1. Basic Equations for Circular Motion
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2. Basic Example for Circular Motion
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3. Centrifugal Force
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4. Exercise Mass on a Rotating Table
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5. Position, Velocity and Acceleration Vectors
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End of Chapter Questions
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End of Chapter Questions (Advanced)
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[{"Name":"1. Basic Equations for Circular Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction to Circular Motion","Duration":"1m ","ChapterTopicVideoID":8967,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.985","Text":"Hello. In this video, I want to talk about circular motion."},{"Start":"00:02.985 ","End":"00:05.280","Text":"We\u0027re going to start with the basics, nothing too complicated."},{"Start":"00:05.280 ","End":"00:08.985","Text":"We\u0027re not going to talk about cylindrical coordinates, for example."},{"Start":"00:08.985 ","End":"00:11.670","Text":"But it\u0027s very important that you listen regardless because this is"},{"Start":"00:11.670 ","End":"00:13.320","Text":"the basic building blocks of what we\u0027re going to"},{"Start":"00:13.320 ","End":"00:15.240","Text":"do and it\u0027s important for the rest of the course."},{"Start":"00:15.240 ","End":"00:17.250","Text":"When we\u0027re talking about circular motion,"},{"Start":"00:17.250 ","End":"00:20.340","Text":"we\u0027re talking about an object that\u0027s moving in an exact circle,"},{"Start":"00:20.340 ","End":"00:24.749","Text":"meaning it has a set radius that doesn\u0027t change, it\u0027s a constant."},{"Start":"00:24.749 ","End":"00:26.700","Text":"If the radius is not constant,"},{"Start":"00:26.700 ","End":"00:28.660","Text":"we\u0027re not talking exactly about circular motion,"},{"Start":"00:28.660 ","End":"00:31.650","Text":"we\u0027re talking about rotational motion or something else like it."},{"Start":"00:31.650 ","End":"00:33.300","Text":"But if our radius is constant,"},{"Start":"00:33.300 ","End":"00:36.300","Text":"we can draw it out and talk about how"},{"Start":"00:36.300 ","End":"00:39.615","Text":"this object is moving with some set formulas we\u0027re going to discuss."},{"Start":"00:39.615 ","End":"00:41.820","Text":"First, let\u0027s give this a radius."},{"Start":"00:41.820 ","End":"00:43.580","Text":"We can draw the line and call it,"},{"Start":"00:43.580 ","End":"00:49.070","Text":"let\u0027s say big R in this case. Here we go."},{"Start":"00:49.070 ","End":"00:53.240","Text":"Now that we have an object moving in an exact circle with a set radius,"},{"Start":"00:53.240 ","End":"00:55.850","Text":"we can talk about a couple of formulas that are"},{"Start":"00:55.850 ","End":"00:59.550","Text":"important to remember when we\u0027re describing this kind of motion."}],"ID":9252},{"Watched":false,"Name":"Linear and Angular Velocity","Duration":"1m 56s","ChapterTopicVideoID":8968,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"The first equation that needs to come to mind when"},{"Start":"00:02.760 ","End":"00:05.325","Text":"you look at circular motion is the following,"},{"Start":"00:05.325 ","End":"00:08.865","Text":"v equals omega r. Now,"},{"Start":"00:08.865 ","End":"00:11.835","Text":"v here equals r linear velocity,"},{"Start":"00:11.835 ","End":"00:18.285","Text":"meaning the velocity at which our object is moving along the circle."},{"Start":"00:18.285 ","End":"00:22.500","Text":"This is going to be a vector and it\u0027s going to be tangential to the circle."},{"Start":"00:22.500 ","End":"00:24.885","Text":"Remember, velocity is always measured in vectors."},{"Start":"00:24.885 ","End":"00:26.970","Text":"Now if we talk about omega for a second,"},{"Start":"00:26.970 ","End":"00:29.475","Text":"omega is the angular velocity."},{"Start":"00:29.475 ","End":"00:34.440","Text":"What does that mean? It means the velocity at which our angle theta moves around."},{"Start":"00:34.440 ","End":"00:36.630","Text":"Meaning how quickly our angle is changing within"},{"Start":"00:36.630 ","End":"00:40.155","Text":"the circle relative to the center of the circle."},{"Start":"00:40.155 ","End":"00:43.970","Text":"1 way to imagine this is imagine that this circle here,"},{"Start":"00:43.970 ","End":"00:47.525","Text":"we\u0027re resting on the x-axis."},{"Start":"00:47.525 ","End":"00:49.400","Text":"Let me draw that out for you. There it is."},{"Start":"00:49.400 ","End":"00:55.235","Text":"The angle theta would be the angle between the line R our radius and the x-axis."},{"Start":"00:55.235 ","End":"00:58.625","Text":"Now, as the object moves,"},{"Start":"00:58.625 ","End":"01:00.305","Text":"the angle theta changes."},{"Start":"01:00.305 ","End":"01:07.190","Text":"So we can think of omega as the change in theta d theta over dt, the change in time."},{"Start":"01:07.190 ","End":"01:09.440","Text":"Or we can write that out as theta dot."},{"Start":"01:09.440 ","End":"01:13.565","Text":"So if we have the angle moving very slowly,"},{"Start":"01:13.565 ","End":"01:16.910","Text":"such as in the following situation where an object moves slowly,"},{"Start":"01:16.910 ","End":"01:20.495","Text":"omega is going to have a small value."},{"Start":"01:20.495 ","End":"01:25.144","Text":"Here\u0027s the example. Of course, in opposition,"},{"Start":"01:25.144 ","End":"01:27.290","Text":"if we have a very quick moving object like that,"},{"Start":"01:27.290 ","End":"01:32.570","Text":"that means that r omega is a higher number and that our theta is changing more quickly."},{"Start":"01:32.570 ","End":"01:36.170","Text":"This should make intuitive sense because the quicker that the angle is changing,"},{"Start":"01:36.170 ","End":"01:38.345","Text":"the quicker that the object should move."},{"Start":"01:38.345 ","End":"01:42.110","Text":"Now, I assume you can see the direct relationship between the 2 terms,"},{"Start":"01:42.110 ","End":"01:44.090","Text":"but of course, that\u0027s only in terms of magnitude."},{"Start":"01:44.090 ","End":"01:45.605","Text":"When we talk about direction,"},{"Start":"01:45.605 ","End":"01:48.110","Text":"for v we\u0027re going to talk about tangents and angles,"},{"Start":"01:48.110 ","End":"01:50.300","Text":"and for omega, we\u0027re going to use the right-hand rule."},{"Start":"01:50.300 ","End":"01:52.890","Text":"But for now, let\u0027s just leave that aside."}],"ID":9253},{"Watched":false,"Name":"Trajectory","Duration":"57s","ChapterTopicVideoID":8969,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.355","Text":"A second formula that\u0027s important to think about is one that deals with position."},{"Start":"00:05.355 ","End":"00:09.720","Text":"Let\u0027s say that we have some sort of object that"},{"Start":"00:09.720 ","End":"00:14.505","Text":"starts along the x-axis there and moves to where our object was before."},{"Start":"00:14.505 ","End":"00:19.995","Text":"Now the trajectory that it is followed is the length of that arc,"},{"Start":"00:19.995 ","End":"00:22.905","Text":"length of the perimeter of that circle basically."},{"Start":"00:22.905 ","End":"00:24.930","Text":"We\u0027ll call that S for now."},{"Start":"00:24.930 ","End":"00:31.020","Text":"Now if we want to think of S and how we can calculate S, let me write that out."},{"Start":"00:31.020 ","End":"00:40.095","Text":"S is going to be equal to the angle of Theta times r, the radius."},{"Start":"00:40.095 ","End":"00:45.080","Text":"Again, we see that there\u0027s a direct relationship here between the circle,"},{"Start":"00:45.080 ","End":"00:48.695","Text":"the angle, and the distance covered."},{"Start":"00:48.695 ","End":"00:52.300","Text":"S is equal to Theta times the radius."},{"Start":"00:52.300 ","End":"00:53.880","Text":"Now you have 2 formulas,"},{"Start":"00:53.880 ","End":"00:57.420","Text":"1 for velocity and 1 for trajectory in position."}],"ID":9254},{"Watched":false,"Name":"Radial Acceleration","Duration":"1m 10s","ChapterTopicVideoID":8970,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8970.jpeg","UploadDate":"2017-03-21T09:11:45.3730000","DurationForVideoObject":"PT1M10S","Description":null,"MetaTitle":"Radial Acceleration: Video + Workbook | Proprep","MetaDescription":"Circular Motion - 1. Basic Equations for Circular Motion. 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/circular-motion/1.-basic-equations-for-circular-motion/vid9255","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"The next equation I want to talk about is one for acceleration."},{"Start":"00:04.080 ","End":"00:06.810","Text":"The moment that you have an object moving around a circle,"},{"Start":"00:06.810 ","End":"00:09.600","Text":"it\u0027s evident that the direction of it\u0027s"},{"Start":"00:09.600 ","End":"00:13.020","Text":"velocity is going to change throughout it\u0027s motion."},{"Start":"00:13.020 ","End":"00:17.730","Text":"Even if the magnitude of velocity stays the same at this point,"},{"Start":"00:17.730 ","End":"00:19.185","Text":"at that point, at any given point,"},{"Start":"00:19.185 ","End":"00:22.725","Text":"it\u0027s going to have a different direction for its velocity vector."},{"Start":"00:22.725 ","End":"00:25.020","Text":"For example, at the point that we\u0027ve drawn on,"},{"Start":"00:25.020 ","End":"00:29.505","Text":"it\u0027s pointing more or less up in a little bit to the left."},{"Start":"00:29.505 ","End":"00:32.400","Text":"But once you go around the circle is going to point in the exact opposite direction."},{"Start":"00:32.400 ","End":"00:34.290","Text":"We have this equation here."},{"Start":"00:34.290 ","End":"00:40.165","Text":"a_ r is equal to v^2 over R equals Omega^2R."},{"Start":"00:40.165 ","End":"00:44.325","Text":"a_ r symbolizes the radial acceleration"},{"Start":"00:44.325 ","End":"00:46.520","Text":"and the direction is towards the center of the circle,"},{"Start":"00:46.520 ","End":"00:48.095","Text":"we\u0027ll write it here in green."},{"Start":"00:48.095 ","End":"00:49.985","Text":"When we\u0027re talking about the magnitude of it,"},{"Start":"00:49.985 ","End":"00:51.890","Text":"we know that the direction is towards the middle of the circle."},{"Start":"00:51.890 ","End":"00:54.350","Text":"The magnitude is v^2 over R, and v,"},{"Start":"00:54.350 ","End":"00:58.130","Text":"we can sometimes replace with Omega and we get"},{"Start":"00:58.130 ","End":"01:02.180","Text":"Omega^2 times R. These are 4 very important equations,"},{"Start":"01:02.180 ","End":"01:04.790","Text":"particularly the first one and the last one, and we\u0027ll use them a lot."},{"Start":"01:04.790 ","End":"01:06.665","Text":"You should remember them and remember that they are"},{"Start":"01:06.665 ","End":"01:09.840","Text":"always correct regardless of the scenario."}],"ID":9255},{"Watched":false,"Name":"Uniform Circular Motion, Frequency and Period Time","Duration":"1m 25s","ChapterTopicVideoID":8971,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.300","Text":"Now I want to split my circular motion into 2 types of circular motion."},{"Start":"00:04.300 ","End":"00:06.685","Text":"First, we have Uniform Circular Motion."},{"Start":"00:06.685 ","End":"00:08.440","Text":"That\u0027s when the velocity is constant."},{"Start":"00:08.440 ","End":"00:12.610","Text":"The magnitude of the velocity is the same throughout the trajectory of the object."},{"Start":"00:12.610 ","End":"00:16.465","Text":"In opposition to that, we have something called Non-Uniform Circular Motion,"},{"Start":"00:16.465 ","End":"00:17.875","Text":"which is the opposite."},{"Start":"00:17.875 ","End":"00:20.770","Text":"What does that mean? It means that the magnitude of"},{"Start":"00:20.770 ","End":"00:24.185","Text":"velocity changes throughout the trajectory of the object."},{"Start":"00:24.185 ","End":"00:26.560","Text":"If we\u0027re dealing with uniform circular motion,"},{"Start":"00:26.560 ","End":"00:30.475","Text":"we can define something called T. T is our period time."},{"Start":"00:30.475 ","End":"00:35.815","Text":"That\u0027s the amount of time it takes for our object to fulfill 1 cycle around the circle."},{"Start":"00:35.815 ","End":"00:37.240","Text":"Now, if we have T,"},{"Start":"00:37.240 ","End":"00:39.010","Text":"we can also define that in a number of ways."},{"Start":"00:39.010 ","End":"00:42.745","Text":"First, we can do 1 over F. F is our frequency."},{"Start":"00:42.745 ","End":"00:47.525","Text":"We can also define it as 2 Pi over omega."},{"Start":"00:47.525 ","End":"00:49.220","Text":"Omega is if you recall,"},{"Start":"00:49.220 ","End":"00:51.725","Text":"our angular velocity that we talked about before."},{"Start":"00:51.725 ","End":"00:54.850","Text":"Now, these 2 are inverse in terms of the relationship T and Omega."},{"Start":"00:54.850 ","End":"00:59.015","Text":"The larger omega is the shorter time we have for T, and vice versa."},{"Start":"00:59.015 ","End":"01:02.750","Text":"F or frequency is the amount of"},{"Start":"01:02.750 ","End":"01:06.665","Text":"rotations that our object performs in a given unit of time."},{"Start":"01:06.665 ","End":"01:11.650","Text":"For example, we could say 10 Hertz is 10 rotations in 1 second."},{"Start":"01:11.650 ","End":"01:14.390","Text":"This formula should also go into your formula sheet."},{"Start":"01:14.390 ","End":"01:16.970","Text":"Now, remember this only works for uniform circular motion."},{"Start":"01:16.970 ","End":"01:18.620","Text":"If we have non-uniform circular motion,"},{"Start":"01:18.620 ","End":"01:20.960","Text":"we can\u0027t talk about a period because nothing is constant,"},{"Start":"01:20.960 ","End":"01:25.650","Text":"nothing is uniform, and things will be changing between each cycle of rotation."}],"ID":9256},{"Watched":false,"Name":"Non-Uniform Circular Motion and Tangential Acceleration","Duration":"1m 32s","ChapterTopicVideoID":8972,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.035","Text":"Moving on, we\u0027re going to talk about non-uniform circular motion."},{"Start":"00:04.035 ","End":"00:06.540","Text":"When you have non-uniform circular motion,"},{"Start":"00:06.540 ","End":"00:08.940","Text":"you gain a second acceleration."},{"Start":"00:08.940 ","End":"00:12.255","Text":"This acceleration is your tangential acceleration"},{"Start":"00:12.255 ","End":"00:16.600","Text":"and we\u0027re going to mark it here with the following formula."},{"Start":"00:16.640 ","End":"00:23.570","Text":"Your tangential acceleration, the symbol is a sub Theta and it\u0027s equal to dv over dt."},{"Start":"00:23.570 ","End":"00:26.585","Text":"Now, remember this is only the magnitude that we\u0027re dealing with,"},{"Start":"00:26.585 ","End":"00:28.940","Text":"so it doesn\u0027t change the direction only the magnitude."},{"Start":"00:28.940 ","End":"00:32.990","Text":"You should remember that. Now we have this because your velocity is changing."},{"Start":"00:32.990 ","End":"00:35.060","Text":"If your velocity was uniform,"},{"Start":"00:35.060 ","End":"00:37.010","Text":"you wouldn\u0027t have a tangential acceleration."},{"Start":"00:37.010 ","End":"00:41.610","Text":"It\u0027s also equal to Alpha R. Alpha R is"},{"Start":"00:41.610 ","End":"00:48.300","Text":"your angular acceleration and Alpha R is equal to your Omega dot."},{"Start":"00:48.300 ","End":"00:52.565","Text":"Omega dot being the derivative of Omega"},{"Start":"00:52.565 ","End":"00:55.860","Text":"and we\u0027re not going to talk about it too much right now"},{"Start":"00:55.860 ","End":"00:59.380","Text":"but it\u0027s basically the change in Omega over the change in time."},{"Start":"00:59.380 ","End":"01:01.380","Text":"Now you have these 6 formulas to use,"},{"Start":"01:01.380 ","End":"01:02.880","Text":"remember v equals Omega R,"},{"Start":"01:02.880 ","End":"01:04.625","Text":"and a_r are the most important."},{"Start":"01:04.625 ","End":"01:07.520","Text":"Uniform circular motion only applies when your velocity"},{"Start":"01:07.520 ","End":"01:10.760","Text":"is constant and non-uniform circular motion."},{"Start":"01:10.760 ","End":"01:14.570","Text":"Your tangential acceleration only applies if your velocity is changing."},{"Start":"01:14.570 ","End":"01:16.945","Text":"Now, remember this is only the magnitude of it,"},{"Start":"01:16.945 ","End":"01:20.090","Text":"for the direction you need to look at the arrow of"},{"Start":"01:20.090 ","End":"01:23.525","Text":"the vector v and that\u0027s going to be the direction of your tangential acceleration."},{"Start":"01:23.525 ","End":"01:25.580","Text":"It could, in theory, be negative v"},{"Start":"01:25.580 ","End":"01:28.070","Text":"but essentially we can just use v and have the same result."},{"Start":"01:28.070 ","End":"01:32.640","Text":"Now you have your 2 acceleration and the equations for them."}],"ID":10776},{"Watched":false,"Name":"How to Approach the Exercises","Duration":"2m 11s","ChapterTopicVideoID":8973,"CourseChapterTopicPlaylistID":85360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.280 ","End":"00:03.780","Text":"These are the formulas that you need to"},{"Start":"00:03.780 ","End":"00:07.170","Text":"remember when we\u0027re talking about circular motion."},{"Start":"00:07.170 ","End":"00:10.500","Text":"What I want to do now is talk with you about how you can"},{"Start":"00:10.500 ","End":"00:13.890","Text":"approach the exercises about circular motion."},{"Start":"00:13.890 ","End":"00:15.510","Text":"It\u0027s nothing too complicated,"},{"Start":"00:15.510 ","End":"00:17.205","Text":"it\u0027s rather simple and short."},{"Start":"00:17.205 ","End":"00:20.400","Text":"What we\u0027re going to do is really with anything having to do with motion,"},{"Start":"00:20.400 ","End":"00:21.750","Text":"we can refer to Newton\u0027s laws,"},{"Start":"00:21.750 ","End":"00:27.265","Text":"but certainly circular motion is going to help us to refer to Newton\u0027s laws."},{"Start":"00:27.265 ","End":"00:31.365","Text":"For circular motion, we\u0027re going to refer to Newton\u0027s second law of motion."},{"Start":"00:31.365 ","End":"00:35.760","Text":"To remind you, Newton\u0027s second law is the sum of the forces"},{"Start":"00:35.760 ","End":"00:39.860","Text":"Sigma F equals the mass times the acceleration vector."},{"Start":"00:39.860 ","End":"00:43.910","Text":"Now remember these are both vectors and the vector is going to be in the same direction."},{"Start":"00:43.910 ","End":"00:49.130","Text":"Now, we can also do once we have that is talk about the radial force."},{"Start":"00:49.130 ","End":"00:51.250","Text":"The force going in towards the center of the circle,"},{"Start":"00:51.250 ","End":"00:56.710","Text":"so the sum of the radial force is also equal to the mass times the radial acceleration."},{"Start":"00:56.710 ","End":"00:58.460","Text":"You have to remember that this is"},{"Start":"00:58.460 ","End":"01:00.608","Text":"the direction going in towards the center of the circle."},{"Start":"01:00.608 ","End":"01:05.660","Text":"That\u0027s going to be your positive vector and you calculate that by summing up"},{"Start":"01:05.660 ","End":"01:08.000","Text":"all the forces pushing in towards the center of"},{"Start":"01:08.000 ","End":"01:12.130","Text":"the circle and that\u0027s going to equal your radial acceleration."},{"Start":"01:12.130 ","End":"01:13.790","Text":"Just another reminder that"},{"Start":"01:13.790 ","End":"01:17.450","Text":"the positive force is going inwards towards the center of the circle."},{"Start":"01:17.450 ","End":"01:19.915","Text":"Now you can do the same thing for Theta."},{"Start":"01:19.915 ","End":"01:23.835","Text":"You can write out Sigma F Theta equals ma Theta."},{"Start":"01:23.835 ","End":"01:27.880","Text":"Now remember, a Theta is your tangential acceleration,"},{"Start":"01:27.880 ","End":"01:32.840","Text":"so only if you have a changing velocity can you use that."},{"Start":"01:32.840 ","End":"01:35.795","Text":"If you don\u0027t have one, it\u0027ll be equal to 0 most likely,"},{"Start":"01:35.795 ","End":"01:41.030","Text":"so it\u0027s not going to be quite as important as Sigma F_r=ma_r,"},{"Start":"01:41.030 ","End":"01:45.955","Text":"but you\u0027re still going to use Sigma F Theta=ma Theta for certain problems."},{"Start":"01:45.955 ","End":"01:52.520","Text":"Of course, if you\u0027re using the ma_r=Sigma F_r of equation,"},{"Start":"01:52.520 ","End":"01:55.445","Text":"you can use your equation down at the bottom here for"},{"Start":"01:55.445 ","End":"02:01.310","Text":"a_r and if you can find the values of the forces on the radial axis,"},{"Start":"02:01.310 ","End":"02:04.180","Text":"you can usually solve the exercise pretty quickly."},{"Start":"02:04.180 ","End":"02:07.190","Text":"That\u0027s the end of the lecture series on this topic."},{"Start":"02:07.190 ","End":"02:11.430","Text":"I\u0027d suggest that you go on to the exercises now. Thanks for listening."}],"ID":9258}],"Thumbnail":null,"ID":85360},{"Name":"2. Basic Example for Circular Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Basic Exercises About Circular Motion","Duration":"4m 59s","ChapterTopicVideoID":9002,"CourseChapterTopicPlaylistID":85361,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this exercise,"},{"Start":"00:01.980 ","End":"00:07.245","Text":"we\u0027re given a pendulum which is on a rod of a given length l and at an angle Theta,"},{"Start":"00:07.245 ","End":"00:10.540","Text":"which is relative to a perpendicular line to the ceiling."},{"Start":"00:10.540 ","End":"00:12.705","Text":"It\u0027s relative to a straight vertical line,"},{"Start":"00:12.705 ","End":"00:17.775","Text":"the angle Theta describes how far the pendulum is from that vertical line."},{"Start":"00:17.775 ","End":"00:21.140","Text":"We know that this pendulum is staying parallel to the ceiling,"},{"Start":"00:21.140 ","End":"00:23.670","Text":"it\u0027s on the same height level compared to the ground into"},{"Start":"00:23.670 ","End":"00:26.925","Text":"the ceiling and it\u0027s going around this vertical axis,"},{"Start":"00:26.925 ","End":"00:31.860","Text":"we can describe its motion like so and this problem is asking"},{"Start":"00:31.860 ","End":"00:33.960","Text":"us to find what happens to that mass"},{"Start":"00:33.960 ","End":"00:37.410","Text":"in terms of the frequency and in terms of the period time."},{"Start":"00:37.410 ","End":"00:40.455","Text":"What we want to do is for the frequency,"},{"Start":"00:40.455 ","End":"00:44.240","Text":"we\u0027re going to use the symbol f and for the period of time,"},{"Start":"00:44.240 ","End":"00:48.020","Text":"we\u0027re going to use the symbol T and if you recall,"},{"Start":"00:48.020 ","End":"00:52.445","Text":"there is a relationship between these 2 things that we need to find."},{"Start":"00:52.445 ","End":"00:56.830","Text":"If you recall, T equals 1 over f."},{"Start":"00:56.830 ","End":"01:02.675","Text":"The time period is 1 over the frequency and that also equals 2 Pi over Omega,"},{"Start":"01:02.675 ","End":"01:05.675","Text":"that means that if we find 1 of these 3 variables,"},{"Start":"01:05.675 ","End":"01:08.230","Text":"we can then find the other 2 very quickly."},{"Start":"01:08.230 ","End":"01:10.550","Text":"In this case, it\u0027s probably easiest for us to"},{"Start":"01:10.550 ","End":"01:13.010","Text":"find Omega and I\u0027ll show you how to do that."},{"Start":"01:13.010 ","End":"01:19.160","Text":"If you recall, we talked about this in the video lecture about approaching the questions."},{"Start":"01:19.160 ","End":"01:21.440","Text":"The best way to approach many of those questions,"},{"Start":"01:21.440 ","End":"01:23.225","Text":"and you\u0027ll see it in many of the video lectures,"},{"Start":"01:23.225 ","End":"01:24.935","Text":"is to start with simple things."},{"Start":"01:24.935 ","End":"01:28.085","Text":"For now, the simplest thing is to start with forces."},{"Start":"01:28.085 ","End":"01:33.050","Text":"We\u0027re going to talk about the force that\u0027s exercise an object from given points or"},{"Start":"01:33.050 ","End":"01:35.360","Text":"towards given points and that\u0027ll help us"},{"Start":"01:35.360 ","End":"01:39.115","Text":"find the variables we need to solve our exercises."},{"Start":"01:39.115 ","End":"01:43.040","Text":"In this problem we have 2 forces that we can talk about."},{"Start":"01:43.040 ","End":"01:48.160","Text":"The first is mg, which of course will be pointing down towards the ground,"},{"Start":"01:48.160 ","End":"01:51.840","Text":"in a straight line like that and the second is mT."},{"Start":"01:51.840 ","End":"01:54.995","Text":"mT Is going to start with the object itself and point"},{"Start":"01:54.995 ","End":"01:59.455","Text":"up along the rod towards the ceiling, something like that."},{"Start":"01:59.455 ","End":"02:04.040","Text":"Now I need to plan out my axes on x axis and y axis and I\u0027m going to have"},{"Start":"02:04.040 ","End":"02:08.565","Text":"the x axis parallel to the ceiling and the y axis perpendicular to the ceiling, why?"},{"Start":"02:08.565 ","End":"02:13.415","Text":"Because I want at least 1 of my axes where I know I\u0027m going to have my acceleration."},{"Start":"02:13.415 ","End":"02:16.310","Text":"I know because the nature of the spinning that some of"},{"Start":"02:16.310 ","End":"02:19.235","Text":"the acceleration is going to be going toward the center on,"},{"Start":"02:19.235 ","End":"02:20.385","Text":"along the x axis."},{"Start":"02:20.385 ","End":"02:23.315","Text":"The y axis, I don\u0027t think he\u0027s going to have any force on it,"},{"Start":"02:23.315 ","End":"02:25.610","Text":"but the x axis certainly will."},{"Start":"02:25.610 ","End":"02:31.290","Text":"Now I can actually calculate my forces for the x and the y axis,"},{"Start":"02:31.540 ","End":"02:34.940","Text":"if you look at angle Theta,"},{"Start":"02:34.940 ","End":"02:36.545","Text":"which is listed there,"},{"Start":"02:36.545 ","End":"02:37.595","Text":"if that\u0027s angle Theta,"},{"Start":"02:37.595 ","End":"02:42.320","Text":"then this is also Theta and we can then calculate"},{"Start":"02:42.320 ","End":"02:47.870","Text":"for the force of x that it is T sine Theta,"},{"Start":"02:47.870 ","End":"02:50.720","Text":"which is equal to mar."},{"Start":"02:50.720 ","End":"02:52.775","Text":"Your radial acceleration, which is the same as"},{"Start":"02:52.775 ","End":"02:57.635","Text":"the acceleration along the x axis and that a_r,"},{"Start":"02:57.635 ","End":"02:58.790","Text":"if we want to solve for Omega,"},{"Start":"02:58.790 ","End":"03:00.925","Text":"we can plug in Omega here,"},{"Start":"03:00.925 ","End":"03:09.830","Text":"a_r is equal to Omega squared times the R. We now know that and for the sum of forces y,"},{"Start":"03:09.830 ","End":"03:16.000","Text":"it\u0027s T cosine Theta minus mg and that equals 0."},{"Start":"03:16.000 ","End":"03:19.700","Text":"At this point I\u0027ve pretty much solved our problem because we"},{"Start":"03:19.700 ","End":"03:22.520","Text":"have 2 equations with 2 unknown variables,"},{"Start":"03:22.520 ","End":"03:27.620","Text":"we have T and we have Omega and you can actually solve pretty quickly for Omega."},{"Start":"03:27.620 ","End":"03:31.790","Text":"The only obstacle here is we don\u0027t yet know the value of R but if you think about it,"},{"Start":"03:31.790 ","End":"03:37.021","Text":"R is the radius from this imaginary vertical line towards our pendulum itself."},{"Start":"03:37.021 ","End":"03:38.210","Text":"Once we solve for that,"},{"Start":"03:38.210 ","End":"03:40.835","Text":"we can solve for T and F as well."},{"Start":"03:40.835 ","End":"03:43.685","Text":"Just to solve R real quick,"},{"Start":"03:43.685 ","End":"03:47.240","Text":"what you can do is if you realize that again,"},{"Start":"03:47.240 ","End":"03:49.175","Text":"R is along the x axis there,"},{"Start":"03:49.175 ","End":"03:52.445","Text":"the line between the vertical pole and the object R equals"},{"Start":"03:52.445 ","End":"03:57.320","Text":"l sine Theta and so if we solve for R,"},{"Start":"03:57.320 ","End":"04:02.465","Text":"we can then solve for Omega quickly and the solution ends up being that Omega"},{"Start":"04:02.465 ","End":"04:10.325","Text":"equals the square root of g over l cosine Theta."},{"Start":"04:10.325 ","End":"04:14.890","Text":"Once we solve for that, we can then get our T and our f as well."},{"Start":"04:14.890 ","End":"04:19.475","Text":"I\u0027m actually going to skip the last step here of solving it, but if you recall,"},{"Start":"04:19.475 ","End":"04:26.850","Text":"f equals Omega over 2 Pi and T equals 2 Pi over Omega."},{"Start":"04:26.850 ","End":"04:29.120","Text":"You can solve that yourself."},{"Start":"04:29.120 ","End":"04:31.955","Text":"I also skip the algebra above."},{"Start":"04:31.955 ","End":"04:34.080","Text":"If you basically isolate T,"},{"Start":"04:34.080 ","End":"04:38.090","Text":"you can then solve for T and you have the whole equation solved at that point."},{"Start":"04:38.090 ","End":"04:39.800","Text":"But the point here,"},{"Start":"04:39.800 ","End":"04:42.095","Text":"what I wanted to focus on this video is the process."},{"Start":"04:42.095 ","End":"04:44.750","Text":"I started with my force equations by"},{"Start":"04:44.750 ","End":"04:47.345","Text":"putting in my x and y axes and finding where the forces are,"},{"Start":"04:47.345 ","End":"04:48.815","Text":"knowing that this is circular motion."},{"Start":"04:48.815 ","End":"04:51.170","Text":"I use my equations for that."},{"Start":"04:51.170 ","End":"04:54.800","Text":"Solve for a_r to find Omega and 1 side Omega found f and"},{"Start":"04:54.800 ","End":"04:59.700","Text":"T. That\u0027s the gist of the problem here. Thanks for listening."}],"ID":9268}],"Thumbnail":null,"ID":85361},{"Name":"3. Centrifugal Force","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Centrifugal Force","Duration":"4m 24s","ChapterTopicVideoID":9004,"CourseChapterTopicPlaylistID":85362,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.315","Text":"Hello, I want to talk for a minute about centrifugal force."},{"Start":"00:03.315 ","End":"00:05.040","Text":"Now when we talk about centrifugal force,"},{"Start":"00:05.040 ","End":"00:08.430","Text":"what it really is is the tool for us to talk about certain equations in"},{"Start":"00:08.430 ","End":"00:13.230","Text":"circular motion differently and to help us solve certain circular motion problems."},{"Start":"00:13.230 ","End":"00:16.710","Text":"The best way to start is with a little recap of what we\u0027ve done before."},{"Start":"00:16.710 ","End":"00:20.880","Text":"If you recall in our first lecture on circular motion, we talked about a_r."},{"Start":"00:20.880 ","End":"00:23.880","Text":"Every object moving in circular motion has a_r,"},{"Start":"00:23.880 ","End":"00:27.255","Text":"which equals v^2 over r,"},{"Start":"00:27.255 ","End":"00:30.075","Text":"or Omega^2 times r. That"},{"Start":"00:30.075 ","End":"00:33.890","Text":"a radial acceleration can also be talked about in terms of radial force,"},{"Start":"00:33.890 ","End":"00:36.650","Text":"the force going towards the middle of a circle."},{"Start":"00:36.650 ","End":"00:41.020","Text":"The radial force equals m times a_r."},{"Start":"00:41.020 ","End":"00:45.515","Text":"Now if this was a stone let\u0027s say on a rope and we\u0027re swinging around in a circle,"},{"Start":"00:45.515 ","End":"00:49.225","Text":"we could call that force rt."},{"Start":"00:49.225 ","End":"00:54.395","Text":"The way we use this is we do a summation of our force,"},{"Start":"00:54.395 ","End":"00:56.930","Text":"we talked about the forces going toward the center of"},{"Start":"00:56.930 ","End":"00:59.555","Text":"the radial force and set that equal to ma_r."},{"Start":"00:59.555 ","End":"01:02.225","Text":"That\u0027s the first way that we can solve these problems."},{"Start":"01:02.225 ","End":"01:04.355","Text":"Is finding our force,"},{"Start":"01:04.355 ","End":"01:06.410","Text":"that is the radial force,"},{"Start":"01:06.410 ","End":"01:08.665","Text":"and setting that equal to ma_r."},{"Start":"01:08.665 ","End":"01:12.980","Text":"With centrifugal force, what we can do is talk about a second way to do this."},{"Start":"01:12.980 ","End":"01:14.720","Text":"We add a second force,"},{"Start":"01:14.720 ","End":"01:18.080","Text":"a force that\u0027s not demonstrated on the object by any given body"},{"Start":"01:18.080 ","End":"01:22.660","Text":"but 1 that we can use nonetheless to help us solve this problem."},{"Start":"01:22.660 ","End":"01:24.290","Text":"If we have t, again,"},{"Start":"01:24.290 ","End":"01:26.720","Text":"in the example of a stone on a rope."},{"Start":"01:26.720 ","End":"01:28.700","Text":"T is a force that\u0027s going inwards,"},{"Start":"01:28.700 ","End":"01:31.685","Text":"we have an equal opposite force going outwards."},{"Start":"01:31.685 ","End":"01:35.930","Text":"That force is going to be m times Omega^2 times"},{"Start":"01:35.930 ","End":"01:42.050","Text":"r. This again is not something that a force that someone is exercising,"},{"Start":"01:42.050 ","End":"01:44.510","Text":"but we\u0027re going to write it out nonetheless."},{"Start":"01:44.510 ","End":"01:46.789","Text":"When we write that into our equation,"},{"Start":"01:46.789 ","End":"01:52.665","Text":"we\u0027re going to do the sum of the radial force."},{"Start":"01:52.665 ","End":"01:57.635","Text":"We\u0027re going to set that equal to t minus"},{"Start":"01:57.635 ","End":"02:00.560","Text":"m Omega^2 R because remember t is"},{"Start":"02:00.560 ","End":"02:04.310","Text":"the positive force so m Omega r must be the negative force."},{"Start":"02:04.310 ","End":"02:07.405","Text":"I\u0027m going to say that equal to 0."},{"Start":"02:07.405 ","End":"02:13.220","Text":"Now, this does make mathematical sense because if a_r equals Omega^2 R,"},{"Start":"02:13.220 ","End":"02:16.100","Text":"then t which if you recall, is ma_r,"},{"Start":"02:16.100 ","End":"02:21.070","Text":"should equal m Omega^2 R. If we subtract the 2 then that should equal 0."},{"Start":"02:21.070 ","End":"02:23.090","Text":"Of course, through both ways,"},{"Start":"02:23.090 ","End":"02:25.790","Text":"whether it\u0027s just finding t or ma_r,"},{"Start":"02:25.790 ","End":"02:27.620","Text":"or finding an equation equal to 0,"},{"Start":"02:27.620 ","End":"02:28.910","Text":"we do come up with the same answer."},{"Start":"02:28.910 ","End":"02:31.115","Text":"There\u0027s only one answer for any problem in physics,"},{"Start":"02:31.115 ","End":"02:34.145","Text":"what this does is it changes our perspective on the problem."},{"Start":"02:34.145 ","End":"02:36.460","Text":"Instead of looking at it from our current perspective,"},{"Start":"02:36.460 ","End":"02:43.475","Text":"if we\u0027re using the example of a stone on a rope,"},{"Start":"02:43.475 ","End":"02:47.300","Text":"we are the boy on the inside of the circle or a girl,"},{"Start":"02:47.300 ","End":"02:48.530","Text":"let\u0027s not be gender biased."},{"Start":"02:48.530 ","End":"02:52.520","Text":"But the person on the inside of the circle throwing the stone,"},{"Start":"02:52.520 ","End":"02:54.115","Text":"holding the rope, and spinning it."},{"Start":"02:54.115 ","End":"02:59.015","Text":"Now, if you\u0027re spinning with the rope you are moving at the angular velocity of Omega."},{"Start":"02:59.015 ","End":"03:00.740","Text":"From the perspective of the child,"},{"Start":"03:00.740 ","End":"03:03.785","Text":"the stone isn\u0027t moving in relation to the child."},{"Start":"03:03.785 ","End":"03:06.590","Text":"What we\u0027ve done here is we\u0027ve added a simulated force."},{"Start":"03:06.590 ","End":"03:10.130","Text":"The stimulated force in this case is m Omega^2 R,"},{"Start":"03:10.130 ","End":"03:14.615","Text":"so that from the child\u0027s perspective nothing is changing we\u0027re in a set system."},{"Start":"03:14.615 ","End":"03:20.405","Text":"What we\u0027re really doing is taking our real force t and subtracting from a simulated force"},{"Start":"03:20.405 ","End":"03:23.600","Text":"to get us to the answer of 0 because it\u0027s the closed system where"},{"Start":"03:23.600 ","End":"03:27.130","Text":"the child seems to see the rock not moving."},{"Start":"03:27.130 ","End":"03:31.400","Text":"This force, the stimulated force that allows us to set the equation equal to 0,"},{"Start":"03:31.400 ","End":"03:33.170","Text":"to look from the child\u0027s perspective,"},{"Start":"03:33.170 ","End":"03:35.870","Text":"is what we\u0027re going to call the centrifugal force."},{"Start":"03:35.870 ","End":"03:41.855","Text":"What\u0027s important to remember is that the centrifugal force is simulated for us."},{"Start":"03:41.855 ","End":"03:43.865","Text":"It\u0027s not a real force,"},{"Start":"03:43.865 ","End":"03:47.705","Text":"but it allows us to look at the problem from the perspective of,"},{"Start":"03:47.705 ","End":"03:51.535","Text":"in our example, the child holding a stone on a rope."},{"Start":"03:51.535 ","End":"03:54.680","Text":"If you don\u0027t understand it now don\u0027t worry we\u0027re going to talk about it in"},{"Start":"03:54.680 ","End":"03:57.800","Text":"much more detail when we talk about relative forces later."},{"Start":"03:57.800 ","End":"04:00.920","Text":"But what\u0027s important for you right now is to understand that this is"},{"Start":"04:00.920 ","End":"04:04.490","Text":"a simulated force and that allows us 2 ways to solve these problems."},{"Start":"04:04.490 ","End":"04:08.480","Text":"One is to find the radial force and set that equal to ma_r,"},{"Start":"04:08.480 ","End":"04:15.185","Text":"and the other is to find that force as well as m Omega^2 R and set that equal to 0."},{"Start":"04:15.185 ","End":"04:17.780","Text":"That it\u0027s useful for us, it\u0027ll help us solve"},{"Start":"04:17.780 ","End":"04:21.630","Text":"certain problems while changing the perspective with which we look at it."}],"ID":9269},{"Watched":false,"Name":"The Centripetal Force","Duration":"58s","ChapterTopicVideoID":9005,"CourseChapterTopicPlaylistID":85362,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"That was centrifugal force."},{"Start":"00:01.935 ","End":"00:04.314","Text":"Remember centrifugal force is assimilated force."},{"Start":"00:04.314 ","End":"00:06.915","Text":"It\u0027s not real and it\u0027s acting towards the outside"},{"Start":"00:06.915 ","End":"00:10.590","Text":"m Omega squared r. A new term for you is this centripetal force."},{"Start":"00:10.590 ","End":"00:13.550","Text":"Now remember this is not the same thing as centrifugal force,"},{"Start":"00:13.550 ","End":"00:14.685","Text":"in fact it\u0027s the opposite,"},{"Start":"00:14.685 ","End":"00:17.040","Text":"even though they\u0027re spelled similarly and look similar,"},{"Start":"00:17.040 ","End":"00:22.140","Text":"Centripetal force is a name for the real force acting towards the center of the circle."},{"Start":"00:22.140 ","End":"00:24.180","Text":"This is what makes something act in an arc."},{"Start":"00:24.180 ","End":"00:27.345","Text":"So if we\u0027re looking at our example from before,"},{"Start":"00:27.345 ","End":"00:29.280","Text":"the centripetal force would be T,"},{"Start":"00:29.280 ","End":"00:33.405","Text":"would be the forces pulling our rock inwards towards the circle."},{"Start":"00:33.405 ","End":"00:36.970","Text":"It\u0027s important to know the difference between centripetal and centrifugal."},{"Start":"00:36.970 ","End":"00:39.400","Text":"Remember, centripetal is a real force acting inwards,"},{"Start":"00:39.400 ","End":"00:42.635","Text":"it\u0027s the opposite of centrifugal force in many ways."},{"Start":"00:42.635 ","End":"00:45.470","Text":"Before we end the lectures on centrifugal and centripetal force,"},{"Start":"00:45.470 ","End":"00:49.865","Text":"I just want to emphasize that once more centripetal force acts inwards and it is real,"},{"Start":"00:49.865 ","End":"00:52.580","Text":"centrifugal force is simulated and it\u0027s outwards."},{"Start":"00:52.580 ","End":"00:58.410","Text":"Now we\u0027ll do an exercise that shows how sometimes centripetal force can be useful for us."}],"ID":9270}],"Thumbnail":null,"ID":85362},{"Name":"4. Exercise Mass on a Rotating Table","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise - Mass on a Rotating Table","Duration":"8m ","ChapterTopicVideoID":12197,"CourseChapterTopicPlaylistID":85363,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.885","Text":"In this exercise we\u0027re given a mass m;"},{"Start":"00:03.885 ","End":"00:08.580","Text":"m, that\u0027s a block resting on a disk in the middle of a table."},{"Start":"00:08.580 ","End":"00:17.235","Text":"Now this disk is rotating at a speed of Omega and our block m is also connected with"},{"Start":"00:17.235 ","End":"00:20.910","Text":"a rope via a pulley that goes through a hole in the middle of"},{"Start":"00:20.910 ","End":"00:26.580","Text":"the disc to another mass called M hanging beneath the table."},{"Start":"00:26.580 ","End":"00:32.010","Text":"Now let\u0027s assume for the sake of the problem that the rope doesn\u0027t fray,"},{"Start":"00:32.010 ","End":"00:35.550","Text":"that there\u0027s no loops in the rope or knots in the rope,"},{"Start":"00:35.550 ","End":"00:38.920","Text":"and that it\u0027s a direct simple connection there."},{"Start":"00:38.920 ","End":"00:44.720","Text":"The given data that I have in this problem is the mass m,"},{"Start":"00:44.720 ","End":"00:46.805","Text":"the mass M,"},{"Start":"00:46.805 ","End":"00:49.985","Text":"the radial velocity Omega,"},{"Start":"00:49.985 ","End":"00:53.920","Text":"and a coefficient of static friction Mu_s."},{"Start":"00:53.920 ","End":"00:55.880","Text":"The question asked what is"},{"Start":"00:55.880 ","End":"01:02.062","Text":"the minimum and maximum radius that I can place m so that it will not move radially?"},{"Start":"01:02.062 ","End":"01:06.380","Text":"What that means is where can I rest my block m so that it"},{"Start":"01:06.380 ","End":"01:11.375","Text":"won\u0027t move inwards towards the hole in the disk or outwards towards the edge of the disk?"},{"Start":"01:11.375 ","End":"01:13.490","Text":"Based on the wording of the question,"},{"Start":"01:13.490 ","End":"01:15.665","Text":"I know that I\u0027m looking for some range,"},{"Start":"01:15.665 ","End":"01:19.460","Text":"some area between radius x and radius y for"},{"Start":"01:19.460 ","End":"01:24.230","Text":"example where I can place block m and it will move around the circle,"},{"Start":"01:24.230 ","End":"01:28.400","Text":"but it won\u0027t move in towards the center or out towards the edge of the disk."},{"Start":"01:28.400 ","End":"01:30.995","Text":"Before we go on to solving the problem,"},{"Start":"01:30.995 ","End":"01:33.155","Text":"I want to talk for a moment about friction."},{"Start":"01:33.155 ","End":"01:34.722","Text":"We have 2 friction;"},{"Start":"01:34.722 ","End":"01:38.015","Text":"we have static friction and we have kinetic friction."},{"Start":"01:38.015 ","End":"01:42.325","Text":"When we talk about static friction; that\u0027s f_s,"},{"Start":"01:42.325 ","End":"01:50.580","Text":"we know that it is always going to be less than or equal to Mu_sN."},{"Start":"01:50.580 ","End":"01:52.400","Text":"Oftentimes it\u0027s unknown."},{"Start":"01:52.400 ","End":"01:56.240","Text":"We don\u0027t know exactly what the value is unless someone"},{"Start":"01:56.240 ","End":"02:00.445","Text":"tells us it\u0027s a maximal which means that it\u0027s going to be equal to MsN."},{"Start":"02:00.445 ","End":"02:04.925","Text":"If we have a maximum static friction it\u0027s going to be equal to MsN,"},{"Start":"02:04.925 ","End":"02:07.850","Text":"otherwise it\u0027s going to be less than or equal to"},{"Start":"02:07.850 ","End":"02:12.205","Text":"MsN and we can\u0027t determine the value without some procedure."},{"Start":"02:12.205 ","End":"02:14.710","Text":"In contrast f_k;"},{"Start":"02:14.710 ","End":"02:16.280","Text":"your kinetic friction,"},{"Start":"02:16.280 ","End":"02:19.305","Text":"is always going to equal M_kN."},{"Start":"02:19.305 ","End":"02:21.280","Text":"It doesn\u0027t matter how fast my object is moving,"},{"Start":"02:21.280 ","End":"02:23.570","Text":"it doesn\u0027t matter the direction or whatever."},{"Start":"02:23.570 ","End":"02:25.723","Text":"It\u0027s always going to equal M_kN,"},{"Start":"02:25.723 ","End":"02:29.720","Text":"so it\u0027s rarely an unknown and we can plug it in automatically."},{"Start":"02:29.720 ","End":"02:33.170","Text":"Now we can start the problem and we\u0027re going to begin it like we do"},{"Start":"02:33.170 ","End":"02:36.115","Text":"any problem by listing out our forces."},{"Start":"02:36.115 ","End":"02:41.140","Text":"First we can talk about the force T that\u0027s pulling m in towards"},{"Start":"02:41.140 ","End":"02:47.015","Text":"the center of the disk and also pulling large M up towards the center of the disk."},{"Start":"02:47.015 ","End":"02:53.415","Text":"In addition we have Mg which is pulling M down."},{"Start":"02:53.415 ","End":"02:59.120","Text":"We also have mg which is pulling m downwards and we"},{"Start":"02:59.120 ","End":"03:04.905","Text":"have a normal force pulling m upwards or away,"},{"Start":"03:04.905 ","End":"03:07.120","Text":"I should say from the center."},{"Start":"03:07.120 ","End":"03:10.370","Text":"We also have our static friction, f_s."},{"Start":"03:10.370 ","End":"03:12.410","Text":"We have a problem with it for now which is we"},{"Start":"03:12.410 ","End":"03:14.510","Text":"don\u0027t know which direction it will be pulling."},{"Start":"03:14.510 ","End":"03:16.445","Text":"We don\u0027t know if it\u0027s going to the left or to the right."},{"Start":"03:16.445 ","End":"03:19.090","Text":"We\u0027ll leave it as it is for now and approach it in a minute."},{"Start":"03:19.090 ","End":"03:20.840","Text":"Now that I\u0027ve listed out the forces"},{"Start":"03:20.840 ","End":"03:22.970","Text":"acting on these objects, I can start writing them out."},{"Start":"03:22.970 ","End":"03:24.710","Text":"This time it needs the second method with"},{"Start":"03:24.710 ","End":"03:27.830","Text":"centrifugal force that we talked about earlier where I"},{"Start":"03:27.830 ","End":"03:33.860","Text":"incorporate the centrifugal force m Omega squared r,"},{"Start":"03:33.860 ","End":"03:38.250","Text":"and use that to set my equation equal to 0."},{"Start":"03:38.250 ","End":"03:41.618","Text":"If you recall m Omega squared r has to equal T,"},{"Start":"03:41.618 ","End":"03:45.170","Text":"and per usual we\u0027re going to set our force going to the inside of the circle."},{"Start":"03:45.170 ","End":"03:46.715","Text":"In this case, that\u0027s the right."},{"Start":"03:46.715 ","End":"03:52.995","Text":"When we write out our force that is the radial force on m,"},{"Start":"03:52.995 ","End":"03:59.700","Text":"and we\u0027re going to write T minus m Omega squared r,"},{"Start":"03:59.700 ","End":"04:05.870","Text":"and then we have to account for the static friction equals 0."},{"Start":"04:05.870 ","End":"04:07.670","Text":"Now when we\u0027re accounting for the static friction,"},{"Start":"04:07.670 ","End":"04:09.605","Text":"we don\u0027t know whether to add it or subtract it yet."},{"Start":"04:09.605 ","End":"04:11.285","Text":"We don\u0027t know which direction it\u0027s acting,"},{"Start":"04:11.285 ","End":"04:13.240","Text":"but we can figure that out actually."},{"Start":"04:13.240 ","End":"04:19.895","Text":"First let\u0027s note that m Omega squared r gets larger as our object m moves farther out."},{"Start":"04:19.895 ","End":"04:23.390","Text":"Now if we look at T for a second; the first T,"},{"Start":"04:23.390 ","End":"04:26.780","Text":"we know that if m isn\u0027t moving then M isn\u0027t moving"},{"Start":"04:26.780 ","End":"04:30.559","Text":"either which means that if M isn\u0027t moving"},{"Start":"04:30.559 ","End":"04:39.095","Text":"then T minus MG has to equal 0 for our equation to work in this situation."},{"Start":"04:39.095 ","End":"04:44.940","Text":"If T minus MG=0, then T=MG."},{"Start":"04:44.940 ","End":"04:46.775","Text":"This tells me that T is constant."},{"Start":"04:46.775 ","End":"04:51.065","Text":"But if m Omega squared r gets larger as the object moves out,"},{"Start":"04:51.065 ","End":"04:55.205","Text":"then we need to have the friction accounting for that to balance things out."},{"Start":"04:55.205 ","End":"04:56.390","Text":"Meaning it\u0027s moving towards the right."},{"Start":"04:56.390 ","End":"05:03.335","Text":"Let\u0027s say for a second that we\u0027re using our maximum radius to make this equal to 0,"},{"Start":"05:03.335 ","End":"05:06.770","Text":"our maximum radius for this problem,"},{"Start":"05:06.770 ","End":"05:12.980","Text":"then we know that m Omega squared r is going to be larger than T"},{"Start":"05:12.980 ","End":"05:15.838","Text":"because again it\u0027s getting larger as it goes farther out and"},{"Start":"05:15.838 ","End":"05:19.527","Text":"our static friction is going to have to account for that and move towards the right."},{"Start":"05:19.527 ","End":"05:21.438","Text":"Meaning move in the positive direction,"},{"Start":"05:21.438 ","End":"05:23.225","Text":"so we\u0027re adding our static friction."},{"Start":"05:23.225 ","End":"05:25.535","Text":"If I know I\u0027m at my maximal radius,"},{"Start":"05:25.535 ","End":"05:26.930","Text":"then I can also know that I\u0027m at"},{"Start":"05:26.930 ","End":"05:31.040","Text":"my maximum static friction because that means it has to be able to account for"},{"Start":"05:31.040 ","End":"05:33.350","Text":"the largest difference between m Omega"},{"Start":"05:33.350 ","End":"05:37.715","Text":"squared r and T. If I\u0027m at my maximum static friction,"},{"Start":"05:37.715 ","End":"05:40.540","Text":"I can set that equal to Mu_sN."},{"Start":"05:40.540 ","End":"05:43.250","Text":"We know N is the normal force and we can actually find"},{"Start":"05:43.250 ","End":"05:46.205","Text":"that by looking at our y-axis, the vertical axis."},{"Start":"05:46.205 ","End":"05:51.630","Text":"We know that on the y-axis and m is not moving, so N minus mg=0."},{"Start":"05:53.210 ","End":"05:57.030","Text":"That\u0027s the other force on the y-axis."},{"Start":"05:57.030 ","End":"05:59.475","Text":"Therefore, N=mg."},{"Start":"05:59.475 ","End":"06:02.470","Text":"If we plug everything back into our equation above;"},{"Start":"06:02.470 ","End":"06:03.550","Text":"our first equation,"},{"Start":"06:03.550 ","End":"06:07.025","Text":"we can talk about things in terms of m more than anything else."},{"Start":"06:07.025 ","End":"06:14.600","Text":"Instead of T we can write this as MG minus m Omega"},{"Start":"06:14.600 ","End":"06:21.780","Text":"squared r plus Mu_smg=0."},{"Start":"06:21.780 ","End":"06:24.520","Text":"Let me write that M a little more clearly."},{"Start":"06:24.520 ","End":"06:27.770","Text":"From here, I can isolate my r,"},{"Start":"06:27.770 ","End":"06:31.310","Text":"my maximal radius, and solve for it."},{"Start":"06:31.310 ","End":"06:37.792","Text":"If we solve, we\u0027ll find that the maximal radius equals large MG"},{"Start":"06:37.792 ","End":"06:46.030","Text":"plus Mu_smg over m Omega squared."},{"Start":"06:46.030 ","End":"06:49.215","Text":"Again, that\u0027s our maximal radius."},{"Start":"06:49.215 ","End":"06:51.564","Text":"When we\u0027re finding the minimal radius,"},{"Start":"06:51.564 ","End":"06:54.550","Text":"we need to think about the exact same problems in the opposite."},{"Start":"06:54.550 ","End":"06:57.130","Text":"The closer that m gets to the center,"},{"Start":"06:57.130 ","End":"07:00.160","Text":"the smaller m Omega squared r will get."},{"Start":"07:00.160 ","End":"07:02.650","Text":"We need the exact same amount of friction just in"},{"Start":"07:02.650 ","End":"07:06.160","Text":"the opposite direction to keep it from falling through the hole."},{"Start":"07:06.160 ","End":"07:08.800","Text":"We\u0027re going to have the exact same amount of friction."},{"Start":"07:08.800 ","End":"07:10.120","Text":"Again, it\u0027s our maximal friction,"},{"Start":"07:10.120 ","End":"07:12.505","Text":"so we can still write that as Mu_sN,"},{"Start":"07:12.505 ","End":"07:15.020","Text":"but it\u0027s just going in the opposite direction."},{"Start":"07:15.020 ","End":"07:17.110","Text":"Remember it\u0027s not when the friction is 0,"},{"Start":"07:17.110 ","End":"07:20.080","Text":"it\u0027s in fact when the friction is going towards the left."},{"Start":"07:20.080 ","End":"07:21.970","Text":"That\u0027s our minimal radius."},{"Start":"07:21.970 ","End":"07:23.222","Text":"When we do minimal radius."},{"Start":"07:23.222 ","End":"07:26.605","Text":"What we need to do is instead of adding that friction,"},{"Start":"07:26.605 ","End":"07:29.060","Text":"we need to simply subtract that friction and we"},{"Start":"07:29.060 ","End":"07:31.670","Text":"can do the exact same process for solving."},{"Start":"07:31.670 ","End":"07:34.040","Text":"It\u0027s just when we go to the bottom to solve"},{"Start":"07:34.040 ","End":"07:36.350","Text":"for our minimal radius instead of our maximal radius,"},{"Start":"07:36.350 ","End":"07:42.210","Text":"we switch the plus to a minus so that it\u0027s MG minus Mu_smg."},{"Start":"07:42.520 ","End":"07:45.095","Text":"That\u0027s our minimal radius."},{"Start":"07:45.095 ","End":"07:46.635","Text":"That\u0027s your solution."},{"Start":"07:46.635 ","End":"07:50.180","Text":"The important thing to remember from this exercise is that we set our forces,"},{"Start":"07:50.180 ","End":"07:51.970","Text":"use our centrifugal force,"},{"Start":"07:51.970 ","End":"07:55.850","Text":"and set the equation equal to 0 and that helps us find the solution we need"},{"Start":"07:55.850 ","End":"08:00.750","Text":"for our maximal and minimal radius and how the friction affects things."}],"ID":12672}],"Thumbnail":null,"ID":85363},{"Name":"5. Position, Velocity and Acceleration Vectors","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Position Vector","Duration":"2m 46s","ChapterTopicVideoID":9011,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.355","Text":"Hello, in this video,"},{"Start":"00:02.355 ","End":"00:08.025","Text":"I wanted to talk about position vectors and how we can describe position vectors,"},{"Start":"00:08.025 ","End":"00:11.550","Text":"velocity and acceleration in circular motion."},{"Start":"00:11.550 ","End":"00:16.200","Text":"The first thing we need to do is we have this circle with some object resting along"},{"Start":"00:16.200 ","End":"00:21.075","Text":"it and we can put this on the x, y-axis."},{"Start":"00:21.075 ","End":"00:22.695","Text":"Notice the x and y-axis,"},{"Start":"00:22.695 ","End":"00:25.440","Text":"that origin is also the middle of the circle."},{"Start":"00:25.440 ","End":"00:28.725","Text":"We label our x and our y axis and we find a point."},{"Start":"00:28.725 ","End":"00:32.310","Text":"Now the position vector in circular motion is always going to"},{"Start":"00:32.310 ","End":"00:36.795","Text":"come from the origin and go out towards the point itself."},{"Start":"00:36.795 ","End":"00:41.055","Text":"We\u0027re going to draw it here, like this,"},{"Start":"00:41.055 ","End":"00:43.410","Text":"and we\u0027re going to label it."},{"Start":"00:43.410 ","End":"00:46.850","Text":"It\u0027s always going to be labeled r vector as"},{"Start":"00:46.850 ","End":"00:51.155","Text":"opposed to the radius which is R. Just remember the difference between the 2."},{"Start":"00:51.155 ","End":"00:55.370","Text":"I want to express this in terms of Cartesian coordinates."},{"Start":"00:55.370 ","End":"00:59.405","Text":"The way I do that is in terms of x and y, of course,"},{"Start":"00:59.405 ","End":"01:05.100","Text":"x hat and y plus y, y hat."},{"Start":"01:05.100 ","End":"01:07.370","Text":"I have both the direction and the magnitude"},{"Start":"01:07.370 ","End":"01:09.880","Text":"represented here in terms of x and x direction,"},{"Start":"01:09.880 ","End":"01:11.670","Text":"y and y direction."},{"Start":"01:11.670 ","End":"01:13.790","Text":"Before I go and find my values,"},{"Start":"01:13.790 ","End":"01:16.300","Text":"we should just remember that this formula is important."},{"Start":"01:16.300 ","End":"01:18.275","Text":"Let\u0027s put in type here,"},{"Start":"01:18.275 ","End":"01:20.690","Text":"r vector equals x,"},{"Start":"01:20.690 ","End":"01:22.130","Text":"x hat plus y,"},{"Start":"01:22.130 ","End":"01:24.985","Text":"y hat. Here it is."},{"Start":"01:24.985 ","End":"01:28.720","Text":"So now we can focus on finding our x and our y values."},{"Start":"01:28.720 ","End":"01:32.680","Text":"The way we do that is take a perpendicular line from the point towards the x-axis,"},{"Start":"01:32.680 ","End":"01:34.270","Text":"and that\u0027ll give us our x value,"},{"Start":"01:34.270 ","End":"01:37.345","Text":"and a perpendicular line towards the y-axis."},{"Start":"01:37.345 ","End":"01:38.765","Text":"That\u0027ll give us our y value."},{"Start":"01:38.765 ","End":"01:44.440","Text":"If we look at the triangle formed by our newly dropped line here,"},{"Start":"01:44.440 ","End":"01:46.495","Text":"we can call that angle Theta."},{"Start":"01:46.495 ","End":"01:49.945","Text":"We see that this is actually a right angle triangle."},{"Start":"01:49.945 ","End":"01:52.180","Text":"With that given data,"},{"Start":"01:52.180 ","End":"01:59.410","Text":"we know that x equals r cosine Theta."},{"Start":"01:59.450 ","End":"02:07.395","Text":"We can write that out, x equals r cosine Theta."},{"Start":"02:07.395 ","End":"02:10.460","Text":"We can do a similar thing for y."},{"Start":"02:10.460 ","End":"02:14.270","Text":"We can find not that y is equal to r cosine Theta,"},{"Start":"02:14.270 ","End":"02:19.300","Text":"but rather y is equal to r sine Theta."},{"Start":"02:19.300 ","End":"02:23.060","Text":"For these problems, it doesn\u0027t matter for using the little r or R,"},{"Start":"02:23.060 ","End":"02:25.160","Text":"because the value there,"},{"Start":"02:25.160 ","End":"02:26.890","Text":"the distance there is the same actually,"},{"Start":"02:26.890 ","End":"02:28.770","Text":"these both represent the radius."},{"Start":"02:28.770 ","End":"02:31.625","Text":"We can put these both in type formulas and we will later."},{"Start":"02:31.625 ","End":"02:37.355","Text":"But for now, what\u0027s important to remember is that we can now express the r vector as"},{"Start":"02:37.355 ","End":"02:46.560","Text":"R cosine Theta x hat plus R sine Theta y hat."}],"ID":9272},{"Watched":false,"Name":"Velocity Vector","Duration":"2m 36s","ChapterTopicVideoID":9012,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Now that we have a position vector,"},{"Start":"00:02.115 ","End":"00:05.040","Text":"the next thing to do is find our velocity vector,"},{"Start":"00:05.040 ","End":"00:08.700","Text":"and we\u0027re going to find the velocity vector, if you recall,"},{"Start":"00:08.700 ","End":"00:12.795","Text":"is the derivative of the position vector."},{"Start":"00:12.795 ","End":"00:15.135","Text":"It\u0027s r vector dot,"},{"Start":"00:15.135 ","End":"00:17.130","Text":"so we know that the R,"},{"Start":"00:17.130 ","End":"00:18.840","Text":"the radius of the circle is constant in time,"},{"Start":"00:18.840 ","End":"00:21.180","Text":"so we\u0027re not doing a derivative based on that."},{"Start":"00:21.180 ","End":"00:23.670","Text":"We\u0027re doing it based on time, so it\u0027s actually Theta"},{"Start":"00:23.670 ","End":"00:26.340","Text":"is something that changes with time, the angle Theta."},{"Start":"00:26.340 ","End":"00:28.530","Text":"What we can do is,"},{"Start":"00:28.530 ","End":"00:33.375","Text":"remember Omega is the change in Theta over the change in time,"},{"Start":"00:33.375 ","End":"00:35.775","Text":"which is the same as saying,"},{"Start":"00:35.775 ","End":"00:38.625","Text":"Omega is the derivative of Theta."},{"Start":"00:38.625 ","End":"00:44.405","Text":"In the same way that v is the derivative of x."},{"Start":"00:44.405 ","End":"00:48.835","Text":"Or we could call that velocity is the derivative of position."},{"Start":"00:48.835 ","End":"00:56.908","Text":"Omega, we can even call that the angular velocity as opposed to the angular position."},{"Start":"00:56.908 ","End":"01:00.440","Text":"The derivative of r is the change in r over"},{"Start":"01:00.440 ","End":"01:04.445","Text":"the change in time and we can use the chain rule to write this differently."},{"Start":"01:04.445 ","End":"01:08.355","Text":"We could write this out as dr over d Theta."},{"Start":"01:08.355 ","End":"01:14.330","Text":"For example, the derivative of cosine Theta is"},{"Start":"01:14.330 ","End":"01:20.915","Text":"negative sine Theta and then we take that and multiply that by d Theta dt,"},{"Start":"01:20.915 ","End":"01:23.090","Text":"or the change in Theta over the change in time,"},{"Start":"01:23.090 ","End":"01:24.215","Text":"or the derivative of Theta,"},{"Start":"01:24.215 ","End":"01:26.330","Text":"which is equal to Omega."},{"Start":"01:26.330 ","End":"01:28.130","Text":"This is how you use the chain rule."},{"Start":"01:28.130 ","End":"01:32.195","Text":"If you\u0027re given an equation with 1 variable,"},{"Start":"01:32.195 ","End":"01:35.225","Text":"but you\u0027d like to do your derivative with another variable."},{"Start":"01:35.225 ","End":"01:38.690","Text":"You can start by doing your derivative with the variable that you have and"},{"Start":"01:38.690 ","End":"01:42.640","Text":"then multiply that by the variable that you need to later do your derivative with."},{"Start":"01:42.640 ","End":"01:46.010","Text":"If for example, I\u0027ve had 2t instead of Theta in there,"},{"Start":"01:46.010 ","End":"01:49.580","Text":"I could have immediately done dr over dt because I didn\u0027t have that."},{"Start":"01:49.580 ","End":"01:54.055","Text":"I did dr over d Theta and then d Theta over dt to get to my desired result."},{"Start":"01:54.055 ","End":"01:55.550","Text":"When we calculate this out,"},{"Start":"01:55.550 ","End":"02:00.620","Text":"we find that dr over d Theta gives us negative R,"},{"Start":"02:00.620 ","End":"02:10.560","Text":"sine Theta in the direction of x plus for a y element."},{"Start":"02:10.560 ","End":"02:19.550","Text":"We get plus R cosine Theta and we have to multiply that by d Theta over dt."},{"Start":"02:19.550 ","End":"02:25.435","Text":"Of course it\u0027s in the y direction and d Theta over dt is Theta dot or Omega,"},{"Start":"02:25.435 ","End":"02:29.060","Text":"so we can even write that right away as Omega and"},{"Start":"02:29.060 ","End":"02:33.780","Text":"that\u0027s our result so far for the velocity vector, that\u0027s the formula."}],"ID":9273},{"Watched":false,"Name":"Acceleration Vector","Duration":"1m 48s","ChapterTopicVideoID":9013,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.810","Text":"The procedure for finding the acceleration vector is very similar."},{"Start":"00:03.810 ","End":"00:09.000","Text":"Of course, the acceleration vector is the derivative of the velocity vector."},{"Start":"00:09.000 ","End":"00:14.655","Text":"We need to do is take the result we just got and do another set of derivatives."},{"Start":"00:14.655 ","End":"00:18.975","Text":"This time we\u0027re doing derivatives on Theta as well as on Omega."},{"Start":"00:18.975 ","End":"00:23.460","Text":"We can think of it as Theta in time as well as the change of Omega in time."},{"Start":"00:23.460 ","End":"00:26.010","Text":"To do the math quickly,"},{"Start":"00:26.010 ","End":"00:29.970","Text":"we can find that the v dot equals,"},{"Start":"00:29.970 ","End":"00:32.250","Text":"for starters, let\u0027s just do the Theta,"},{"Start":"00:32.250 ","End":"00:38.160","Text":"it would be negative R cosine Theta instead of sine Theta,"},{"Start":"00:38.160 ","End":"00:43.190","Text":"cosine Theta in the direction of x, and instead of plus,"},{"Start":"00:43.190 ","End":"00:50.875","Text":"we end up with a minus R sine Theta in the direction of y."},{"Start":"00:50.875 ","End":"00:54.020","Text":"Now we have our initial Omega to add in and"},{"Start":"00:54.020 ","End":"00:56.570","Text":"because we\u0027re multiplying again by d Theta over dt,"},{"Start":"00:56.570 ","End":"00:58.490","Text":"I\u0027m going to put another Omega,"},{"Start":"00:58.490 ","End":"01:01.650","Text":"so instead of just 1 Omega, it\u0027s Omega^2."},{"Start":"01:01.760 ","End":"01:04.770","Text":"Now for the second part of our procedure,"},{"Start":"01:04.770 ","End":"01:08.385","Text":"we\u0027ll do the change in Omega over the change in time."},{"Start":"01:08.385 ","End":"01:10.105","Text":"Because we\u0027re not dealing with Theta,"},{"Start":"01:10.105 ","End":"01:12.215","Text":"the initial part of our equation stays the same."},{"Start":"01:12.215 ","End":"01:15.065","Text":"Those are our findings of the velocity vector."},{"Start":"01:15.065 ","End":"01:21.790","Text":"We can write negative R sine Theta x hat plus R cosine Theta y hat"},{"Start":"01:21.790 ","End":"01:26.505","Text":"and the derivative of Omega is going to be Alpha."},{"Start":"01:26.505 ","End":"01:28.690","Text":"Alpha is the derivative of Omega,"},{"Start":"01:28.690 ","End":"01:30.230","Text":"which is the derivative of Theta."},{"Start":"01:30.230 ","End":"01:34.490","Text":"Therefore, Alpha is the second derivative of Theta."},{"Start":"01:34.490 ","End":"01:38.390","Text":"You go from the position of the angle to"},{"Start":"01:38.390 ","End":"01:42.605","Text":"the angular velocity to the angular acceleration."},{"Start":"01:42.605 ","End":"01:44.420","Text":"That\u0027s the relationship between Alpha,"},{"Start":"01:44.420 ","End":"01:46.560","Text":"Omega, and Theta."}],"ID":9274},{"Watched":false,"Name":"The Radial Acceleration Formula","Duration":"1m 18s","ChapterTopicVideoID":9014,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.770 ","End":"00:03.060","Text":"If you look at this for a second,"},{"Start":"00:03.060 ","End":"00:04.950","Text":"you might see something very interesting."},{"Start":"00:04.950 ","End":"00:08.115","Text":"Let\u0027s assume that Alpha ends up equaling 0,"},{"Start":"00:08.115 ","End":"00:12.675","Text":"which means that the second half of our acceleration vector equation equals 0."},{"Start":"00:12.675 ","End":"00:14.910","Text":"Then what you\u0027re left with is"},{"Start":"00:14.910 ","End":"00:20.875","Text":"negative R cosine Theta x-hat minus R sine Theta y-hat times Omega^2."},{"Start":"00:20.875 ","End":"00:24.810","Text":"That part in the parentheses is actually the negative of our R vector."},{"Start":"00:24.810 ","End":"00:28.470","Text":"If we took the negative and put on the outside, we\u0027d have the same amount."},{"Start":"00:28.470 ","End":"00:37.725","Text":"The acceleration vector equals the negative of the position vector times Omega^2."},{"Start":"00:37.725 ","End":"00:39.780","Text":"This should make some sense because"},{"Start":"00:39.780 ","End":"00:43.205","Text":"the R vector is going from the origin outwards and the a,"},{"Start":"00:43.205 ","End":"00:47.425","Text":"the acceleration vector is going inwards from that point towards the origin."},{"Start":"00:47.425 ","End":"00:49.455","Text":"It\u0027s going in that direction."},{"Start":"00:49.455 ","End":"00:51.845","Text":"Omega^2 should give us a quantity."},{"Start":"00:51.845 ","End":"00:53.690","Text":"Now, if we were to take the magnitudes,"},{"Start":"00:53.690 ","End":"00:57.125","Text":"the absolute value of the acceleration vector,"},{"Start":"00:57.125 ","End":"01:02.615","Text":"it would equal Omega^2 times the absolute value of the R vector."},{"Start":"01:02.615 ","End":"01:05.420","Text":"That should be a very familiar formula that we"},{"Start":"01:05.420 ","End":"01:08.150","Text":"just talked about a couple of video lectures ago."},{"Start":"01:08.150 ","End":"01:10.700","Text":"This is our circular motion formula for"},{"Start":"01:10.700 ","End":"01:13.775","Text":"the acceleration vector when we have a constant speed."},{"Start":"01:13.775 ","End":"01:16.070","Text":"There you have it, that\u0027s where it comes from."},{"Start":"01:16.070 ","End":"01:18.690","Text":"Now, for one more thing."}],"ID":9275},{"Watched":false,"Name":"The Velocity Formula","Duration":"1m 58s","ChapterTopicVideoID":9015,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.645","Text":"If we want to find the magnitude of the velocity vector,"},{"Start":"00:03.645 ","End":"00:07.410","Text":"what we need to do is take the square and then"},{"Start":"00:07.410 ","End":"00:12.510","Text":"the square root of both of these elements of the formula."},{"Start":"00:12.510 ","End":"00:15.750","Text":"We know that Omega^2 and square root is"},{"Start":"00:15.750 ","End":"00:20.820","Text":"Omega and if we take the other bit and put it inside a square root,"},{"Start":"00:20.820 ","End":"00:31.025","Text":"find square root of R^2 sine^2 Theta plus R^2 cosine^2 Theta."},{"Start":"00:31.025 ","End":"00:38.295","Text":"Now the sine^2 plus cosine^2 is going to equal 1 and that\u0027s going to drop out."},{"Start":"00:38.295 ","End":"00:43.650","Text":"What we\u0027re left with is Omega times R and if you recall,"},{"Start":"00:43.650 ","End":"00:49.285","Text":"V equals Omega R is one of our other formulas from circular motion."},{"Start":"00:49.285 ","End":"00:53.870","Text":"That\u0027s our magnitude and if we want to find the direction of the V vector,"},{"Start":"00:53.870 ","End":"00:58.670","Text":"we can do a scalar multiplication with the R vector."},{"Start":"00:58.670 ","End":"01:01.775","Text":"If we do a scalar multiplication here,"},{"Start":"01:01.775 ","End":"01:08.535","Text":"that\u0027ll be v.r or r.v and that would equal."},{"Start":"01:08.535 ","End":"01:10.950","Text":"Let\u0027s see the x element first."},{"Start":"01:10.950 ","End":"01:16.145","Text":"R^2 cosine Theta sine Theta,"},{"Start":"01:16.145 ","End":"01:21.100","Text":"and that\u0027ll be negative because we have negative sine Theta with our V element."},{"Start":"01:21.100 ","End":"01:23.980","Text":"We don\u0027t have directions anymore, just values."},{"Start":"01:23.980 ","End":"01:32.260","Text":"That\u0027s going to be also times Omega plus R^2 sine Theta,"},{"Start":"01:32.260 ","End":"01:35.840","Text":"cosine Theta and those 2 are like terms,"},{"Start":"01:35.840 ","End":"01:38.090","Text":"it\u0027s 1 minus the other is they equal 0."},{"Start":"01:38.090 ","End":"01:43.610","Text":"What that means for us is that v is perpendicular to r. If it\u0027s perpendicular to r,"},{"Start":"01:43.610 ","End":"01:46.265","Text":"it means it\u0027s going tangentially to the circle."},{"Start":"01:46.265 ","End":"01:48.890","Text":"It could be tangential in either direction really."},{"Start":"01:48.890 ","End":"01:51.950","Text":"But what we know now is that v is perpendicular or"},{"Start":"01:51.950 ","End":"01:56.430","Text":"orthogonal to r and tangential to our circle."}],"ID":9276},{"Watched":false,"Name":"The Tangential Acceleration Formula","Duration":"1m 48s","ChapterTopicVideoID":9016,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"The last thing I want to talk about has to do with the acceleration vector and how"},{"Start":"00:04.500 ","End":"00:08.730","Text":"the last portion of the formula here, the end segment,"},{"Start":"00:08.730 ","End":"00:19.395","Text":"is actually equal to Alpha times the value R in the direction of the v vector."},{"Start":"00:19.395 ","End":"00:23.205","Text":"Now we\u0027re only talking about the last part of the equation."},{"Start":"00:23.205 ","End":"00:26.705","Text":"We know that the first part of the acceleration vector equation is"},{"Start":"00:26.705 ","End":"00:30.200","Text":"equal to Omega^2*R in the direction of"},{"Start":"00:30.200 ","End":"00:33.140","Text":"negative R. You\u0027ll see that the portion that I"},{"Start":"00:33.140 ","End":"00:37.565","Text":"highlighted is the same as the v vector without an Omega."},{"Start":"00:37.565 ","End":"00:40.970","Text":"We can say that that\u0027s equal to the v vector divided by Omega."},{"Start":"00:40.970 ","End":"00:46.515","Text":"We write that out below we have v vector divided by Omega times Alpha,"},{"Start":"00:46.515 ","End":"00:48.945","Text":"which is remaining outside of the parentheses."},{"Start":"00:48.945 ","End":"00:52.700","Text":"We can rewrite the v vector as the magnitude and the direction,"},{"Start":"00:52.700 ","End":"00:54.230","Text":"the 2 elements of a vector."},{"Start":"00:54.230 ","End":"01:01.860","Text":"We can write v magnitude times v direction over Omega times Alpha."},{"Start":"01:01.860 ","End":"01:05.210","Text":"We happen to already know that the magnitude of the v vector from"},{"Start":"01:05.210 ","End":"01:09.125","Text":"before is equal to Omega R."},{"Start":"01:09.125 ","End":"01:12.570","Text":"Omega R over Omega times Alpha"},{"Start":"01:12.570 ","End":"01:16.560","Text":"times the direction of the v vector and if we cancel out the Omegas,"},{"Start":"01:16.560 ","End":"01:21.010","Text":"we end up with Alpha R in the direction of the v vector."},{"Start":"01:21.010 ","End":"01:23.590","Text":"You could do this a different way of finding the magnitude by doing"},{"Start":"01:23.590 ","End":"01:26.320","Text":"a squaring and square root to find that that actually"},{"Start":"01:26.320 ","End":"01:29.005","Text":"equals R Alpha and that the direction is"},{"Start":"01:29.005 ","End":"01:32.425","Text":"also the direction of the v vector and tangential."},{"Start":"01:32.425 ","End":"01:34.510","Text":"We know that this is"},{"Start":"01:34.510 ","End":"01:38.860","Text":"the tangential acceleration where the first part is the radial acceleration."},{"Start":"01:38.860 ","End":"01:43.000","Text":"This shows us very clearly that the highlighted part is tangential acceleration and"},{"Start":"01:43.000 ","End":"01:48.470","Text":"the upper part is radial acceleration and the tangential acceleration equals Alpha R."}],"ID":9277},{"Watched":false,"Name":"Summary","Duration":"1m 33s","ChapterTopicVideoID":9017,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Let\u0027s take a moment to summarize."},{"Start":"00:02.025 ","End":"00:06.360","Text":"In circular motion, we\u0027re moving around a circle with a constant radius."},{"Start":"00:06.360 ","End":"00:10.110","Text":"The way we can describe the position is the way we did below"},{"Start":"00:10.110 ","End":"00:14.215","Text":"in red xx-hat yy-hat in Cartesian coordinates."},{"Start":"00:14.215 ","End":"00:19.875","Text":"We know that x is r cosine Theta and y is r sine Theta."},{"Start":"00:19.875 ","End":"00:21.660","Text":"The way that we write the r vector,"},{"Start":"00:21.660 ","End":"00:24.675","Text":"the position vector is like this."},{"Start":"00:24.675 ","End":"00:27.645","Text":"That\u0027s really the most important thing to remember is that this"},{"Start":"00:27.645 ","End":"00:30.855","Text":"is the position vector in Cartesian coordinates."},{"Start":"00:30.855 ","End":"00:34.245","Text":"Then we took a derivative of that to find the velocity vector."},{"Start":"00:34.245 ","End":"00:36.200","Text":"If you recall the velocity vector,"},{"Start":"00:36.200 ","End":"00:39.050","Text":"we had to use the chain rule to do a derivative over time."},{"Start":"00:39.050 ","End":"00:43.835","Text":"What we ended up with was this whole thing multiplied by Omega^2."},{"Start":"00:43.835 ","End":"00:46.160","Text":"Omega, of course, being the difference in the angle over"},{"Start":"00:46.160 ","End":"00:49.715","Text":"the difference in time or the angular velocity."},{"Start":"00:49.715 ","End":"00:54.410","Text":"If you recall, the magnitude of the velocity vector was"},{"Start":"00:54.410 ","End":"00:58.835","Text":"equal to Omega r and the angle was tangential,"},{"Start":"00:58.835 ","End":"01:01.325","Text":"was orthogonal to the circle itself."},{"Start":"01:01.325 ","End":"01:04.475","Text":"Then we did another derivative to find our acceleration vector."},{"Start":"01:04.475 ","End":"01:06.065","Text":"The first part which ended up being"},{"Start":"01:06.065 ","End":"01:09.110","Text":"our radial acceleration is in the direction towards the origin,"},{"Start":"01:09.110 ","End":"01:10.985","Text":"towards the center of the circle."},{"Start":"01:10.985 ","End":"01:16.720","Text":"Its magnitude was Omega^2r."},{"Start":"01:16.720 ","End":"01:20.945","Text":"The second part, the tangential acceleration was"},{"Start":"01:20.945 ","End":"01:25.565","Text":"in the same tangential direction or the same orthogonal direction."},{"Start":"01:25.565 ","End":"01:29.090","Text":"The tangential acceleration a Theta was in the direction of"},{"Start":"01:29.090 ","End":"01:34.710","Text":"the v vector with the value of Alpha r. That\u0027s the end of the lecture. Thank you."}],"ID":9278},{"Watched":false,"Name":"Exercise 1","Duration":"9m 27s","ChapterTopicVideoID":10025,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this question,"},{"Start":"00:01.830 ","End":"00:06.390","Text":"we\u0027re being told that a body travels in a circle of radius R,"},{"Start":"00:06.390 ","End":"00:09.615","Text":"with constant tangential acceleration,"},{"Start":"00:09.615 ","End":"00:13.395","Text":"a_t, and no initial velocity."},{"Start":"00:13.395 ","End":"00:17.760","Text":"What does that mean? That means that our acceleration in"},{"Start":"00:17.760 ","End":"00:23.415","Text":"the tangential direction is equal to a constant,"},{"Start":"00:23.415 ","End":"00:26.255","Text":"and we have no initial velocity,"},{"Start":"00:26.255 ","End":"00:32.975","Text":"which means that our velocity at t=0 is equal to 0."},{"Start":"00:32.975 ","End":"00:39.365","Text":"What we\u0027re being told is to calculate our radial acceleration."},{"Start":"00:39.365 ","End":"00:41.510","Text":"This is what we\u0027re trying to find."},{"Start":"00:41.510 ","End":"00:47.550","Text":"In question 1, we\u0027re trying to find this as a function of time."},{"Start":"00:49.250 ","End":"00:51.505","Text":"For question number 1,"},{"Start":"00:51.505 ","End":"00:53.195","Text":"let\u0027s see how to solve this."},{"Start":"00:53.195 ","End":"01:00.470","Text":"Now, we have 2 equations or we know 2 equations for the tangential acceleration."},{"Start":"01:00.470 ","End":"01:05.525","Text":"The first one is that our tangential acceleration is equal to"},{"Start":"01:05.525 ","End":"01:12.025","Text":"angular acceleration multiplied by the radius of the circle."},{"Start":"01:12.025 ","End":"01:22.650","Text":"The next equation is equal to the change in the size of the velocity by time."},{"Start":"01:23.320 ","End":"01:27.500","Text":"Here we\u0027re dealing with our magnitude of"},{"Start":"01:27.500 ","End":"01:32.190","Text":"the velocity and we\u0027re not taking into account its direction."},{"Start":"01:32.350 ","End":"01:38.120","Text":"Now, what\u0027s important to know is that our radial acceleration is"},{"Start":"01:38.120 ","End":"01:46.500","Text":"perpendicular to the velocity and is therefore only changing its direction,"},{"Start":"01:46.520 ","End":"01:54.860","Text":"whereas our tangential acceleration is from the name tangential to the velocity,"},{"Start":"01:54.860 ","End":"02:00.250","Text":"and so it\u0027s only changing its size or the magnitude."},{"Start":"02:00.250 ","End":"02:06.965","Text":"Any change in the magnitude has to do with our tangential acceleration."},{"Start":"02:06.965 ","End":"02:14.280","Text":"Now I have my tangential acceleration and it\u0027s equal to d(V) by dt."},{"Start":"02:14.280 ","End":"02:16.770","Text":"In order to find my V,"},{"Start":"02:16.770 ","End":"02:18.570","Text":"so here we have a magnitude,"},{"Start":"02:18.570 ","End":"02:24.695","Text":"because my v doesn\u0027t have the arrow on top denoting that it\u0027s not a vector."},{"Start":"02:24.695 ","End":"02:30.180","Text":"So that is going to be equal to the integral of a_t dt."},{"Start":"02:30.790 ","End":"02:36.665","Text":"Now we\u0027re given in the question that we have constant tangential acceleration,"},{"Start":"02:36.665 ","End":"02:39.260","Text":"which means that it\u0027s not a function of t,"},{"Start":"02:39.260 ","End":"02:43.400","Text":"which means that when we do this integral,"},{"Start":"02:43.400 ","End":"02:49.260","Text":"we\u0027ll get that it\u0027s equal to a_t(t)."},{"Start":"02:49.630 ","End":"02:55.017","Text":"Now, of course, we\u0027re meant to add here plus some integrating constant."},{"Start":"02:55.017 ","End":"03:01.530","Text":"But because we\u0027re told that our velocity at t=0 is equal to 0."},{"Start":"03:01.530 ","End":"03:04.130","Text":"Once we sub in t is equal to 0,"},{"Start":"03:04.130 ","End":"03:05.990","Text":"we\u0027ll get that v is equal to c,"},{"Start":"03:05.990 ","End":"03:07.670","Text":"then it has to be equal to 0."},{"Start":"03:07.670 ","End":"03:11.610","Text":"That means that our c is equal to 0."},{"Start":"03:11.930 ","End":"03:19.210","Text":"Now we have our equation for our velocity as a function of time."},{"Start":"03:19.210 ","End":"03:25.030","Text":"Now what we want to do is we want to find our radial acceleration."},{"Start":"03:25.030 ","End":"03:30.490","Text":"Now we know that our radial acceleration is equal"},{"Start":"03:30.490 ","End":"03:35.560","Text":"to V squared divided by R. Now we have our V,"},{"Start":"03:35.560 ","End":"03:37.150","Text":"So we can just substitute it in."},{"Start":"03:37.150 ","End":"03:41.710","Text":"So V^2 is tangential acceleration multiplied by"},{"Start":"03:41.710 ","End":"03:48.995","Text":"t^2 divided by R. That\u0027s our answer to question number 1."},{"Start":"03:48.995 ","End":"03:53.470","Text":"Of course, if we want to know the direction of this radial acceleration,"},{"Start":"03:53.470 ","End":"03:58.175","Text":"it\u0027s obviously into the center of the circle."},{"Start":"03:58.175 ","End":"04:02.395","Text":"In the next video, we\u0027re going to answer question number 2."},{"Start":"04:02.395 ","End":"04:04.630","Text":"Now, in question number 2,"},{"Start":"04:04.630 ","End":"04:08.315","Text":"we\u0027re trying to find the radial acceleration,"},{"Start":"04:08.315 ","End":"04:12.595","Text":"and this time as a function of the angle Theta."},{"Start":"04:12.595 ","End":"04:16.490","Text":"What we\u0027re going to do in order to solve this is we\u0027re going to"},{"Start":"04:16.490 ","End":"04:21.390","Text":"find what the angle Theta is as a function of time,"},{"Start":"04:21.390 ","End":"04:24.290","Text":"then we\u0027re going to isolate out the time,"},{"Start":"04:24.290 ","End":"04:27.005","Text":"and then we\u0027re going to plug in"},{"Start":"04:27.005 ","End":"04:32.670","Text":"our equation that has Theta as its variable rather than t,"},{"Start":"04:32.670 ","End":"04:39.540","Text":"and then we\u0027ll get our radial acceleration as a function of the angle."},{"Start":"04:39.540 ","End":"04:42.030","Text":"How are we going to do this?"},{"Start":"04:42.030 ","End":"04:46.370","Text":"What we\u0027re going to need to do is we\u0027re going to have to find out"},{"Start":"04:46.370 ","End":"04:51.470","Text":"what our angular velocity is equal to, our Omega."},{"Start":"04:51.470 ","End":"04:53.375","Text":"How are we going to do that?"},{"Start":"04:53.375 ","End":"04:57.065","Text":"Now, we have to remember that our Omega is equal to"},{"Start":"04:57.065 ","End":"05:00.995","Text":"the time derivative of our angle Theta."},{"Start":"05:00.995 ","End":"05:04.045","Text":"What we need is our Theta."},{"Start":"05:04.045 ","End":"05:07.620","Text":"Therefore, we can work out what our Omega is,"},{"Start":"05:07.620 ","End":"05:10.185","Text":"what our angular velocity is equal to,"},{"Start":"05:10.185 ","End":"05:12.245","Text":"and then we can integrate,"},{"Start":"05:12.245 ","End":"05:19.450","Text":"and we\u0027ll get that our Theta is equal to the integral of Omega dt."},{"Start":"05:19.790 ","End":"05:23.930","Text":"This is what we\u0027re going to do for our first step."},{"Start":"05:23.930 ","End":"05:31.340","Text":"In circular motion, we know that our Omega is always going to be equal to"},{"Start":"05:31.340 ","End":"05:39.260","Text":"V divided by R. We know from the previous question that our V,"},{"Start":"05:39.260 ","End":"05:45.830","Text":"our velocity, is equal to our tangential acceleration multiplied by t,"},{"Start":"05:45.830 ","End":"05:51.645","Text":"and then we divide it by R. This is our Omega."},{"Start":"05:51.645 ","End":"05:54.690","Text":"Now, in order to find our Theta,"},{"Start":"05:54.690 ","End":"05:57.805","Text":"we\u0027re going to plug this in over here,"},{"Start":"05:57.805 ","End":"06:01.820","Text":"our Theta is going to be the integral of Omega,"},{"Start":"06:01.820 ","End":"06:06.900","Text":"which is a_t t divided by R dt."},{"Start":"06:07.400 ","End":"06:13.700","Text":"That is simply going to be equal to a_ t tangential acceleration"},{"Start":"06:13.700 ","End":"06:19.805","Text":"multiplied by t^2 divided by 2 plus some integrating constant."},{"Start":"06:19.805 ","End":"06:25.205","Text":"Now, if we define that our initial angle,"},{"Start":"06:25.205 ","End":"06:29.120","Text":"so Theta at t=0, is equal to 0,"},{"Start":"06:29.120 ","End":"06:33.420","Text":"if we use this over here,"},{"Start":"06:33.420 ","End":"06:36.890","Text":"then in this time over here,"},{"Start":"06:36.890 ","End":"06:39.490","Text":"t will be 0^2."},{"Start":"06:39.490 ","End":"06:42.205","Text":"This term over here will be equal to 0."},{"Start":"06:42.205 ","End":"06:46.040","Text":"Then we\u0027ll have that our Theta is equal to c,"},{"Start":"06:46.040 ","End":"06:47.315","Text":"which is equal to 0."},{"Start":"06:47.315 ","End":"06:51.940","Text":"Therefore, our c will be equal to 0 and we can cancel it out."},{"Start":"06:51.940 ","End":"06:54.855","Text":"Now we have our equation for Theta."},{"Start":"06:54.855 ","End":"07:03.780","Text":"Theta is equal to the tangential acceleration divided by 2 multiplied by t^2."},{"Start":"07:03.780 ","End":"07:07.430","Text":"Now what we want to do is we want to isolate out"},{"Start":"07:07.430 ","End":"07:13.305","Text":"our t and then sub it in to this equation over here."},{"Start":"07:13.305 ","End":"07:17.610","Text":"Let\u0027s scroll down to give us a little bit of space."},{"Start":"07:17.860 ","End":"07:22.270","Text":"What we\u0027ll get is that t^2,"},{"Start":"07:22.270 ","End":"07:26.720","Text":"let\u0027s leave it as t^2 because we can see that in this equation we have a t^2,"},{"Start":"07:26.720 ","End":"07:35.470","Text":"t^2 is going to be equal to 2 times Theta divided by a_t,"},{"Start":"07:35.470 ","End":"07:39.825","Text":"divided by the tangential acceleration."},{"Start":"07:39.825 ","End":"07:43.570","Text":"Now we can plug it into this equation over here."},{"Start":"07:43.570 ","End":"07:47.645","Text":"We have that our radial acceleration is equal to,"},{"Start":"07:47.645 ","End":"07:52.700","Text":"so let\u0027s square each item in the brackets individually."},{"Start":"07:52.700 ","End":"08:00.915","Text":"We\u0027ll have that our tangential acceleration squared divided by R multiplied by t squared."},{"Start":"08:00.915 ","End":"08:03.935","Text":"That\u0027s why I kept this over here as t squared."},{"Start":"08:03.935 ","End":"08:06.680","Text":"Obviously, I could have taken the square root and plugged it in and"},{"Start":"08:06.680 ","End":"08:09.320","Text":"I would have gotten the same answer."},{"Start":"08:09.320 ","End":"08:13.370","Text":"This is just a little shorthand for this specific example over here."},{"Start":"08:13.370 ","End":"08:20.390","Text":"Then we\u0027ll have our tangential acceleration squared divided by R multiplied by t squared,"},{"Start":"08:20.390 ","End":"08:30.185","Text":"which is equal to 2 times Theta divided by our tangential acceleration."},{"Start":"08:30.185 ","End":"08:32.357","Text":"Notice important, I made a mistake,"},{"Start":"08:32.357 ","End":"08:35.255","Text":"I forgot to carry my R through over here."},{"Start":"08:35.255 ","End":"08:38.355","Text":"There\u0027s an R over here from here."},{"Start":"08:38.355 ","End":"08:41.960","Text":"Because of my Omega, it\u0027s divided by R. Then that"},{"Start":"08:41.960 ","End":"08:46.130","Text":"means that here I\u0027m going to be multiplying by R,"},{"Start":"08:46.130 ","End":"08:49.065","Text":"because I\u0027m isolating out my t^2."},{"Start":"08:49.065 ","End":"08:51.390","Text":"I multiply my Theta by R,"},{"Start":"08:51.390 ","End":"08:54.410","Text":"and that means that also over here,"},{"Start":"08:54.410 ","End":"08:59.775","Text":"I\u0027m going to have an R. Now we can cancel out this R with this R,"},{"Start":"08:59.775 ","End":"09:04.275","Text":"and 1 a_t over here with this a_t down here."},{"Start":"09:04.275 ","End":"09:09.795","Text":"Then we\u0027ll therefore get that our radial acceleration is equal to"},{"Start":"09:09.795 ","End":"09:16.905","Text":"2 Theta multiplied by the tangential acceleration."},{"Start":"09:16.905 ","End":"09:19.960","Text":"This is our answer to question 2."},{"Start":"09:19.960 ","End":"09:25.690","Text":"This is our radial acceleration as a function of Theta."},{"Start":"09:25.690 ","End":"09:28.390","Text":"That\u0027s the end of this lesson."}],"ID":9801},{"Watched":false,"Name":"Exercise 2","Duration":"7m 7s","ChapterTopicVideoID":10026,"CourseChapterTopicPlaylistID":85364,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.440","Text":"Hello. In this question we\u0027re being told that the angular position of"},{"Start":"00:04.440 ","End":"00:09.495","Text":"a point on the edge of a spinning wheel is given by this equation over here."},{"Start":"00:09.495 ","End":"00:13.439","Text":"Now, the fact that this is the position on the edge of a spinning wheel isn\u0027t important."},{"Start":"00:13.439 ","End":"00:18.180","Text":"All that\u0027s important to realize is that we\u0027re dealing with circular motion."},{"Start":"00:18.180 ","End":"00:21.345","Text":"Question Number 1 is asking us,"},{"Start":"00:21.345 ","End":"00:28.815","Text":"what is the angular velocity at t is equal to 2 seconds and at t is equal to 4 seconds."},{"Start":"00:28.815 ","End":"00:36.490","Text":"We know that the angular velocity Omega is given by the time derivative of our angle."},{"Start":"00:36.490 ","End":"00:40.205","Text":"Similarly, we get that our linear velocity,"},{"Start":"00:40.205 ","End":"00:48.410","Text":"v is equal to our position let\u0027s say in the x-direction at the time derivative of it."},{"Start":"00:48.410 ","End":"00:51.170","Text":"This is analogous."},{"Start":"00:51.170 ","End":"00:57.125","Text":"What we can do is we can take the time derivative of our angle fee."},{"Start":"00:57.125 ","End":"01:00.170","Text":"The derivative of 5t is 5,"},{"Start":"01:00.170 ","End":"01:03.920","Text":"the derivative of 3t^2 is 6t,"},{"Start":"01:03.920 ","End":"01:11.940","Text":"and the derivative of negative 2t^3 is negative 6t^2."},{"Start":"01:13.280 ","End":"01:19.070","Text":"Now what we can do is we can simply plug"},{"Start":"01:19.070 ","End":"01:25.530","Text":"in when our t is equal to 2 and when our t is equal to 4."},{"Start":"01:25.530 ","End":"01:31.425","Text":"At t is equal to 2 we have 5 plus"},{"Start":"01:31.425 ","End":"01:38.860","Text":"6 times 2 minus 6 times 2^2."},{"Start":"01:39.020 ","End":"01:48.770","Text":"That is going to be equal to negative 7 radians per second."},{"Start":"01:48.770 ","End":"01:55.325","Text":"Now, this minus simply denotes that we\u0027re moving anticlockwise."},{"Start":"01:55.325 ","End":"01:59.240","Text":"Now similarly, at t is equal to 4,"},{"Start":"01:59.240 ","End":"02:09.335","Text":"so we\u0027ll have 5 plus 6 times 4 minus 6 times 4^2."},{"Start":"02:09.335 ","End":"02:18.665","Text":"This is simply going to be equal to negative 67 radians per second."},{"Start":"02:18.665 ","End":"02:22.385","Text":"Again, we have this minus sign denoting that we\u0027re going"},{"Start":"02:22.385 ","End":"02:27.360","Text":"against or in the anticlockwise direction."},{"Start":"02:27.360 ","End":"02:29.920","Text":"That\u0027s the end of question Number 1."},{"Start":"02:29.920 ","End":"02:33.850","Text":"In the next video we\u0027re going to be answering question Number 2."},{"Start":"02:33.850 ","End":"02:36.515","Text":"Question Number 2 is,"},{"Start":"02:36.515 ","End":"02:41.825","Text":"what is the average angular acceleration between these 2 times?"},{"Start":"02:41.825 ","End":"02:47.635","Text":"Where the 2 times is a t is equal to 2 and a t is equal to 4."},{"Start":"02:47.635 ","End":"02:52.160","Text":"Angular acceleration is denoted by the Greek letter Alpha."},{"Start":"02:52.160 ","End":"02:59.150","Text":"It is equal to the time derivative of our angular velocity Omega which is also"},{"Start":"02:59.150 ","End":"03:06.665","Text":"in turn equal to the second time derivative of our position Phi."},{"Start":"03:06.665 ","End":"03:09.740","Text":"This is if we\u0027re going to liken this to linear motion,"},{"Start":"03:09.740 ","End":"03:14.750","Text":"this is the same as our linear acceleration being equal to the time derivative of"},{"Start":"03:14.750 ","End":"03:22.005","Text":"our velocity which is in turn equal to the second time derivative of our position."},{"Start":"03:22.005 ","End":"03:24.080","Text":"Let\u0027s answer this question."},{"Start":"03:24.080 ","End":"03:29.005","Text":"We\u0027re being asked to find the average angular acceleration between these 2 times."},{"Start":"03:29.005 ","End":"03:32.390","Text":"We know that the average linear acceleration,"},{"Start":"03:32.390 ","End":"03:40.310","Text":"so denoted by this bar on top is equal to Delta v divided by Delta"},{"Start":"03:40.310 ","End":"03:49.860","Text":"t. That\u0027s the final velocity minus the initial velocity divided by the time difference."},{"Start":"03:49.860 ","End":"03:56.390","Text":"Similarly, we can write that the average angular acceleration,"},{"Start":"03:56.390 ","End":"04:00.620","Text":"so denoted by a bar on top because it\u0027s average is equal to"},{"Start":"04:00.620 ","End":"04:06.560","Text":"Delta Omega divided by Delta t. This is going to"},{"Start":"04:06.560 ","End":"04:11.000","Text":"be equal to our angular velocity at our final time which"},{"Start":"04:11.000 ","End":"04:16.520","Text":"is at t is equal to 4 minus our angular velocity"},{"Start":"04:16.520 ","End":"04:20.685","Text":"at our initial time which is at t is equal to"},{"Start":"04:20.685 ","End":"04:27.100","Text":"2 divided by the change in time which is 4 minus 2."},{"Start":"04:27.100 ","End":"04:29.505","Text":"Let\u0027s plug in these values."},{"Start":"04:29.505 ","End":"04:35.314","Text":"Our Omega t is equal to 4 is equal to negative 67 minus"},{"Start":"04:35.314 ","End":"04:43.370","Text":"our Omega t equals 2 which is equal to negative 7 divided by 4 minus 2 which is 2."},{"Start":"04:43.370 ","End":"04:51.665","Text":"Then this is going to be equal to 30 and then it\u0027s in units of acceleration"},{"Start":"04:51.665 ","End":"05:00.690","Text":"which is going to be radians per second squared and of course it\u0027s negative 30."},{"Start":"05:00.690 ","End":"05:03.750","Text":"This is the answer to question Number 2."},{"Start":"05:03.750 ","End":"05:08.750","Text":"In the next video we\u0027re going to be answering question Number 3."},{"Start":"05:09.130 ","End":"05:11.405","Text":"Question Number 3 is,"},{"Start":"05:11.405 ","End":"05:17.105","Text":"what is the instantaneous angular acceleration at these 2 times?"},{"Start":"05:17.105 ","End":"05:20.755","Text":"Our angular acceleration, Alpha,"},{"Start":"05:20.755 ","End":"05:27.320","Text":"as we said before is equal to the first derivative of our angular velocity."},{"Start":"05:27.320 ","End":"05:29.855","Text":"Just like in linear acceleration,"},{"Start":"05:29.855 ","End":"05:35.650","Text":"our acceleration is equal to the first derivative of our linear velocity."},{"Start":"05:35.650 ","End":"05:38.510","Text":"Therefore, going from this,"},{"Start":"05:38.510 ","End":"05:40.655","Text":"the first derivative of"},{"Start":"05:40.655 ","End":"05:48.065","Text":"our angular velocity is going to simply be to take the derivative of this."},{"Start":"05:48.065 ","End":"05:54.255","Text":"Our 5 will go to 0 and then our derivative of 60"},{"Start":"05:54.255 ","End":"06:01.740","Text":"is 6 and our derivative of 60^2 is going to be equal to negative 12t."},{"Start":"06:02.300 ","End":"06:05.810","Text":"Now this is going to be equal to,"},{"Start":"06:05.810 ","End":"06:08.900","Text":"so we\u0027re looking for the instantaneous angular acceleration"},{"Start":"06:08.900 ","End":"06:11.845","Text":"at the 2 times t is equal to 2."},{"Start":"06:11.845 ","End":"06:15.530","Text":"All we do is we substitute in 2 over here."},{"Start":"06:15.530 ","End":"06:21.914","Text":"We\u0027ll have that this is equal to 6 minus 12 times"},{"Start":"06:21.914 ","End":"06:28.715","Text":"2 which is simply going to be equal to negative 18."},{"Start":"06:28.715 ","End":"06:31.280","Text":"Of course it\u0027s units for acceleration,"},{"Start":"06:31.280 ","End":"06:36.955","Text":"so it\u0027s going to be radians per second squared."},{"Start":"06:36.955 ","End":"06:44.165","Text":"Now, for t is equal to 4 seconds we\u0027ll have that this is equal to 6 minus"},{"Start":"06:44.165 ","End":"06:51.695","Text":"12 times 4 which is simply going to be equal to negative 42."},{"Start":"06:51.695 ","End":"06:56.405","Text":"Again, the units for angular acceleration,"},{"Start":"06:56.405 ","End":"07:00.980","Text":"radians per second square."},{"Start":"07:00.980 ","End":"07:07.680","Text":"This is the answer to question Number 3 and now we\u0027ve finished this exercise."}],"ID":9802}],"Thumbnail":null,"ID":85364},{"Name":"End of Chapter Questions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"11m 1s","ChapterTopicVideoID":8989,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.405","Text":"Hello. In this exercise,"},{"Start":"00:03.405 ","End":"00:08.280","Text":"we\u0027re given a hoop of radius R with a bead threaded onto the hoop."},{"Start":"00:08.280 ","End":"00:13.440","Text":"The hoop is rotating along the axis over here,"},{"Start":"00:13.440 ","End":"00:15.630","Text":"this main thick line in the hoop,"},{"Start":"00:15.630 ","End":"00:20.170","Text":"and it\u0027s rotating in this direction."},{"Start":"00:20.210 ","End":"00:25.995","Text":"Now, as the hoop rotates at speed Omega,"},{"Start":"00:25.995 ","End":"00:31.215","Text":"the bead that\u0027s threaded on the hoop moves upwards until it reaches"},{"Start":"00:31.215 ","End":"00:36.360","Text":"an angle of Alpha between this dotted line and the hoop."},{"Start":"00:36.360 ","End":"00:39.450","Text":"The first question that we\u0027re being asked is,"},{"Start":"00:39.450 ","End":"00:42.420","Text":"what is the frequency of rotation?"},{"Start":"00:42.420 ","End":"00:46.265","Text":"Which means that we\u0027re being asked to find what Omega is."},{"Start":"00:46.265 ","End":"00:48.680","Text":"Let\u0027s begin answering this question."},{"Start":"00:48.680 ","End":"00:54.080","Text":"We\u0027re going to start by drawing a free body diagram for the bead."},{"Start":"00:54.080 ","End":"00:57.260","Text":"So this is the bead and as usual,"},{"Start":"00:57.260 ","End":"01:00.900","Text":"it has mg pointing downwards,"},{"Start":"01:00.900 ","End":"01:05.275","Text":"and then it also has some normal force,"},{"Start":"01:05.275 ","End":"01:10.370","Text":"which is obviously applied by the hoop because the bead is touching the hoop and"},{"Start":"01:10.370 ","End":"01:17.435","Text":"the normal force is at 90 degrees to the tangent of the hoop."},{"Start":"01:17.435 ","End":"01:20.750","Text":"Now of course, we don\u0027t know what this normal force is,"},{"Start":"01:20.750 ","End":"01:23.860","Text":"it\u0027s an unknown, so we\u0027ll put a tilde on top."},{"Start":"01:23.860 ","End":"01:26.450","Text":"Now let\u0027s redraw this diagram,"},{"Start":"01:26.450 ","End":"01:32.105","Text":"but we\u0027re going to split the normal force into its separate components."},{"Start":"01:32.105 ","End":"01:39.815","Text":"The mg is still pointing downwards and then we have this component of the normal force,"},{"Start":"01:39.815 ","End":"01:44.405","Text":"and we have this component of the normal force."},{"Start":"01:44.405 ","End":"01:48.290","Text":"Now, if we go back to the diagram of the hoop itself,"},{"Start":"01:48.290 ","End":"01:51.545","Text":"we can see that the angle Alpha is here,"},{"Start":"01:51.545 ","End":"01:56.885","Text":"so we have to move the angle so that we can work with it,"},{"Start":"01:56.885 ","End":"02:02.490","Text":"which works out that the angle Alpha is over here."},{"Start":"02:02.800 ","End":"02:08.960","Text":"Now we can split into the separate components and we can say that here is"},{"Start":"02:08.960 ","End":"02:16.985","Text":"N cosine of Alpha because it\u0027s the adjacent to the angle."},{"Start":"02:16.985 ","End":"02:24.905","Text":"Here, the force is N sine of Alpha because it\u0027s the opposite."},{"Start":"02:24.905 ","End":"02:26.300","Text":"Instead of drawing it here,"},{"Start":"02:26.300 ","End":"02:30.175","Text":"I could have also drawn it here if it makes it more obvious to you."},{"Start":"02:30.175 ","End":"02:34.370","Text":"Now what we\u0027re going to do is write down what all of our forces"},{"Start":"02:34.370 ","End":"02:38.465","Text":"are on each axis separately."},{"Start":"02:38.465 ","End":"02:44.470","Text":"First, we have the forces on the z-axis."},{"Start":"02:44.470 ","End":"02:50.750","Text":"It\u0027s sum of the forces on the Z and the"},{"Start":"02:50.750 ","End":"02:57.515","Text":"next is the sum of the forces on the radial axis."},{"Start":"02:57.515 ","End":"03:03.275","Text":"Now here it\u0027s a bit confusing because this is the radial axis,"},{"Start":"03:03.275 ","End":"03:09.830","Text":"not in the direction of this line, not that."},{"Start":"03:09.830 ","End":"03:12.005","Text":"It\u0027s this. This is the radius."},{"Start":"03:12.005 ","End":"03:17.870","Text":"Now we have on the z-axis, it\u0027s really easy."},{"Start":"03:17.870 ","End":"03:20.570","Text":"We have mg pointing downwards,"},{"Start":"03:20.570 ","End":"03:26.520","Text":"so write minus mg plus"},{"Start":"03:26.520 ","End":"03:32.879","Text":"N cosine of Alpha"},{"Start":"03:32.960 ","End":"03:38.230","Text":"and we know that from the equation F equals ma."},{"Start":"03:38.230 ","End":"03:43.750","Text":"We have ma on the z-axis."},{"Start":"03:43.750 ","End":"03:47.440","Text":"Now, if we look at the bead on the hoop,"},{"Start":"03:47.440 ","End":"03:50.875","Text":"we know that in the z-axis, it doesn\u0027t move."},{"Start":"03:50.875 ","End":"03:54.565","Text":"It stays exactly in the same place."},{"Start":"03:54.565 ","End":"04:03.970","Text":"We know that this equals 0 and now into the radial direction, the radial axis."},{"Start":"04:03.970 ","End":"04:08.780","Text":"The only thing we have is this N sine Alpha."},{"Start":"04:08.780 ","End":"04:19.350","Text":"Let\u0027s write it in and we again know that F equals ma on the radial axis and we"},{"Start":"04:19.350 ","End":"04:25.819","Text":"know that on the radial axis equals"},{"Start":"04:25.819 ","End":"04:34.280","Text":"Omega^2 R. Now let\u0027s rewrite the equations."},{"Start":"04:34.280 ","End":"04:36.110","Text":"Let\u0027s do it in a different color."},{"Start":"04:36.110 ","End":"04:46.320","Text":"We have that minus mg plus N cosine of Alpha equals 0"},{"Start":"04:46.320 ","End":"04:51.195","Text":"and we also have the N sine Alpha"},{"Start":"04:51.195 ","End":"04:58.410","Text":"equals Omega^2 R. Now let\u0027s look at all of our unknowns."},{"Start":"04:58.410 ","End":"05:00.920","Text":"M we\u0027re given in the question, we know g,"},{"Start":"05:00.920 ","End":"05:03.725","Text":"we know N we don\u0027t know,"},{"Start":"05:03.725 ","End":"05:06.600","Text":"so we mark it with a tilde,"},{"Start":"05:06.600 ","End":"05:08.238","Text":"Alpha, we know,"},{"Start":"05:08.238 ","End":"05:12.470","Text":"it\u0027s given in the question and we\u0027ve already said that we don\u0027t know Alpha,"},{"Start":"05:12.470 ","End":"05:16.345","Text":"we know Omega we don\u0027t know,"},{"Start":"05:16.345 ","End":"05:21.660","Text":"so mark it and also R we don\u0027t know because it\u0027s not given."},{"Start":"05:21.660 ","End":"05:24.720","Text":"Currently we have 2 equations,"},{"Start":"05:24.720 ","End":"05:27.440","Text":"but 3 unknowns, which as we know,"},{"Start":"05:27.440 ","End":"05:28.580","Text":"if we have 3 unknowns,"},{"Start":"05:28.580 ","End":"05:30.605","Text":"then we have to have 3 equations."},{"Start":"05:30.605 ","End":"05:37.570","Text":"We\u0027re going to use trigonometry in order to find what the 3rd equation will be."},{"Start":"05:37.570 ","End":"05:39.950","Text":"Let\u0027s take a look at this."},{"Start":"05:39.950 ","End":"05:44.465","Text":"This red arrow is a i,"},{"Start":"05:44.465 ","End":"05:46.625","Text":"and here if we look,"},{"Start":"05:46.625 ","End":"05:49.525","Text":"we have a triangle."},{"Start":"05:49.525 ","End":"05:52.490","Text":"Let\u0027s draw this triangle over here."},{"Start":"05:52.490 ","End":"05:56.990","Text":"We have this, we have this line and this line."},{"Start":"05:56.990 ","End":"05:59.950","Text":"We have here that this is Alpha."},{"Start":"05:59.950 ","End":"06:06.510","Text":"This is i the radial axis and"},{"Start":"06:06.510 ","End":"06:14.655","Text":"here we know that this is this line and we know that this is I,"},{"Start":"06:14.655 ","End":"06:16.840","Text":"the radius of the hoop."},{"Start":"06:16.840 ","End":"06:24.545","Text":"Then we can write our 3rd equation as sine of Alpha,"},{"Start":"06:24.545 ","End":"06:25.760","Text":"which as we know,"},{"Start":"06:25.760 ","End":"06:30.050","Text":"is opposite over hypotenuse."},{"Start":"06:30.050 ","End":"06:38.625","Text":"The opposite is i and the hypotenuse is the I."},{"Start":"06:38.625 ","End":"06:41.115","Text":"This is our 3rd equation."},{"Start":"06:41.115 ","End":"06:45.965","Text":"Now, in order to find what Omega is,"},{"Start":"06:45.965 ","End":"06:47.735","Text":"the frequency of rotation,"},{"Start":"06:47.735 ","End":"06:51.170","Text":"we\u0027re just going to rearrange all of the equations that we have"},{"Start":"06:51.170 ","End":"06:55.442","Text":"here and sub in for the unknowns."},{"Start":"06:55.442 ","End":"06:58.805","Text":"One of the unknowns is the normal force,"},{"Start":"06:58.805 ","End":"07:04.685","Text":"so we\u0027re going to going to rearrange it to N cosine of Alpha equals mg."},{"Start":"07:04.685 ","End":"07:08.255","Text":"I just moved mg to the other side of the equal sign,"},{"Start":"07:08.255 ","End":"07:21.235","Text":"which means that N"},{"Start":"07:21.235 ","End":"07:26.025","Text":"equals mg over cosine of Alpha."},{"Start":"07:26.025 ","End":"07:33.770","Text":"Then we can sub that into this equation,"},{"Start":"07:33.770 ","End":"07:37.950","Text":"which will mean that instead of m sine Alpha,"},{"Start":"07:37.950 ","End":"07:44.180","Text":"we have mg over cosine Alpha multiplied by sine Alpha,"},{"Start":"07:44.180 ","End":"07:50.210","Text":"but we know that sine Alpha is r over R,"},{"Start":"07:50.210 ","End":"07:54.540","Text":"which equals Omega^2 i."},{"Start":"07:54.540 ","End":"07:58.765","Text":"Now we can cross out both i\u0027s from either side of the equation."},{"Start":"07:58.765 ","End":"08:02.780","Text":"Then we just have mg over"},{"Start":"08:02.780 ","End":"08:10.500","Text":"R cosine Alpha equals Omega^2,"},{"Start":"08:10.500 ","End":"08:17.515","Text":"so we\u0027ll square root both sides and then we have that Omega."},{"Start":"08:17.515 ","End":"08:26.735","Text":"The frequency of rotation is the square root of mg divided by R cosine of Alpha."},{"Start":"08:26.735 ","End":"08:29.310","Text":"Now I\u0027m just going to rub out all of"},{"Start":"08:29.310 ","End":"08:37.825","Text":"the equations to make some room in order to answer the second question."},{"Start":"08:37.825 ","End":"08:40.315","Text":"Now, in the second question,"},{"Start":"08:40.315 ","End":"08:43.915","Text":"we\u0027re being asked what should be the frequency of rotation in"},{"Start":"08:43.915 ","End":"08:48.510","Text":"order for the angle Alpha to be at 90 degrees."},{"Start":"08:48.510 ","End":"08:58.885","Text":"What they\u0027re asking from us is to find Alpha as a function of Omega."},{"Start":"08:58.885 ","End":"09:05.250","Text":"What they\u0027re asking us is how Alpha will change as Omega changes and they\u0027re asking"},{"Start":"09:05.250 ","End":"09:12.150","Text":"us what Omega should be in order for Alpha to be equal to 90 degrees?"},{"Start":"09:12.150 ","End":"09:18.385","Text":"What we\u0027re going to do in order to answer this question is to isolate cosine of Alpha."},{"Start":"09:18.385 ","End":"09:21.685","Text":"All we have to do is from the previous question,"},{"Start":"09:21.685 ","End":"09:27.160","Text":"just rearrange the equation in order to have cosine Alpha isolated."},{"Start":"09:27.160 ","End":"09:32.830","Text":"What we\u0027re going to do is we\u0027re going to square both sides to give us mg"},{"Start":"09:32.830 ","End":"09:39.433","Text":"over r cosine of Alpha equals Omega^2,"},{"Start":"09:39.433 ","End":"09:42.830","Text":"and then we\u0027re going to divide both sides by Omega^2"},{"Start":"09:42.830 ","End":"09:47.180","Text":"and we\u0027re going to multiply both sides by cosine of Alpha,"},{"Start":"09:47.180 ","End":"09:53.780","Text":"which will give us that cosine of Alpha equals mg over,"},{"Start":"09:53.780 ","End":"10:00.220","Text":"sorry, that\u0027s a R, over R Omega^2."},{"Start":"10:00.220 ","End":"10:08.525","Text":"Now we\u0027re being asked what Omega^2 needs to be in order for Alpha to be at 90 degrees."},{"Start":"10:08.525 ","End":"10:13.145","Text":"Now, if Alpha is at 90 degrees,"},{"Start":"10:13.145 ","End":"10:17.075","Text":"then cosine of Alpha will have to equal 0."},{"Start":"10:17.075 ","End":"10:21.590","Text":"Now because mg and R are all constants,"},{"Start":"10:21.590 ","End":"10:26.690","Text":"it means that in order for it to equal this side of the equation,"},{"Start":"10:26.690 ","End":"10:28.700","Text":"for it to equal 0,"},{"Start":"10:28.700 ","End":"10:33.995","Text":"it means that Omega^2 has to be equal to infinity."},{"Start":"10:33.995 ","End":"10:35.390","Text":"Now, if you think about it,"},{"Start":"10:35.390 ","End":"10:36.890","Text":"this actually makes sense."},{"Start":"10:36.890 ","End":"10:42.680","Text":"Why? Because in order to lift this bead to be at 90 degrees,"},{"Start":"10:42.680 ","End":"10:45.955","Text":"to be perpendicular to the z-axis,"},{"Start":"10:45.955 ","End":"10:51.230","Text":"it means that we have to spin the hoop around so fast,"},{"Start":"10:51.230 ","End":"10:55.970","Text":"which means so many times in order for that to happen that it"},{"Start":"10:55.970 ","End":"11:02.100","Text":"turns out that Omega will have to be equal to infinity and that\u0027s it."}],"ID":9259},{"Watched":false,"Name":"Exercise 2","Duration":"8m 46s","ChapterTopicVideoID":8990,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.360","Text":"In this question, we\u0027re being asked what the distance is of the mass from the main pole?"},{"Start":"00:07.670 ","End":"00:16.170","Text":"They\u0027re talking about this distance and let\u0027s call it r. In the 2nd question,"},{"Start":"00:16.170 ","End":"00:19.920","Text":"we\u0027re being asked, how will the height of the mass be"},{"Start":"00:19.920 ","End":"00:24.630","Text":"affected if the length of the string d were to be doubled."},{"Start":"00:24.630 ","End":"00:34.770","Text":"In other words, this is the height and if the length d were to be doubled,"},{"Start":"00:34.770 ","End":"00:38.590","Text":"how would this height be affected?"},{"Start":"00:38.750 ","End":"00:41.075","Text":"Let\u0027s see how we do this."},{"Start":"00:41.075 ","End":"00:43.640","Text":"Let\u0027s start off with the first question."},{"Start":"00:43.640 ","End":"00:47.345","Text":"If we look at drawing the free body diagram,"},{"Start":"00:47.345 ","End":"00:50.645","Text":"so if this is the mass,"},{"Start":"00:50.645 ","End":"00:56.465","Text":"then we have some force T from the rope going in this direction."},{"Start":"00:56.465 ","End":"01:00.589","Text":"I\u0027ll draw a tilde on top because it\u0027s unknown and going downwards,"},{"Start":"01:00.589 ","End":"01:05.275","Text":"we have mg. Now,"},{"Start":"01:05.275 ","End":"01:09.050","Text":"because the mass isn\u0027t moving in any direction,"},{"Start":"01:09.050 ","End":"01:11.435","Text":"is just spinning around in a circle,"},{"Start":"01:11.435 ","End":"01:14.270","Text":"it\u0027s not being pulled to left, right, up,"},{"Start":"01:14.270 ","End":"01:20.445","Text":"or down so it has acceleration to be precise radial acceleration,"},{"Start":"01:20.445 ","End":"01:24.995","Text":"which as we know, is Omega squared."},{"Start":"01:24.995 ","End":"01:30.840","Text":"Omega is the angular velocity multiplied by r. Now,"},{"Start":"01:30.840 ","End":"01:38.250","Text":"this equation is very important for you to know in general."},{"Start":"01:38.690 ","End":"01:45.470","Text":"Now, let\u0027s talk about all of the forces working in the radial direction."},{"Start":"01:45.470 ","End":"01:50.854","Text":"What we can do is we can split up the T into"},{"Start":"01:50.854 ","End":"01:58.560","Text":"its separate components and call this T_z because it\u0027s on the z-axis and this,"},{"Start":"01:58.560 ","End":"02:02.760","Text":"we can call it T_r because it\u0027s in the radial axis."},{"Start":"02:02.760 ","End":"02:08.880","Text":"Let\u0027s see what T_r equals."},{"Start":"02:08.880 ","End":"02:11.880","Text":"T_r equals t,"},{"Start":"02:11.880 ","End":"02:19.390","Text":"the unknown that we have multiplied by cosine of Alpha."},{"Start":"02:19.390 ","End":"02:21.455","Text":"Let\u0027s call this angle Alpha."},{"Start":"02:21.455 ","End":"02:31.135","Text":"This is also an unknown and so this component is t cosine of Alpha."},{"Start":"02:31.135 ","End":"02:36.275","Text":"As we know that the sum of all the forces is equal to,"},{"Start":"02:36.275 ","End":"02:38.570","Text":"hold on. Let\u0027s just draw."},{"Start":"02:38.570 ","End":"02:42.125","Text":"This is like saying,"},{"Start":"02:42.125 ","End":"02:48.185","Text":"the sum of all the forces in the radial direction is equal to,"},{"Start":"02:48.185 ","End":"02:54.205","Text":"mass times acceleration ma."},{"Start":"02:54.205 ","End":"03:01.520","Text":"But here, our mass is m and our a is what\u0027s our radial acceleration,"},{"Start":"03:01.520 ","End":"03:11.745","Text":"which is written above Omega squared times r. Let\u0027s label who our unknowns are."},{"Start":"03:11.745 ","End":"03:13.320","Text":"We have Alpha. Also,"},{"Start":"03:13.320 ","End":"03:15.995","Text":"unknown T is already been marked as unknown,"},{"Start":"03:15.995 ","End":"03:18.350","Text":"and also our r is unknown."},{"Start":"03:18.350 ","End":"03:23.550","Text":"This is another equation that we have."},{"Start":"03:24.530 ","End":"03:31.925","Text":"As we can see, we have 3 unknowns in this equation and only 1 equation,"},{"Start":"03:31.925 ","End":"03:36.340","Text":"which means that I need at least another 2 equations."},{"Start":"03:36.340 ","End":"03:40.200","Text":"Let\u0027s now take a look at what our 2nd equation is going to be."},{"Start":"03:40.200 ","End":"03:42.750","Text":"Our 2nd equation involves T_z,"},{"Start":"03:42.750 ","End":"03:53.460","Text":"T on the z-axis and we\u0027re going to take the t component over here."},{"Start":"03:53.650 ","End":"03:58.470","Text":"We have t sine of Alpha,"},{"Start":"03:58.470 ","End":"04:02.970","Text":"so T_z equals T sine of"},{"Start":"04:02.970 ","End":"04:09.035","Text":"Alpha and because the sum of all the forces on the z-axis is equal to 0,"},{"Start":"04:09.035 ","End":"04:15.040","Text":"we can say that T sine of Alpha is equal to mg."},{"Start":"04:15.040 ","End":"04:19.740","Text":"This is 2nd equation."},{"Start":"04:19.740 ","End":"04:21.750","Text":"My next equation,"},{"Start":"04:21.750 ","End":"04:25.050","Text":"I\u0027m going to take from basic trigonometry."},{"Start":"04:25.050 ","End":"04:32.315","Text":"What do we do? We can say that the angle here is equal to Alpha."},{"Start":"04:32.315 ","End":"04:34.370","Text":"Now, of course, I don\u0027t know what Alpha is."},{"Start":"04:34.370 ","End":"04:35.915","Text":"It\u0027s the same Alpha,"},{"Start":"04:35.915 ","End":"04:44.690","Text":"but I know that the cosine of an angle is adjacent over hypotenuse."},{"Start":"04:44.690 ","End":"04:48.500","Text":"My adjacent angle is r,"},{"Start":"04:48.500 ","End":"04:58.900","Text":"this is my edge angle and the hypotenuse is d. I can say that"},{"Start":"04:59.330 ","End":"05:06.695","Text":"the cosine of Alpha equals adjacent r over"},{"Start":"05:06.695 ","End":"05:14.275","Text":"hypotenuse d. This is going to be our next equation."},{"Start":"05:14.275 ","End":"05:18.185","Text":"Here, we\u0027ve practically finished the first question."},{"Start":"05:18.185 ","End":"05:20.330","Text":"All you have to do is because they\u0027re"},{"Start":"05:20.330 ","End":"05:23.105","Text":"asking what the distance of the masses from the main pole,"},{"Start":"05:23.105 ","End":"05:27.035","Text":"we use all of these equations and isolate out"},{"Start":"05:27.035 ","End":"05:31.375","Text":"what r is because that\u0027s the distance and there you have your answer."},{"Start":"05:31.375 ","End":"05:35.490","Text":"Now, we\u0027re going to move on to the next question, the 2nd one."},{"Start":"05:35.490 ","End":"05:37.820","Text":"I\u0027m going to make all of these equations slightly"},{"Start":"05:37.820 ","End":"05:41.740","Text":"smaller and move them a bit so that we have slightly more room."},{"Start":"05:41.740 ","End":"05:45.185","Text":"Now, in the 2nd question, we\u0027re being asked,"},{"Start":"05:45.185 ","End":"05:47.570","Text":"how will the height of the mass be affected if"},{"Start":"05:47.570 ","End":"05:51.120","Text":"the length of the string d were to be doubled?"},{"Start":"05:51.170 ","End":"05:54.630","Text":"If we double this,"},{"Start":"05:54.630 ","End":"05:56.700","Text":"how will this change?"},{"Start":"05:56.700 ","End":"06:00.950","Text":"What we have to do is we have to find out what h is,"},{"Start":"06:00.950 ","End":"06:04.850","Text":"and then we have to find out what d is and what its effect is,"},{"Start":"06:04.850 ","End":"06:07.235","Text":"what its relationship is to each."},{"Start":"06:07.235 ","End":"06:09.550","Text":"Let\u0027s see how we do this."},{"Start":"06:09.550 ","End":"06:15.020","Text":"We\u0027re going to start by trying to find out what h is."},{"Start":"06:15.020 ","End":"06:19.925","Text":"Now, h equals d sine of Alpha."},{"Start":"06:19.925 ","End":"06:24.725","Text":"As we can see in the diagram here,"},{"Start":"06:24.725 ","End":"06:31.735","Text":"the y component or the z component for d is d sine of Alpha."},{"Start":"06:31.735 ","End":"06:35.400","Text":"Now, we don\u0027t know what sine of Alpha is."},{"Start":"06:35.400 ","End":"06:39.530","Text":"However, if we look at our equation over here for T_z,"},{"Start":"06:39.530 ","End":"06:43.175","Text":"we can see that T sine of Alpha is equal to mg,"},{"Start":"06:43.175 ","End":"06:49.740","Text":"which means that sine of Alpha is equal to mg over T. We can write here,"},{"Start":"06:49.740 ","End":"06:54.795","Text":"it\u0027s dmg over T,"},{"Start":"06:54.795 ","End":"06:58.040","Text":"but we also don\u0027t know what T is."},{"Start":"06:58.040 ","End":"07:03.980","Text":"However, if we look in our equation for T_r over here,"},{"Start":"07:03.980 ","End":"07:09.725","Text":"we can see that T cosine of Alpha is m Omega squared times r,"},{"Start":"07:09.725 ","End":"07:17.435","Text":"which means that, T is equal to m Omega squared times r divided by cosine of Alpha."},{"Start":"07:17.435 ","End":"07:18.895","Text":"Let\u0027s write that."},{"Start":"07:18.895 ","End":"07:23.415","Text":"T equals m Omega squared r,"},{"Start":"07:23.415 ","End":"07:27.095","Text":"divided by cosine of Alpha."},{"Start":"07:27.095 ","End":"07:31.355","Text":"But we also don\u0027t know what cosine of Alpha is."},{"Start":"07:31.355 ","End":"07:34.204","Text":"However, if we look at this equation,"},{"Start":"07:34.204 ","End":"07:41.390","Text":"we know that cosine of Alpha is r over d. This becomes T equals m Omega"},{"Start":"07:41.390 ","End":"07:49.225","Text":"squared r over r over d. Then we can cancel out the rs,"},{"Start":"07:49.225 ","End":"07:56.840","Text":"which means that T equals m Omega squared d."},{"Start":"07:58.740 ","End":"08:04.060","Text":"We have here h equals"},{"Start":"08:04.060 ","End":"08:13.770","Text":"dmg over m Omega squared d. Now,"},{"Start":"08:13.770 ","End":"08:15.825","Text":"this d can cross out, this d you can cross out,"},{"Start":"08:15.825 ","End":"08:17.340","Text":"m and m crosses out,"},{"Start":"08:17.340 ","End":"08:22.965","Text":"which means that h equals g over Omega squared."},{"Start":"08:22.965 ","End":"08:25.750","Text":"As we can see, h,"},{"Start":"08:25.750 ","End":"08:32.240","Text":"the starting height is only dependent on g divided by Omega squared,"},{"Start":"08:32.240 ","End":"08:33.875","Text":"which means that,"},{"Start":"08:33.875 ","End":"08:35.705","Text":"if we double d,"},{"Start":"08:35.705 ","End":"08:39.150","Text":"because there\u0027s no relation between d and h,"},{"Start":"08:39.150 ","End":"08:41.565","Text":"h will remain exactly the same."},{"Start":"08:41.565 ","End":"08:44.070","Text":"That\u0027s the answer to that question and here,"},{"Start":"08:44.070 ","End":"08:47.170","Text":"we\u0027ve finished the exercise."}],"ID":9260},{"Watched":false,"Name":"Exercise 3","Duration":"11m 11s","ChapterTopicVideoID":8991,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"In the diagram below,"},{"Start":"00:02.055 ","End":"00:08.220","Text":"we\u0027re given three masses which are rotating around an axis while attached to rods."},{"Start":"00:08.220 ","End":"00:13.935","Text":"What we have is we have one rod,"},{"Start":"00:13.935 ","End":"00:18.915","Text":"a second rod, and masses attached to these rods."},{"Start":"00:18.915 ","End":"00:24.050","Text":"There\u0027s a central pole and around the central pole is the third mass."},{"Start":"00:24.050 ","End":"00:27.470","Text":"The third mass can move up and down depending on"},{"Start":"00:27.470 ","End":"00:30.980","Text":"the speed at which the system is rotating."},{"Start":"00:30.980 ","End":"00:37.115","Text":"Meaning that the faster the system rotates M will move upwards."},{"Start":"00:37.115 ","End":"00:40.130","Text":"However, if the system slows down,"},{"Start":"00:40.130 ","End":"00:43.790","Text":"then M will move downwards because of centrifugal force."},{"Start":"00:43.790 ","End":"00:49.305","Text":"We are told that the rods and the pole in the middle have no mass."},{"Start":"00:49.305 ","End":"00:52.400","Text":"We\u0027re also given the small mass m,"},{"Start":"00:52.400 ","End":"00:55.355","Text":"we\u0027re given the large mass here in the middle,"},{"Start":"00:55.355 ","End":"00:59.885","Text":"which is M. This mass is also M. We\u0027re given d,"},{"Start":"00:59.885 ","End":"01:02.655","Text":"which is this length,"},{"Start":"01:02.655 ","End":"01:06.075","Text":"d and this is also d,"},{"Start":"01:06.075 ","End":"01:11.760","Text":"and the current length of D, which is this."},{"Start":"01:11.760 ","End":"01:16.355","Text":"Webbing asked in this question to find the angular velocity Omega."},{"Start":"01:16.355 ","End":"01:18.930","Text":"Let\u0027s begin answering this question."},{"Start":"01:18.930 ","End":"01:23.160","Text":"Of course, the first thing that we\u0027re going to do is draw a free body diagram."},{"Start":"01:23.160 ","End":"01:25.395","Text":"This is mass m,"},{"Start":"01:25.395 ","End":"01:28.970","Text":"both of the masses m obviously equal to each other."},{"Start":"01:28.970 ","End":"01:35.080","Text":"There\u0027s some T_1, the tension, here."},{"Start":"01:35.080 ","End":"01:38.345","Text":"Then there\u0027s in this direction,"},{"Start":"01:38.345 ","End":"01:44.755","Text":"some T_2, which is the tangent here, T_1, T_2."},{"Start":"01:44.755 ","End":"01:51.020","Text":"Then we have also mg pointing downwards."},{"Start":"01:51.020 ","End":"01:55.895","Text":"Now, as in the video with the two masses,"},{"Start":"01:55.895 ","End":"01:59.420","Text":"if you don\u0027t understand why this angle is Alpha and also"},{"Start":"01:59.420 ","End":"02:03.145","Text":"this angle is Alpha please watch that because I explain it there."},{"Start":"02:03.145 ","End":"02:08.645","Text":"Now we have to spill up the T_1 and the T_2 into their separate components."},{"Start":"02:08.645 ","End":"02:13.115","Text":"Now let\u0027s draw this system when it\u0027s split up into its separate components."},{"Start":"02:13.115 ","End":"02:21.535","Text":"Here\u0027s the mass and then this pointing upwards is T_1 sine of Alpha."},{"Start":"02:21.535 ","End":"02:27.480","Text":"Pointing downwards, we have T_2 sine of Alpha,"},{"Start":"02:27.480 ","End":"02:34.925","Text":"and also we have mg. Then we have two forces pointing in the leftwards direction,"},{"Start":"02:34.925 ","End":"02:42.655","Text":"which is T_1 cosine of Alpha and T_2 cosine of Alpha."},{"Start":"02:42.655 ","End":"02:44.840","Text":"Now, it\u0027s important to know,"},{"Start":"02:44.840 ","End":"02:46.700","Text":"we don\u0027t know what Alpha is,"},{"Start":"02:46.700 ","End":"02:53.675","Text":"so I\u0027ll put a tilde on top and we also don\u0027t know what Omega is so that\u0027s another tilde."},{"Start":"02:53.675 ","End":"02:57.470","Text":"Now we can begin writing down the separate equations."},{"Start":"02:57.470 ","End":"03:05.100","Text":"I\u0027m going to split this up into the sum of all the forces on the z-axis."},{"Start":"03:06.290 ","End":"03:14.075","Text":"This is the z-axis and then the sum of all the forces on the radial axis,"},{"Start":"03:14.075 ","End":"03:18.050","Text":"not going to call it x I\u0027m going to call it the radial axis because it\u0027s going in"},{"Start":"03:18.050 ","End":"03:22.900","Text":"a radial direction which also reminds me that this is circular motion."},{"Start":"03:22.900 ","End":"03:26.740","Text":"Let\u0027s begin by doing on the z-axis."},{"Start":"03:26.740 ","End":"03:30.725","Text":"We have in the positive direction,"},{"Start":"03:30.725 ","End":"03:38.645","Text":"we have T_1 sine of Alpha because this is the positive direction going in this way."},{"Start":"03:38.645 ","End":"03:46.215","Text":"Then we have minus T_2 sine of Alpha and minus mg,"},{"Start":"03:46.215 ","End":"03:49.160","Text":"because these are in the negative direction."},{"Start":"03:49.160 ","End":"03:53.390","Text":"This equals to the sum of all the accelerations,"},{"Start":"03:53.390 ","End":"03:57.110","Text":"which is mass times the acceleration on the z-axis."},{"Start":"03:57.110 ","End":"03:59.840","Text":"Now because there is no acceleration on the z-axis,"},{"Start":"03:59.840 ","End":"04:02.465","Text":"we can say that az is equal to 0,"},{"Start":"04:02.465 ","End":"04:06.170","Text":"which means that this whole equation is equal to 0."},{"Start":"04:06.170 ","End":"04:10.400","Text":"Now let\u0027s take a look at the sum of all the forces and the radial axis."},{"Start":"04:10.400 ","End":"04:15.185","Text":"We\u0027re going to make that this direction is the negative direction."},{"Start":"04:15.185 ","End":"04:19.940","Text":"Why? Because it\u0027s going inwards into the center,"},{"Start":"04:19.940 ","End":"04:22.895","Text":"which we\u0027re going to call the negative direction."},{"Start":"04:22.895 ","End":"04:28.170","Text":"We\u0027re going to have minus cosine of Alpha."},{"Start":"04:28.170 ","End":"04:34.145","Text":"I\u0027m just putting all the common factors outside of the brackets and then we have"},{"Start":"04:34.145 ","End":"04:39.230","Text":"T_1 plus T_2 and"},{"Start":"04:39.230 ","End":"04:45.080","Text":"this equals mass times acceleration in the radial direction."},{"Start":"04:45.080 ","End":"04:49.340","Text":"We know that the acceleration in the radial direction is equal to,"},{"Start":"04:49.340 ","End":"04:52.895","Text":"in this case minus Omega^2r,"},{"Start":"04:52.895 ","End":"04:55.370","Text":"r we also don\u0027t know."},{"Start":"04:55.370 ","End":"05:00.605","Text":"Now let\u0027s write down the equations in a slightly more readable way."},{"Start":"05:00.605 ","End":"05:03.455","Text":"We have on the z-axis,"},{"Start":"05:03.455 ","End":"05:08.685","Text":"we just have that T_1 sine of Alpha is"},{"Start":"05:08.685 ","End":"05:14.745","Text":"equal to mg plus T_2 sine of Alpha."},{"Start":"05:14.745 ","End":"05:21.200","Text":"I just moved the minus factors to the other side of the equal sign."},{"Start":"05:21.200 ","End":"05:23.600","Text":"Then on the radial axis,"},{"Start":"05:23.600 ","End":"05:27.135","Text":"we have a minus cosine of"},{"Start":"05:27.135 ","End":"05:35.810","Text":"Alpha T1 plus T2 equals negative m Omega^2r."},{"Start":"05:35.810 ","End":"05:39.605","Text":"Now, of course, let\u0027s label our unknowns."},{"Start":"05:39.605 ","End":"05:44.960","Text":"T_1 is unknown, T_2 is unknown, Alphas is unknown."},{"Start":"05:44.960 ","End":"05:46.580","Text":"In the second equation,"},{"Start":"05:46.580 ","End":"05:52.830","Text":"we also have that Omega is unknown and r is unknown."},{"Start":"05:52.830 ","End":"05:54.825","Text":"We have two equations,"},{"Start":"05:54.825 ","End":"05:56.555","Text":"but we have five unknowns,"},{"Start":"05:56.555 ","End":"06:01.435","Text":"which means that we need to find another three equations in order to solve this question."},{"Start":"06:01.435 ","End":"06:05.285","Text":"Let\u0027s see where I can get the other equations from."},{"Start":"06:05.285 ","End":"06:08.075","Text":"Now up until now,"},{"Start":"06:08.075 ","End":"06:11.390","Text":"I\u0027ve been working on the force diagram for this mass."},{"Start":"06:11.390 ","End":"06:15.395","Text":"Now, if I drew another force diagram and equations for this mass,"},{"Start":"06:15.395 ","End":"06:18.579","Text":"I\u0027ll get the exact same thing because it\u0027s symmetrical."},{"Start":"06:18.579 ","End":"06:23.570","Text":"I\u0027m going to now start working on this mass and finding all the equations."},{"Start":"06:23.570 ","End":"06:26.930","Text":"Now I am going to rub out all the writing in blue in order to"},{"Start":"06:26.930 ","End":"06:31.350","Text":"make a little bit more room and to make the screen a little bit clear."},{"Start":"06:31.610 ","End":"06:41.000","Text":"Let\u0027s begin now by drawing the free body diagram for the mass M. Here\u0027s"},{"Start":"06:41.000 ","End":"06:46.050","Text":"our mass M and then we have going in"},{"Start":"06:46.050 ","End":"06:52.830","Text":"this direction T_2 and then we also have going in this direction T2."},{"Start":"06:52.830 ","End":"06:55.700","Text":"Remember, because this whole system is symmetrical around"},{"Start":"06:55.700 ","End":"06:59.750","Text":"the z-axis and going down, of course,"},{"Start":"06:59.750 ","End":"07:04.570","Text":"we have Mg. Now,"},{"Start":"07:04.570 ","End":"07:08.360","Text":"you\u0027ll notice that if I draw here,"},{"Start":"07:08.360 ","End":"07:11.225","Text":"this angle here is Alpha,"},{"Start":"07:11.225 ","End":"07:13.355","Text":"similar to the previous diagrams."},{"Start":"07:13.355 ","End":"07:17.810","Text":"You can go back in the video if you want to take a look at it and why that is."},{"Start":"07:17.810 ","End":"07:23.370","Text":"But I\u0027ll quickly explain it here just in case to save some time."},{"Start":"07:23.830 ","End":"07:27.605","Text":"Because this is D,"},{"Start":"07:27.605 ","End":"07:30.710","Text":"and this is D. If we split it in half,"},{"Start":"07:30.710 ","End":"07:33.755","Text":"we know that this is Alpha and this is Alpha."},{"Start":"07:33.755 ","End":"07:36.440","Text":"When we\u0027re looking at this M,"},{"Start":"07:36.440 ","End":"07:40.775","Text":"we\u0027re looking at this section and then this is Alpha,"},{"Start":"07:40.775 ","End":"07:45.450","Text":"which is exactly the same as this."},{"Start":"07:45.450 ","End":"07:51.175","Text":"Now, let\u0027s split this up into the separate components."},{"Start":"07:51.175 ","End":"07:59.020","Text":"Here is our mass M. Going in the upwards direction we have T_2 sine Alpha."},{"Start":"07:59.020 ","End":"08:05.000","Text":"Now because we have here T_2 and here T_2 so it means we have 2T_2s."},{"Start":"08:05.000 ","End":"08:09.240","Text":"We have 2T_2 sine of Alpha."},{"Start":"08:09.240 ","End":"08:15.660","Text":"In the downwards direction of course we have mg. Now going left and right,"},{"Start":"08:15.660 ","End":"08:17.045","Text":"I\u0027m not going to write anything."},{"Start":"08:17.045 ","End":"08:19.640","Text":"Why? Because as you\u0027ll notice,"},{"Start":"08:19.640 ","End":"08:22.429","Text":"going in the rightward direction,"},{"Start":"08:22.429 ","End":"08:25.130","Text":"I\u0027d have T_2 cosine of Alpha,"},{"Start":"08:25.130 ","End":"08:30.080","Text":"but also going in the leftward direction I\u0027d have T_2 cosine of Alpha,"},{"Start":"08:30.080 ","End":"08:35.110","Text":"which means that these two forces will balance each other out and cancel."},{"Start":"08:35.110 ","End":"08:38.150","Text":"Now, because in this question,"},{"Start":"08:38.150 ","End":"08:40.220","Text":"we know that m currently in"},{"Start":"08:40.220 ","End":"08:43.415","Text":"the current state isn\u0027t moving up and isn\u0027t moving down, sorry,"},{"Start":"08:43.415 ","End":"08:50.390","Text":"M. It means that the acceleration in the z-axis is equal to 0,"},{"Start":"08:50.390 ","End":"08:54.030","Text":"which means that the equation will equal to 0."},{"Start":"08:54.030 ","End":"08:57.785","Text":"That means that if I just move things along on the equal side,"},{"Start":"08:57.785 ","End":"09:03.650","Text":"I can say that the sum of all the forces on the z-axis is equal"},{"Start":"09:03.650 ","End":"09:12.695","Text":"to 2T_2 sine of Alpha minus mg equals 0,"},{"Start":"09:12.695 ","End":"09:22.700","Text":"which means that 2T_2 sine of Alpha is equal to mg. Now,"},{"Start":"09:22.700 ","End":"09:24.290","Text":"this is going to be our third equation,"},{"Start":"09:24.290 ","End":"09:26.915","Text":"but we\u0027re still missing two equations because we have"},{"Start":"09:26.915 ","End":"09:30.755","Text":"5 unknowns and right now we only have three equations."},{"Start":"09:30.755 ","End":"09:34.699","Text":"Now that I found all the forces working on each of the masses,"},{"Start":"09:34.699 ","End":"09:38.240","Text":"the only way that I\u0027m going to find the next two equations to"},{"Start":"09:38.240 ","End":"09:42.925","Text":"complete my list of five will be by looking at simple trigonometry."},{"Start":"09:42.925 ","End":"09:47.330","Text":"Now let\u0027s use our favorite equation from back"},{"Start":"09:47.330 ","End":"09:52.995","Text":"in our high school days, which is Pythagoras."},{"Start":"09:52.995 ","End":"09:56.770","Text":"Here we have r,"},{"Start":"09:56.770 ","End":"10:03.140","Text":"here we have d/2."},{"Start":"10:03.140 ","End":"10:04.740","Text":"With Pythagoras, we know"},{"Start":"10:04.740 ","End":"10:13.735","Text":"that r^2 plus (d/2)^2 equals d^2."},{"Start":"10:13.735 ","End":"10:17.650","Text":"It\u0027s important to note this angle here is 90 degrees."},{"Start":"10:17.650 ","End":"10:21.500","Text":"Another one of our favorites is our SOHCAHTOA,"},{"Start":"10:21.500 ","End":"10:23.600","Text":"which states that the SOH,"},{"Start":"10:23.600 ","End":"10:26.390","Text":"sine is opposite over hypotenuse."},{"Start":"10:26.390 ","End":"10:30.395","Text":"We know that sine of Alpha is, now,"},{"Start":"10:30.395 ","End":"10:36.020","Text":"we know that this is Alpha the opposite angle is D divided by 2,"},{"Start":"10:36.020 ","End":"10:41.720","Text":"and our hypotenuse is d. To sum up,"},{"Start":"10:41.720 ","End":"10:45.550","Text":"we have one equation,"},{"Start":"10:45.550 ","End":"10:52.000","Text":"two equations, three equations,"},{"Start":"10:52.000 ","End":"10:55.970","Text":"a fourth equation, and our fifth equation."},{"Start":"10:55.970 ","End":"10:58.400","Text":"Now we have five equations and five unknowns,"},{"Start":"10:58.400 ","End":"11:00.010","Text":"so we can solve the question."},{"Start":"11:00.010 ","End":"11:04.250","Text":"Now all that\u0027s left to do in order to solve this question till the end,"},{"Start":"11:04.250 ","End":"11:08.825","Text":"is to simply solve these five equations with five unknowns."},{"Start":"11:08.825 ","End":"11:11.670","Text":"Here we finish this lesson."}],"ID":9261},{"Watched":false,"Name":"Exercise 4","Duration":"15m 29s","ChapterTopicVideoID":8992,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.320","Text":"In the question, we are told that a mass is tied to a rope of length small d,"},{"Start":"00:07.320 ","End":"00:10.275","Text":"and the rope is tied to a pole,"},{"Start":"00:10.275 ","End":"00:17.770","Text":"and the pole is of length capital D. The pole is rotating at a speed Omega,"},{"Start":"00:17.770 ","End":"00:20.900","Text":"and at some stage the rope disconnects."},{"Start":"00:20.900 ","End":"00:26.030","Text":"What will be the displacement of the mass by the time it hits the ground."},{"Start":"00:26.030 ","End":"00:27.920","Text":"When the rope disconnects,"},{"Start":"00:27.920 ","End":"00:30.350","Text":"the mass flies outwards,"},{"Start":"00:30.350 ","End":"00:37.114","Text":"and they\u0027re asking how far the mass will fly from the pole by the time it\u0027s motion stops,"},{"Start":"00:37.114 ","End":"00:39.340","Text":"which is when it hits the ground."},{"Start":"00:39.340 ","End":"00:46.020","Text":"Over here, we have the diagrams."},{"Start":"00:46.150 ","End":"00:52.760","Text":"This one is from this side and this one is from up top."},{"Start":"00:52.760 ","End":"00:55.700","Text":"Let\u0027s look at the diagram from the side."},{"Start":"00:55.700 ","End":"00:59.700","Text":"This here is a pole of"},{"Start":"00:59.700 ","End":"01:06.140","Text":"D. This is the length of the rope,"},{"Start":"01:06.140 ","End":"01:09.455","Text":"small d. This is the mass,"},{"Start":"01:09.455 ","End":"01:12.245","Text":"and then here we have r,"},{"Start":"01:12.245 ","End":"01:18.455","Text":"which represents the distance of the mass from the pole when it\u0027s attached to a rope D,"},{"Start":"01:18.455 ","End":"01:20.335","Text":"whilst the pole is spinning."},{"Start":"01:20.335 ","End":"01:24.980","Text":"This is angle Alpha, some unknown angle."},{"Start":"01:24.980 ","End":"01:29.885","Text":"Our unknowns are the angle Alpha and the radius."},{"Start":"01:29.885 ","End":"01:34.985","Text":"Now let\u0027s take a look at what\u0027s happening from the angle above."},{"Start":"01:34.985 ","End":"01:37.715","Text":"As we can see, we\u0027re looking at this diagram."},{"Start":"01:37.715 ","End":"01:40.560","Text":"Here, from the angle above,"},{"Start":"01:40.560 ","End":"01:45.110","Text":"we can see that the mass is going in a circular motion"},{"Start":"01:45.110 ","End":"01:50.365","Text":"and that this radius is actually also this radius."},{"Start":"01:50.365 ","End":"01:54.590","Text":"Notice that as the mass travels around in the circle,"},{"Start":"01:54.590 ","End":"01:58.175","Text":"it has a tangential velocity,"},{"Start":"01:58.175 ","End":"02:00.950","Text":"which goes in this direction."},{"Start":"02:00.950 ","End":"02:05.255","Text":"At every moment, whilst the mass is traveling along the circle,"},{"Start":"02:05.255 ","End":"02:09.020","Text":"it has a tangential velocity, and of course,"},{"Start":"02:09.020 ","End":"02:13.265","Text":"the equation for the tangential velocity is written over here."},{"Start":"02:13.265 ","End":"02:21.050","Text":"V equals Omega multiplied by r. Now let\u0027s go over the forces acting on the mass."},{"Start":"02:21.050 ","End":"02:23.480","Text":"We\u0027ll draw it on a free body diagram."},{"Start":"02:23.480 ","End":"02:24.830","Text":"Now I\u0027ve already drawn it,"},{"Start":"02:24.830 ","End":"02:26.000","Text":"but let\u0027s go over this."},{"Start":"02:26.000 ","End":"02:33.515","Text":"Here we have the mass and then we have mg pointing downwards as per usual."},{"Start":"02:33.515 ","End":"02:39.335","Text":"Then we have the tension of the string D going in this direction."},{"Start":"02:39.335 ","End":"02:40.860","Text":"T for tension."},{"Start":"02:40.860 ","End":"02:45.635","Text":"Obviously we have to split up the tension into its separate components."},{"Start":"02:45.635 ","End":"02:50.265","Text":"We can do this upper row as T_z."},{"Start":"02:50.265 ","End":"02:53.970","Text":"I\u0027ve just labeled it z, it doesn\u0027t really matter."},{"Start":"02:53.970 ","End":"02:59.790","Text":"This t is T_r, why r?"},{"Start":"02:59.790 ","End":"03:02.420","Text":"Because it\u0027s going in the direction of the radius."},{"Start":"03:02.420 ","End":"03:05.075","Text":"It\u0027s always going to be in the direction of the radius."},{"Start":"03:05.075 ","End":"03:08.255","Text":"It just makes it easier for me to label it like this."},{"Start":"03:08.255 ","End":"03:11.390","Text":"Also we can label the axis."},{"Start":"03:11.390 ","End":"03:20.045","Text":"The axis going in this direction is the y axis,"},{"Start":"03:20.045 ","End":"03:26.330","Text":"and the axis going in this direction is the x axis,"},{"Start":"03:26.330 ","End":"03:28.925","Text":"but of course, as I already said,"},{"Start":"03:28.925 ","End":"03:32.420","Text":"this is also the radius axis,"},{"Start":"03:32.420 ","End":"03:34.055","Text":"the axis of the radius."},{"Start":"03:34.055 ","End":"03:37.355","Text":"I can also call it the r axis."},{"Start":"03:37.355 ","End":"03:39.935","Text":"It doesn\u0027t really matter, it\u0027s just to label it."},{"Start":"03:39.935 ","End":"03:42.965","Text":"We can say that the sum of all the forces is"},{"Start":"03:42.965 ","End":"03:45.890","Text":"equal to the sum of all of the accelerations."},{"Start":"03:45.890 ","End":"03:48.170","Text":"In the z axis direction,"},{"Start":"03:48.170 ","End":"03:52.410","Text":"remember, we said, this direction,"},{"Start":"03:53.210 ","End":"03:56.075","Text":"we have no motion,"},{"Start":"03:56.075 ","End":"03:59.450","Text":"which also means that the sum of"},{"Start":"03:59.450 ","End":"04:04.975","Text":"the forces is 0 and that also means that the sum of the accelerations is also 0."},{"Start":"04:04.975 ","End":"04:14.840","Text":"We can say that T_z equals mg. Let\u0027s label this equation number 1, T_z=mg."},{"Start":"04:14.840 ","End":"04:17.615","Text":"Now let\u0027s take a look at the other axis."},{"Start":"04:17.615 ","End":"04:19.955","Text":"We have this here, T_r."},{"Start":"04:19.955 ","End":"04:22.465","Text":"Now let\u0027s take a look at the other axis."},{"Start":"04:22.465 ","End":"04:24.070","Text":"Now we\u0027re looking at T_r."},{"Start":"04:24.070 ","End":"04:28.820","Text":"Again, the sum of all the forces equals the sum of all the accelerations."},{"Start":"04:28.820 ","End":"04:31.385","Text":"The sum of all the forces is T_r."},{"Start":"04:31.385 ","End":"04:38.635","Text":"We write that here, and the sum of all the accelerations is of course ma,"},{"Start":"04:38.635 ","End":"04:42.560","Text":"where my a over here is in the r direction,"},{"Start":"04:42.560 ","End":"04:46.160","Text":"just like my T is in the r direction."},{"Start":"04:46.160 ","End":"04:50.180","Text":"The sum of all the forces in the direction of r is the sum of"},{"Start":"04:50.180 ","End":"04:56.080","Text":"all the forces of all the accelerations in the direction of r. Now, as we know,"},{"Start":"04:56.080 ","End":"05:01.025","Text":"a_r is equal to Omega squared r. This is"},{"Start":"05:01.025 ","End":"05:07.460","Text":"my radial acceleration and it\u0027s very very important to learn this equation of by heart."},{"Start":"05:07.460 ","End":"05:12.425","Text":"Radial acceleration equals Omega squared multiplied by r. Now,"},{"Start":"05:12.425 ","End":"05:17.630","Text":"what exactly is this radial acceleration that I\u0027m speaking about? Let\u0027s explain."},{"Start":"05:17.630 ","End":"05:19.325","Text":"Let\u0027s go back to this diagram."},{"Start":"05:19.325 ","End":"05:21.275","Text":"We have our mass over here,"},{"Start":"05:21.275 ","End":"05:26.045","Text":"and it\u0027s going in a circular motion around and around and around."},{"Start":"05:26.045 ","End":"05:30.350","Text":"Now it\u0027s obvious to everyone that if the rope here"},{"Start":"05:30.350 ","End":"05:36.200","Text":"wasn\u0027t holding the mass whilst it was going around in a circular motion."},{"Start":"05:36.200 ","End":"05:39.530","Text":"The mass would just fly off in this direction,"},{"Start":"05:39.530 ","End":"05:40.580","Text":"in this direction,"},{"Start":"05:40.580 ","End":"05:43.070","Text":"in this direction, and whatever."},{"Start":"05:43.070 ","End":"05:46.220","Text":"However, because we have the rope holding it in,"},{"Start":"05:46.220 ","End":"05:49.100","Text":"there is some force acting on the mass."},{"Start":"05:49.100 ","End":"05:50.780","Text":"What is this force called?"},{"Start":"05:50.780 ","End":"05:52.925","Text":"It\u0027s called the radial acceleration."},{"Start":"05:52.925 ","End":"06:00.345","Text":"It\u0027s the force that\u0027s causing the mass to accelerate inwards towards the center,"},{"Start":"06:00.345 ","End":"06:06.535","Text":"and this force keeps the mass from flying out at a tangent."},{"Start":"06:06.535 ","End":"06:14.510","Text":"This radial acceleration is really what keeps the mass going in the circular motion."},{"Start":"06:14.510 ","End":"06:19.010","Text":"It\u0027s important to remember that the equation for radial acceleration"},{"Start":"06:19.010 ","End":"06:23.945","Text":"is Omega squared r. Right now,"},{"Start":"06:23.945 ","End":"06:26.735","Text":"I have 3 unknowns."},{"Start":"06:26.735 ","End":"06:29.445","Text":"I have T_z as unknown,"},{"Start":"06:29.445 ","End":"06:31.650","Text":"I have T_r is unknown,"},{"Start":"06:31.650 ","End":"06:34.920","Text":"and I also have that r is unknown."},{"Start":"06:34.920 ","End":"06:39.715","Text":"I have 3 unknowns and only 1, 2 equations."},{"Start":"06:39.715 ","End":"06:41.870","Text":"I\u0027m missing 1 equation."},{"Start":"06:41.870 ","End":"06:45.695","Text":"Now in a second we\u0027re going to see how we find this missing equation."},{"Start":"06:45.695 ","End":"06:47.015","Text":"But in the meantime,"},{"Start":"06:47.015 ","End":"06:48.935","Text":"let\u0027s take a look at this."},{"Start":"06:48.935 ","End":"06:55.290","Text":"The first thing that we\u0027re going to have a look at is this equation."},{"Start":"06:55.430 ","End":"07:00.875","Text":"We have cosine of Alpha equals r over d. Now,"},{"Start":"07:00.875 ","End":"07:04.985","Text":"this equation connects r the radius to d,"},{"Start":"07:04.985 ","End":"07:07.400","Text":"the length of the rope holding the mass in."},{"Start":"07:07.400 ","End":"07:11.160","Text":"Let\u0027s go to this diagram."},{"Start":"07:11.160 ","End":"07:13.575","Text":"It\u0027s this, as you know,"},{"Start":"07:13.575 ","End":"07:17.475","Text":"SOHCAHTOA, SOHCAH,"},{"Start":"07:17.475 ","End":"07:22.365","Text":"cosine is adjacent over hypotenuse,"},{"Start":"07:22.365 ","End":"07:23.730","Text":"adjacent being the r,"},{"Start":"07:23.730 ","End":"07:32.060","Text":"Hypotenuse being the d. Then the next equation that we can look at is this one."},{"Start":"07:32.060 ","End":"07:36.020","Text":"Tan of Alpha is, and again,"},{"Start":"07:36.020 ","End":"07:42.450","Text":"SOHCAHTOA, Tan is opposite over adjacent."},{"Start":"07:42.450 ","End":"07:44.370","Text":"If we look over here,"},{"Start":"07:44.370 ","End":"07:47.085","Text":"T_z is the Opposite,"},{"Start":"07:47.085 ","End":"07:49.175","Text":"because even though it\u0027s here,"},{"Start":"07:49.175 ","End":"07:51.740","Text":"we could also draw it here."},{"Start":"07:51.740 ","End":"07:53.985","Text":"As you remember with vectors,"},{"Start":"07:53.985 ","End":"07:58.715","Text":"as long as it\u0027s the same size and pointing in the same direction,"},{"Start":"07:58.715 ","End":"08:01.100","Text":"you can move it around and it doesn\u0027t matter,"},{"Start":"08:01.100 ","End":"08:07.235","Text":"and the Adjacent is the T_r, which is this,"},{"Start":"08:07.235 ","End":"08:09.755","Text":"because obviously this is Alpha,"},{"Start":"08:09.755 ","End":"08:12.530","Text":"because if you look in this diagram,"},{"Start":"08:12.530 ","End":"08:15.645","Text":"here\u0027s the mass, and here\u0027s this."},{"Start":"08:15.645 ","End":"08:17.055","Text":"If this is Alpha,"},{"Start":"08:17.055 ","End":"08:20.475","Text":"then this is Alpha."},{"Start":"08:20.475 ","End":"08:23.975","Text":"Now, just in case you\u0027re unfamiliar with SOHCAHTOA,"},{"Start":"08:23.975 ","End":"08:26.439","Text":"Let\u0027s just write this."},{"Start":"08:26.439 ","End":"08:33.865","Text":"Is sine of Alpha equals opposite over hypotenuse?"},{"Start":"08:33.865 ","End":"08:43.030","Text":"Cosine of the angle equals adjacent over hypotenuse,"},{"Start":"08:43.030 ","End":"08:52.880","Text":"and tan of the angle equals opposite over adjacent."},{"Start":"08:53.370 ","End":"08:56.140","Text":"You can copy this down."},{"Start":"08:56.140 ","End":"08:58.360","Text":"All of these equations,"},{"Start":"08:58.360 ","End":"09:01.270","Text":"this one, this one,"},{"Start":"09:01.270 ","End":"09:03.625","Text":"this, and this are"},{"Start":"09:03.625 ","End":"09:07.075","Text":"all the equations that we need right now in order to solve this question."},{"Start":"09:07.075 ","End":"09:09.520","Text":"Now after we find all of our unknowns,"},{"Start":"09:09.520 ","End":"09:14.920","Text":"we will also have a value for r. Now why do I need r?"},{"Start":"09:14.920 ","End":"09:17.095","Text":"I need to know what r is,"},{"Start":"09:17.095 ","End":"09:20.590","Text":"because once I know what r is I can known also what"},{"Start":"09:20.590 ","End":"09:25.240","Text":"the d is and I can also know this up until here."},{"Start":"09:25.240 ","End":"09:27.385","Text":"Now, in other words,"},{"Start":"09:27.385 ","End":"09:32.275","Text":"what I need to know actually is this value here."},{"Start":"09:32.275 ","End":"09:39.445","Text":"The height that the mass is from the ground."},{"Start":"09:39.445 ","End":"09:42.355","Text":"Now at the moment I know the height of the mass,"},{"Start":"09:42.355 ","End":"09:47.485","Text":"I know that the mass will disconnect from the rope at"},{"Start":"09:47.485 ","End":"09:54.250","Text":"velocity 0 on the z-axis."},{"Start":"09:54.250 ","End":"10:01.465","Text":"The fact that the mask has some tangential velocity doesn\u0027t matter to me."},{"Start":"10:01.465 ","End":"10:08.260","Text":"Just to reiterate, with regards to how long it will take for the mass to hit the ground."},{"Start":"10:08.260 ","End":"10:11.905","Text":"It doesn\u0027t matter if the mass is released from a resting position,"},{"Start":"10:11.905 ","End":"10:19.390","Text":"or via this tangential velocity by flying around in circular motion."},{"Start":"10:19.390 ","End":"10:20.755","Text":"In order to do this,"},{"Start":"10:20.755 ","End":"10:25.195","Text":"I need to understand what this height is."},{"Start":"10:25.195 ","End":"10:28.165","Text":"In order to know what this height is,"},{"Start":"10:28.165 ","End":"10:36.775","Text":"I need to know what this is and what this is,"},{"Start":"10:36.775 ","End":"10:39.609","Text":"and once I know what d is,"},{"Start":"10:39.609 ","End":"10:44.530","Text":"so I need to know what this value is."},{"Start":"10:44.530 ","End":"10:45.910","Text":"Let\u0027s call it height."},{"Start":"10:45.910 ","End":"10:47.995","Text":"Now how I do this is,"},{"Start":"10:47.995 ","End":"10:54.265","Text":"I know that the total length of the pole is d."},{"Start":"10:54.265 ","End":"11:01.870","Text":"I have to minus from the D, d sine Alpha."},{"Start":"11:01.870 ","End":"11:07.450","Text":"D minus this section will give me the height."},{"Start":"11:07.450 ","End":"11:15.985","Text":"My height is D minus d sine Alpha and my question is,"},{"Start":"11:15.985 ","End":"11:21.475","Text":"how long does it take the mass to fall this distance?"},{"Start":"11:21.475 ","End":"11:25.045","Text":"In order to find out when the mass will hit the ground,"},{"Start":"11:25.045 ","End":"11:28.360","Text":"we\u0027re going to use this equation."},{"Start":"11:28.360 ","End":"11:31.893","Text":"Now, this x is my position,"},{"Start":"11:31.893 ","End":"11:34.165","Text":"the position that I\u0027m trying to find."},{"Start":"11:34.165 ","End":"11:39.024","Text":"The x_0 is my starting position,"},{"Start":"11:39.024 ","End":"11:43.030","Text":"v_0 is my velocity."},{"Start":"11:43.030 ","End":"11:47.620","Text":"T is the time plus half multiplied by a,"},{"Start":"11:47.620 ","End":"11:49.765","Text":"which is my acceleration downwards,"},{"Start":"11:49.765 ","End":"11:53.290","Text":"multiplied by t squared, my time squared."},{"Start":"11:53.290 ","End":"11:55.420","Text":"Let\u0027s plug in our values."},{"Start":"11:55.420 ","End":"11:58.802","Text":"The position I wanted to know of the question that we\u0027re asking,"},{"Start":"11:58.802 ","End":"12:04.090","Text":"when the mass hits the ground is 0 because it\u0027s a 0 height,"},{"Start":"12:04.090 ","End":"12:07.930","Text":"my current position, sorry, this isn\u0027t correct."},{"Start":"12:07.930 ","End":"12:14.980","Text":"This is meant to be D minus d sine Alpha."},{"Start":"12:14.980 ","End":"12:19.000","Text":"Put this here, sorry about that mistake, so 0,"},{"Start":"12:19.000 ","End":"12:22.540","Text":"the position I want to know of equals the height,"},{"Start":"12:22.540 ","End":"12:24.850","Text":"my starting position, the starting height,"},{"Start":"12:24.850 ","End":"12:27.820","Text":"which is D minus d sine Alpha."},{"Start":"12:27.820 ","End":"12:37.750","Text":"This D minus d sine Alpha plus 0 velocity because as we said before,"},{"Start":"12:37.750 ","End":"12:40.420","Text":"our velocity as the ball falls down,"},{"Start":"12:40.420 ","End":"12:44.695","Text":"we have 0 velocity on the z-axis and 0 times t is 0,"},{"Start":"12:44.695 ","End":"12:50.710","Text":"so that\u0027s that plus half multiplied by the acceleration times t squared."},{"Start":"12:50.710 ","End":"12:54.510","Text":"Now, our acceleration is the negative value of"},{"Start":"12:54.510 ","End":"12:58.605","Text":"gravity because this is going in a downward motion."},{"Start":"12:58.605 ","End":"13:03.515","Text":"Negative g multiplied by t squared."},{"Start":"13:03.515 ","End":"13:07.300","Text":"Now, because my question right now is"},{"Start":"13:07.300 ","End":"13:10.885","Text":"I\u0027m trying to find the time that it takes the mass to hit the ground,"},{"Start":"13:10.885 ","End":"13:18.055","Text":"I need to isolate out my t. After doing very simple algebraic steps,"},{"Start":"13:18.055 ","End":"13:22.270","Text":"I get to that t is the square root of"},{"Start":"13:22.270 ","End":"13:30.010","Text":"2d sine Alpha divided by g. This is my value for t. Now what do I have to do with this t?"},{"Start":"13:30.010 ","End":"13:36.160","Text":"Now they\u0027re asking me what the distance is that the mass moves from the poll,"},{"Start":"13:36.160 ","End":"13:38.530","Text":"so I have the velocity,"},{"Start":"13:38.530 ","End":"13:43.330","Text":"and this tangential velocity is written here in the diagram,"},{"Start":"13:43.330 ","End":"13:49.465","Text":"v equals Omega times r. Here as well, it\u0027s written."},{"Start":"13:49.465 ","End":"13:54.895","Text":"This is the velocity that it\u0027s traveling in."},{"Start":"13:54.895 ","End":"13:57.700","Text":"Then from the equation that we just came up with,"},{"Start":"13:57.700 ","End":"14:00.280","Text":"I also have my value for t,"},{"Start":"14:00.280 ","End":"14:03.475","Text":"the amount of time that it is traveling."},{"Start":"14:03.475 ","End":"14:06.190","Text":"Now, as you know, displacement,"},{"Start":"14:06.190 ","End":"14:10.197","Text":"let\u0027s call displacement x,"},{"Start":"14:10.197 ","End":"14:13.345","Text":"equals velocity times time."},{"Start":"14:13.345 ","End":"14:14.890","Text":"We already know this."},{"Start":"14:14.890 ","End":"14:20.950","Text":"Now, we have here an equation for velocity,"},{"Start":"14:20.950 ","End":"14:26.485","Text":"which is this and we have here our equation for time,"},{"Start":"14:26.485 ","End":"14:30.535","Text":"so we can do displacement equals v,"},{"Start":"14:30.535 ","End":"14:35.845","Text":"which is Omega times r multiplied by t,"},{"Start":"14:35.845 ","End":"14:40.765","Text":"which here from our equation is root"},{"Start":"14:40.765 ","End":"14:49.660","Text":"2d sine Alpha over g and we know what our r value is,"},{"Start":"14:49.660 ","End":"14:54.640","Text":"because this we currently don\u0027t know."},{"Start":"14:54.640 ","End":"14:57.955","Text":"Let\u0027s do it here. Let\u0027s isolate the r,"},{"Start":"14:57.955 ","End":"15:03.115","Text":"r equals d cos Alpha."},{"Start":"15:03.115 ","End":"15:06.580","Text":"This actually is displacement equals"},{"Start":"15:06.580 ","End":"15:12.970","Text":"omega times d cos Alpha multiplied by the square root of"},{"Start":"15:12.970 ","End":"15:19.570","Text":"2d sine Alpha over g. This is"},{"Start":"15:19.570 ","End":"15:27.130","Text":"the distance that our mass will fly away from the pole when the rope disconnects,"},{"Start":"15:27.130 ","End":"15:30.530","Text":"and here we finished our question."}],"ID":9262},{"Watched":false,"Name":"Exercise 5","Duration":"21m 21s","ChapterTopicVideoID":8993,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.950 ","End":"00:04.155","Text":"Hello. In this question,"},{"Start":"00:04.155 ","End":"00:06.340","Text":"we\u0027re given a carrousel."},{"Start":"00:06.340 ","End":"00:09.300","Text":"The height of the seat from the ground is H,"},{"Start":"00:09.300 ","End":"00:13.275","Text":"so the distance here"},{"Start":"00:13.275 ","End":"00:19.965","Text":"is H. The length of the rope is D,"},{"Start":"00:19.965 ","End":"00:26.140","Text":"so this is D."},{"Start":"00:27.800 ","End":"00:32.700","Text":"The distance of the seat from the center of the pole is R,"},{"Start":"00:32.700 ","End":"00:38.260","Text":"so this is R."},{"Start":"00:39.260 ","End":"00:43.080","Text":"A boy of mass M sits on the seat,"},{"Start":"00:43.080 ","End":"00:49.745","Text":"so this is M. Now the first question that we\u0027re being asked is,"},{"Start":"00:49.745 ","End":"00:56.370","Text":"what will the boy\u0027s height be off the ground when the carrousel will begin to rotate?"},{"Start":"00:56.740 ","End":"00:58.955","Text":"What does this mean?"},{"Start":"00:58.955 ","End":"01:02.375","Text":"Let\u0027s take this page a little bit to the side for a second,"},{"Start":"01:02.375 ","End":"01:06.050","Text":"this means that as the carrousel begins to spin,"},{"Start":"01:06.050 ","End":"01:12.480","Text":"the swing will move slightly in this direction and will suddenly be here."},{"Start":"01:13.820 ","End":"01:20.390","Text":"This is the rope, which means that its height is slightly different because it\u0027s up here."},{"Start":"01:20.390 ","End":"01:26.720","Text":"Then this will be a new height,"},{"Start":"01:26.720 ","End":"01:28.335","Text":"height tag for instance."},{"Start":"01:28.335 ","End":"01:32.490","Text":"That\u0027s what we\u0027re being asked."},{"Start":"01:32.490 ","End":"01:39.330","Text":"Right now we\u0027re looking at the diagram as if it\u0027s spinning."},{"Start":"01:39.330 ","End":"01:42.330","Text":"Here we know that this is Omega,"},{"Start":"01:42.330 ","End":"01:48.935","Text":"and this is the carrousel and it\u0027s spinning and the seats are out to the side."},{"Start":"01:48.935 ","End":"01:51.895","Text":"Now, it doesn\u0027t matter if there\u0027s just 1 seat,"},{"Start":"01:51.895 ","End":"01:54.365","Text":"2 seats, a 1,000 seats, it doesn\u0027t matter."},{"Start":"01:54.365 ","End":"01:57.785","Text":"The question, the working out remains the same."},{"Start":"01:57.785 ","End":"02:03.545","Text":"Now let\u0027s begin by drawing a free body diagram for the boy."},{"Start":"02:03.545 ","End":"02:09.585","Text":"This is the boy, and the mass,"},{"Start":"02:09.585 ","End":"02:12.570","Text":"mg it\u0027s pointing downwards,"},{"Start":"02:12.570 ","End":"02:16.535","Text":"and then over here from the rope of the swing,"},{"Start":"02:16.535 ","End":"02:23.750","Text":"we have T. Now at this stage of where we are in mechanics,"},{"Start":"02:23.750 ","End":"02:29.525","Text":"we can say that there\u0027s only a force acting on an object,"},{"Start":"02:29.525 ","End":"02:31.535","Text":"if it is in contact with objects."},{"Start":"02:31.535 ","End":"02:33.995","Text":"Aside from, of course, gravity,"},{"Start":"02:33.995 ","End":"02:35.690","Text":"but for instance, tension,"},{"Start":"02:35.690 ","End":"02:38.500","Text":"a pulling force, a pushing force,"},{"Start":"02:38.500 ","End":"02:41.115","Text":"anything else that there might be,"},{"Start":"02:41.115 ","End":"02:42.800","Text":"there\u0027s only a force,"},{"Start":"02:42.800 ","End":"02:46.910","Text":"right now as far as we\u0027re concerned at this level,"},{"Start":"02:46.910 ","End":"02:51.330","Text":"only if it\u0027s touching the body."},{"Start":"02:51.980 ","End":"02:58.710","Text":"Now let\u0027s draw some axes that will suit our question,"},{"Start":"02:58.710 ","End":"03:00.780","Text":"so we\u0027ll have this as the y,"},{"Start":"03:00.780 ","End":"03:04.745","Text":"and this is the x, and then I\u0027ll just call this angle Alpha."},{"Start":"03:04.745 ","End":"03:13.105","Text":"Now I\u0027m going to write on the side the things that we don\u0027t know with a tilde on top,"},{"Start":"03:13.105 ","End":"03:16.025","Text":"T we don\u0027t know, the tension in the rope we don\u0027t know,"},{"Start":"03:16.025 ","End":"03:19.970","Text":"and we also don\u0027t know Alpha, the angle."},{"Start":"03:19.970 ","End":"03:21.530","Text":"I\u0027ve written them here,"},{"Start":"03:21.530 ","End":"03:25.290","Text":"so that I don\u0027t forget whether tilde on top."},{"Start":"03:26.090 ","End":"03:31.145","Text":"Sorry, I called this the y-axis, but actually,"},{"Start":"03:31.145 ","End":"03:33.110","Text":"let\u0027s call it because it\u0027s circular motion,"},{"Start":"03:33.110 ","End":"03:36.195","Text":"the z axis and this isn\u0027t the x-axis,"},{"Start":"03:36.195 ","End":"03:38.720","Text":"it\u0027s in fact the radial axis."},{"Start":"03:38.720 ","End":"03:46.550","Text":"Because these names are better suited to our question here because it\u0027s circular motion."},{"Start":"03:46.550 ","End":"03:50.185","Text":"Let\u0027s write down our equations."},{"Start":"03:50.185 ","End":"03:55.445","Text":"We have the sum of all of the forces on the z axis,"},{"Start":"03:55.445 ","End":"04:01.750","Text":"and we have to write an equation for the sum of all the forces on the radial axis."},{"Start":"04:01.750 ","End":"04:04.715","Text":"Let\u0027s begin, on the z axis,"},{"Start":"04:04.715 ","End":"04:07.900","Text":"we have mg in the downwards direction,"},{"Start":"04:07.900 ","End":"04:12.830","Text":"now we\u0027re going to take the upwards direction to be the positive direction,"},{"Start":"04:12.830 ","End":"04:15.980","Text":"so we have minus mg,"},{"Start":"04:15.980 ","End":"04:26.425","Text":"and then we have plus T Cosine of Alpha."},{"Start":"04:26.425 ","End":"04:27.755","Text":"Why cosine of Alpha?"},{"Start":"04:27.755 ","End":"04:31.115","Text":"Because we\u0027re trying to find here,"},{"Start":"04:31.115 ","End":"04:33.560","Text":"which is the adjacent, so it\u0027s cosine."},{"Start":"04:33.560 ","End":"04:40.430","Text":"Now, we know that the sum of all the forces is equal to mass times acceleration."},{"Start":"04:40.430 ","End":"04:41.600","Text":"Now we\u0027re on the z-axis,"},{"Start":"04:41.600 ","End":"04:44.375","Text":"so that\u0027s the acceleration in the z axis."},{"Start":"04:44.375 ","End":"04:47.960","Text":"Now, because our boy isn\u0027t moving up and down,"},{"Start":"04:47.960 ","End":"04:51.740","Text":"we can say that the acceleration in the z axis is equal to 0,"},{"Start":"04:51.740 ","End":"04:55.195","Text":"which means that this is equal to 0."},{"Start":"04:55.195 ","End":"05:00.330","Text":"Now the sum of all the forces on the radial axis."},{"Start":"05:00.680 ","End":"05:07.000","Text":"The only force that is acting on this is T sine of Alpha."},{"Start":"05:07.000 ","End":"05:08.585","Text":"Why sine of Alpha?"},{"Start":"05:08.585 ","End":"05:12.650","Text":"Because we\u0027re looking for this component over here,"},{"Start":"05:12.650 ","End":"05:14.695","Text":"which is the opposite so it\u0027s sine,"},{"Start":"05:14.695 ","End":"05:20.180","Text":"which equals mass times acceleration in the radial direction."},{"Start":"05:20.180 ","End":"05:26.135","Text":"Obviously, we know that ar is equal to mass times ar,"},{"Start":"05:26.135 ","End":"05:32.620","Text":"which is Omega^2 r. We know this already."},{"Start":"05:32.930 ","End":"05:39.935","Text":"Now I\u0027m going to make these equations slightly smaller so that the page is tidier,"},{"Start":"05:39.935 ","End":"05:42.380","Text":"and I\u0027m going to rearrange the equations in"},{"Start":"05:42.380 ","End":"05:46.590","Text":"order to make it slightly more readable and understandable."},{"Start":"05:47.000 ","End":"05:50.075","Text":"From the first equation that we have,"},{"Start":"05:50.075 ","End":"05:56.475","Text":"we have mg equals T cosine of Alpha."},{"Start":"05:56.475 ","End":"05:58.295","Text":"In the second equation,"},{"Start":"05:58.295 ","End":"06:03.635","Text":"we have that T sine of Alpha is equal to"},{"Start":"06:03.635 ","End":"06:10.270","Text":"m Omega^2 r. Let\u0027s talk about the r\u0027s."},{"Start":"06:10.270 ","End":"06:15.819","Text":"As we\u0027ve noticed, here we have R. Why am I writing here r?"},{"Start":"06:15.819 ","End":"06:19.980","Text":"Because if you think back to here,"},{"Start":"06:19.980 ","End":"06:24.455","Text":"R is this distance over here."},{"Start":"06:24.455 ","End":"06:26.045","Text":"But right now we\u0027re talking,"},{"Start":"06:26.045 ","End":"06:28.190","Text":"which is when the carrousel is at rest."},{"Start":"06:28.190 ","End":"06:30.695","Text":"When the carrousel is moving though,"},{"Start":"06:30.695 ","End":"06:33.397","Text":"then the r increases,"},{"Start":"06:33.397 ","End":"06:40.590","Text":"so this is R and this is the r of whilst the carrousel is rotating,"},{"Start":"06:40.590 ","End":"06:42.480","Text":"so don\u0027t get confused here."},{"Start":"06:42.480 ","End":"06:44.915","Text":"Right now we don\u0027t need to use the R,"},{"Start":"06:44.915 ","End":"06:47.975","Text":"maybe later in 1 of the equations will come in handy,"},{"Start":"06:47.975 ","End":"06:51.480","Text":"but right now, we\u0027re just going to leave it."},{"Start":"06:51.700 ","End":"06:56.075","Text":"Now, let\u0027s begin labeling what are unknowns r?"},{"Start":"06:56.075 ","End":"06:58.715","Text":"As we\u0027ve already spoken,"},{"Start":"06:58.715 ","End":"07:00.705","Text":"T and Alpha unknown,"},{"Start":"07:00.705 ","End":"07:03.000","Text":"so T and Alpha unknown, T I\u0027ve already,"},{"Start":"07:03.000 ","End":"07:05.940","Text":"Alpha I\u0027ve already written."},{"Start":"07:05.940 ","End":"07:10.230","Text":"M we know, Omega we know, r we don\u0027t know."},{"Start":"07:10.230 ","End":"07:13.369","Text":"We have 2 equations but 3 unknowns,"},{"Start":"07:13.369 ","End":"07:18.030","Text":"which means that we have to find another equation in order to solve this."},{"Start":"07:18.350 ","End":"07:22.835","Text":"What can we do in order to find the third equation?"},{"Start":"07:22.835 ","End":"07:26.030","Text":"You see this triangle over here,"},{"Start":"07:26.030 ","End":"07:30.660","Text":"let\u0027s enlarge it, we\u0027ll draw it over here."},{"Start":"07:30.660 ","End":"07:33.140","Text":"Here\u0027s the seats, this is how it goes."},{"Start":"07:33.140 ","End":"07:36.305","Text":"Now, when we were drawing originally,"},{"Start":"07:36.305 ","End":"07:41.000","Text":"we said that this angle here is Alpha,"},{"Start":"07:41.000 ","End":"07:45.535","Text":"which means that if this is 90 degrees,"},{"Start":"07:45.535 ","End":"07:47.485","Text":"this is also Alpha,"},{"Start":"07:47.485 ","End":"07:53.118","Text":"and we know that the distance over here is equal to D,"},{"Start":"07:53.118 ","End":"07:59.215","Text":"because it says the length of the rope is D. That\u0027s D. Now,"},{"Start":"07:59.215 ","End":"08:04.970","Text":"here is the center pole of the carrousel."},{"Start":"08:04.970 ","End":"08:14.470","Text":"We know that the distance to here is R and that the total distance up until here is"},{"Start":"08:14.470 ","End":"08:18.110","Text":"r. Then by using"},{"Start":"08:18.110 ","End":"08:27.040","Text":"sine Alpha is opposite over hypotenuse,"},{"Start":"08:28.220 ","End":"08:33.955","Text":"we can say that the opposite is here,"},{"Start":"08:33.955 ","End":"08:37.075","Text":"we only need this distance,"},{"Start":"08:37.075 ","End":"08:40.645","Text":"so it\u0027s r minus the R,"},{"Start":"08:40.645 ","End":"08:43.520","Text":"so that\u0027s this distance."},{"Start":"08:43.520 ","End":"08:51.250","Text":"Opposite is r minus R divided by hypotenuse,"},{"Start":"08:51.250 ","End":"08:55.809","Text":"which is this, which is the D. Now we have a third equation."},{"Start":"08:55.809 ","End":"09:06.105","Text":"We have sine of Alpha equals r minus R over D. Of course we know what R is,"},{"Start":"09:06.105 ","End":"09:08.160","Text":"we know what D is,"},{"Start":"09:08.160 ","End":"09:10.280","Text":"and so now we can solve it,"},{"Start":"09:10.280 ","End":"09:12.370","Text":"we have 3 equation."},{"Start":"09:12.560 ","End":"09:15.845","Text":"Now, if we go back to the original question,"},{"Start":"09:15.845 ","End":"09:20.825","Text":"what will the boy\u0027s height off the ground be when the carrousel will rotate?"},{"Start":"09:20.825 ","End":"09:25.230","Text":"We\u0027re trying to figure out what r is."},{"Start":"09:25.390 ","End":"09:32.503","Text":"What we have to do is we have to isolate out r. Now how do we do that?"},{"Start":"09:32.503 ","End":"09:37.100","Text":"For every 1 of the equations we can substitute n,"},{"Start":"09:37.100 ","End":"09:42.080","Text":"into another equation until we find out what r is,"},{"Start":"09:42.080 ","End":"09:48.020","Text":"as an expression of only the quantities that"},{"Start":"09:48.020 ","End":"09:53.840","Text":"we know such as mg Omega."},{"Start":"09:53.840 ","End":"09:56.695","Text":"Let\u0027s begin."},{"Start":"09:56.695 ","End":"09:59.605","Text":"Let\u0027s use this equation first."},{"Start":"09:59.605 ","End":"10:02.110","Text":"We\u0027re isolating out the r,"},{"Start":"10:02.110 ","End":"10:05.229","Text":"which equals multiply this,"},{"Start":"10:05.229 ","End":"10:13.240","Text":"so it\u0027s D sine of Alpha plus R. Very simple algebra until now."},{"Start":"10:13.240 ","End":"10:15.205","Text":"Now that I\u0027ve used this equation,"},{"Start":"10:15.205 ","End":"10:21.424","Text":"I\u0027ll put a tick next to it in order to show myself so that I don\u0027t go round in circles."},{"Start":"10:21.424 ","End":"10:26.920","Text":"In this equation, now we have sine Alpha and we don\u0027t know what sine Alpha is,"},{"Start":"10:26.920 ","End":"10:29.920","Text":"I can get sine Alpha from this equation though."},{"Start":"10:29.920 ","End":"10:35.380","Text":"From here, I can say that sine of Alpha equals to m Omega"},{"Start":"10:35.380 ","End":"10:40.960","Text":"squared r divided by T. Now I sub that in,"},{"Start":"10:40.960 ","End":"10:44.500","Text":"so I have that r equals D sine of Alpha,"},{"Start":"10:44.500 ","End":"10:50.995","Text":"which is equal to this m Omega squared r divided by T"},{"Start":"10:50.995 ","End":"11:00.220","Text":"plus R. But now my unknown is T. From this equation,"},{"Start":"11:00.220 ","End":"11:09.625","Text":"I can extract T and I can say that T equals mg divided by Cosine of Alpha."},{"Start":"11:09.625 ","End":"11:15.580","Text":"Then this equation now becomes Dm Omega"},{"Start":"11:15.580 ","End":"11:22.225","Text":"squared r divided by T which is divided by mg,"},{"Start":"11:22.225 ","End":"11:23.995","Text":"divided by cosine Alpha,"},{"Start":"11:23.995 ","End":"11:27.520","Text":"which brings it up here,"},{"Start":"11:27.520 ","End":"11:34.330","Text":"multiplied by cosine Alpha plus R. But now,"},{"Start":"11:34.330 ","End":"11:38.020","Text":"I don\u0027t know what cosine Alpha is equal to."},{"Start":"11:38.020 ","End":"11:42.010","Text":"How can I work out what cosine of Alpha is equal to?"},{"Start":"11:42.010 ","End":"11:45.220","Text":"So I\u0027ll remind you of this triangle over here."},{"Start":"11:45.220 ","End":"11:48.820","Text":"Let\u0027s draw it again and if you remember,"},{"Start":"11:48.820 ","End":"11:51.550","Text":"we said that this was Alpha."},{"Start":"11:51.550 ","End":"11:53.245","Text":"This is 90,"},{"Start":"11:53.245 ","End":"11:56.170","Text":"this is the hypotenuse which is equal to D,"},{"Start":"11:56.170 ","End":"11:58.105","Text":"it\u0027s the length of the string,"},{"Start":"11:58.105 ","End":"12:03.580","Text":"and then we\u0027re trying to find what cosine of Alpha is,"},{"Start":"12:03.580 ","End":"12:06.805","Text":"and that\u0027s this side."},{"Start":"12:06.805 ","End":"12:08.515","Text":"Now we don\u0027t know what that is,"},{"Start":"12:08.515 ","End":"12:11.590","Text":"but it\u0027s the adjacent side to the angle."},{"Start":"12:11.590 ","End":"12:16.180","Text":"Let\u0027s call it a, a for adjacent because it\u0027s the adjacent side."},{"Start":"12:16.180 ","End":"12:24.070","Text":"We know that cosine of Alpha is equal to adjacent over hypotenuse."},{"Start":"12:24.070 ","End":"12:27.370","Text":"But we don\u0027t know what this a is."},{"Start":"12:27.370 ","End":"12:37.945","Text":"But we can use Pythagoras because we know that this is r minus R remember this sect here."},{"Start":"12:37.945 ","End":"12:43.675","Text":"We can say that D squared is equal to"},{"Start":"12:43.675 ","End":"12:49.555","Text":"a squared plus r minus R squared,"},{"Start":"12:49.555 ","End":"12:55.420","Text":"and then we can say that a is equal to"},{"Start":"12:55.420 ","End":"13:03.410","Text":"the square root of D squared minus r minus R squared."},{"Start":"13:04.470 ","End":"13:08.485","Text":"Then we just plug it into this equation."},{"Start":"13:08.485 ","End":"13:15.235","Text":"Let\u0027s this instead of the cosine Alpha,"},{"Start":"13:15.235 ","End":"13:19.705","Text":"and that is our equation for R. But notice"},{"Start":"13:19.705 ","End":"13:25.479","Text":"that the first question is asking what the boy\u0027s height will be off the ground?"},{"Start":"13:25.479 ","End":"13:27.850","Text":"If we look in the picture,"},{"Start":"13:27.850 ","End":"13:30.985","Text":"we can see that it\u0027s going to be H,"},{"Start":"13:30.985 ","End":"13:36.685","Text":"which was the starting height plus this a."},{"Start":"13:36.685 ","End":"13:43.675","Text":"The total height is meant to be H plus a."},{"Start":"13:43.675 ","End":"13:47.665","Text":"Now a, we worked out to be this,"},{"Start":"13:47.665 ","End":"13:52.330","Text":"and in order to find out what this is,"},{"Start":"13:52.330 ","End":"13:54.684","Text":"because we have an r which is unknown,"},{"Start":"13:54.684 ","End":"13:59.320","Text":"then we can sub in this r into here,"},{"Start":"13:59.320 ","End":"14:02.960","Text":"and that is the answer to the question."},{"Start":"14:03.570 ","End":"14:08.230","Text":"Now in the second question we\u0027re being asked that if"},{"Start":"14:08.230 ","End":"14:12.940","Text":"a coin would drop out of the boy\u0027s pocket whilst the carousel spitting,"},{"Start":"14:12.940 ","End":"14:17.290","Text":"how far away from the center of the pole will the coin land?"},{"Start":"14:17.290 ","End":"14:20.440","Text":"Let\u0027s see how we answer this question."},{"Start":"14:20.440 ","End":"14:24.310","Text":"Now in this question because in the previous question we worked out"},{"Start":"14:24.310 ","End":"14:27.850","Text":"what r, a, and Alpha."},{"Start":"14:27.850 ","End":"14:32.320","Text":"We\u0027re, so here we can consider them as given."},{"Start":"14:32.320 ","End":"14:36.475","Text":"What we\u0027re going to do in order to answer question number 2,"},{"Start":"14:36.475 ","End":"14:40.405","Text":"we\u0027re going to look at the carousel from a bird\u0027s eye view."},{"Start":"14:40.405 ","End":"14:41.830","Text":"We have the center,"},{"Start":"14:41.830 ","End":"14:43.600","Text":"this is the rotating pole,"},{"Start":"14:43.600 ","End":"14:47.439","Text":"and it goes around in a circle."},{"Start":"14:47.439 ","End":"14:52.030","Text":"Now here we have the y-axis,"},{"Start":"14:52.030 ","End":"14:57.265","Text":"and here we have the x-axis."},{"Start":"14:57.265 ","End":"15:02.575","Text":"Now here is the swing with the boy going around in circles,"},{"Start":"15:02.575 ","End":"15:08.080","Text":"and this is all the way to here is"},{"Start":"15:08.080 ","End":"15:11.860","Text":"r. It\u0027s not the R because the R just"},{"Start":"15:11.860 ","End":"15:15.955","Text":"represents the distance when the carousel is at rest."},{"Start":"15:15.955 ","End":"15:20.545","Text":"The r represents when the carousel is rotating."},{"Start":"15:20.545 ","End":"15:27.160","Text":"Now we know that the carousel is rotating at a speed of Omega,"},{"Start":"15:27.160 ","End":"15:32.530","Text":"and we\u0027re being asked at what distance the coin will fall."},{"Start":"15:32.530 ","End":"15:34.840","Text":"To do that, let\u0027s first work out"},{"Start":"15:34.840 ","End":"15:39.130","Text":"the velocity that the coin will be at when it drops out of the boy\u0027s pocket."},{"Start":"15:39.130 ","End":"15:40.825","Text":"If this is Omega,"},{"Start":"15:40.825 ","End":"15:45.100","Text":"then we can know that the velocity is going to be equal"},{"Start":"15:45.100 ","End":"15:49.720","Text":"to Omega r. Now that we know the speed,"},{"Start":"15:49.720 ","End":"15:51.940","Text":"we know that the distance,"},{"Start":"15:51.940 ","End":"15:57.040","Text":"let\u0027s call it d is equal to velocity times time,"},{"Start":"15:57.040 ","End":"16:04.614","Text":"so it\u0027s Omega r multiplied by t. Now we have unknowns in this equation,"},{"Start":"16:04.614 ","End":"16:07.975","Text":"which is the d and the t. We don\u0027t know them."},{"Start":"16:07.975 ","End":"16:13.135","Text":"Now it\u0027s going to be very easy to find out the time that the coin is in the air,"},{"Start":"16:13.135 ","End":"16:14.620","Text":"the amount of time."},{"Start":"16:14.620 ","End":"16:16.855","Text":"But we\u0027re going to get to that in a second."},{"Start":"16:16.855 ","End":"16:19.450","Text":"Let\u0027s go back to the question for 1 moment."},{"Start":"16:19.450 ","End":"16:21.085","Text":"Whilst the carousel\u0027s spinning,"},{"Start":"16:21.085 ","End":"16:25.240","Text":"how far away from the center pole will the coin land?"},{"Start":"16:25.240 ","End":"16:27.325","Text":"From the center pole."},{"Start":"16:27.325 ","End":"16:32.330","Text":"They\u0027re asking me, say that this is d,"},{"Start":"16:34.800 ","End":"16:38.095","Text":"this is the distance that the coin travels."},{"Start":"16:38.095 ","End":"16:40.990","Text":"But they\u0027re not asking from this point."},{"Start":"16:40.990 ","End":"16:47.425","Text":"They\u0027re asking from here all of this,"},{"Start":"16:47.425 ","End":"16:51.355","Text":"this is the distance that we\u0027re being asked to find."},{"Start":"16:51.355 ","End":"16:54.355","Text":"Now this means that we need to use Pythagoras."},{"Start":"16:54.355 ","End":"16:58.615","Text":"Now this length is d as we know."},{"Start":"16:58.615 ","End":"17:01.375","Text":"This length is r as we know."},{"Start":"17:01.375 ","End":"17:04.060","Text":"Now by using Pythagoras, we can say,"},{"Start":"17:04.060 ","End":"17:10.900","Text":"let\u0027s call this hype because it\u0027s the hypotenuse."},{"Start":"17:10.900 ","End":"17:18.940","Text":"Hypotenuse, this is equal to the square root of r squared."},{"Start":"17:18.940 ","End":"17:23.650","Text":"This side plus this side squared,"},{"Start":"17:23.650 ","End":"17:32.900","Text":"which is Omega rt squared square root of all of that."},{"Start":"17:33.150 ","End":"17:39.055","Text":"Now in order to find the time that it takes for the coin to hit the ground,"},{"Start":"17:39.055 ","End":"17:45.715","Text":"we have to look at the free fall velocity in the y direction."},{"Start":"17:45.715 ","End":"17:49.690","Text":"We have that, this is the coin."},{"Start":"17:49.690 ","End":"17:54.924","Text":"Now obviously, it\u0027s also flying in this direction with a certain velocity."},{"Start":"17:54.924 ","End":"17:57.340","Text":"But for now it doesn\u0027t matter because we\u0027re"},{"Start":"17:57.340 ","End":"18:00.145","Text":"trying to find out the amount of time until it hits the ground."},{"Start":"18:00.145 ","End":"18:04.015","Text":"Now obviously the faster this velocity will move forwards"},{"Start":"18:04.015 ","End":"18:07.945","Text":"more in the time that it takes to hit the ground. But it doesn\u0027t matter."},{"Start":"18:07.945 ","End":"18:11.245","Text":"We\u0027re just trying to find the time that it takes to hit the ground."},{"Start":"18:11.245 ","End":"18:17.095","Text":"In order to find that we need to know what the height is from the ground."},{"Start":"18:17.095 ","End":"18:20.110","Text":"Now we worked it out here."},{"Start":"18:20.110 ","End":"18:25.550","Text":"The height is H plus a."},{"Start":"18:27.900 ","End":"18:33.670","Text":"Now this we can consider as a given number,"},{"Start":"18:33.670 ","End":"18:38.185","Text":"given amount, and so we can use this in our calculation."},{"Start":"18:38.185 ","End":"18:42.370","Text":"Now there\u0027s an equation for this, and it goes,"},{"Start":"18:42.370 ","End":"18:49.915","Text":"y equals y_0 plus v_0 t plus half at squared."},{"Start":"18:49.915 ","End":"18:52.090","Text":"Now I\u0027m going to write it in z."},{"Start":"18:52.090 ","End":"18:56.035","Text":"Because we\u0027re talking about the z axis."},{"Start":"18:56.035 ","End":"19:01.780","Text":"We have z equals z_0 plus"},{"Start":"19:01.780 ","End":"19:09.490","Text":"v_0 t plus 1/2 at squared."},{"Start":"19:09.490 ","End":"19:11.575","Text":"Now here I said,"},{"Start":"19:11.575 ","End":"19:15.790","Text":"this is the end distance that we wanted to find out."},{"Start":"19:15.790 ","End":"19:20.074","Text":"That\u0027s when it equals to 0 because it\u0027s on the ground, it\u0027s at height 0."},{"Start":"19:20.074 ","End":"19:24.864","Text":"Z_0 is our starting position,"},{"Start":"19:24.864 ","End":"19:32.920","Text":"which is H plus a plus v_0 t,"},{"Start":"19:32.920 ","End":"19:40.315","Text":"which here is equal to 0 because our starting speed in the z direction is equal to 0,"},{"Start":"19:40.315 ","End":"19:45.820","Text":"we have a velocity in the radial direction or in the x direction,"},{"Start":"19:45.820 ","End":"19:47.275","Text":"however you want to look at it."},{"Start":"19:47.275 ","End":"19:50.500","Text":"But in the z direction, it\u0027s 0."},{"Start":"19:50.500 ","End":"19:55.310","Text":"So 0 plus, now here it\u0027s going to be minus"},{"Start":"19:55.310 ","End":"19:59.930","Text":"because our acceleration is actually equal to g pointing downwards."},{"Start":"19:59.930 ","End":"20:06.850","Text":"So it\u0027s going to be minus because it\u0027s pointing downwards, minus 1/2gt squared."},{"Start":"20:06.850 ","End":"20:08.765","Text":"Then if we rearrange it,"},{"Start":"20:08.765 ","End":"20:17.250","Text":"we can say that H plus a equals 1/2gt squared."},{"Start":"20:17.350 ","End":"20:21.530","Text":"Now in order to find out what t is,"},{"Start":"20:21.530 ","End":"20:22.830","Text":"we can isolate it,"},{"Start":"20:22.830 ","End":"20:28.640","Text":"and we can say that t is equal to H"},{"Start":"20:28.640 ","End":"20:35.090","Text":"plus a multiplied by 2 divided by g,"},{"Start":"20:35.090 ","End":"20:36.845","Text":"and because this is a squared,"},{"Start":"20:36.845 ","End":"20:39.120","Text":"it\u0027s the square root of all of this."},{"Start":"20:39.120 ","End":"20:44.620","Text":"Then we just sub in t into here."},{"Start":"20:44.620 ","End":"20:47.120","Text":"Then we get that the hypotenuse,"},{"Start":"20:47.120 ","End":"20:51.530","Text":"which is the distance that we\u0027re trying to find that the coin falls from"},{"Start":"20:51.530 ","End":"21:00.425","Text":"the center pole is equal to the square root of r squared plus Omega rt,"},{"Start":"21:00.425 ","End":"21:04.490","Text":"which is square root of 2H plus"},{"Start":"21:04.490 ","End":"21:12.590","Text":"a divided by g. All of this squared."},{"Start":"21:12.590 ","End":"21:17.600","Text":"Then you just finish it off with some algebra to make it a bit neater."},{"Start":"21:17.600 ","End":"21:21.390","Text":"Here, we have finished this question."}],"ID":9263},{"Watched":false,"Name":"Exercise 6","Duration":"7m 12s","ChapterTopicVideoID":8994,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.500 ","End":"00:08.080","Text":"Hello. We\u0027re given a rotating disk with center here"},{"Start":"00:08.080 ","End":"00:12.450","Text":"and we\u0027re being asked what the furthest distance that"},{"Start":"00:12.450 ","End":"00:17.070","Text":"the ladybird can be from the center of the disk without it sliding off."},{"Start":"00:17.070 ","End":"00:26.595","Text":"We\u0027re being asked what this distance is between the center and the ladybird."},{"Start":"00:26.595 ","End":"00:32.115","Text":"We\u0027re given the coefficient of friction Mu over here."},{"Start":"00:32.115 ","End":"00:35.045","Text":"How do I deal with this question?"},{"Start":"00:35.045 ","End":"00:39.170","Text":"Well, we\u0027ve already learned that sometimes it\u0027s easier to just"},{"Start":"00:39.170 ","End":"00:46.265","Text":"pretend that the disk isn\u0027t moving and add in a frame of reference."},{"Start":"00:46.265 ","End":"00:49.295","Text":"If we look at the disk from above,"},{"Start":"00:49.295 ","End":"00:53.360","Text":"we can see that here is the ladybird,"},{"Start":"00:53.360 ","End":"00:55.865","Text":"here\u0027s the center,"},{"Start":"00:55.865 ","End":"01:03.005","Text":"and the only thing we\u0027re interested in is the radius between the center and the ladybird."},{"Start":"01:03.005 ","End":"01:07.310","Text":"We\u0027re not interested in the radius of the circle itself."},{"Start":"01:07.310 ","End":"01:11.880","Text":"Then if I add here the frame of reference."},{"Start":"01:14.170 ","End":"01:18.740","Text":"Here again, the r over here is referring to"},{"Start":"01:18.740 ","End":"01:22.610","Text":"the radius between the center of the circle and the ladybird."},{"Start":"01:22.610 ","End":"01:25.715","Text":"Then because I\u0027ve added in this frame of reference,"},{"Start":"01:25.715 ","End":"01:27.800","Text":"then as far as I\u0027m concerned,"},{"Start":"01:27.800 ","End":"01:30.385","Text":"it\u0027s as if the disk isn\u0027t moving at all."},{"Start":"01:30.385 ","End":"01:34.265","Text":"Now if we look at this disk and the ladybird from the side,"},{"Start":"01:34.265 ","End":"01:35.945","Text":"now here of course,"},{"Start":"01:35.945 ","End":"01:38.255","Text":"is the center of the disk."},{"Start":"01:38.255 ","End":"01:45.385","Text":"Here again is our r. The radius between the center of the disk and the ladybird."},{"Start":"01:45.385 ","End":"01:50.175","Text":"Then there\u0027s some force here,"},{"Start":"01:50.175 ","End":"01:52.035","Text":"which is, as we know,"},{"Start":"01:52.035 ","End":"01:54.885","Text":"M Omega squared r,"},{"Start":"01:54.885 ","End":"01:59.475","Text":"where r is the center till the ladybird."},{"Start":"01:59.475 ","End":"02:05.210","Text":"My question is, what is the furthest that there\u0027s the largest number that is r"},{"Start":"02:05.210 ","End":"02:11.680","Text":"can be such that the ladybird doesn\u0027t slide and fly off the disk."},{"Start":"02:11.680 ","End":"02:19.580","Text":"Now let\u0027s go over all of the forces that are working on the ladybird. Let\u0027s draw it here."},{"Start":"02:19.580 ","End":"02:23.500","Text":"Going downwards, we have mg,"},{"Start":"02:23.500 ","End":"02:27.574","Text":"going upwards, we have the normal force."},{"Start":"02:27.574 ","End":"02:30.650","Text":"Then we have this force,"},{"Start":"02:30.650 ","End":"02:36.305","Text":"and in the opposite direction to balance out this frame of reference,"},{"Start":"02:36.305 ","End":"02:39.060","Text":"we have the frictional force."},{"Start":"02:39.060 ","End":"02:41.150","Text":"Now, why do we know that there\u0027s"},{"Start":"02:41.150 ","End":"02:45.065","Text":"this frictional force that balances out the frame of reference."},{"Start":"02:45.065 ","End":"02:51.945","Text":"It\u0027s because we know in the question we\u0027re being told that the ladybird isn\u0027t moving,"},{"Start":"02:51.945 ","End":"02:55.715","Text":"that the ladybird is static at rest and we don\u0027t want her to move."},{"Start":"02:55.715 ","End":"02:58.280","Text":"That\u0027s the whole point. That\u0027s why we know that there\u0027s"},{"Start":"02:58.280 ","End":"03:02.765","Text":"this frictional force countering the other force."},{"Start":"03:02.765 ","End":"03:09.620","Text":"Now, our unknowns are the normal force and the frictional force."},{"Start":"03:09.620 ","End":"03:17.320","Text":"We\u0027re being asked what the maximum r can be such that the ladybird doesn\u0027t slide."},{"Start":"03:17.320 ","End":"03:22.250","Text":"In other words, that this expression"},{"Start":"03:22.250 ","End":"03:29.220","Text":"here does not become larger than this, than the friction."},{"Start":"03:29.930 ","End":"03:33.425","Text":"Let\u0027s write the forces that are working here."},{"Start":"03:33.425 ","End":"03:40.760","Text":"We have the normal force is equal to mg because the ladybird isn\u0027t moving up or down,"},{"Start":"03:40.760 ","End":"03:42.380","Text":"which means that she\u0027s at rest."},{"Start":"03:42.380 ","End":"03:43.985","Text":"The forces are equal,"},{"Start":"03:43.985 ","End":"03:49.205","Text":"and we have that the frictional force equals"},{"Start":"03:49.205 ","End":"03:56.690","Text":"m Omega squared r. That\u0027s because the ladybird also isn\u0027t moving side to side."},{"Start":"03:56.690 ","End":"04:05.265","Text":"That\u0027s the whole point. Now, we also know that the frictional force equals Mu N,"},{"Start":"04:05.265 ","End":"04:10.170","Text":"and we know that N is mg from the first equation."},{"Start":"04:10.170 ","End":"04:18.335","Text":"We know that the frictional force equals Mu mg. Now the frictional force, we know."},{"Start":"04:18.335 ","End":"04:19.805","Text":"Now my question is,"},{"Start":"04:19.805 ","End":"04:25.970","Text":"what is the largest frictional force that I can get which will hold this?"},{"Start":"04:25.970 ","End":"04:32.040","Text":"The largest frictional force which will hold Mu mg. Now this is"},{"Start":"04:32.040 ","End":"04:37.880","Text":"clear because the larger the radius,"},{"Start":"04:37.880 ","End":"04:42.575","Text":"the further away the ladybird is from the center of the disk,"},{"Start":"04:42.575 ","End":"04:47.630","Text":"the larger the frictional force will have to be in order to hold the ladybird in."},{"Start":"04:47.630 ","End":"04:53.120","Text":"Now, obviously, if the ladybird was standing at the center of the disk,"},{"Start":"04:53.120 ","End":"04:55.610","Text":"then the friction could be 0."},{"Start":"04:55.610 ","End":"04:57.425","Text":"This is why, as you know,"},{"Start":"04:57.425 ","End":"05:05.060","Text":"the equation generally is frictional force is smaller or equal to Mu N. But here,"},{"Start":"05:05.060 ","End":"05:14.330","Text":"we can rub that out because we\u0027re being asked what the maximum is?"},{"Start":"05:14.330 ","End":"05:21.500","Text":"The maximum is when the frictional force is equal to Mu N. If it was smaller or equal to,"},{"Start":"05:21.500 ","End":"05:23.060","Text":"then again it could be 0,"},{"Start":"05:23.060 ","End":"05:26.615","Text":"it could be 1, it could be 2,"},{"Start":"05:26.615 ","End":"05:31.010","Text":"but the maximum is when the frictional force is equal to Mu N,"},{"Start":"05:31.010 ","End":"05:35.090","Text":"which here is Mu mg. Now,"},{"Start":"05:35.090 ","End":"05:43.935","Text":"all that\u0027s left for me to ask is when does f equal m Omega squared r?"},{"Start":"05:43.935 ","End":"05:53.289","Text":"When does Mu mg equals m Omega squared?"},{"Start":"05:53.289 ","End":"05:55.340","Text":"This is my question."},{"Start":"05:55.340 ","End":"05:57.505","Text":"This is what I need to know."},{"Start":"05:57.505 ","End":"06:01.805","Text":"They\u0027re asking me to find out what the radius is."},{"Start":"06:01.805 ","End":"06:08.400","Text":"Because all the rest of the variables I already know."},{"Start":"06:08.480 ","End":"06:11.390","Text":"Now what I need to do is I can look at"},{"Start":"06:11.390 ","End":"06:14.465","Text":"this equation and I see that there\u0027s M\u0027s on both sides."},{"Start":"06:14.465 ","End":"06:16.295","Text":"I can cancel them out,"},{"Start":"06:16.295 ","End":"06:19.735","Text":"and then because I want to know what r is,"},{"Start":"06:19.735 ","End":"06:23.370","Text":"I\u0027ll just put my r on 1 side,"},{"Start":"06:23.370 ","End":"06:28.410","Text":"equals and then I\u0027ll just have"},{"Start":"06:28.410 ","End":"06:34.395","Text":"Mu g divided by Omega squared."},{"Start":"06:34.395 ","End":"06:40.350","Text":"This is the largest radius that the ladybird can be"},{"Start":"06:40.350 ","End":"06:46.815","Text":"found on the disk such that the ladybird doesn\u0027t fly off the disk."},{"Start":"06:46.815 ","End":"06:49.869","Text":"This is the largest radius."},{"Start":"06:50.140 ","End":"06:54.260","Text":"Now, obviously the radius can be smaller than this."},{"Start":"06:54.260 ","End":"07:02.430","Text":"But if the radius exceeds the limit of Mu g divided by Omega squared,"},{"Start":"07:02.430 ","End":"07:07.605","Text":"then the ladybird will fly off and slide off the disk."},{"Start":"07:07.605 ","End":"07:09.830","Text":"Now we\u0027ve finished this question."},{"Start":"07:09.830 ","End":"07:13.470","Text":"Let\u0027s move on to the rest of the examples."}],"ID":9264},{"Watched":false,"Name":"Exercise 7","Duration":"7m 59s","ChapterTopicVideoID":8995,"CourseChapterTopicPlaylistID":85365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"In the question in front of us,"},{"Start":"00:01.890 ","End":"00:06.450","Text":"we\u0027re given a diagram where there are 2 strings attached to a pole."},{"Start":"00:06.450 ","End":"00:11.475","Text":"This is 1 string and this is the other string."},{"Start":"00:11.475 ","End":"00:15.180","Text":"At the midpoint of each string, we have a mass."},{"Start":"00:15.180 ","End":"00:16.395","Text":"This is mass number 1,"},{"Start":"00:16.395 ","End":"00:18.730","Text":"and this is mass number 2."},{"Start":"00:18.950 ","End":"00:24.795","Text":"Then pull in the strings have no masses and we\u0027re given Omega,"},{"Start":"00:24.795 ","End":"00:27.885","Text":"which is the frequency of rotation."},{"Start":"00:27.885 ","End":"00:32.975","Text":"We\u0027re being ask, what is the tension in the string T?"},{"Start":"00:32.975 ","End":"00:36.965","Text":"The system in front of us is symmetrical, so in that case,"},{"Start":"00:36.965 ","End":"00:39.245","Text":"what\u0027s happening on the right side of the diagram"},{"Start":"00:39.245 ","End":"00:42.520","Text":"and on the left side of the diagram is identical."},{"Start":"00:42.520 ","End":"00:47.935","Text":"This tension, we can call it T_1 because it\u0027s attached to a mass."},{"Start":"00:47.935 ","End":"00:50.810","Text":"This side of the string,"},{"Start":"00:50.810 ","End":"00:55.117","Text":"the tension can be called T_2 because it\u0027s also attached to a mass,"},{"Start":"00:55.117 ","End":"00:57.890","Text":"so they might have different tensions."},{"Start":"00:57.890 ","End":"01:01.540","Text":"Now, we\u0027re given the length of small d,"},{"Start":"01:01.540 ","End":"01:06.845","Text":"which is half the length of the string and we\u0027re given the length of the capital D,"},{"Start":"01:06.845 ","End":"01:10.645","Text":"which is this length."},{"Start":"01:10.645 ","End":"01:14.980","Text":"What is unknown is T_1 and T_2,"},{"Start":"01:14.980 ","End":"01:17.885","Text":"and that is what we\u0027re being asked to find."},{"Start":"01:17.885 ","End":"01:20.665","Text":"Now let\u0027s see how we answer this question."},{"Start":"01:20.665 ","End":"01:25.525","Text":"Because we already know that we need to find what T_1 and T_2 is,"},{"Start":"01:25.525 ","End":"01:27.730","Text":"we know that we have 2 unknowns,"},{"Start":"01:27.730 ","End":"01:31.830","Text":"which means that we\u0027re looking for 2 equations in order to solve this."},{"Start":"01:31.830 ","End":"01:36.070","Text":"Let\u0027s begin by drawing a free body diagram because it\u0027s symmetrical,"},{"Start":"01:36.070 ","End":"01:40.910","Text":"we only have to draw a free body diagram for just one of the masses."},{"Start":"01:41.330 ","End":"01:47.230","Text":"We have on the mass mg pointing downwards."},{"Start":"01:47.230 ","End":"01:49.975","Text":"Then we have in this direction,"},{"Start":"01:49.975 ","End":"01:57.270","Text":"we have T_1 and in this direction, we have T_2."},{"Start":"01:57.270 ","End":"02:02.345","Text":"Now, we\u0027re going to split into components."},{"Start":"02:02.345 ","End":"02:08.195","Text":"We can say that this angle is Alpha and also this angle is Alpha."},{"Start":"02:08.195 ","End":"02:10.565","Text":"How do I know that both of them are equal?"},{"Start":"02:10.565 ","End":"02:12.965","Text":"Let\u0027s look at this side of the triangle for a second."},{"Start":"02:12.965 ","End":"02:14.990","Text":"Now, on this side of the diagram,"},{"Start":"02:14.990 ","End":"02:19.445","Text":"we\u0027ll notice that d represents half the length of the string,"},{"Start":"02:19.445 ","End":"02:22.070","Text":"which means that this length over here is also equal"},{"Start":"02:22.070 ","End":"02:25.790","Text":"to small d. The large D if we split it in the mid point here,"},{"Start":"02:25.790 ","End":"02:28.985","Text":"this section will be half large D,"},{"Start":"02:28.985 ","End":"02:35.390","Text":"and this is also half large capital D. Which means that the angle here,"},{"Start":"02:35.390 ","End":"02:40.910","Text":"and angle here are identical because this is an isosceles triangle."},{"Start":"02:40.910 ","End":"02:45.755","Text":"Now what do we have to do is we have to split up everything into the components."},{"Start":"02:45.755 ","End":"02:47.810","Text":"Let\u0027s draw the mass again."},{"Start":"02:47.810 ","End":"02:57.185","Text":"Now, pointing upwards, we have T_2 sine of Alpha and pointing downwards,"},{"Start":"02:57.185 ","End":"03:02.920","Text":"we have T_1 sine of Alpha."},{"Start":"03:02.920 ","End":"03:06.915","Text":"We also have mg pointing downwards."},{"Start":"03:06.915 ","End":"03:12.790","Text":"Then to the sides we have 2 forces,"},{"Start":"03:12.790 ","End":"03:20.630","Text":"which is T_2 cosine of Alpha and T_1 cosine of Alpha."},{"Start":"03:20.630 ","End":"03:24.320","Text":"Whoever doesn\u0027t know how I split this up into the separate components,"},{"Start":"03:24.320 ","End":"03:27.790","Text":"please go back to the lesson where explain how you do that."},{"Start":"03:27.790 ","End":"03:32.340","Text":"Now let\u0027s work out the sum of all the forces."},{"Start":"03:32.340 ","End":"03:37.065","Text":"We can call this axis,"},{"Start":"03:37.065 ","End":"03:43.470","Text":"the z-axis and this will be the positive direction pointing upwards."},{"Start":"03:43.470 ","End":"03:49.130","Text":"We can say that the sum of all the forces on the z-axis is equal"},{"Start":"03:49.130 ","End":"03:55.085","Text":"to T_2 sine Alpha because that\u0027s pointing upwards,"},{"Start":"03:55.085 ","End":"04:03.125","Text":"minus T_1 sine Alpha minus mg. Now this is"},{"Start":"04:03.125 ","End":"04:12.065","Text":"equal to the sum of the accelerations mass times the acceleration in the z-direction."},{"Start":"04:12.065 ","End":"04:15.725","Text":"Now the acceleration in the z-direction is equal to 0,"},{"Start":"04:15.725 ","End":"04:21.190","Text":"which means that this whole equation is equal to 0."},{"Start":"04:21.320 ","End":"04:25.275","Text":"We can just not even look at this section."},{"Start":"04:25.275 ","End":"04:31.235","Text":"This is going to be our first equation that we\u0027re going to use in order to solve this."},{"Start":"04:31.235 ","End":"04:34.725","Text":"Now, what we\u0027re going to do is we\u0027re going to work out"},{"Start":"04:34.725 ","End":"04:41.010","Text":"the sum of all the forces on the radial axis."},{"Start":"04:41.010 ","End":"04:51.165","Text":"On the radial axis we\u0027re going to take out cos Alpha because it appears in both."},{"Start":"04:51.165 ","End":"04:56.320","Text":"Then we\u0027re going to do T_2 plus T_1."},{"Start":"04:56.320 ","End":"05:02.080","Text":"This equals mass times the acceleration in the radial axis."},{"Start":"05:02.080 ","End":"05:09.145","Text":"The acceleration in the radial axis is equal to Omega squared times I."},{"Start":"05:09.145 ","End":"05:14.875","Text":"This is going to be equal to mass Omega squared times r."},{"Start":"05:14.875 ","End":"05:21.230","Text":"Then this is going to be our second equation that we use."},{"Start":"05:21.230 ","End":"05:25.110","Text":"In this section we can just ignore because I don\u0027t need it right now."},{"Start":"05:25.110 ","End":"05:28.725","Text":"Let\u0027s see how many unknowns we have."},{"Start":"05:28.725 ","End":"05:33.795","Text":"We know that T_2 is an unknown,"},{"Start":"05:33.795 ","End":"05:35.700","Text":"T_1 is an unknown,"},{"Start":"05:35.700 ","End":"05:38.040","Text":"Alpha is an unknown."},{"Start":"05:38.040 ","End":"05:41.700","Text":"We know that also r is an unknown."},{"Start":"05:41.700 ","End":"05:45.160","Text":"We have 4 unknowns and we\u0027ve only got 2 equations,"},{"Start":"05:45.160 ","End":"05:49.000","Text":"which means that we need to find another 2 equations in order to solve this."},{"Start":"05:49.000 ","End":"05:51.970","Text":"Now, I feel like it will be a lot easier to find"},{"Start":"05:51.970 ","End":"05:55.525","Text":"the other 2 equations from using simple trigonometry."},{"Start":"05:55.525 ","End":"05:57.670","Text":"Let\u0027s look what we can do."},{"Start":"05:57.670 ","End":"05:59.859","Text":"Let\u0027s go back to our diagram."},{"Start":"05:59.859 ","End":"06:03.545","Text":"Now this is r,"},{"Start":"06:03.545 ","End":"06:09.515","Text":"and this section here is capital D/2,"},{"Start":"06:09.515 ","End":"06:15.645","Text":"and this is small d. Through Pythagoras,"},{"Start":"06:15.645 ","End":"06:24.725","Text":"we know that r squared plus capital D/2 squared equals the hypotenuse,"},{"Start":"06:24.725 ","End":"06:27.680","Text":"which here is small d squared."},{"Start":"06:27.680 ","End":"06:34.970","Text":"This is another equation that we can use in order to find out what T_1 and T_2 is."},{"Start":"06:34.970 ","End":"06:43.200","Text":"The next equation still using trigonometry is that I can use sine Alpha."},{"Start":"06:43.200 ","End":"06:46.780","Text":"Because now we\u0027re trying to find what Alpha equals,"},{"Start":"06:46.780 ","End":"06:53.893","Text":"which equals according to the SOHCAHTOA sine opposite over hypotenuse,"},{"Start":"06:53.893 ","End":"06:59.340","Text":"so our opposite is d/2 divided by our hypotenuse,"},{"Start":"06:59.340 ","End":"07:02.505","Text":"which is small d, and that\u0027s it."},{"Start":"07:02.505 ","End":"07:05.145","Text":"This is our fourth equation."},{"Start":"07:05.145 ","End":"07:08.795","Text":"Now we have 4 unknowns and 4 equations."},{"Start":"07:08.795 ","End":"07:11.815","Text":"Here it\u0027s just now a question of solving"},{"Start":"07:11.815 ","End":"07:16.555","Text":"these 4 equations to find what T_2 and T_1 equals."},{"Start":"07:16.555 ","End":"07:23.785","Text":"Now, a warning a lot of students decide, instead of writing,"},{"Start":"07:23.785 ","End":"07:28.530","Text":"let\u0027s say for sine Alpha capital D/2/d,"},{"Start":"07:28.530 ","End":"07:34.165","Text":"instead of the small though d they\u0027ll write T_2."},{"Start":"07:34.165 ","End":"07:36.560","Text":"Because it\u0027s equal to that."},{"Start":"07:36.560 ","End":"07:41.300","Text":"However, this is incorrect because small d represents a length."},{"Start":"07:41.300 ","End":"07:43.475","Text":"That\u0027s what we\u0027re trying to use."},{"Start":"07:43.475 ","End":"07:46.115","Text":"The t\u0027s represent a vector."},{"Start":"07:46.115 ","End":"07:48.680","Text":"Because we don\u0027t know the size of the vector,"},{"Start":"07:48.680 ","End":"07:52.070","Text":"we cannot use that in this calculation."},{"Start":"07:52.070 ","End":"07:57.050","Text":"Be aware when you\u0027re working out distances, only use distances."},{"Start":"07:57.050 ","End":"07:59.700","Text":"That\u0027s the end of this lesson."}],"ID":9265}],"Thumbnail":null,"ID":85365},{"Name":"End of Chapter Questions (Advanced)","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 8","Duration":"22m 57s","ChapterTopicVideoID":9001,"CourseChapterTopicPlaylistID":85366,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.625","Text":"Hello. In this video,"},{"Start":"00:02.625 ","End":"00:06.345","Text":"we have a bird, which is this in red,"},{"Start":"00:06.345 ","End":"00:10.050","Text":"which is flying from a rotating disk in the direction"},{"Start":"00:10.050 ","End":"00:14.205","Text":"of the x-axis at a given velocity V_0."},{"Start":"00:14.205 ","End":"00:20.805","Text":"The radius of the disc is R. At that exact moment,"},{"Start":"00:20.805 ","End":"00:26.220","Text":"a policeman mounts the disk and rotates with the disk towards the bird."},{"Start":"00:26.220 ","End":"00:30.470","Text":"What velocity will the policeman speedometer measure if it is"},{"Start":"00:30.470 ","End":"00:35.465","Text":"known that the speedometer only measures the velocity in his direction?"},{"Start":"00:35.465 ","End":"00:41.660","Text":"What does this mean? This means that if the policeman is standing here and the bird is"},{"Start":"00:41.660 ","End":"00:50.375","Text":"stationary here and flying instead at V_0 in this direction."},{"Start":"00:50.375 ","End":"00:53.780","Text":"That means that only if the bird is flying towards"},{"Start":"00:53.780 ","End":"00:59.090","Text":"the policeman will his speedometer be able to notice the velocity."},{"Start":"00:59.090 ","End":"01:00.440","Text":"If the bird however,"},{"Start":"01:00.440 ","End":"01:02.869","Text":"is flying in this direction,"},{"Start":"01:02.869 ","End":"01:05.600","Text":"because the bird isn\u0027t flying towards"},{"Start":"01:05.600 ","End":"01:12.709","Text":"the police officer then the velocity that the speedometer will record will be 0."},{"Start":"01:12.709 ","End":"01:16.970","Text":"Only if the bird is flying directly towards"},{"Start":"01:16.970 ","End":"01:20.945","Text":"the police officer will there be a velocity recorded."},{"Start":"01:20.945 ","End":"01:25.250","Text":"Otherwise, if the velocity is perpendicular to"},{"Start":"01:25.250 ","End":"01:31.320","Text":"the direction of the police officer in this direction then the velocity will be 0."},{"Start":"01:31.540 ","End":"01:36.680","Text":"Now that we\u0027ve understood the question let\u0027s see how we answer this."},{"Start":"01:36.680 ","End":"01:41.270","Text":"Now, in this diagram over here I\u0027ve drawn it a moment"},{"Start":"01:41.270 ","End":"01:47.105","Text":"after the bird has started to fly and the policeman has rotated slightly."},{"Start":"01:47.105 ","End":"01:51.200","Text":"Here is the policeman and here is the bird that\u0027s moved."},{"Start":"01:51.200 ","End":"01:54.770","Text":"Now, the policeman is recording the velocity from"},{"Start":"01:54.770 ","End":"01:59.450","Text":"the speedometer in this direction in this line because the component"},{"Start":"01:59.450 ","End":"02:02.600","Text":"of velocity that\u0027s going in"},{"Start":"02:02.600 ","End":"02:11.954","Text":"this direction cannot be recorded by the policeman so we\u0027re recording this."},{"Start":"02:11.954 ","End":"02:16.400","Text":"To solve this we have to find the relative velocity."},{"Start":"02:16.400 ","End":"02:22.250","Text":"Now, in order to find the relative velocity we have to use Cartesian coordinates."},{"Start":"02:22.250 ","End":"02:27.110","Text":"Because if we\u0027re going to use polar coordinates we can\u0027t do a polar coordinate minus"},{"Start":"02:27.110 ","End":"02:32.785","Text":"another polar coordinate in order to find the relative distance."},{"Start":"02:32.785 ","End":"02:39.450","Text":"Whenever working with circular motion we have to use Cartesian coordinates."},{"Start":"02:39.450 ","End":"02:44.405","Text":"Now let\u0027s write down an equation for the position of the bird,"},{"Start":"02:44.405 ","End":"02:48.237","Text":"the position of the policeman,"},{"Start":"02:48.237 ","End":"02:51.105","Text":"and let\u0027s see what we do from that case."},{"Start":"02:51.105 ","End":"02:56.990","Text":"Let\u0027s write the coordinates for the position of the policeman."},{"Start":"02:56.990 ","End":"03:02.420","Text":"We\u0027ll write it x_p for the policeman and then for"},{"Start":"03:02.420 ","End":"03:08.570","Text":"the bird we\u0027ll write x_b."},{"Start":"03:08.570 ","End":"03:10.820","Text":"Let\u0027s look at the policeman."},{"Start":"03:10.820 ","End":"03:14.555","Text":"When the policeman starts here."},{"Start":"03:14.555 ","End":"03:16.430","Text":"Let\u0027s draw it."},{"Start":"03:16.430 ","End":"03:17.810","Text":"Okay, fine here."},{"Start":"03:17.810 ","End":"03:26.100","Text":"This is our y-axis and this is our x-axis with this being the origin."},{"Start":"03:26.210 ","End":"03:31.280","Text":"The bird flies and the police man rotates,"},{"Start":"03:31.280 ","End":"03:35.910","Text":"say he gets to this position here."},{"Start":"03:35.950 ","End":"03:42.375","Text":"His coordinate will be R. Sorry."},{"Start":"03:42.375 ","End":"03:47.730","Text":"Then this angle here is"},{"Start":"03:47.730 ","End":"03:55.895","Text":"Omega t. Because it\u0027s here and we want this distance,"},{"Start":"03:55.895 ","End":"04:01.390","Text":"this position, because it\u0027s the opposite angle it will be"},{"Start":"04:01.390 ","End":"04:07.655","Text":"R sine of Omega t. Now,"},{"Start":"04:07.655 ","End":"04:17.010","Text":"as for the y coordinates of the policeman it will be R cosine Omega t,"},{"Start":"04:17.010 ","End":"04:19.565","Text":"for the exact same reason,"},{"Start":"04:19.565 ","End":"04:22.440","Text":"but now we want this."},{"Start":"04:23.600 ","End":"04:26.100","Text":"Now, as for the bird."},{"Start":"04:26.100 ","End":"04:31.595","Text":"We know that the bird is flying along the x-axis,"},{"Start":"04:31.595 ","End":"04:37.460","Text":"in which case that the equation that\u0027s obvious to use is V_0,"},{"Start":"04:37.460 ","End":"04:44.105","Text":"the velocity multiplied by t. It\u0027s the usual equation,"},{"Start":"04:44.105 ","End":"04:45.871","Text":"velocity is distance over time."},{"Start":"04:45.871 ","End":"04:46.970","Text":"But we\u0027re looking for distance,"},{"Start":"04:46.970 ","End":"04:48.980","Text":"so it\u0027s velocity times time."},{"Start":"04:48.980 ","End":"04:54.545","Text":"The y equation for the bird is equal to,"},{"Start":"04:54.545 ","End":"04:59.765","Text":"of course, 0 because it\u0027s not moving up or down in the y direction at all."},{"Start":"04:59.765 ","End":"05:06.380","Text":"Now, using the locations we have to work out what their velocities are."},{"Start":"05:06.380 ","End":"05:11.840","Text":"We can say that the velocity of the police man is equal"},{"Start":"05:11.840 ","End":"05:17.585","Text":"to the differentiated position equation,"},{"Start":"05:17.585 ","End":"05:23.170","Text":"x dots, which equals to, let\u0027s differentiate."},{"Start":"05:23.170 ","End":"05:31.145","Text":"We\u0027re doing it with regards to t so it will be Omega r cosine of"},{"Start":"05:31.145 ","End":"05:39.695","Text":"Omega t. The differentiated location of the bird,"},{"Start":"05:39.695 ","End":"05:42.620","Text":"which is the velocity of the bird, sorry,"},{"Start":"05:42.620 ","End":"05:44.555","Text":"this is p and this is b,"},{"Start":"05:44.555 ","End":"05:47.930","Text":"will be again with regards to t V_0."},{"Start":"05:47.930 ","End":"05:55.880","Text":"In the y direction the y dot the velocity in the y-direction of"},{"Start":"05:55.880 ","End":"06:06.140","Text":"the policeman will be negative Omega R,"},{"Start":"06:06.140 ","End":"06:16.775","Text":"sine Omega t and the y velocity of the bird will equal 0 still."},{"Start":"06:16.775 ","End":"06:24.200","Text":"Now what we want to do is to find the relative velocity. How do we do that?"},{"Start":"06:24.200 ","End":"06:28.685","Text":"We do the velocity of the policeman"},{"Start":"06:28.685 ","End":"06:36.890","Text":"minus the velocity of the bird in both the x and y direction."},{"Start":"06:36.890 ","End":"06:42.785","Text":"We have v relative is equal to"},{"Start":"06:42.785 ","End":"06:53.475","Text":"Omega R cos Omega t minus V_0."},{"Start":"06:53.475 ","End":"07:04.280","Text":"Then in the y-direction it\u0027s minus Omega R sine Omega t minus 0,"},{"Start":"07:04.280 ","End":"07:06.110","Text":"which is just that."},{"Start":"07:06.110 ","End":"07:14.315","Text":"Now what we have here is we have the relative velocity with regards to the x and y-axis."},{"Start":"07:14.315 ","End":"07:16.850","Text":"Now, this isn\u0027t what we want because these axes"},{"Start":"07:16.850 ","End":"07:20.995","Text":"aren\u0027t comfortable for us in order to do these calculations."},{"Start":"07:20.995 ","End":"07:26.210","Text":"We want to change the axes into an axis that is more uncomfortable for us."},{"Start":"07:26.210 ","End":"07:33.080","Text":"If we look at this diagram over here we wanted an axis going in this direction."},{"Start":"07:33.080 ","End":"07:39.304","Text":"We can call it the parallel axis and going in this direction,"},{"Start":"07:39.304 ","End":"07:44.800","Text":"which is perpendicular and we can call that the normal component."},{"Start":"07:44.800 ","End":"07:47.300","Text":"Let\u0027s see how we can do this."},{"Start":"07:47.300 ","End":"07:53.360","Text":"What we can do is we can look and use angles in order to find"},{"Start":"07:53.360 ","End":"08:00.200","Text":"the projection of the y-axis on here and the x-axis on here."},{"Start":"08:00.200 ","End":"08:04.385","Text":"We can do that, but it\u0027s a bit messy and a bit chaotic."},{"Start":"08:04.385 ","End":"08:07.205","Text":"Let\u0027s find a different way that we can do this."},{"Start":"08:07.205 ","End":"08:12.980","Text":"There\u0027s an equation to find the projection with the projection of vectors."},{"Start":"08:12.980 ","End":"08:15.020","Text":"Let\u0027s see how we do this."},{"Start":"08:15.020 ","End":"08:19.835","Text":"There\u0027s a simple equation for vector projection."},{"Start":"08:19.835 ","End":"08:23.045","Text":"Let\u0027s see how it goes."},{"Start":"08:23.045 ","End":"08:25.385","Text":"This is our equation,"},{"Start":"08:25.385 ","End":"08:27.305","Text":"and let\u0027s see how we use it."},{"Start":"08:27.305 ","End":"08:35.855","Text":"This is the velocity of the vector in the direction of the perpendicular component."},{"Start":"08:35.855 ","End":"08:41.239","Text":"I forgot to write the position vector relative."},{"Start":"08:41.239 ","End":"08:43.505","Text":"You work out in the exact same way."},{"Start":"08:43.505 ","End":"08:49.415","Text":"Just start here with the x_ p minus x_ b and y_ p minus y_ b."},{"Start":"08:49.415 ","End":"08:51.965","Text":"It works out to"},{"Start":"08:51.965 ","End":"09:00.050","Text":"R sine Omega t minus v_ 0t and then the y-direction,"},{"Start":"09:00.050 ","End":"09:06.805","Text":"it\u0027s R cosine of Omega t minus 0,"},{"Start":"09:06.805 ","End":"09:10.310","Text":"which is just like that."},{"Start":"09:10.450 ","End":"09:17.610","Text":"Let\u0027s go back to our equation of velocity in the parallel direction."},{"Start":"09:18.040 ","End":"09:25.145","Text":"If I have some vector and I would like to project it onto another vector."},{"Start":"09:25.145 ","End":"09:29.720","Text":"What I do is I take the vector and I scale or"},{"Start":"09:29.720 ","End":"09:33.920","Text":"multiply it with the vector that I would like to project it"},{"Start":"09:33.920 ","End":"09:38.630","Text":"onto and then I divide it by"},{"Start":"09:38.630 ","End":"09:44.610","Text":"the size of the vector which I want this vector to be projected onto."},{"Start":"09:44.800 ","End":"09:48.440","Text":"What is in fact happening here is that"},{"Start":"09:48.440 ","End":"09:51.470","Text":"because I\u0027m multiplying by this vector but then dividing"},{"Start":"09:51.470 ","End":"09:57.825","Text":"by the size I haven\u0027t actually changed the size of this vector."},{"Start":"09:57.825 ","End":"10:06.530","Text":"All I\u0027ve done is found this vector in the direction that I want it to go in."},{"Start":"10:06.530 ","End":"10:11.420","Text":"In Other words, I\u0027m getting the component of"},{"Start":"10:11.420 ","End":"10:17.780","Text":"v rel in the direction of r rel."},{"Start":"10:17.780 ","End":"10:22.105","Text":"Let\u0027s see how we figure this out."},{"Start":"10:22.105 ","End":"10:24.415","Text":"We have v relative,"},{"Start":"10:24.415 ","End":"10:26.835","Text":"which is this equation,"},{"Start":"10:26.835 ","End":"10:29.840","Text":"with a scalar multiple of r relative,"},{"Start":"10:29.840 ","End":"10:32.075","Text":"which is this equation here."},{"Start":"10:32.075 ","End":"10:34.460","Text":"Let\u0027s do this."},{"Start":"10:34.460 ","End":"10:38.180","Text":"What we do is because it\u0027s the dot-product we"},{"Start":"10:38.180 ","End":"10:43.085","Text":"do x multiplied by x plus y multiplied by y."},{"Start":"10:43.085 ","End":"10:45.830","Text":"In the x-component of each,"},{"Start":"10:45.830 ","End":"10:48.275","Text":"we have this multiplied by this,"},{"Start":"10:48.275 ","End":"10:50.045","Text":"this multiplied by this,"},{"Start":"10:50.045 ","End":"10:51.365","Text":"this multiplied by this,"},{"Start":"10:51.365 ","End":"10:52.894","Text":"and this multiplied by this."},{"Start":"10:52.894 ","End":"10:54.770","Text":"Let\u0027s do this."},{"Start":"10:54.770 ","End":"10:56.465","Text":"Let\u0027s start with the end."},{"Start":"10:56.465 ","End":"11:01.715","Text":"We have negative v_ 0 multiplied by v_ 0t."},{"Start":"11:01.715 ","End":"11:08.150","Text":"It works out to v_ 0^2 t. The minuses cancel out."},{"Start":"11:08.150 ","End":"11:11.555","Text":"The next thing that we can look at when we multiply is"},{"Start":"11:11.555 ","End":"11:19.430","Text":"the Omega R cos Omega t multiplied by R sine Omega t. If you notice in the y components,"},{"Start":"11:19.430 ","End":"11:23.510","Text":"we have negative Omega R Sine Omega t and"},{"Start":"11:23.510 ","End":"11:30.200","Text":"R cosine of Omega t. As we know in the dot product, it\u0027s the x."},{"Start":"11:30.200 ","End":"11:34.745","Text":"This plus this, and because this is a minus,"},{"Start":"11:34.745 ","End":"11:36.215","Text":"this will cross out."},{"Start":"11:36.215 ","End":"11:41.975","Text":"Because we\u0027ll end up with R Omega^2 cosine Omega t sine Omega t"},{"Start":"11:41.975 ","End":"11:48.800","Text":"minus OmegaR^2 cosine Omega t sine Omega t. That will cross out."},{"Start":"11:48.800 ","End":"11:50.690","Text":"Because of that move,"},{"Start":"11:50.690 ","End":"11:54.650","Text":"we cannot look at the y components at all,"},{"Start":"11:54.650 ","End":"11:59.810","Text":"because we\u0027ve already figured out that this times this cancels out with this times this."},{"Start":"11:59.810 ","End":"12:07.595","Text":"All that\u0027s left to do is to multiply negative v_ 0 by R Sine Omega t. Then"},{"Start":"12:07.595 ","End":"12:11.015","Text":"to multiply Omega R cosine Omega t by"},{"Start":"12:11.015 ","End":"12:15.395","Text":"negative v_ 0 t. If we take out the common multiples,"},{"Start":"12:15.395 ","End":"12:18.200","Text":"it will work out to,"},{"Start":"12:18.200 ","End":"12:22.580","Text":"we can see that in each we\u0027ll have v_ 0,"},{"Start":"12:22.580 ","End":"12:33.490","Text":"and in each we have R. We can take out negative v_ 0 R. Then"},{"Start":"12:33.490 ","End":"12:36.595","Text":"we have sine of"},{"Start":"12:36.595 ","End":"12:39.500","Text":"Omega t"},{"Start":"12:46.480 ","End":"12:52.370","Text":"plus Omega t cosine"},{"Start":"12:52.370 ","End":"12:59.305","Text":"of Omega t. Let\u0027s just go over how we did that."},{"Start":"12:59.305 ","End":"13:02.350","Text":"Because we\u0027re doing this times this and this times this."},{"Start":"13:02.350 ","End":"13:08.390","Text":"We can notice that in both multiplications we have a v_ 0 and an R. We take out the v_"},{"Start":"13:08.390 ","End":"13:14.945","Text":"0 and the R. Then we have v_ 0 times R of sine Omega t,"},{"Start":"13:14.945 ","End":"13:18.350","Text":"and obviously it\u0027s a negative because here there\u0027s a negative."},{"Start":"13:18.350 ","End":"13:22.835","Text":"Then we have to get a negative because here\u0027s a negative multiplied by this."},{"Start":"13:22.835 ","End":"13:27.335","Text":"Here we put a positive so that the negative and the positive equal a negative."},{"Start":"13:27.335 ","End":"13:33.169","Text":"Then here we have v_ 0t multiplied by Omega R, cosine."},{"Start":"13:33.169 ","End":"13:42.200","Text":"We have t and Omega multiplied by cosine of Omega t. That\u0027s for that."},{"Start":"13:42.200 ","End":"13:46.820","Text":"Then we have to multiply by the size of our rel."},{"Start":"13:46.820 ","End":"13:49.040","Text":"Let\u0027s see how we do that one."},{"Start":"13:49.040 ","End":"13:51.110","Text":"How do we find the size?"},{"Start":"13:51.110 ","End":"13:59.030","Text":"We do the square root of the x component squared,"},{"Start":"13:59.030 ","End":"14:02.940","Text":"plus of the y component squared."},{"Start":"14:02.950 ","End":"14:08.465","Text":"I\u0027m not going to write down each of the steps because it\u0027s just a waste of time."},{"Start":"14:08.465 ","End":"14:11.150","Text":"Either try and do it in your head or follow"},{"Start":"14:11.150 ","End":"14:14.610","Text":"on a piece of paper and pause the video as you go along."},{"Start":"14:14.680 ","End":"14:19.445","Text":"We have R sine Omega t minus v_ 0t,"},{"Start":"14:19.445 ","End":"14:21.620","Text":"which is our x-component squared."},{"Start":"14:21.620 ","End":"14:22.850","Text":"What does that mean?"},{"Start":"14:22.850 ","End":"14:32.165","Text":"We have this squared minus 2 times this times this plus this squared."},{"Start":"14:32.165 ","End":"14:37.490","Text":"Let\u0027s do that. That means that we have this squared."},{"Start":"14:37.490 ","End":"14:39.695","Text":"We have R^2 sine squared"},{"Start":"14:39.695 ","End":"14:47.930","Text":"Omega t. Here notice in the y section because it\u0027s x^2 plus y^2,"},{"Start":"14:47.930 ","End":"14:54.995","Text":"we\u0027re going to have R^2 cosine squared Omega t. Remember the equation"},{"Start":"14:54.995 ","End":"15:02.030","Text":"of sin^2 x plus cos^2 x=1."},{"Start":"15:02.030 ","End":"15:05.405","Text":"Here we\u0027re going to have cosine squared and sine squared,"},{"Start":"15:05.405 ","End":"15:07.940","Text":"which means that if we take out the common multiple,"},{"Start":"15:07.940 ","End":"15:11.060","Text":"which here is the R,"},{"Start":"15:11.060 ","End":"15:15.240","Text":"it\u0027s going to cancel down to R^2 multiplied by 1,"},{"Start":"15:15.760 ","End":"15:18.965","Text":"which equals just R^2."},{"Start":"15:18.965 ","End":"15:22.683","Text":"We can put that right over there."},{"Start":"15:22.683 ","End":"15:25.870","Text":"Now we\u0027ve finished the multiplication of"},{"Start":"15:25.870 ","End":"15:29.545","Text":"this squared and the multiplication of the y component,"},{"Start":"15:29.545 ","End":"15:33.850","Text":"because together they formed into an identity that we"},{"Start":"15:33.850 ","End":"15:38.680","Text":"already know to make this answer slightly more readable."},{"Start":"15:38.680 ","End":"15:40.675","Text":"After we\u0027ve done all of that,"},{"Start":"15:40.675 ","End":"15:44.605","Text":"what we have left is,"},{"Start":"15:44.605 ","End":"15:47.605","Text":"sorry, negative (v_0t)^2,"},{"Start":"15:47.605 ","End":"15:51.490","Text":"which just equals positive (v_0)^2 t^2"},{"Start":"15:51.490 ","End":"16:00.460","Text":"and negative 2 times this times this,"},{"Start":"16:00.460 ","End":"16:08.290","Text":"which is negative 2(v_0t) are sine"},{"Start":"16:08.290 ","End":"16:16.615","Text":"of Omega t. This is the size of the vector R relative."},{"Start":"16:16.615 ","End":"16:19.540","Text":"Now this whole thing over"},{"Start":"16:19.540 ","End":"16:27.160","Text":"here is in fact the answer that we were looking for."},{"Start":"16:27.160 ","End":"16:30.550","Text":"Now this question is slightly harder because you\u0027re asked"},{"Start":"16:30.550 ","End":"16:33.865","Text":"to do quite a few different things,"},{"Start":"16:33.865 ","End":"16:36.220","Text":"but now you can see it\u0027s not actually that"},{"Start":"16:36.220 ","End":"16:39.295","Text":"difficult and it\u0027s pretty simple once you break it down."},{"Start":"16:39.295 ","End":"16:44.365","Text":"Now what we\u0027re going to do is we\u0027re going to look at this whole situation from"},{"Start":"16:44.365 ","End":"16:49.945","Text":"certain angles in order to understand exactly what\u0027s going on and to get some intuition."},{"Start":"16:49.945 ","End":"16:52.960","Text":"Now let\u0027s take a look at the first scenario."},{"Start":"16:52.960 ","End":"16:54.211","Text":"Now we\u0027re looking here."},{"Start":"16:54.211 ","End":"16:58.900","Text":"If I input into the angle right at the beginning, t equals 0,"},{"Start":"16:58.900 ","End":"17:01.915","Text":"which means that the whole angle is equal to 0,"},{"Start":"17:01.915 ","End":"17:03.265","Text":"because it\u0027s Omega times t,"},{"Start":"17:03.265 ","End":"17:05.365","Text":"which is equal to 0."},{"Start":"17:05.365 ","End":"17:13.870","Text":"Now I meant to get that the velocity that the speedometer measures also has to equal 0."},{"Start":"17:13.870 ","End":"17:17.650","Text":"Let\u0027s see how we do this and if this works out correct."},{"Start":"17:17.650 ","End":"17:20.920","Text":"We said that t equals 0."},{"Start":"17:20.920 ","End":"17:28.645","Text":"Here we can see (v_0)^2 multiplied by 0, that\u0027s 0."},{"Start":"17:28.645 ","End":"17:36.400","Text":"Now here the angle will be 0 and sine of 0 is 0, so this falls."},{"Start":"17:36.400 ","End":"17:38.875","Text":"Here, we have a t over here,"},{"Start":"17:38.875 ","End":"17:40.378","Text":"which means that this is 0."},{"Start":"17:40.378 ","End":"17:43.345","Text":"Now the whole numerator is equal to 0,"},{"Start":"17:43.345 ","End":"17:46.240","Text":"which means that our v is equal to 0,"},{"Start":"17:46.240 ","End":"17:48.520","Text":"which is exactly what we wanted."},{"Start":"17:48.520 ","End":"17:53.541","Text":"Now, I can look at this moment,"},{"Start":"17:53.541 ","End":"17:58.210","Text":"when the policeman reaches this point over here."},{"Start":"17:58.210 ","End":"18:03.370","Text":"At this point, what will the policeman\u0027s speedometer measure?"},{"Start":"18:03.370 ","End":"18:06.220","Text":"It\u0027s obvious that it will measure v_0"},{"Start":"18:06.220 ","End":"18:10.585","Text":"because directly to the policeman at a velocity of v_0."},{"Start":"18:10.585 ","End":"18:13.960","Text":"The speedometer, as we mentioned at the beginning how it works,"},{"Start":"18:13.960 ","End":"18:19.105","Text":"should measure v_0. Let\u0027s check ourselves."},{"Start":"18:19.105 ","End":"18:22.150","Text":"It\u0027s also important to note that even though"},{"Start":"18:22.150 ","End":"18:25.389","Text":"the police officer is still moving in this direction,"},{"Start":"18:25.389 ","End":"18:32.420","Text":"it doesn\u0027t matter because the speedometer is only recording this relative velocity."},{"Start":"18:32.760 ","End":"18:35.275","Text":"If the police man is here,"},{"Start":"18:35.275 ","End":"18:42.400","Text":"then the angle between the starting position and getting here is equal to 90 degrees."},{"Start":"18:42.400 ","End":"18:46.285","Text":"Or better yet, instead of calling it 90 degrees,"},{"Start":"18:46.285 ","End":"18:50.965","Text":"we can call it Pi over 4."},{"Start":"18:50.965 ","End":"18:53.410","Text":"Now let\u0027s substitute this end,"},{"Start":"18:53.410 ","End":"18:54.700","Text":"so the angle is Pi over 4,"},{"Start":"18:54.700 ","End":"19:00.070","Text":"which means that it equals Omega t. Omega t is Pi over 4,"},{"Start":"19:00.070 ","End":"19:05.110","Text":"cosine of Pi over 4 is 0,so this cancels out."},{"Start":"19:05.110 ","End":"19:11.305","Text":"When we substitute Pi divided by 4 into the sine expression, we\u0027re going to get 1,"},{"Start":"19:11.305 ","End":"19:17.875","Text":"which means that this whole section is equal to 1."},{"Start":"19:17.875 ","End":"19:23.245","Text":"Then we have (v_0)^2 multiplied by t minus"},{"Start":"19:23.245 ","End":"19:30.220","Text":"v_0 r. This is the expression that we have right now to use."},{"Start":"19:30.220 ","End":"19:35.230","Text":"Now let\u0027s look at what is happening in the denominator."},{"Start":"19:35.230 ","End":"19:37.135","Text":"Now, as for the denominator,"},{"Start":"19:37.135 ","End":"19:44.130","Text":"I\u0027m going to write it over here because it\u0027s slightly more complicated."},{"Start":"19:44.130 ","End":"19:48.750","Text":"If we substitute in Pi over 4 into the sign,"},{"Start":"19:48.750 ","End":"19:50.721","Text":"we already know that this becomes 1,"},{"Start":"19:50.721 ","End":"19:54.425","Text":"which means that we have the square root of"},{"Start":"19:54.425 ","End":"20:01.955","Text":"r^2 plus (v_0)^2 t^2 minus 2 v_0 t,"},{"Start":"20:01.955 ","End":"20:07.975","Text":"which is equal to the square root of"},{"Start":"20:07.975 ","End":"20:14.070","Text":"r minus (v_0 t)^2."},{"Start":"20:14.070 ","End":"20:17.670","Text":"Now, the squared and the square root cancel each other out,"},{"Start":"20:17.670 ","End":"20:19.740","Text":"which equals r minus"},{"Start":"20:19.740 ","End":"20:28.165","Text":"v_0 t. Then if you notice going back to the numerator here,"},{"Start":"20:28.165 ","End":"20:34.540","Text":"we have an equation that goes (v_0)^2 t minus"},{"Start":"20:34.540 ","End":"20:42.855","Text":"v_0 r divided by r minus v_0 t. Now,"},{"Start":"20:42.855 ","End":"20:49.794","Text":"if we take the common multiple from the numerator,"},{"Start":"20:49.794 ","End":"20:58.960","Text":"we\u0027ll have v_0 multiplied by v_0 t minus r over, sorry,"},{"Start":"20:58.960 ","End":"21:05.635","Text":"this is a capital R, I keep forgetting over capital R minus v_0 t. Now,"},{"Start":"21:05.635 ","End":"21:08.965","Text":"this is the minus of this."},{"Start":"21:08.965 ","End":"21:12.630","Text":"These cross out, leaving a negative over here."},{"Start":"21:12.630 ","End":"21:19.195","Text":"Then we see that the speedometer is measuring a velocity of minus v. Now,"},{"Start":"21:19.195 ","End":"21:22.735","Text":"we expect it already this v_0, which is perfect."},{"Start":"21:22.735 ","End":"21:25.240","Text":"Why the minus I hear you asking?"},{"Start":"21:25.240 ","End":"21:28.060","Text":"Because the police man is standing right over here"},{"Start":"21:28.060 ","End":"21:31.915","Text":"and he sees something coming towards him at v_0."},{"Start":"21:31.915 ","End":"21:37.985","Text":"Now, notice from the starting point,"},{"Start":"21:37.985 ","End":"21:42.833","Text":"sorry, the bird is flying in this direction at v_0,"},{"Start":"21:42.833 ","End":"21:46.615","Text":"which means that if we\u0027re looking from this direction,"},{"Start":"21:46.615 ","End":"21:49.555","Text":"because it\u0027s coming towards the object,"},{"Start":"21:49.555 ","End":"21:53.539","Text":"the police man, it will be minus v_0."},{"Start":"21:53.539 ","End":"21:57.280","Text":"Out from the starting point it\u0027s a positive,"},{"Start":"21:57.280 ","End":"22:00.235","Text":"but looking from this direction,"},{"Start":"22:00.235 ","End":"22:03.160","Text":"the relative direction it\u0027s negative v_0,"},{"Start":"22:03.160 ","End":"22:06.470","Text":"which is exactly what we expected to get."},{"Start":"22:06.780 ","End":"22:12.700","Text":"Another way that you can think of if you don\u0027t want to accept the answer of"},{"Start":"22:12.700 ","End":"22:18.250","Text":"negative v_0 is that you can remember that when you take the square root of something,"},{"Start":"22:18.250 ","End":"22:20.620","Text":"you get a plus or a minus."},{"Start":"22:20.620 ","End":"22:23.800","Text":"Now, obviously here the square root canceled out,"},{"Start":"22:23.800 ","End":"22:26.200","Text":"but it doesn\u0027t matter, the idea is still remains."},{"Start":"22:26.200 ","End":"22:27.460","Text":"You can have a plus or a minus,"},{"Start":"22:27.460 ","End":"22:30.070","Text":"which means that you can choose which sign you want to take."},{"Start":"22:30.070 ","End":"22:35.875","Text":"Then you can make it so that this turns out to minus r plus v_0 t,"},{"Start":"22:35.875 ","End":"22:40.075","Text":"and then it just crosses out and there\u0027s not the minus over here."},{"Start":"22:40.075 ","End":"22:41.425","Text":"It doesn\u0027t really matter."},{"Start":"22:41.425 ","End":"22:46.690","Text":"It just checks that what we would expect to get that here,"},{"Start":"22:46.690 ","End":"22:50.230","Text":"it would be 0 and here it would be v_0 actually"},{"Start":"22:50.230 ","End":"22:54.700","Text":"happens and that our equation over here is correct."},{"Start":"22:54.700 ","End":"22:57.980","Text":"This is the end of the lesson."}],"ID":9597}],"Thumbnail":null,"ID":85366}]

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