"02:14.160 ","End":"02:21.080","Text":"now we just have to add the moments of inertia of the small disk."},{"Start":"02:21.080 ","End":"02:25.085","Text":"Now because this small disk isn\u0027t a small disk, it\u0027s actually hole,"},{"Start":"02:25.085 ","End":"02:27.260","Text":"so we\u0027re going to say that the mass is negative,"},{"Start":"02:27.260 ","End":"02:31.440","Text":"so we can in fact just minus all of this."},{"Start":"02:31.440 ","End":"02:36.310","Text":"Now here comes in the Steiner because this hole,"},{"Start":"02:36.310 ","End":"02:41.690","Text":"isn\u0027t located in the center of the disk where axis of rotation is,"},{"Start":"02:41.690 ","End":"02:45.620","Text":"it\u0027s located, R divide it by 2 away from there."},{"Start":"02:45.620 ","End":"02:49.280","Text":"Then what we\u0027re going to do is we\u0027re going to add in"},{"Start":"02:49.280 ","End":"02:53.375","Text":"the equation for the moment of inertia of the hole,"},{"Start":"02:53.375 ","End":"02:56.585","Text":"which is again going to be the m,"},{"Start":"02:56.585 ","End":"03:00.185","Text":"the mass of the disk multiplied by its radius."},{"Start":"03:00.185 ","End":"03:05.875","Text":"It\u0027s R over 4^2 divided by 2."},{"Start":"03:05.875 ","End":"03:09.305","Text":"It\u0027s the same equation just with different numbers substituted in."},{"Start":"03:09.305 ","End":"03:13.220","Text":"Then we have to use Steiner."},{"Start":"03:13.220 ","End":"03:19.470","Text":"Its distance is R divided by 2 so we add plus m,"},{"Start":"03:19.470 ","End":"03:23.160","Text":"the mass multiplied by its distance squared and"},{"Start":"03:23.160 ","End":"03:27.750","Text":"its distance is R divided by 2 and then we square that."},{"Start":"03:27.750 ","End":"03:32.420","Text":"Now we just have to substitute in our values for"},{"Start":"03:32.420 ","End":"03:38.150","Text":"our m and rearrange and cancel out terms."},{"Start":"03:38.150 ","End":"03:41.675","Text":"In the end, our answer is going to be"},{"Start":"03:41.675 ","End":"03:49.975","Text":"247 divided by 512 MR^2."},{"Start":"03:49.975 ","End":"03:53.360","Text":"I\u0027m not going to do this now because it\u0027s just going to be a waste of time,"},{"Start":"03:53.360 ","End":"03:54.860","Text":"but do it on a piece of paper."},{"Start":"03:54.860 ","End":"03:59.130","Text":"You substitute in the value for m,"},{"Start":"03:59.150 ","End":"04:03.300","Text":"into here and into here."},{"Start":"04:03.300 ","End":"04:07.590","Text":"Then you rearrange everything and you\u0027ll get this answer."},{"Start":"04:08.260 ","End":"04:12.995","Text":"Now we\u0027re going to look at Question number 2."},{"Start":"04:12.995 ","End":"04:14.900","Text":"In Question number 2,"},{"Start":"04:14.900 ","End":"04:17.590","Text":"we can see that we\u0027re being asked to find"},{"Start":"04:17.590 ","End":"04:21.705","Text":"the moment of inertia of the disk rotating about its center of mass,"},{"Start":"04:21.705 ","End":"04:23.700","Text":"not at the center of the disk."},{"Start":"04:23.700 ","End":"04:27.650","Text":"The center of mass is somewhere along here."},{"Start":"04:27.650 ","End":"04:31.810","Text":"We\u0027ve had a lesson where we find again back in the unit"},{"Start":"04:31.810 ","End":"04:35.065","Text":"for a center of mass and we found that the"},{"Start":"04:35.065 ","End":"04:41.660","Text":"distance from the center of this disk or the hole until the center of mass so"},{"Start":"04:41.660 ","End":"04:50.765","Text":"this distance d is going to be R divided by 30,"},{"Start":"04:50.765 ","End":"04:54.175","Text":"that is this distance."},{"Start":"04:54.175 ","End":"04:58.605","Text":"Now we\u0027re just going to use our rearranged equation for a Steiner."},{"Start":"04:58.605 ","End":"05:02.495","Text":"Before we knew that our I that we wanted to find,"},{"Start":"05:02.495 ","End":"05:09.835","Text":"it\u0027s going to be equal to the I center of mass plus mass times distance squared."},{"Start":"05:09.835 ","End":"05:13.760","Text":"But now here we just worked our outer I,"},{"Start":"05:13.760 ","End":"05:17.980","Text":"this is our I_0 and what we want to find is our I center of mass."},{"Start":"05:17.980 ","End":"05:20.120","Text":"We\u0027re going to rearrange this equation,"},{"Start":"05:20.120 ","End":"05:26.510","Text":"we\u0027ll have I_0 arrow minus md^2 is going to be equal to our I center of mass,"},{"Start":"05:26.510 ","End":"05:29.930","Text":"which will be the answer to Question number 2."},{"Start":"05:29.930 ","End":"05:35.820","Text":"Now we substitute in what our I_0 is,"},{"Start":"05:35.820 ","End":"05:41.860","Text":"which we found to be 247 over 512 MR^2."},{"Start":"05:42.680 ","End":"05:46.040","Text":"Then we\u0027re going to have minus our mass."},{"Start":"05:46.040 ","End":"05:47.195","Text":"Now, what is our mass?"},{"Start":"05:47.195 ","End":"05:52.220","Text":"We\u0027re dealing with our total mass minus the mass of the drilled hole."},{"Start":"05:52.220 ","End":"05:57.394","Text":"We found that the mass of the drill hole is 116th M. So we\u0027re going to have minus"},{"Start":"05:57.394 ","End":"06:03.830","Text":"15 over 16 M multiplied by d^2,"},{"Start":"06:03.830 ","End":"06:09.620","Text":"and our d^2 is R over 30^2."},{"Start":"06:09.620 ","End":"06:15.150","Text":"This will give us I center of mass."},{"Start":"06:15.590 ","End":"06:17.600","Text":"Then the final answer,"},{"Start":"06:17.600 ","End":"06:21.080","Text":"you can do it with a calculator or just on your page."},{"Start":"06:21.080 ","End":"06:29.230","Text":"It will be MR^2 multiplied by 3,697 divided by 7,680."},{"Start":"06:29.230 ","End":"06:32.760","Text":"That is the final answer."},{"Start":"06:32.760 ","End":"06:34.695","Text":"It\u0027s very simple,"},{"Start":"06:34.695 ","End":"06:37.580","Text":"once you use the Steiner theorem and the trick for"},{"Start":"06:37.580 ","End":"06:41.480","Text":"additivity that we spoke about right at the beginning of the unit."},{"Start":"06:41.480 ","End":"06:44.490","Text":"That\u0027s the end of this lesson."}],"ID":9384},{"Watched":false,"Name":"I of Sphere","Duration":"7m 1s","ChapterTopicVideoID":9112,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"Hello. In this lesson,"},{"Start":"00:02.490 ","End":"00:08.055","Text":"we\u0027re going to be showing how to work out the moment of inertia of a full sphere."},{"Start":"00:08.055 ","End":"00:17.775","Text":"As we know, the equation for the moment of inertia is equal to the sum of m_i r_i^2."},{"Start":"00:17.775 ","End":"00:20.085","Text":"Now this r_i^2,"},{"Start":"00:20.085 ","End":"00:25.905","Text":"I\u0027m going to put a tilde on top because this is the dist"},{"Start":"00:25.905 ","End":"00:32.980","Text":"from axis of rotation."},{"Start":"00:33.350 ","End":"00:41.760","Text":"Now, because we\u0027re going to be working on a full sphere and we\u0027re taking little squares."},{"Start":"00:41.760 ","End":"00:51.680","Text":"We\u0027re going to say that this is equal to the integral on i^2dm."},{"Start":"00:51.680 ","End":"00:58.595","Text":"Let\u0027s see. Our r tilde here means because our axes of rotation is the z-axis,"},{"Start":"00:58.595 ","End":"01:01.700","Text":"so this is our r tilde."},{"Start":"01:01.700 ","End":"01:07.425","Text":"We\u0027re going to see very soon that it\u0027s very different to our i."},{"Start":"01:07.425 ","End":"01:11.415","Text":"Let\u0027s see what our dm is equal to."},{"Start":"01:11.415 ","End":"01:14.359","Text":"Our dm because we\u0027re dealing with volume,"},{"Start":"01:14.359 ","End":"01:18.520","Text":"we\u0027re going to be using Rho dv."},{"Start":"01:18.520 ","End":"01:21.945","Text":"Now we\u0027re going to see what is Rho."},{"Start":"01:21.945 ","End":"01:28.415","Text":"Rho is equal to the total mass of the body divided by the total volume of the body."},{"Start":"01:28.415 ","End":"01:33.350","Text":"That\u0027s going to be the mass of the sphere divided by the volume of"},{"Start":"01:33.350 ","End":"01:39.540","Text":"a sphere is 4PiR^3 divided by 3."},{"Start":"01:39.540 ","End":"01:43.160","Text":"Then we can just rearrange a spring the 3 up,"},{"Start":"01:43.160 ","End":"01:47.550","Text":"so we\u0027ll have 3M divided by 4Pi^3."},{"Start":"01:49.040 ","End":"01:53.600","Text":"Now let\u0027s see what our dv is equal to."},{"Start":"01:53.600 ","End":"02:00.500","Text":"Now, our dv because we\u0027re speaking about a sphere and we\u0027re speaking about volume."},{"Start":"02:00.500 ","End":"02:06.080","Text":"The easiest thing to do is to use spherical coordinates because we\u0027re using a sphere."},{"Start":"02:06.080 ","End":"02:10.850","Text":"Remembering the Jacobian in spherical coordinates,"},{"Start":"02:10.850 ","End":"02:20.385","Text":"the Jacobian is r^2 sine of Phi dr d Theta d Phi."},{"Start":"02:20.385 ","End":"02:23.570","Text":"Remember that d Theta and d Phi are"},{"Start":"02:23.570 ","End":"02:28.760","Text":"interchangeable within different textbooks and with different professors."},{"Start":"02:28.760 ","End":"02:34.970","Text":"But just remember that your sine Phi or sine Theta,"},{"Start":"02:34.970 ","End":"02:42.270","Text":"whatever it might be has to equate to the bounds that you put in, in the integral."},{"Start":"02:42.440 ","End":"02:44.910","Text":"Now let\u0027s set up this equation."},{"Start":"02:44.910 ","End":"02:48.065","Text":"We have I is equal to and now because we\u0027re dealing with volume,"},{"Start":"02:48.065 ","End":"02:57.390","Text":"it\u0027s going to be a triple integral of r tilde squared multiplied by r [inaudible],"},{"Start":"02:57.390 ","End":"03:06.090","Text":"which is going to be 3M divided by 4PiR^3 multiplied by our dv,"},{"Start":"03:06.090 ","End":"03:14.260","Text":"which is going to be r^2 sine Phi dr d Theta d Phi."},{"Start":"03:15.560 ","End":"03:18.180","Text":"Now as we can see,"},{"Start":"03:18.180 ","End":"03:20.395","Text":"r tilde and our r,"},{"Start":"03:20.395 ","End":"03:24.510","Text":"we have separate variables in our equation."},{"Start":"03:24.510 ","End":"03:29.750","Text":"Now there\u0027s too many variables and we have to find the relation between the 2."},{"Start":"03:29.750 ","End":"03:35.330","Text":"As we know, this is going to be r. We know that this angle,"},{"Start":"03:35.330 ","End":"03:37.070","Text":"the definition of Phi,"},{"Start":"03:37.070 ","End":"03:41.235","Text":"is that it\u0027s the angle between the radius and the z axis."},{"Start":"03:41.235 ","End":"03:50.005","Text":"Therefore we can say that our r tilde is equal to r sine of Phi."},{"Start":"03:50.005 ","End":"03:52.770","Text":"Basic trigonometry."},{"Start":"03:54.140 ","End":"03:56.645","Text":"Now, we can rewrite this."},{"Start":"03:56.645 ","End":"03:58.340","Text":"I\u0027ll keep going in the pink pen."},{"Start":"03:58.340 ","End":"04:06.665","Text":"We can say that our I is equal to the triple integral of r tilde squared,"},{"Start":"04:06.665 ","End":"04:16.080","Text":"which is going to be r squared sine squared Phi multiplied by 3M divided by"},{"Start":"04:16.080 ","End":"04:22.140","Text":"4PiR^3 multiplied by i^2 sine"},{"Start":"04:22.140 ","End":"04:27.435","Text":"of Phi dr d Theta d Phi."},{"Start":"04:27.435 ","End":"04:30.700","Text":"Let\u0027s talk about our bounds our r,"},{"Start":"04:30.700 ","End":"04:36.305","Text":"is of course going to be from 0 until capital R. Our Theta is going to"},{"Start":"04:36.305 ","End":"04:44.380","Text":"be from 0 until 2Pi and our Phi is going to be from 0 until Pi."},{"Start":"04:49.370 ","End":"04:59.760","Text":"Let\u0027s take a look. We have r^4 and sine Phi^3."},{"Start":"04:59.760 ","End":"05:06.185","Text":"Notice that our i\u0027s are Thetas and Phi\u0027s don\u0027t rely on 1 another."},{"Start":"05:06.185 ","End":"05:08.750","Text":"It makes the integral slightly easier."},{"Start":"05:08.750 ","End":"05:12.350","Text":"But what does make it slightly more complicated is the fact"},{"Start":"05:12.350 ","End":"05:16.465","Text":"that we have a sine cubed Phi over here."},{"Start":"05:16.465 ","End":"05:19.040","Text":"Let\u0027s see how we do this."},{"Start":"05:19.040 ","End":"05:25.895","Text":"We can rewrite the integral of sine cubed Phi d Phi"},{"Start":"05:25.895 ","End":"05:35.460","Text":"into the integral of sine squared Phi multiplied by sine Phi d Phi,"},{"Start":"05:35.460 ","End":"05:37.625","Text":"I haven\u0027t changed anything over here."},{"Start":"05:37.625 ","End":"05:47.915","Text":"Then we can say that sine squared Phi is equal to 1 minus cosine squared Phi."},{"Start":"05:47.915 ","End":"05:51.800","Text":"Then we have sine Phi d Phi."},{"Start":"05:51.800 ","End":"05:54.620","Text":"Then if I name another variable,"},{"Start":"05:54.620 ","End":"05:59.615","Text":"say t, and I say that that equals to cosine of Phi."},{"Start":"05:59.615 ","End":"06:03.740","Text":"Then I\u0027ll have that"},{"Start":"06:03.740 ","End":"06:12.780","Text":"my dt is equal to negative of sine Phi d Phi."},{"Start":"06:12.880 ","End":"06:20.460","Text":"Then I can say that this whole thing over here is equal to negative dt."},{"Start":"06:22.160 ","End":"06:29.495","Text":"Then what I can do is I can just substitute that in, into this integral."},{"Start":"06:29.495 ","End":"06:32.700","Text":"All I have to do is do the integral."},{"Start":"06:33.680 ","End":"06:41.130","Text":"We\u0027ll scroll up. My final answer for what my I will be, will be 2/5MR^2."},{"Start":"06:44.150 ","End":"06:50.000","Text":"This is the final answer for the moment of inertia of a full sphere,"},{"Start":"06:50.000 ","End":"06:53.840","Text":"you\u0027re more than welcome and in fact encouraged to"},{"Start":"06:53.840 ","End":"06:58.345","Text":"do this integration alone on a piece of paper."},{"Start":"06:58.345 ","End":"07:01.210","Text":"That\u0027s the end of this lesson."}],"ID":9385},{"Watched":false,"Name":"I of Electric Gate","Duration":"9m 13s","ChapterTopicVideoID":12207,"CourseChapterTopicPlaylistID":5389,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:02.790","Text":"Hello. In this question,"},{"Start":"00:02.790 ","End":"00:05.310","Text":"we\u0027re being told to calculate the moment of inertia of"},{"Start":"00:05.310 ","End":"00:09.435","Text":"an electric gate of mass m and length l. At the end of the gate,"},{"Start":"00:09.435 ","End":"00:13.125","Text":"a mass of M and of length L is attached,"},{"Start":"00:13.125 ","End":"00:16.995","Text":"and then we\u0027re told that the gate rotates about the center of mass."},{"Start":"00:16.995 ","End":"00:19.620","Text":"As we can see, the center of mass will be over here,"},{"Start":"00:19.620 ","End":"00:24.480","Text":"which is why this triangle is located over here."},{"Start":"00:24.480 ","End":"00:28.515","Text":"Now, as you remember from the units about center of mass,"},{"Start":"00:28.515 ","End":"00:31.860","Text":"we spoke about a very similar question"},{"Start":"00:31.860 ","End":"00:34.995","Text":"and we showed why it\u0027s important that this triangle is"},{"Start":"00:34.995 ","End":"00:38.190","Text":"exactly where the center of mass is between the center of"},{"Start":"00:38.190 ","End":"00:42.465","Text":"mass of the rod and the center of mass of this mass."},{"Start":"00:42.465 ","End":"00:46.175","Text":"Because it means that the least amount of"},{"Start":"00:46.175 ","End":"00:50.375","Text":"energy or work has to be done in order to open the gate,"},{"Start":"00:50.375 ","End":"00:54.955","Text":"shut the gates, and also to keep the gate open when it is open."},{"Start":"00:54.955 ","End":"00:58.760","Text":"We spoke about it in terms of the center of mass,"},{"Start":"00:58.760 ","End":"01:01.535","Text":"but right now, because we are speaking about the moment of inertia,"},{"Start":"01:01.535 ","End":"01:04.295","Text":"we\u0027re going to speak about the exact same question,"},{"Start":"01:04.295 ","End":"01:10.440","Text":"but in terms of moments of inertia and using Steiner\u0027s theorem."},{"Start":"01:11.180 ","End":"01:16.355","Text":"What we\u0027re going to do is first we\u0027re going to remember what Steiner\u0027s theorem says."},{"Start":"01:16.355 ","End":"01:21.780","Text":"It says that the I_new, let\u0027s call it,"},{"Start":"01:21.780 ","End":"01:30.625","Text":"is equal to the I_cm plus m multiplied ny distance squared."},{"Start":"01:30.625 ","End":"01:35.165","Text":"So that means that the moment of inertia of the new system,"},{"Start":"01:35.165 ","End":"01:38.000","Text":"of the system not rotating about the center of mass,"},{"Start":"01:38.000 ","End":"01:44.900","Text":"but about a different axis of rotation is equal to the moment of inertia of the center"},{"Start":"01:44.900 ","End":"01:47.855","Text":"of mass plus the mass of the object"},{"Start":"01:47.855 ","End":"01:52.675","Text":"multiplied by its distance squared from the axis of rotation."},{"Start":"01:52.675 ","End":"01:58.455","Text":"The distance of the center of mass from the axis of rotation,"},{"Start":"01:58.455 ","End":"02:04.715","Text":"and we\u0027re also going to remember that we have the additivity of I,"},{"Start":"02:04.715 ","End":"02:11.530","Text":"meaning that I total is equal to I_1 plus I_2,"},{"Start":"02:11.530 ","End":"02:15.150","Text":"remember these 2 things."},{"Start":"02:15.150 ","End":"02:17.745","Text":"Let\u0027s see, so I_1,"},{"Start":"02:17.745 ","End":"02:26.775","Text":"let\u0027s call that the rod and I_2 will refer to this mass M. So let\u0027s see,"},{"Start":"02:26.775 ","End":"02:30.130","Text":"I_1 is equal to."},{"Start":"02:30.460 ","End":"02:34.535","Text":"Now, this is something handy to put into"},{"Start":"02:34.535 ","End":"02:41.330","Text":"your notes for taking into the exam that the moment of inertia of"},{"Start":"02:41.330 ","End":"02:46.770","Text":"a rod spinning about its center of mass is going to be 1"},{"Start":"02:46.770 ","End":"02:53.090","Text":"over 12 multiplied by ml^2."},{"Start":"02:53.090 ","End":"02:59.595","Text":"This is important to remember this I of rod,"},{"Start":"02:59.595 ","End":"03:03.230","Text":"and remember this is specifically rotating about its center of mass,"},{"Start":"03:03.230 ","End":"03:06.875","Text":"and then we\u0027re going to add on the section of Steiner,"},{"Start":"03:06.875 ","End":"03:11.370","Text":"which is its mass multiplied by,"},{"Start":"03:11.370 ","End":"03:16.600","Text":"let\u0027s see, so the center of mass is over here."},{"Start":"03:16.600 ","End":"03:24.755","Text":"This distance from the center of mass of the rod until the center of mass of the system,"},{"Start":"03:24.755 ","End":"03:27.965","Text":"which is also going to be its axis of rotation."},{"Start":"03:27.965 ","End":"03:35.140","Text":"We\u0027re just going to call this c. So we\u0027ll say plus mc^2."},{"Start":"03:35.390 ","End":"03:42.845","Text":"The distance between the center of mass of the entire system of the electric gate to"},{"Start":"03:42.845 ","End":"03:50.675","Text":"the center of mass of the rod itself is going to be c, that\u0027s our I1 want."},{"Start":"03:50.675 ","End":"03:54.010","Text":"Let\u0027s see what I2 equals."},{"Start":"03:54.010 ","End":"03:57.200","Text":"Now, this is another handy equation."},{"Start":"03:57.200 ","End":"04:03.600","Text":"What is the moment of inertia of some kind of box shape."},{"Start":"04:03.600 ","End":"04:09.160","Text":"What it\u0027s going to be is again I over 12,"},{"Start":"04:09.160 ","End":"04:12.785","Text":"this has to do with the symmetry on the z axis,"},{"Start":"04:12.785 ","End":"04:14.000","Text":"so it\u0027s not exactly,"},{"Start":"04:14.000 ","End":"04:17.450","Text":"but it takes that trick into account,"},{"Start":"04:17.450 ","End":"04:19.954","Text":"remember the lesson symmetry on the z-axis."},{"Start":"04:19.954 ","End":"04:23.765","Text":"It might give you a clue and might be easy to remember."},{"Start":"04:23.765 ","End":"04:26.720","Text":"Then what we do,"},{"Start":"04:26.720 ","End":"04:35.550","Text":"it\u0027s going to be I over 12 of this length^2 plus this length^2."},{"Start":"04:35.650 ","End":"04:42.150","Text":"Now we\u0027re going to say that the box is of length L,"},{"Start":"04:42.250 ","End":"04:45.920","Text":"L by L, which means that it\u0027s a square,"},{"Start":"04:45.920 ","End":"04:48.830","Text":"it\u0027s not a rectangle, which is what\u0027s being drawn over here."},{"Start":"04:48.830 ","End":"04:50.480","Text":"Just so that you remember this,"},{"Start":"04:50.480 ","End":"04:52.970","Text":"I\u0027m going to say l^2 plus l^2,"},{"Start":"04:52.970 ","End":"04:58.580","Text":"which is meant to be the width^2 plus the length^2, for instance,"},{"Start":"04:58.580 ","End":"05:01.310","Text":"of the box, and then of course,"},{"Start":"05:01.310 ","End":"05:11.495","Text":"multiplied by m. This is the I of some rectangle."},{"Start":"05:11.495 ","End":"05:17.045","Text":"Then again, we\u0027re going to add on the section of the Steiner theorem,"},{"Start":"05:17.045 ","End":"05:18.440","Text":"which as we know,"},{"Start":"05:18.440 ","End":"05:22.775","Text":"is going to be M multiplied by the distance squared."},{"Start":"05:22.775 ","End":"05:24.185","Text":"What is the distance squared?"},{"Start":"05:24.185 ","End":"05:27.860","Text":"We\u0027re going to want to go from the midpoint of"},{"Start":"05:27.860 ","End":"05:33.155","Text":"the box until the center of mass of the entire system."},{"Start":"05:33.155 ","End":"05:35.015","Text":"what is this distance?"},{"Start":"05:35.015 ","End":"05:44.585","Text":"We know that this distance over here is simply going to be L divided by 2"},{"Start":"05:44.585 ","End":"05:54.145","Text":"minus c. Because we know that the distance from here until here is L divided by 2."},{"Start":"05:54.145 ","End":"05:58.100","Text":"Then we know that this distance over here is C,"},{"Start":"05:58.100 ","End":"06:03.260","Text":"so L divided by 2 minus c will give us this distance,"},{"Start":"06:03.260 ","End":"06:08.705","Text":"the distance to the edge of the rod from the center of mass of the entire system."},{"Start":"06:08.705 ","End":"06:13.375","Text":"Then we also have to add the center of mass of the box."},{"Start":"06:13.375 ","End":"06:17.850","Text":"That\u0027s going to be simply at L over 2,"},{"Start":"06:17.850 ","End":"06:19.440","Text":"L over 2,"},{"Start":"06:19.440 ","End":"06:23.640","Text":"so plus L over 2."},{"Start":"06:23.640 ","End":"06:25.920","Text":"This distance over here,"},{"Start":"06:25.920 ","End":"06:28.870","Text":"this is our I2."},{"Start":"06:29.750 ","End":"06:33.740","Text":"Now the final thing that is left to do is"},{"Start":"06:33.740 ","End":"06:37.570","Text":"to find out what the c is because we don\u0027t know what that means."},{"Start":"06:37.570 ","End":"06:41.430","Text":"Leaving in the c won\u0027t give you all of the points,"},{"Start":"06:41.430 ","End":"06:45.395","Text":"how are we going to find what this distance c is equal to?"},{"Start":"06:45.395 ","End":"06:48.275","Text":"As we know from one of the previous units,"},{"Start":"06:48.275 ","End":"06:51.605","Text":"we can use the concept of the center of mass."},{"Start":"06:51.605 ","End":"06:54.470","Text":"What does the center of mass mean,"},{"Start":"06:54.470 ","End":"06:57.985","Text":"it means its location on the rod over here,"},{"Start":"06:57.985 ","End":"07:02.285","Text":"it will give us the location of where this triangle stand is meant to be,"},{"Start":"07:02.285 ","End":"07:09.625","Text":"which will just give us also the distance c if we choose our axis correctly."},{"Start":"07:09.625 ","End":"07:14.030","Text":"Let\u0027s see how we can do this. If we say that over here,"},{"Start":"07:14.030 ","End":"07:18.125","Text":"just for the sake of this calculation"},{"Start":"07:18.125 ","End":"07:23.465","Text":"that our origin goes through the center of mass of the rod,"},{"Start":"07:23.465 ","End":"07:25.100","Text":"right in the middle of the rod,"},{"Start":"07:25.100 ","End":"07:28.545","Text":"the center of mass of the rod is our origin."},{"Start":"07:28.545 ","End":"07:33.395","Text":"Now we can do the calculation for the x center of mass,"},{"Start":"07:33.395 ","End":"07:36.740","Text":"which will also equal to C that"},{"Start":"07:36.740 ","End":"07:41.745","Text":"we are looking for because that will give us the location."},{"Start":"07:41.745 ","End":"07:44.600","Text":"As we know for the center of mass,"},{"Start":"07:44.600 ","End":"07:49.204","Text":"it\u0027s the sum of the masses multiplied by their distance"},{"Start":"07:49.204 ","End":"07:54.890","Text":"from specific axis divided by the sum of all of the masses."},{"Start":"07:54.890 ","End":"08:00.140","Text":"We know that the mass of the rod is small m, but its location,"},{"Start":"08:00.140 ","End":"08:04.100","Text":"so we take its location to be at the center of mass of the rod,"},{"Start":"08:04.100 ","End":"08:06.500","Text":"which is going to be here at the origin."},{"Start":"08:06.500 ","End":"08:14.925","Text":"It\u0027s going to be multiplied by 0 because it\u0027s a distance 0 away plus mass M,"},{"Start":"08:14.925 ","End":"08:20.180","Text":"and its distance from the center of the rod is"},{"Start":"08:20.180 ","End":"08:26.100","Text":"going to be l divided by 2 plus L divided by 2,"},{"Start":"08:26.100 ","End":"08:29.945","Text":"because we\u0027re going to have to travel this distance"},{"Start":"08:29.945 ","End":"08:36.045","Text":"plus this distance in order to get to our origin."},{"Start":"08:36.045 ","End":"08:39.890","Text":"Then we divide all of this by the sum of all of the masses,"},{"Start":"08:39.890 ","End":"08:47.015","Text":"which is m plus M. This is obviously equal to 0,"},{"Start":"08:47.015 ","End":"08:51.980","Text":"and we can cancel this out to equal to M(l"},{"Start":"08:51.980 ","End":"08:59.430","Text":"plus L) over m plus M. This is equal to c,"},{"Start":"08:59.430 ","End":"09:02.240","Text":"then all we have to do is we substitute"},{"Start":"09:02.240 ","End":"09:06.350","Text":"this back into where we have our c\u0027s and also over"},{"Start":"09:06.350 ","End":"09:14.140","Text":"here and then that\u0027s it we\u0027ve solved the question. That\u0027s the end."}],"ID":12677}],"Thumbnail":null,"ID":5389}]

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