Equation and Laws of Conservation
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- 1.1 Equation
- 1.2 Angular Momentum of Body Moving in a Straight Line
- 1.3 Connection Between Angular Momentum and Torque Conservation
- 1.4 How to Answer Questions
- Exercise- Ball Rotating
- Exercise- Ball In A Cone
- Exercise- Ball Attached To Hanging Mass
- 2.1 Angular Momentum of CM Intro
- 2.2 Using the Equation
- 2.3 Multiple Bodies- Same Equation
- 2.4 Center of Mass Doesnt Need to Move in a Straight Line
- 2.5 Deriving the Equation

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[{"Name":"Equation and Laws of Conservation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1.1 Equation","Duration":"2m 50s","ChapterTopicVideoID":12222,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"Hello. In this lesson,"},{"Start":"00:01.740 ","End":"00:04.050","Text":"we\u0027re going to be learning about angular momentum."},{"Start":"00:04.050 ","End":"00:09.030","Text":"Now, the equation for angular momentum is like this."},{"Start":"00:09.030 ","End":"00:12.555","Text":"L, this is the angular momentum and it\u0027s a vector"},{"Start":"00:12.555 ","End":"00:16.860","Text":"which is equal to our r vector which has our position,"},{"Start":"00:16.860 ","End":"00:21.869","Text":"cross multiplied with our P vector,"},{"Start":"00:21.869 ","End":"00:23.894","Text":"which if you remember,"},{"Start":"00:23.894 ","End":"00:26.264","Text":"is our linear momentum."},{"Start":"00:26.264 ","End":"00:29.459","Text":"This is the equation and of course,"},{"Start":"00:29.459 ","End":"00:35.438","Text":"a lot of the time we want to know the size of the angular momentum."},{"Start":"00:35.438 ","End":"00:42.385","Text":"This is going to be equal to the size of our vector multiplied by"},{"Start":"00:42.385 ","End":"00:52.425","Text":"the sine of our linear momentum vector multiplied by sine of the angle between the two."},{"Start":"00:52.425 ","End":"00:55.310","Text":"So let\u0027s give a simple example so that it"},{"Start":"00:55.310 ","End":"00:58.025","Text":"will be slightly easier to understand what\u0027s happening."},{"Start":"00:58.025 ","End":"01:04.400","Text":"Say we have some kind of body of mass m and it\u0027s traveling over here."},{"Start":"01:04.400 ","End":"01:06.245","Text":"It has its v_0,"},{"Start":"01:06.245 ","End":"01:13.215","Text":"and it\u0027s traveling around in a circle."},{"Start":"01:13.215 ","End":"01:16.880","Text":"The axis of rotation is going to be the center of"},{"Start":"01:16.880 ","End":"01:20.946","Text":"the circle and its radius capital R. Now,"},{"Start":"01:20.946 ","End":"01:25.655","Text":"notice that the radius is perpendicular to the velocity."},{"Start":"01:25.655 ","End":"01:28.625","Text":"Now, to know the size of our angular momentum,"},{"Start":"01:28.625 ","End":"01:30.800","Text":"that\u0027s what matters at the moment."},{"Start":"01:30.800 ","End":"01:34.415","Text":"It\u0027s going to be equal to our position,"},{"Start":"01:34.415 ","End":"01:40.565","Text":"which is located at R multiplied by our linear momentum vector,"},{"Start":"01:40.565 ","End":"01:46.385","Text":"which is going to be our mass times R v_0."},{"Start":"01:46.385 ","End":"01:50.915","Text":"Then it\u0027s going to be multiplied by sine of the angle between them."},{"Start":"01:50.915 ","End":"01:52.580","Text":"The angle between them is 90."},{"Start":"01:52.580 ","End":"01:56.970","Text":"Now, for the direction,"},{"Start":"01:56.970 ","End":"02:02.000","Text":"we can see that the direction is going in this direction."},{"Start":"02:02.000 ","End":"02:03.680","Text":"It\u0027s clockwise."},{"Start":"02:03.680 ","End":"02:05.700","Text":"It\u0027s traveling like this."},{"Start":"02:05.700 ","End":"02:10.985","Text":"Because it\u0027s clockwise here our direction will be either plus or minus,"},{"Start":"02:10.985 ","End":"02:14.860","Text":"depending on where we said our positive direction was."},{"Start":"02:14.860 ","End":"02:19.820","Text":"If we say ahead of time that our positive direction is clockwise,"},{"Start":"02:19.820 ","End":"02:23.855","Text":"then this will be a positive direction and if it\u0027s anticlockwise,"},{"Start":"02:23.855 ","End":"02:26.510","Text":"then this will have a negative in front."},{"Start":"02:26.510 ","End":"02:29.735","Text":"But also what we can do is if we use the right-hand rule."},{"Start":"02:29.735 ","End":"02:32.930","Text":"If you bend your fingers in the direction that"},{"Start":"02:32.930 ","End":"02:36.245","Text":"the mass is traveling in and it\u0027s traveling in the clockwise direction,"},{"Start":"02:36.245 ","End":"02:40.745","Text":"then you\u0027ll see that your thumb is pointing into the screen."},{"Start":"02:40.745 ","End":"02:47.450","Text":"We can say that the z-axis or whatever the axis of rotation is pointing into the page."},{"Start":"02:47.450 ","End":"02:50.730","Text":"But it doesn\u0027t really matter at this stage."}],"ID":12698},{"Watched":false,"Name":"1.2 Angular Momentum of Body Moving in a Straight Line","Duration":"4m 27s","ChapterTopicVideoID":12223,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"Now let\u0027s give another example."},{"Start":"00:04.140 ","End":"00:06.150","Text":"We\u0027ll say that again."},{"Start":"00:06.150 ","End":"00:07.680","Text":"We have our body,"},{"Start":"00:07.680 ","End":"00:09.255","Text":"which is of mass m,"},{"Start":"00:09.255 ","End":"00:10.410","Text":"and it\u0027s traveling,"},{"Start":"00:10.410 ","End":"00:12.225","Text":"again here we have our V_0."},{"Start":"00:12.225 ","End":"00:16.365","Text":"Now this time it\u0027s traveling in a straight line,"},{"Start":"00:16.365 ","End":"00:20.100","Text":"and the axis of rotation is going to be over here."},{"Start":"00:20.100 ","End":"00:25.115","Text":"Let\u0027s say that the distance to the axis of rotation,"},{"Start":"00:25.115 ","End":"00:29.480","Text":"that this distance over here is going to be equal to d,"},{"Start":"00:29.480 ","End":"00:33.200","Text":"and of course we can see these lines are perpendicular."},{"Start":"00:33.200 ","End":"00:36.860","Text":"The first thing I have to do is I have to draw its location vector."},{"Start":"00:36.860 ","End":"00:40.745","Text":"The location vector will be going from here until here."},{"Start":"00:40.745 ","End":"00:43.850","Text":"This is our r vector."},{"Start":"00:44.000 ","End":"00:48.740","Text":"now what we have to do to find the angular momentum is we\u0027re going to"},{"Start":"00:48.740 ","End":"00:52.985","Text":"write L our angular momentum is equal to,"},{"Start":"00:52.985 ","End":"00:54.560","Text":"we\u0027re doing the size right now,"},{"Start":"00:54.560 ","End":"00:59.278","Text":"so our linear momentum is mass times velocity,"},{"Start":"00:59.278 ","End":"01:04.785","Text":"so MV_0 multiplied by the size of our I vector,"},{"Start":"01:04.785 ","End":"01:10.145","Text":"which is r, and then multiply by the sine of the angle between them."},{"Start":"01:10.145 ","End":"01:13.985","Text":"This angle over here is going to be equal to Alpha."},{"Start":"01:13.985 ","End":"01:15.920","Text":"We don\u0027t know what Alpha is currently,"},{"Start":"01:15.920 ","End":"01:18.860","Text":"we\u0027re going to say that is multiplied by sine of Alpha."},{"Start":"01:18.860 ","End":"01:22.610","Text":"But what can we do in order to find out what this is?"},{"Start":"01:22.610 ","End":"01:27.045","Text":"We\u0027ll, notice that we know what the size of this length is,"},{"Start":"01:27.045 ","End":"01:31.850","Text":"what this side is, and we know what the size of this side is."},{"Start":"01:31.850 ","End":"01:36.470","Text":"Now, what we can do is we can use the handy SOHCAHTOA,"},{"Start":"01:36.470 ","End":"01:43.500","Text":"remember, SOH CAH TOA."},{"Start":"01:43.500 ","End":"01:46.275","Text":"What we have is the opposite."},{"Start":"01:46.275 ","End":"01:49.055","Text":"This is the opposite side to the angle,"},{"Start":"01:49.055 ","End":"01:50.870","Text":"and this is the longest side."},{"Start":"01:50.870 ","End":"01:53.570","Text":"This is the hypotenuse side."},{"Start":"01:53.570 ","End":"01:58.865","Text":"We can see that sine is"},{"Start":"01:58.865 ","End":"02:03.950","Text":"sine of the angle is equal to opposite over hypotenuse."},{"Start":"02:03.950 ","End":"02:07.370","Text":"We know that that means therefore that"},{"Start":"02:07.370 ","End":"02:11.600","Text":"our sine Alpha is equal to our opposite over hypotenuse,"},{"Start":"02:11.600 ","End":"02:16.475","Text":"which is d over the size of our r vector,"},{"Start":"02:16.475 ","End":"02:21.980","Text":"d over r. That means that now we can just say that this is equal"},{"Start":"02:21.980 ","End":"02:31.365","Text":"to MV_0r multiplied by d over r. These 2 rs cross out."},{"Start":"02:31.365 ","End":"02:40.655","Text":"Then therefore we can say that our angular momentum is equal to MV_0 multiplied by d,"},{"Start":"02:40.655 ","End":"02:44.795","Text":"where d is the distance from the axis of rotation."},{"Start":"02:44.795 ","End":"02:48.500","Text":"If we draw a line perpendicular to the line of travel,"},{"Start":"02:48.500 ","End":"02:53.360","Text":"that is our distance d. What this is is"},{"Start":"02:53.360 ","End":"02:59.165","Text":"the angular momentum of a body traveling in a straight line always."},{"Start":"02:59.165 ","End":"03:03.895","Text":"Every single body traveling in straight line has this angular momentum."},{"Start":"03:03.895 ","End":"03:07.970","Text":"Now, this is also known as r effective."},{"Start":"03:07.970 ","End":"03:11.330","Text":"The most important thing to remember about this,"},{"Start":"03:11.330 ","End":"03:13.985","Text":"is that when you\u0027ve got your axis of rotation,"},{"Start":"03:13.985 ","End":"03:15.710","Text":"you draw a straight line,"},{"Start":"03:15.710 ","End":"03:19.945","Text":"which is at 90 degrees to the direction of travel,"},{"Start":"03:19.945 ","End":"03:21.920","Text":"that\u0027s the most important."},{"Start":"03:21.920 ","End":"03:23.405","Text":"I hear you asking,"},{"Start":"03:23.405 ","End":"03:24.860","Text":"or some of you asking,"},{"Start":"03:24.860 ","End":"03:27.245","Text":"why does a body traveling in"},{"Start":"03:27.245 ","End":"03:31.850","Text":"a straight line have an axis of rotation and it\u0027s traveling in a straight line?"},{"Start":"03:31.850 ","End":"03:34.985","Text":"How does it have angular momentum? Let\u0027s go over this."},{"Start":"03:34.985 ","End":"03:39.960","Text":"If you imagine that this is you standing over here and this body is in an airplane,"},{"Start":"03:39.960 ","End":"03:43.730","Text":"let\u0027s say, and you\u0027re standing over here and you\u0027re looking at the airplane."},{"Start":"03:43.730 ","End":"03:47.330","Text":"The angle of where the airplane is,"},{"Start":"03:47.330 ","End":"03:50.240","Text":"is changing relative to you all the time."},{"Start":"03:50.240 ","End":"03:53.255","Text":"You can see that when the airplane is here,"},{"Start":"03:53.255 ","End":"03:55.960","Text":"you\u0027re looking like this,"},{"Start":"03:55.960 ","End":"03:59.255","Text":"when the airplane moves over here,"},{"Start":"03:59.255 ","End":"04:01.505","Text":"you\u0027ll be looking in this direction."},{"Start":"04:01.505 ","End":"04:04.730","Text":"When the airplane is over here,"},{"Start":"04:04.730 ","End":"04:07.010","Text":"you\u0027ll be looking in this direction,"},{"Start":"04:07.010 ","End":"04:09.000","Text":"and so on and so forth."},{"Start":"04:09.000 ","End":"04:14.255","Text":"You\u0027ll carry on, rotating about this axis of rotation."},{"Start":"04:14.255 ","End":"04:20.569","Text":"As you can see, this is going in an angular direction,"},{"Start":"04:20.569 ","End":"04:23.960","Text":"and that\u0027s why there is angular momentum relative"},{"Start":"04:23.960 ","End":"04:28.380","Text":"to the axis of rotation relative to you."}],"ID":12699},{"Watched":false,"Name":"1.3 Connection Between Angular Momentum and Torque Conservation","Duration":"3m 16s","ChapterTopicVideoID":12224,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.195","Text":"Now another very important equation to add to your formula sheet is that"},{"Start":"00:06.195 ","End":"00:12.730","Text":"the sum of all of the torques is equal to dL,"},{"Start":"00:12.730 ","End":"00:17.220","Text":"the change in angular momentum by dt."},{"Start":"00:17.220 ","End":"00:20.490","Text":"This comes from similar thing,"},{"Start":"00:20.490 ","End":"00:29.985","Text":"such as the sum of all of the forces is equal to the change in the linear momentum by dt."},{"Start":"00:29.985 ","End":"00:35.189","Text":"Now, another important fact is if the sum of all of the torques,"},{"Start":"00:35.189 ","End":"00:38.205","Text":"but the external torques."},{"Start":"00:38.205 ","End":"00:42.545","Text":"External is equal to 0,"},{"Start":"00:42.545 ","End":"00:45.170","Text":"then angular momentum is conserved."},{"Start":"00:45.170 ","End":"00:50.060","Text":"The sum of all of the external torques is equal to"},{"Start":"00:50.060 ","End":"00:55.495","Text":"0 that equates to conservation of angular momentum."},{"Start":"00:55.495 ","End":"00:58.445","Text":"Now I just want to show that this equation,"},{"Start":"00:58.445 ","End":"01:03.950","Text":"L is equal to mv0d is an equation that we can use all the time."},{"Start":"01:03.950 ","End":"01:11.325","Text":"Let\u0027s go back to this example over here and we\u0027ll work out now once the body has moved."},{"Start":"01:11.325 ","End":"01:13.535","Text":"Again from the axis of rotation,"},{"Start":"01:13.535 ","End":"01:16.805","Text":"I\u0027m going to draw its position vector,"},{"Start":"01:16.805 ","End":"01:18.875","Text":"which is going to be r tag."},{"Start":"01:18.875 ","End":"01:24.320","Text":"Then I\u0027m going to label this angle over here as Alpha tag."},{"Start":"01:24.320 ","End":"01:27.525","Text":"Now if we again write out our equation."},{"Start":"01:27.525 ","End":"01:33.170","Text":"We\u0027ll have L or angular momentum is equal to linear momentum,"},{"Start":"01:33.170 ","End":"01:34.825","Text":"which is mv_0,"},{"Start":"01:34.825 ","End":"01:38.930","Text":"and then multiplied by the position vector."},{"Start":"01:38.930 ","End":"01:42.365","Text":"It\u0027s r tag, the size of it,"},{"Start":"01:42.365 ","End":"01:45.510","Text":"which is just going to be r tag,"},{"Start":"01:45.830 ","End":"01:50.615","Text":"multiplied by sine of Alpha tag."},{"Start":"01:50.615 ","End":"01:53.105","Text":"Now, again, we don\u0027t know what Alpha tag is."},{"Start":"01:53.105 ","End":"01:59.425","Text":"However, we can take a look that if this is our angle Alpha,"},{"Start":"01:59.425 ","End":"02:04.070","Text":"then again our d is going to be our opposite,"},{"Start":"02:04.070 ","End":"02:07.925","Text":"and r tag is going to be our hypotenuse."},{"Start":"02:07.925 ","End":"02:12.140","Text":"Again, sine of Alpha is going to be opposite over hypotenuse,"},{"Start":"02:12.140 ","End":"02:14.840","Text":"which is going to be d/r tag."},{"Start":"02:14.840 ","End":"02:19.235","Text":"This is going to be equal to d/r tag."},{"Start":"02:19.235 ","End":"02:22.160","Text":"Then you can see that our r tags are going to cross out."},{"Start":"02:22.160 ","End":"02:25.265","Text":"Which means that once again we\u0027ll get mv_0."},{"Start":"02:25.265 ","End":"02:30.650","Text":"The I tags cross off multiplied by d. We can see"},{"Start":"02:30.650 ","End":"02:35.990","Text":"that it doesn\u0027t matter where our body is in relation to our axes of rotation,"},{"Start":"02:35.990 ","End":"02:39.160","Text":"the same equation applies."},{"Start":"02:39.160 ","End":"02:41.465","Text":"This is important to note."},{"Start":"02:41.465 ","End":"02:46.115","Text":"If a body is traveling in a straight line at a constant velocity,"},{"Start":"02:46.115 ","End":"02:52.405","Text":"then that means that the sum of all of the forces is going to be equal to 0."},{"Start":"02:52.405 ","End":"02:55.070","Text":"If the sum of all of the forces is equal to 0,"},{"Start":"02:55.070 ","End":"02:56.825","Text":"then the sum of the moments,"},{"Start":"02:56.825 ","End":"03:00.065","Text":"the sum of the torques, is also going to be equal to 0,"},{"Start":"03:00.065 ","End":"03:02.930","Text":"which means that our angular momentum is conserved."},{"Start":"03:02.930 ","End":"03:07.955","Text":"That means that the angular momentum of the body over here and over here,"},{"Start":"03:07.955 ","End":"03:10.190","Text":"and at every single point in-between,"},{"Start":"03:10.190 ","End":"03:17.490","Text":"and also further out of the page is also going to be exactly the same."}],"ID":12700},{"Watched":false,"Name":"1.4 How to Answer Questions","Duration":"2m 6s","ChapterTopicVideoID":12225,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.950","Text":"Now let\u0027s give a little list that will help us to solve some of the questions."},{"Start":"00:04.950 ","End":"00:09.105","Text":"A lot of times solving questions is easier if we know what is being conserved."},{"Start":"00:09.105 ","End":"00:12.405","Text":"In order to know if our momentum is being conserved,"},{"Start":"00:12.405 ","End":"00:18.375","Text":"we have to ask ourselves if the sum of all of the external forces,"},{"Start":"00:18.375 ","End":"00:22.035","Text":"external forces very important, is equal to 0."},{"Start":"00:22.035 ","End":"00:25.875","Text":"If the sum of all of the external forces is not equal to 0,"},{"Start":"00:25.875 ","End":"00:28.125","Text":"then momentum is not being conserved."},{"Start":"00:28.125 ","End":"00:29.715","Text":"What are the external forces?"},{"Start":"00:29.715 ","End":"00:37.320","Text":"If the body is accelerating then the external forces are not equal to 0."},{"Start":"00:37.320 ","End":"00:43.790","Text":"Inside the body, the forces might be accelerating when traveling at a constant velocity,"},{"Start":"00:43.790 ","End":"00:44.920","Text":"whatever it might be."},{"Start":"00:44.920 ","End":"00:47.645","Text":"Let\u0027s say in the body of a human,"},{"Start":"00:47.645 ","End":"00:49.790","Text":"the blood is flowing through the body."},{"Start":"00:49.790 ","End":"00:52.640","Text":"That doesn\u0027t make a difference if I\u0027m running down"},{"Start":"00:52.640 ","End":"00:55.940","Text":"the pitch and I\u0027m accelerating because I\u0027m in a 100-meter sprint."},{"Start":"00:55.940 ","End":"00:58.175","Text":"It\u0027s the external forces."},{"Start":"00:58.175 ","End":"01:01.655","Text":"What\u0027s happening to the cells inside your body doesn\u0027t matter,"},{"Start":"01:01.655 ","End":"01:05.735","Text":"it\u0027s outside if your body is moving at a constant velocity,"},{"Start":"01:05.735 ","End":"01:08.330","Text":"is stationary, or is accelerating."},{"Start":"01:08.330 ","End":"01:11.750","Text":"Where the energy is a conservation of forces,"},{"Start":"01:11.750 ","End":"01:13.535","Text":"for energy to be conserved,"},{"Start":"01:13.535 ","End":"01:15.605","Text":"we have to have all of the forces,"},{"Start":"01:15.605 ","End":"01:20.095","Text":"external and internal forces being conserved."},{"Start":"01:20.095 ","End":"01:22.660","Text":"Now of course, for angular momentum,"},{"Start":"01:22.660 ","End":"01:24.460","Text":"for it to be conserved,"},{"Start":"01:24.460 ","End":"01:33.030","Text":"we have to check if the external moments are equal to 0."},{"Start":"01:33.030 ","End":"01:37.285","Text":"The sum of all of the external moments are equal to 0 and this corresponds"},{"Start":"01:37.285 ","End":"01:44.380","Text":"to what is written in this box over here."},{"Start":"01:44.380 ","End":"01:47.190","Text":"A tip for answering questions."},{"Start":"01:47.190 ","End":"01:49.900","Text":"In order to get all of your equations and in order to"},{"Start":"01:49.900 ","End":"01:52.885","Text":"find out all of your unknowns, first,"},{"Start":"01:52.885 ","End":"01:56.140","Text":"you can take a look and see if anything is conserved,"},{"Start":"01:56.140 ","End":"01:59.300","Text":"whether it\u0027s momentum, energy or angular momentum."},{"Start":"01:59.300 ","End":"02:03.770","Text":"Then that gives you an extra equation which will help you solve it."},{"Start":"02:03.870 ","End":"02:07.160","Text":"That\u0027s the end of this lesson."}],"ID":12701},{"Watched":false,"Name":"Exercise- Ball Rotating","Duration":"19m 1s","ChapterTopicVideoID":9123,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.010 ","End":"00:06.435","Text":"In this lesson, we\u0027re being given an example to solve."},{"Start":"00:06.435 ","End":"00:08.610","Text":"Now we\u0027re being told that we have a ball of"},{"Start":"00:08.610 ","End":"00:13.470","Text":"mass m. This is the ball of mass m and it\u0027s attached"},{"Start":"00:13.470 ","End":"00:16.410","Text":"to a string of length l. This is"},{"Start":"00:16.410 ","End":"00:20.855","Text":"the string and it\u0027s rotating in a circle perpendicular to the ground."},{"Start":"00:20.855 ","End":"00:24.535","Text":"This over here is the ground."},{"Start":"00:24.535 ","End":"00:29.355","Text":"The velocity of the ball at its maximum height is v0,"},{"Start":"00:29.355 ","End":"00:32.220","Text":"and now we\u0027re being told to find the torque."},{"Start":"00:32.220 ","End":"00:33.675","Text":"In question number 1,"},{"Start":"00:33.675 ","End":"00:40.205","Text":"we\u0027re being told to find the torque acting on the ball as a function of the angle Alpha."},{"Start":"00:40.205 ","End":"00:43.500","Text":"We can see that this is the angle Alpha."},{"Start":"00:43.670 ","End":"00:47.690","Text":"Let\u0027s see how we are going to solve this."},{"Start":"00:47.690 ","End":"00:53.340","Text":"The first thing I\u0027m going to do is I\u0027m going to label my forces."},{"Start":"00:53.340 ","End":"00:59.475","Text":"Because this is the string so the ball has just moved down the circle."},{"Start":"00:59.475 ","End":"01:01.995","Text":"This is also of length l,"},{"Start":"01:01.995 ","End":"01:05.675","Text":"and we have over here in the string,"},{"Start":"01:05.675 ","End":"01:15.840","Text":"we have T for tension and we have acting downwards our mass times our gravity."},{"Start":"01:16.430 ","End":"01:19.475","Text":"Before we start working this out,"},{"Start":"01:19.475 ","End":"01:26.790","Text":"we\u0027re going to say that the positive direction is going to be anticlockwise."},{"Start":"01:27.170 ","End":"01:34.060","Text":"Now we know that our equation for torque is equal to,"},{"Start":"01:34.060 ","End":"01:36.995","Text":"if we\u0027re going to do the size,"},{"Start":"01:36.995 ","End":"01:43.820","Text":"so it equals to the size of our I vector multiplied by the size of our force vector,"},{"Start":"01:43.820 ","End":"01:48.300","Text":"multiplied by sine of the angle between the 2."},{"Start":"01:48.560 ","End":"01:51.135","Text":"Let\u0027s see what this is over here."},{"Start":"01:51.135 ","End":"01:52.700","Text":"We\u0027re going to write that the torque,"},{"Start":"01:52.700 ","End":"01:55.880","Text":"so it\u0027s actually going to be the sum of the torques."},{"Start":"01:55.880 ","End":"01:58.165","Text":"Is going to be equal to what?"},{"Start":"01:58.165 ","End":"02:00.510","Text":"We have to find our I vector."},{"Start":"02:00.510 ","End":"02:05.645","Text":"Our I vector is going from the center because this is our axis of rotation."},{"Start":"02:05.645 ","End":"02:13.220","Text":"It\u0027s going to go all the way up until it gets to the ball in this direction."},{"Start":"02:13.220 ","End":"02:18.695","Text":"Now it\u0027s size is l. We know that because that\u0027s the length of the string,"},{"Start":"02:18.695 ","End":"02:27.205","Text":"and it cancels out with our T because our T is also going all the way until here."},{"Start":"02:27.205 ","End":"02:30.210","Text":"This is our T and it\u0027s equal and opposite."},{"Start":"02:30.210 ","End":"02:32.070","Text":"Their directions cancel out,"},{"Start":"02:32.070 ","End":"02:35.645","Text":"which means that when I\u0027m working out this equation,"},{"Start":"02:35.645 ","End":"02:38.215","Text":"when I\u0027m going to now draw my vector diagram,"},{"Start":"02:38.215 ","End":"02:42.710","Text":"in order to find out the direction of the torque,"},{"Start":"02:42.710 ","End":"02:45.635","Text":"I don\u0027t need to add in T because it\u0027s canceled out"},{"Start":"02:45.635 ","End":"02:49.370","Text":"by my I because they are on equal opposite directions."},{"Start":"02:49.370 ","End":"02:56.370","Text":"I have my T multiplied by my force. What is my force?"},{"Start":"02:56.370 ","End":"03:03.710","Text":"It\u0027s mg, so multiplied by mg multiplied by sine of the angle between the 2,"},{"Start":"03:03.710 ","End":"03:05.270","Text":"which, as we can see,"},{"Start":"03:05.270 ","End":"03:09.090","Text":"because these 2 lines are parallel,"},{"Start":"03:09.090 ","End":"03:11.505","Text":"so we know that this is also Alpha."},{"Start":"03:11.505 ","End":"03:13.800","Text":"This is alpha, and this is Alpha."},{"Start":"03:13.800 ","End":"03:18.900","Text":"It\u0027s going to be multiplied by sine of Alpha."},{"Start":"03:18.900 ","End":"03:22.160","Text":"Now in order for us to know which direction it\u0027s going in,"},{"Start":"03:22.160 ","End":"03:25.130","Text":"so if I draw my I vector,"},{"Start":"03:25.130 ","End":"03:28.250","Text":"so we can see it\u0027s going in this direction,"},{"Start":"03:28.250 ","End":"03:30.780","Text":"and mg vector,"},{"Start":"03:30.780 ","End":"03:33.690","Text":"my force vector is pointing downwards."},{"Start":"03:33.690 ","End":"03:35.570","Text":"To get from my I and my mg,"},{"Start":"03:35.570 ","End":"03:36.590","Text":"I have to go this way,"},{"Start":"03:36.590 ","End":"03:38.365","Text":"which is clockwise,"},{"Start":"03:38.365 ","End":"03:40.910","Text":"which is the negative direction because we said"},{"Start":"03:40.910 ","End":"03:43.310","Text":"that anticlockwise was the positive direction."},{"Start":"03:43.310 ","End":"03:46.590","Text":"I added a negative over here."},{"Start":"03:48.980 ","End":"03:54.340","Text":"Now we\u0027re going to take a look at the second question."},{"Start":"03:54.340 ","End":"03:57.110","Text":"Here we\u0027re being told that we have to find"},{"Start":"03:57.110 ","End":"04:00.755","Text":"the angular momentum of the ball as a function of Alpha."},{"Start":"04:00.755 ","End":"04:07.085","Text":"As we know that the size of the angular momentum is equal to"},{"Start":"04:07.085 ","End":"04:14.675","Text":"I position vector multiplied by I linear momentum vector,"},{"Start":"04:14.675 ","End":"04:19.680","Text":"and again multiplied by sine of the angle between the 2."},{"Start":"04:21.110 ","End":"04:26.840","Text":"As we know, our velocity vector is always going to be in the direction which"},{"Start":"04:26.840 ","End":"04:32.040","Text":"is tangential and perpendicular to our radius."},{"Start":"04:32.040 ","End":"04:34.755","Text":"Here our radius in this direction."},{"Start":"04:34.755 ","End":"04:38.440","Text":"Our velocity vector will be perpendicular to it,"},{"Start":"04:38.440 ","End":"04:41.795","Text":"and it will also be at a tangent to the circle."},{"Start":"04:41.795 ","End":"04:46.965","Text":"This is our velocity at some angle Alpha."},{"Start":"04:46.965 ","End":"04:50.855","Text":"Now we\u0027re told in the question that at maximum height,"},{"Start":"04:50.855 ","End":"04:52.700","Text":"the velocity is v0."},{"Start":"04:52.700 ","End":"04:55.835","Text":"As you know, because of conservation of energy,"},{"Start":"04:55.835 ","End":"05:02.765","Text":"so here it will have the most potential energy and the least kinetic energy,"},{"Start":"05:02.765 ","End":"05:04.550","Text":"and as it goes down,"},{"Start":"05:04.550 ","End":"05:08.810","Text":"the potential energy and the kinetic energy will eventually balance each other"},{"Start":"05:08.810 ","End":"05:12.650","Text":"out until it reaches the bottom where there\u0027ll be the least potential energy,"},{"Start":"05:12.650 ","End":"05:14.300","Text":"but the most kinetic energy."},{"Start":"05:14.300 ","End":"05:17.160","Text":"Then go back in the circle like that."},{"Start":"05:20.150 ","End":"05:27.850","Text":"The velocity is changing as a function of the angle depending on where it is."},{"Start":"05:29.120 ","End":"05:31.250","Text":"We\u0027re going to write this here,"},{"Start":"05:31.250 ","End":"05:35.690","Text":"and this is at 90 degrees to the radius."},{"Start":"05:35.690 ","End":"05:38.045","Text":"Now we\u0027ll write down that"},{"Start":"05:38.045 ","End":"05:46.280","Text":"our angular momentum is equal to the size of our position vector,"},{"Start":"05:46.280 ","End":"05:50.825","Text":"which again the length of the string is l. It\u0027s multiplied by l,"},{"Start":"05:50.825 ","End":"05:55.475","Text":"multiplied by the size of the momentum vector,"},{"Start":"05:55.475 ","End":"06:04.670","Text":"which is going to be multiplied by m multiplied by velocity,"},{"Start":"06:04.670 ","End":"06:08.585","Text":"which is changing as a function of the angle."},{"Start":"06:08.585 ","End":"06:13.610","Text":"Then it\u0027s going to be multiplied by sine of the angle between them,"},{"Start":"06:13.610 ","End":"06:15.785","Text":"which has sine of 90 degrees,"},{"Start":"06:15.785 ","End":"06:17.060","Text":"which as we know,"},{"Start":"06:17.060 ","End":"06:19.765","Text":"is equal to 1."},{"Start":"06:19.765 ","End":"06:26.885","Text":"I\u0027ve put here a little reminder that the linear momentum is equal to mass times velocity."},{"Start":"06:26.885 ","End":"06:34.820","Text":"This is going to be equal to lmv as a function of Alpha multiplied by 1,"},{"Start":"06:34.820 ","End":"06:37.505","Text":"so it\u0027s just lmv as a function of Alpha."},{"Start":"06:37.505 ","End":"06:42.655","Text":"Now, let\u0027s see which direction it\u0027s going to be going in."},{"Start":"06:42.655 ","End":"06:49.175","Text":"If we say that our positive direction as the anticlockwise direction,"},{"Start":"06:49.175 ","End":"06:51.830","Text":"so if you lift up your right hand and"},{"Start":"06:51.830 ","End":"06:56.750","Text":"you curl your fingers in the anticlockwise direction,"},{"Start":"06:56.750 ","End":"06:59.840","Text":"you\u0027ll see that the positive direction is when"},{"Start":"06:59.840 ","End":"07:04.205","Text":"the z-axis is coming out of the screen and towards you."},{"Start":"07:04.205 ","End":"07:11.210","Text":"Now what we can see it is because the velocity is moving in this direction,"},{"Start":"07:11.210 ","End":"07:17.074","Text":"so we can see that the ball is moving in the clockwise direction,"},{"Start":"07:17.074 ","End":"07:19.145","Text":"which is the negative direction."},{"Start":"07:19.145 ","End":"07:24.060","Text":"That means that we\u0027re going in the negative z-direction."},{"Start":"07:24.060 ","End":"07:28.125","Text":"I\u0027ll just write here negative z hat."},{"Start":"07:28.125 ","End":"07:34.060","Text":"Here you see that if you curl your fingers in the anticlockwise direction,"},{"Start":"07:34.060 ","End":"07:37.300","Text":"then your thumb is pointing in the negative z-direction."},{"Start":"07:37.300 ","End":"07:38.980","Text":"That\u0027s where this comes from."},{"Start":"07:38.980 ","End":"07:40.825","Text":"Now another way of doing this"},{"Start":"07:40.825 ","End":"07:46.060","Text":"is you can write out that your angular momentum as you know,"},{"Start":"07:46.060 ","End":"07:52.735","Text":"it\u0027s also equal to the r vector cross product with the momentum vector."},{"Start":"07:52.735 ","End":"07:57.280","Text":"We always use this equation for the sizes and then afterwards using"},{"Start":"07:57.280 ","End":"08:01.615","Text":"the right-hand rule to find the direction because a lot of the time it\u0027s actually easier."},{"Start":"08:01.615 ","End":"08:06.860","Text":"But let\u0027s go over the other way because it\u0027s good that you know how to use both ways."},{"Start":"08:06.860 ","End":"08:12.205","Text":"Our r vector is actually going to be, so what is it?"},{"Start":"08:12.205 ","End":"08:17.320","Text":"We know that we have our r hat going in this direction,"},{"Start":"08:17.320 ","End":"08:18.879","Text":"in the radial direction."},{"Start":"08:18.879 ","End":"08:23.710","Text":"Then always perpendicular to the radial direction will have the Theta hat,"},{"Start":"08:23.710 ","End":"08:25.255","Text":"in the Theta direction."},{"Start":"08:25.255 ","End":"08:29.725","Text":"We\u0027re working this out in polar coordinates with r, Theta and z."},{"Start":"08:29.725 ","End":"08:33.400","Text":"Specifically in this question we\u0027re only dealing with r and Theta."},{"Start":"08:33.400 ","End":"08:36.220","Text":"Theta is always going to be perpendicular to our"},{"Start":"08:36.220 ","End":"08:40.170","Text":"r. Now we just have to see in which direction,"},{"Start":"08:40.170 ","End":"08:43.560","Text":"if it\u0027s going to be pointing in this direction or in this direction."},{"Start":"08:43.560 ","End":"08:48.082","Text":"Now, because we\u0027ve said that our positive direction is in the clockwise direction,"},{"Start":"08:48.082 ","End":"08:51.895","Text":"so our Theta is going to be pointing in this direction."},{"Start":"08:51.895 ","End":"08:54.130","Text":"This is our Theta hat."},{"Start":"08:54.130 ","End":"08:56.665","Text":"Then when we\u0027re going from r to Theta,"},{"Start":"08:56.665 ","End":"08:59.350","Text":"it\u0027s going in the anticlockwise direction"},{"Start":"08:59.350 ","End":"09:02.005","Text":"which here we\u0027ve said is the positive direction."},{"Start":"09:02.005 ","End":"09:04.825","Text":"It always goes from r to Theta."},{"Start":"09:04.825 ","End":"09:13.300","Text":"Our r vector is going to be rl multiplied in the direction of our r hat."},{"Start":"09:13.300 ","End":"09:22.585","Text":"Then with a cross product with our momentum which is going to be our p vector,"},{"Start":"09:22.585 ","End":"09:28.435","Text":"which is going to be multiplied by mv as a function of Alpha."},{"Start":"09:28.435 ","End":"09:31.675","Text":"What direction is this?"},{"Start":"09:31.675 ","End":"09:36.970","Text":"We can see that it\u0027s perpendicular to the radius so that we know."},{"Start":"09:36.970 ","End":"09:41.095","Text":"But we can see that it\u0027s going in the opposite direction to our Theta hat."},{"Start":"09:41.095 ","End":"09:42.850","Text":"Our Theta hat is pointing upwards in"},{"Start":"09:42.850 ","End":"09:46.795","Text":"this direction and our velocity is pointing downwards in this direction."},{"Start":"09:46.795 ","End":"09:51.445","Text":"It\u0027s going to be in the negative Theta hat direction."},{"Start":"09:51.445 ","End":"09:53.230","Text":"What does that mean?"},{"Start":"09:53.230 ","End":"09:56.800","Text":"That means that in polar coordinates,"},{"Start":"09:56.800 ","End":"10:01.435","Text":"in the r direction we have l and then in the Theta direction,"},{"Start":"10:01.435 ","End":"10:03.535","Text":"and the z-direction we have 0."},{"Start":"10:03.535 ","End":"10:09.490","Text":"Then we have this as a cross product with our mv Alpha in the negative Theta direction."},{"Start":"10:09.490 ","End":"10:16.705","Text":"In this r direction we have 0 then we have negative mv as a function of Alpha."},{"Start":"10:16.705 ","End":"10:19.900","Text":"Then in the z-direction again we have 0."},{"Start":"10:19.900 ","End":"10:23.590","Text":"Then when we do the cross products,"},{"Start":"10:23.590 ","End":"10:26.620","Text":"then what we will be left with is"},{"Start":"10:26.620 ","End":"10:35.140","Text":"negative mvl in the z-direction which is equal to,"},{"Start":"10:35.140 ","End":"10:36.970","Text":"if you want to look at it in vector format,"},{"Start":"10:36.970 ","End":"10:41.270","Text":"it\u0027s going to be 0, 0 and then negative mvl."},{"Start":"10:43.530 ","End":"10:49.880","Text":"Obviously this v over here and also over here is as a function of Alpha."},{"Start":"10:50.160 ","End":"10:55.270","Text":"Now to those of you that don\u0027t remember how to do the cross product,"},{"Start":"10:55.270 ","End":"10:58.690","Text":"please go back to the lesson where I go through this."},{"Start":"10:58.690 ","End":"11:05.935","Text":"It\u0027s really not very complicated and it\u0027s very important that you go over."},{"Start":"11:05.935 ","End":"11:09.775","Text":"Now how are we going to find our v?"},{"Start":"11:09.775 ","End":"11:14.170","Text":"What can do is through the conservation of energy,"},{"Start":"11:14.170 ","End":"11:19.940","Text":"we can say that our energy is always going to be equal to."},{"Start":"11:20.070 ","End":"11:27.520","Text":"When the ball is right at the top it\u0027s going to be half m its velocity at the top,"},{"Start":"11:27.520 ","End":"11:31.480","Text":"which is v_0^2 plus its potential energy,"},{"Start":"11:31.480 ","End":"11:33.745","Text":"which is going to be multiplied by"},{"Start":"11:33.745 ","End":"11:39.970","Text":"mg. Then we said because the height is equal to 0 at the axis of rotation,"},{"Start":"11:39.970 ","End":"11:47.380","Text":"so this height right at the top is going to be l, so plus mgl."},{"Start":"11:47.380 ","End":"11:54.055","Text":"This is going to be equal to the energy that it will have when it is at this point."},{"Start":"11:54.055 ","End":"11:56.485","Text":"Because of the conservation of energy."},{"Start":"11:56.485 ","End":"12:01.520","Text":"This is going to be equal to half mv^2."},{"Start":"12:02.100 ","End":"12:10.720","Text":"This is the v. This is the v that we want to find out plus,"},{"Start":"12:10.720 ","End":"12:15.640","Text":"and now we have to find out what our potential energy over here will be."},{"Start":"12:15.640 ","End":"12:20.710","Text":"The potential energy is going to be mg. Then"},{"Start":"12:20.710 ","End":"12:26.040","Text":"we have to find what this height is over here."},{"Start":"12:26.040 ","End":"12:31.000","Text":"This is our height. How do we do this?"},{"Start":"12:31.000 ","End":"12:34.120","Text":"We know that the hypotenuse is r radius,"},{"Start":"12:34.120 ","End":"12:40.315","Text":"which is l. This is the angle and we\u0027re trying to find the adjacent angle."},{"Start":"12:40.315 ","End":"12:47.590","Text":"That\u0027s will be cos of the angle is equal to the height,"},{"Start":"12:47.590 ","End":"12:52.060","Text":"h that we\u0027re trying to find over our hypotenuse,"},{"Start":"12:52.060 ","End":"12:57.280","Text":"which is our l. Therefore,"},{"Start":"12:57.280 ","End":"13:02.770","Text":"l cosine of Alpha will equal to the height that we\u0027re trying to find."},{"Start":"13:02.770 ","End":"13:08.330","Text":"Then we\u0027ll have here l cosine of Alpha."},{"Start":"13:08.580 ","End":"13:12.250","Text":"Now what we can do is from both sides,"},{"Start":"13:12.250 ","End":"13:17.650","Text":"we can just divide through by this m because it\u0027s a common factor, it doesn\u0027t matter."},{"Start":"13:17.650 ","End":"13:22.285","Text":"Then we can also multiply everything by 2."},{"Start":"13:22.285 ","End":"13:25.719","Text":"Let\u0027s scroll down to give us a little bit more space."},{"Start":"13:25.719 ","End":"13:34.600","Text":"Then we\u0027ll have that our v_0^2 plus 2gl is going to be equal to v^2,"},{"Start":"13:34.600 ","End":"13:42.625","Text":"which is unknown, plus 2gl cosine of Alpha."},{"Start":"13:42.625 ","End":"13:45.910","Text":"Now what we\u0027re going to do is we\u0027re going to isolate out"},{"Start":"13:45.910 ","End":"13:49.090","Text":"this v because that\u0027s where we want to do."},{"Start":"13:49.090 ","End":"13:57.010","Text":"We\u0027re going to say that v^2 is equal to v_0^2 plus 2gl,"},{"Start":"13:57.010 ","End":"13:59.290","Text":"which is a common factor,"},{"Start":"13:59.290 ","End":"14:03.895","Text":"1 minus cosine of Alpha."},{"Start":"14:03.895 ","End":"14:07.330","Text":"I literally just did very easy algebra."},{"Start":"14:07.330 ","End":"14:10.540","Text":"I isolated out my v^2."},{"Start":"14:10.540 ","End":"14:12.775","Text":"Then in order to just get v,"},{"Start":"14:12.775 ","End":"14:19.465","Text":"I\u0027ll just rub out this squared and I\u0027ll just square root both sides."},{"Start":"14:19.465 ","End":"14:24.190","Text":"This is our v. Now,"},{"Start":"14:24.190 ","End":"14:27.370","Text":"because in the question they asked me to find out what"},{"Start":"14:27.370 ","End":"14:31.270","Text":"my angular momentum was as a function of Alpha."},{"Start":"14:31.270 ","End":"14:33.985","Text":"Here we have my v as a function of Alpha."},{"Start":"14:33.985 ","End":"14:38.365","Text":"I\u0027m going to substitute it in to this equation over here."},{"Start":"14:38.365 ","End":"14:41.215","Text":"Then let\u0027s scroll down a little bit more."},{"Start":"14:41.215 ","End":"14:51.265","Text":"We\u0027ll have that our angular momentum is equal to negative lm v as a function of Alpha."},{"Start":"14:51.265 ","End":"14:57.220","Text":"that is going to be v_0^2 plus 2gl,"},{"Start":"14:57.220 ","End":"15:02.190","Text":"1 minus cosine of Alpha."},{"Start":"15:02.190 ","End":"15:10.260","Text":"Then this is going to be the square root of all of that and in the direction of z hat."},{"Start":"15:10.500 ","End":"15:14.000","Text":"That is the end of this exercise."},{"Start":"15:14.000 ","End":"15:18.125","Text":"But let\u0027s just go back to question number 1."},{"Start":"15:18.125 ","End":"15:20.675","Text":"Let\u0027s just move along here."},{"Start":"15:20.675 ","End":"15:26.540","Text":"Now, what we\u0027ll do is going back to question number 1,"},{"Start":"15:26.540 ","End":"15:30.335","Text":"we\u0027ll see that our equation for Sigma Tau,"},{"Start":"15:30.335 ","End":"15:37.685","Text":"the sum of the torque is actually equal to what we know it is,"},{"Start":"15:37.685 ","End":"15:42.090","Text":"which is dl by dt."},{"Start":"15:42.390 ","End":"15:44.980","Text":"What is dl by dt?"},{"Start":"15:44.980 ","End":"15:55.495","Text":"It equals to dl by d Alpha multiplied by d Alpha by dt."},{"Start":"15:55.495 ","End":"15:57.895","Text":"That is the chain rule."},{"Start":"15:57.895 ","End":"16:02.480","Text":"I did the chain rule like this because I know that in my example"},{"Start":"16:02.480 ","End":"16:08.720","Text":"specifically that my angular momentum is with the variable Alpha."},{"Start":"16:09.330 ","End":"16:17.420","Text":"Let\u0027s do this. My dl by d Alpha is going to be equal to this."},{"Start":"16:17.420 ","End":"16:22.730","Text":"This is my l and I differentiated according to Alpha."},{"Start":"16:22.730 ","End":"16:30.145","Text":"This is going to be equal to negative ml divided by 2."},{"Start":"16:30.145 ","End":"16:36.545","Text":"Then instead of rewriting out this entire expression over here because it\u0027s really long."},{"Start":"16:36.545 ","End":"16:39.035","Text":"It equals to my v as a function of Alpha."},{"Start":"16:39.035 ","End":"16:42.650","Text":"I\u0027m just going to write v because it\u0027s this."},{"Start":"16:42.650 ","End":"16:46.625","Text":"Then that\u0027s going to be multiplied by"},{"Start":"16:46.625 ","End":"16:53.400","Text":"2gl sine of Alpha"},{"Start":"16:53.400 ","End":"16:58.620","Text":"and then multiplied by the derivative of Alpha."},{"Start":"16:59.370 ","End":"17:03.695","Text":"This whole expression of negative ml over"},{"Start":"17:03.695 ","End":"17:10.295","Text":"2v multiplied by 2gl sine of Alpha is our dl by d Alpha."},{"Start":"17:10.295 ","End":"17:14.390","Text":"This is the derivative of"},{"Start":"17:14.390 ","End":"17:19.160","Text":"the outside expression and this is the derivative of the inside expression."},{"Start":"17:19.160 ","End":"17:22.880","Text":"Then my Alpha dot is my d Alpha by dt."},{"Start":"17:22.880 ","End":"17:27.310","Text":"Now let\u0027s take a look at what Alpha dot means."},{"Start":"17:27.310 ","End":"17:29.540","Text":"What does Alpha dot mean?"},{"Start":"17:29.540 ","End":"17:32.560","Text":"Alpha dot is the change in angle."},{"Start":"17:32.560 ","End":"17:36.665","Text":"What does that mean? That\u0027s my angular velocity, my Omega."},{"Start":"17:36.665 ","End":"17:39.590","Text":"Because the angle is changing as a function of time,"},{"Start":"17:39.590 ","End":"17:42.125","Text":"which is exactly my angular velocity."},{"Start":"17:42.125 ","End":"17:48.795","Text":"Now, Omega is also equal to the velocity divided by the radius."},{"Start":"17:48.795 ","End":"17:53.515","Text":"Here we know that the radius is l. We\u0027ll substitute that in."},{"Start":"17:53.515 ","End":"17:57.575","Text":"It\u0027s v divided by l. Now we\u0027ll go back to"},{"Start":"17:57.575 ","End":"18:02.310","Text":"our expression and we\u0027ll get the sum of the torque is equal to"},{"Start":"18:02.310 ","End":"18:10.735","Text":"negative 2mlgl sine Alpha"},{"Start":"18:10.735 ","End":"18:15.580","Text":"divided by 2v multiplied by my Alpha dot,"},{"Start":"18:15.580 ","End":"18:22.120","Text":"so multiplied by v over l. Now we can cross off all like terms."},{"Start":"18:22.120 ","End":"18:24.160","Text":"We have 2 and 2 will cross out,"},{"Start":"18:24.160 ","End":"18:29.350","Text":"v and v will cross out. Here\u0027s an Alpha."},{"Start":"18:29.350 ","End":"18:33.175","Text":"Then 1 of the l\u0027s and this l will cross off."},{"Start":"18:33.175 ","End":"18:42.360","Text":"Then in the end, we\u0027ll be left with negative mlg sine of Alpha."},{"Start":"18:42.360 ","End":"18:46.610","Text":"Now, if we look to our answer of what we got in number 1,"},{"Start":"18:46.610 ","End":"18:52.760","Text":"we got that the sum of torque was equal to negative lmg sine Alpha."},{"Start":"18:52.760 ","End":"18:54.900","Text":"We got the exact same answer."},{"Start":"18:54.900 ","End":"18:58.480","Text":"Now you see that it works."},{"Start":"18:58.480 ","End":"19:01.400","Text":"That\u0027s the end of this lesson."}],"ID":12702},{"Watched":false,"Name":"Exercise- Ball In A Cone","Duration":"27m 48s","ChapterTopicVideoID":9124,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"Hello, in this lesson,"},{"Start":"00:02.370 ","End":"00:06.525","Text":"we\u0027re going to be speaking about a small ball which rolls through a cone,"},{"Start":"00:06.525 ","End":"00:09.060","Text":"which is attached to the ground via its tip."},{"Start":"00:09.060 ","End":"00:11.535","Text":"It\u0027s an upside-down cone,"},{"Start":"00:11.535 ","End":"00:13.260","Text":"and there\u0027s a ball inside."},{"Start":"00:13.260 ","End":"00:18.434","Text":"The initial velocity of the ball is v_0 in the horizontal direction,"},{"Start":"00:18.434 ","End":"00:20.520","Text":"tangent to the side of the cone."},{"Start":"00:20.520 ","End":"00:24.090","Text":"Right now, it\u0027s going to be into the page."},{"Start":"00:24.090 ","End":"00:29.419","Text":"We\u0027re being told that the initial height of the ball H is given over here,"},{"Start":"00:29.419 ","End":"00:34.605","Text":"and we\u0027re being asked to find the maximum height which the ball will reach."},{"Start":"00:34.605 ","End":"00:38.490","Text":"Now, we\u0027re being told that the cone is stationary,"},{"Start":"00:38.490 ","End":"00:42.105","Text":"so this whole system isn\u0027t moving, just the ball is,"},{"Start":"00:42.105 ","End":"00:46.690","Text":"and that a cubic equation will be accepted as an answer."},{"Start":"00:46.690 ","End":"00:52.160","Text":"How we\u0027re going to solve this is by using the idea of conservation of energy."},{"Start":"00:52.160 ","End":"00:56.060","Text":"However, let\u0027s speak about the dynamics of the situation so that you can"},{"Start":"00:56.060 ","End":"01:01.230","Text":"understand also another way to solve this. How can I do this?"},{"Start":"01:01.230 ","End":"01:06.020","Text":"What I could do is a free body diagram."},{"Start":"01:06.020 ","End":"01:10.940","Text":"I\u0027ll have my mg pointing downwards as per usual."},{"Start":"01:10.940 ","End":"01:12.935","Text":"Then I\u0027ll have my normal,"},{"Start":"01:12.935 ","End":"01:18.005","Text":"which is going to be at 90 degrees to the wall of the cone."},{"Start":"01:18.005 ","End":"01:26.970","Text":"Now, I can say that either the sum of all of my forces is going to be equal to mw^2r,"},{"Start":"01:26.970 ","End":"01:29.010","Text":"but that could be a little bit more complicated."},{"Start":"01:29.010 ","End":"01:36.870","Text":"What I\u0027m going to do is I\u0027m going to use the idea of a fictitious force,"},{"Start":"01:36.870 ","End":"01:41.975","Text":"and what we\u0027re going to do is we\u0027re going to use the centrifugal force,"},{"Start":"01:41.975 ","End":"01:49.470","Text":"so then I can write at 90 degrees to my mg over here."},{"Start":"01:49.470 ","End":"01:53.330","Text":"I\u0027ll write my equation for the centrifugal force,"},{"Start":"01:53.330 ","End":"02:00.560","Text":"which is going to be mw^2 multiplied by r. What, in fact,"},{"Start":"02:00.560 ","End":"02:02.360","Text":"we\u0027ve done over here is,"},{"Start":"02:02.360 ","End":"02:05.555","Text":"instead of saying that the sum of all of the forces is equal to mw^2r,"},{"Start":"02:05.555 ","End":"02:09.050","Text":"and then I have to work that out."},{"Start":"02:09.050 ","End":"02:11.915","Text":"What I\u0027ve said instead is I\u0027ve said,"},{"Start":"02:11.915 ","End":"02:18.375","Text":"imagine that the system is rotating in the direction of the ball,"},{"Start":"02:18.375 ","End":"02:20.030","Text":"so that\u0027s the system."},{"Start":"02:20.030 ","End":"02:25.595","Text":"Then instead, I can say that the acceleration is going to be equal to 0,"},{"Start":"02:25.595 ","End":"02:31.105","Text":"and then my centrifugal force is going to be mw^2r."},{"Start":"02:31.105 ","End":"02:32.900","Text":"It\u0027s a different way of looking at it."},{"Start":"02:32.900 ","End":"02:34.770","Text":"I\u0027m saying that my acceleration is 0,"},{"Start":"02:34.770 ","End":"02:37.525","Text":"so the sum of all my forces is equal to 0,"},{"Start":"02:37.525 ","End":"02:44.910","Text":"and then that I have my centrifugal force acting and that my entire system is rotating."},{"Start":"02:45.800 ","End":"02:52.040","Text":"Now the next thing that we can do is we can split this up into x and y components."},{"Start":"02:52.040 ","End":"02:58.640","Text":"The easiest thing to say is that along the wall of the cone,"},{"Start":"02:58.640 ","End":"03:01.700","Text":"we can say that this is our x-axis,"},{"Start":"03:01.700 ","End":"03:05.210","Text":"like this, and that going in this direction,"},{"Start":"03:05.210 ","End":"03:07.590","Text":"we have our y-axis."},{"Start":"03:08.090 ","End":"03:12.405","Text":"It doesn\u0027t matter right now which direction this is in."},{"Start":"03:12.405 ","End":"03:19.100","Text":"Now what we can say is that if our omega is very large, so what does that mean?"},{"Start":"03:19.100 ","End":"03:23.030","Text":"We know that the equation for velocity is equal"},{"Start":"03:23.030 ","End":"03:27.905","Text":"to omega multiplied by r. If my omega is very large,"},{"Start":"03:27.905 ","End":"03:30.275","Text":"then my velocity will be very large."},{"Start":"03:30.275 ","End":"03:34.930","Text":"Here we have an equation with omega in our centrifugal equation."},{"Start":"03:34.930 ","End":"03:37.655","Text":"We can say that if omega is very large,"},{"Start":"03:37.655 ","End":"03:39.860","Text":"then our velocity is going to be very large."},{"Start":"03:39.860 ","End":"03:43.640","Text":"If, when we split this up into x and y components,"},{"Start":"03:43.640 ","End":"03:52.025","Text":"so if the force which will represent our mw^2r in the x- direction,"},{"Start":"03:52.025 ","End":"03:58.495","Text":"and this direction is greater than the force that gravity is applying in this direction."},{"Start":"03:58.495 ","End":"04:01.115","Text":"If this force over here is greater,"},{"Start":"04:01.115 ","End":"04:06.835","Text":"then we can see that the ball is going to slowly rotate upwards."},{"Start":"04:06.835 ","End":"04:13.505","Text":"It\u0027s going to be spinning around in a spiral up the cone."},{"Start":"04:13.505 ","End":"04:17.150","Text":"Can you see that? That\u0027s what\u0027s going to happen."},{"Start":"04:17.150 ","End":"04:20.105","Text":"If our omega is small enough,"},{"Start":"04:20.105 ","End":"04:21.470","Text":"then our force,"},{"Start":"04:21.470 ","End":"04:24.440","Text":"the gravity will apply along the axis,"},{"Start":"04:24.440 ","End":"04:27.005","Text":"but in the opposite direction will overpower that,"},{"Start":"04:27.005 ","End":"04:29.060","Text":"and then the ball will drop down."},{"Start":"04:29.060 ","End":"04:31.880","Text":"That\u0027s how you can look at this."},{"Start":"04:31.880 ","End":"04:34.670","Text":"Now, in this question, we\u0027re being told to find"},{"Start":"04:34.670 ","End":"04:37.295","Text":"the maximum height which the ball will reach."},{"Start":"04:37.295 ","End":"04:40.385","Text":"Because we know that our starting height is already H,"},{"Start":"04:40.385 ","End":"04:43.610","Text":"we can assume from the clue in the question that the ball"},{"Start":"04:43.610 ","End":"04:46.860","Text":"isn\u0027t going to actually be going downwards."},{"Start":"04:46.860 ","End":"04:49.920","Text":"In fact, its height is going to be increasing."},{"Start":"04:49.920 ","End":"04:55.550","Text":"Now we can know that from the conservation of energy or angular momentum,"},{"Start":"04:55.550 ","End":"05:00.565","Text":"either way you want to look at it that as the ball gains height,"},{"Start":"05:00.565 ","End":"05:03.565","Text":"so as it moves up the cone,"},{"Start":"05:03.565 ","End":"05:09.320","Text":"so we will see that the angular velocity of the ball will become smaller."},{"Start":"05:09.320 ","End":"05:12.210","Text":"Why will this happen?"},{"Start":"05:12.420 ","End":"05:18.205","Text":"We can see that if we rearrange this equation to get our angular velocity,"},{"Start":"05:18.205 ","End":"05:24.985","Text":"so we\u0027ll have omega is equal to v over r. Then as the ball goes up,"},{"Start":"05:24.985 ","End":"05:27.805","Text":"that we can see that the radius is increasing,"},{"Start":"05:27.805 ","End":"05:32.770","Text":"which means that our denominator will be increasing as well,"},{"Start":"05:32.770 ","End":"05:37.400","Text":"which means that our angular velocity will decrease."},{"Start":"05:39.320 ","End":"05:46.840","Text":"What we\u0027ll see that will happen is that our ball will move up in spiral fashion"},{"Start":"05:46.840 ","End":"05:54.010","Text":"up the cone until it reaches some point where it\u0027s potential energy,"},{"Start":"05:54.010 ","End":"05:57.525","Text":"and its kinetic energy will be equal,"},{"Start":"05:57.525 ","End":"05:59.340","Text":"so equal in opposite,"},{"Start":"05:59.340 ","End":"06:01.106","Text":"so they will cancel each other out."},{"Start":"06:01.106 ","End":"06:06.080","Text":"The ball though, won\u0027t stop over here"},{"Start":"06:06.080 ","End":"06:11.975","Text":"because it\u0027s going to have a component of velocity carrying on in this direction."},{"Start":"06:11.975 ","End":"06:16.115","Text":"Even though the energies will be equal,"},{"Start":"06:16.115 ","End":"06:21.680","Text":"so what we\u0027re going to have is that the ball will still have this component of velocity."},{"Start":"06:21.680 ","End":"06:25.160","Text":"It will carry on moving until it reaches some point."},{"Start":"06:25.160 ","End":"06:27.020","Text":"Let\u0027s say it\u0027s here,"},{"Start":"06:27.020 ","End":"06:30.905","Text":"and then when its velocity will be equal to 0,"},{"Start":"06:30.905 ","End":"06:34.325","Text":"only then would start traveling back down,"},{"Start":"06:34.325 ","End":"06:42.730","Text":"because then the force acting due to the gravity will be obviously greater."},{"Start":"06:42.890 ","End":"06:46.950","Text":"Even when the energy is balanced,"},{"Start":"06:46.950 ","End":"06:48.425","Text":"and cancel each other out,"},{"Start":"06:48.425 ","End":"06:51.890","Text":"there\u0027s still a component of velocity which will push the ball,"},{"Start":"06:51.890 ","End":"06:54.530","Text":"still make it carry on, moving upwards,"},{"Start":"06:54.530 ","End":"06:56.360","Text":"still in a spiral fashion,"},{"Start":"06:56.360 ","End":"06:59.255","Text":"until its velocity will become 0,"},{"Start":"06:59.255 ","End":"07:01.405","Text":"and then it will start moving down."},{"Start":"07:01.405 ","End":"07:07.090","Text":"We have 2 sections to look at when working out this equation."},{"Start":"07:07.090 ","End":"07:09.880","Text":"Then once again, of course,"},{"Start":"07:09.880 ","End":"07:11.525","Text":"once the ball reaches this point,"},{"Start":"07:11.525 ","End":"07:15.800","Text":"it will go back down until they\u0027re same height over here,"},{"Start":"07:15.800 ","End":"07:19.115","Text":"H and then the same exact thing will repeat."},{"Start":"07:19.115 ","End":"07:21.860","Text":"It will move up until this point and then again,"},{"Start":"07:21.860 ","End":"07:24.360","Text":"until this point, and back down."},{"Start":"07:25.340 ","End":"07:33.955","Text":"Let\u0027s begin by solving this using conservation of energy and momentum."},{"Start":"07:33.955 ","End":"07:40.835","Text":"What we can say is that because we see that our mw^2r is a fictitious force,"},{"Start":"07:40.835 ","End":"07:50.345","Text":"we can say that the force is actually currently acting is going to be mg and N_1."},{"Start":"07:50.345 ","End":"07:57.980","Text":"Then that is going"},{"Start":"07:57.980 ","End":"08:03.970","Text":"to mean that our energy is equal to a constant."},{"Start":"08:03.970 ","End":"08:09.410","Text":"Now what we\u0027re going to want to do is to find the momentum."},{"Start":"08:09.410 ","End":"08:16.855","Text":"What we can say is that if our z-axis is our axis of rotation,"},{"Start":"08:16.855 ","End":"08:19.080","Text":"so here we want to find our momentum,"},{"Start":"08:19.080 ","End":"08:24.090","Text":"so we have our axis going down, like so."},{"Start":"08:24.100 ","End":"08:31.340","Text":"If we have a look, we have our r vector perpendicular to the z-axis going like this,"},{"Start":"08:31.340 ","End":"08:33.810","Text":"and this is some kind of r."},{"Start":"08:34.610 ","End":"08:44.195","Text":"We can draw a force diagram,"},{"Start":"08:44.195 ","End":"08:47.920","Text":"and we can say that if our r is going in this direction,"},{"Start":"08:47.920 ","End":"08:53.610","Text":"and then we can see that our force our mg is going in this direction."},{"Start":"08:53.610 ","End":"08:57.840","Text":"We know that our equation is r cross F,"},{"Start":"08:57.840 ","End":"09:02.773","Text":"so we\u0027ll get that our moment is going, sorry,"},{"Start":"09:02.773 ","End":"09:05.780","Text":"is going in the anticlockwise direction,"},{"Start":"09:05.780 ","End":"09:13.835","Text":"which means that the moment is coming towards us,"},{"Start":"09:13.835 ","End":"09:17.820","Text":"so from the screen towards our face."},{"Start":"09:17.860 ","End":"09:24.965","Text":"That means that we can see that our moment is going to be coming out of the page,"},{"Start":"09:24.965 ","End":"09:27.845","Text":"and let\u0027s call this x Tilda."},{"Start":"09:27.845 ","End":"09:31.640","Text":"Now another way that we can look at this is if we"},{"Start":"09:31.640 ","End":"09:35.599","Text":"write down that our force due to gravity,"},{"Start":"09:35.599 ","End":"09:43.260","Text":"let\u0027s call it w, is equal to mg in the negative z-direction."},{"Start":"09:43.260 ","End":"09:49.955","Text":"Then we can see that our radius is going to be some r,"},{"Start":"09:49.955 ","End":"09:57.855","Text":"the radius of our mg where this is starting is going to be"},{"Start":"09:57.855 ","End":"10:06.460","Text":"some r and it\u0027s going to be in the negative y-direction, like we saw before."},{"Start":"10:06.460 ","End":"10:09.710","Text":"Remember, because we said that along the wall of the cone,"},{"Start":"10:09.710 ","End":"10:11.075","Text":"this is our x-direction,"},{"Start":"10:11.075 ","End":"10:13.490","Text":"and then perpendicular to it,"},{"Start":"10:13.490 ","End":"10:15.125","Text":"it\u0027s going to be our y-direction."},{"Start":"10:15.125 ","End":"10:17.825","Text":"We can see that our r is going to be something,"},{"Start":"10:17.825 ","End":"10:21.400","Text":"in the negative y direction."},{"Start":"10:21.400 ","End":"10:26.015","Text":"Then if we do the cross-product between z and y,"},{"Start":"10:26.015 ","End":"10:28.100","Text":"and because it\u0027s negative z and negative y,"},{"Start":"10:28.100 ","End":"10:29.570","Text":"so the negatives will cross out."},{"Start":"10:29.570 ","End":"10:30.785","Text":"Between z and y,"},{"Start":"10:30.785 ","End":"10:33.125","Text":"we\u0027re going to get something in the x-direction,"},{"Start":"10:33.125 ","End":"10:37.680","Text":"which we\u0027ll say that this is this red arrow."},{"Start":"10:38.080 ","End":"10:43.915","Text":"Now we can see that our moment or our torque is in the x-direction."},{"Start":"10:43.915 ","End":"10:50.370","Text":"What we can say is that tau of mg is equal"},{"Start":"10:50.370 ","End":"10:58.600","Text":"to our mg multiplied by r in the x-direction."},{"Start":"10:59.540 ","End":"11:02.880","Text":"This could be a positive or a negative,"},{"Start":"11:02.880 ","End":"11:04.635","Text":"It doesn\u0027t matter right now."},{"Start":"11:04.635 ","End":"11:09.210","Text":"The important thing to note from what we\u0027ve written here is that it\u0027s in"},{"Start":"11:09.210 ","End":"11:15.000","Text":"the x-direction and not in the z-direction."},{"Start":"11:15.000 ","End":"11:20.490","Text":"Now similarly, if we want to find the torque for our N vector,"},{"Start":"11:20.490 ","End":"11:23.340","Text":"so it\u0027s going to be the exact same r,"},{"Start":"11:23.340 ","End":"11:29.370","Text":"the same r vector because we\u0027re still speaking about the same point, the bowl."},{"Start":"11:29.370 ","End":"11:31.605","Text":"Then we can say,"},{"Start":"11:31.605 ","End":"11:34.080","Text":"so what is our normal made up of?"},{"Start":"11:34.080 ","End":"11:40.110","Text":"Our normal is going to be our normal in the z-direction."},{"Start":"11:40.110 ","End":"11:43.110","Text":"We can split them up into components,"},{"Start":"11:43.110 ","End":"11:52.395","Text":"and we\u0027re going to have plus our normal in our y direction."},{"Start":"11:52.395 ","End":"11:54.870","Text":"We can split them up into the components,"},{"Start":"11:54.870 ","End":"11:57.000","Text":"and then the r is the same,"},{"Start":"11:57.000 ","End":"12:00.510","Text":"so when you do the cross-product between y and y,"},{"Start":"12:00.510 ","End":"12:02.835","Text":"we\u0027ll see that these 2 cancel out."},{"Start":"12:02.835 ","End":"12:04.410","Text":"Between y and z, again,"},{"Start":"12:04.410 ","End":"12:07.275","Text":"it\u0027s going to be in the x-direction."},{"Start":"12:07.275 ","End":"12:11.520","Text":"Just like when we were speaking about with our mg. Because again,"},{"Start":"12:11.520 ","End":"12:14.160","Text":"we\u0027re having a y crossed with a z."},{"Start":"12:14.160 ","End":"12:17.040","Text":"Again, the signs don\u0027t matter, this could be a negative."},{"Start":"12:17.040 ","End":"12:19.050","Text":"There\u0027s a negative, there\u0027s a positive, there\u0027s a negative."},{"Start":"12:19.050 ","End":"12:20.985","Text":"It doesn\u0027t matter right now,"},{"Start":"12:20.985 ","End":"12:23.625","Text":"we just want to show the direction that it\u0027s in,"},{"Start":"12:23.625 ","End":"12:26.640","Text":"which again, also for the normal vector,"},{"Start":"12:26.640 ","End":"12:30.580","Text":"the torque is going to be in the x-direction."},{"Start":"12:30.980 ","End":"12:38.460","Text":"We can say that our torque for our normal vector is also going to equal plus or minus,"},{"Start":"12:38.460 ","End":"12:40.875","Text":"because right now we don\u0027t care about the sign."},{"Start":"12:40.875 ","End":"12:45.690","Text":"Some normal in the z-direction,"},{"Start":"12:45.690 ","End":"12:48.210","Text":"the component in the z-direction,"},{"Start":"12:48.210 ","End":"12:55.060","Text":"multiplied by our radius and this is going to be in the x-direction as well."},{"Start":"12:55.100 ","End":"13:00.945","Text":"Now the next thing to note is if you remember the equation,"},{"Start":"13:00.945 ","End":"13:02.895","Text":"I\u0027m going to write it up over here,"},{"Start":"13:02.895 ","End":"13:05.745","Text":"for the sum of all of the torques,"},{"Start":"13:05.745 ","End":"13:07.545","Text":"so what is this equal to?"},{"Start":"13:07.545 ","End":"13:14.340","Text":"It equals to L by dt,"},{"Start":"13:14.340 ","End":"13:20.290","Text":"which means that it\u0027s the rate of change of angular momentum."},{"Start":"13:20.300 ","End":"13:26.840","Text":"Just like we can separate out our angular momentum into our x, y,"},{"Start":"13:26.840 ","End":"13:27.920","Text":"and z components,"},{"Start":"13:27.920 ","End":"13:33.200","Text":"so we can do this also with the sum of all of the torques."},{"Start":"13:33.200 ","End":"13:37.310","Text":"We can say that the sum of the torques, let\u0027s say,"},{"Start":"13:37.310 ","End":"13:46.485","Text":"in the x-direction is going to be equal to by dL in our x-direction divided by dt."},{"Start":"13:46.485 ","End":"13:50.025","Text":"We can do the same thing for x,"},{"Start":"13:50.025 ","End":"13:53.230","Text":"y, and z similarly."},{"Start":"13:53.480 ","End":"14:02.775","Text":"Now, usually we\u0027re dealing with some plane which is going to be in the x, y axis,"},{"Start":"14:02.775 ","End":"14:06.045","Text":"and then we\u0027re dealing with the z-axis,"},{"Start":"14:06.045 ","End":"14:08.550","Text":"because that is perpendicular to both the x and"},{"Start":"14:08.550 ","End":"14:13.350","Text":"the y-axis and we\u0027ll be dealing with the torque in that axis."},{"Start":"14:13.350 ","End":"14:17.505","Text":"Now, in the x-axis,"},{"Start":"14:17.505 ","End":"14:20.520","Text":"we can see that the torque is in conserved,"},{"Start":"14:20.520 ","End":"14:23.865","Text":"and we can see that also in the y-axis,"},{"Start":"14:23.865 ","End":"14:25.455","Text":"because it\u0027s on the same plane."},{"Start":"14:25.455 ","End":"14:29.085","Text":"Now, because the z-axis is perpendicular to that,"},{"Start":"14:29.085 ","End":"14:32.565","Text":"and we can see that there\u0027s no torque on the z-axis,"},{"Start":"14:32.565 ","End":"14:39.090","Text":"therefore we can say"},{"Start":"14:39.090 ","End":"14:45.540","Text":"that our angular momentum in our z-axis is a constant."},{"Start":"14:45.540 ","End":"14:50.775","Text":"It\u0027s conserved. Now,"},{"Start":"14:50.775 ","End":"14:53.670","Text":"that was a bit of an explanation of why the energy and"},{"Start":"14:53.670 ","End":"14:56.805","Text":"why the angular momentum is conserved."},{"Start":"14:56.805 ","End":"15:01.125","Text":"However, when you get a question like this,"},{"Start":"15:01.125 ","End":"15:04.830","Text":"you should know that in the test most likely it will be conserved anyway,"},{"Start":"15:04.830 ","End":"15:08.625","Text":"because it\u0027s going to be really difficult to solve this if it\u0027s not."},{"Start":"15:08.625 ","End":"15:11.940","Text":"You could have just said immediately, energy conserved,"},{"Start":"15:11.940 ","End":"15:15.165","Text":"angular momentum conserved, so constant and constant,"},{"Start":"15:15.165 ","End":"15:17.235","Text":"but it doesn\u0027t matter."},{"Start":"15:17.235 ","End":"15:19.155","Text":"That was the explanation."},{"Start":"15:19.155 ","End":"15:25.240","Text":"Now, we\u0027re going to plug everything into the equations and see what we get."},{"Start":"15:25.700 ","End":"15:32.010","Text":"What we\u0027re going to do now is we\u0027re going to start with our equation for our energy."},{"Start":"15:32.010 ","End":"15:34.275","Text":"Here we go."},{"Start":"15:34.275 ","End":"15:37.350","Text":"We know that at the start our potential energy is going to"},{"Start":"15:37.350 ","End":"15:40.680","Text":"be mg multiplied by its initial height,"},{"Start":"15:40.680 ","End":"15:45.630","Text":"which is H. Then we also are going to add in its kinetic energy,"},{"Start":"15:45.630 ","End":"15:47.070","Text":"so it\u0027s going to be plus."},{"Start":"15:47.070 ","End":"15:50.590","Text":"The equation for kinetic energy is 1/2MV^2."},{"Start":"15:51.590 ","End":"15:57.360","Text":"We know that our initial velocity is V_0, so V_0^2."},{"Start":"15:57.360 ","End":"15:58.770","Text":"That\u0027s right at the beginning."},{"Start":"15:58.770 ","End":"16:00.255","Text":"Now at the end,"},{"Start":"16:00.255 ","End":"16:03.785","Text":"so when maximum height is achieved,"},{"Start":"16:03.785 ","End":"16:05.510","Text":"so we\u0027re going to have, again,"},{"Start":"16:05.510 ","End":"16:07.130","Text":"our potential energy, which is, again,"},{"Start":"16:07.130 ","End":"16:11.760","Text":"going to be m and g multiplied by H. We don\u0027t know what this H is,"},{"Start":"16:11.760 ","End":"16:13.590","Text":"is going to be H max,"},{"Start":"16:13.590 ","End":"16:15.465","Text":"some maximum height,"},{"Start":"16:15.465 ","End":"16:18.525","Text":"let\u0027s mark it,"},{"Start":"16:18.525 ","End":"16:20.740","Text":"we don\u0027t know what this is."},{"Start":"16:20.870 ","End":"16:26.370","Text":"Then what we\u0027re going to do is we\u0027re going to add onto that by kinetic energy,"},{"Start":"16:26.370 ","End":"16:29.625","Text":"which is again going to be 1/2 multiplied by m"},{"Start":"16:29.625 ","End":"16:33.570","Text":"and then it\u0027s going to be multiplied by some V^2,"},{"Start":"16:33.570 ","End":"16:36.674","Text":"which is going to be our minimum velocity."},{"Start":"16:36.674 ","End":"16:39.945","Text":"Now, this we also don\u0027t know."},{"Start":"16:39.945 ","End":"16:43.613","Text":"All we know is that it\u0027s going to be some minimum velocity,"},{"Start":"16:43.613 ","End":"16:45.270","Text":"but again, it doesn\u0027t matter."},{"Start":"16:45.270 ","End":"16:51.180","Text":"You just know it\u0027s multiplied by some V. Now,"},{"Start":"16:51.180 ","End":"16:59.055","Text":"what we\u0027re going to do is we\u0027re going to write the equation for our angular momentum."},{"Start":"16:59.055 ","End":"17:02.805","Text":"At the beginning we know that it\u0027s going to be equal to"},{"Start":"17:02.805 ","End":"17:06.480","Text":"m multiplied by its initial velocity,"},{"Start":"17:06.480 ","End":"17:11.565","Text":"which is going to be V_0, and then multiplied by the radius,"},{"Start":"17:11.565 ","End":"17:16.410","Text":"which is the distance to the axis of rotation."},{"Start":"17:16.410 ","End":"17:19.830","Text":"Here we can see that it\u0027s the I."},{"Start":"17:19.830 ","End":"17:21.885","Text":"Then at the end,"},{"Start":"17:21.885 ","End":"17:27.135","Text":"so it\u0027s again going to be m multiplied by its final velocity,"},{"Start":"17:27.135 ","End":"17:29.715","Text":"so this is again going to be V_min,"},{"Start":"17:29.715 ","End":"17:32.910","Text":"the same velocity over here."},{"Start":"17:32.910 ","End":"17:37.440","Text":"Then it\u0027s going to be multiplied by the radius."},{"Start":"17:37.440 ","End":"17:40.680","Text":"Now, we know that with finding"},{"Start":"17:40.680 ","End":"17:45.555","Text":"the greatest height that it will be and coincidentally at the greatest height,"},{"Start":"17:45.555 ","End":"17:48.990","Text":"its radius will also be the greatest, because it\u0027s a cone."},{"Start":"17:48.990 ","End":"17:53.010","Text":"The radius is increasing as we go up the cone."},{"Start":"17:53.010 ","End":"17:57.525","Text":"Then it\u0027s going to be multiplied by r_max."},{"Start":"17:57.525 ","End":"18:00.510","Text":"But we don\u0027t have to really write r_max it doesn\u0027t matter."},{"Start":"18:00.510 ","End":"18:05.685","Text":"It will just explain what\u0027s happening to us a little bit easier,"},{"Start":"18:05.685 ","End":"18:10.240","Text":"but it\u0027s just multiplied by some r, which is over here."},{"Start":"18:10.340 ","End":"18:13.020","Text":"Now, an important note over here,"},{"Start":"18:13.020 ","End":"18:15.495","Text":"because we\u0027re dealing with the angular momentum,"},{"Start":"18:15.495 ","End":"18:17.700","Text":"the z component of it."},{"Start":"18:17.700 ","End":"18:20.790","Text":"It so happens that in this question specifically,"},{"Start":"18:20.790 ","End":"18:24.060","Text":"we\u0027re being told that our V_0 is in the horizontal direction,"},{"Start":"18:24.060 ","End":"18:28.020","Text":"which means that it\u0027s perpendicular to the z-axis."},{"Start":"18:28.020 ","End":"18:31.110","Text":"That\u0027s why we can write V_0 here and V_0 here,"},{"Start":"18:31.110 ","End":"18:34.470","Text":"and it\u0027s the exact same velocities."},{"Start":"18:34.470 ","End":"18:39.480","Text":"However, if our V_0 wasn\u0027t on the plane,"},{"Start":"18:39.480 ","End":"18:42.225","Text":"which is perpendicular to our z-axis,"},{"Start":"18:42.225 ","End":"18:46.080","Text":"then we would have to take components, and then the V_0,"},{"Start":"18:46.080 ","End":"18:52.665","Text":"our velocity that we would plug into this equation would be different to this velocity."},{"Start":"18:52.665 ","End":"18:55.140","Text":"This one would stay the same."},{"Start":"18:55.140 ","End":"19:00.690","Text":"It would include our velocity in the x,"},{"Start":"19:00.690 ","End":"19:02.610","Text":"y, and z directions."},{"Start":"19:02.610 ","End":"19:08.220","Text":"It will be V_x^2 plus V_y^2 plus V_z^2."},{"Start":"19:08.220 ","End":"19:10.665","Text":"That will be that. But then over here,"},{"Start":"19:10.665 ","End":"19:12.630","Text":"we\u0027d have to take the components."},{"Start":"19:12.630 ","End":"19:14.010","Text":"Now, how would we take the components?"},{"Start":"19:14.010 ","End":"19:16.665","Text":"We\u0027d have to do the cross-product."},{"Start":"19:16.665 ","End":"19:18.315","Text":"What we would have to do,"},{"Start":"19:18.315 ","End":"19:20.460","Text":"I\u0027m just going to do this in a different color and I\u0027m going to rub it"},{"Start":"19:20.460 ","End":"19:23.835","Text":"out soon because for this question,"},{"Start":"19:23.835 ","End":"19:27.165","Text":"it is irrelevant, but it\u0027s something to look out for."},{"Start":"19:27.165 ","End":"19:35.720","Text":"We have your r cross your MV."},{"Start":"19:35.720 ","End":"19:37.675","Text":"That will be equal to,"},{"Start":"19:37.675 ","End":"19:41.770","Text":"let\u0027s say we had a velocity in the x and the y direction,"},{"Start":"19:41.770 ","End":"19:47.540","Text":"we\u0027d have x plus y."},{"Start":"19:48.240 ","End":"19:57.565","Text":"This is y, and this will be cross and then it will be"},{"Start":"19:57.565 ","End":"20:01.420","Text":"the x component of the velocity in"},{"Start":"20:01.420 ","End":"20:07.450","Text":"the x direction plus the y component of the velocity in the y direction."},{"Start":"20:07.450 ","End":"20:10.060","Text":"The cross product we would get,"},{"Start":"20:10.060 ","End":"20:14.260","Text":"so therefore L_ z would equal to this cross product,"},{"Start":"20:14.260 ","End":"20:23.710","Text":"which will equal to xV_y in minus yV_x."},{"Start":"20:23.710 ","End":"20:28.345","Text":"Doing the cross product you can also see this via taking the determinant."},{"Start":"20:28.345 ","End":"20:33.250","Text":"You could have L_x, L_y,"},{"Start":"20:33.250 ","End":"20:37.098","Text":"L_z, xyz,"},{"Start":"20:37.098 ","End":"20:40.330","Text":"V_x, V_y, V_z."},{"Start":"20:40.330 ","End":"20:43.045","Text":"Then when you take the determinant, if you want for L_z,"},{"Start":"20:43.045 ","End":"20:44.935","Text":"because that\u0027s what\u0027s relevant over here,"},{"Start":"20:44.935 ","End":"20:46.870","Text":"then we can see that it\u0027s this times this,"},{"Start":"20:46.870 ","End":"20:49.085","Text":"minus this times this."},{"Start":"20:49.085 ","End":"20:57.040","Text":"We\u0027ll just get this. This is super important that you know this difference."},{"Start":"20:57.040 ","End":"20:58.360","Text":"In this question right now,"},{"Start":"20:58.360 ","End":"21:01.090","Text":"it is irrelevant because we\u0027re dealing"},{"Start":"21:01.090 ","End":"21:05.920","Text":"with things in which are perpendicular to one another."},{"Start":"21:05.920 ","End":"21:07.975","Text":"We don\u0027t have to take all the components."},{"Start":"21:07.975 ","End":"21:10.105","Text":"If it wasn\u0027t like this,"},{"Start":"21:10.105 ","End":"21:13.060","Text":"how it\u0027s set up in this question specifically than we would have"},{"Start":"21:13.060 ","End":"21:17.300","Text":"to work out the cross product."},{"Start":"21:17.640 ","End":"21:24.445","Text":"Another note, how come these V_mins, they\u0027re equal?"},{"Start":"21:24.445 ","End":"21:27.595","Text":"Because I\u0027m trying to find my maximum height."},{"Start":"21:27.595 ","End":"21:29.020","Text":"When does that happen?"},{"Start":"21:29.020 ","End":"21:30.130","Text":"What we said before,"},{"Start":"21:30.130 ","End":"21:36.415","Text":"when the z component of the velocity over here is equal to 0,"},{"Start":"21:36.415 ","End":"21:38.380","Text":"which means that\u0027s it."},{"Start":"21:38.380 ","End":"21:41.330","Text":"It\u0027s reached its maximum height."},{"Start":"21:42.000 ","End":"21:48.310","Text":"We know that our V_min,"},{"Start":"21:48.310 ","End":"21:52.525","Text":"the z component of it is going to be equal to 0."},{"Start":"21:52.525 ","End":"21:59.570","Text":"That\u0027s why these 2 V_mins will equal to 0 because at any other points,"},{"Start":"21:59.570 ","End":"22:02.670","Text":"the ball is going to have velocity in the x,"},{"Start":"22:02.670 ","End":"22:04.365","Text":"y, and z components,"},{"Start":"22:04.365 ","End":"22:06.765","Text":"which means that we haven\u0027t reached maximum height,"},{"Start":"22:06.765 ","End":"22:10.125","Text":"and maximum height, our z components are equal to 0,"},{"Start":"22:10.125 ","End":"22:11.490","Text":"which is what we want."},{"Start":"22:11.490 ","End":"22:14.150","Text":"That\u0027s what proves that we\u0027re at maximum height."},{"Start":"22:14.150 ","End":"22:18.565","Text":"Now what we can do using the trigonometric identities,"},{"Start":"22:18.565 ","End":"22:23.815","Text":"we can say that the angle of our cone over here,"},{"Start":"22:23.815 ","End":"22:26.185","Text":"we\u0027ll call this angle Theta."},{"Start":"22:26.185 ","End":"22:32.845","Text":"Now we can find the relationship between our radius and our height throughout the cone."},{"Start":"22:32.845 ","End":"22:37.810","Text":"What we can see, is that we know that our tan for my SOHCAHTOA,"},{"Start":"22:37.810 ","End":"22:42.400","Text":"TOA tan of Theta is opposite over adjacent."},{"Start":"22:42.400 ","End":"22:49.630","Text":"Tan of Theta is equal to opposite over adjacent."},{"Start":"22:49.630 ","End":"22:52.525","Text":"What is our opposite side?"},{"Start":"22:52.525 ","End":"22:58.310","Text":"It\u0027s our radius R. What is our adjacent? It\u0027s our height."},{"Start":"22:58.830 ","End":"23:04.600","Text":"We have this, so then we can arrange this such that we"},{"Start":"23:04.600 ","End":"23:10.600","Text":"know that R is equal to h tan of Theta."},{"Start":"23:10.600 ","End":"23:15.015","Text":"This is the relationship,"},{"Start":"23:15.015 ","End":"23:21.220","Text":"and then we can also say that our r_max is"},{"Start":"23:21.220 ","End":"23:28.300","Text":"equal to h_max multiplied by tan of Theta."},{"Start":"23:28.300 ","End":"23:31.870","Text":"We can see that through a simple trigonometric identity"},{"Start":"23:31.870 ","End":"23:35.575","Text":"we can find the relationship between the radius and the height."},{"Start":"23:35.575 ","End":"23:41.680","Text":"Now what we can do is we can substitute this R over here,"},{"Start":"23:41.680 ","End":"23:46.630","Text":"and we can substitute our r_max over into here."},{"Start":"23:46.630 ","End":"23:48.865","Text":"Let\u0027s rewrite this."},{"Start":"23:48.865 ","End":"23:56.710","Text":"We have mV_0 multiplied by h tan of Theta,"},{"Start":"23:56.710 ","End":"24:04.300","Text":"which is going to be equal to mV_min multiplied by r_max,"},{"Start":"24:04.300 ","End":"24:10.505","Text":"which is multiplied by h_max tan of Theta."},{"Start":"24:10.505 ","End":"24:13.005","Text":"In both sides, we have m\u0027s. We can cross them off."},{"Start":"24:13.005 ","End":"24:15.675","Text":"On both sides we have tan of Theta so we can cross them off."},{"Start":"24:15.675 ","End":"24:20.125","Text":"Notice that nothing depends on the angle of the cone itself."},{"Start":"24:20.125 ","End":"24:23.440","Text":"It doesn\u0027t matter that we don\u0027t know the angle in the question."},{"Start":"24:23.440 ","End":"24:26.020","Text":"Then we can see that we\u0027re left with V_0,"},{"Start":"24:26.020 ","End":"24:35.790","Text":"H is equal to V_min multiplied by h_max."},{"Start":"24:35.790 ","End":"24:38.470","Text":"Now what we can do,"},{"Start":"24:38.470 ","End":"24:41.850","Text":"is these are exactly the 2 things that we don\u0027t"},{"Start":"24:41.850 ","End":"24:45.775","Text":"know what they are V_min and h_max, there are unknowns."},{"Start":"24:45.775 ","End":"24:49.360","Text":"What we can do first is we can isolate out."},{"Start":"24:49.360 ","End":"24:52.060","Text":"Let\u0027s say our V_min."},{"Start":"24:52.060 ","End":"25:01.780","Text":"Then we can say that our V_min is equal to V_0 H divided by our h_max."},{"Start":"25:01.780 ","End":"25:07.885","Text":"Then we can substitute this in over here."},{"Start":"25:07.885 ","End":"25:10.690","Text":"Let\u0027s rewrite this."},{"Start":"25:10.690 ","End":"25:20.210","Text":"We have that our equation for energy is going to be equal to mgH plus 1/2m(V_0)^2,"},{"Start":"25:21.120 ","End":"25:23.530","Text":"which is going to be equal to"},{"Start":"25:23.530 ","End":"25:28.260","Text":"mgh_max"},{"Start":"25:28.620 ","End":"25:34.720","Text":"plus 1/2m(V_min)^2,"},{"Start":"25:34.720 ","End":"25:41.930","Text":"which is going to be V_0^2 H^2 over h_max^2."},{"Start":"25:42.990 ","End":"25:49.300","Text":"We can see that the common factor over here is our m. We"},{"Start":"25:49.300 ","End":"25:55.060","Text":"can divide everything by M. Then,"},{"Start":"25:55.060 ","End":"25:58.450","Text":"because we have halves, we can multiply everything by 2."},{"Start":"25:58.450 ","End":"26:06.640","Text":"I get 2gH plus V_0^2 is equal to"},{"Start":"26:06.640 ","End":"26:12.130","Text":"2gh_max"},{"Start":"26:12.130 ","End":"26:17.950","Text":"plus V_0^2 H^2"},{"Start":"26:17.950 ","End":"26:25.090","Text":"over (h_max)^2."},{"Start":"26:25.090 ","End":"26:28.554","Text":"Now, all that\u0027s left is the algebra."},{"Start":"26:28.554 ","End":"26:31.285","Text":"If you want to watch, then that\u0027s fine."},{"Start":"26:31.285 ","End":"26:35.680","Text":"Otherwise, fast forward to the end of the video where you\u0027ll just see the answer which is"},{"Start":"26:35.680 ","End":"26:41.390","Text":"going to be a cubic equation involving our h_max."},{"Start":"26:41.640 ","End":"26:44.080","Text":"Right now go to the end of the video,"},{"Start":"26:44.080 ","End":"26:49.525","Text":"unless you want to see how I do this. How we do this?"},{"Start":"26:49.525 ","End":"26:54.580","Text":"Let\u0027s multiply both sides by (h_max)^2."},{"Start":"26:54.580 ","End":"27:01.870","Text":"We\u0027ll have 2gH plus V_0^2."},{"Start":"27:01.870 ","End":"27:06.085","Text":"This is going to be multiplied by h_max^2."},{"Start":"27:06.085 ","End":"27:09.530","Text":"Then this is going to be equal to"},{"Start":"27:12.150 ","End":"27:21.010","Text":"2gh_max plus V_0^2 H^2."},{"Start":"27:21.010 ","End":"27:27.085","Text":"Over here, it\u0027s going to be cubed because we multiplied by H^2."},{"Start":"27:27.085 ","End":"27:28.405","Text":"This is the final answer,"},{"Start":"27:28.405 ","End":"27:32.110","Text":"we have our h_max cubed."},{"Start":"27:32.110 ","End":"27:34.645","Text":"A cubic equation will be accepted."},{"Start":"27:34.645 ","End":"27:36.430","Text":"You can further simplify it."},{"Start":"27:36.430 ","End":"27:40.150","Text":"I\u0027m not going to do it now. This would be finding a test,"},{"Start":"27:40.150 ","End":"27:41.890","Text":"but to further simplify it,"},{"Start":"27:41.890 ","End":"27:43.375","Text":"you can do this on your own,"},{"Start":"27:43.375 ","End":"27:45.760","Text":"but this is the answer you\u0027re meant again."},{"Start":"27:45.760 ","End":"27:48.500","Text":"That\u0027s the end of this lesson."}],"ID":12703},{"Watched":false,"Name":"Exercise- Ball Attached To Hanging Mass","Duration":"21m 47s","ChapterTopicVideoID":9125,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this question,"},{"Start":"00:02.100 ","End":"00:06.600","Text":"we\u0027re being told that a mass m moves on a frictionless table."},{"Start":"00:06.600 ","End":"00:09.870","Text":"We\u0027re told that the mass is attached via a string of"},{"Start":"00:09.870 ","End":"00:13.485","Text":"length L. This whole string is of length L,"},{"Start":"00:13.485 ","End":"00:18.690","Text":"which is threaded through the center of the table to another mass M,"},{"Start":"00:18.690 ","End":"00:20.347","Text":"which hangs in the air."},{"Start":"00:20.347 ","End":"00:23.460","Text":"We\u0027re being told at the time t=0,"},{"Start":"00:23.460 ","End":"00:25.500","Text":"our mass M,"},{"Start":"00:25.500 ","End":"00:27.870","Text":"this one over here is at rest and"},{"Start":"00:27.870 ","End":"00:32.610","Text":"our mass m is a distance R from the center of the table."},{"Start":"00:32.610 ","End":"00:37.270","Text":"This distance is R."},{"Start":"00:38.080 ","End":"00:45.080","Text":"We\u0027re being told also at this time that our m is traveling at a velocity v_0,"},{"Start":"00:45.080 ","End":"00:49.270","Text":"which is tangent or perpendicular to the radius."},{"Start":"00:49.270 ","End":"00:52.605","Text":"There\u0027s a 90-degree angle over here."},{"Start":"00:52.605 ","End":"00:56.420","Text":"We\u0027re being asked to write an equation for the conservation of energy and"},{"Start":"00:56.420 ","End":"01:02.150","Text":"angular momentum that means the equation of motion of this system."},{"Start":"01:02.150 ","End":"01:05.005","Text":"Then to find the differential equation,"},{"Start":"01:05.005 ","End":"01:07.595","Text":"which is dependent only on the size R,"},{"Start":"01:07.595 ","End":"01:11.825","Text":"only on the distance of m to the center of the table,"},{"Start":"01:11.825 ","End":"01:14.260","Text":"on this distance over here."},{"Start":"01:15.440 ","End":"01:17.865","Text":"How are we going to solve this?"},{"Start":"01:17.865 ","End":"01:23.060","Text":"The first thing we want to do is we know that the string is of length L,"},{"Start":"01:23.060 ","End":"01:24.305","Text":"that\u0027s the total length,"},{"Start":"01:24.305 ","End":"01:30.815","Text":"and we want to find how that\u0027s dependent on this distance over here,"},{"Start":"01:30.815 ","End":"01:37.325","Text":"which we\u0027ll call r. That means that we have some height change."},{"Start":"01:37.325 ","End":"01:41.210","Text":"Our mass M will go up and down in height depending"},{"Start":"01:41.210 ","End":"01:45.510","Text":"on the distance of our m from the center of the table."},{"Start":"01:46.790 ","End":"01:51.405","Text":"First of all, we\u0027ll pick our axis."},{"Start":"01:51.405 ","End":"01:58.395","Text":"Let\u0027s say that our z-axis is going to be in the downwards direction."},{"Start":"01:58.395 ","End":"02:01.140","Text":"This is our z-axis."},{"Start":"02:01.140 ","End":"02:03.810","Text":"We can write the equation for the strings."},{"Start":"02:03.810 ","End":"02:05.950","Text":"We can say that our z,"},{"Start":"02:06.250 ","End":"02:10.490","Text":"our height, the height of this mass, M,"},{"Start":"02:10.490 ","End":"02:14.375","Text":"is going to be equal to the total length of the string,"},{"Start":"02:14.375 ","End":"02:15.620","Text":"which is L,"},{"Start":"02:15.620 ","End":"02:18.060","Text":"minus this distance over here,"},{"Start":"02:18.060 ","End":"02:23.080","Text":"so minus r. Now r is a variable, it\u0027s changing."},{"Start":"02:24.080 ","End":"02:29.375","Text":"Now what we can do is we can write our equation for our energy."},{"Start":"02:29.375 ","End":"02:32.810","Text":"We know that our energy is going to be conserved."},{"Start":"02:32.810 ","End":"02:35.660","Text":"Lets call this equation number 1."},{"Start":"02:35.660 ","End":"02:39.980","Text":"It\u0027s going to be our first important equation for energy."},{"Start":"02:39.980 ","End":"02:41.555","Text":"What is this equal to?"},{"Start":"02:41.555 ","End":"02:44.195","Text":"Our initial potential energy,"},{"Start":"02:44.195 ","End":"02:46.750","Text":"which is going to be our M,"},{"Start":"02:46.750 ","End":"02:48.265","Text":"which is in the air,"},{"Start":"02:48.265 ","End":"02:51.425","Text":"Mg multiplied by its height,"},{"Start":"02:51.425 ","End":"02:53.585","Text":"which is going to be L minus,"},{"Start":"02:53.585 ","End":"02:55.160","Text":"and because it\u0027s at the beginning,"},{"Start":"02:55.160 ","End":"03:00.032","Text":"it\u0027s R because that\u0027s the location of our m at the beginning,"},{"Start":"03:00.032 ","End":"03:01.650","Text":"plus our kinetic energy,"},{"Start":"03:01.650 ","End":"03:03.915","Text":"so that\u0027s going to be 1/2."},{"Start":"03:03.915 ","End":"03:10.340","Text":"We only need the kinetic energy of our m,"},{"Start":"03:10.340 ","End":"03:13.129","Text":"because our m is moving at a velocity"},{"Start":"03:13.129 ","End":"03:17.630","Text":"at v_0 and our M is at rest as given in the question."},{"Start":"03:17.630 ","End":"03:23.075","Text":"Plus 1/2m v_0^2. Also another note,"},{"Start":"03:23.075 ","End":"03:27.294","Text":"I didn\u0027t have to add in the potential energy of my mass m"},{"Start":"03:27.294 ","End":"03:32.195","Text":"because we can just say that it\u0027s at 0 on the z-axis,"},{"Start":"03:32.195 ","End":"03:36.770","Text":"which means that it will be Mg multiplied by 0, which is 0."},{"Start":"03:36.770 ","End":"03:41.735","Text":"Then the final energy distribution"},{"Start":"03:41.735 ","End":"03:48.125","Text":"between potential and kinetic is going to be the kinetic energy of our mass m,"},{"Start":"03:48.125 ","End":"03:51.275","Text":"which is going to be the kinetic energy, sorry,"},{"Start":"03:51.275 ","End":"03:55.620","Text":"1/2m, and then we\u0027ll just say some v_1^2,"},{"Start":"03:55.620 ","End":"03:56.767","Text":"we don\u0027t know what this is yet,"},{"Start":"03:56.767 ","End":"04:01.160","Text":"plus the kinetic energy of our mass M,"},{"Start":"04:01.160 ","End":"04:03.445","Text":"which is going to be 1/2M,"},{"Start":"04:03.445 ","End":"04:06.840","Text":"and then some velocity v_2^2,"},{"Start":"04:06.840 ","End":"04:09.270","Text":"and then the potential energy."},{"Start":"04:09.270 ","End":"04:16.535","Text":"Again, our m is still on this plane where z is equal to 0."},{"Start":"04:16.535 ","End":"04:18.560","Text":"It\u0027s going to have no potential energy,"},{"Start":"04:18.560 ","End":"04:23.645","Text":"and it\u0027s only this mass M which can move up and down along the z-axis."},{"Start":"04:23.645 ","End":"04:27.450","Text":"So only this will have potential energy."},{"Start":"04:27.450 ","End":"04:34.780","Text":"It\u0027s going to be plus Mg multiplied by our z,"},{"Start":"04:34.780 ","End":"04:37.405","Text":"which we worked out over here."},{"Start":"04:37.405 ","End":"04:47.340","Text":"Multiplied by L minus r. Multiplied by its position relative to where our m is."},{"Start":"04:47.480 ","End":"04:50.300","Text":"Now, the next thing to look out for,"},{"Start":"04:50.300 ","End":"04:51.635","Text":"which is a bit confusing."},{"Start":"04:51.635 ","End":"04:56.585","Text":"Now, even though I said that my z-axis was pointing in the downwards direction,"},{"Start":"04:56.585 ","End":"05:05.255","Text":"the height of my mass M is going to be moving up or down,"},{"Start":"05:05.255 ","End":"05:08.540","Text":"where the downwards direction is a smaller height,"},{"Start":"05:08.540 ","End":"05:11.095","Text":"it\u0027s a reduction in the height."},{"Start":"05:11.095 ","End":"05:14.675","Text":"We\u0027re going to say that that\u0027s a minus when it\u0027s going down."},{"Start":"05:14.675 ","End":"05:16.250","Text":"Everywhere where I have z,"},{"Start":"05:16.250 ","End":"05:18.590","Text":"I\u0027m going to have to add a minus."},{"Start":"05:18.590 ","End":"05:24.715","Text":"That is over here and over here also."},{"Start":"05:24.715 ","End":"05:30.140","Text":"Now, the next thing that we have to do is we have to take a look at our equation."},{"Start":"05:30.140 ","End":"05:33.125","Text":"Now. We have our v_1 and our v_2,"},{"Start":"05:33.125 ","End":"05:34.775","Text":"which we don\u0027t know what that is,"},{"Start":"05:34.775 ","End":"05:38.370","Text":"or do we? Let\u0027s see how we do this."},{"Start":"05:38.370 ","End":"05:41.475","Text":"Let\u0027s take a look at our v_1."},{"Start":"05:41.475 ","End":"05:43.685","Text":"What can we say about this?"},{"Start":"05:43.685 ","End":"05:45.080","Text":"We don\u0027t really know what\u0027s happening,"},{"Start":"05:45.080 ","End":"05:53.160","Text":"but we know that our m is rotating around in a circle."},{"Start":"05:53.160 ","End":"05:57.710","Text":"The general formula for the velocity of"},{"Start":"05:57.710 ","End":"06:02.030","Text":"a body and the movement of a body when it\u0027s traveling around in a circle,"},{"Start":"06:02.030 ","End":"06:05.785","Text":"we\u0027re going to call this right now v_1,"},{"Start":"06:05.785 ","End":"06:08.325","Text":"and it\u0027s a vector quantity."},{"Start":"06:08.325 ","End":"06:14.190","Text":"What is this? It\u0027s going to be equal r dot in"},{"Start":"06:14.190 ","End":"06:22.635","Text":"the r direction plus r Theta dot in the Theta direction."},{"Start":"06:22.635 ","End":"06:27.560","Text":"Then I\u0027m going to do this in cylindrical coordinates just so that you have this,"},{"Start":"06:27.560 ","End":"06:29.840","Text":"but in this example,"},{"Start":"06:29.840 ","End":"06:32.525","Text":"I could have done it in polar coordinates or cylindrical coordinates,"},{"Start":"06:32.525 ","End":"06:33.875","Text":"it wouldn\u0027t have made a difference,"},{"Start":"06:33.875 ","End":"06:37.384","Text":"and you\u0027ll see why, plus in the z direction."},{"Start":"06:37.384 ","End":"06:45.390","Text":"Now, because my small mass is at z=0 throughout the entire movement,"},{"Start":"06:45.390 ","End":"06:48.500","Text":"everything, z is always equal to 0,"},{"Start":"06:48.500 ","End":"06:52.665","Text":"so we\u0027re going to have 0 in the z direction."},{"Start":"06:52.665 ","End":"06:55.790","Text":"Right now it\u0027s in cylindrical coordinates because I"},{"Start":"06:55.790 ","End":"06:58.669","Text":"also have included this z direction, however,"},{"Start":"06:58.669 ","End":"07:01.490","Text":"I could have done it in this example also just in"},{"Start":"07:01.490 ","End":"07:05.770","Text":"polar coordinates with just r hat and Theta hat without the z,"},{"Start":"07:05.770 ","End":"07:11.760","Text":"because anyway, we don\u0027t have any movement of our small mass in the z direction."},{"Start":"07:12.290 ","End":"07:18.585","Text":"What I\u0027ve done is I\u0027ve put a square around this and I haven\u0027t included the z,"},{"Start":"07:18.585 ","End":"07:20.930","Text":"because if we\u0027re writing"},{"Start":"07:20.930 ","End":"07:24.440","Text":"a general equation to describe the movement in polar coordinates,"},{"Start":"07:24.440 ","End":"07:27.600","Text":"just with r and Theta,"},{"Start":"07:27.600 ","End":"07:30.150","Text":"this is the general equation,"},{"Start":"07:30.150 ","End":"07:32.750","Text":"that\u0027s why I\u0027ve put it in a square because this might be"},{"Start":"07:32.750 ","End":"07:36.005","Text":"useful for you in other questions."},{"Start":"07:36.005 ","End":"07:40.555","Text":"That\u0027s v_1, and now let\u0027s take a look at v_2."},{"Start":"07:40.555 ","End":"07:43.225","Text":"v_2 as you\u0027ll notice,"},{"Start":"07:43.225 ","End":"07:47.390","Text":"it\u0027s in the equation together with our M. We\u0027re describing"},{"Start":"07:47.390 ","End":"07:54.175","Text":"the velocity with which this M moves along up and down the z-axis."},{"Start":"07:54.175 ","End":"07:58.010","Text":"What equation do we have written on the screen that we already spoke"},{"Start":"07:58.010 ","End":"08:02.030","Text":"about which is referring to this movement along the z-axis?"},{"Start":"08:02.030 ","End":"08:04.460","Text":"It\u0027s this, z is equal to L minus"},{"Start":"08:04.460 ","End":"08:13.650","Text":"r. Our L is a constant and our r is the position."},{"Start":"08:13.650 ","End":"08:15.965","Text":"How are we going to find our velocity?"},{"Start":"08:15.965 ","End":"08:19.600","Text":"Our velocity is the derivative of the position."},{"Start":"08:19.600 ","End":"08:25.080","Text":"That means that in order to find what our v_2 is equal to,"},{"Start":"08:25.080 ","End":"08:29.300","Text":"we\u0027re just going to differentiate this equation."},{"Start":"08:29.300 ","End":"08:33.170","Text":"We\u0027re going to say that it\u0027s z dot,"},{"Start":"08:33.170 ","End":"08:36.080","Text":"and of course that in the z direction."},{"Start":"08:36.080 ","End":"08:40.535","Text":"Always add in the direction because we\u0027re dealing with vectors."},{"Start":"08:40.535 ","End":"08:43.190","Text":"It\u0027s z dot in the z direction."},{"Start":"08:43.190 ","End":"08:44.975","Text":"Let\u0027s differentiate this."},{"Start":"08:44.975 ","End":"08:47.134","Text":"As we said, L is a constant."},{"Start":"08:47.134 ","End":"08:52.090","Text":"So that\u0027s just going to become 0 when we differentiate it."},{"Start":"08:52.090 ","End":"08:55.441","Text":"Then negative r when we differentiate this,"},{"Start":"08:55.441 ","End":"08:57.484","Text":"because this is the position,"},{"Start":"08:57.484 ","End":"09:01.520","Text":"it\u0027s going to become negative r dot,"},{"Start":"09:01.520 ","End":"09:08.210","Text":"and this is of course in the z direction because we\u0027re dealing with vectors."},{"Start":"09:09.530 ","End":"09:11.990","Text":"Now what we want to do is,"},{"Start":"09:11.990 ","End":"09:16.235","Text":"because we see in our equations we have v_1^2 and v_2^2,"},{"Start":"09:16.235 ","End":"09:19.685","Text":"let\u0027s just write this out."},{"Start":"09:19.685 ","End":"09:21.815","Text":"v_1^2."},{"Start":"09:21.815 ","End":"09:26.960","Text":"You\u0027ll see that when you multiply this out,"},{"Start":"09:26.960 ","End":"09:30.260","Text":"you can do this using a pen and paper,"},{"Start":"09:30.260 ","End":"09:38.805","Text":"you\u0027ll get r dot squared plus r Theta dot squared."},{"Start":"09:38.805 ","End":"09:43.700","Text":"This is what you\u0027ll get when you square your v_1."},{"Start":"09:43.700 ","End":"09:48.575","Text":"Then when we square our v_2^2,"},{"Start":"09:48.575 ","End":"09:54.930","Text":"what we will get is just simply r dot squared."},{"Start":"09:54.930 ","End":"09:56.610","Text":"Try it yourself,"},{"Start":"09:56.610 ","End":"09:59.140","Text":"it\u0027s the dot product, you\u0027ll see."},{"Start":"09:59.720 ","End":"10:01.850","Text":"Now in our question,"},{"Start":"10:01.850 ","End":"10:04.595","Text":"we\u0027re being asked to write our equation,"},{"Start":"10:04.595 ","End":"10:10.655","Text":"which is dependent only on the size r. But here we see that we have r Theta dot."},{"Start":"10:10.655 ","End":"10:13.490","Text":"What do we do? We have to try and get rid of this somehow."},{"Start":"10:13.490 ","End":"10:15.170","Text":"How are we going to do this?"},{"Start":"10:15.170 ","End":"10:21.560","Text":"We\u0027re going to write an equation using the conservation of angular momentum."},{"Start":"10:21.560 ","End":"10:26.480","Text":"Using that, hopefully we\u0027ll be able to get rid of this Theta dot."},{"Start":"10:26.480 ","End":"10:28.640","Text":"Let\u0027s see how we do this."},{"Start":"10:28.640 ","End":"10:32.180","Text":"The reason that we can say that we have a conservation of"},{"Start":"10:32.180 ","End":"10:35.780","Text":"angular momentum is because there\u0027s no friction,"},{"Start":"10:35.780 ","End":"10:38.660","Text":"so the only external force acting in"},{"Start":"10:38.660 ","End":"10:42.590","Text":"the system is going to be our Mg pointing in this direction,"},{"Start":"10:42.590 ","End":"10:47.450","Text":"which means that we have a conservation of angular momentum."},{"Start":"10:47.450 ","End":"10:51.140","Text":"Another way of looking at this is to look at"},{"Start":"10:51.140 ","End":"10:57.250","Text":"our mass m. We can see that the only force acting on our m is"},{"Start":"10:57.250 ","End":"11:01.520","Text":"acting into the center"},{"Start":"11:04.370 ","End":"11:07.500","Text":"and also it\u0027s Mg,"},{"Start":"11:07.500 ","End":"11:09.000","Text":"but into the center,"},{"Start":"11:09.000 ","End":"11:12.154","Text":"which means that it\u0027s not changing the angular momentum."},{"Start":"11:12.154 ","End":"11:15.320","Text":"It\u0027s not in the axis of the angular momentum,"},{"Start":"11:15.320 ","End":"11:17.915","Text":"which means the angular momentum is conserved."},{"Start":"11:17.915 ","End":"11:22.675","Text":"Now we have to write our equations for angular momentum."},{"Start":"11:22.675 ","End":"11:25.755","Text":"Let\u0027s do that."},{"Start":"11:25.755 ","End":"11:30.890","Text":"Then we will hopefully be able to get rid of r Theta dot."},{"Start":"11:30.890 ","End":"11:38.135","Text":"We know that the total angular momentum is going to be the angular momentum of"},{"Start":"11:38.135 ","End":"11:46.200","Text":"our m plus the angular momentum of our M."},{"Start":"11:47.160 ","End":"11:52.675","Text":"Now, when we\u0027re looking at the angular momentum of our M,"},{"Start":"11:52.675 ","End":"11:56.230","Text":"we can say that its angular momentum is equal to 0."},{"Start":"11:56.230 ","End":"11:59.860","Text":"Why? Because it\u0027s only moving up and"},{"Start":"11:59.860 ","End":"12:03.730","Text":"down the z-axis and it\u0027s not moving around in a circle."},{"Start":"12:03.730 ","End":"12:07.570","Text":"It has no angular velocity or anything."},{"Start":"12:07.570 ","End":"12:11.710","Text":"We can say that this is equal to 0."},{"Start":"12:11.710 ","End":"12:20.000","Text":"We can say that our L total is equal to just our L_m."},{"Start":"12:20.000 ","End":"12:27.040","Text":"Now what we\u0027re going to want to do is find out what exactly our equation is for our L_m."},{"Start":"12:28.650 ","End":"12:31.750","Text":"The equation in its simplest form,"},{"Start":"12:31.750 ","End":"12:36.850","Text":"so we know that it\u0027s going to be r cross p,"},{"Start":"12:36.850 ","End":"12:40.135","Text":"the radius cross the momentum."},{"Start":"12:40.135 ","End":"12:42.205","Text":"What does that equal to?"},{"Start":"12:42.205 ","End":"12:44.635","Text":"That\u0027s going to be the radius,"},{"Start":"12:44.635 ","End":"12:48.460","Text":"the distance, cross our mass,"},{"Start":"12:48.460 ","End":"12:53.725","Text":"which is small m multiplied by our v_1 vector,"},{"Start":"12:53.725 ","End":"12:57.565","Text":"mass times velocity, which is going to be equal to,"},{"Start":"12:57.565 ","End":"13:03.985","Text":"the radius in the most general form is going to be r in the r direction so r times"},{"Start":"13:03.985 ","End":"13:10.765","Text":"r hat and then cross multiply by m. Then what is our v_1?"},{"Start":"13:10.765 ","End":"13:12.280","Text":"Let\u0027s take a look."},{"Start":"13:12.280 ","End":"13:14.860","Text":"We see here in our polar coordinates,"},{"Start":"13:14.860 ","End":"13:20.890","Text":"because this is just going to cancel out to 0 multiplied by r dot in"},{"Start":"13:20.890 ","End":"13:28.220","Text":"the r direction plus r Theta dot in the Theta direction."},{"Start":"13:29.190 ","End":"13:33.025","Text":"Now when we cross multiply,"},{"Start":"13:33.025 ","End":"13:34.825","Text":"you can do this on a piece of paper."},{"Start":"13:34.825 ","End":"13:39.595","Text":"You\u0027ll see that we\u0027ll be left with mr^2,"},{"Start":"13:39.595 ","End":"13:44.690","Text":"Theta dot in the z direction."},{"Start":"13:46.140 ","End":"13:49.210","Text":"Through the conservation of angular momentum,"},{"Start":"13:49.210 ","End":"13:52.600","Text":"we know that the angular momentum at the beginning is going to"},{"Start":"13:52.600 ","End":"13:56.050","Text":"be equal to the angular momentum at the end,"},{"Start":"13:56.050 ","End":"14:00.310","Text":"and also at every single moment throughout."},{"Start":"14:00.310 ","End":"14:05.380","Text":"Now because we know that the angular momentum of M=0,"},{"Start":"14:05.380 ","End":"14:10.270","Text":"the angular momentum, this is always going to be the same."},{"Start":"14:10.270 ","End":"14:12.950","Text":"We can say at the beginning,"},{"Start":"14:12.950 ","End":"14:16.154","Text":"we know that our angular momentum,"},{"Start":"14:16.154 ","End":"14:23.910","Text":"let\u0027s write this down L_m at t=0 so right at the beginning is going to be equal"},{"Start":"14:23.910 ","End":"14:34.730","Text":"to our r cross our m v_0 in the Theta direction."},{"Start":"14:35.490 ","End":"14:40.285","Text":"Now we know that our r is equal to R,"},{"Start":"14:40.285 ","End":"14:48.580","Text":"at t=0 and then we\u0027re just going to multiply it by simply m v_0 in the z direction."},{"Start":"14:48.580 ","End":"14:55.300","Text":"Why is this? Because our v_0 is perpendicular to our r vector,"},{"Start":"14:55.300 ","End":"14:57.700","Text":"so this is going to be our answer."},{"Start":"14:57.700 ","End":"15:00.430","Text":"Then we know that at the end and throughout"},{"Start":"15:00.430 ","End":"15:04.615","Text":"every moment that this is going to be equal to L_m."},{"Start":"15:04.615 ","End":"15:13.000","Text":"Then we can say that our R mv_0 in the z direction is going to be equal to"},{"Start":"15:13.000 ","End":"15:23.080","Text":"mr^2 Theta dot in the z direction because of conservation of angular momentum."},{"Start":"15:23.080 ","End":"15:27.085","Text":"Then what we can do is we can isolate out"},{"Start":"15:27.085 ","End":"15:32.050","Text":"our Theta dot because that was our problem in the first place."},{"Start":"15:32.050 ","End":"15:39.265","Text":"We\u0027re just going to divide both sides by mr^2 z hat."},{"Start":"15:39.265 ","End":"15:44.110","Text":"These 2 cancel out and then we\u0027re going to get that our Theta dot is going"},{"Start":"15:44.110 ","End":"15:49.270","Text":"to be equal to R mv_0 divided"},{"Start":"15:49.270 ","End":"15:55.390","Text":"by mr^2 and our m can also cross out so we\u0027re going to be"},{"Start":"15:55.390 ","End":"16:02.990","Text":"left with our Theta dot being equal to R v_0 divided by r^2."},{"Start":"16:03.960 ","End":"16:09.175","Text":"Now what we can do is we have our Theta dot over here,"},{"Start":"16:09.175 ","End":"16:16.210","Text":"and we have our v_2 and we have our v_1 and our v_1^2 and our v_2^2."},{"Start":"16:16.210 ","End":"16:19.630","Text":"Now we can just substitute in our Theta dot into our v_1"},{"Start":"16:19.630 ","End":"16:24.620","Text":"squared and then substitute everything into our equation for energy."},{"Start":"16:24.840 ","End":"16:27.070","Text":"Let\u0027s see how we do this."},{"Start":"16:27.070 ","End":"16:30.070","Text":"We\u0027re going to have that our energy is equal to"},{"Start":"16:30.070 ","End":"16:37.300","Text":"negative Mg L minus R. I\u0027m literally just copying this out."},{"Start":"16:37.300 ","End":"16:42.200","Text":"Plus 1/2 mv_0^2,"},{"Start":"16:42.200 ","End":"16:48.625","Text":"which is going to be equal to 1/2 multiplied by m multiplied by v_1^2."},{"Start":"16:48.625 ","End":"16:50.110","Text":"What is v_1^2?"},{"Start":"16:50.110 ","End":"16:57.090","Text":"It\u0027s r dot squared plus r multiplied by Theta dot,"},{"Start":"16:57.090 ","End":"17:04.025","Text":"which is going to be multiplied by R v_0 divided by r^2."},{"Start":"17:04.025 ","End":"17:06.309","Text":"All of this squared,"},{"Start":"17:06.309 ","End":"17:16.165","Text":"because this is v_1^2 so mv_1^2 plus 1/2 M v^2^2,"},{"Start":"17:16.165 ","End":"17:21.820","Text":"which is going to be r dot squared minus"},{"Start":"17:21.820 ","End":"17:26.960","Text":"Mg L minus r."},{"Start":"17:28.110 ","End":"17:32.590","Text":"Now what we can do is we can simplify this equation."},{"Start":"17:32.590 ","End":"17:34.810","Text":"Notice we have negative MgL on"},{"Start":"17:34.810 ","End":"17:38.635","Text":"this side and negative MgL on this side so they can cross off."},{"Start":"17:38.635 ","End":"17:47.935","Text":"We\u0027ll be left with MgR plus 1/2 mv_0^2,"},{"Start":"17:47.935 ","End":"17:57.460","Text":"which will equal to 1/2 m r dot squared plus 1/2 m and then"},{"Start":"17:57.460 ","End":"18:01.705","Text":"this r will cross off with the squared over here plus 1/2"},{"Start":"18:01.705 ","End":"18:07.795","Text":"R v_0 over r^2 because remember there\u0027s a squared here,"},{"Start":"18:07.795 ","End":"18:13.660","Text":"plus 1/2 m r dot squared."},{"Start":"18:13.660 ","End":"18:22.555","Text":"Then our minus MgL was crossed off with this one, so plus Mgr."},{"Start":"18:22.555 ","End":"18:28.555","Text":"Now, all we have to do is rearrange this a little bit."},{"Start":"18:28.555 ","End":"18:30.820","Text":"What we can do is in the end,"},{"Start":"18:30.820 ","End":"18:35.080","Text":"we\u0027ll multiply everything by 2 to get rid of the halves."},{"Start":"18:35.080 ","End":"18:40.390","Text":"We\u0027ll have 2 MgR plus"},{"Start":"18:40.390 ","End":"18:46.645","Text":"mv_0^2 minus Mgr,"},{"Start":"18:46.645 ","End":"18:48.700","Text":"so we\u0027re minusing this,"},{"Start":"18:48.700 ","End":"18:54.460","Text":"minus Mgr and then minus,"},{"Start":"18:54.460 ","End":"19:02.270","Text":"let\u0027s see, mv_0 R^2."},{"Start":"19:04.260 ","End":"19:10.780","Text":"This will be multiplied by 1/r^2."},{"Start":"19:10.780 ","End":"19:13.390","Text":"Then this will equal to"},{"Start":"19:13.390 ","End":"19:22.525","Text":"m plus m multiplied by r dot squared."},{"Start":"19:22.525 ","End":"19:25.870","Text":"All I\u0027ve done is rearrange this equation."},{"Start":"19:25.870 ","End":"19:28.750","Text":"You can do this on your own on a piece of"},{"Start":"19:28.750 ","End":"19:32.630","Text":"paper now to make sure that you understand the algebra."},{"Start":"19:32.850 ","End":"19:39.745","Text":"What we will get in the end is an equation in the form of,"},{"Start":"19:39.745 ","End":"19:44.245","Text":"let\u0027s see, we\u0027ll have to 2 MgR,"},{"Start":"19:44.245 ","End":"19:47.330","Text":"we\u0027ll divide both sides by m plus"},{"Start":"19:47.340 ","End":"19:59.000","Text":"m minus and then we\u0027ll have 2Mg."},{"Start":"19:59.310 ","End":"20:04.135","Text":"This is also going to be divided by m plus"},{"Start":"20:04.135 ","End":"20:12.070","Text":"m multiplied by r and then,"},{"Start":"20:12.070 ","End":"20:22.420","Text":"sorry, this is also going to be a minus mv_0 R^2"},{"Start":"20:22.420 ","End":"20:28.120","Text":"divided by m plus"},{"Start":"20:28.120 ","End":"20:33.775","Text":"m. Then this is going to be equal to r dot squared."},{"Start":"20:33.775 ","End":"20:41.170","Text":"Now all we\u0027ll do is we will square root both sides to get rid of the squared over here."},{"Start":"20:41.170 ","End":"20:43.730","Text":"We can even do this now."},{"Start":"20:45.800 ","End":"20:49.300","Text":"It\u0027s a bit messy. I\u0027m sorry."},{"Start":"20:50.510 ","End":"20:59.665","Text":"Now you\u0027ll notice that our r dot is equal to dr by dt."},{"Start":"20:59.665 ","End":"21:03.970","Text":"We can see that everything here is a known,"},{"Start":"21:03.970 ","End":"21:09.940","Text":"M and m and R and g. We know what these values are,"},{"Start":"21:09.940 ","End":"21:11.110","Text":"they were given in the question."},{"Start":"21:11.110 ","End":"21:17.380","Text":"Our only unknown is this r over here and our dr by dt."},{"Start":"21:17.380 ","End":"21:21.930","Text":"This is exactly what we were asked to find in the question."},{"Start":"21:21.930 ","End":"21:23.805","Text":"This is the differential equation."},{"Start":"21:23.805 ","End":"21:29.430","Text":"Now, all we have to do is separate out our unknowns and do the integration."},{"Start":"21:29.430 ","End":"21:32.460","Text":"Now the integration is a little bit complicated and"},{"Start":"21:32.460 ","End":"21:35.880","Text":"also that\u0027s something that you should be doing in your next quizzes."},{"Start":"21:35.880 ","End":"21:37.290","Text":"I\u0027m not going to go into it now,"},{"Start":"21:37.290 ","End":"21:41.175","Text":"but this is the differential equation that you were asked to find."},{"Start":"21:41.175 ","End":"21:43.560","Text":"If you would like to attempt to solve this,"},{"Start":"21:43.560 ","End":"21:45.170","Text":"by all means, please do."},{"Start":"21:45.170 ","End":"21:48.350","Text":"But this is the end of this lesson."}],"ID":12704},{"Watched":false,"Name":"2.1 Angular Momentum of CM Intro","Duration":"1m 18s","ChapterTopicVideoID":12226,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/12226.jpeg","UploadDate":"2021-11-19T08:39:45.5770000","DurationForVideoObject":"PT1M18S","Description":null,"MetaTitle":"2.1 Angular Momentum of CM Intro: Video + Workbook | Proprep","MetaDescription":"Angular Momentum - Equation and Laws of Conservation. 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/angular-momentum/equation-and-laws-of-conservation/vid12705","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"In this lesson, we\u0027re going to derive an equation which is"},{"Start":"00:03.210 ","End":"00:07.010","Text":"very important and will be very useful in the future."},{"Start":"00:07.010 ","End":"00:10.200","Text":"The equation is going to be speaking about a system,"},{"Start":"00:10.200 ","End":"00:13.230","Text":"where we have for instance these 2 balls,"},{"Start":"00:13.230 ","End":"00:17.865","Text":"which is both rotating around its center of mass,"},{"Start":"00:17.865 ","End":"00:21.850","Text":"and its center of mass is moving."},{"Start":"00:22.910 ","End":"00:27.690","Text":"Our goal is to work out the angular momentum of"},{"Start":"00:27.690 ","End":"00:32.040","Text":"this type of system relative to some point of reference."},{"Start":"00:32.040 ","End":"00:34.449","Text":"Let\u0027s say over here."},{"Start":"00:34.910 ","End":"00:39.210","Text":"The angular momentum of this whole system is going to be equal"},{"Start":"00:39.210 ","End":"00:43.205","Text":"to the angular momentum of the center of mass,"},{"Start":"00:43.205 ","End":"00:46.650","Text":"plus the internal angular momentum."},{"Start":"00:46.650 ","End":"00:53.545","Text":"What is that mean? The angular momentum relative to the center of mass."},{"Start":"00:53.545 ","End":"00:59.330","Text":"Obviously, this equation isn\u0027t just correct to assist them with just 2 balls,"},{"Start":"00:59.330 ","End":"01:02.990","Text":"but is rather correct for every system no matter"},{"Start":"01:02.990 ","End":"01:09.210","Text":"what shape the bodies and the system are and how many bodies there are in the system."},{"Start":"01:09.210 ","End":"01:12.860","Text":"1st, what we\u0027re going to do is we\u0027re going"},{"Start":"01:12.860 ","End":"01:16.565","Text":"to see how we use the equation and then afterwards,"},{"Start":"01:16.565 ","End":"01:19.050","Text":"we\u0027ll see how to derive it."}],"ID":12705},{"Watched":false,"Name":"2.2 Using the Equation","Duration":"9m 17s","ChapterTopicVideoID":12227,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Right now, what we\u0027re going to do is we\u0027re"},{"Start":"00:03.030 ","End":"00:06.240","Text":"going to take a look at this example right over here."},{"Start":"00:06.240 ","End":"00:10.770","Text":"We\u0027re going to say that our 2 balls are rotating about the center of"},{"Start":"00:10.770 ","End":"00:15.795","Text":"mass with an angular velocity of Omega 0."},{"Start":"00:15.795 ","End":"00:19.196","Text":"Then also, our center of mass,"},{"Start":"00:19.196 ","End":"00:22.710","Text":"the whole system is also moving forwards, so in the meantime,"},{"Start":"00:22.710 ","End":"00:26.070","Text":"let\u0027s just say it\u0027s moving forwards in a straight line with"},{"Start":"00:26.070 ","End":"00:30.885","Text":"a linear velocity of V. Let\u0027s call it V_c.m."},{"Start":"00:30.885 ","End":"00:40.035","Text":"The whole system is going to be moving something like this in this kind of situation."},{"Start":"00:40.035 ","End":"00:44.330","Text":"For someone standing in the lab watching this,"},{"Start":"00:44.330 ","End":"00:47.970","Text":"it\u0027s going to be pretty hard to calculate what\u0027s going on."},{"Start":"00:47.970 ","End":"00:51.050","Text":"Let\u0027s see how we can do it and let\u0027s just write out"},{"Start":"00:51.050 ","End":"00:54.650","Text":"the equation again using our equation,"},{"Start":"00:54.650 ","End":"00:59.435","Text":"which is the total angular momentum is going to equal to"},{"Start":"00:59.435 ","End":"01:06.875","Text":"the angular momentum of the center of mass plus the angular momentum internal."},{"Start":"01:06.875 ","End":"01:12.590","Text":"The first things that we can do is we can see that in the system, there are 2 balls,"},{"Start":"01:12.590 ","End":"01:19.545","Text":"so we\u0027ll say that this ball is of mass m1 and this ball is of mass m2."},{"Start":"01:19.545 ","End":"01:22.930","Text":"Now, what we can do is we can start by"},{"Start":"01:22.930 ","End":"01:26.890","Text":"working out the angular momentum of the center of mass."},{"Start":"01:26.890 ","End":"01:30.040","Text":"Now, instead of considering this a large system,"},{"Start":"01:30.040 ","End":"01:34.975","Text":"we\u0027re going to consider it a point system over here. How do we do this?"},{"Start":"01:34.975 ","End":"01:43.030","Text":"We know that our equation for angular momentum is going to be equal to r"},{"Start":"01:43.030 ","End":"01:52.600","Text":"cross r momentum and this is going to be of cm and this is also going to be of cm."},{"Start":"01:52.600 ","End":"01:54.545","Text":"Now, what is our P?"},{"Start":"01:54.545 ","End":"01:56.835","Text":"What\u0027s our momentum?"},{"Start":"01:56.835 ","End":"02:00.760","Text":"We know that it\u0027s equal to the total mass of"},{"Start":"02:00.760 ","End":"02:06.035","Text":"the system multiplied by the velocity of the center of mass."},{"Start":"02:06.035 ","End":"02:12.185","Text":"Then we can say that it\u0027s r_cm cross-product with this,"},{"Start":"02:12.185 ","End":"02:14.695","Text":"which is equal to the total mass,"},{"Start":"02:14.695 ","End":"02:25.040","Text":"so m1 plus m2 and the system multiplied by the velocity of the center of mass."},{"Start":"02:25.040 ","End":"02:27.995","Text":"Then this m1 plus m2,"},{"Start":"02:27.995 ","End":"02:29.435","Text":"we can multiply it here,"},{"Start":"02:29.435 ","End":"02:31.889","Text":"or we can move it to the beginning of the equation,"},{"Start":"02:31.889 ","End":"02:34.925","Text":"it doesn\u0027t matter because it\u0027s a scalar number."},{"Start":"02:34.925 ","End":"02:38.300","Text":"Then what we can do is in order to find"},{"Start":"02:38.300 ","End":"02:44.555","Text":"what the size of our angular momentum is, remember this equation."},{"Start":"02:44.555 ","End":"02:51.125","Text":"Let\u0027s call this over here M to shorten this,"},{"Start":"02:51.125 ","End":"02:55.655","Text":"so we can say that it\u0027s the total mass of the system,"},{"Start":"02:55.655 ","End":"03:00.660","Text":"multiplied by the size of the r_cm,"},{"Start":"03:01.040 ","End":"03:06.800","Text":"multiplied by the size of our momentum,"},{"Start":"03:06.800 ","End":"03:08.585","Text":"which is just going to be right now our"},{"Start":"03:08.585 ","End":"03:18.025","Text":"V. Then this is multiplied by sine of the angle between them."},{"Start":"03:18.025 ","End":"03:21.750","Text":"Now let\u0027s take a look at what our r is."},{"Start":"03:21.750 ","End":"03:28.130","Text":"Our r is going to go from here all the way until the center of mass,"},{"Start":"03:28.130 ","End":"03:31.355","Text":"because right now we\u0027re dealing with the center of mass,"},{"Start":"03:31.355 ","End":"03:39.200","Text":"so I\u0027ll write this r_cm and now if for instance we know what this height is over here,"},{"Start":"03:39.200 ","End":"03:41.100","Text":"so if we know that,"},{"Start":"03:41.100 ","End":"03:43.290","Text":"let\u0027s call this height d,"},{"Start":"03:43.290 ","End":"03:46.770","Text":"then where we have sine of Alpha,"},{"Start":"03:46.770 ","End":"03:48.095","Text":"so as we know,"},{"Start":"03:48.095 ","End":"03:54.575","Text":"sine of Alpha is equal to the opposite over the hypotenuse."},{"Start":"03:54.575 ","End":"03:57.400","Text":"What is the opposite and the hypotenuse over here?"},{"Start":"03:57.400 ","End":"04:00.470","Text":"We can see that our angle Alpha will be over here,"},{"Start":"04:00.470 ","End":"04:03.178","Text":"so the opposite side will be the d,"},{"Start":"04:03.178 ","End":"04:06.975","Text":"because this is the side opposite to the angle and"},{"Start":"04:06.975 ","End":"04:12.125","Text":"the hypotenuse is obviously going to be the r divided by r_cm."},{"Start":"04:12.125 ","End":"04:14.975","Text":"When we plug all of this into the equation,"},{"Start":"04:14.975 ","End":"04:22.460","Text":"so we\u0027ll have our m multiplied by our V_cm and then our r and our r will"},{"Start":"04:22.460 ","End":"04:30.630","Text":"cancel out multiplied by our d. All of this is just our L_cm."},{"Start":"04:30.630 ","End":"04:33.050","Text":"It\u0027s very important, sorry, I didn\u0027t write it."},{"Start":"04:33.050 ","End":"04:34.985","Text":"This is our L_cm."},{"Start":"04:34.985 ","End":"04:38.605","Text":"Now we want to find out what our L internal is."},{"Start":"04:38.605 ","End":"04:40.875","Text":"What is the trick over here?"},{"Start":"04:40.875 ","End":"04:43.425","Text":"Now I\u0027m trying to find out the L internal."},{"Start":"04:43.425 ","End":"04:47.600","Text":"That means that I\u0027m not going to be looking at the center of mass,"},{"Start":"04:47.600 ","End":"04:49.715","Text":"I\u0027m looking relative to the center of mass."},{"Start":"04:49.715 ","End":"04:53.575","Text":"That means that I can rub out everything."},{"Start":"04:53.575 ","End":"04:55.070","Text":"I can rub out this,"},{"Start":"04:55.070 ","End":"04:59.945","Text":"I\u0027m not looking to this point and I can rub out even this point of reference,"},{"Start":"04:59.945 ","End":"05:06.284","Text":"it\u0027s not relative to me and now I\u0027m a person standing on top of the center of mass."},{"Start":"05:06.284 ","End":"05:11.550","Text":"I\u0027m standing at the center of mass and I\u0027m looking at the balls and their movement,"},{"Start":"05:11.550 ","End":"05:13.790","Text":"so I\u0027ll draw it in purple just to make it more clear."},{"Start":"05:13.790 ","End":"05:19.340","Text":"I can see this ball going in this direction at Omega 0 and this 1 going like"},{"Start":"05:19.340 ","End":"05:22.730","Text":"this with angular velocity Omega 0 and now"},{"Start":"05:22.730 ","End":"05:26.930","Text":"I\u0027m trying to find when I\u0027m standing here at the center of mass,"},{"Start":"05:26.930 ","End":"05:32.420","Text":"the angular momentum of the bodies that are spinning around me."},{"Start":"05:32.420 ","End":"05:34.625","Text":"This is super simple."},{"Start":"05:34.625 ","End":"05:40.475","Text":"All I have to do is I have to know the distance each body is from the center of mass."},{"Start":"05:40.475 ","End":"05:43.025","Text":"In this example, specifically,"},{"Start":"05:43.025 ","End":"05:45.740","Text":"both of the balls are a distance"},{"Start":"05:45.740 ","End":"05:50.330","Text":"R away from the center of mass and it\u0027s the same over here."},{"Start":"05:50.330 ","End":"05:52.505","Text":"Now all I have to do,"},{"Start":"05:52.505 ","End":"05:55.400","Text":"so I can just say that my L and let\u0027s just say,"},{"Start":"05:55.400 ","End":"05:58.190","Text":"let\u0027s call it L int instead of L internal,"},{"Start":"05:58.190 ","End":"06:00.515","Text":"this is going to equal to,"},{"Start":"06:00.515 ","End":"06:02.290","Text":"we\u0027ll do for the first body,"},{"Start":"06:02.290 ","End":"06:06.950","Text":"so m1 multiplied by its distance from the center of mass,"},{"Start":"06:06.950 ","End":"06:11.520","Text":"which is R, multiplied by its angular velocity."},{"Start":"06:12.020 ","End":"06:20.300","Text":"This is when we\u0027re dealing with circular motion."},{"Start":"06:20.300 ","End":"06:23.990","Text":"Obviously, if these shapes aren\u0027t moving in circular motion,"},{"Start":"06:23.990 ","End":"06:27.785","Text":"then we would just write our velocity accordingly."},{"Start":"06:27.785 ","End":"06:30.380","Text":"Then I do the exact same thing for the second body,"},{"Start":"06:30.380 ","End":"06:32.165","Text":"so that\u0027s the mass of the second body,"},{"Start":"06:32.165 ","End":"06:37.115","Text":"m2 multiplied by its distance from the center of mass,"},{"Start":"06:37.115 ","End":"06:39.020","Text":"because we are relative to the center of mass."},{"Start":"06:39.020 ","End":"06:40.895","Text":"That\u0027s R and then again,"},{"Start":"06:40.895 ","End":"06:43.880","Text":"multiplied by the velocity which is"},{"Start":"06:43.880 ","End":"06:51.924","Text":"Omega 0 R. That is for our angular momentum of the internal."},{"Start":"06:51.924 ","End":"06:56.975","Text":"Now, all I have to do is I write finally that"},{"Start":"06:56.975 ","End":"07:04.400","Text":"therefore my L is equal to and then I just rewrite this out,"},{"Start":"07:04.400 ","End":"07:10.495","Text":"so I\u0027ll have mv center of mass d plus"},{"Start":"07:10.495 ","End":"07:20.120","Text":"m1 plus m2R^2 Omega 0."},{"Start":"07:20.120 ","End":"07:22.010","Text":"Now you can see how easy that was,"},{"Start":"07:22.010 ","End":"07:25.055","Text":"which is much easier than working out for each body,"},{"Start":"07:25.055 ","End":"07:31.500","Text":"moving with this type of movement and working out its angular momentum."},{"Start":"07:32.120 ","End":"07:36.445","Text":"An important thing to note about this equation"},{"Start":"07:36.445 ","End":"07:40.840","Text":"might make it easier to remember is that this equation is"},{"Start":"07:40.840 ","End":"07:49.435","Text":"very similar to the equation for velocity of a body relative to some point of reference."},{"Start":"07:49.435 ","End":"07:53.620","Text":"If you remember, if we have some body here, let\u0027s call it 1,"},{"Start":"07:53.620 ","End":"08:00.640","Text":"and then we can say that our velocity of body number 1 relative to it\u0027s center of"},{"Start":"08:00.640 ","End":"08:04.270","Text":"mass will be equal to the velocity of"},{"Start":"08:04.270 ","End":"08:08.470","Text":"body number 1 minus the velocity of the center of mass."},{"Start":"08:08.470 ","End":"08:12.385","Text":"Then we can also rewrite this as the velocity of 1"},{"Start":"08:12.385 ","End":"08:16.625","Text":"is equal to the velocity of the center of mass,"},{"Start":"08:16.625 ","End":"08:22.295","Text":"plus the velocity of 1 relative to the center of mass."},{"Start":"08:22.295 ","End":"08:24.055","Text":"This is the exact same thing,"},{"Start":"08:24.055 ","End":"08:30.875","Text":"so we can look at our total angular momentum as equaling to this,"},{"Start":"08:30.875 ","End":"08:34.030","Text":"the velocity relative to the center of mass."},{"Start":"08:34.030 ","End":"08:36.755","Text":"If I draw in this system,"},{"Start":"08:36.755 ","End":"08:39.890","Text":"again with the person standing here,"},{"Start":"08:39.890 ","End":"08:42.320","Text":"so we can say that my L_cm,"},{"Start":"08:42.320 ","End":"08:48.530","Text":"this L of the center of mass is the same as my V_cm,"},{"Start":"08:48.530 ","End":"08:52.830","Text":"so the person standing here relative to the person standing here."},{"Start":"08:53.110 ","End":"08:57.945","Text":"Then this is like the tag that\u0027s over here,"},{"Start":"08:57.945 ","End":"09:01.220","Text":"so it\u0027s the L internal when we\u0027re standing inside"},{"Start":"09:01.220 ","End":"09:04.700","Text":"the system and we\u0027re seeing the angular momentum going on around it,"},{"Start":"09:04.700 ","End":"09:06.380","Text":"so it\u0027s like when we\u0027re standing in the system"},{"Start":"09:06.380 ","End":"09:08.860","Text":"and we\u0027re seeing the velocity going around it."},{"Start":"09:08.860 ","End":"09:13.160","Text":"Now we\u0027re going to see how we derive this equation and it\u0027s"},{"Start":"09:13.160 ","End":"09:18.030","Text":"going to be in a very similar way to how we derive this equation."}],"ID":12706},{"Watched":false,"Name":"2.3 Multiple Bodies- Same Equation","Duration":"2m 57s","ChapterTopicVideoID":12228,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.560","Text":"Let\u0027s before we do that,"},{"Start":"00:01.560 ","End":"00:06.045","Text":"speak about a system that has more bodies moving around."},{"Start":"00:06.045 ","End":"00:07.410","Text":"We have a body here,"},{"Start":"00:07.410 ","End":"00:08.760","Text":"a body here, a body here,"},{"Start":"00:08.760 ","End":"00:12.850","Text":"a body here, it doesn\u0027t really matter, lots of bodies."},{"Start":"00:13.340 ","End":"00:18.540","Text":"We look at the equation in the exact same way."},{"Start":"00:18.540 ","End":"00:24.600","Text":"Our L_ cm will be the exact same thing because we\u0027re just speaking about the distance"},{"Start":"00:24.600 ","End":"00:30.480","Text":"from the origin until the center of mass cross-product with the total mass of the system."},{"Start":"00:30.480 ","End":"00:36.620","Text":"We\u0027ll just say that the total mass of the system is this capital M, V_CM."},{"Start":"00:36.620 ","End":"00:38.765","Text":"It\u0027s going to be the exact same thing."},{"Start":"00:38.765 ","End":"00:43.085","Text":"Then when we\u0027re just working out the size of our L_CM,"},{"Start":"00:43.085 ","End":"00:46.100","Text":"so we just take the size of the radius,"},{"Start":"00:46.100 ","End":"00:47.705","Text":"the size of the velocity,"},{"Start":"00:47.705 ","End":"00:50.010","Text":"multiplied by sine of the angle."},{"Start":"00:50.010 ","End":"00:52.850","Text":"We can use the same geometry to find out"},{"Start":"00:52.850 ","End":"00:57.000","Text":"what sine of the angle is even if we do not have the angle included."},{"Start":"00:57.000 ","End":"01:00.425","Text":"The L_CM calculation will remain exactly the same."},{"Start":"01:00.425 ","End":"01:03.100","Text":"But when we\u0027re dealing with the L internal equation,"},{"Start":"01:03.100 ","End":"01:06.270","Text":"so here the equation could be slightly different."},{"Start":"01:06.270 ","End":"01:07.910","Text":"We\u0027d have to work out,"},{"Start":"01:07.910 ","End":"01:11.795","Text":"instead of just doing m_1 multiplied by"},{"Start":"01:11.795 ","End":"01:15.845","Text":"the distance from the center of mass and the velocity."},{"Start":"01:15.845 ","End":"01:18.680","Text":"Notice here this is in circular motion,"},{"Start":"01:18.680 ","End":"01:24.225","Text":"and then m_2 so we would also have add in our m_3,"},{"Start":"01:24.225 ","End":"01:28.790","Text":"m_4, and so on and so forth multiplied by the distance,"},{"Start":"01:28.790 ","End":"01:30.410","Text":"which could be different."},{"Start":"01:30.410 ","End":"01:33.970","Text":"This could be r,"},{"Start":"01:33.970 ","End":"01:36.405","Text":"and this is r_1."},{"Start":"01:36.405 ","End":"01:38.760","Text":"All of these distances are different."},{"Start":"01:38.760 ","End":"01:45.255","Text":"This could be changing by distance and then R velocity."},{"Start":"01:45.255 ","End":"01:49.410","Text":"Here it was Omega_ 0 multiplied by R because we\u0027re in circular motion."},{"Start":"01:49.410 ","End":"01:51.305","Text":"If we\u0027re not traveling in circular motion,"},{"Start":"01:51.305 ","End":"01:54.575","Text":"as I said, these bodies could be moving not in circular motion."},{"Start":"01:54.575 ","End":"01:56.330","Text":"How would we do this?"},{"Start":"01:56.330 ","End":"01:59.365","Text":"Let\u0027s just clear this out."},{"Start":"01:59.365 ","End":"02:02.765","Text":"How would we do this if they\u0027re not moving in circular motion?"},{"Start":"02:02.765 ","End":"02:09.450","Text":"Here, notice that we have it moving at 90 degrees."},{"Start":"02:10.720 ","End":"02:16.675","Text":"Sine 90 is going to equal 1."},{"Start":"02:16.675 ","End":"02:22.040","Text":"That\u0027s why this is as if we did Omega_ 0 multiplied by R,"},{"Start":"02:22.040 ","End":"02:26.165","Text":"multiplied by sine(90) degrees,"},{"Start":"02:26.165 ","End":"02:28.490","Text":"which is equal to 1."},{"Start":"02:28.490 ","End":"02:31.460","Text":"If this isn\u0027t moving in circular motion,"},{"Start":"02:31.460 ","End":"02:34.325","Text":"then this angle won\u0027t be 90,"},{"Start":"02:34.325 ","End":"02:36.875","Text":"it will be some kind of Alpha."},{"Start":"02:36.875 ","End":"02:40.135","Text":"Then we\u0027ll multiply it by sine(Alpha)."},{"Start":"02:40.135 ","End":"02:45.005","Text":"Then we\u0027ll just get the exact same answer but with some angle."},{"Start":"02:45.005 ","End":"02:47.300","Text":"Again, if there\u0027s a way to find out what"},{"Start":"02:47.300 ","End":"02:52.160","Text":"sine(Alpha) is just like we did here with geometry and trigonometry,"},{"Start":"02:52.160 ","End":"02:57.700","Text":"we can do the same thing with the each individual body."}],"ID":12707},{"Watched":false,"Name":"2.4 Center of Mass Doesnt Need to Move in a Straight Line","Duration":"1m ","ChapterTopicVideoID":12229,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:06.203","Text":"Now, similarly, if our center of mass isn\u0027t moving in a straight line,"},{"Start":"00:06.203 ","End":"00:10.905","Text":"if for instance our center of mass is moving in a more complicated formation,"},{"Start":"00:10.905 ","End":"00:14.445","Text":"say in a circle or whatever it might be."},{"Start":"00:14.445 ","End":"00:17.010","Text":"Our center of mass now is moving."},{"Start":"00:17.010 ","End":"00:20.460","Text":"Then our equation for our L_CM will be different."},{"Start":"00:20.460 ","End":"00:24.810","Text":"Because our sine Alpha won\u0027t be d over the radius."},{"Start":"00:24.810 ","End":"00:26.520","Text":"It will be something else."},{"Start":"00:26.520 ","End":"00:30.149","Text":"But we can still use the exact same equation."},{"Start":"00:30.149 ","End":"00:36.165","Text":"That means that this equation is correct for however many shapes"},{"Start":"00:36.165 ","End":"00:42.290","Text":"we have moving around the center of mass and for any movement that they have,"},{"Start":"00:42.290 ","End":"00:48.400","Text":"be it circular motion or whatever it might be around the center of mass."},{"Start":"00:48.400 ","End":"00:53.585","Text":"It\u0027s also correct for the center of mass moving in any type of formation,"},{"Start":"00:53.585 ","End":"00:56.870","Text":"in a straight line, in a circle, whatever it might be."},{"Start":"00:56.870 ","End":"01:00.930","Text":"This equation works for absolutely everything."}],"ID":12708},{"Watched":false,"Name":"2.5 Deriving the Equation","Duration":"15m 47s","ChapterTopicVideoID":12230,"CourseChapterTopicPlaylistID":5331,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.960","Text":"Right now, I\u0027m actually going to derive the equation."},{"Start":"00:03.960 ","End":"00:06.525","Text":"For those of you who this isn\u0027t relevant to,"},{"Start":"00:06.525 ","End":"00:09.195","Text":"then this is the end of this video."},{"Start":"00:09.195 ","End":"00:12.585","Text":"For starters, I want to work out"},{"Start":"00:12.585 ","End":"00:18.990","Text":"the angular momentum for these 2 bodies relative to this point over here."},{"Start":"00:18.990 ","End":"00:21.435","Text":"Let\u0027s do this without using the center of mass."},{"Start":"00:21.435 ","End":"00:23.025","Text":"Imagine this isn\u0027t here."},{"Start":"00:23.025 ","End":"00:26.445","Text":"Let\u0027s call this body 1 and body 2."},{"Start":"00:26.445 ","End":"00:29.210","Text":"Let\u0027s work out for each ball what\u0027s happening."},{"Start":"00:29.210 ","End":"00:30.875","Text":"This ball, number 1,"},{"Start":"00:30.875 ","End":"00:34.070","Text":"isn\u0027t moving exactly in a circle because"},{"Start":"00:34.070 ","End":"00:37.675","Text":"we know that this whole thing is moving with some velocity."},{"Start":"00:37.675 ","End":"00:39.470","Text":"v_0, let\u0027s call it."},{"Start":"00:39.470 ","End":"00:43.055","Text":"If we look just at the movement of this body over here,"},{"Start":"00:43.055 ","End":"00:47.390","Text":"we\u0027ll see that it\u0027s moving like this."},{"Start":"00:47.390 ","End":"00:53.730","Text":"Let\u0027s work out the angular momentum of this. Let\u0027s write this down."},{"Start":"00:53.730 ","End":"00:59.120","Text":"We have the angular momentum of body number 1 is, as we know,"},{"Start":"00:59.120 ","End":"01:02.360","Text":"the basic equation is going to be my r vector,"},{"Start":"01:02.360 ","End":"01:06.770","Text":"cross multiplied with my momentum vector."},{"Start":"01:06.770 ","End":"01:08.360","Text":"Let\u0027s see what this is."},{"Start":"01:08.360 ","End":"01:11.600","Text":"My r vector is going to be from here,"},{"Start":"01:11.600 ","End":"01:17.015","Text":"my point of reference, until the body itself."},{"Start":"01:17.015 ","End":"01:23.310","Text":"Let\u0027s call this my r. Obviously over here,"},{"Start":"01:23.310 ","End":"01:26.015","Text":"this is r_1 and this is our momentum."},{"Start":"01:26.015 ","End":"01:30.920","Text":"Now, my momentum is going something like this."},{"Start":"01:30.920 ","End":"01:33.815","Text":"Now, let\u0027s see what our momentum is equal to."},{"Start":"01:33.815 ","End":"01:38.915","Text":"My P_1 is going to be equal to the mass of my first body,"},{"Start":"01:38.915 ","End":"01:40.340","Text":"let\u0027s call it m_1,"},{"Start":"01:40.340 ","End":"01:44.500","Text":"multiplied by the velocity."},{"Start":"01:44.500 ","End":"01:47.095","Text":"Now we\u0027re looking at this v_1."},{"Start":"01:47.095 ","End":"01:50.030","Text":"We don\u0027t really know what this is."},{"Start":"01:50.030 ","End":"01:52.520","Text":"How do we work this out?"},{"Start":"01:52.520 ","End":"01:55.265","Text":"This isn\u0027t given in the question. I don\u0027t know what this is."},{"Start":"01:55.265 ","End":"01:59.940","Text":"What we\u0027re going to do is now we\u0027re going to start working with our center of mass."},{"Start":"01:59.940 ","End":"02:04.415","Text":"We\u0027re going to imagine that we\u0027re standing right over here at the center of mass."},{"Start":"02:04.415 ","End":"02:06.005","Text":"A few minutes ago,"},{"Start":"02:06.005 ","End":"02:10.160","Text":"we spoke about the equation for the velocity of"},{"Start":"02:10.160 ","End":"02:15.095","Text":"a body relative to some frame of reference."},{"Start":"02:15.095 ","End":"02:25.585","Text":"I know that the velocity of this ball number 1 relative to the center of mass,"},{"Start":"02:25.585 ","End":"02:28.039","Text":"relative to my center of mass,"},{"Start":"02:28.039 ","End":"02:34.970","Text":"is going to be equal to the velocity of the ball relative to my point of"},{"Start":"02:34.970 ","End":"02:42.710","Text":"reference minus the velocity of my center of mass."},{"Start":"02:42.710 ","End":"02:46.175","Text":"Now notice that in our equation,"},{"Start":"02:46.175 ","End":"02:49.940","Text":"we are trying to find v_1, which is this."},{"Start":"02:49.940 ","End":"02:56.820","Text":"What we\u0027re going to do now is we are going to isolate out our v_1. Let\u0027s write this."},{"Start":"02:56.820 ","End":"03:04.665","Text":"I\u0027ll have that our v_1 is going to be equal to our v tag,"},{"Start":"03:04.665 ","End":"03:07.185","Text":"which is this,"},{"Start":"03:07.185 ","End":"03:12.500","Text":"plus our velocity of the center of mass."},{"Start":"03:12.500 ","End":"03:17.350","Text":"Now, we\u0027ll substitute this into our equation for momentum,"},{"Start":"03:17.350 ","End":"03:20.840","Text":"and then we will get that, therefore,"},{"Start":"03:20.840 ","End":"03:24.850","Text":"our momentum for our first body is going to be"},{"Start":"03:24.850 ","End":"03:28.570","Text":"equal to the mass of the first body multiplied"},{"Start":"03:28.570 ","End":"03:37.250","Text":"by our v tag 1cm plus our v_cm."},{"Start":"03:37.250 ","End":"03:39.759","Text":"Now notice because I\u0027m dealing with vectors,"},{"Start":"03:39.759 ","End":"03:44.065","Text":"so we don\u0027t have to worry about splitting this up into the different components."},{"Start":"03:44.065 ","End":"03:50.170","Text":"Then, therefore, if we substitute this into our equation for angular momentum."},{"Start":"03:50.170 ","End":"03:53.800","Text":"We\u0027ll get that the angular momentum of the first body is equal"},{"Start":"03:53.800 ","End":"03:57.775","Text":"to our radius 1, our position,"},{"Start":"03:57.775 ","End":"04:00.770","Text":"cross multiplied by our momentum,"},{"Start":"04:00.770 ","End":"04:03.425","Text":"which is m_1,"},{"Start":"04:03.425 ","End":"04:06.310","Text":"and then this,"},{"Start":"04:06.310 ","End":"04:10.280","Text":"and then also with the center of mass."},{"Start":"04:10.280 ","End":"04:14.705","Text":"Now, notice that our m_1 is a scalar multiple,"},{"Start":"04:14.705 ","End":"04:17.390","Text":"which means that we can just put it also over"},{"Start":"04:17.390 ","End":"04:20.560","Text":"here as the coefficient of r. It doesn\u0027t really matter."},{"Start":"04:20.560 ","End":"04:24.485","Text":"Then we\u0027re left with what\u0027s going on here inside the brackets."},{"Start":"04:24.485 ","End":"04:26.600","Text":"Now when we\u0027re doing cross multiplication,"},{"Start":"04:26.600 ","End":"04:32.600","Text":"we can multiply each item in the bracket as if we were doing normal multiplication."},{"Start":"04:32.600 ","End":"04:34.340","Text":"Let\u0027s see how we do this."},{"Start":"04:34.340 ","End":"04:36.785","Text":"Now let\u0027s open up the brackets."},{"Start":"04:36.785 ","End":"04:43.770","Text":"Therefore, we\u0027ll get that our m_1 multiplied by r_1."},{"Start":"04:43.770 ","End":"04:46.365","Text":"Then it\u0027s going to be cross."},{"Start":"04:46.365 ","End":"04:47.900","Text":"First, we\u0027ll do this."},{"Start":"04:47.900 ","End":"04:50.870","Text":"You\u0027ll see why. We\u0027ll just make the explanation easier."},{"Start":"04:50.870 ","End":"04:59.710","Text":"Cross v_cm plus m_1 r_1 cross"},{"Start":"04:59.710 ","End":"05:05.150","Text":"our v_1 tag cm."},{"Start":"05:05.370 ","End":"05:09.880","Text":"This is going to equal our L_1."},{"Start":"05:09.880 ","End":"05:13.790","Text":"Then we can say that our L_2 Is going"},{"Start":"05:13.790 ","End":"05:17.350","Text":"to equal the exact same thing except instead of r_1,"},{"Start":"05:17.350 ","End":"05:20.610","Text":"it will be r_2, and instead of m_1, it will be m_2."},{"Start":"05:20.610 ","End":"05:22.700","Text":"Therefore, we\u0027ll just take it from here."},{"Start":"05:22.700 ","End":"05:28.765","Text":"We\u0027ll have m_2 multiplied by r_2 cross"},{"Start":"05:28.765 ","End":"05:38.160","Text":"v_cm plus m_2 r_2 cross,"},{"Start":"05:38.160 ","End":"05:42.795","Text":"and then here it will be the v tag,"},{"Start":"05:42.795 ","End":"05:45.180","Text":"but then here it will be 2cm."},{"Start":"05:45.180 ","End":"05:50.810","Text":"Now we can say that our total angular momentum will"},{"Start":"05:50.810 ","End":"05:57.085","Text":"just equal to our L_1 plus our L_2."},{"Start":"05:57.085 ","End":"05:59.395","Text":"Let\u0027s write that out."},{"Start":"05:59.395 ","End":"06:03.355","Text":"We can see that our L_1 and L_2 here,"},{"Start":"06:03.355 ","End":"06:06.670","Text":"we\u0027re multiplying cross-product by our v_cm."},{"Start":"06:06.670 ","End":"06:09.580","Text":"Here it\u0027s also a cross product by the v_cm."},{"Start":"06:09.580 ","End":"06:15.715","Text":"We can take that out as a common factor and we\u0027ll be left with brackets,"},{"Start":"06:15.715 ","End":"06:25.330","Text":"m_1 r_1 plus m_2 r_2 cross"},{"Start":"06:25.330 ","End":"06:30.910","Text":"multiplied with our v_cm."},{"Start":"06:30.910 ","End":"06:33.220","Text":"Now, with the next factors,"},{"Start":"06:33.220 ","End":"06:35.500","Text":"they don\u0027t have any common factors,"},{"Start":"06:35.500 ","End":"06:38.425","Text":"so we can just add them in."},{"Start":"06:38.425 ","End":"06:39.910","Text":"Let\u0027s just write that."},{"Start":"06:39.910 ","End":"06:47.490","Text":"Then we have plus m_1 r_1 cross multiplied with our v_1 tag."},{"Start":"06:47.490 ","End":"06:49.650","Text":"Instead of writing this cm,"},{"Start":"06:49.650 ","End":"06:53.250","Text":"we\u0027ll just shorten it, but that\u0027s the same thing."},{"Start":"06:53.250 ","End":"07:00.130","Text":"Then plus m_2 r_2 and then cross again,"},{"Start":"07:00.130 ","End":"07:04.600","Text":"we\u0027ll just do v_2 tag instead of the cm."},{"Start":"07:04.600 ","End":"07:08.305","Text":"Now let\u0027s look at what is going on over here."},{"Start":"07:08.305 ","End":"07:11.020","Text":"Notice we have something in brackets cross"},{"Start":"07:11.020 ","End":"07:15.070","Text":"multiplied with the velocity of the center of mass."},{"Start":"07:15.070 ","End":"07:18.385","Text":"Let\u0027s look at what is inside the brackets and think"},{"Start":"07:18.385 ","End":"07:22.315","Text":"how it can be connected to the velocity of the center of mass,"},{"Start":"07:22.315 ","End":"07:25.375","Text":"something with the center of mass. Let\u0027s try and remember."},{"Start":"07:25.375 ","End":"07:30.550","Text":"Our equation for center of mass, if you remember,"},{"Start":"07:30.550 ","End":"07:37.780","Text":"is equal to m_1 r_1 plus m_2"},{"Start":"07:37.780 ","End":"07:45.355","Text":"r_2 divided by the total mass of the system."},{"Start":"07:45.355 ","End":"07:50.425","Text":"Now, obviously, this equation for the center of mass is just for 2 bodies,"},{"Start":"07:50.425 ","End":"07:52.735","Text":"which right here we have 2 bodies."},{"Start":"07:52.735 ","End":"07:57.145","Text":"But remember that if we had 3 bodies or a 10 bodies or however it would be,"},{"Start":"07:57.145 ","End":"07:59.980","Text":"it would just be plus m_3 r_3 plus,"},{"Start":"07:59.980 ","End":"08:03.130","Text":"plus, plus, plus up until m_10 r_10."},{"Start":"08:03.130 ","End":"08:07.135","Text":"It\u0027s the same thing. But right here, we have 2 bodies."},{"Start":"08:07.135 ","End":"08:12.625","Text":"As you can see, this m_1 r1 plus m_2 r_2 is exactly what is in the brackets."},{"Start":"08:12.625 ","End":"08:15.730","Text":"That will mean that what is in the brackets is equal to"},{"Start":"08:15.730 ","End":"08:19.615","Text":"our r_cm multiplied by our total mass."},{"Start":"08:19.615 ","End":"08:22.809","Text":"Now we can rewrite this."},{"Start":"08:22.809 ","End":"08:24.580","Text":"We can say, therefore,"},{"Start":"08:24.580 ","End":"08:28.420","Text":"that our L total is equal to,"},{"Start":"08:28.420 ","End":"08:30.880","Text":"so inside the brackets,"},{"Start":"08:30.880 ","End":"08:34.180","Text":"which will be our total mass of the system"},{"Start":"08:34.180 ","End":"08:38.800","Text":"multiplied by the position of the center of mass,"},{"Start":"08:38.800 ","End":"08:43.630","Text":"cross multiplied by our V of the center of mass."},{"Start":"08:43.630 ","End":"08:45.790","Text":"That\u0027s what\u0027s happening over here."},{"Start":"08:45.790 ","End":"08:50.660","Text":"Then we\u0027ll add on this over here."},{"Start":"08:51.060 ","End":"08:53.290","Text":"I added in this,"},{"Start":"08:53.290 ","End":"08:55.720","Text":"which is just simply this from over here."},{"Start":"08:55.720 ","End":"09:04.550","Text":"Then we can see that this over here is the exact equation for L_cm."},{"Start":"09:05.100 ","End":"09:07.945","Text":"Now we have this."},{"Start":"09:07.945 ","End":"09:11.950","Text":"We can see how our equation is already starting to come together."},{"Start":"09:11.950 ","End":"09:16.015","Text":"Now what we want to do is we want to deal with this equation,"},{"Start":"09:16.015 ","End":"09:18.670","Text":"just like what we did for our velocity,"},{"Start":"09:18.670 ","End":"09:23.950","Text":"where we said over here that our v_1 is going to be equal"},{"Start":"09:23.950 ","End":"09:26.770","Text":"to the velocity of the body"},{"Start":"09:26.770 ","End":"09:30.490","Text":"relative to the center of mass plus the velocity of the center of mass."},{"Start":"09:30.490 ","End":"09:33.925","Text":"We can do the same thing for our r\u0027s."},{"Start":"09:33.925 ","End":"09:36.970","Text":"We can say that our r_1,"},{"Start":"09:36.970 ","End":"09:40.495","Text":"let\u0027s just scroll up to give you what this is."},{"Start":"09:40.495 ","End":"09:47.215","Text":"Remember r_1, the distance of the body relative to our point of reference,"},{"Start":"09:47.215 ","End":"09:54.160","Text":"which is over here, is going to be equal to our r_1 tag."},{"Start":"09:54.160 ","End":"10:03.085","Text":"That is the body relative to our center of mass plus our r center of mass."},{"Start":"10:03.085 ","End":"10:06.235","Text":"Now that\u0027s the same for r_2."},{"Start":"10:06.235 ","End":"10:14.590","Text":"We\u0027ll have that our r_2 is equal to r_2 tag plus r_cm."},{"Start":"10:14.590 ","End":"10:19.105","Text":"Now what we\u0027re going to do is we\u0027re going to substitute this in"},{"Start":"10:19.105 ","End":"10:24.955","Text":"everywhere we have this r_1 and this r_2 so over into here."},{"Start":"10:24.955 ","End":"10:29.770","Text":"This is r_1. Into here,"},{"Start":"10:29.770 ","End":"10:33.925","Text":"and here we\u0027re going to substitute in these values."},{"Start":"10:33.925 ","End":"10:37.060","Text":"Just as a note,"},{"Start":"10:37.060 ","End":"10:45.460","Text":"this is going to be our r_cm."},{"Start":"10:45.460 ","End":"10:50.450","Text":"Now I\u0027m going to scroll back down so that we have a little bit more space."},{"Start":"10:50.760 ","End":"10:55.930","Text":"Then we\u0027ll get that our L total is going to equal to this,"},{"Start":"10:55.930 ","End":"10:57.250","Text":"which is our L_cm."},{"Start":"10:57.250 ","End":"11:01.105","Text":"I\u0027m just going to save some time and write L_cm"},{"Start":"11:01.105 ","End":"11:06.820","Text":"plus m_1 multiplied by r_1, which is this."},{"Start":"11:06.820 ","End":"11:08.155","Text":"We\u0027ll have our brackets."},{"Start":"11:08.155 ","End":"11:12.790","Text":"It\u0027s r_1 tag plus r_cm,"},{"Start":"11:12.790 ","End":"11:17.770","Text":"and this cross multiplied by our v_1 tag, sorry,"},{"Start":"11:17.770 ","End":"11:22.585","Text":"and then plus our m_2 multiplied by r_2,"},{"Start":"11:22.585 ","End":"11:28.915","Text":"which is going to be r_2 tag plus r_cm,"},{"Start":"11:28.915 ","End":"11:33.625","Text":"and this is going to be multiplied by our v_2 tag."},{"Start":"11:33.625 ","End":"11:40.585","Text":"Now what we\u0027re going to do is we will open up the brackets and we can see"},{"Start":"11:40.585 ","End":"11:43.840","Text":"that we\u0027re going to have r_cm with"},{"Start":"11:43.840 ","End":"11:47.740","Text":"some expression over here and r_cm with some expression over here."},{"Start":"11:47.740 ","End":"11:51.250","Text":"We\u0027ll take out our r_cm is our common factor."},{"Start":"11:51.250 ","End":"11:54.145","Text":"Then we\u0027ll have here our L_cm,"},{"Start":"11:54.145 ","End":"11:55.720","Text":"which as we remember,"},{"Start":"11:55.720 ","End":"11:57.310","Text":"is this over here."},{"Start":"11:57.310 ","End":"12:04.360","Text":"Plus, then we\u0027ll have our m_1 r_1 tag cross multiplied by"},{"Start":"12:04.360 ","End":"12:07.300","Text":"our v_1 tag plus"},{"Start":"12:07.300 ","End":"12:14.600","Text":"our m_2 r_2 tag."},{"Start":"12:14.600 ","End":"12:18.010","Text":"Sorry, this is also cross multiplied."},{"Start":"12:18.010 ","End":"12:22.090","Text":"Cross multiplied by our v_2 tag plus,"},{"Start":"12:22.090 ","End":"12:24.100","Text":"and then we\u0027ll take out a common factor,"},{"Start":"12:24.100 ","End":"12:25.240","Text":"which is our r_cm."},{"Start":"12:25.240 ","End":"12:30.235","Text":"We\u0027ll have r_cm cross multiplied with"},{"Start":"12:30.235 ","End":"12:40.030","Text":"our m_1 multiplied by"},{"Start":"12:40.030 ","End":"12:49.435","Text":"v_1 tag plus our m_2 multiplied by our v_2 tag."},{"Start":"12:49.435 ","End":"12:55.270","Text":"Now we can see that our m_1 r_1 tag cross v_1 tag plus m_2"},{"Start":"12:55.270 ","End":"13:01.225","Text":"r_2 tag plus cross multiplied with our v_2 tag is going to be our angular momentum,"},{"Start":"13:01.225 ","End":"13:04.585","Text":"but our internal, this is our L internal."},{"Start":"13:04.585 ","End":"13:07.525","Text":"I\u0027ll call it L_int."},{"Start":"13:07.525 ","End":"13:11.500","Text":"Now we\u0027ve gotten to this, but what is this equation?"},{"Start":"13:11.500 ","End":"13:14.560","Text":"It doesn\u0027t really make much sense. Let\u0027s take a look at it."},{"Start":"13:14.560 ","End":"13:18.370","Text":"As we remember, for the position of the center of mass,"},{"Start":"13:18.370 ","End":"13:20.185","Text":"this is our equation."},{"Start":"13:20.185 ","End":"13:23.215","Text":"If we differentiate this equation,"},{"Start":"13:23.215 ","End":"13:26.320","Text":"we\u0027ll get velocity for our center of mass."},{"Start":"13:26.320 ","End":"13:34.510","Text":"I\u0027ll just get that our v_cm is equal to m_1 v_1 plus m_2 v_2 divided by the total mass."},{"Start":"13:34.510 ","End":"13:37.165","Text":"Let\u0027s just write this over here."},{"Start":"13:37.165 ","End":"13:43.855","Text":"So our v_cm multiplied by our mass total is going to be equal to"},{"Start":"13:43.855 ","End":"13:52.285","Text":"our m_1 v_1 plus our m_2 v_2."},{"Start":"13:52.285 ","End":"13:55.540","Text":"I\u0027ve added in tags just because right now we\u0027re"},{"Start":"13:55.540 ","End":"13:59.305","Text":"using in the tags to represent relative to the center of mass."},{"Start":"13:59.305 ","End":"14:02.200","Text":"Now, the trick here is to"},{"Start":"14:02.200 ","End":"14:07.045","Text":"notice that we\u0027re dealing with here the velocity of the center of mass."},{"Start":"14:07.045 ","End":"14:11.575","Text":"This is therefore going to equal to"},{"Start":"14:11.575 ","End":"14:19.109","Text":"r_cm cross multiplied with our total mass."},{"Start":"14:19.109 ","End":"14:22.880","Text":"Then here there\u0027s also missing a tag,"},{"Start":"14:22.880 ","End":"14:28.375","Text":"cross multiplied with our v tag cm. What does that mean?"},{"Start":"14:28.375 ","End":"14:30.325","Text":"What\u0027s our v tag cm?"},{"Start":"14:30.325 ","End":"14:34.640","Text":"It\u0027s our center of mass relative to our center of mass."},{"Start":"14:34.640 ","End":"14:36.410","Text":"That\u0027s what this tag means."},{"Start":"14:36.410 ","End":"14:41.750","Text":"We\u0027re finding that the velocity of the center of mass relative to the center of mass."},{"Start":"14:41.750 ","End":"14:43.160","Text":"What does that going to be?"},{"Start":"14:43.160 ","End":"14:46.670","Text":"That is always going to be equal to 0."},{"Start":"14:46.670 ","End":"14:51.875","Text":"The velocity of something relative to itself is always equal to 0."},{"Start":"14:51.875 ","End":"14:53.960","Text":"Here this is going to be equal to 0,"},{"Start":"14:53.960 ","End":"14:58.640","Text":"which means that this whole expression is therefore going to be equal to 0."},{"Start":"14:58.640 ","End":"15:02.660","Text":"Therefore, we can just rub it out."},{"Start":"15:03.270 ","End":"15:05.920","Text":"Therefore, the final thing,"},{"Start":"15:05.920 ","End":"15:07.510","Text":"I\u0027ll write in red."},{"Start":"15:07.510 ","End":"15:15.680","Text":"Our L total is going to be equal to our L_cm, which we saw,"},{"Start":"15:15.680 ","End":"15:17.210","Text":"it\u0027s equal to this,"},{"Start":"15:17.210 ","End":"15:21.405","Text":"plus our L internal,"},{"Start":"15:21.405 ","End":"15:23.515","Text":"which is equal to this,"},{"Start":"15:23.515 ","End":"15:25.405","Text":"and this is equal to 0."},{"Start":"15:25.405 ","End":"15:27.860","Text":"That, if we scroll back up,"},{"Start":"15:27.860 ","End":"15:32.460","Text":"is exactly the equation that we needed to derive."},{"Start":"15:32.610 ","End":"15:35.060","Text":"This was the end of the lesson."},{"Start":"15:35.060 ","End":"15:37.310","Text":"We\u0027ve seen how to derive this equation."},{"Start":"15:37.310 ","End":"15:39.440","Text":"I hope it also makes a bit more sense and"},{"Start":"15:39.440 ","End":"15:41.690","Text":"that now you will understand the equation and how to"},{"Start":"15:41.690 ","End":"15:47.130","Text":"use it further and also why it\u0027s correct. That\u0027s it."}],"ID":12709}],"Thumbnail":null,"ID":5331},{"Name":"Precession","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Precession Size and Direction","Duration":"22m 7s","ChapterTopicVideoID":9132,"CourseChapterTopicPlaylistID":9397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:02.835","Text":"Hello. In this lesson,"},{"Start":"00:02.835 ","End":"00:05.955","Text":"we\u0027re going to be speaking about precession."},{"Start":"00:05.955 ","End":"00:08.970","Text":"In order to try and explain what this is,"},{"Start":"00:08.970 ","End":"00:11.805","Text":"let\u0027s start with an example."},{"Start":"00:11.805 ","End":"00:17.520","Text":"Imagine that this is my axis of rotation over here."},{"Start":"00:17.520 ","End":"00:24.130","Text":"I have some body going in this shape, a Z shape."},{"Start":"00:24.770 ","End":"00:30.455","Text":"This is 1 bar over here and then another 1 over here,"},{"Start":"00:30.455 ","End":"00:35.655","Text":"and here I have 2 masses joined."},{"Start":"00:35.655 ","End":"00:40.325","Text":"This is the midpoint of the bar here and the height,"},{"Start":"00:40.325 ","End":"00:45.150","Text":"this is z and similarly,"},{"Start":"00:45.150 ","End":"00:47.905","Text":"this is also z."},{"Start":"00:47.905 ","End":"00:58.265","Text":"Let\u0027s say that this over here is length a and similarly over here,"},{"Start":"00:58.265 ","End":"01:01.110","Text":"this is also of length a."},{"Start":"01:01.790 ","End":"01:07.350","Text":"Now if we say that this is our axis of rotation."},{"Start":"01:07.390 ","End":"01:09.755","Text":"If we save that now,"},{"Start":"01:09.755 ","End":"01:15.085","Text":"this body is rotating around this axis of rotation,"},{"Start":"01:15.085 ","End":"01:18.195","Text":"and it\u0027s going in this direction,"},{"Start":"01:18.195 ","End":"01:22.895","Text":"you\u0027ll have this mass and obviously these 2 masses are identical."},{"Start":"01:22.895 ","End":"01:26.750","Text":"It will go into the page and then come back out of"},{"Start":"01:26.750 ","End":"01:31.445","Text":"the page and return to this and this mass will do the exact same thing."},{"Start":"01:31.445 ","End":"01:35.095","Text":"Out of the page and then into the page."},{"Start":"01:35.095 ","End":"01:38.130","Text":"They\u0027re both doing the same movement."},{"Start":"01:38.130 ","End":"01:42.695","Text":"What we\u0027ll see is that up until now,"},{"Start":"01:42.695 ","End":"01:50.015","Text":"we\u0027ve seen that our angular momentum will be in the same direction as our Omega."},{"Start":"01:50.015 ","End":"01:51.875","Text":"That\u0027s what we\u0027ve seen till now."},{"Start":"01:51.875 ","End":"01:56.493","Text":"If our Omega was in this direction along the z-axis,"},{"Start":"01:56.493 ","End":"01:59.705","Text":"so will our L. However here,"},{"Start":"01:59.705 ","End":"02:01.445","Text":"when we\u0027re dealing with precession,"},{"Start":"02:01.445 ","End":"02:08.420","Text":"so we\u0027ll see that our angular momentum is actually in a different direction."},{"Start":"02:08.420 ","End":"02:11.630","Text":"We\u0027ll see why, and we\u0027ll see how we work it out."},{"Start":"02:11.630 ","End":"02:14.390","Text":"But we\u0027ll see that in this example, specifically,"},{"Start":"02:14.390 ","End":"02:20.231","Text":"the angular momentum is going to be in this direction over here."},{"Start":"02:20.231 ","End":"02:25.550","Text":"In this lesson, what I\u0027m actually going to do is I\u0027m"},{"Start":"02:25.550 ","End":"02:31.505","Text":"going to show you how to work out the angular momentum of an example like this."},{"Start":"02:31.505 ","End":"02:34.055","Text":"I\u0027m going to do it in 2 ways."},{"Start":"02:34.055 ","End":"02:37.940","Text":"One via vector multiplication and the other"},{"Start":"02:37.940 ","End":"02:40.340","Text":"way by working out the size and"},{"Start":"02:40.340 ","End":"02:43.790","Text":"then by working out the direction of the angular momentum."},{"Start":"02:43.790 ","End":"02:48.920","Text":"The first way I\u0027m going to start with is by working out the size"},{"Start":"02:48.920 ","End":"02:54.380","Text":"and then the direction. Let\u0027s see."},{"Start":"02:54.380 ","End":"03:00.895","Text":"First what we\u0027re going to do,"},{"Start":"03:00.895 ","End":"03:03.875","Text":"is we\u0027re going to work out the velocity."},{"Start":"03:03.875 ","End":"03:09.710","Text":"As we know, the linear velocity is going"},{"Start":"03:09.710 ","End":"03:15.815","Text":"to be equal to the angular velocity Omega multiplied by a."},{"Start":"03:15.815 ","End":"03:17.750","Text":"This is going to be for each ball,"},{"Start":"03:17.750 ","End":"03:22.070","Text":"because each ball has"},{"Start":"03:22.070 ","End":"03:27.680","Text":"the same distance a over here and is traveling at the same angular velocity."},{"Start":"03:27.680 ","End":"03:30.640","Text":"That means that our momentum,"},{"Start":"03:30.640 ","End":"03:33.345","Text":"we\u0027ll say I for each ball."},{"Start":"03:33.345 ","End":"03:34.630","Text":"This will be 1 or 2,"},{"Start":"03:34.630 ","End":"03:38.150","Text":"for example, as we know,"},{"Start":"03:38.150 ","End":"03:44.880","Text":"it\u0027s going to be equal to m multiplied by v_i because they both have the same mass."},{"Start":"03:45.050 ","End":"03:53.640","Text":"Let\u0027s say that this is v_i. That is going to be m Omega multiplied by a."},{"Start":"03:53.640 ","End":"03:58.264","Text":"Now in order to find the angular momentum,"},{"Start":"03:58.264 ","End":"04:01.100","Text":"so let\u0027s write down the equation first."},{"Start":"04:01.100 ","End":"04:08.305","Text":"We know that it\u0027s r cross multiplied with the momentum."},{"Start":"04:08.305 ","End":"04:11.730","Text":"Now this is where the catch is,"},{"Start":"04:11.730 ","End":"04:14.260","Text":"how to work this out."},{"Start":"04:14.930 ","End":"04:18.485","Text":"You\u0027ll remember that in previous examples,"},{"Start":"04:18.485 ","End":"04:21.275","Text":"when we were finding out what our r was,"},{"Start":"04:21.275 ","End":"04:25.370","Text":"we just took its distance from the axis of rotation."},{"Start":"04:25.370 ","End":"04:28.805","Text":"That would have been this distance here, say a."},{"Start":"04:28.805 ","End":"04:36.170","Text":"In this example, we cannot do that because we have to take it from the origins."},{"Start":"04:36.170 ","End":"04:38.645","Text":"If we say that this is the origin,"},{"Start":"04:38.645 ","End":"04:43.760","Text":"so this would be our r vector as opposed to what we would have chosen in"},{"Start":"04:43.760 ","End":"04:45.980","Text":"our previous examples when we weren\u0027t dealing with"},{"Start":"04:45.980 ","End":"04:50.300","Text":"precession to taking a vector from here."},{"Start":"04:50.300 ","End":"04:52.580","Text":"Now, in the previous examples,"},{"Start":"04:52.580 ","End":"04:56.735","Text":"we were also meant to take our r vector from the origin."},{"Start":"04:56.735 ","End":"04:58.220","Text":"However, in those examples,"},{"Start":"04:58.220 ","End":"05:00.200","Text":"it didn\u0027t matter how we did it,"},{"Start":"05:00.200 ","End":"05:02.615","Text":"if we did it that way or if we do it this way."},{"Start":"05:02.615 ","End":"05:06.005","Text":"Later in this video,"},{"Start":"05:06.005 ","End":"05:09.395","Text":"I\u0027ll explain in which cases we have to take"},{"Start":"05:09.395 ","End":"05:12.920","Text":"from the origin or in which cases it doesn\u0027t really make a difference if we"},{"Start":"05:12.920 ","End":"05:22.490","Text":"take from the origin or from the the closest displacement from the axis of rotation."},{"Start":"05:22.490 ","End":"05:24.605","Text":"But here in precession,"},{"Start":"05:24.605 ","End":"05:27.380","Text":"it\u0027s very important to remember to take it from the origin,"},{"Start":"05:27.380 ","End":"05:30.755","Text":"which will be the center. This is our r vector."},{"Start":"05:30.755 ","End":"05:37.820","Text":"Now let\u0027s take a look at how we will do this vector cross-product."},{"Start":"05:37.820 ","End":"05:42.470","Text":"This r cross p. As we can see,"},{"Start":"05:42.470 ","End":"05:44.116","Text":"our r is going in this direction."},{"Start":"05:44.116 ","End":"05:48.605","Text":"Then we\u0027re doing cross p. Now what is P?"},{"Start":"05:48.605 ","End":"05:50.555","Text":"It\u0027s m Omega a,"},{"Start":"05:50.555 ","End":"05:54.360","Text":"which is our mv."},{"Start":"05:54.360 ","End":"05:59.460","Text":"At this point, when our ball or masses over here."},{"Start":"05:59.460 ","End":"06:05.224","Text":"We can see, because our angular velocity is going in this direction,"},{"Start":"06:05.224 ","End":"06:08.600","Text":"the ball is right now going into the page,"},{"Start":"06:08.600 ","End":"06:12.380","Text":"so I\u0027ll say that r velocity vector,"},{"Start":"06:12.380 ","End":"06:14.945","Text":"is into the page,"},{"Start":"06:14.945 ","End":"06:17.755","Text":"that\u0027s what this x represents."},{"Start":"06:17.755 ","End":"06:19.775","Text":"If it\u0027s into the page,"},{"Start":"06:19.775 ","End":"06:23.374","Text":"so now we have to find our angular momentum,"},{"Start":"06:23.374 ","End":"06:29.875","Text":"which is going to be perpendicular to both our r vector and to our v,"},{"Start":"06:29.875 ","End":"06:31.700","Text":"because if it\u0027s perpendicular to our v,"},{"Start":"06:31.700 ","End":"06:35.150","Text":"then it\u0027s also going to be perpendicular to our p,"},{"Start":"06:35.150 ","End":"06:37.865","Text":"which is exactly what we\u0027re looking for."},{"Start":"06:37.865 ","End":"06:43.055","Text":"If we draw this, let\u0027s see,"},{"Start":"06:43.055 ","End":"06:47.385","Text":"so these are axes,"},{"Start":"06:47.385 ","End":"06:53.655","Text":"and then we have going in over here at the origin."},{"Start":"06:53.655 ","End":"07:02.205","Text":"That is our v and then also we have going in this direction our r vector."},{"Start":"07:02.205 ","End":"07:03.440","Text":"I\u0027m just copying that out."},{"Start":"07:03.440 ","End":"07:04.850","Text":"Instead of drawing the v over here,"},{"Start":"07:04.850 ","End":"07:07.475","Text":"I\u0027ve drawn it over here. There is no difference."},{"Start":"07:07.475 ","End":"07:09.560","Text":"Now what do we have to do,"},{"Start":"07:09.560 ","End":"07:11.525","Text":"is we have to find our angular momentum,"},{"Start":"07:11.525 ","End":"07:15.910","Text":"which is going to be perpendicular to both of these vectors."},{"Start":"07:15.910 ","End":"07:26.285","Text":"As we can see, it\u0027s going to be therefore this and over here,"},{"Start":"07:26.285 ","End":"07:28.490","Text":"this angle is going to be 90 degrees."},{"Start":"07:28.490 ","End":"07:32.810","Text":"I know that this picture isn\u0027t 100 percent, this diagram."},{"Start":"07:32.810 ","End":"07:36.700","Text":"But imagine that this is a 90 degree angle over here."},{"Start":"07:36.700 ","End":"07:39.950","Text":"As you can see, if this is our angular momentum,"},{"Start":"07:39.950 ","End":"07:42.110","Text":"this is exactly what we said at the beginning."},{"Start":"07:42.110 ","End":"07:45.740","Text":"Because we can see that it\u0027s going to be 90 degrees to"},{"Start":"07:45.740 ","End":"07:50.310","Text":"the vector that\u0027s going into the page over here."},{"Start":"07:50.310 ","End":"07:57.420","Text":"It has to be at 90 degrees to this r. It\u0027s going to be this diagonal over here."},{"Start":"07:58.370 ","End":"08:04.160","Text":"Up until now, we found the direction of the angular momentum."},{"Start":"08:04.160 ","End":"08:06.575","Text":"Now we\u0027re going to find the size."},{"Start":"08:06.575 ","End":"08:08.000","Text":"The size as we know,"},{"Start":"08:08.000 ","End":"08:09.530","Text":"we symbolize it like this."},{"Start":"08:09.530 ","End":"08:13.880","Text":"It\u0027s going to be equal to the size of"},{"Start":"08:13.880 ","End":"08:18.650","Text":"our r vector multiplied by the size of our p vector,"},{"Start":"08:18.650 ","End":"08:24.065","Text":"momentum vector multiplied by sine of the angle between them."},{"Start":"08:24.065 ","End":"08:30.600","Text":"Now the angle between them is 90 degrees,"},{"Start":"08:30.600 ","End":"08:32.735","Text":"as we can see over here."},{"Start":"08:32.735 ","End":"08:34.715","Text":"If you can imagine,"},{"Start":"08:34.715 ","End":"08:36.200","Text":"let\u0027s say for example,"},{"Start":"08:36.200 ","End":"08:39.233","Text":"that this is your y and this is your x-axis,"},{"Start":"08:39.233 ","End":"08:41.930","Text":"so the R is in the plane of the xy,"},{"Start":"08:41.930 ","End":"08:43.355","Text":"on the xy plane,"},{"Start":"08:43.355 ","End":"08:46.730","Text":"and rp, which is rv,"},{"Start":"08:46.730 ","End":"08:50.765","Text":"so rp is the z axis, going into the page."},{"Start":"08:50.765 ","End":"08:53.660","Text":"Obviously it\u0027s something that\u0027s on the z axis is going to be"},{"Start":"08:53.660 ","End":"08:57.755","Text":"perpendicular to something on the xy plane."},{"Start":"08:57.755 ","End":"09:03.860","Text":"As we know, sine of 90 is equal to 1."},{"Start":"09:03.860 ","End":"09:09.200","Text":"It\u0027s going to be the size of r multiplied by the size of p multiplied by 1."},{"Start":"09:09.200 ","End":"09:11.255","Text":"What is the size of r?"},{"Start":"09:11.255 ","End":"09:19.080","Text":"It\u0027s hypotenuse is going to be the square root of a^2 plus z^2,"},{"Start":"09:20.770 ","End":"09:26.405","Text":"a^2 plus z^2 is the size of our r vector"},{"Start":"09:26.405 ","End":"09:31.775","Text":"and then multiplied by the size of our p vector and what is our p vector?"},{"Start":"09:31.775 ","End":"09:38.490","Text":"It\u0027s going to be mass multiplied by Omega, multiplied by a."},{"Start":"09:39.550 ","End":"09:47.510","Text":"This is the size and this is the direction of the angular momentum going to this shape."},{"Start":"09:47.510 ","End":"09:51.665","Text":"But now let\u0027s see to this shape, to this body."},{"Start":"09:51.665 ","End":"09:56.125","Text":"Now this is going to be our r vector."},{"Start":"09:56.125 ","End":"09:59.600","Text":"Again, exactly like what we did over here,"},{"Start":"09:59.600 ","End":"10:01.085","Text":"at this exact point,"},{"Start":"10:01.085 ","End":"10:02.825","Text":"when our r vector is here,"},{"Start":"10:02.825 ","End":"10:06.620","Text":"according to what we know about our angular velocity,"},{"Start":"10:06.620 ","End":"10:11.705","Text":"and its direction, the ball will be exactly heading out of the page."},{"Start":"10:11.705 ","End":"10:18.000","Text":"Our velocity vector will be coming out of the page."},{"Start":"10:18.860 ","End":"10:21.155","Text":"If it\u0027s like that,"},{"Start":"10:21.155 ","End":"10:24.930","Text":"so it\u0027s going to be exactly opposite to what we have here."},{"Start":"10:24.930 ","End":"10:29.310","Text":"Here we have some axis,"},{"Start":"10:29.310 ","End":"10:32.415","Text":"it doesn\u0027t really matter the labels right now."},{"Start":"10:32.415 ","End":"10:39.335","Text":"Now, we can see that we\u0027re going to have our r vector going in this direction."},{"Start":"10:39.335 ","End":"10:45.530","Text":"This is our r and because it\u0027s the first item in our cross multiplication,"},{"Start":"10:45.530 ","End":"10:50.375","Text":"so we have to rotate it in the direction of our velocity vector,"},{"Start":"10:50.375 ","End":"10:54.230","Text":"which will be in the direction of our momentum."},{"Start":"10:54.230 ","End":"10:59.520","Text":"As we can see, rv is coming out, so this is rv."},{"Start":"11:01.020 ","End":"11:04.420","Text":"Here our v is coming out of the page."},{"Start":"11:04.420 ","End":"11:09.160","Text":"If I rotate my eye to coming out of the page,"},{"Start":"11:09.160 ","End":"11:19.315","Text":"then you\u0027ll see that the angular momentum is still going to be in this direction."},{"Start":"11:19.315 ","End":"11:25.540","Text":"Let\u0027s take a look at what this means and to the logic behind it."},{"Start":"11:25.540 ","End":"11:28.104","Text":"As we saw in our first example,"},{"Start":"11:28.104 ","End":"11:31.525","Text":"the radius vector was pointing in this direction."},{"Start":"11:31.525 ","End":"11:36.625","Text":"The opposite direction to it will be this direction so we flip this around."},{"Start":"11:36.625 ","End":"11:40.060","Text":"However, also our velocity vector has been flipped"},{"Start":"11:40.060 ","End":"11:43.555","Text":"around before it was going into the page and now it\u0027s here,"},{"Start":"11:43.555 ","End":"11:45.340","Text":"it\u0027s coming out of the page."},{"Start":"11:45.340 ","End":"11:49.030","Text":"Which means that we\u0027re looking at the exact same system,"},{"Start":"11:49.030 ","End":"11:52.765","Text":"just basically from the exact opposite angle,"},{"Start":"11:52.765 ","End":"11:56.140","Text":"which means that our angular momentum is still going to"},{"Start":"11:56.140 ","End":"11:59.515","Text":"be in the exact same direction as before."},{"Start":"11:59.515 ","End":"12:05.110","Text":"Also we can once again see that if we rotate our eye over here,"},{"Start":"12:05.110 ","End":"12:10.510","Text":"so we can see that the angular momentum is going in this direction."},{"Start":"12:10.510 ","End":"12:17.470","Text":"We can see that it\u0027s perpendicular both to velocity which is coming out of the page,"},{"Start":"12:17.470 ","End":"12:21.820","Text":"and to radius which is going in this direction."},{"Start":"12:21.820 ","End":"12:28.400","Text":"Now we have the direction for this shape of the angular momentum."},{"Start":"12:28.920 ","End":"12:35.260","Text":"Now what we can do is we can work out the angular momentum of this shape."},{"Start":"12:35.260 ","End":"12:39.610","Text":"However, we know that it\u0027s going to turn out to be this exact same thing."},{"Start":"12:39.610 ","End":"12:43.435","Text":"What we can say is that our angular momentum,"},{"Start":"12:43.435 ","End":"12:45.070","Text":"our total angular momentum,"},{"Start":"12:45.070 ","End":"12:50.275","Text":"the size of it, is just going to be equal to 2 times what we have here."},{"Start":"12:50.275 ","End":"12:58.030","Text":"It\u0027s going to be 2 times a^2 plus z^2 multiplied by"},{"Start":"12:58.030 ","End":"13:07.240","Text":"m Omega e. Now we just have to find how to describe the direction."},{"Start":"13:07.240 ","End":"13:10.735","Text":"We can say that this angle over here is called Alpha,"},{"Start":"13:10.735 ","End":"13:16.030","Text":"which is going to be the same angle over here, which is Alpha,"},{"Start":"13:16.030 ","End":"13:23.330","Text":"and the same over here if we were to draw a radius vector."},{"Start":"13:24.630 ","End":"13:32.470","Text":"Then we can say that Cosine of Alpha is going to be equal to,"},{"Start":"13:32.470 ","End":"13:35.995","Text":"it\u0027s going to be a over the size of r,"},{"Start":"13:35.995 ","End":"13:44.910","Text":"which is just going to be equal to a divided by root a^2 plus z^2."},{"Start":"13:44.910 ","End":"13:53.995","Text":"We can say that sine of Alpha is going to be equal to z over our r,"},{"Start":"13:53.995 ","End":"14:03.715","Text":"which is going to be a^2 plus z^2."},{"Start":"14:03.715 ","End":"14:05.095","Text":"Now we\u0027ve figured that out."},{"Start":"14:05.095 ","End":"14:09.370","Text":"Also you can see that as the system rotates,"},{"Start":"14:09.370 ","End":"14:13.615","Text":"angular momentum is going to remain constant. It\u0027s not going to change."},{"Start":"14:13.615 ","End":"14:21.505","Text":"If I have my system like this and it\u0027s rotated a bit in this direction,"},{"Start":"14:21.505 ","End":"14:25.045","Text":"this is still going to be my angular momentum over here."},{"Start":"14:25.045 ","End":"14:28.345","Text":"Because then instead of looking at it from this angle straight on,"},{"Start":"14:28.345 ","End":"14:31.960","Text":"I can rotate myself and stand over here and"},{"Start":"14:31.960 ","End":"14:36.460","Text":"look and the working out will be the exact same thing."},{"Start":"14:36.460 ","End":"14:38.200","Text":"As the system rotates,"},{"Start":"14:38.200 ","End":"14:42.280","Text":"it doesn\u0027t matter that angular momentum remains constant throughout."},{"Start":"14:42.280 ","End":"14:45.234","Text":"Now this is going to be slightly more complicated."},{"Start":"14:45.234 ","End":"14:50.440","Text":"What we\u0027re going to do is we\u0027re going to draw axes onto this system over here."},{"Start":"14:50.440 ","End":"14:54.340","Text":"Let\u0027s say that this is our x-axis,"},{"Start":"14:54.340 ","End":"14:57.580","Text":"this will be our y-axis,"},{"Start":"14:57.580 ","End":"15:01.010","Text":"and this will be z-axis."},{"Start":"15:01.170 ","End":"15:04.150","Text":"Now, we can work with x, y,"},{"Start":"15:04.150 ","End":"15:06.610","Text":"and z, or we can work with polar coordinates."},{"Start":"15:06.610 ","End":"15:10.760","Text":"But in the meantime, let\u0027s work with x, y, and z."},{"Start":"15:11.130 ","End":"15:15.415","Text":"As you can see, if I draw this out,"},{"Start":"15:15.415 ","End":"15:24.115","Text":"so this is my z axis and my l begins over here."},{"Start":"15:24.115 ","End":"15:28.405","Text":"As my system rotates, you\u0027ll see that my l,"},{"Start":"15:28.405 ","End":"15:33.505","Text":"if it, once in its starting position was on the y-axis."},{"Start":"15:33.505 ","End":"15:36.655","Text":"It would only have a y-component."},{"Start":"15:36.655 ","End":"15:38.800","Text":"As my whole body rotates,"},{"Start":"15:38.800 ","End":"15:43.210","Text":"eventually, my l will be above the x-axis."},{"Start":"15:43.210 ","End":"15:44.800","Text":"Right above here."},{"Start":"15:44.800 ","End":"15:46.315","Text":"As it rotates,"},{"Start":"15:46.315 ","End":"15:49.855","Text":"it will go again to over here."},{"Start":"15:49.855 ","End":"15:52.600","Text":"Then it will be on the y-axis again,"},{"Start":"15:52.600 ","End":"15:54.460","Text":"and it\u0027s just rotating."},{"Start":"15:54.460 ","End":"15:56.590","Text":"It can take any value around,"},{"Start":"15:56.590 ","End":"15:58.960","Text":"rotating around the z axis."},{"Start":"15:58.960 ","End":"16:03.220","Text":"You can imagine it as something like this."},{"Start":"16:03.220 ","End":"16:06.950","Text":"As it\u0027s rotating, it will look like this."},{"Start":"16:09.200 ","End":"16:14.080","Text":"This is what\u0027s really complicated over here because we can"},{"Start":"16:14.080 ","End":"16:18.100","Text":"see that everything is changing direction all the time."},{"Start":"16:18.100 ","End":"16:20.829","Text":"That\u0027s the thing of the procession."},{"Start":"16:20.829 ","End":"16:24.700","Text":"The angular momentum is constantly changing direction."},{"Start":"16:24.700 ","End":"16:28.540","Text":"Let\u0027s see how we can deal with this and how we can write it"},{"Start":"16:28.540 ","End":"16:33.320","Text":"out on the different components for the different axes."},{"Start":"16:33.780 ","End":"16:39.670","Text":"Now what I\u0027m going to do is I\u0027m going to choose one random arrow,"},{"Start":"16:39.670 ","End":"16:44.020","Text":"which is going to be my most general and my most random example for where"},{"Start":"16:44.020 ","End":"16:50.500","Text":"my angular momentum will be located relative to all the other axes."},{"Start":"16:50.500 ","End":"16:54.235","Text":"Through that, I\u0027m going to work out the separate components."},{"Start":"16:54.235 ","End":"16:57.310","Text":"Let\u0027s choose this one."},{"Start":"16:57.310 ","End":"17:01.600","Text":"I\u0027m just going to rub out everything and draw this again."},{"Start":"17:01.600 ","End":"17:04.255","Text":"This is my z-axis and I chose"},{"Start":"17:04.255 ","End":"17:13.160","Text":"this to be my angular momentum."},{"Start":"17:16.620 ","End":"17:19.585","Text":"Let\u0027s see how we can draw this."},{"Start":"17:19.585 ","End":"17:21.820","Text":"As we know over here,"},{"Start":"17:21.820 ","End":"17:29.260","Text":"I\u0027ll have the projection l_z on the z-axis."},{"Start":"17:29.260 ","End":"17:32.180","Text":"Sorry, without the arrow."},{"Start":"17:35.430 ","End":"17:38.530","Text":"My projection on the z-axis."},{"Start":"17:38.530 ","End":"17:49.250","Text":"Then similarly, I\u0027ll have over here my x axes and over here my y-axis."},{"Start":"17:49.350 ","End":"17:54.040","Text":"Then I will have my projection on."},{"Start":"17:54.040 ","End":"18:00.235","Text":"Therefore, the x, y plane will be like this."},{"Start":"18:00.235 ","End":"18:03.565","Text":"This will be my L, x, y."},{"Start":"18:03.565 ","End":"18:06.265","Text":"Because that\u0027s my projection on the x, y plane."},{"Start":"18:06.265 ","End":"18:10.705","Text":"Now we can call this angle over here Alpha."},{"Start":"18:10.705 ","End":"18:15.310","Text":"Now, we can see how to solve this."},{"Start":"18:15.310 ","End":"18:20.530","Text":"We can say that my l_z because it\u0027s my projection on the z-axis."},{"Start":"18:20.530 ","End":"18:25.240","Text":"This is also going to be my z component of my L,"},{"Start":"18:25.240 ","End":"18:26.965","Text":"of my angular momentum."},{"Start":"18:26.965 ","End":"18:37.580","Text":"It\u0027s going to be equal to the size of my L multiplied by cosine of the angle."},{"Start":"18:37.830 ","End":"18:41.860","Text":"As we know, cosine of the angle is going to be this."},{"Start":"18:41.860 ","End":"18:48.820","Text":"This is the size of my L. V\u0027s root signs are going to cancel and I\u0027m"},{"Start":"18:48.820 ","End":"18:56.229","Text":"going to be left with m Omega a^2 multiplied,"},{"Start":"18:56.229 ","End":"19:02.630","Text":"sorry, and then multiplied by 2 because we have 2 bodies."},{"Start":"19:02.730 ","End":"19:06.550","Text":"Then similarly, my l x, y,"},{"Start":"19:06.550 ","End":"19:09.850","Text":"notice this is not yet my x-component and my y component."},{"Start":"19:09.850 ","End":"19:13.540","Text":"This is a general x-y component on this plane,"},{"Start":"19:13.540 ","End":"19:18.400","Text":"is again going to be equal to the size of my angular momentum,"},{"Start":"19:18.400 ","End":"19:22.090","Text":"but this time multiplied by sine of the angle."},{"Start":"19:22.090 ","End":"19:26.350","Text":"Notice again that here there\u0027s a bit of a mess."},{"Start":"19:26.350 ","End":"19:31.450","Text":"I see that these square roots are going to cancel."},{"Start":"19:31.450 ","End":"19:34.825","Text":"We\u0027re just going to get 2,"},{"Start":"19:34.825 ","End":"19:42.520","Text":"because we have 2 bodies multiplied by m Omega a multiplied by z."},{"Start":"19:42.650 ","End":"19:45.285","Text":"Because the square roots cancel out."},{"Start":"19:45.285 ","End":"19:49.180","Text":"Then we have this and multiply by 2 because we have the 2 bodies."},{"Start":"19:49.380 ","End":"19:56.020","Text":"Now what we want to do is we want to find the actual x and the actual y components."},{"Start":"19:56.020 ","End":"20:00.145","Text":"What we can do is we can say that"},{"Start":"20:00.145 ","End":"20:06.250","Text":"here this angle is going to be equal to Theta. What is Theta?"},{"Start":"20:06.250 ","End":"20:11.365","Text":"It\u0027s the angle between the x axis until my vector."},{"Start":"20:11.365 ","End":"20:14.450","Text":"Until my vector projection."},{"Start":"20:14.730 ","End":"20:21.310","Text":"Therefore, I can say that my l_x is going to be equal to my l,"},{"Start":"20:21.310 ","End":"20:27.555","Text":"x, y multiplied by cosine of Theta."},{"Start":"20:27.555 ","End":"20:31.255","Text":"Because it\u0027s always, or usually,"},{"Start":"20:31.255 ","End":"20:33.940","Text":"depending on how you set up your system."},{"Start":"20:33.940 ","End":"20:35.650","Text":"But on the x-axis it\u0027s always going to be"},{"Start":"20:35.650 ","End":"20:39.145","Text":"cosine Theta and the y-axis is always going to be sine Theta."},{"Start":"20:39.145 ","End":"20:41.890","Text":"Again, if you set up your axes in a different way,"},{"Start":"20:41.890 ","End":"20:44.620","Text":"but generally speaking, it\u0027s going to be like this."},{"Start":"20:44.620 ","End":"20:47.530","Text":"What is that going to be equal to?"},{"Start":"20:47.530 ","End":"20:57.145","Text":"It\u0027s going to be equal to 2 Omega a_z multiplied by m,"},{"Start":"20:57.145 ","End":"21:03.080","Text":"sorry, multiplied by cosine of theta."},{"Start":"21:03.660 ","End":"21:08.230","Text":"Then my l_y is just going to be l,"},{"Start":"21:08.230 ","End":"21:11.485","Text":"x, y multiplied by my sine Theta."},{"Start":"21:11.485 ","End":"21:21.710","Text":"That\u0027s just going to equal to 2m omega a_z and then multiplied by sine of Theta."},{"Start":"21:22.650 ","End":"21:27.835","Text":"Now to write out the total angular momentum."},{"Start":"21:27.835 ","End":"21:36.680","Text":"We\u0027ll write it out in vector form so we can see that our common factors are 2m omega a."},{"Start":"21:36.680 ","End":"21:40.065","Text":"Then we\u0027ll put it into our axes."},{"Start":"21:40.065 ","End":"21:43.890","Text":"Then we have on our x-axis,"},{"Start":"21:43.890 ","End":"21:47.155","Text":"we have Z cosine of Theta."},{"Start":"21:47.155 ","End":"21:52.015","Text":"On our y-axis, we have sine of Theta."},{"Start":"21:52.015 ","End":"21:54.955","Text":"Then I know z axes,"},{"Start":"21:54.955 ","End":"21:57.430","Text":"we have just multiplied by a."},{"Start":"21:57.430 ","End":"22:03.385","Text":"In the next video we\u0027re going to speak how to do this via vector multiplication."},{"Start":"22:03.385 ","End":"22:07.860","Text":"That\u0027s the end for this section of the video."}],"ID":9523},{"Watched":false,"Name":"Precession Vector Multiplication","Duration":"13m 53s","ChapterTopicVideoID":9133,"CourseChapterTopicPlaylistID":9397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.520 ","End":"00:03.645","Text":"In the last video,"},{"Start":"00:03.645 ","End":"00:06.764","Text":"we saw how to work out the angular momentum"},{"Start":"00:06.764 ","End":"00:10.590","Text":"by working out its size and then by working out its direction,"},{"Start":"00:10.590 ","End":"00:11.820","Text":"and putting it together."},{"Start":"00:11.820 ","End":"00:13.140","Text":"Now, in this lesson,"},{"Start":"00:13.140 ","End":"00:17.010","Text":"we\u0027re going to work it out through vector multiplication."},{"Start":"00:17.010 ","End":"00:20.580","Text":"For some of you, this way will be harder and more complicated."},{"Start":"00:20.580 ","End":"00:22.260","Text":"For others, it will be easier."},{"Start":"00:22.260 ","End":"00:25.455","Text":"It\u0027s very useful to know how to do it either way."},{"Start":"00:25.455 ","End":"00:28.410","Text":"We\u0027re going to go over it now."},{"Start":"00:28.410 ","End":"00:33.465","Text":"The first thing that we\u0027re going to do is over here on our original diagram,"},{"Start":"00:33.465 ","End":"00:37.575","Text":"we\u0027re looking at the system from a sideways view."},{"Start":"00:37.575 ","End":"00:42.095","Text":"Now, if we want to look at it from a bird\u0027s-eye view."},{"Start":"00:42.095 ","End":"00:44.805","Text":"Let\u0027s draw this right now."},{"Start":"00:44.805 ","End":"00:49.575","Text":"If we draw our y-axis going in"},{"Start":"00:49.575 ","End":"00:56.875","Text":"this direction and our x-axis pointing downwards like so."},{"Start":"00:56.875 ","End":"01:01.060","Text":"Now we can draw our system on this axis."},{"Start":"01:01.060 ","End":"01:07.515","Text":"Then we have our z-axis coming out of the page, in 3 dimensions."},{"Start":"01:07.515 ","End":"01:10.320","Text":"We can say that the ball that\u0027s higher,"},{"Start":"01:10.320 ","End":"01:13.870","Text":"for instance, this ball is located over here."},{"Start":"01:13.870 ","End":"01:15.590","Text":"That would mean that the lower ball,"},{"Start":"01:15.590 ","End":"01:16.910","Text":"this one over here,"},{"Start":"01:16.910 ","End":"01:20.795","Text":"would be located somewhere over here."},{"Start":"01:20.795 ","End":"01:22.520","Text":"This is what it would look like."},{"Start":"01:22.520 ","End":"01:26.104","Text":"Then this ball would be going in this direction,"},{"Start":"01:26.104 ","End":"01:28.550","Text":"and this one will be going in this direction,"},{"Start":"01:28.550 ","End":"01:31.880","Text":"because this is going into the page over here,"},{"Start":"01:31.880 ","End":"01:33.635","Text":"and this one is coming out of the page."},{"Start":"01:33.635 ","End":"01:36.550","Text":"That translates to something like this."},{"Start":"01:36.550 ","End":"01:41.230","Text":"Then after that this system has rotated."},{"Start":"01:42.620 ","End":"01:47.000","Text":"This will be the rotated system where this over here will be"},{"Start":"01:47.000 ","End":"01:51.340","Text":"the upper ball and this over here will be the lower one."},{"Start":"01:51.340 ","End":"01:56.420","Text":"Now, we can write out the position vector."},{"Start":"01:56.420 ","End":"01:59.569","Text":"Here over to here,"},{"Start":"01:59.569 ","End":"02:01.520","Text":"we\u0027ll have our r_1."},{"Start":"02:01.520 ","End":"02:03.865","Text":"It is a function of x,"},{"Start":"02:03.865 ","End":"02:05.940","Text":"y, and z."},{"Start":"02:05.940 ","End":"02:07.350","Text":"It has x, y,"},{"Start":"02:07.350 ","End":"02:10.875","Text":"and z components. Let\u0027s write it out."},{"Start":"02:10.875 ","End":"02:15.680","Text":"We\u0027ll say that our r_1 is going to be equal to?"},{"Start":"02:15.680 ","End":"02:17.705","Text":"Let\u0027s see. In the x-direction,"},{"Start":"02:17.705 ","End":"02:21.500","Text":"we know that this length over here is going to be a."},{"Start":"02:21.500 ","End":"02:23.310","Text":"In the x-direction,"},{"Start":"02:23.310 ","End":"02:27.540","Text":"we\u0027ll have a cosine of Theta."},{"Start":"02:27.540 ","End":"02:33.120","Text":"Sorry, let me just mark over here, this is Theta."},{"Start":"02:33.230 ","End":"02:43.874","Text":"I have a cosine Theta in the x-direction plus a sine of Theta in the y-direction,"},{"Start":"02:43.874 ","End":"02:48.359","Text":"and then plus z in the z-direction."},{"Start":"02:48.359 ","End":"02:52.960","Text":"Its height over here that we said was z."},{"Start":"02:52.960 ","End":"02:55.000","Text":"Then similarly for r_2,"},{"Start":"02:55.000 ","End":"02:57.265","Text":"which will be the vector going over here."},{"Start":"02:57.265 ","End":"02:59.900","Text":"Let\u0027s draw this as well."},{"Start":"03:00.200 ","End":"03:02.895","Text":"This is r_2."},{"Start":"03:02.895 ","End":"03:06.190","Text":"Notice r_1 and r_2 are meant to be the same length."},{"Start":"03:06.190 ","End":"03:08.515","Text":"I know you can\u0027t really see that in the picture."},{"Start":"03:08.515 ","End":"03:14.620","Text":"Then we can see that r_2 is just the exact opposite of r_1."},{"Start":"03:14.620 ","End":"03:18.610","Text":"It\u0027s a diagonal going downward instead of upwards."},{"Start":"03:18.610 ","End":"03:27.035","Text":"We can say therefore that our r_2 is going to be equal to negative our r_1."},{"Start":"03:27.035 ","End":"03:31.395","Text":"That means that we\u0027ll have negative a cosine of Theta in the x-direction,"},{"Start":"03:31.395 ","End":"03:33.915","Text":"negative a sine of Theta and the y-direction,"},{"Start":"03:33.915 ","End":"03:38.520","Text":"and also negative z in the z-direction."},{"Start":"03:38.520 ","End":"03:40.755","Text":"These are our positions."},{"Start":"03:40.755 ","End":"03:43.455","Text":"Now we\u0027re going to speak about the velocities."},{"Start":"03:43.455 ","End":"03:46.770","Text":"This is going to be slightly more complicated."},{"Start":"03:46.770 ","End":"03:49.455","Text":"As we can see this arrow over here,"},{"Start":"03:49.455 ","End":"03:51.900","Text":"that represents the velocity."},{"Start":"03:51.900 ","End":"03:57.240","Text":"This will be v_1 and it\u0027s a vector."},{"Start":"03:57.240 ","End":"04:01.035","Text":"Let\u0027s see how we work this out."},{"Start":"04:01.035 ","End":"04:07.035","Text":"Notice that v_1 is perpendicular to our r vector,"},{"Start":"04:07.035 ","End":"04:10.245","Text":"to the xy coordinates of our r vector."},{"Start":"04:10.245 ","End":"04:15.030","Text":"Notice that our v is just on the xy plane,"},{"Start":"04:15.030 ","End":"04:18.345","Text":"and doesn\u0027t have any components in the x-direction,"},{"Start":"04:18.345 ","End":"04:20.820","Text":"as opposed to our r vector."},{"Start":"04:20.820 ","End":"04:22.380","Text":"As you can see in this diagram,"},{"Start":"04:22.380 ","End":"04:24.930","Text":"our r has components in the x, y,"},{"Start":"04:24.930 ","End":"04:27.660","Text":"and z planes,"},{"Start":"04:27.660 ","End":"04:31.420","Text":"so here, our v is only in the xy."},{"Start":"04:33.530 ","End":"04:37.260","Text":"Our v is perpendicular to our r vector."},{"Start":"04:37.260 ","End":"04:40.200","Text":"I can copy, as you know with vectors,"},{"Start":"04:40.200 ","End":"04:45.410","Text":"I can move them, as long as I keep their size and direction, anywhere I want."},{"Start":"04:45.410 ","End":"04:49.230","Text":"I can move this to be over here."},{"Start":"04:49.670 ","End":"04:52.745","Text":"If I\u0027ve moved v_1 over here,"},{"Start":"04:52.745 ","End":"04:55.849","Text":"so you\u0027ll notice that this angle,"},{"Start":"04:55.849 ","End":"04:57.515","Text":"this is also Theta."},{"Start":"04:57.515 ","End":"05:01.025","Text":"If this is Theta, then this is also Theta."},{"Start":"05:01.025 ","End":"05:07.170","Text":"Now, because it\u0027s at 90 degrees to my r vector,"},{"Start":"05:07.170 ","End":"05:11.185","Text":"we can see that when I\u0027m writing out my v_1 and"},{"Start":"05:11.185 ","End":"05:16.370","Text":"because it only has components in the x and y planes,"},{"Start":"05:16.370 ","End":"05:21.530","Text":"so I\u0027m just going to switch over my x and y components."},{"Start":"05:21.530 ","End":"05:23.800","Text":"I\u0027ll show you how I do this."},{"Start":"05:23.800 ","End":"05:27.045","Text":"My v is going to be negative."},{"Start":"05:27.045 ","End":"05:29.100","Text":"We know that it\u0027s Omega a."},{"Start":"05:29.100 ","End":"05:31.800","Text":"It\u0027s going to be negative Omega a."},{"Start":"05:31.800 ","End":"05:36.350","Text":"Now instead of writing cosine Theta in the x-direction,"},{"Start":"05:36.350 ","End":"05:38.300","Text":"because it\u0027s perpendicular to my r,"},{"Start":"05:38.300 ","End":"05:41.755","Text":"so I\u0027m going to switch it to my sine Theta."},{"Start":"05:41.755 ","End":"05:51.600","Text":"Sine Theta in the x-direction and then plus Omega a."},{"Start":"05:51.600 ","End":"05:54.830","Text":"Now, instead of writing sine Theta in the y-direction,"},{"Start":"05:54.830 ","End":"06:01.710","Text":"I\u0027m going to write cosine Theta in the y-direction."},{"Start":"06:01.710 ","End":"06:05.450","Text":"Let\u0027s just give this a little explanation again."},{"Start":"06:05.450 ","End":"06:08.435","Text":"We had my position vector,"},{"Start":"06:08.435 ","End":"06:12.370","Text":"my r. Now we\u0027re dealing with my velocity vector."},{"Start":"06:12.370 ","End":"06:18.857","Text":"Now because my velocity vector is perpendicular to my position vector"},{"Start":"06:18.857 ","End":"06:25.790","Text":"and this angle Theta is the same angle Theta over here because it\u0027s perpendicular,"},{"Start":"06:25.790 ","End":"06:29.795","Text":"I\u0027m switching, instead of writing cosine Theta in the x direction,"},{"Start":"06:29.795 ","End":"06:32.705","Text":"I\u0027m writing sine Theta in the x-direction,"},{"Start":"06:32.705 ","End":"06:34.670","Text":"because it\u0027s the opposite."},{"Start":"06:34.670 ","End":"06:37.700","Text":"Then where does this negative come from?"},{"Start":"06:37.700 ","End":"06:39.505","Text":"It comes from,"},{"Start":"06:39.505 ","End":"06:41.500","Text":"if we look at our r vector,"},{"Start":"06:41.500 ","End":"06:44.210","Text":"we can see that it\u0027s in the downward diagonal,"},{"Start":"06:44.210 ","End":"06:48.170","Text":"which means that if we split it up into components,"},{"Start":"06:48.170 ","End":"06:52.070","Text":"we can see that it\u0027s going downwards in"},{"Start":"06:52.070 ","End":"06:57.560","Text":"the negative x-direction and then in the positive y-direction."},{"Start":"06:57.560 ","End":"06:58.775","Text":"As opposed to this,"},{"Start":"06:58.775 ","End":"07:03.260","Text":"our v_1 vector is in the positive diagonal. What does that mean?"},{"Start":"07:03.260 ","End":"07:05.480","Text":"If we split it up into components,"},{"Start":"07:05.480 ","End":"07:11.240","Text":"it\u0027s going in the negative x-direction."},{"Start":"07:11.240 ","End":"07:12.395","Text":"In the y-direction,"},{"Start":"07:12.395 ","End":"07:15.225","Text":"it\u0027s still going in the positive y-direction."},{"Start":"07:15.225 ","End":"07:17.945","Text":"Because our x directions have switched,"},{"Start":"07:17.945 ","End":"07:20.555","Text":"so here comes this minus."},{"Start":"07:20.555 ","End":"07:26.610","Text":"Here, it stays the plus because it\u0027s still going in the positive y-direction."},{"Start":"07:28.430 ","End":"07:30.855","Text":"As for v_2,"},{"Start":"07:30.855 ","End":"07:32.715","Text":"let\u0027s write it over here."},{"Start":"07:32.715 ","End":"07:37.450","Text":"Just like with r_2,"},{"Start":"07:37.450 ","End":"07:41.300","Text":"it\u0027s going to be equal to negative v_1,"},{"Start":"07:41.300 ","End":"07:43.910","Text":"because it\u0027s the exact same thing,"},{"Start":"07:43.910 ","End":"07:46.420","Text":"just in the opposite direction."},{"Start":"07:46.420 ","End":"07:49.875","Text":"Now we\u0027re going to do the cross multiplication."},{"Start":"07:49.875 ","End":"07:55.700","Text":"We\u0027re going to find the angular momentum of body number 1. How are we going to do this?"},{"Start":"07:55.700 ","End":"07:59.965","Text":"We\u0027re going to work it out through the determinant."},{"Start":"07:59.965 ","End":"08:04.245","Text":"We\u0027re going to have our x components,"},{"Start":"08:04.245 ","End":"08:09.520","Text":"our y components, and z components."},{"Start":"08:09.520 ","End":"08:13.460","Text":"Then we\u0027re going to write down our r\u0027s."},{"Start":"08:13.460 ","End":"08:18.525","Text":"For our x, we have a cosine of Theta."},{"Start":"08:18.525 ","End":"08:22.410","Text":"For our y, we have a sine of Theta."},{"Start":"08:22.410 ","End":"08:25.335","Text":"For our z, we just have z."},{"Start":"08:25.335 ","End":"08:27.574","Text":"Then for our velocity,"},{"Start":"08:27.574 ","End":"08:33.975","Text":"we have negative Omega a sine of Theta."},{"Start":"08:33.975 ","End":"08:39.750","Text":"For our y, we have Omega a cosine of Theta."},{"Start":"08:39.750 ","End":"08:43.450","Text":"For or z, we have 0."},{"Start":"08:43.610 ","End":"08:47.220","Text":"Now let\u0027s work out the determinant."},{"Start":"08:47.220 ","End":"08:49.570","Text":"In the x-direction,"},{"Start":"08:49.570 ","End":"08:53.960","Text":"we\u0027re going to have a sine of Theta multiplied by 0,"},{"Start":"08:53.960 ","End":"08:58.820","Text":"which is 0, minus Omega a cosine of Theta times z."},{"Start":"08:58.820 ","End":"09:06.320","Text":"We\u0027ll have z Omega a cosine of Theta in the x-direction."},{"Start":"09:06.320 ","End":"09:07.790","Text":"Now in the y direction,"},{"Start":"09:07.790 ","End":"09:09.650","Text":"that means that we\u0027re crossing out this,"},{"Start":"09:09.650 ","End":"09:12.995","Text":"so we\u0027ll have a cosine of Theta multiplied by 0,"},{"Start":"09:12.995 ","End":"09:17.000","Text":"negative, negative Omega a sine of Theta multiplied by z."},{"Start":"09:17.000 ","End":"09:18.530","Text":"That will be a positive."},{"Start":"09:18.530 ","End":"09:22.470","Text":"We\u0027ll have positive Omega a sine of"},{"Start":"09:22.470 ","End":"09:29.965","Text":"Theta in the y-direction with the z here."},{"Start":"09:29.965 ","End":"09:32.624","Text":"Then in the z direction,"},{"Start":"09:32.624 ","End":"09:38.650","Text":"we\u0027re going to have a cosine of Theta multiplied by Omega a cosine of Theta,"},{"Start":"09:38.900 ","End":"09:45.640","Text":"negative, negative Omega a sine of Theta multiplied by a sine of Theta."},{"Start":"09:45.710 ","End":"09:50.410","Text":"That\u0027s going to equal to plus."},{"Start":"09:50.720 ","End":"09:53.750","Text":"You\u0027ll notice that in both multiplications,"},{"Start":"09:53.750 ","End":"09:56.810","Text":"we\u0027re going to have Omega multiplied by a^2,"},{"Start":"09:56.810 ","End":"09:58.860","Text":"so Omega a^2,"},{"Start":"09:58.860 ","End":"10:00.490","Text":"and then open brackets."},{"Start":"10:00.490 ","End":"10:05.685","Text":"Then here, we\u0027ll have cosine squared of Theta."},{"Start":"10:05.685 ","End":"10:07.790","Text":"Then over here, because we have negative,"},{"Start":"10:07.790 ","End":"10:09.350","Text":"negative, so it will be a positive,"},{"Start":"10:09.350 ","End":"10:14.270","Text":"so plus sine squared Theta."},{"Start":"10:14.270 ","End":"10:18.350","Text":"As we know, cosine squared Theta plus sine squared Theta is equal to 1,"},{"Start":"10:18.350 ","End":"10:20.720","Text":"and obviously in the z direction,"},{"Start":"10:20.720 ","End":"10:26.165","Text":"which makes sense because as we can see when it comes to our z component,"},{"Start":"10:26.165 ","End":"10:28.415","Text":"it has nothing to do with our Theta angle."},{"Start":"10:28.415 ","End":"10:34.680","Text":"Our Theta angle is only to do with the components in x and y."},{"Start":"10:34.680 ","End":"10:37.590","Text":"That makes a lot of sense."},{"Start":"10:37.590 ","End":"10:41.380","Text":"Of course, I forgot that we have to add in our m"},{"Start":"10:41.380 ","End":"10:45.685","Text":"because this is a cross multiplication by m Omega a,"},{"Start":"10:45.685 ","End":"10:50.680","Text":"so everything is going to be multiplied by m. Also here,"},{"Start":"10:50.680 ","End":"10:53.565","Text":"we\u0027ll have an m over here,"},{"Start":"10:53.565 ","End":"10:56.475","Text":"and also an m over here."},{"Start":"10:56.475 ","End":"10:59.490","Text":"Now with our L_2,"},{"Start":"10:59.490 ","End":"11:02.505","Text":"for our body number 2 of angular momentum."},{"Start":"11:02.505 ","End":"11:09.810","Text":"If we look, it\u0027s going to be our r_2 cross multiplied with v_2,"},{"Start":"11:09.810 ","End":"11:18.010","Text":"which is equal to negative r_1 cross multiplied with negative v_1,"},{"Start":"11:20.330 ","End":"11:25.980","Text":"because we have the negative and the negative and they cancel out,"},{"Start":"11:25.980 ","End":"11:29.755","Text":"it\u0027s just going to be equal to our L_1,"},{"Start":"11:29.755 ","End":"11:31.865","Text":"which is also what we expected."},{"Start":"11:31.865 ","End":"11:36.290","Text":"We wanted to get our angular momentum in the same direction."},{"Start":"11:36.290 ","End":"11:38.600","Text":"Then our L total,"},{"Start":"11:38.600 ","End":"11:41.540","Text":"the total angular momentum of the system,"},{"Start":"11:41.540 ","End":"11:47.600","Text":"is simply going to be L_1 plus our L_2,"},{"Start":"11:47.600 ","End":"11:51.720","Text":"which is simply going to be equal to 2L_1."},{"Start":"11:53.720 ","End":"12:00.830","Text":"Sorry, this is where the whole mistake comes in because here,"},{"Start":"12:00.830 ","End":"12:05.450","Text":"we\u0027re meant to have a p. I\u0027m just going to say that"},{"Start":"12:05.450 ","End":"12:10.835","Text":"this is p because we have to multiply."},{"Start":"12:10.835 ","End":"12:14.090","Text":"The cross multiplication is between r and p,"},{"Start":"12:14.090 ","End":"12:21.860","Text":"and p is m times the v. That\u0027s going to be just multiplying m on the line in"},{"Start":"12:21.860 ","End":"12:30.215","Text":"the determinant where we put in our v. I\u0027ll just add an m over here and an m over here."},{"Start":"12:30.215 ","End":"12:33.735","Text":"Then we get the onset."},{"Start":"12:33.735 ","End":"12:40.550","Text":"Because we have that r total angular momentum of the whole system is 2 times our L_1,"},{"Start":"12:40.550 ","End":"12:43.320","Text":"I\u0027ll rise up here."},{"Start":"12:45.700 ","End":"12:55.305","Text":"Therefore, our total angular momentum is going to be equal to 2."},{"Start":"12:55.305 ","End":"12:59.970","Text":"Then we have our common factors,"},{"Start":"12:59.970 ","End":"13:03.315","Text":"so it\u0027s going to be m Omega a."},{"Start":"13:03.315 ","End":"13:06.795","Text":"Then for our x-component,"},{"Start":"13:06.795 ","End":"13:11.535","Text":"we have z cosine of Theta."},{"Start":"13:11.535 ","End":"13:16.750","Text":"For our y, we\u0027re going to have z sine of Theta."},{"Start":"13:16.970 ","End":"13:20.310","Text":"Then for our z,"},{"Start":"13:20.310 ","End":"13:23.620","Text":"we\u0027re just going to have an a over here."},{"Start":"13:24.440 ","End":"13:33.625","Text":"That\u0027s how we work out the total angular momentum of a system by using the cross-product."},{"Start":"13:33.625 ","End":"13:39.950","Text":"It\u0027s also very important that when you\u0027re working out the velocity vector,"},{"Start":"13:39.950 ","End":"13:45.025","Text":"that you remember that the cross multiplication is between p and not v,"},{"Start":"13:45.025 ","End":"13:46.625","Text":"which means that in the end,"},{"Start":"13:46.625 ","End":"13:50.070","Text":"you have to multiply everything by m over here."},{"Start":"13:50.380 ","End":"13:53.970","Text":"That\u0027s it. That\u0027s the end of this lesson."}],"ID":9524},{"Watched":false,"Name":"Torque and Precession","Duration":"7m 32s","ChapterTopicVideoID":9134,"CourseChapterTopicPlaylistID":9397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:05.655","Text":"Hello. Now we\u0027re going to speak a little bit about our results."},{"Start":"00:05.655 ","End":"00:08.895","Text":"I\u0027m going to move a little bit to the side over here."},{"Start":"00:08.895 ","End":"00:12.659","Text":"We can see that our Omega,"},{"Start":"00:12.659 ","End":"00:15.345","Text":"what is our Omega equal to?"},{"Start":"00:15.345 ","End":"00:20.295","Text":"Our Omega is our angular velocity,"},{"Start":"00:20.295 ","End":"00:24.645","Text":"which means that it\u0027s how r Theta changes with time."},{"Start":"00:24.645 ","End":"00:27.970","Text":"It\u0027s equal to Theta dot."},{"Start":"00:29.630 ","End":"00:32.400","Text":"That is what our Omega is,"},{"Start":"00:32.400 ","End":"00:36.060","Text":"and we can see that our Theta is the angle."},{"Start":"00:36.060 ","End":"00:39.300","Text":"We\u0027re constantly changing our direction."},{"Start":"00:39.300 ","End":"00:42.555","Text":"We can see that our a is a constant,"},{"Start":"00:42.555 ","End":"00:46.990","Text":"so our z component is always going to be constant."},{"Start":"00:46.990 ","End":"00:50.450","Text":"However, our x and y components are"},{"Start":"00:50.450 ","End":"00:55.470","Text":"changing with time because they are dependent on Theta."},{"Start":"00:55.520 ","End":"00:58.715","Text":"This is something a little bit new to us because"},{"Start":"00:58.715 ","End":"01:02.240","Text":"before we saw that our angular momentum is always constant."},{"Start":"01:02.240 ","End":"01:04.895","Text":"However, here when we\u0027re dealing with precession,"},{"Start":"01:04.895 ","End":"01:08.420","Text":"because everything is changing,"},{"Start":"01:08.420 ","End":"01:11.670","Text":"the angle is changing."},{"Start":"01:12.160 ","End":"01:18.335","Text":"We can see that angular momentum when dealing with precession,"},{"Start":"01:18.335 ","End":"01:20.405","Text":"is also constantly changing."},{"Start":"01:20.405 ","End":"01:23.375","Text":"If we find the size of our angular momentum,"},{"Start":"01:23.375 ","End":"01:26.855","Text":"we will see that we will get rid of our Cosine Theta over here,"},{"Start":"01:26.855 ","End":"01:29.165","Text":"so the size is always constant."},{"Start":"01:29.165 ","End":"01:32.240","Text":"However, the direction is always changing."},{"Start":"01:32.240 ","End":"01:37.595","Text":"That means that because our angular momentum is a vector,"},{"Start":"01:37.595 ","End":"01:40.650","Text":"it means that it is also changing."},{"Start":"01:41.540 ","End":"01:44.690","Text":"That means that as we said before,"},{"Start":"01:44.690 ","End":"01:47.449","Text":"if this is the axis of rotation,"},{"Start":"01:47.449 ","End":"01:49.684","Text":"and this is our angular momentum,"},{"Start":"01:49.684 ","End":"01:54.495","Text":"so its size is constantly the same."},{"Start":"01:54.495 ","End":"01:57.405","Text":"This is just a three-dimensional drawing."},{"Start":"01:57.405 ","End":"02:04.085","Text":"But its direction is constantly changing because it\u0027s rotating around the axes."},{"Start":"02:04.085 ","End":"02:06.240","Text":"If you look at it like that."},{"Start":"02:07.160 ","End":"02:11.150","Text":"Now we\u0027re going to be reminded by this law that"},{"Start":"02:11.150 ","End":"02:15.530","Text":"says that if the sum of the external moments is equal to 0,"},{"Start":"02:15.530 ","End":"02:19.210","Text":"then that means that there\u0027s conservation of angular momentum."},{"Start":"02:19.210 ","End":"02:22.684","Text":"Now, we\u0027ve already seen that our angular momentum"},{"Start":"02:22.684 ","End":"02:26.885","Text":"is changing with time because its direction is changing."},{"Start":"02:26.885 ","End":"02:32.460","Text":"That means that we know"},{"Start":"02:32.780 ","End":"02:41.080","Text":"that our angular momentum dl by dt,"},{"Start":"02:41.080 ","End":"02:43.655","Text":"because it is changing with time."},{"Start":"02:43.655 ","End":"02:48.950","Text":"That means that there\u0027s some of external moments,"},{"Start":"02:48.950 ","End":"02:51.570","Text":"which means external torques."},{"Start":"02:56.270 ","End":"03:00.790","Text":"That this is not equal to 0."},{"Start":"03:01.490 ","End":"03:07.335","Text":"Now we can see that we have some external torque."},{"Start":"03:07.335 ","End":"03:10.520","Text":"External moment or an external moment of force."},{"Start":"03:10.520 ","End":"03:14.040","Text":"Remember, all of these words are the same."},{"Start":"03:14.360 ","End":"03:18.320","Text":"We don\u0027t know exactly what is providing this."},{"Start":"03:18.320 ","End":"03:22.905","Text":"However, we can find out the torque itself."},{"Start":"03:22.905 ","End":"03:26.595","Text":"As we can see, our dl by dt will equal that,"},{"Start":"03:26.595 ","End":"03:33.180","Text":"so all we have to do is we have to take the derivative of this,"},{"Start":"03:33.180 ","End":"03:37.410","Text":"and then we will find the sum of the external torques."},{"Start":"03:38.030 ","End":"03:41.270","Text":"Now what we\u0027re going to do is we\u0027re going to derive"},{"Start":"03:41.270 ","End":"03:45.955","Text":"the angular momentum with respect to time."},{"Start":"03:45.955 ","End":"03:49.965","Text":"Let\u0027s be reminded of the chain rule,"},{"Start":"03:49.965 ","End":"03:56.340","Text":"so dl by dt is going to be equal to dl by"},{"Start":"03:56.340 ","End":"04:01.275","Text":"d Theta with respect to Theta"},{"Start":"04:01.275 ","End":"04:08.325","Text":"multiplied by dTheta with respect to time, by dt."},{"Start":"04:08.325 ","End":"04:10.490","Text":"Let\u0027s see what this equals to."},{"Start":"04:10.490 ","End":"04:14.960","Text":"So first we\u0027re going to take the derivative of L with respect to Theta."},{"Start":"04:14.960 ","End":"04:21.480","Text":"What we\u0027re going to have is 2 multiplied by m omega a,"},{"Start":"04:21.480 ","End":"04:28.875","Text":"and then this will be negative z Sine of Theta."},{"Start":"04:28.875 ","End":"04:33.820","Text":"This will be z Cosine of Theta."},{"Start":"04:35.840 ","End":"04:41.630","Text":"This a with regards to Theta is just going to be equal to 0."},{"Start":"04:41.630 ","End":"04:43.400","Text":"There\u0027s no z component."},{"Start":"04:43.400 ","End":"04:48.680","Text":"Now, we have to multiply it by d Theta with respect to t. Now as we know,"},{"Start":"04:48.680 ","End":"04:51.425","Text":"our d Theta with respect to t,"},{"Start":"04:51.425 ","End":"04:55.030","Text":"that\u0027s what is this equal to."},{"Start":"04:55.030 ","End":"04:58.575","Text":"Is equal to Omega."},{"Start":"04:58.575 ","End":"05:01.450","Text":"Because it equals to Theta."},{"Start":"05:02.450 ","End":"05:09.910","Text":"Now I just have to multiply this whole expression by Omega, so Omega squared."},{"Start":"05:09.910 ","End":"05:18.900","Text":"This is going to be the size of our external torque."},{"Start":"05:19.300 ","End":"05:22.550","Text":"Now 2 things to notice is that one,"},{"Start":"05:22.550 ","End":"05:24.905","Text":"it has no z component,"},{"Start":"05:24.905 ","End":"05:27.920","Text":"it\u0027s only on the x, y plane,"},{"Start":"05:27.920 ","End":"05:31.135","Text":"and that number 2,"},{"Start":"05:31.135 ","End":"05:37.535","Text":"we can see that it\u0027s going to be perpendicular to our angular momentum."},{"Start":"05:37.535 ","End":"05:40.505","Text":"How do we know that it\u0027s perpendicular to the angular momentum?"},{"Start":"05:40.505 ","End":"05:44.975","Text":"Here, our z component in the Angular Momentum is Cosine of Theta,"},{"Start":"05:44.975 ","End":"05:47.110","Text":"and here its Sine of Theta."},{"Start":"05:47.110 ","End":"05:49.545","Text":"The y component is Sine of Theta,"},{"Start":"05:49.545 ","End":"05:51.810","Text":"and here it\u0027s Cosine of Theta."},{"Start":"05:51.810 ","End":"05:55.310","Text":"It\u0027s the opposite, which means that it\u0027s perpendicular."},{"Start":"05:55.310 ","End":"06:01.905","Text":"Similarly, if you were to do the dot product of the moments,"},{"Start":"06:01.905 ","End":"06:06.905","Text":"the external torque, dot-product with our angular momentum,"},{"Start":"06:06.905 ","End":"06:11.435","Text":"you\u0027ll get 0 because they\u0027re perpendicular one to another."},{"Start":"06:11.435 ","End":"06:14.450","Text":"That means if I take this diagram again,"},{"Start":"06:14.450 ","End":"06:17.450","Text":"and I just take it at one moment,"},{"Start":"06:17.450 ","End":"06:19.400","Text":"at the first instance,"},{"Start":"06:19.400 ","End":"06:23.270","Text":"when the angular momentum is over here."},{"Start":"06:23.270 ","End":"06:32.610","Text":"Then we will see that our external torque will be perpendicular to its x y components,"},{"Start":"06:32.610 ","End":"06:35.420","Text":"to the projection on the x y plane."},{"Start":"06:35.420 ","End":"06:37.490","Text":"Notice that here it\u0027s going in"},{"Start":"06:37.490 ","End":"06:43.470","Text":"the negative x-direction to here the negative perpendicular direction."},{"Start":"06:43.470 ","End":"06:47.860","Text":"We\u0027ll be going something like in this direction."},{"Start":"06:47.860 ","End":"06:55.635","Text":"This will be the torque and when the angular momentum moves along to its next stage,"},{"Start":"06:55.635 ","End":"07:00.465","Text":"this will be over here."},{"Start":"07:00.465 ","End":"07:05.330","Text":"We know that the perpendicular one to another because their x and"},{"Start":"07:05.330 ","End":"07:10.595","Text":"y components are the opposite with the Cosine and Sine until they\u0027ve been flipped."},{"Start":"07:10.595 ","End":"07:13.390","Text":"Because there\u0027s a negative over here,"},{"Start":"07:13.390 ","End":"07:15.515","Text":"it\u0027s going in the negative direction."},{"Start":"07:15.515 ","End":"07:21.000","Text":"The arrow is pointing in the opposite direction to this arrow over here."},{"Start":"07:21.370 ","End":"07:32.399","Text":"This external torque is what is rotating the whole system around its axis of rotation.c"}],"ID":9525},{"Watched":false,"Name":"4.1 Point On Wheel","Duration":"21m 43s","ChapterTopicVideoID":12233,"CourseChapterTopicPlaylistID":9397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"Hello in this question we\u0027re being told that we have a"},{"Start":"00:05.610 ","End":"00:11.505","Text":"solid filled-in wheel of radius small r and of width a."},{"Start":"00:11.505 ","End":"00:20.124","Text":"We\u0027re being told that at some point on the edge of the wheel we have some point mass m,"},{"Start":"00:20.124 ","End":"00:22.920","Text":"and the wheel is spinning."},{"Start":"00:22.920 ","End":"00:29.030","Text":"In question a, we\u0027re being told to find the equation for"},{"Start":"00:29.030 ","End":"00:32.300","Text":"the angular momentum of this system and to"},{"Start":"00:32.300 ","End":"00:36.410","Text":"show also that it\u0027s going to be dependent on time."},{"Start":"00:36.410 ","End":"00:40.175","Text":"We\u0027re finding, l of the mass,"},{"Start":"00:40.175 ","End":"00:46.860","Text":"which is dependent on time and we\u0027re trying to find out what the equation for this is."},{"Start":"00:47.540 ","End":"00:50.550","Text":"Let\u0027s take a look at how we do this."},{"Start":"00:50.550 ","End":"00:55.595","Text":"Also we have to note that the origin,"},{"Start":"00:55.595 ","End":"00:58.860","Text":"is going to be right in the middle of the wheel."},{"Start":"01:00.140 ","End":"01:05.545","Text":"The way that we\u0027re going to solve this is we\u0027re going to use the idea of precession."},{"Start":"01:05.545 ","End":"01:08.500","Text":"Why do we know that we\u0027re going to be using precession?"},{"Start":"01:08.500 ","End":"01:12.385","Text":"Because we can see that there\u0027s no symmetry in the system."},{"Start":"01:12.385 ","End":"01:15.460","Text":"Usually when we\u0027re dealing with a whole wheel,"},{"Start":"01:15.460 ","End":"01:18.200","Text":"there\u0027s no point mass over here."},{"Start":"01:18.200 ","End":"01:20.265","Text":"The whole thing is symmetrical."},{"Start":"01:20.265 ","End":"01:22.672","Text":"Then you can just work out the angular momentum."},{"Start":"01:22.672 ","End":"01:28.013","Text":"Here because we have 1 mass over here and the whole system is out of balance,"},{"Start":"01:28.013 ","End":"01:31.610","Text":"we can see that we\u0027re going to be using procession."},{"Start":"01:32.030 ","End":"01:36.280","Text":"Let\u0027s begin. The first thing that we\u0027re going to"},{"Start":"01:36.280 ","End":"01:39.955","Text":"start is by writing the equation for angular momentum."},{"Start":"01:39.955 ","End":"01:45.650","Text":"We\u0027re going to say that our angular momentum is equal to our r vector,"},{"Start":"01:45.650 ","End":"01:47.195","Text":"which is our position vector,"},{"Start":"01:47.195 ","End":"01:51.225","Text":"cross multiplied with our p vector,"},{"Start":"01:51.225 ","End":"01:54.160","Text":"which is our momentum vector."},{"Start":"01:54.170 ","End":"01:58.185","Text":"Now let\u0027s write down what our r is."},{"Start":"01:58.185 ","End":"01:59.855","Text":"Let\u0027s see how we do this."},{"Start":"01:59.855 ","End":"02:06.340","Text":"Now, we\u0027re going to say that our wheel is rotating at an angular velocity of Omega 0."},{"Start":"02:06.340 ","End":"02:09.140","Text":"Now, because we\u0027re dealing with a wheel and we can"},{"Start":"02:09.140 ","End":"02:11.573","Text":"see that it\u0027s in a cylindrical shape,"},{"Start":"02:11.573 ","End":"02:14.495","Text":"we\u0027re going to work in cylindrical coordinates."},{"Start":"02:14.495 ","End":"02:19.495","Text":"Before we start writing out what our r vector is going to be,"},{"Start":"02:19.495 ","End":"02:23.070","Text":"we have to decide on our axes."},{"Start":"02:23.070 ","End":"02:25.155","Text":"We\u0027re working in cylindrical coordinates,"},{"Start":"02:25.155 ","End":"02:30.535","Text":"so we can say that the axis of rotation is around the z-axis."},{"Start":"02:30.535 ","End":"02:37.740","Text":"Then we have to find the radial direction so that\u0027s going to be in this direction."},{"Start":"02:38.510 ","End":"02:42.965","Text":"Then we have to find the Theta direction,"},{"Start":"02:42.965 ","End":"02:47.135","Text":"which is going to be perpendicular to both the z and the r axis."},{"Start":"02:47.135 ","End":"02:52.435","Text":"We\u0027re going to say that the Theta direction is going to be into the page."},{"Start":"02:52.435 ","End":"02:55.035","Text":"This is Theta hat."},{"Start":"02:55.035 ","End":"02:59.465","Text":"Now in order to write down our position vector,"},{"Start":"02:59.465 ","End":"03:03.230","Text":"we can just simply look at what it is."},{"Start":"03:03.230 ","End":"03:09.000","Text":"We\u0027re starting from the origin until a point where our mass is."},{"Start":"03:09.000 ","End":"03:12.740","Text":"This is our radius."},{"Start":"03:12.740 ","End":"03:16.595","Text":"Now let\u0027s take a look what that means in coordinates."},{"Start":"03:16.595 ","End":"03:21.215","Text":"We can see that along the radial direction it\u0027s a distance of"},{"Start":"03:21.215 ","End":"03:26.760","Text":"r. We said it will be r_1 to make it more clear,"},{"Start":"03:26.760 ","End":"03:31.840","Text":"so we have r_1 in the radial direction."},{"Start":"03:32.810 ","End":"03:40.003","Text":"If this whole width is a and our origin is right in the center of the entire wheel,"},{"Start":"03:40.003 ","End":"03:46.375","Text":"we know that the origin is going to be in the z-axis at a/2."},{"Start":"03:46.375 ","End":"03:54.755","Text":"Then we can say that this is a plus a/2 in the z direction."},{"Start":"03:54.755 ","End":"03:56.690","Text":"Again, why is it a/2?"},{"Start":"03:56.690 ","End":"03:58.535","Text":"Because it\u0027s in the center of the wheel."},{"Start":"03:58.535 ","End":"04:06.625","Text":"We have to find the midpoint of the width of the wheel, which is a/2."},{"Start":"04:06.625 ","End":"04:11.100","Text":"Now we have to do is to work out what our p is."},{"Start":"04:11.150 ","End":"04:17.595","Text":"Our p is equal to mass times our velocity."},{"Start":"04:17.595 ","End":"04:19.885","Text":"Let\u0027s take a look at what our velocity is."},{"Start":"04:19.885 ","End":"04:27.340","Text":"We can see that at this point because our angular velocity is going in this direction."},{"Start":"04:27.340 ","End":"04:31.090","Text":"We can see that our v over here is going to be into"},{"Start":"04:31.090 ","End":"04:35.005","Text":"the page at this exact point over here,"},{"Start":"04:35.005 ","End":"04:37.120","Text":"when the mass is right here at the top."},{"Start":"04:37.120 ","End":"04:40.900","Text":"Then we can say that we have m and then what is our v?"},{"Start":"04:40.900 ","End":"04:43.240","Text":"It\u0027s going to be Omega,"},{"Start":"04:43.240 ","End":"04:46.450","Text":"the angular velocity multiplied by the radius."},{"Start":"04:46.450 ","End":"04:52.800","Text":"Here it\u0027s going to be the angular velocity is Omega 0 and the radius is r1."},{"Start":"04:52.800 ","End":"04:55.790","Text":"Now because our v is a vector quantity,"},{"Start":"04:55.790 ","End":"04:57.695","Text":"we have to add in the direction."},{"Start":"04:57.695 ","End":"05:02.030","Text":"As we can see, the v is going into the page over here"},{"Start":"05:02.030 ","End":"05:07.520","Text":"which is exactly the positive Theta direction that we spoke about over here."},{"Start":"05:07.520 ","End":"05:12.095","Text":"Then we can just write in the Theta direction."},{"Start":"05:12.095 ","End":"05:16.835","Text":"Now alternatively, another way to solve this is"},{"Start":"05:16.835 ","End":"05:21.770","Text":"if we were going to be using the polar coordinate system."},{"Start":"05:21.770 ","End":"05:27.835","Text":"What we can do is use the usual recipe which is going to be m multiplied by"},{"Start":"05:27.835 ","End":"05:33.405","Text":"r dot in the r direction plus r Theta dot"},{"Start":"05:33.405 ","End":"05:38.130","Text":"in the Theta direction plus if there\u0027s some z component,"},{"Start":"05:38.130 ","End":"05:41.715","Text":"so z dot in the z direction."},{"Start":"05:41.715 ","End":"05:46.054","Text":"Now, because we can see that our radius is remaining constant,"},{"Start":"05:46.054 ","End":"05:49.880","Text":"so this isn\u0027t changing r dot will be 0."},{"Start":"05:49.880 ","End":"05:52.250","Text":"Because we see that in our z direction,"},{"Start":"05:52.250 ","End":"05:53.645","Text":"we\u0027re also being constant,"},{"Start":"05:53.645 ","End":"05:56.390","Text":"so z dot also doesn\u0027t exist."},{"Start":"05:56.390 ","End":"05:59.155","Text":"It\u0027s 0 because it\u0027s a constant."},{"Start":"05:59.155 ","End":"06:02.835","Text":"Then we\u0027re left with r Theta dot."},{"Start":"06:02.835 ","End":"06:03.975","Text":"Now what is Theta dot?"},{"Start":"06:03.975 ","End":"06:06.465","Text":"As we know, that equals to Omega."},{"Start":"06:06.465 ","End":"06:10.820","Text":"Again, we\u0027re left with m multiplied by the radius multiplied by Omega"},{"Start":"06:10.820 ","End":"06:15.990","Text":"in the Theta direction which is exactly what we got over here."},{"Start":"06:16.780 ","End":"06:21.320","Text":"Now what\u0027s left for us to do is to work out what our Lm is."},{"Start":"06:21.320 ","End":"06:24.430","Text":"We\u0027re going to do the cross product."},{"Start":"06:24.430 ","End":"06:28.020","Text":"We\u0027re going to have that our L is equal to our r,"},{"Start":"06:28.020 ","End":"06:38.530","Text":"which is r1 in the r direction plus a/2 in the z direction cross multiplied with rp,"},{"Start":"06:38.530 ","End":"06:46.200","Text":"which is going to be m Omega 0 r1 Theta hat."},{"Start":"06:46.200 ","End":"06:53.370","Text":"Now when you are doing cross multiplication and you have in your brackets sum or minus,"},{"Start":"06:53.370 ","End":"07:01.300","Text":"so you can just open up the brackets like you would with a regular multiplication."},{"Start":"07:01.300 ","End":"07:09.215","Text":"You can move all of the constants to one side so you can have r1 m Omega 0,"},{"Start":"07:09.215 ","End":"07:11.915","Text":"and then another r1 so squared."},{"Start":"07:11.915 ","End":"07:18.465","Text":"Then you have r hat cross multiplied with this Theta hat."},{"Start":"07:18.465 ","End":"07:22.020","Text":"Then again plus and then all of the constants,"},{"Start":"07:22.020 ","End":"07:27.945","Text":"so a/2 m Omega 0 r1."},{"Start":"07:27.945 ","End":"07:34.685","Text":"Then we\u0027re doing z hat cross multiplied with Theta hat."},{"Start":"07:34.685 ","End":"07:40.125","Text":"Then if you look at the drawing and through the right-hand rule,"},{"Start":"07:40.125 ","End":"07:44.820","Text":"so we will get this answer m Omega"},{"Start":"07:44.820 ","End":"07:50.580","Text":"0 r1 squared in the z direction."},{"Start":"07:50.580 ","End":"08:00.690","Text":"Then plus a/2 Omega 0 r1 in the negative r direction."},{"Start":"08:00.690 ","End":"08:04.230","Text":"Notice that here there\u0027s a negative it\u0027s important."},{"Start":"08:04.230 ","End":"08:08.090","Text":"Now another way if you don\u0027t want to use the right-hand rule is"},{"Start":"08:08.090 ","End":"08:11.780","Text":"to write this out as vector quantities."},{"Start":"08:11.780 ","End":"08:17.760","Text":"That would mean you\u0027ll have your r component over here,"},{"Start":"08:17.760 ","End":"08:19.860","Text":"your Theta component over here,"},{"Start":"08:19.860 ","End":"08:23.810","Text":"and your z component over here and then cross"},{"Start":"08:23.810 ","End":"08:28.885","Text":"multiplied with the same thing for this over here."},{"Start":"08:28.885 ","End":"08:31.605","Text":"Substitute it in,"},{"Start":"08:31.605 ","End":"08:38.440","Text":"and then just cross multiply instead of doing it like this."},{"Start":"08:38.620 ","End":"08:44.578","Text":"Now, this is how we notice another way of saying that this is actually precession,"},{"Start":"08:44.578 ","End":"08:48.470","Text":"and not just normal angular momentum. Why is that?"},{"Start":"08:48.470 ","End":"08:52.460","Text":"Because we can see that we have 2 components for our angular momentum."},{"Start":"08:52.460 ","End":"08:58.255","Text":"We have a component in the z direction and then the other 1 in the negative r direction."},{"Start":"08:58.255 ","End":"09:00.470","Text":"If we\u0027re looking over here,"},{"Start":"09:00.470 ","End":"09:05.595","Text":"we can see that we\u0027ll have here our L,"},{"Start":"09:05.595 ","End":"09:09.060","Text":"with the z component and then we\u0027ll have over"},{"Start":"09:09.060 ","End":"09:14.685","Text":"here our angular momentum but in the radial component."},{"Start":"09:14.685 ","End":"09:19.455","Text":"Which means that our total so let\u0027s call it Lt,"},{"Start":"09:19.455 ","End":"09:22.065","Text":"which is just all of this."},{"Start":"09:22.065 ","End":"09:25.570","Text":"It\u0027s going in this diagonal direction."},{"Start":"09:25.570 ","End":"09:29.180","Text":"This is with regards to what\u0027s going on over here."},{"Start":"09:29.180 ","End":"09:35.375","Text":"That means that our L total is going from the center in this direction."},{"Start":"09:35.375 ","End":"09:39.520","Text":"However, if we notice if our point mass was over here,"},{"Start":"09:39.520 ","End":"09:41.308","Text":"instead of where the blue is,"},{"Start":"09:41.308 ","End":"09:48.905","Text":"so we would see that our angular momentum would be in this direction."},{"Start":"09:48.905 ","End":"09:52.850","Text":"What we can see actually is that we have procession."},{"Start":"09:52.850 ","End":"09:56.730","Text":"Our angular momentum is rotating."},{"Start":"09:57.600 ","End":"10:01.960","Text":"It\u0027s rotating itself around the z axis,"},{"Start":"10:01.960 ","End":"10:05.270","Text":"so it\u0027s constantly changing direction."},{"Start":"10:06.240 ","End":"10:11.260","Text":"We found the equation for our angular momentum."},{"Start":"10:11.260 ","End":"10:13.930","Text":"However, we\u0027re being told to show how it\u0027s dependent on"},{"Start":"10:13.930 ","End":"10:18.955","Text":"time and also to write out the equation when it\u0027s dependent on time."},{"Start":"10:18.955 ","End":"10:20.875","Text":"Let\u0027s talk about this."},{"Start":"10:20.875 ","End":"10:26.200","Text":"We can see that our angular momentum is changing its direction,"},{"Start":"10:26.200 ","End":"10:30.890","Text":"it\u0027s r, so it\u0027s constantly changing in the radial direction."},{"Start":"10:31.650 ","End":"10:35.739","Text":"Our r, we know is always dependent on time."},{"Start":"10:35.739 ","End":"10:38.590","Text":"We can see that because it\u0027s changing,"},{"Start":"10:38.590 ","End":"10:40.645","Text":"our r direction is always changing,"},{"Start":"10:40.645 ","End":"10:43.795","Text":"so we know that as the time moves on,"},{"Start":"10:43.795 ","End":"10:47.710","Text":"the size of this vector will remain the same however,"},{"Start":"10:47.710 ","End":"10:49.165","Text":"its direction will be changing,"},{"Start":"10:49.165 ","End":"10:51.595","Text":"so it\u0027s changing with the time."},{"Start":"10:51.595 ","End":"10:54.160","Text":"How can we convert this?"},{"Start":"10:54.160 ","End":"11:01.030","Text":"We know that our r hat is going to be equal to cosine of Theta in"},{"Start":"11:01.030 ","End":"11:07.705","Text":"the x direction plus sine of Theta in the y-direction."},{"Start":"11:07.705 ","End":"11:14.960","Text":"Notice that there\u0027s no coefficients over here because r hat is the unit vector."},{"Start":"11:15.690 ","End":"11:22.119","Text":"Then the final thing is that we know that our Theta is what is dependent in time."},{"Start":"11:22.119 ","End":"11:26.725","Text":"Now Theta is equal to our angular velocity."},{"Start":"11:26.725 ","End":"11:31.810","Text":"Here it\u0027s Omega_0 multiplied by t, by the time."},{"Start":"11:31.810 ","End":"11:35.815","Text":"Now if we substitute all of this into our equation,"},{"Start":"11:35.815 ","End":"11:38.245","Text":"we\u0027ll get that it\u0027s dependent with time."},{"Start":"11:38.245 ","End":"11:40.550","Text":"We can write this out."},{"Start":"11:41.160 ","End":"11:46.375","Text":"I wrote this out now, I just copied over the z and then I put in the minus,"},{"Start":"11:46.375 ","End":"11:50.650","Text":"because I moved it from over here to over here and then all of the coefficients."},{"Start":"11:50.650 ","End":"11:53.440","Text":"Then instead of having in the r direction,"},{"Start":"11:53.440 ","End":"11:54.910","Text":"because the minus already moved,"},{"Start":"11:54.910 ","End":"11:56.800","Text":"so now it becomes the r direction."},{"Start":"11:56.800 ","End":"12:01.225","Text":"I just substituted in this and for Theta I substituted in"},{"Start":"12:01.225 ","End":"12:06.040","Text":"Omega_0 t. In the x-direction and in the y direction."},{"Start":"12:06.040 ","End":"12:10.105","Text":"Now we can see that we got the equation dependent on time."},{"Start":"12:10.105 ","End":"12:14.270","Text":"Let\u0027s go on to Section B."},{"Start":"12:15.600 ","End":"12:21.159","Text":"Question B is to show that the torque which changes"},{"Start":"12:21.159 ","End":"12:26.500","Text":"the angular momentum is given by the force which rotates the mass."},{"Start":"12:26.500 ","End":"12:28.540","Text":"This is a bit of a complicated question,"},{"Start":"12:28.540 ","End":"12:31.390","Text":"but let\u0027s take a look at what this means."},{"Start":"12:31.390 ","End":"12:38.475","Text":"We know that our mass is being rotated because we have angular velocity,"},{"Start":"12:38.475 ","End":"12:42.615","Text":"which means that we have some force in the meantime,"},{"Start":"12:42.615 ","End":"12:44.204","Text":"some of the forces,"},{"Start":"12:44.204 ","End":"12:47.244","Text":"and we\u0027ll call this the force,"},{"Start":"12:47.244 ","End":"12:49.390","Text":"and it\u0027s a vector quantity."},{"Start":"12:49.390 ","End":"12:55.360","Text":"The sum of the forces is going to be equal to mass multiplied by acceleration."},{"Start":"12:55.360 ","End":"12:58.938","Text":"Now because we\u0027re working with angular velocity,"},{"Start":"12:58.938 ","End":"13:04.870","Text":"our acceleration is going to be equal to Omega_0 squared,"},{"Start":"13:04.870 ","End":"13:12.055","Text":"because we know that this is our angular velocity multiplied by our radius r_1."},{"Start":"13:12.055 ","End":"13:15.310","Text":"Of course, because this is a vector quantity,"},{"Start":"13:15.310 ","End":"13:18.145","Text":"so we have to add in the vector quantities."},{"Start":"13:18.145 ","End":"13:21.159","Text":"One, we know it\u0027s going to be in the radial direction."},{"Start":"13:21.159 ","End":"13:26.620","Text":"Now let\u0027s see if it\u0027s going to be in the positive or a negative radial direction."},{"Start":"13:26.620 ","End":"13:28.675","Text":"Let\u0027s take a look over here,"},{"Start":"13:28.675 ","End":"13:36.115","Text":"we can see that it\u0027s rotating because we\u0027re dealing with angular velocity."},{"Start":"13:36.115 ","End":"13:40.000","Text":"The force is going to be an inwards force,"},{"Start":"13:40.000 ","End":"13:41.830","Text":"which means that it\u0027s in"},{"Start":"13:41.830 ","End":"13:46.435","Text":"the negative radial direction because this is the positive radial direction."},{"Start":"13:46.435 ","End":"13:51.020","Text":"Then we\u0027re just going to add a negative over here."},{"Start":"13:52.410 ","End":"13:57.625","Text":"We\u0027ve already established that our force is going to be towards the center."},{"Start":"13:57.625 ","End":"14:02.200","Text":"However, now we can see that it has some moment, why?"},{"Start":"14:02.200 ","End":"14:04.840","Text":"Because it has the same r vector."},{"Start":"14:04.840 ","End":"14:06.970","Text":"This red arrow over here,"},{"Start":"14:06.970 ","End":"14:09.370","Text":"which you can\u0027t really see right now."},{"Start":"14:09.370 ","End":"14:12.475","Text":"This is the vector."},{"Start":"14:12.475 ","End":"14:14.455","Text":"It\u0027s acting from over here."},{"Start":"14:14.455 ","End":"14:17.650","Text":"However, the force is in this direction."},{"Start":"14:17.650 ","End":"14:21.310","Text":"It\u0027s not in the same direction as our radial vector,"},{"Start":"14:21.310 ","End":"14:27.920","Text":"which means that we have some torque."},{"Start":"14:29.010 ","End":"14:31.075","Text":"Now we have the force,"},{"Start":"14:31.075 ","End":"14:33.175","Text":"let\u0027s work out the torque."},{"Start":"14:33.175 ","End":"14:37.540","Text":"We\u0027re going to find the torque of this force over here."},{"Start":"14:37.540 ","End":"14:46.140","Text":"As we know, it\u0027s going to be the r vector cross multiplied with the F vector."},{"Start":"14:46.140 ","End":"14:49.125","Text":"Now I know in this example,"},{"Start":"14:49.125 ","End":"14:52.600","Text":"I worked it out just through"},{"Start":"14:52.600 ","End":"14:57.100","Text":"identity as of what r cross Theta was and what z cross Theta was."},{"Start":"14:57.100 ","End":"14:59.680","Text":"However, it\u0027s come to my attention that some people might"},{"Start":"14:59.680 ","End":"15:02.320","Text":"find that a little bit more complicated,"},{"Start":"15:02.320 ","End":"15:08.035","Text":"and also that it\u0027s easier to make mistakes using this way how we did it over here."},{"Start":"15:08.035 ","End":"15:11.320","Text":"I\u0027m going to write it out in a very clear way,"},{"Start":"15:11.320 ","End":"15:19.090","Text":"which in an exam will hopefully guarantee no calculation errors."},{"Start":"15:19.090 ","End":"15:21.445","Text":"How do we do our r cross F?"},{"Start":"15:21.445 ","End":"15:23.710","Text":"As we know, our r vector is going to be"},{"Start":"15:23.710 ","End":"15:28.555","Text":"the exact same r vector that we had in Section A. That\u0027s this over here."},{"Start":"15:28.555 ","End":"15:33.130","Text":"What is this in vector form?"},{"Start":"15:33.130 ","End":"15:34.825","Text":"In our vector form,"},{"Start":"15:34.825 ","End":"15:38.394","Text":"we\u0027ll have r_1 in the r direction."},{"Start":"15:38.394 ","End":"15:40.975","Text":"We have no Theta components, so it\u0027s 0."},{"Start":"15:40.975 ","End":"15:46.220","Text":"Then in the z direction it\u0027s going to be a divided by 2."},{"Start":"15:46.320 ","End":"15:50.290","Text":"We can write this over here as our r vector."},{"Start":"15:50.290 ","End":"15:55.000","Text":"We\u0027ll have r_1 0 a over 2,"},{"Start":"15:55.000 ","End":"15:59.605","Text":"then cross multiplied with our force vector."},{"Start":"15:59.605 ","End":"16:01.345","Text":"What does this equal to?"},{"Start":"16:01.345 ","End":"16:03.640","Text":"It equals to in the r direction,"},{"Start":"16:03.640 ","End":"16:10.105","Text":"negative m Omega_0 squared r_1 in the r direction."},{"Start":"16:10.105 ","End":"16:15.280","Text":"Then 0 in the Theta direction because we have no Theta component and the same with a z."},{"Start":"16:15.280 ","End":"16:16.930","Text":"We have no z components,"},{"Start":"16:16.930 ","End":"16:18.298","Text":"so it\u0027s also a 0."},{"Start":"16:18.298 ","End":"16:20.665","Text":"Now we\u0027ll just substitute that in,"},{"Start":"16:20.665 ","End":"16:26.005","Text":"m Omega_0 squared r_1 0 0."},{"Start":"16:26.005 ","End":"16:29.155","Text":"Now let\u0027s do this multiplication."},{"Start":"16:29.155 ","End":"16:31.690","Text":"The first time that we do this multiplication,"},{"Start":"16:31.690 ","End":"16:36.925","Text":"we\u0027re going to, you can put your finger over the first row,"},{"Start":"16:36.925 ","End":"16:43.720","Text":"so cross this out and then you\u0027ll take the 0 multiplied by this 0,"},{"Start":"16:43.720 ","End":"16:48.655","Text":"which is 0, negative a over 2 multiplied by 0."},{"Start":"16:48.655 ","End":"16:52.310","Text":"That is equal to 0."},{"Start":"16:52.560 ","End":"16:59.110","Text":"Then what we\u0027ll do is we\u0027ll move to the next component."},{"Start":"16:59.110 ","End":"17:04.495","Text":"We cross out this line over here."},{"Start":"17:04.495 ","End":"17:08.650","Text":"Then now when we\u0027re in the middle component, we do the opposite."},{"Start":"17:08.650 ","End":"17:10.315","Text":"We take a over 2,"},{"Start":"17:10.315 ","End":"17:13.570","Text":"before we would have go on top like this."},{"Start":"17:13.570 ","End":"17:16.810","Text":"Now we\u0027re doing the x a bit different like this."},{"Start":"17:16.810 ","End":"17:23.530","Text":"A over 2 multiplied by negative m Omega_0 squared r_1,"},{"Start":"17:23.530 ","End":"17:26.815","Text":"so it\u0027s just going to be a over 2,"},{"Start":"17:26.815 ","End":"17:33.250","Text":"with a negative and then m Omega_0 squared r_1,"},{"Start":"17:33.250 ","End":"17:36.955","Text":"negative r_1 multiplied by 0,"},{"Start":"17:36.955 ","End":"17:39.685","Text":"which is 0, so negative 0."},{"Start":"17:39.685 ","End":"17:46.015","Text":"Now we\u0027ll cross off this row and do the same thing."},{"Start":"17:46.015 ","End":"17:48.445","Text":"R_1 times 0 is 0,"},{"Start":"17:48.445 ","End":"17:52.150","Text":"minus minus 0 times this,"},{"Start":"17:52.150 ","End":"17:54.775","Text":"which is going to be 0."},{"Start":"17:54.775 ","End":"17:56.650","Text":"This is then equal to,"},{"Start":"17:56.650 ","End":"17:58.630","Text":"if we\u0027re going to put it back into this type of"},{"Start":"17:58.630 ","End":"18:01.300","Text":"format where we write down the components."},{"Start":"18:01.300 ","End":"18:04.165","Text":"We\u0027ll have negative a over 2,"},{"Start":"18:04.165 ","End":"18:11.995","Text":"m Omega_0 squared r_1 in the r direction."},{"Start":"18:11.995 ","End":"18:16.375","Text":"This is our torque for this force F over here."},{"Start":"18:16.375 ","End":"18:17.980","Text":"Before I\u0027m going to continue,"},{"Start":"18:17.980 ","End":"18:20.065","Text":"just notice over here and over here,"},{"Start":"18:20.065 ","End":"18:22.480","Text":"I forgot to multiply by m,"},{"Start":"18:22.480 ","End":"18:27.445","Text":"here there\u0027s an m and I forgot to write it down in these two lines so I corrected it."},{"Start":"18:27.445 ","End":"18:31.810","Text":"Back to our question. Now, we\u0027re being asked to show that"},{"Start":"18:31.810 ","End":"18:33.190","Text":"the torque which changes"},{"Start":"18:33.190 ","End":"18:36.925","Text":"the angular momentum is given by the force which rotates the mass."},{"Start":"18:36.925 ","End":"18:40.015","Text":"We found the force that rotates the mass."},{"Start":"18:40.015 ","End":"18:45.210","Text":"We found the torque that comes with this force."},{"Start":"18:45.210 ","End":"18:48.685","Text":"Now we want to check that this is the torque"},{"Start":"18:48.685 ","End":"18:52.865","Text":"which changes the angular momentum. How do we do that?"},{"Start":"18:52.865 ","End":"18:56.010","Text":"Which changes the angular momentum."},{"Start":"18:56.010 ","End":"18:57.830","Text":"That means dl by dt."},{"Start":"18:57.830 ","End":"19:02.770","Text":"The change of angular momentum in time"},{"Start":"19:02.770 ","End":"19:08.185","Text":"is the changes of the angular momentum. Let\u0027s do this."},{"Start":"19:08.185 ","End":"19:12.080","Text":"Let\u0027s differentiate what we got over here."},{"Start":"19:12.080 ","End":"19:16.920","Text":"Let\u0027s take it over here instead of using this over here."},{"Start":"19:16.920 ","End":"19:23.985","Text":"Now an expression that is useful to remember is that r hat,"},{"Start":"19:23.985 ","End":"19:33.000","Text":"the derivative of this is going to be equal to Theta dot in the direction of Theta."},{"Start":"19:33.000 ","End":"19:35.595","Text":"Theta dot, as we know, is Omega,"},{"Start":"19:35.595 ","End":"19:41.740","Text":"which over here it\u0027s going to be Omega_0 in the Theta direction."},{"Start":"19:42.720 ","End":"19:46.315","Text":"Let\u0027s see how we differentiate this over here."},{"Start":"19:46.315 ","End":"19:49.990","Text":"As we saw, a z component also over here,"},{"Start":"19:49.990 ","End":"19:53.440","Text":"a z component has nothing to do with t,"},{"Start":"19:53.440 ","End":"19:57.280","Text":"it\u0027s a constant, so when we differentiate it, it just disappears."},{"Start":"19:57.280 ","End":"20:02.845","Text":"In the z-direction we\u0027re going to have 0 and then plus our r component."},{"Start":"20:02.845 ","End":"20:05.200","Text":"Now we can see that our r component,"},{"Start":"20:05.200 ","End":"20:07.390","Text":"when we discussed in this question,"},{"Start":"20:07.390 ","End":"20:09.690","Text":"we saw that it was changing in time,"},{"Start":"20:09.690 ","End":"20:16.570","Text":"that\u0027s where our r is equal to this and our Theta was dependent on time."},{"Start":"20:16.570 ","End":"20:19.390","Text":"Also over here we can see that it\u0027s dependent on time."},{"Start":"20:19.390 ","End":"20:22.315","Text":"Now, instead of taking the derivative of this,"},{"Start":"20:22.315 ","End":"20:23.455","Text":"which will be slightly longer,"},{"Start":"20:23.455 ","End":"20:26.495","Text":"I\u0027m going to take the derivative of this expression over here,"},{"Start":"20:26.495 ","End":"20:29.570","Text":"which is the exact same thing as this expression, just shorter."},{"Start":"20:29.570 ","End":"20:35.320","Text":"I\u0027m going to use the knowledge that my r vector,"},{"Start":"20:35.320 ","End":"20:38.580","Text":"the derivative of it is equal to this."},{"Start":"20:39.490 ","End":"20:43.500","Text":"Plus a over 2,"},{"Start":"20:43.640 ","End":"20:46.055","Text":"a over 2 as a constant,"},{"Start":"20:46.055 ","End":"20:48.425","Text":"Omega_0 is a constant,"},{"Start":"20:48.425 ","End":"20:50.990","Text":"r_1 is a constant,"},{"Start":"20:50.990 ","End":"20:58.645","Text":"and then it\u0027s multiplied by a negative over here so negative."},{"Start":"20:58.645 ","End":"21:01.620","Text":"Then our r-hat,"},{"Start":"21:01.620 ","End":"21:08.520","Text":"the derivative of it is going to be Omega_0 2 in the direction of Theta hat."},{"Start":"21:08.520 ","End":"21:12.545","Text":"Another Omega_0 in the direction of Theta hat."},{"Start":"21:12.545 ","End":"21:15.905","Text":"Then I can just rub out the 0 because it makes no difference."},{"Start":"21:15.905 ","End":"21:21.015","Text":"You\u0027ll see that I got the exact same expression over here."},{"Start":"21:21.015 ","End":"21:24.800","Text":"Sorry, again, I forgot to multiply by the mass,"},{"Start":"21:24.800 ","End":"21:26.750","Text":"I forgot to carry it in over here."},{"Start":"21:26.750 ","End":"21:29.420","Text":"We got the exact same expression."},{"Start":"21:29.420 ","End":"21:33.050","Text":"We actually got that this torque for"},{"Start":"21:33.050 ","End":"21:37.575","Text":"this force is what is changing the direction of the angular momentum."},{"Start":"21:37.575 ","End":"21:40.630","Text":"That\u0027s the answer to Question number b."},{"Start":"21:40.630 ","End":"21:44.100","Text":"Now we have finished this question."}],"ID":12712},{"Watched":false,"Name":"4.2 Rod Rotates At An Angle","Duration":"25m 15s","ChapterTopicVideoID":12234,"CourseChapterTopicPlaylistID":9397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:03.285","Text":"Hello. In this question,"},{"Start":"00:03.285 ","End":"00:06.990","Text":"we have a rod of length L and of mass M,"},{"Start":"00:06.990 ","End":"00:10.200","Text":"which is resting at an angle of Phi relative to"},{"Start":"00:10.200 ","End":"00:15.255","Text":"the Z axis and the angle Phi is never changing."},{"Start":"00:15.255 ","End":"00:21.615","Text":"The rod rotates about the Z axis at a constant angular momentum Omega."},{"Start":"00:21.615 ","End":"00:25.870","Text":"What moment of force is acting on the rod?"},{"Start":"00:26.990 ","End":"00:31.530","Text":"As we can see, our system about the Z axis,"},{"Start":"00:31.530 ","End":"00:33.734","Text":"this is the Z axis."},{"Start":"00:33.734 ","End":"00:37.630","Text":"We can see that there isn\u0027t exactly symmetry."},{"Start":"00:38.420 ","End":"00:43.000","Text":"What this means is that we\u0027re dealing with precession."},{"Start":"00:43.400 ","End":"00:49.030","Text":"Now what we want to do is we want to find the moment of force,"},{"Start":"00:49.030 ","End":"00:51.400","Text":"or the torque acting on the rod."},{"Start":"00:51.400 ","End":"00:53.080","Text":"So what are we going to do?"},{"Start":"00:53.080 ","End":"00:58.085","Text":"We\u0027re going to find the angular momentum and then"},{"Start":"00:58.085 ","End":"01:03.370","Text":"we\u0027re going to take the derivative of it and that will give us the moment of force."},{"Start":"01:03.370 ","End":"01:06.020","Text":"Let\u0027s see how we do this."},{"Start":"01:07.390 ","End":"01:10.660","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"01:10.660 ","End":"01:13.840","Text":"split the rod up into small lengths."},{"Start":"01:13.840 ","End":"01:18.350","Text":"We\u0027re going to say that each length is of length dL."},{"Start":"01:19.410 ","End":"01:24.200","Text":"Now we can say that dM."},{"Start":"01:24.470 ","End":"01:32.375","Text":"Each length has some small mass because the rod is a uniform rod."},{"Start":"01:32.375 ","End":"01:41.285","Text":"As we know, it\u0027s going to be equal to Lambda multiplied by its small length, dL."},{"Start":"01:41.285 ","End":"01:43.385","Text":"Now what is Lambda?"},{"Start":"01:43.385 ","End":"01:46.760","Text":"It\u0027s like density, per unit length."},{"Start":"01:46.760 ","End":"01:56.105","Text":"We can say we know that our Lambda is equal to mass divided by the length always."},{"Start":"01:56.105 ","End":"01:59.030","Text":"When we\u0027re dealing with questions and mechanics,"},{"Start":"01:59.030 ","End":"02:02.065","Text":"Lambda is going to be mass."},{"Start":"02:02.065 ","End":"02:06.870","Text":"Sorry our mass here is capital M divided by our total length."},{"Start":"02:06.870 ","End":"02:12.180","Text":"Total mass divided by total length."},{"Start":"02:12.180 ","End":"02:19.230","Text":"Now what we want to do is we want to find what our dL is,"},{"Start":"02:19.230 ","End":"02:22.125","Text":"but also relative to the Z axis."},{"Start":"02:22.125 ","End":"02:29.220","Text":"We want to find our projection of every dL on the Z axis."},{"Start":"02:29.220 ","End":"02:36.200","Text":"We\u0027re going to do this exactly how we do this when we\u0027re dealing with our X and Y axes."},{"Start":"02:36.200 ","End":"02:42.530","Text":"The fact that this is the Z axis and this could be the x-axis and this the Y,"},{"Start":"02:42.530 ","End":"02:45.085","Text":"it doesn\u0027t really matter, or vice versa."},{"Start":"02:45.085 ","End":"02:47.565","Text":"We do this the exact same way."},{"Start":"02:47.565 ","End":"02:52.895","Text":"Now, notice that our angle is over here."},{"Start":"02:52.895 ","End":"02:56.180","Text":"Usually, we\u0027re used to seeing the angle over here."},{"Start":"02:56.180 ","End":"02:59.690","Text":"We can see that our dL is going to be"},{"Start":"02:59.690 ","End":"03:07.245","Text":"our hypotenuse and then our adjacent is going to be our Z axis."},{"Start":"03:07.245 ","End":"03:10.100","Text":"If we go back to our SOHCAHTOA,"},{"Start":"03:10.100 ","End":"03:13.835","Text":"we have sine is opposite over hypotenuse,"},{"Start":"03:13.835 ","End":"03:21.315","Text":"cos is adjacent over hypotenuse and tan is opposite over adjacent."},{"Start":"03:21.315 ","End":"03:26.850","Text":"We can see that because we have our Z is our adjacent and our dL is our hypotenuse,"},{"Start":"03:26.850 ","End":"03:34.925","Text":"we\u0027re going to be using cos in order to find the projection of our rod on the Z axis."},{"Start":"03:34.925 ","End":"03:39.120","Text":"We\u0027re going to say that therefore,"},{"Start":"03:39.350 ","End":"03:44.255","Text":"cos of Phi is going to be equal to adjacent,"},{"Start":"03:44.255 ","End":"03:46.610","Text":"which our dZ,"},{"Start":"03:46.610 ","End":"03:48.365","Text":"because we\u0027re dealing with dL,"},{"Start":"03:48.365 ","End":"03:50.750","Text":"divided by our hypotenuse."},{"Start":"03:50.750 ","End":"03:55.050","Text":"Each infinite has more length, dL."},{"Start":"03:55.790 ","End":"04:00.915","Text":"Of course, our Phi is constant, it\u0027s never changing."},{"Start":"04:00.915 ","End":"04:06.679","Text":"Then we can say that our dL therefore,"},{"Start":"04:06.679 ","End":"04:09.260","Text":"because remember we want to plug it into this equation"},{"Start":"04:09.260 ","End":"04:12.485","Text":"so we have Lambda over here and now we\u0027re working our dL."},{"Start":"04:12.485 ","End":"04:20.740","Text":"Our dL is just going to be dZ divided by cosine of Phi."},{"Start":"04:21.170 ","End":"04:24.135","Text":"Now let\u0027s substitute this in."},{"Start":"04:24.135 ","End":"04:26.835","Text":"Our dM is going to be Lambda,"},{"Start":"04:26.835 ","End":"04:31.995","Text":"which is our M divided by l multiplied by our dL,"},{"Start":"04:31.995 ","End":"04:39.390","Text":"which is going to be dZ divided by cosine of Phi."},{"Start":"04:40.340 ","End":"04:45.370","Text":"Now I\u0027m going to find the angular momentum now because the symbol for"},{"Start":"04:45.370 ","End":"04:49.720","Text":"angular momentum is a L and I see that in the question I\u0027ve said"},{"Start":"04:49.720 ","End":"04:54.080","Text":"that the length of the rod is L. I\u0027m just going to change all the capital L\u0027s"},{"Start":"04:54.080 ","End":"05:00.410","Text":"representing the length to a l and then they\u0027ll be less confusion."},{"Start":"05:01.400 ","End":"05:09.145","Text":"Now, all of the L\u0027s representing the length of the rod have been changed to l. Now,"},{"Start":"05:09.145 ","End":"05:13.074","Text":"we can speak about the angular momentum,"},{"Start":"05:13.074 ","End":"05:17.680","Text":"which is L. Because we\u0027re currently dealing"},{"Start":"05:17.680 ","End":"05:22.650","Text":"with a small segment of length dL and a small segment of mass dM."},{"Start":"05:22.650 ","End":"05:28.555","Text":"We\u0027ll speak about the angular momentum also in it\u0027s dL form."},{"Start":"05:28.555 ","End":"05:35.310","Text":"This is the angular momentum of each tiny section of the rod."},{"Start":"05:35.620 ","End":"05:41.270","Text":"As we know, its equation is going to be its position vector"},{"Start":"05:41.270 ","End":"05:46.975","Text":"r cross multiplied with its momentum."},{"Start":"05:46.975 ","End":"05:52.450","Text":"What is its momentum? Here we\u0027re dealing with it\u0027s dp,"},{"Start":"05:52.450 ","End":"05:56.910","Text":"not it\u0027s p, but the momentum of each individual piece."},{"Start":"05:57.220 ","End":"06:00.655","Text":"Let\u0027s see what this equals to."},{"Start":"06:00.655 ","End":"06:04.325","Text":"Our r vector, our position vector."},{"Start":"06:04.325 ","End":"06:07.685","Text":"Now, because we\u0027re dealing with cylindrical coordinates."},{"Start":"06:07.685 ","End":"06:08.900","Text":"Why cylindrical?"},{"Start":"06:08.900 ","End":"06:11.450","Text":"Because it\u0027s going to be easier if we take"},{"Start":"06:11.450 ","End":"06:15.095","Text":"the radius each time from this section over here,"},{"Start":"06:15.095 ","End":"06:18.919","Text":"rather than taking the radius from the origin,"},{"Start":"06:18.919 ","End":"06:23.410","Text":"because then it will complicate this calculation over here."},{"Start":"06:23.410 ","End":"06:27.515","Text":"Always in cylindrical coordinates,"},{"Start":"06:27.515 ","End":"06:37.335","Text":"our position vector is going to be at r in the r direction plus z in the z direction."},{"Start":"06:37.335 ","End":"06:42.095","Text":"This is always in the cylindrical coordinates and the Theta is incorporated into here."},{"Start":"06:42.095 ","End":"06:46.230","Text":"You just don\u0027t include the Theta component."},{"Start":"06:46.790 ","End":"06:53.010","Text":"Next, we\u0027re going to deal with r momentum dp."},{"Start":"06:53.010 ","End":"06:55.230","Text":"What is this going to be equal to?"},{"Start":"06:55.230 ","End":"06:57.930","Text":"Our normal momentum is mass times velocity."},{"Start":"06:57.930 ","End":"07:01.315","Text":"Here, we\u0027re dealing with a small amount of mass so"},{"Start":"07:01.315 ","End":"07:07.050","Text":"dM is our mass multiplied by our velocity vector."},{"Start":"07:07.050 ","End":"07:09.265","Text":"What is our velocity vector?"},{"Start":"07:09.265 ","End":"07:10.690","Text":"Seeing as we\u0027re dealing with"},{"Start":"07:10.690 ","End":"07:15.879","Text":"cylindrical coordinates and we\u0027re dealing with circular motion,"},{"Start":"07:15.879 ","End":"07:21.310","Text":"our velocity is going to be equal to the angular velocity,"},{"Start":"07:21.310 ","End":"07:24.385","Text":"which is Omega multiplied by the radius."},{"Start":"07:24.385 ","End":"07:31.240","Text":"I just want to remind everyone that this is our radius in cylindrical coordinates."},{"Start":"07:31.240 ","End":"07:33.515","Text":"Not from the origin,"},{"Start":"07:33.515 ","End":"07:37.785","Text":"but from the specific points on the Z axis."},{"Start":"07:37.785 ","End":"07:40.450","Text":"Then we\u0027re going to have dM,"},{"Start":"07:40.450 ","End":"07:49.950","Text":"the mass multiplied by Omega r so the length of the radius."},{"Start":"07:49.950 ","End":"07:52.250","Text":"Then because it\u0027s rotating,"},{"Start":"07:52.250 ","End":"07:59.160","Text":"it\u0027s going to be in the Theta direction because there\u0027s circular motion."},{"Start":"07:59.540 ","End":"08:02.210","Text":"Now, as we would have expected,"},{"Start":"08:02.210 ","End":"08:06.350","Text":"we\u0027re going to cross-multiply like so."},{"Start":"08:06.350 ","End":"08:07.865","Text":"Now I want to remind you,"},{"Start":"08:07.865 ","End":"08:10.760","Text":"that there\u0027s an easier way to do it where you just open up"},{"Start":"08:10.760 ","End":"08:14.525","Text":"the multiplication and you just take"},{"Start":"08:14.525 ","End":"08:16.820","Text":"all the constants to 1 side and then you just cross"},{"Start":"08:16.820 ","End":"08:21.815","Text":"multiply r cross multiplied with Theta and Z cross multiplied with Theta."},{"Start":"08:21.815 ","End":"08:25.160","Text":"However, it can sometimes if you get confused,"},{"Start":"08:25.160 ","End":"08:26.450","Text":"lead to calculation error."},{"Start":"08:26.450 ","End":"08:30.290","Text":"I\u0027m going to show you the sure-fire ways to do this."},{"Start":"08:30.290 ","End":"08:35.280","Text":"We\u0027re going to say that our dL is going to be equal to."},{"Start":"08:35.350 ","End":"08:38.585","Text":"So our position vector, r vector,"},{"Start":"08:38.585 ","End":"08:41.600","Text":"which has r in the r direction,"},{"Start":"08:41.600 ","End":"08:43.715","Text":"0 in the Theta direction,"},{"Start":"08:43.715 ","End":"08:46.880","Text":"and Z in the Z direction,"},{"Start":"08:46.880 ","End":"08:51.830","Text":"cross multiplied with our dp,"},{"Start":"08:51.830 ","End":"08:56.830","Text":"which has 0 in the r direction,"},{"Start":"08:56.830 ","End":"09:01.584","Text":"dM multiplied by Omega r in the Theta direction,"},{"Start":"09:01.584 ","End":"09:07.062","Text":"and 0 in the Z direction."},{"Start":"09:07.062 ","End":"09:11.565","Text":"Now, we\u0027re just going to do the cross-multiplication."},{"Start":"09:11.565 ","End":"09:17.700","Text":"First component, we cross off the top line and we have 0 multiplied by 0 which is 0."},{"Start":"09:17.700 ","End":"09:24.135","Text":"Negative z multiplied by dM Omega r. We\u0027re going to have"},{"Start":"09:24.135 ","End":"09:32.235","Text":"negative Omega r z multiplied by dM."},{"Start":"09:32.235 ","End":"09:36.840","Text":"Then we\u0027re going to cross off the middle and then we flip it,"},{"Start":"09:36.840 ","End":"09:42.255","Text":"so z times 0 is 0 and r times 0 is 0."},{"Start":"09:42.255 ","End":"09:50.715","Text":"Then we\u0027re going to cross off the last line and we\u0027re going to have r times dM Omega r,"},{"Start":"09:50.715 ","End":"09:54.630","Text":"so we\u0027re going to have dM,"},{"Start":"09:54.630 ","End":"09:57.885","Text":"and then Omega r^2,"},{"Start":"09:57.885 ","End":"10:04.035","Text":"and then minus z times 0, which is 0."},{"Start":"10:04.035 ","End":"10:07.950","Text":"Then if we want to put it into this format,"},{"Start":"10:07.950 ","End":"10:18.800","Text":"we will have Omega rzdM"},{"Start":"10:18.800 ","End":"10:23.858","Text":"in the negative r direction,"},{"Start":"10:23.858 ","End":"10:33.810","Text":"plus Omega r^2 dM in the z direction."},{"Start":"10:35.300 ","End":"10:40.185","Text":"Now, what we want to do is we want to get rid of this d over here."},{"Start":"10:40.185 ","End":"10:44.505","Text":"We want to find the total angular momentum."},{"Start":"10:44.505 ","End":"10:49.450","Text":"Obviously, what we\u0027re going to have to do is an integral."},{"Start":"10:50.270 ","End":"10:53.685","Text":"Let\u0027s scroll down a little bit."},{"Start":"10:53.685 ","End":"10:57.420","Text":"In order to find our L,"},{"Start":"10:57.420 ","End":"11:02.310","Text":"we\u0027re going to have to integrate on dL,"},{"Start":"11:02.310 ","End":"11:05.280","Text":"which is going to be an integral on,"},{"Start":"11:05.280 ","End":"11:08.350","Text":"let\u0027s do our z component first,"},{"Start":"11:08.350 ","End":"11:16.875","Text":"so on Omega r^2 dM in the z direction."},{"Start":"11:16.875 ","End":"11:26.470","Text":"Then plus an integral on our Omega r z dM."},{"Start":"11:29.060 ","End":"11:33.570","Text":"Then let\u0027s have a negative over here,"},{"Start":"11:33.570 ","End":"11:37.630","Text":"for the negative r direction."},{"Start":"11:39.530 ","End":"11:42.980","Text":"Let\u0027s see. Now,"},{"Start":"11:42.980 ","End":"11:44.715","Text":"if we look back at our dM,"},{"Start":"11:44.715 ","End":"11:47.925","Text":"our dM has a, dZ in it."},{"Start":"11:47.925 ","End":"11:50.430","Text":"We have the total mass divided by"},{"Start":"11:50.430 ","End":"11:57.420","Text":"the total length multiplied by dZ divided by cosine of Phi."},{"Start":"11:57.420 ","End":"11:59.925","Text":"When we have this r over here,"},{"Start":"11:59.925 ","End":"12:04.125","Text":"we don\u0027t really know what this means,"},{"Start":"12:04.125 ","End":"12:08.408","Text":"but we know that our radius is dependent on our angle and on our height,"},{"Start":"12:08.408 ","End":"12:13.125","Text":"on our z value and on our Phi or Phi,"},{"Start":"12:13.125 ","End":"12:15.060","Text":"which is something that we know."},{"Start":"12:15.060 ","End":"12:19.905","Text":"We know what our Phi is. Let\u0027s see what this is."},{"Start":"12:19.905 ","End":"12:22.755","Text":"If we look at our SOHCAHTOA,"},{"Start":"12:22.755 ","End":"12:27.150","Text":"we want to find this over here, the opposite angle."},{"Start":"12:27.150 ","End":"12:34.245","Text":"In order to do that we can do tan is equal to opposite over adjacent."},{"Start":"12:34.245 ","End":"12:40.785","Text":"That means that we have tan of Phi is equal to our opposite,"},{"Start":"12:40.785 ","End":"12:47.530","Text":"which is our r divided by our adjacent, which is our z."},{"Start":"12:47.540 ","End":"12:52.950","Text":"Then that means that if we just get our r out,"},{"Start":"12:52.950 ","End":"12:54.240","Text":"isolate out the r,"},{"Start":"12:54.240 ","End":"13:01.660","Text":"we\u0027ll have that our r is equal to z multiplied by tan of Phi."},{"Start":"13:02.780 ","End":"13:07.815","Text":"Now what we\u0027re going to do is write out our integral,"},{"Start":"13:07.815 ","End":"13:11.265","Text":"but substituting in everywhere we see r with"},{"Start":"13:11.265 ","End":"13:16.065","Text":"z tan Phi and our dM with what we have over here."},{"Start":"13:16.065 ","End":"13:23.745","Text":"Our L is going to be equal to the integral of Omega, then r^2,"},{"Start":"13:23.745 ","End":"13:32.805","Text":"so that\u0027s going to be z^2 tan^2 Phi multiplied by dM."},{"Start":"13:32.805 ","End":"13:36.645","Text":"Then we\u0027re going to have multiplied by M divided by L,"},{"Start":"13:36.645 ","End":"13:44.250","Text":"multiplied by dZ divided by cosine of Phi."},{"Start":"13:44.250 ","End":"13:51.045","Text":"That\u0027s for that and this is going to be in the z direction."},{"Start":"13:51.045 ","End":"13:57.405","Text":"Then we have here minus the integral of Omega."},{"Start":"13:57.405 ","End":"13:58.710","Text":"Again, we have an r,"},{"Start":"13:58.710 ","End":"14:06.270","Text":"so it\u0027s going to be z tan Phi multiplied by z,"},{"Start":"14:06.270 ","End":"14:08.237","Text":"so we\u0027ll have a squared here,"},{"Start":"14:08.237 ","End":"14:10.070","Text":"multiplied by dM,"},{"Start":"14:10.070 ","End":"14:13.505","Text":"which is M divided by L,"},{"Start":"14:13.505 ","End":"14:19.830","Text":"dZ divided by cosine of Phi,"},{"Start":"14:19.830 ","End":"14:25.990","Text":"and then all of these is going to be in the r direction."},{"Start":"14:26.360 ","End":"14:28.847","Text":"Now let\u0027s put in our borders,"},{"Start":"14:28.847 ","End":"14:30.390","Text":"so they\u0027re going to be the same,"},{"Start":"14:30.390 ","End":"14:32.565","Text":"in the z and in the r direction."},{"Start":"14:32.565 ","End":"14:42.930","Text":"It\u0027s going to go from negative L cosine of Phi until L cosine of Phi."},{"Start":"14:42.930 ","End":"14:53.145","Text":"Same over here, negative L cosine of Phi until L cosine of Phi."},{"Start":"14:53.145 ","End":"14:55.410","Text":"What we\u0027re doing, in fact,"},{"Start":"14:55.410 ","End":"15:00.885","Text":"is we\u0027re integrating along the z-axis in both integrals."},{"Start":"15:00.885 ","End":"15:07.720","Text":"Both of our integrals share the same boundaries, so along here."},{"Start":"15:09.320 ","End":"15:14.080","Text":"These are the lengths that we\u0027re integrating along."},{"Start":"15:14.510 ","End":"15:19.185","Text":"From negative L cosine Phi until L cosine of Phi."},{"Start":"15:19.185 ","End":"15:22.233","Text":"Because we\u0027re integrating along dZ,"},{"Start":"15:22.233 ","End":"15:23.640","Text":"everything, our Phi,"},{"Start":"15:23.640 ","End":"15:25.740","Text":"our Omega, our Ms,"},{"Start":"15:25.740 ","End":"15:30.000","Text":"our Ls, everything is going to be constant."},{"Start":"15:30.000 ","End":"15:33.195","Text":"The only thing were just changing is our z."},{"Start":"15:33.195 ","End":"15:36.300","Text":"Now notice that in both integrals,"},{"Start":"15:36.300 ","End":"15:38.550","Text":"we\u0027re integrating by z,"},{"Start":"15:38.550 ","End":"15:42.915","Text":"both of them are z^2 and their bounds are the same."},{"Start":"15:42.915 ","End":"15:46.650","Text":"Now what we can do is we can put all of"},{"Start":"15:46.650 ","End":"15:51.060","Text":"our like terms together and make this integration a lot easier."},{"Start":"15:51.060 ","End":"15:59.354","Text":"Then we can say that this is equal to Omega tan of Phi,"},{"Start":"15:59.354 ","End":"16:04.605","Text":"multiplied by M divided by L cosine of Phi."},{"Start":"16:04.605 ","End":"16:08.820","Text":"I\u0027m literally just rewriting this out like terms,"},{"Start":"16:08.820 ","End":"16:12.120","Text":"so then we\u0027re going to integrate our z^2,"},{"Start":"16:12.120 ","End":"16:13.330","Text":"which are common factors,"},{"Start":"16:13.330 ","End":"16:19.120","Text":"so that\u0027s going to be z^3 divided by 3."},{"Start":"16:19.250 ","End":"16:30.590","Text":"Then our bounds are between negative L cosine Phi until L cosine Phi."},{"Start":"16:30.590 ","End":"16:37.360","Text":"Then multiplied by we have"},{"Start":"16:37.360 ","End":"16:46.340","Text":"another tan Phi in the Z direction and negative r hat."},{"Start":"16:47.490 ","End":"16:55.000","Text":"Now, let\u0027s substitute this in and then simplify this a little bit,"},{"Start":"16:55.000 ","End":"17:00.640","Text":"so we\u0027ll have 2/3 Omega M"},{"Start":"17:00.640 ","End":"17:07.400","Text":"multiplied by sine Phi over cos Phi,"},{"Start":"17:08.250 ","End":"17:17.825","Text":"and then this is going to be multiplied by L^2 cos squared Phi."},{"Start":"17:17.825 ","End":"17:23.740","Text":"Then we\u0027re going to have here tan Phi"},{"Start":"17:23.740 ","End":"17:29.722","Text":"in the Z direction, negative r hat."},{"Start":"17:29.722 ","End":"17:34.375","Text":"Now, we can simplify this a little bit more,"},{"Start":"17:34.375 ","End":"17:38.035","Text":"by crossing out our cosine Phi\u0027s."},{"Start":"17:38.035 ","End":"17:41.970","Text":"Canceling everything out and our tan Phi and everything over here."},{"Start":"17:41.970 ","End":"17:45.135","Text":"So we\u0027re just going to simplify this."},{"Start":"17:45.135 ","End":"17:53.610","Text":"Again, so Omega M L^2 sine 2 Phi."},{"Start":"17:53.610 ","End":"18:01.150","Text":"This comes from the tan and the cosine identity,"},{"Start":"18:01.150 ","End":"18:09.910","Text":"so this is 2 Phi not sine squared Phi sine 2 Phi divided by 3,"},{"Start":"18:09.910 ","End":"18:12.340","Text":"and this is also where this 2,"},{"Start":"18:12.340 ","End":"18:14.965","Text":"into the sine 2 Phi."},{"Start":"18:14.965 ","End":"18:22.360","Text":"Then it\u0027s going to be multiplied by tan Phi in"},{"Start":"18:22.360 ","End":"18:30.685","Text":"the Z direction minus r. This is exactly the same as what I have over here."},{"Start":"18:30.685 ","End":"18:34.040","Text":"We\u0027ve just simplified it a little bit."},{"Start":"18:34.800 ","End":"18:41.065","Text":"Now what I have to do in order to find the torque or the moment of force."},{"Start":"18:41.065 ","End":"18:42.714","Text":"Let\u0027s go back to the question."},{"Start":"18:42.714 ","End":"18:45.640","Text":"What moment of force is acting on the rod?"},{"Start":"18:45.640 ","End":"18:49.255","Text":"What torque? In order to find this out,"},{"Start":"18:49.255 ","End":"18:54.200","Text":"I\u0027m going to have to take the derivative of this expression over here."},{"Start":"18:55.710 ","End":"18:59.080","Text":"Now, because we already said at the beginning,"},{"Start":"18:59.080 ","End":"19:02.710","Text":"I\u0027m going to scroll up again that what\u0027s"},{"Start":"19:02.710 ","End":"19:07.480","Text":"happening here is that everything is changing direction,"},{"Start":"19:07.480 ","End":"19:12.460","Text":"there\u0027s a lack of symmetry over here because we have a horizontal line."},{"Start":"19:12.460 ","End":"19:14.830","Text":"We said that we\u0027re dealing with precession."},{"Start":"19:14.830 ","End":"19:17.005","Text":"What does precession mean?"},{"Start":"19:17.005 ","End":"19:23.755","Text":"It means that our angular momentum is constantly changing with time."},{"Start":"19:23.755 ","End":"19:28.285","Text":"As time goes on, the direction of our angular momentum is changing."},{"Start":"19:28.285 ","End":"19:29.980","Text":"Its size is constant,"},{"Start":"19:29.980 ","End":"19:32.185","Text":"but its direction is changing."},{"Start":"19:32.185 ","End":"19:36.910","Text":"That means that we can find this moment,"},{"Start":"19:36.910 ","End":"19:41.890","Text":"so the sum of the torques or the sum of the moments or moments of force."},{"Start":"19:41.890 ","End":"19:43.885","Text":"They all mean the same thing,"},{"Start":"19:43.885 ","End":"19:48.145","Text":"is equal to the change in our angular momentum,"},{"Start":"19:48.145 ","End":"19:51.730","Text":"dt with regards to time."},{"Start":"19:51.730 ","End":"19:54.970","Text":"Let\u0027s take a look at what this means."},{"Start":"19:54.970 ","End":"19:58.165","Text":"We already spoke about in our previous video,"},{"Start":"19:58.165 ","End":"20:01.810","Text":"in the previous question dealing with precession that"},{"Start":"20:01.810 ","End":"20:06.445","Text":"our time is only when we\u0027re dealing with the radial direction."},{"Start":"20:06.445 ","End":"20:11.530","Text":"That\u0027s what\u0027s changing with time as that direction is constant with time."},{"Start":"20:11.530 ","End":"20:14.305","Text":"When we derive this,"},{"Start":"20:14.305 ","End":"20:17.800","Text":"this will go to 0 because it\u0027s in the Z direction."},{"Start":"20:17.800 ","End":"20:21.790","Text":"Now, we\u0027re just dealing with r hat."},{"Start":"20:21.790 ","End":"20:25.760","Text":"Something that is good to remember,"},{"Start":"20:27.060 ","End":"20:33.970","Text":"and it\u0027s very important is that the derivative of r hat is"},{"Start":"20:33.970 ","End":"20:40.405","Text":"going to be equal to Theta dot in the Theta direction,"},{"Start":"20:40.405 ","End":"20:42.490","Text":"which is Theta dot, as we know,"},{"Start":"20:42.490 ","End":"20:47.690","Text":"that\u0027s Omega, so it\u0027s going to be Omega in the Theta direction."},{"Start":"20:49.440 ","End":"20:51.670","Text":"Let\u0027s derive this,"},{"Start":"20:51.670 ","End":"20:56.375","Text":"so this means that we\u0027re going to be taking Omega."},{"Start":"20:56.375 ","End":"20:58.830","Text":"Remember this term cancels out,"},{"Start":"20:58.830 ","End":"21:01.905","Text":"so we\u0027re just multiplying for this direction,"},{"Start":"21:01.905 ","End":"21:08.005","Text":"so it\u0027s going to be omega ML^2 sine"},{"Start":"21:08.005 ","End":"21:13.135","Text":"of 2Phi divided by 3,"},{"Start":"21:13.135 ","End":"21:18.205","Text":"and then in the negative r hat derivative."},{"Start":"21:18.205 ","End":"21:21.220","Text":"The only thing that we\u0027re deriving here is the r hat,"},{"Start":"21:21.220 ","End":"21:25.900","Text":"because everything else is a constant and this is the only thing dependent on time."},{"Start":"21:25.900 ","End":"21:31.060","Text":"Now we know what our r hat dot is going to be equal to,"},{"Start":"21:31.060 ","End":"21:34.300","Text":"so we\u0027ll just multiply everything already by Omega,"},{"Start":"21:34.300 ","End":"21:41.215","Text":"so we will have Omega ML^2 sine"},{"Start":"21:41.215 ","End":"21:46.335","Text":"2Phi divided by 3,"},{"Start":"21:46.335 ","End":"21:48.690","Text":"and then we multiply by Omega,"},{"Start":"21:48.690 ","End":"21:50.340","Text":"so this will become a squared,"},{"Start":"21:50.340 ","End":"21:52.590","Text":"and you\u0027ll notice that there\u0027s a negative over here,"},{"Start":"21:52.590 ","End":"21:54.180","Text":"so we\u0027ll put that over here."},{"Start":"21:54.180 ","End":"21:58.210","Text":"Then in the Theta direction."},{"Start":"21:59.310 ","End":"22:02.755","Text":"This is our final answer to the question,"},{"Start":"22:02.755 ","End":"22:04.885","Text":"so let\u0027s take a look at what this means."},{"Start":"22:04.885 ","End":"22:13.435","Text":"Remember that it\u0027s some expression in the negative theta direction. What does that mean?"},{"Start":"22:13.435 ","End":"22:15.580","Text":"Let\u0027s take a look at the diagram,"},{"Start":"22:15.580 ","End":"22:17.710","Text":"so we\u0027re dealing with a negative Theta direction."},{"Start":"22:17.710 ","End":"22:20.215","Text":"Now, at this exact point over here,"},{"Start":"22:20.215 ","End":"22:24.085","Text":"so let\u0027s flip this around for a second."},{"Start":"22:24.085 ","End":"22:30.985","Text":"We can say, if this is the y-axis and this is the x-axis for example,"},{"Start":"22:30.985 ","End":"22:33.085","Text":"and at this specific point,"},{"Start":"22:33.085 ","End":"22:39.140","Text":"we\u0027re dealing with our dL in the z, y plane."},{"Start":"22:39.690 ","End":"22:47.980","Text":"Because our angular velocity is going around the z-axis in the anticlockwise direction,"},{"Start":"22:47.980 ","End":"22:53.620","Text":"so we can say that Theta is currently in this direction."},{"Start":"22:53.620 ","End":"22:59.690","Text":"This is our Theta direction going like this."},{"Start":"23:00.360 ","End":"23:06.024","Text":"Now, we can see that because our torque is in the negative Theta direction,"},{"Start":"23:06.024 ","End":"23:15.100","Text":"it\u0027s going to be currently pointing in the x-direction,"},{"Start":"23:15.100 ","End":"23:18.775","Text":"in the opposite direction because it\u0027s negative Theta."},{"Start":"23:18.775 ","End":"23:22.220","Text":"This is the direction that our torque is pointing in."},{"Start":"23:22.590 ","End":"23:26.200","Text":"Now, I\u0027ve scrolled to the side because I want to go over something"},{"Start":"23:26.200 ","End":"23:29.320","Text":"because right now we were dealing with precession,"},{"Start":"23:29.320 ","End":"23:30.790","Text":"and why was a precession?"},{"Start":"23:30.790 ","End":"23:36.145","Text":"Because our angular momentum is constantly changing direction as time goes by."},{"Start":"23:36.145 ","End":"23:41.800","Text":"I want to remind you if we\u0027re dealing with something like this,"},{"Start":"23:41.800 ","End":"23:44.050","Text":"so this is the z-axis,"},{"Start":"23:44.050 ","End":"23:45.580","Text":"this is the y-axis,"},{"Start":"23:45.580 ","End":"23:47.605","Text":"and this is the x-axis."},{"Start":"23:47.605 ","End":"23:52.105","Text":"If we have some kind of shape that\u0027s on the x y plane, only,"},{"Start":"23:52.105 ","End":"23:56.800","Text":"then its angular momentum is always going to be in the z direction,"},{"Start":"23:56.800 ","End":"23:59.155","Text":"solely in the z direction,"},{"Start":"23:59.155 ","End":"24:01.720","Text":"and that is for every single shape,"},{"Start":"24:01.720 ","End":"24:03.835","Text":"however weird it is, it doesn\u0027t matter."},{"Start":"24:03.835 ","End":"24:05.875","Text":"That is in the x, y plane."},{"Start":"24:05.875 ","End":"24:10.750","Text":"However, if we\u0027re dealing with some shape which is not in the x, y plane,"},{"Start":"24:10.750 ","End":"24:12.130","Text":"but has or is,"},{"Start":"24:12.130 ","End":"24:14.500","Text":"but also has some z component,"},{"Start":"24:14.500 ","End":"24:19.390","Text":"such as this shape,"},{"Start":"24:19.390 ","End":"24:22.015","Text":"whatever it might be."},{"Start":"24:22.015 ","End":"24:26.740","Text":"Then we\u0027re dealing with precession because our angular momentum is"},{"Start":"24:26.740 ","End":"24:31.825","Text":"going to have a z component and some other components."},{"Start":"24:31.825 ","End":"24:37.000","Text":"Usually, it will be some component in the r direction."},{"Start":"24:37.000 ","End":"24:40.150","Text":"But not always, so we\u0027re"},{"Start":"24:40.150 ","End":"24:43.465","Text":"dealing with precession when there\u0027s another component aside from the z,"},{"Start":"24:43.465 ","End":"24:45.670","Text":"and it doesn\u0027t cancel out."},{"Start":"24:45.670 ","End":"24:48.070","Text":"When we have something like this,"},{"Start":"24:48.070 ","End":"24:52.940","Text":"then we\u0027re going to be dealing with precession."},{"Start":"24:53.820 ","End":"24:56.770","Text":"This is the end of this lesson."},{"Start":"24:56.770 ","End":"25:01.270","Text":"Now, the next lesson I\u0027m going to briefly explain a little bit"},{"Start":"25:01.270 ","End":"25:06.335","Text":"more about our dL and what it means and how we can find it in general."},{"Start":"25:06.335 ","End":"25:08.925","Text":"Our dL is this over here,"},{"Start":"25:08.925 ","End":"25:12.771","Text":"so a brief explanation on that and how we use it in general."},{"Start":"25:12.771 ","End":"25:15.550","Text":"That\u0027s the end of this video."}],"ID":12713}],"Thumbnail":null,"ID":9397}]

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