Examples of Exam Questions
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[{"Name":"Examples of Exam Questions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Cycloid Motion","Duration":"29m 19s","ChapterTopicVideoID":9203,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9203.jpeg","UploadDate":"2017-03-24T12:52:46.4370000","DurationForVideoObject":"PT29M19S","Description":null,"MetaTitle":"Cycloid Motion: Video + Workbook | Proprep","MetaDescription":"Exam Questions - Examples of Exam Questions. 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/exam-questions/examples-of-exam-questions/vid9473","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"Hello, Let\u0027s take a look at this question."},{"Start":"00:03.090 ","End":"00:06.180","Text":"It\u0027s a slightly more complicated question,"},{"Start":"00:06.180 ","End":"00:08.280","Text":"also just to understand."},{"Start":"00:08.280 ","End":"00:11.325","Text":"We\u0027re being told that we have a point mass m,"},{"Start":"00:11.325 ","End":"00:17.730","Text":"which is moving along a cycloid course as given by this parametric equation."},{"Start":"00:17.730 ","End":"00:19.890","Text":"Cycloid course, it doesn\u0027t really matter,"},{"Start":"00:19.890 ","End":"00:22.815","Text":"you can look it up on the Internet if you want to know what it is."},{"Start":"00:22.815 ","End":"00:26.160","Text":"But basically we have our mass moving along this course over here."},{"Start":"00:26.160 ","End":"00:31.215","Text":"Its motion is given in vector form using a parametric equation,"},{"Start":"00:31.215 ","End":"00:36.855","Text":"which means that our x and y components are being expressed by one variable,"},{"Start":"00:36.855 ","End":"00:39.225","Text":"which here is our variable Phi."},{"Start":"00:39.225 ","End":"00:40.890","Text":"These Phi\u0027s are the same,"},{"Start":"00:40.890 ","End":"00:42.325","Text":"they just look different over here."},{"Start":"00:42.325 ","End":"00:44.510","Text":"It\u0027s the same letter."},{"Start":"00:44.510 ","End":"00:48.875","Text":"We\u0027re being told that a is a constant and our Phi is the variable."},{"Start":"00:48.875 ","End":"00:54.325","Text":"We\u0027re being told that the body begins its motion from rest from the point (0,0),"},{"Start":"00:54.325 ","End":"00:55.610","Text":"so from the origin,"},{"Start":"00:55.610 ","End":"00:57.980","Text":"and we\u0027re being told that it moves through"},{"Start":"00:57.980 ","End":"01:01.100","Text":"gravities force field as shown in the diagram,"},{"Start":"01:01.100 ","End":"01:02.615","Text":"so it moves down."},{"Start":"01:02.615 ","End":"01:06.440","Text":"The point of reference for our potential energy is at"},{"Start":"01:06.440 ","End":"01:10.745","Text":"the bottom of the trajectory, where y=2a."},{"Start":"01:10.745 ","End":"01:15.655","Text":"This is our point of reference for our potential energy."},{"Start":"01:15.655 ","End":"01:19.230","Text":"Just to explain what I just said so our body"},{"Start":"01:19.230 ","End":"01:22.875","Text":"begins from rest at our point (0,0) the origin."},{"Start":"01:22.875 ","End":"01:24.210","Text":"That\u0027s right over here."},{"Start":"01:24.210 ","End":"01:26.745","Text":"This is our point (0,0)."},{"Start":"01:26.745 ","End":"01:30.200","Text":"Then we\u0027re told that it moves through gravities force field,"},{"Start":"01:30.200 ","End":"01:32.840","Text":"which means that we have mg pointing downwards,"},{"Start":"01:32.840 ","End":"01:36.770","Text":"as shown in the diagram, and our point of reference for our potential energy"},{"Start":"01:36.770 ","End":"01:40.775","Text":"is at the bottom of their trajectory, where y=2a."},{"Start":"01:40.775 ","End":"01:43.880","Text":"At this dotted line over here where y=2a,"},{"Start":"01:43.880 ","End":"01:47.420","Text":"so our point of reference for potential energy."},{"Start":"01:47.420 ","End":"01:52.090","Text":"What that means is that our h over here is equal to 0."},{"Start":"01:52.090 ","End":"01:56.024","Text":"For our equation for potential energy is mgh,"},{"Start":"01:56.024 ","End":"01:57.560","Text":"so here our h is 0."},{"Start":"01:57.560 ","End":"01:59.330","Text":"Our first question is,"},{"Start":"01:59.330 ","End":"02:03.590","Text":"what is the velocity of the mass at the lowest point of the trajectory?"},{"Start":"02:03.590 ","End":"02:06.185","Text":"That\u0027s where our y=2a."},{"Start":"02:06.185 ","End":"02:10.300","Text":"We\u0027re being asked to find the velocity of our mass over here."},{"Start":"02:10.300 ","End":"02:14.000","Text":"Let\u0027s see how we look at this. There\u0027s two options."},{"Start":"02:14.000 ","End":"02:17.450","Text":"We can look at solving this via energy or via forces."},{"Start":"02:17.450 ","End":"02:20.840","Text":"Now, via forces is going to be a little bit complicated because we can see"},{"Start":"02:20.840 ","End":"02:24.455","Text":"that we have quite a weird trajectory,"},{"Start":"02:24.455 ","End":"02:30.185","Text":"a weird course that our mass or our body is traveling through."},{"Start":"02:30.185 ","End":"02:32.120","Text":"We don\u0027t really know at this stage,"},{"Start":"02:32.120 ","End":"02:34.670","Text":"or it also makes it very much more complicated,"},{"Start":"02:34.670 ","End":"02:37.055","Text":"how to solve this type of thing with forces."},{"Start":"02:37.055 ","End":"02:41.855","Text":"However, a clue is that whenever we\u0027re dealing with a change in height,"},{"Start":"02:41.855 ","End":"02:47.855","Text":"the general rule is to work by solving with energy."},{"Start":"02:47.855 ","End":"02:50.465","Text":"Change in height means solving with energy."},{"Start":"02:50.465 ","End":"02:55.625","Text":"Also another clue is that we\u0027re being asked about velocity and in our energy equations,"},{"Start":"02:55.625 ","End":"02:59.075","Text":"we have our variable for velocity in there,"},{"Start":"02:59.075 ","End":"03:02.540","Text":"however in our forces it\u0027s more dealing with acceleration."},{"Start":"03:02.540 ","End":"03:05.615","Text":"We\u0027re going to be working with energy."},{"Start":"03:05.615 ","End":"03:09.080","Text":"Now because we have our gravitational force field,"},{"Start":"03:09.080 ","End":"03:12.365","Text":"that means that we\u0027re going to have energy conservation."},{"Start":"03:12.365 ","End":"03:14.389","Text":"With conservation of energy,"},{"Start":"03:14.389 ","End":"03:17.225","Text":"our energy before so our initial energy,"},{"Start":"03:17.225 ","End":"03:19.820","Text":"has to be equal to our energy after,"},{"Start":"03:19.820 ","End":"03:20.965","Text":"so our final energy."},{"Start":"03:20.965 ","End":"03:25.620","Text":"Let\u0027s take a look at what our initial energy is going to be equal to."},{"Start":"03:25.810 ","End":"03:28.970","Text":"With kinetic energy, we\u0027re being told in"},{"Start":"03:28.970 ","End":"03:32.825","Text":"the question that the body begins its motion from rest,"},{"Start":"03:32.825 ","End":"03:35.300","Text":"from the origin, from over here."},{"Start":"03:35.300 ","End":"03:41.300","Text":"Its kinetic energy is going to be 0 because it\u0027s from rest, it starts stationary."},{"Start":"03:41.300 ","End":"03:43.265","Text":"Then it\u0027s going to have potential energy,"},{"Start":"03:43.265 ","End":"03:47.305","Text":"which is going to be equal to mgh."},{"Start":"03:47.305 ","End":"03:50.510","Text":"Now, let\u0027s take a look at what our h is going to be."},{"Start":"03:50.510 ","End":"03:54.770","Text":"Now, because we said that our point of reference when dealing with our h value,"},{"Start":"03:54.770 ","End":"03:56.495","Text":"for our potential energy,"},{"Start":"03:56.495 ","End":"04:00.850","Text":"our 0 is over here at y=2a."},{"Start":"04:00.850 ","End":"04:04.140","Text":"That means that our h is going to be 2a,"},{"Start":"04:04.140 ","End":"04:08.315","Text":"and it\u0027s going to be positive 2a because it\u0027s above the 0 line."},{"Start":"04:08.315 ","End":"04:11.840","Text":"Or also another way of looking at it is that we\u0027re getting further away"},{"Start":"04:11.840 ","End":"04:15.630","Text":"from the Earth so it\u0027s going to be positive."},{"Start":"04:15.630 ","End":"04:22.605","Text":"We\u0027re just going to have 2mga, because our h=2a."},{"Start":"04:22.605 ","End":"04:27.690","Text":"Now let\u0027s look at our final energy, so our E_f."},{"Start":"04:27.690 ","End":"04:31.865","Text":"That\u0027s going to be at this point over here."},{"Start":"04:31.865 ","End":"04:35.795","Text":"Let\u0027s take a look. We\u0027re definitely going to have kinetic energy."},{"Start":"04:35.795 ","End":"04:42.225","Text":"Our kinetic energy is going to be 1/2mv_f^2."},{"Start":"04:42.225 ","End":"04:45.290","Text":"Then our potential energy is going to be mgh,"},{"Start":"04:45.290 ","End":"04:48.457","Text":"but our h over here at this lowest point is equal to 0,"},{"Start":"04:48.457 ","End":"04:51.935","Text":"so plus our potential energy,"},{"Start":"04:51.935 ","End":"04:54.425","Text":"which over here is going to be equal to 0,"},{"Start":"04:54.425 ","End":"05:02.035","Text":"that means that our E_f=1/2mv_f^2."},{"Start":"05:02.035 ","End":"05:04.505","Text":"Now because we have conservation of energy,"},{"Start":"05:04.505 ","End":"05:09.140","Text":"we equate these and all we have to do is isolate out our v_f."},{"Start":"05:09.140 ","End":"05:19.020","Text":"What we\u0027ll get is that a 1/2mv_f^2=2mga."},{"Start":"05:19.020 ","End":"05:23.915","Text":"We can divide both sides by m. Then via simple algebra,"},{"Start":"05:23.915 ","End":"05:27.660","Text":"we\u0027ll get that our v_f=2 times the square root of ga."},{"Start":"05:31.790 ","End":"05:36.600","Text":"This is the final answer for the question."},{"Start":"05:37.100 ","End":"05:41.300","Text":"This is the easiest way to work it out. It\u0027s pretty simple."},{"Start":"05:41.300 ","End":"05:45.710","Text":"However, there\u0027s another way which is slightly more complicated maybe,"},{"Start":"05:45.710 ","End":"05:47.630","Text":"however, if we understand it,"},{"Start":"05:47.630 ","End":"05:50.669","Text":"it might make understanding the question a"},{"Start":"05:50.669 ","End":"05:55.100","Text":"little bit easier and also if they ask you to find the velocity,"},{"Start":"05:55.100 ","End":"05:56.570","Text":"not at our bottom point,"},{"Start":"05:56.570 ","End":"06:01.280","Text":"but at any point throughout the course of the motion,"},{"Start":"06:01.280 ","End":"06:04.790","Text":"it will also help you to work that out."},{"Start":"06:04.790 ","End":"06:07.190","Text":"Let\u0027s go over this way."},{"Start":"06:07.190 ","End":"06:10.210","Text":"Let\u0027s scroll down and we\u0027re still in question 1."},{"Start":"06:10.210 ","End":"06:15.785","Text":"Another way to do it is to say that we\u0027re going to choose"},{"Start":"06:15.785 ","End":"06:24.310","Text":"some point P and its coordinates are going to be some general x and some general y."},{"Start":"06:24.310 ","End":"06:28.710","Text":"This is some point P in this course."},{"Start":"06:28.710 ","End":"06:33.245","Text":"Then we\u0027re told in the question that we\u0027re trying to find"},{"Start":"06:33.245 ","End":"06:38.540","Text":"the velocity of the mass at the lowest point of the trajectory."},{"Start":"06:38.540 ","End":"06:44.959","Text":"We\u0027re trying to just find the velocity of the mass at some point on the trajectory,"},{"Start":"06:44.959 ","End":"06:48.560","Text":"some point P. This is where we\u0027re trying to find the velocity."},{"Start":"06:48.560 ","End":"06:52.115","Text":"Now we can say that from conservation of energy,"},{"Start":"06:52.115 ","End":"07:00.560","Text":"we can say that kinetic energy at our point P plus our potential energy at our point P is"},{"Start":"07:00.560 ","End":"07:04.700","Text":"going to be equal to our kinetic energy at"},{"Start":"07:04.700 ","End":"07:09.275","Text":"the origin plus our potential energy at the origin."},{"Start":"07:09.275 ","End":"07:11.930","Text":"Now we can write out our equations."},{"Start":"07:11.930 ","End":"07:16.985","Text":"Our kinetic energy at point P is going to be equal to a 1/2mv^2"},{"Start":"07:16.985 ","End":"07:23.780","Text":"and our potential energy is going to be equal to,"},{"Start":"07:23.780 ","End":"07:26.280","Text":"let\u0027s take a look, it\u0027s going to be mg times our height."},{"Start":"07:27.850 ","End":"07:30.395","Text":"What is our height going to be?"},{"Start":"07:30.395 ","End":"07:34.400","Text":"Because we know that this total length is going to be 2a and"},{"Start":"07:34.400 ","End":"07:38.900","Text":"our body is always moving between over here in this dotted line,"},{"Start":"07:38.900 ","End":"07:42.115","Text":"in this area of length 2a."},{"Start":"07:42.115 ","End":"07:44.990","Text":"Let\u0027s say we choose our P to be over here."},{"Start":"07:44.990 ","End":"07:48.545","Text":"In order to find what this height is over here,"},{"Start":"07:48.545 ","End":"07:50.840","Text":"if this is what we call our h,"},{"Start":"07:50.840 ","End":"07:55.565","Text":"we can say that it\u0027s equal to 2a negative,"},{"Start":"07:55.565 ","End":"07:58.370","Text":"this distance over here,"},{"Start":"07:58.370 ","End":"08:01.395","Text":"y, so negative y."},{"Start":"08:01.395 ","End":"08:04.480","Text":"Because then we can see that y+h=2a."},{"Start":"08:04.790 ","End":"08:15.135","Text":"Our h over here is going to be equal to 2a-y that we\u0027ve gone down."},{"Start":"08:15.135 ","End":"08:18.890","Text":"Our y is positive because, first of all,"},{"Start":"08:18.890 ","End":"08:21.230","Text":"our positive y direction is going down,"},{"Start":"08:21.230 ","End":"08:23.120","Text":"but also it\u0027s above our h=0."},{"Start":"08:23.120 ","End":"08:28.710","Text":"As a reminder, this is our h,"},{"Start":"08:28.710 ","End":"08:30.515","Text":"for our potential energy."},{"Start":"08:30.515 ","End":"08:34.880","Text":"Then this is going to be equal to our kinetic energy at the origin,"},{"Start":"08:34.880 ","End":"08:39.200","Text":"which here is specifically because they told us that we start from a stationary position."},{"Start":"08:39.200 ","End":"08:43.340","Text":"It\u0027s going to be 0 plus our potential energy at the origin,"},{"Start":"08:43.340 ","End":"08:46.400","Text":"which is going to be plus mg,"},{"Start":"08:46.400 ","End":"08:48.290","Text":"and then we know that the height,"},{"Start":"08:48.290 ","End":"08:51.370","Text":"our h, at the beginning is 2a."},{"Start":"08:54.490 ","End":"08:59.295","Text":"Then we have to substitute n what our y is equal to."},{"Start":"08:59.295 ","End":"09:05.180","Text":"Here our y=2a because we\u0027re being asked at the bottom of our trajectory."},{"Start":"09:05.180 ","End":"09:08.254","Text":"Because we substitute n where our P point is."},{"Start":"09:08.254 ","End":"09:11.100","Text":"Then we\u0027ll get mg2a-2a0."},{"Start":"09:11.110 ","End":"09:18.000","Text":"Then we\u0027re left with our 1/2mv^2= mg(2a)."},{"Start":"09:18.000 ","End":"09:20.885","Text":"That\u0027s exactly this equation over here,"},{"Start":"09:20.885 ","End":"09:24.980","Text":"so we can just skip this and we can just go straight"},{"Start":"09:24.980 ","End":"09:30.110","Text":"to that our v=2 times the square root of ga,"},{"Start":"09:30.110 ","End":"09:33.280","Text":"which is the exact same solution that we got over here."},{"Start":"09:33.280 ","End":"09:36.140","Text":"This is a different way to look at this,"},{"Start":"09:36.140 ","End":"09:42.575","Text":"to find the general equation for the conservation of energy"},{"Start":"09:42.575 ","End":"09:46.655","Text":"throughout the course of the movement and then to substitute in"},{"Start":"09:46.655 ","End":"09:50.900","Text":"our point specifically and then we can find the velocity over here,"},{"Start":"09:50.900 ","End":"09:54.085","Text":"over here, wherever we want."},{"Start":"09:54.085 ","End":"09:56.790","Text":"That\u0027s the end of question number 1."},{"Start":"09:56.790 ","End":"10:00.030","Text":"Now let\u0027s go to question number 2."},{"Start":"10:00.030 ","End":"10:08.265","Text":"Question number 2 is, what is the equation of motion of the body throughout?"},{"Start":"10:08.265 ","End":"10:11.285","Text":"When we have to write an equation of motion,"},{"Start":"10:11.285 ","End":"10:16.430","Text":"what do we have to do is we have to find some connection between our position,"},{"Start":"10:16.430 ","End":"10:19.130","Text":"our velocity, our acceleration."},{"Start":"10:19.130 ","End":"10:22.145","Text":"Just like with harmonic motion where we have,"},{"Start":"10:22.145 ","End":"10:27.125","Text":"let\u0027s say negative kx= mx double dot."},{"Start":"10:27.125 ","End":"10:30.080","Text":"We can see a relationship between our x and our x double-dot."},{"Start":"10:30.080 ","End":"10:34.030","Text":"Its derivative. It\u0027s the same variable, just it\u0027s derivative."},{"Start":"10:34.030 ","End":"10:40.785","Text":"Here, because our variable is our Phi,"},{"Start":"10:40.785 ","End":"10:46.970","Text":"we\u0027re going to try and get to some equation with Phi and its derivatives."},{"Start":"10:46.970 ","End":"10:49.510","Text":"So Phi dot or Phi double-dot."},{"Start":"10:49.510 ","End":"10:53.540","Text":"We\u0027re going to have to play with a lot of mathematics in order to answer this."},{"Start":"10:53.540 ","End":"11:00.095","Text":"Now I see that I answered my first question by using energy so let\u0027s start with that."},{"Start":"11:00.095 ","End":"11:04.465","Text":"Let\u0027s start answering our question 2."},{"Start":"11:04.465 ","End":"11:07.220","Text":"The first thing that we\u0027re going to do is we\u0027re going"},{"Start":"11:07.220 ","End":"11:09.590","Text":"to use our idea of conservation of energy."},{"Start":"11:09.590 ","End":"11:13.450","Text":"We\u0027re going to start by writing our energy equations."},{"Start":"11:13.450 ","End":"11:16.025","Text":"We know that the total energy of the system,"},{"Start":"11:16.025 ","End":"11:17.810","Text":"at any point,"},{"Start":"11:17.810 ","End":"11:22.895","Text":"is going to be equal to its kinetic energy plus its potential energy."},{"Start":"11:22.895 ","End":"11:26.080","Text":"Now, another thing to remember is that"},{"Start":"11:26.080 ","End":"11:31.765","Text":"its velocity squared because it\u0027s a vector because we have our x and y components."},{"Start":"11:31.765 ","End":"11:34.330","Text":"It\u0027s going to be our velocity in the x-direction"},{"Start":"11:34.330 ","End":"11:38.620","Text":"squared plus the velocity in our y-direction squared."},{"Start":"11:38.620 ","End":"11:42.850","Text":"That is equal to our velocity in our x-direction can be"},{"Start":"11:42.850 ","End":"11:49.465","Text":"called x dot squared and the velocity and our y-direction can be called y dot squared."},{"Start":"11:49.465 ","End":"11:55.900","Text":"They\u0027re the same. Now let\u0027s write down first of all our kinetic energy."},{"Start":"11:55.900 ","End":"11:59.995","Text":"Our E_k kinetic energy is going to be equal to"},{"Start":"11:59.995 ","End":"12:08.590","Text":"1/2m x dot squared plus 1/2my dot squared."},{"Start":"12:08.590 ","End":"12:12.670","Text":"We\u0027re taking the kinetic energy for each component,"},{"Start":"12:12.670 ","End":"12:16.660","Text":"both in the x-direction and the y-direction."},{"Start":"12:16.660 ","End":"12:19.795","Text":"Just to remind you what our vector was."},{"Start":"12:19.795 ","End":"12:24.745","Text":"We had our x, y and then we had a parametric equation."},{"Start":"12:24.745 ","End":"12:32.545","Text":"Our x was equal to a Phi plus sine Phi and"},{"Start":"12:32.545 ","End":"12:41.155","Text":"our y was equal to a1 minus cosine of Phi."},{"Start":"12:41.155 ","End":"12:44.500","Text":"We have 1/2 and we have our ms which are given to us."},{"Start":"12:44.500 ","End":"12:48.625","Text":"Now we just have to see what our x dot and our y dot is equal to."},{"Start":"12:48.625 ","End":"12:55.240","Text":"In this case, our x dot is going to be equal to and we\u0027re going to use the chain rule."},{"Start":"12:55.240 ","End":"12:58.525","Text":"Our a is a constant so it remains outside."},{"Start":"12:58.525 ","End":"13:00.745","Text":"Then we have our Phi."},{"Start":"13:00.745 ","End":"13:03.445","Text":"The derivative of our Phi is 1,"},{"Start":"13:03.445 ","End":"13:06.910","Text":"the derivative of sine Phi,"},{"Start":"13:06.910 ","End":"13:09.445","Text":"sorry, this is meant to be a negative,"},{"Start":"13:09.445 ","End":"13:11.425","Text":"so negative sine Phi."},{"Start":"13:11.425 ","End":"13:17.200","Text":"We\u0027re going to have a and then we have 1 minus and then the derivative of sine Phi is"},{"Start":"13:17.200 ","End":"13:22.689","Text":"cosine Phi and then multiplied by the inner derivative."},{"Start":"13:22.689 ","End":"13:25.000","Text":"That\u0027s going to be Phi dot."},{"Start":"13:25.000 ","End":"13:26.965","Text":"We\u0027re using the chain rule."},{"Start":"13:26.965 ","End":"13:31.075","Text":"Then that means therefore here that"},{"Start":"13:31.075 ","End":"13:36.670","Text":"our x dot squared is simply going to be the square root."},{"Start":"13:36.670 ","End":"13:40.255","Text":"It\u0027s going to be equal to a^2 multiplied by"},{"Start":"13:40.255 ","End":"13:48.760","Text":"1 minus cosine of Phi squared multiplied by Phi dot squared."},{"Start":"13:48.760 ","End":"13:50.245","Text":"Then we can do the same for y."},{"Start":"13:50.245 ","End":"13:52.525","Text":"We\u0027ll have that our y dot is equal to, again,"},{"Start":"13:52.525 ","End":"13:55.825","Text":"a is a constant multiplied by,"},{"Start":"13:55.825 ","End":"13:58.885","Text":"a 1 is going to cancel out."},{"Start":"13:58.885 ","End":"14:01.615","Text":"We\u0027re going to lose that so we don\u0027t need the brackets in fact,"},{"Start":"14:01.615 ","End":"14:05.290","Text":"then multiply it by negative cosine Phi."},{"Start":"14:05.290 ","End":"14:09.760","Text":"The derivative of negative cosine Phi is going to be sine Phi"},{"Start":"14:09.760 ","End":"14:14.620","Text":"and then multiplied by the inner derivative which is Phi dot."},{"Start":"14:14.620 ","End":"14:19.390","Text":"Then therefore our y dot squared is going to be equal to"},{"Start":"14:19.390 ","End":"14:26.950","Text":"a^2 Phi dot squared multiplied by sine^2 of Phi."},{"Start":"14:26.950 ","End":"14:32.470","Text":"Now let\u0027s substitute all of this into our equation for kinetic energy."},{"Start":"14:32.470 ","End":"14:37.225","Text":"Therefore, kinetic energy is going to be equal to,"},{"Start":"14:37.225 ","End":"14:42.400","Text":"our common factors are 1/2m and a."},{"Start":"14:42.400 ","End":"14:45.490","Text":"Then we can open our brackets,"},{"Start":"14:45.490 ","End":"14:51.250","Text":"and then we can see that in our a^2 even, so x dot squared."},{"Start":"14:51.250 ","End":"14:54.370","Text":"We\u0027ve already put our x dot."},{"Start":"14:54.370 ","End":"15:00.430","Text":"Now we\u0027ll have 1 minus cosine of Phi."},{"Start":"15:00.430 ","End":"15:09.685","Text":"This is going to be squared and then we\u0027re going to have plus our sine^2 of Phi."},{"Start":"15:09.685 ","End":"15:12.985","Text":"Then we can close the brackets and multiply by"},{"Start":"15:12.985 ","End":"15:16.660","Text":"our Phi dot squared which we also have in both terms."},{"Start":"15:16.660 ","End":"15:18.985","Text":"Then by using some trig identities,"},{"Start":"15:18.985 ","End":"15:21.355","Text":"we can simplify this even more."},{"Start":"15:21.355 ","End":"15:25.269","Text":"I\u0027ll write it as half multiplied by m multiplied"},{"Start":"15:25.269 ","End":"15:32.875","Text":"by 2 a^2 multiplied by 1 minus cosine of Phi,"},{"Start":"15:32.875 ","End":"15:36.370","Text":"then multiplied by our Phi dot squared."},{"Start":"15:36.370 ","End":"15:41.260","Text":"All I\u0027ve done is simplify this via some trig identities."},{"Start":"15:41.260 ","End":"15:42.550","Text":"You can do this alone."},{"Start":"15:42.550 ","End":"15:44.065","Text":"You\u0027ll see that we get to this."},{"Start":"15:44.065 ","End":"15:47.830","Text":"Now for our potential energy, our U,"},{"Start":"15:47.830 ","End":"15:52.145","Text":"as we know, this is equal to our mgh."},{"Start":"15:52.145 ","End":"15:58.555","Text":"Just like how before we defined our h for any point randomly,"},{"Start":"15:58.555 ","End":"16:01.390","Text":"wherever we want. Where was it?"},{"Start":"16:01.390 ","End":"16:05.815","Text":"Was down here, where it was 2a minus y."},{"Start":"16:05.815 ","End":"16:11.860","Text":"Let\u0027s go back and we\u0027re going to do it the exact same way."},{"Start":"16:11.860 ","End":"16:18.625","Text":"We\u0027re going to have mg multiplied by 2a minus y."},{"Start":"16:18.625 ","End":"16:25.150","Text":"This is for any random point where our h at the bottom is set to equal 0."},{"Start":"16:25.150 ","End":"16:28.675","Text":"Then again, we can substitute in for our y,"},{"Start":"16:28.675 ","End":"16:36.895","Text":"we\u0027ll get that therefore our potential energy is equal to mg multiplied by 2a minus"},{"Start":"16:36.895 ","End":"16:47.240","Text":"our y which is a plus a cosine of Phi."},{"Start":"16:47.240 ","End":"16:51.595","Text":"I just opened up the brackets over here, so 2a minus,"},{"Start":"16:51.595 ","End":"16:54.085","Text":"and then we have a minus minus,"},{"Start":"16:54.085 ","End":"16:56.665","Text":"that\u0027s plus, a cosine of Phi."},{"Start":"16:56.665 ","End":"17:02.830","Text":"That\u0027s going to be equal to mg and then 2a minus a is going to be a."},{"Start":"17:02.830 ","End":"17:08.995","Text":"We can take that out as a common factor,"},{"Start":"17:08.995 ","End":"17:16.795","Text":"mga, and then it\u0027s going to be 1 minus cosine of Phi."},{"Start":"17:16.795 ","End":"17:19.915","Text":"Now we have our kinetic energy and our potential energy."},{"Start":"17:19.915 ","End":"17:25.060","Text":"That means that the total energy of the system is going"},{"Start":"17:25.060 ","End":"17:30.610","Text":"to be a combination of this for our potential energy and this for our kinetic energy."},{"Start":"17:30.610 ","End":"17:34.375","Text":"Let\u0027s write that out. We\u0027re going to have that our energy is equal to"},{"Start":"17:34.375 ","End":"17:38.350","Text":"1/2m multiplied by 2 a^2 multiplied by"},{"Start":"17:38.350 ","End":"17:47.545","Text":"1 minus cosine of Phi multiplied by Phi dot squared plus,"},{"Start":"17:47.545 ","End":"17:57.565","Text":"over here, our mga 1 minus cosine of Phi."},{"Start":"17:57.565 ","End":"18:03.025","Text":"Then this is going to equal our energy at the beginning."},{"Start":"18:03.025 ","End":"18:05.650","Text":"This is in general and then our initial energy,"},{"Start":"18:05.650 ","End":"18:07.750","Text":"which is what we worked out in question number 1,"},{"Start":"18:07.750 ","End":"18:13.420","Text":"we saw that we only had potential energy because it starts from rest."},{"Start":"18:13.420 ","End":"18:20.680","Text":"There is no kinetic energy and we saw that it was equal to 2mga."},{"Start":"18:20.680 ","End":"18:30.760","Text":"Now what\u0027s left to do is to isolate out our Phi dot by playing around with it."},{"Start":"18:30.760 ","End":"18:31.960","Text":"To save time, I won\u0027t do it,"},{"Start":"18:31.960 ","End":"18:35.080","Text":"but if you want to you can try it on your own."},{"Start":"18:35.080 ","End":"18:40.770","Text":"Our Phi dot squared is going to be equal to our g divided by a."},{"Start":"18:40.770 ","End":"18:47.485","Text":"That means that we have this differential equation."},{"Start":"18:47.485 ","End":"18:56.170","Text":"Then we can just simply solve this by saying that our Phi therefore is going to be"},{"Start":"18:56.170 ","End":"19:04.810","Text":"equal to the square root of g over a multiplied by t plus some constant."},{"Start":"19:04.810 ","End":"19:06.565","Text":"We\u0027re simply going to,"},{"Start":"19:06.565 ","End":"19:08.350","Text":"I say that our Phi-dot,"},{"Start":"19:08.350 ","End":"19:10.975","Text":"so square root this and then we\u0027re just going to"},{"Start":"19:10.975 ","End":"19:14.230","Text":"integrate it to get what our Phi is equal to."},{"Start":"19:14.230 ","End":"19:19.585","Text":"We have this constant over here and we need to find out what this is."},{"Start":"19:19.585 ","End":"19:23.110","Text":"It\u0027s a constant from integration. How do we find this out?"},{"Start":"19:23.110 ","End":"19:25.690","Text":"We find this out from initial conditions."},{"Start":"19:25.690 ","End":"19:31.645","Text":"What we were told in the question was that at a time of t is equal to 0,"},{"Start":"19:31.645 ","End":"19:35.710","Text":"our velocity is equal to 0 because we are released from rest"},{"Start":"19:35.710 ","End":"19:41.125","Text":"and our position is at the origin 0, 0."},{"Start":"19:41.125 ","End":"19:46.315","Text":"I\u0027m going to find it hard to fit in my velocity into this equation."},{"Start":"19:46.315 ","End":"19:49.795","Text":"I can set in my positions."},{"Start":"19:49.795 ","End":"19:52.869","Text":"In my t is equal to 0."},{"Start":"19:52.869 ","End":"19:56.410","Text":"Let\u0027s take a look up over here."},{"Start":"19:56.410 ","End":"20:01.660","Text":"Let\u0027s use my equation for my y because it will be easier to substitute this in."},{"Start":"20:01.660 ","End":"20:03.970","Text":"I\u0027m reminding you that my 0,"},{"Start":"20:03.970 ","End":"20:08.650","Text":"0 is my x, y point coordinate."},{"Start":"20:08.650 ","End":"20:11.965","Text":"Now I\u0027m using this parametric equation over here."},{"Start":"20:11.965 ","End":"20:14.620","Text":"If I said in that my x is equal to 0,"},{"Start":"20:14.620 ","End":"20:17.245","Text":"we can see it\u0027s a little bit more complicated to use."},{"Start":"20:17.245 ","End":"20:19.210","Text":"But if my y I said that equal to 0,"},{"Start":"20:19.210 ","End":"20:21.265","Text":"so I can easily find my Phi."},{"Start":"20:21.265 ","End":"20:25.225","Text":"Let\u0027s say that my y is equal to 0,"},{"Start":"20:25.225 ","End":"20:28.255","Text":"that means 0 is equal to,"},{"Start":"20:28.255 ","End":"20:30.595","Text":"let\u0027s see what our equation for y is."},{"Start":"20:30.595 ","End":"20:34.180","Text":"It\u0027s a 1 minus cosine of Phi."},{"Start":"20:34.180 ","End":"20:36.685","Text":"This is our occasion."},{"Start":"20:36.685 ","End":"20:40.255","Text":"We know that our a is a constant and can\u0027t be equal to 0."},{"Start":"20:40.255 ","End":"20:45.955","Text":"That means that what I have over here has to be equal to 0."},{"Start":"20:45.955 ","End":"20:48.205","Text":"In order for this to equal to 0,"},{"Start":"20:48.205 ","End":"20:53.305","Text":"I have to say that my cosine of Phi is equal to 1,"},{"Start":"20:53.305 ","End":"20:56.110","Text":"because then I have 1 minus 1 and that\u0027s equal to 0."},{"Start":"20:56.110 ","End":"21:01.335","Text":"That means cosine of 0 is equal to 1."},{"Start":"21:01.335 ","End":"21:05.145","Text":"That means that my Phi is equal to 0."},{"Start":"21:05.145 ","End":"21:08.715","Text":"Then I can substitute that into this equation."},{"Start":"21:08.715 ","End":"21:13.285","Text":"I can say that my Phi is equal to 0,"},{"Start":"21:13.285 ","End":"21:19.780","Text":"which is equal to the square root of g divided by a multiplied by my t,"},{"Start":"21:19.780 ","End":"21:24.340","Text":"which is also equal to 0 plus my c. I have 0 is equal"},{"Start":"21:24.340 ","End":"21:29.875","Text":"to 0 plus c. Therefore my c is equal to 0."},{"Start":"21:29.875 ","End":"21:33.865","Text":"Then that means therefore that my equation"},{"Start":"21:33.865 ","End":"21:38.740","Text":"for my motion is Phi is equal to the square root of"},{"Start":"21:38.740 ","End":"21:48.205","Text":"g divided by a multiplied by t. Now we have our equation Phi as function of time."},{"Start":"21:48.205 ","End":"21:56.335","Text":"Then we can take that Phi and substitute it into our parametric equations for x and y."},{"Start":"21:56.335 ","End":"21:59.590","Text":"Then we can find its position, its x-coordinate,"},{"Start":"21:59.590 ","End":"22:04.510","Text":"and its y-coordinate according to our time with using our Phi."},{"Start":"22:04.510 ","End":"22:09.610","Text":"Now question 3 is solve the equation from 2 according to the initial conditions."},{"Start":"22:09.610 ","End":"22:12.205","Text":"In actual fact, we\u0027ve already solved that."},{"Start":"22:12.205 ","End":"22:16.960","Text":"That\u0027s how we got rid of our constant c. That\u0027s how we found it."},{"Start":"22:16.960 ","End":"22:19.720","Text":"What do we really needed to do in 2 was just to come"},{"Start":"22:19.720 ","End":"22:22.795","Text":"to our differential equation of Phi- dot,"},{"Start":"22:22.795 ","End":"22:25.825","Text":"and then in 3 using our initial conditions."},{"Start":"22:25.825 ","End":"22:28.090","Text":"We\u0027ve already solved 2 and 3."},{"Start":"22:28.090 ","End":"22:30.250","Text":"Now let\u0027s look at question 4."},{"Start":"22:30.250 ","End":"22:35.365","Text":"Question number 4 is to show that the body moves in some harmonic motion."},{"Start":"22:35.365 ","End":"22:39.925","Text":"Periodical with a value for t,"},{"Start":"22:39.925 ","End":"22:44.965","Text":"the time per period matching that of a mathematical pendulum."},{"Start":"22:44.965 ","End":"22:50.452","Text":"They\u0027re not saying that the body is moving in a similar way to a mathematical pendulum,"},{"Start":"22:50.452 ","End":"22:54.640","Text":"they\u0027re wanting us to show that the time between each period or"},{"Start":"22:54.640 ","End":"22:59.860","Text":"the time taken for each period is the same as the time taken for"},{"Start":"22:59.860 ","End":"23:03.100","Text":"each period when looking at the mathematical pendulum of"},{"Start":"23:03.100 ","End":"23:06.685","Text":"length l. Then we\u0027re also being asked"},{"Start":"23:06.685 ","End":"23:09.880","Text":"to find what the appropriate value for l would be"},{"Start":"23:09.880 ","End":"23:13.840","Text":"here if we were to equate this to a mathematical pendulum."},{"Start":"23:13.840 ","End":"23:16.990","Text":"We know what our equation for Phi is."},{"Start":"23:16.990 ","End":"23:24.670","Text":"It\u0027s equal to the square root of g divided by a multiplied by t. Now,"},{"Start":"23:24.670 ","End":"23:28.750","Text":"we can substitute in our Phi over here and get our values for x and y."},{"Start":"23:28.750 ","End":"23:32.710","Text":"Let\u0027s do it 1 second for our y because it\u0027s a bit easier to calculate."},{"Start":"23:32.710 ","End":"23:35.410","Text":"I\u0027ll get that our y is equal to a multiplied by"},{"Start":"23:35.410 ","End":"23:40.680","Text":"1 minus cosine of the square root of g divided"},{"Start":"23:40.680 ","End":"23:49.240","Text":"by a multiplied by t. Now we can see that our cosine is a periodical function."},{"Start":"23:49.240 ","End":"23:51.175","Text":"It has a period,"},{"Start":"23:51.175 ","End":"23:54.625","Text":"which means that our y is also going to have"},{"Start":"23:54.625 ","End":"23:59.815","Text":"some period because it\u0027s dependent on this cosine."},{"Start":"23:59.815 ","End":"24:04.570","Text":"Now we want to find its frequency, its Omega."},{"Start":"24:04.570 ","End":"24:09.565","Text":"What do we have here is in fact Cosine of Omega multiplied by t."},{"Start":"24:09.565 ","End":"24:16.675","Text":"That means that our Omega is going to be equal to the square root of g divided by a."},{"Start":"24:16.675 ","End":"24:22.990","Text":"As we know, our Omega is equal to 2 Pi divided by t,"},{"Start":"24:22.990 ","End":"24:25.870","Text":"and t is our time per period."},{"Start":"24:25.870 ","End":"24:28.195","Text":"This is what we want to find."},{"Start":"24:28.195 ","End":"24:32.920","Text":"We can isolate out our t through algebra and we\u0027ll get that our t is equal"},{"Start":"24:32.920 ","End":"24:38.935","Text":"to 2 Pi multiplied by the square root of g divided by a."},{"Start":"24:38.935 ","End":"24:42.895","Text":"Let\u0027s explain what this movement is 1 second."},{"Start":"24:42.895 ","End":"24:45.685","Text":"If we\u0027re going to just look at the y again,"},{"Start":"24:45.685 ","End":"24:48.400","Text":"we can see that when our cosine,"},{"Start":"24:48.400 ","End":"24:51.370","Text":"this expression over here is equal to 0,"},{"Start":"24:51.370 ","End":"24:53.965","Text":"we\u0027re located right over here at the start."},{"Start":"24:53.965 ","End":"24:57.655","Text":"When our cosine function is equal to 1,"},{"Start":"24:57.655 ","End":"25:00.700","Text":"then we\u0027re located right over here at the bottom."},{"Start":"25:00.700 ","End":"25:07.030","Text":"Then when our Cosine completes its period,"},{"Start":"25:07.030 ","End":"25:09.715","Text":"because it has a period of 2Pi,"},{"Start":"25:09.715 ","End":"25:13.690","Text":"so it\u0027s again equal to 0."},{"Start":"25:13.690 ","End":"25:17.110","Text":"Again, we\u0027re going to be up here."},{"Start":"25:17.110 ","End":"25:23.695","Text":"That means that we\u0027ve done a whole period with our cosine."},{"Start":"25:23.695 ","End":"25:26.050","Text":"Our cosine function has had a whole period."},{"Start":"25:26.050 ","End":"25:28.705","Text":"However, our movement over here,"},{"Start":"25:28.705 ","End":"25:31.270","Text":"our trajectory hasn\u0027t had a whole period,"},{"Start":"25:31.270 ","End":"25:34.495","Text":"because a whole period is starting from some point,"},{"Start":"25:34.495 ","End":"25:37.330","Text":"going to its maximum and returning."},{"Start":"25:37.330 ","End":"25:41.290","Text":"But we\u0027ve only gone from one point to the other side."},{"Start":"25:41.290 ","End":"25:49.540","Text":"In actual fact, we\u0027ve done half a period of our cycloid course."},{"Start":"25:49.540 ","End":"25:57.745","Text":"This t that we\u0027ve actually found is actually our t tag and it\u0027s equal to 1/2 of"},{"Start":"25:57.745 ","End":"26:06.970","Text":"our t that we were actually trying to find 1/2 of our actual periods in here."},{"Start":"26:06.970 ","End":"26:15.880","Text":"Then what we get is that our actual time per period is going to be equal to 2 times this."},{"Start":"26:15.880 ","End":"26:21.955","Text":"4Pi multiplied by the square root of g divided by a."},{"Start":"26:21.955 ","End":"26:27.880","Text":"Now I found my t and now I have to show that it\u0027s matching the time per"},{"Start":"26:27.880 ","End":"26:34.300","Text":"period of that of a mathematical pendulum of length l. Let\u0027s scroll down a little bit."},{"Start":"26:34.300 ","End":"26:39.130","Text":"We know that the time per period of a mathematical pendulum of"},{"Start":"26:39.130 ","End":"26:44.109","Text":"length l is going to be equal to 2Pi multiplied"},{"Start":"26:44.109 ","End":"26:48.745","Text":"by the square root of l divided by g. This"},{"Start":"26:48.745 ","End":"26:53.545","Text":"is the time period T-tilde for a mathematical pendulum."},{"Start":"26:53.545 ","End":"26:56.170","Text":"I\u0027m just reminding you that the Omega for"},{"Start":"26:56.170 ","End":"26:59.320","Text":"a mathematical pendulum is the square root of g divided by"},{"Start":"26:59.320 ","End":"27:06.640","Text":"l. If we substitute it in to the equation that we had over here for Omega,"},{"Start":"27:06.640 ","End":"27:11.530","Text":"we\u0027ll get that time period for a mathematical pendulum is 2Pi multiplied"},{"Start":"27:11.530 ","End":"27:17.590","Text":"by the square root of l divided by g. You can substitute this in yourself."},{"Start":"27:17.590 ","End":"27:20.650","Text":"With your Omega for mathematical pendulum being"},{"Start":"27:20.650 ","End":"27:24.620","Text":"this into this type of equation and you\u0027ll get this answer."},{"Start":"27:26.100 ","End":"27:31.855","Text":"There is a similarity and now we\u0027re going to check it out."},{"Start":"27:31.855 ","End":"27:36.580","Text":"If we say that these 2 are equal to one another, that they\u0027re matching,"},{"Start":"27:36.580 ","End":"27:46.420","Text":"then we will get that 4Pi multiplied by the square root of g divided by a is equal"},{"Start":"27:46.420 ","End":"27:50.470","Text":"to 2Pi multiplied by the square root of"},{"Start":"27:50.470 ","End":"27:56.365","Text":"l divided by g. Then we can cross out this and this and this,"},{"Start":"27:56.365 ","End":"27:59.065","Text":"and then we\u0027ll have a 2 over here instead."},{"Start":"27:59.065 ","End":"28:02.800","Text":"Now we can square both sides to get rid of our square root sign."},{"Start":"28:02.800 ","End":"28:08.500","Text":"We\u0027ll have 4 multiplied by g divided by a is"},{"Start":"28:08.500 ","End":"28:14.245","Text":"going to be equal to l divided by g. Now,"},{"Start":"28:14.245 ","End":"28:19.855","Text":"all we have to do because we\u0027re being asked what is an appropriate value for l over here."},{"Start":"28:19.855 ","End":"28:24.115","Text":"We have to do is we have to isolate out this l."},{"Start":"28:24.115 ","End":"28:29.830","Text":"I made a calculation error over here and then to have"},{"Start":"28:29.830 ","End":"28:40.765","Text":"a over g. That means also here a over g. Then also over here a over g,"},{"Start":"28:40.765 ","End":"28:47.395","Text":"and here a over g. Now I can cancel off my gs on both sides,"},{"Start":"28:47.395 ","End":"28:50.170","Text":"and then when I look at my l,"},{"Start":"28:50.170 ","End":"28:54.715","Text":"I\u0027ll have them l is equal to 4 multiplied by a."},{"Start":"28:54.715 ","End":"28:56.665","Text":"This is our onset."},{"Start":"28:56.665 ","End":"29:03.745","Text":"Now we\u0027ve seen that the period of the movement of our body in this system,"},{"Start":"29:03.745 ","End":"29:05.770","Text":"in this diagram, in this course,"},{"Start":"29:05.770 ","End":"29:11.780","Text":"is the same as the period of a mathematical pendulum of length 4a."},{"Start":"29:12.330 ","End":"29:19.660","Text":"Now we\u0027ve answered also our questioned 4 and that\u0027s the end of our question."}],"ID":9473},{"Watched":false,"Name":"Center Of Mass Full Cone And Sphere","Duration":"52m 7s","ChapterTopicVideoID":9204,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"Hello. In this question,"},{"Start":"00:02.490 ","End":"00:05.910","Text":"we\u0027re being told that we have a body consisting of a cone with"},{"Start":"00:05.910 ","End":"00:10.694","Text":"an apex angle of Alpha. What does that mean?"},{"Start":"00:10.694 ","End":"00:14.280","Text":"Our apex angle of Alpha is going to be all of this."},{"Start":"00:14.280 ","End":"00:16.770","Text":"This angle here is Alpha."},{"Start":"00:16.770 ","End":"00:22.185","Text":"We see that our base radius is going to be a."},{"Start":"00:22.185 ","End":"00:27.959","Text":"This radius over here is going to be a and it\u0027s of"},{"Start":"00:27.959 ","End":"00:33.405","Text":"height h. It means that from the tip all the way to here,"},{"Start":"00:33.405 ","End":"00:35.850","Text":"this is our height h,"},{"Start":"00:35.850 ","End":"00:39.874","Text":"and our cone is sitting on a half sphere,"},{"Start":"00:39.874 ","End":"00:42.085","Text":"which is also a radius a."},{"Start":"00:42.085 ","End":"00:44.254","Text":"They share the same radius."},{"Start":"00:44.254 ","End":"00:48.980","Text":"The cone and half sphere have the same uniform density of Rho."},{"Start":"00:48.980 ","End":"00:52.435","Text":"Their density is uniform, it\u0027s Rho."},{"Start":"00:52.435 ","End":"00:57.624","Text":"Also, both the cone and the half sphere are full."},{"Start":"00:57.624 ","End":"01:00.604","Text":"In question number 1, we\u0027re being asked,"},{"Start":"01:00.604 ","End":"01:03.754","Text":"what is the center of mass of the cone,"},{"Start":"01:03.754 ","End":"01:06.844","Text":"just of the cone, relative to the origin,"},{"Start":"01:06.844 ","End":"01:11.355","Text":"which is located on the plane connecting the cone and the half sphere?"},{"Start":"01:11.355 ","End":"01:15.514","Text":"We can see where x and y axes cross."},{"Start":"01:15.514 ","End":"01:18.965","Text":"That there is a circular plane over here,"},{"Start":"01:18.965 ","End":"01:20.929","Text":"where half sphere and our cone meet,"},{"Start":"01:20.929 ","End":"01:28.275","Text":"and we can see that we have our point over here where our origin is."},{"Start":"01:28.275 ","End":"01:32.000","Text":"We\u0027re being asked, what is the center of mass of the cone?"},{"Start":"01:32.000 ","End":"01:34.175","Text":"Just of the cone without the half sphere,"},{"Start":"01:34.175 ","End":"01:37.020","Text":"relative to this point over here."},{"Start":"01:37.580 ","End":"01:41.224","Text":"Before we start answering the question,"},{"Start":"01:41.224 ","End":"01:44.254","Text":"we can see that because we have our angle over here, Alpha,"},{"Start":"01:44.254 ","End":"01:48.549","Text":"and we can see that this is a right-angled triangle,"},{"Start":"01:48.549 ","End":"01:55.585","Text":"we can write a connection between our angle and our a and our height."},{"Start":"01:55.585 ","End":"01:58.770","Text":"That\u0027s going to be this,"},{"Start":"01:58.770 ","End":"02:02.715","Text":"Tan of Alpha divided by 2,"},{"Start":"02:02.715 ","End":"02:06.860","Text":"because Alpha represents all of this and we only want half."},{"Start":"02:06.860 ","End":"02:12.615","Text":"That is going to be equal to a divided by h,"},{"Start":"02:12.615 ","End":"02:20.180","Text":"because we have Tan is the opposite side divided by the adjacent."},{"Start":"02:20.330 ","End":"02:24.924","Text":"We can see that if our angle Alpha divided by 2,"},{"Start":"02:24.924 ","End":"02:30.995","Text":"this is our Alpha divided by 2, this over here."},{"Start":"02:30.995 ","End":"02:36.114","Text":"Our opposite side is our a and our adjacent side is this,"},{"Start":"02:36.114 ","End":"02:42.230","Text":"which is of length h. That\u0027s how we get this."},{"Start":"02:42.950 ","End":"02:46.015","Text":"Because we\u0027re dealing with a cone,"},{"Start":"02:46.015 ","End":"02:49.615","Text":"and it\u0027s a regular cone with a circular base,"},{"Start":"02:49.615 ","End":"02:52.480","Text":"that means that we have symmetry along the x,"},{"Start":"02:52.480 ","End":"02:57.670","Text":"y axes because our origin is located right in the middle."},{"Start":"02:57.670 ","End":"03:00.700","Text":"The only thing that we have to work out in order to"},{"Start":"03:00.700 ","End":"03:03.715","Text":"find our center of mass is our z center of mass."},{"Start":"03:03.715 ","End":"03:04.870","Text":"Where\u0027s the center of mass?"},{"Start":"03:04.870 ","End":"03:07.985","Text":"On our z-axis, somewhere on this line."},{"Start":"03:07.985 ","End":"03:12.830","Text":"Let\u0027s label, this is question number 1."},{"Start":"03:12.830 ","End":"03:19.259","Text":"Our equation for Z_CM is going to be equal to an integral on"},{"Start":"03:19.259 ","End":"03:26.580","Text":"z_dm divided by an integral on dm."},{"Start":"03:26.980 ","End":"03:29.449","Text":"Now this is, of course,"},{"Start":"03:29.449 ","End":"03:33.124","Text":"our total mass, M total."},{"Start":"03:33.124 ","End":"03:34.819","Text":"Let\u0027s, for the meantime,"},{"Start":"03:34.819 ","End":"03:37.295","Text":"imagine that we know this value,"},{"Start":"03:37.295 ","End":"03:40.650","Text":"and we\u0027ll see at the end it will cancel out, anyway."},{"Start":"03:40.910 ","End":"03:44.090","Text":"The first thing that we have to figure out is,"},{"Start":"03:44.090 ","End":"03:45.940","Text":"what is our dm?"},{"Start":"03:45.940 ","End":"03:47.655","Text":"Let\u0027s take a look at that."},{"Start":"03:47.655 ","End":"03:53.704","Text":"Now our dm is always going to be equal to our Rho dv,"},{"Start":"03:53.704 ","End":"03:57.620","Text":"where our Rho is our density per"},{"Start":"03:57.620 ","End":"04:04.910","Text":"unit of volume,"},{"Start":"04:04.910 ","End":"04:08.299","Text":"and our dv is going to be our unit of volume,"},{"Start":"04:08.299 ","End":"04:10.279","Text":"it\u0027s a small unit of volume."},{"Start":"04:10.279 ","End":"04:14.429","Text":"Then we also know what our Rho is."},{"Start":"04:14.429 ","End":"04:17.470","Text":"Our Rho, because its density per unit volume,"},{"Start":"04:17.470 ","End":"04:21.469","Text":"so the way that we work it out is we take the total mass of"},{"Start":"04:21.469 ","End":"04:26.609","Text":"the system and divide it by the total volume."},{"Start":"04:26.720 ","End":"04:29.355","Text":"Then our volume,"},{"Start":"04:29.355 ","End":"04:31.145","Text":"when dealing with a cone,"},{"Start":"04:31.145 ","End":"04:33.589","Text":"you can look in your equation sheet,"},{"Start":"04:33.589 ","End":"04:37.180","Text":"it\u0027s going to be equal to Pi a^2."},{"Start":"04:37.180 ","End":"04:39.750","Text":"This is the area of the base."},{"Start":"04:39.750 ","End":"04:42.679","Text":"It\u0027s circular area where a is our radius,"},{"Start":"04:42.679 ","End":"04:45.004","Text":"multiplied by the height of the cone,"},{"Start":"04:45.004 ","End":"04:48.630","Text":"and then divided by 3."},{"Start":"04:49.220 ","End":"04:54.080","Text":"The last thing that we need is to see what our dv is."},{"Start":"04:54.080 ","End":"04:56.735","Text":"Let\u0027s write over here dv."},{"Start":"04:56.735 ","End":"05:03.190","Text":"Our dv represents the units or the axes that we\u0027re going to be integrating on."},{"Start":"05:03.190 ","End":"05:05.685","Text":"Here we need to have our Z,"},{"Start":"05:05.685 ","End":"05:08.760","Text":"because we\u0027re finding our Z_CM."},{"Start":"05:08.760 ","End":"05:11.480","Text":"Here, we can see we can have our x and our y."},{"Start":"05:11.480 ","End":"05:13.295","Text":"But because we\u0027re dealing with a circle,"},{"Start":"05:13.295 ","End":"05:14.900","Text":"usually, when dealing with a circle,"},{"Start":"05:14.900 ","End":"05:23.790","Text":"it\u0027s easier to deal with our radius and our angle Theta, using polar coordinates."},{"Start":"05:23.790 ","End":"05:28.150","Text":"Then we\u0027re going to have dr d Theta,"},{"Start":"05:28.150 ","End":"05:31.340","Text":"which are polar coordinates, and then dz."},{"Start":"05:31.340 ","End":"05:33.245","Text":"Whenever we have our d Theta,"},{"Start":"05:33.245 ","End":"05:36.575","Text":"we have to multiply it by the Jacobian,"},{"Start":"05:36.575 ","End":"05:37.789","Text":"which is our r,"},{"Start":"05:37.789 ","End":"05:42.900","Text":"so r dr d Theta dz is our dv."},{"Start":"05:43.940 ","End":"05:50.000","Text":"Now we have everything that we need in order to write out our equation."},{"Start":"05:50.000 ","End":"05:53.880","Text":"Let\u0027s substitute everything in. Let\u0027s see."},{"Start":"05:53.880 ","End":"05:57.495","Text":"We have that our Z_CM is going to be equal to."},{"Start":"05:57.495 ","End":"06:01.395","Text":"This integral on dm is our M total."},{"Start":"06:01.395 ","End":"06:05.790","Text":"We\u0027re going to have 1 divided by M total,"},{"Start":"06:05.790 ","End":"06:12.920","Text":"and then we have to multiply it by the integral on our z_dm."},{"Start":"06:12.920 ","End":"06:16.189","Text":"In case we\u0027re going to have our integral here, then let\u0027s see."},{"Start":"06:16.189 ","End":"06:18.160","Text":"What is our z?"},{"Start":"06:18.160 ","End":"06:19.980","Text":"In the meantime,"},{"Start":"06:19.980 ","End":"06:28.115","Text":"our Z this is our variable that we\u0027re integrating on and then multiplied by our dm."},{"Start":"06:28.115 ","End":"06:30.290","Text":"Let\u0027s take a look of what our dm is?"},{"Start":"06:30.290 ","End":"06:32.495","Text":"Our dm is our Rho, dv."},{"Start":"06:32.495 ","End":"06:37.814","Text":"Our Rho is M total divided by V,"},{"Start":"06:37.814 ","End":"06:44.549","Text":"where our V is going to be Pi a^2 h divided by 3,"},{"Start":"06:44.549 ","End":"06:46.950","Text":"so the 3 comes up here."},{"Start":"06:46.950 ","End":"06:49.855","Text":"That was our Rho,"},{"Start":"06:49.855 ","End":"06:51.770","Text":"which was our empty divided by v,"},{"Start":"06:51.770 ","End":"06:54.650","Text":"where this is our V and now our dv."},{"Start":"06:54.650 ","End":"07:04.185","Text":"That\u0027s going to be our r multiplied by dr d Theta, and then dz."},{"Start":"07:04.185 ","End":"07:09.545","Text":"Because we have our integration according to r, Theta, and z,"},{"Start":"07:09.545 ","End":"07:14.959","Text":"so we have to have 3 integration signs."},{"Start":"07:14.959 ","End":"07:19.945","Text":"Let\u0027s see what our limits for integration are."},{"Start":"07:19.945 ","End":"07:22.855","Text":"Let\u0027s just rewrite our Z_CM."},{"Start":"07:22.855 ","End":"07:25.360","Text":"We\u0027re going to have that our Z_CM is equal to,"},{"Start":"07:25.360 ","End":"07:30.430","Text":"now notice, here we have 1 divided by M_T and here we\u0027re multiplying by M_T."},{"Start":"07:30.430 ","End":"07:32.034","Text":"Just like I said before,"},{"Start":"07:32.034 ","End":"07:35.710","Text":"that we can take this as something that\u0027s given and it will cancel out in the end,"},{"Start":"07:35.710 ","End":"07:36.999","Text":"so here we see it."},{"Start":"07:36.999 ","End":"07:44.154","Text":"In case we can cancel that out and then we can see that our 3 divided by Pi a^2 h are"},{"Start":"07:44.154 ","End":"07:48.070","Text":"constants because we\u0027re not integrating with"},{"Start":"07:48.070 ","End":"07:53.280","Text":"variables a or h. I can take them out of the integration sign,"},{"Start":"07:53.280 ","End":"08:02.319","Text":"so I have 3 divided by Pi a^2 h. Now I\u0027m entering in my integral sign,"},{"Start":"08:02.319 ","End":"08:06.520","Text":"so I\u0027m leaving some space so that I can write in the limits and then I\u0027m"},{"Start":"08:06.520 ","End":"08:14.046","Text":"integrating zr dr d Theta dz."},{"Start":"08:14.046 ","End":"08:18.955","Text":"Now let\u0027s take a look at what our limits have to be."},{"Start":"08:18.955 ","End":"08:26.259","Text":"Let\u0027s draw over here our cone and we know that our radius over"},{"Start":"08:26.259 ","End":"08:30.729","Text":"here is a and then our height over"},{"Start":"08:30.729 ","End":"08:37.040","Text":"here is h. Let\u0027s leave r for 1 second and let\u0027s look at our Theta."},{"Start":"08:37.040 ","End":"08:41.619","Text":"Now, because we have to integrate along this entire circle,"},{"Start":"08:41.619 ","End":"08:43.450","Text":"it\u0027s a circular base,"},{"Start":"08:43.450 ","End":"08:45.730","Text":"I have to integrate along the entire circle."},{"Start":"08:45.730 ","End":"08:51.310","Text":"I know that it\u0027s going to be from 0 and up until 2Pi."},{"Start":"08:51.310 ","End":"08:56.420","Text":"I write this in the middle because my d Theta is between my dr and my dz."},{"Start":"08:57.750 ","End":"09:03.679","Text":"Now let\u0027s take a look at our integration with our dr and our dz."},{"Start":"09:03.810 ","End":"09:07.719","Text":"We\u0027ll notice that if we go along our dz,"},{"Start":"09:07.719 ","End":"09:12.339","Text":"we\u0027re integrating along this entire length."},{"Start":"09:12.339 ","End":"09:15.625","Text":"However, if we go according to my dr,"},{"Start":"09:15.625 ","End":"09:17.485","Text":"also from 0 to a,"},{"Start":"09:17.485 ","End":"09:21.460","Text":"then we\u0027ll see that I\u0027ll go up some length in"},{"Start":"09:21.460 ","End":"09:25.780","Text":"z and then I\u0027ll go my radius a and again up in"},{"Start":"09:25.780 ","End":"09:29.830","Text":"z and again my radius a and then we\u0027ll eventually see that"},{"Start":"09:29.830 ","End":"09:34.915","Text":"I\u0027m integrating along a cylinder."},{"Start":"09:34.915 ","End":"09:38.424","Text":"I don\u0027t want a cylinder, I want a cone."},{"Start":"09:38.424 ","End":"09:41.515","Text":"Let\u0027s see how I do this."},{"Start":"09:41.515 ","End":"09:46.300","Text":"I need to find some way of connecting going"},{"Start":"09:46.300 ","End":"09:52.299","Text":"up this certain amount of heights and then getting to this point over here."},{"Start":"09:52.299 ","End":"09:56.689","Text":"If we call this distance r_max,"},{"Start":"09:57.150 ","End":"10:01.584","Text":"and each time depending on my height, it changes."},{"Start":"10:01.584 ","End":"10:06.475","Text":"If this point over here is z and this entire length is h,"},{"Start":"10:06.475 ","End":"10:13.839","Text":"so that means that this length from"},{"Start":"10:13.839 ","End":"10:21.730","Text":"the tip until my z is going to be h minus z because the whole length is h,"},{"Start":"10:21.730 ","End":"10:23.590","Text":"up until here is z,"},{"Start":"10:23.590 ","End":"10:28.219","Text":"so this length is going to be h minus z."},{"Start":"10:28.980 ","End":"10:32.560","Text":"Then I know that this,"},{"Start":"10:32.560 ","End":"10:35.320","Text":"if I know that this angle, remember,"},{"Start":"10:35.320 ","End":"10:39.759","Text":"we said that this angle was Alpha divided by 2,"},{"Start":"10:39.759 ","End":"10:44.770","Text":"so I can write that same equation that we"},{"Start":"10:44.770 ","End":"10:49.795","Text":"had before for my tangents of Alpha divided by 2."},{"Start":"10:49.795 ","End":"10:51.718","Text":"Let\u0027s just remind you."},{"Start":"10:51.718 ","End":"11:01.359","Text":"I had that my Tan of Alpha divided by 2 was equal to my a divided by h. Because it was"},{"Start":"11:01.359 ","End":"11:06.190","Text":"my opposite side to my angle divided by"},{"Start":"11:06.190 ","End":"11:11.959","Text":"my adjacent side to the angle because that\u0027s the equation when dealing with Tan."},{"Start":"11:11.959 ","End":"11:18.820","Text":"Now we can see that my adjacent angle is now h minus z,"},{"Start":"11:18.820 ","End":"11:23.919","Text":"it\u0027s this green 1 and my opposite angle is now r_max."},{"Start":"11:23.919 ","End":"11:29.049","Text":"I\u0027m looking at this as if it\u0027s some triangle formation."},{"Start":"11:29.049 ","End":"11:35.124","Text":"First, I have this triangle and then aside from that,"},{"Start":"11:35.124 ","End":"11:42.130","Text":"I have the bigger triangle and the relationship between the angles is the same."},{"Start":"11:42.130 ","End":"11:47.829","Text":"Then I can say that this is also equal to this adjacent angles,"},{"Start":"11:47.829 ","End":"11:57.370","Text":"it\u0027s divided by adjacent angle,"},{"Start":"11:57.370 ","End":"12:05.089","Text":"so divided by h minus z and then our opposite side it is our r_max."},{"Start":"12:06.990 ","End":"12:09.880","Text":"Then through this relationship,"},{"Start":"12:09.880 ","End":"12:14.690","Text":"we can work out our limits for integration."},{"Start":"12:16.040 ","End":"12:21.734","Text":"Now what I want to isolate out is my r_max because I want to know what I\u0027m"},{"Start":"12:21.734 ","End":"12:28.285","Text":"integrating on with my r. We can just simply rearrange."},{"Start":"12:28.285 ","End":"12:36.769","Text":"I\u0027ll have the my r_max is going to be a divided by h multiplied by h minus z."},{"Start":"12:38.010 ","End":"12:42.460","Text":"Now we can substitute this in to our integration."},{"Start":"12:42.460 ","End":"12:44.155","Text":"Let\u0027s do our r,"},{"Start":"12:44.155 ","End":"12:48.205","Text":"so r, our radius starts at 0."},{"Start":"12:48.205 ","End":"12:54.010","Text":"Then we integrate up until this, for a certain z."},{"Start":"12:54.010 ","End":"12:57.130","Text":"This is our upper bound,"},{"Start":"12:57.130 ","End":"13:05.299","Text":"it\u0027s going to be a divided by h multiplied by h minus z."},{"Start":"13:06.030 ","End":"13:10.735","Text":"This is our upper limit and then with our z,"},{"Start":"13:10.735 ","End":"13:17.082","Text":"we\u0027re integrating and cross the whole heights of the cone."},{"Start":"13:17.082 ","End":"13:21.955","Text":"From here, from our 0 until our h,"},{"Start":"13:21.955 ","End":"13:25.449","Text":"which is our highest value for z."},{"Start":"13:25.449 ","End":"13:29.349","Text":"We can put in here from 0 until z."},{"Start":"13:29.349 ","End":"13:35.979","Text":"Now, all that\u0027s left to do is to actually do the integration."},{"Start":"13:35.979 ","End":"13:39.175","Text":"Let\u0027s begin doing the integral."},{"Start":"13:39.175 ","End":"13:43.689","Text":"We\u0027ll get that our Z_CM is equal to our"},{"Start":"13:43.689 ","End":"13:50.890","Text":"3 divided by Pi a^2 h. Then let\u0027s start off with our Theta."},{"Start":"13:50.890 ","End":"13:54.925","Text":"Because we don\u0027t have any variables over here for Theta,"},{"Start":"13:54.925 ","End":"13:56.439","Text":"when I integrate on that,"},{"Start":"13:56.439 ","End":"14:00.099","Text":"it\u0027s like integrating on 1 d Theta,"},{"Start":"14:00.099 ","End":"14:04.839","Text":"which is just going to be Theta and then when we send in our bounds of 0 and 2Pi,"},{"Start":"14:04.839 ","End":"14:07.269","Text":"it\u0027s just going to be multiplied by 2Pi."},{"Start":"14:07.269 ","End":"14:07.600","Text":"Then we can integrate on our dr,"},{"Start":"14:07.600 ","End":"14:07.601","Text":"we\u0027ll leave"},{"Start":"14:07.601 ","End":"14:07.602","Text":"our dz"},{"Start":"14:07.602 ","End":"14:19.296","Text":"last."},{"Start":"14:19.296 ","End":"14:23.259","Text":"We\u0027re integrating up until h our maximum height"},{"Start":"14:23.259 ","End":"14:27.670","Text":"is h. Why are we leaving our dz till last?"},{"Start":"14:27.670 ","End":"14:29.154","Text":"Because as we can see,"},{"Start":"14:29.154 ","End":"14:32.725","Text":"when we integrate on our r in our bound,"},{"Start":"14:32.725 ","End":"14:35.950","Text":"we have our variable z over here."},{"Start":"14:35.950 ","End":"14:39.325","Text":"That means that we also have to integrate on this,"},{"Start":"14:39.325 ","End":"14:43.905","Text":"that\u0027s why we can\u0027t leave our r last because if we integrate first on our z."},{"Start":"14:43.905 ","End":"14:47.454","Text":"Then we have our z and r integral,"},{"Start":"14:47.454 ","End":"14:49.585","Text":"then we haven\u0027t done this correctly."},{"Start":"14:49.585 ","End":"14:53.889","Text":"Whenever you have your another variable in 1 of the limits,"},{"Start":"14:53.889 ","End":"14:55.749","Text":"be an upper or lower,"},{"Start":"14:55.749 ","End":"15:00.290","Text":"then you have to leave that integration until last."},{"Start":"15:00.450 ","End":"15:07.000","Text":"Now we\u0027re going to do dr. Let\u0027s see."},{"Start":"15:07.000 ","End":"15:08.755","Text":"When we\u0027re integrating an r,"},{"Start":"15:08.755 ","End":"15:16.345","Text":"it\u0027s going to become r^2 divided by 2 and then we set in our bounds."},{"Start":"15:16.345 ","End":"15:24.740","Text":"That\u0027s going to be 0 until a over h multiplied by h minus z."},{"Start":"15:24.740 ","End":"15:29.729","Text":"Then, we also have to still integrate"},{"Start":"15:29.729 ","End":"15:38.470","Text":"from 0 until h on our d, z."},{"Start":"15:39.290 ","End":"15:43.174","Text":"Let\u0027s see what we do."},{"Start":"15:43.174 ","End":"15:46.059","Text":"If we set in our limits,"},{"Start":"15:46.059 ","End":"15:54.835","Text":"it\u0027s going to be equal to 3 divided by Pi a^2 h multiplied by 2 Pi,"},{"Start":"15:54.835 ","End":"15:57.669","Text":"we\u0027ll cancel things out soon."},{"Start":"15:57.669 ","End":"16:01.000","Text":"Then the integral from 0 to h(z)."},{"Start":"16:01.000 ","End":"16:04.599","Text":"Then when we\u0027ve substituted in our limits,"},{"Start":"16:04.599 ","End":"16:10.989","Text":"we\u0027ll get a divided by h multiplied by h minus"},{"Start":"16:10.989 ","End":"16:20.440","Text":"z^2 minus 0 divided by 2,"},{"Start":"16:20.440 ","End":"16:23.575","Text":"and then d, z."},{"Start":"16:23.575 ","End":"16:26.500","Text":"Let\u0027s try and solve this really quickly."},{"Start":"16:26.500 ","End":"16:28.390","Text":"We have our 2 over here,"},{"Start":"16:28.390 ","End":"16:30.715","Text":"our whole expression is divided by 2."},{"Start":"16:30.715 ","End":"16:32.529","Text":"But here we\u0027re multiplying by 2."},{"Start":"16:32.529 ","End":"16:35.064","Text":"Our 2 and our 2 can cancel out,"},{"Start":"16:35.064 ","End":"16:37.449","Text":"and our Pi and our Pi can cancel out."},{"Start":"16:37.449 ","End":"16:41.065","Text":"Then we have 3 divided by"},{"Start":"16:41.065 ","End":"16:48.475","Text":"a^2 h and our integration between 0 to h(z)."},{"Start":"16:48.475 ","End":"16:51.685","Text":"Now let\u0027s open up these brackets over here."},{"Start":"16:51.685 ","End":"16:55.300","Text":"We\u0027re going to have z multiplied by a"},{"Start":"16:55.300 ","End":"16:59.455","Text":"divided by h and then let\u0027s open up these square brackets."},{"Start":"16:59.455 ","End":"17:03.309","Text":"We\u0027re going to have h^2"},{"Start":"17:03.309 ","End":"17:12.415","Text":"minus 2zh plus z^2 and then of course d, z."},{"Start":"17:12.415 ","End":"17:20.200","Text":"Again, 3 divided by a^2 h from 0 until h, z."},{"Start":"17:20.200 ","End":"17:25.450","Text":"Now we can divide everything by h and multiply everything by a."},{"Start":"17:25.450 ","End":"17:29.290","Text":"We\u0027re going to have z is equal to a,"},{"Start":"17:29.290 ","End":"17:39.920","Text":"h minus 2z a plus z squared a divided by h, d z."},{"Start":"17:42.570 ","End":"17:47.665","Text":"Now we can just multiply everything in here by z."},{"Start":"17:47.665 ","End":"17:51.295","Text":"Let\u0027s rub this out and then we can have a z here,"},{"Start":"17:51.295 ","End":"17:55.720","Text":"here, and this will become to the third."},{"Start":"17:55.720 ","End":"17:58.870","Text":"Now there\u0027s something pretty easy to integrate."},{"Start":"17:58.870 ","End":"18:06.310","Text":"We\u0027re going to have 3 divided by a^2 h. Then we\u0027re going to have,"},{"Start":"18:06.310 ","End":"18:10.319","Text":"this is going to be z^2 divided by 2 a,"},{"Start":"18:10.319 ","End":"18:15.715","Text":"h minus 2 z^3 divided by 3,"},{"Start":"18:15.715 ","End":"18:22.190","Text":"a plus z to the 4a divided by 4h."},{"Start":"18:22.440 ","End":"18:26.845","Text":"Then we have to set in our integration limits."},{"Start":"18:26.845 ","End":"18:29.139","Text":"When our z is equal to 0,"},{"Start":"18:29.139 ","End":"18:30.984","Text":"notice everything cancels out."},{"Start":"18:30.984 ","End":"18:34.585","Text":"It\u0027s enough to just set in our upper bound."},{"Start":"18:34.585 ","End":"18:39.849","Text":"Then we\u0027ll get that we have 3 divided by a^2 h multiplied"},{"Start":"18:39.849 ","End":"18:44.890","Text":"by h^2 multiplied h^3 a divided by"},{"Start":"18:44.890 ","End":"18:49.959","Text":"2 minus 2 h^3 a divided"},{"Start":"18:49.959 ","End":"18:55.885","Text":"by 3 plus h to the 4a divided by 4h."},{"Start":"18:55.885 ","End":"19:02.155","Text":"Then we can rub out this h and say that this is to the 3."},{"Start":"19:02.155 ","End":"19:07.780","Text":"Then we can divide each term by h. This will become squared, squared,"},{"Start":"19:07.780 ","End":"19:12.880","Text":"squared and also by an a,"},{"Start":"19:12.880 ","End":"19:15.640","Text":"so get rid of all of these a\u0027s."},{"Start":"19:15.640 ","End":"19:19.644","Text":"Now, if we try and cancel these out and solve,"},{"Start":"19:19.644 ","End":"19:24.715","Text":"we will get that our final answer is going to be h divided by 4."},{"Start":"19:24.715 ","End":"19:27.294","Text":"This is our final answer,"},{"Start":"19:27.294 ","End":"19:31.840","Text":"that our z_CM is located at h divided by 4."},{"Start":"19:31.840 ","End":"19:34.705","Text":"Now let\u0027s go to question number 2."},{"Start":"19:34.705 ","End":"19:37.834","Text":"Question number 2 is saying,"},{"Start":"19:37.834 ","End":"19:40.904","Text":"if the center of mass of the half sphere,"},{"Start":"19:40.904 ","End":"19:42.674","Text":"that\u0027s this over here,"},{"Start":"19:42.674 ","End":"19:45.135","Text":"not including the cone, just of the half sphere,"},{"Start":"19:45.135 ","End":"19:48.105","Text":"is located at this Z center mass,"},{"Start":"19:48.105 ","End":"19:50.795","Text":"negative 3/8 of a,"},{"Start":"19:50.795 ","End":"19:54.850","Text":"relative to the connecting plane from question number 1,"},{"Start":"19:54.850 ","End":"19:57.355","Text":"that\u0027s our x, y plane."},{"Start":"19:57.355 ","End":"19:59.814","Text":"That\u0027s relative to our origin."},{"Start":"19:59.814 ","End":"20:03.550","Text":"What is the center of mass of the entire body?"},{"Start":"20:03.550 ","End":"20:07.750","Text":"We already saw that our answer to question 1 for"},{"Start":"20:07.750 ","End":"20:13.454","Text":"our Z_CM of the cone is equal to h divided by 4."},{"Start":"20:13.454 ","End":"20:20.625","Text":"That means that somewhere around here is our Z_CM of the code."},{"Start":"20:20.625 ","End":"20:23.340","Text":"Now they\u0027re giving us the Z_CM of"},{"Start":"20:23.340 ","End":"20:28.855","Text":"the half sphere and they\u0027re asking us what is the center of mass of everything together?"},{"Start":"20:28.855 ","End":"20:31.404","Text":"All we have to do is we have to find"},{"Start":"20:31.404 ","End":"20:40.449","Text":"the center of mass between the center of mass of the half sphere,"},{"Start":"20:40.449 ","End":"20:42.745","Text":"and the center of mass of the cone."},{"Start":"20:42.745 ","End":"20:45.430","Text":"We\u0027re looking at every body."},{"Start":"20:45.430 ","End":"20:55.105","Text":"The cone and the half sphere as point masses where they\u0027re located at each one\u0027s Z_CM."},{"Start":"20:55.105 ","End":"20:58.750","Text":"Now I have to find the center of mass between these 2 points."},{"Start":"20:58.750 ","End":"21:02.635","Text":"Now again, because we are symmetrical about the origin,"},{"Start":"21:02.635 ","End":"21:05.455","Text":"with regards to our x and our y."},{"Start":"21:05.455 ","End":"21:14.650","Text":"Again, we only have to find our Z_CM between the 2 bodies. Let\u0027s answer this."},{"Start":"21:14.650 ","End":"21:16.465","Text":"We\u0027re on question number 2."},{"Start":"21:16.465 ","End":"21:21.910","Text":"Our Z_CM, now because we only have 2 bodies before we did an integration,"},{"Start":"21:21.910 ","End":"21:27.565","Text":"because we had 1 whole body that was a rigid body, a large body."},{"Start":"21:27.565 ","End":"21:33.295","Text":"Now we only have 2 bodies which are each being considered as 2 point masses."},{"Start":"21:33.295 ","End":"21:40.134","Text":"We don\u0027t have to integrate, we just have to do our usual equation. What is that?"},{"Start":"21:40.134 ","End":"21:42.850","Text":"We\u0027re going to multiply mass 1,"},{"Start":"21:42.850 ","End":"21:45.444","Text":"which is the mass of the cone,"},{"Start":"21:45.444 ","End":"21:51.400","Text":"multiplied by its distance from the origin,"},{"Start":"21:51.400 ","End":"21:55.839","Text":"which is our Z_CM of the cone that we found in our previous question."},{"Start":"21:55.839 ","End":"21:59.900","Text":"That\u0027s multiplied by h divided by 4."},{"Start":"22:00.060 ","End":"22:03.355","Text":"Then the same for our M_2,"},{"Start":"22:03.355 ","End":"22:07.449","Text":"which is going to be the mass of the half sphere and multiply"},{"Start":"22:07.449 ","End":"22:11.994","Text":"it by its distance away from the origin."},{"Start":"22:11.994 ","End":"22:17.989","Text":"Its distance away is given to us negative 3 divided by 8."},{"Start":"22:17.989 ","End":"22:23.879","Text":"Notice the negative sign because it\u0027s below 0."},{"Start":"22:24.190 ","End":"22:32.210","Text":"That\u0027s that and then divide it by the total mass of the system so m_1 plus m_2."},{"Start":"22:32.210 ","End":"22:36.095","Text":"We\u0027ve taken some origin to point masses"},{"Start":"22:36.095 ","End":"22:40.474","Text":"a certain distance away and we\u0027re just doing this equation."},{"Start":"22:40.474 ","End":"22:42.934","Text":"Now, obviously in my question,"},{"Start":"22:42.934 ","End":"22:45.440","Text":"I wasn\u0027t given my values for m_1 or m_2,"},{"Start":"22:45.440 ","End":"22:47.060","Text":"so I have to work them out."},{"Start":"22:47.060 ","End":"22:50.930","Text":"My m_1 is going to be equal to Rho,"},{"Start":"22:50.930 ","End":"22:55.100","Text":"my density multiplied by my volume."},{"Start":"22:55.100 ","End":"22:57.965","Text":"My volume when dealing with a cone,"},{"Start":"22:57.965 ","End":"23:02.585","Text":"as we saw before, it\u0027s the same volume as a cylinder divided by 3."},{"Start":"23:02.585 ","End":"23:06.635","Text":"It\u0027s going to be Rho multiplied by Pi times the radius squared."},{"Start":"23:06.635 ","End":"23:13.400","Text":"Here our radius is a multiplied by the height and then divided by 3."},{"Start":"23:13.400 ","End":"23:17.614","Text":"Similarly, a second mass is also going to be"},{"Start":"23:17.614 ","End":"23:21.965","Text":"a Rho multiplied by the volume v_2 of our half sphere"},{"Start":"23:21.965 ","End":"23:25.924","Text":"so it\u0027s going to be Rho and then the volume of a full sphere"},{"Start":"23:25.924 ","End":"23:30.170","Text":"is 3 divided by 4 multiplied by Pi r^3,"},{"Start":"23:30.170 ","End":"23:35.499","Text":"cubed because it\u0027s volume and here our radius again is a,"},{"Start":"23:35.499 ","End":"23:40.719","Text":"so a^3 and this is the volume for an entire whole sphere but we want half a sphere,"},{"Start":"23:40.719 ","End":"23:42.820","Text":"so we multiply it by 1/2."},{"Start":"23:42.820 ","End":"23:51.439","Text":"Once you plug in your values for your m_1 and your m_2 into here, here and here,"},{"Start":"23:51.439 ","End":"23:59.495","Text":"then what you should get for your z center of mass of the 2 bodies so it should be"},{"Start":"23:59.495 ","End":"24:05.240","Text":"8h^2 minus 27 divided by"},{"Start":"24:05.240 ","End":"24:15.425","Text":"2 a^2 and all of this divided by 8h plus 9a."},{"Start":"24:15.425 ","End":"24:18.424","Text":"It\u0027s a bit of an ugly expression."},{"Start":"24:18.424 ","End":"24:22.504","Text":"You can just do the algebra on your own."},{"Start":"24:22.504 ","End":"24:26.450","Text":"This is the answer to question number 2."},{"Start":"24:26.450 ","End":"24:32.389","Text":"Question number 3 is telling us that the body is tilted by an angle of Theta"},{"Start":"24:32.389 ","End":"24:39.950","Text":"relative to the perpendicular axis so relative to this axis, z over here."},{"Start":"24:39.950 ","End":"24:45.050","Text":"What is the potential energy as a function of this angle?"},{"Start":"24:45.050 ","End":"24:49.515","Text":"This question is relating to this diagram."},{"Start":"24:49.515 ","End":"24:52.975","Text":"As we can see here on this diagram,"},{"Start":"24:52.975 ","End":"24:55.345","Text":"this over here is our angle Theta."},{"Start":"24:55.345 ","End":"24:57.220","Text":"They\u0027ve showed us where it was."},{"Start":"24:57.220 ","End":"25:02.604","Text":"We can see that this line before it was going straight vertically upwards,"},{"Start":"25:02.604 ","End":"25:06.039","Text":"90 degrees to this surface over here."},{"Start":"25:06.039 ","End":"25:09.060","Text":"However now it\u0027s been tilted at some angle"},{"Start":"25:09.060 ","End":"25:13.430","Text":"and they\u0027re helping us by saying that this line over here is our a,"},{"Start":"25:13.430 ","End":"25:15.630","Text":"this is our radius."},{"Start":"25:15.850 ","End":"25:22.829","Text":"Now what we have to do is to find the potential energy as a function of this angle Theta."},{"Start":"25:23.260 ","End":"25:30.334","Text":"What we\u0027re trying to find is our u as a function of Theta over here."},{"Start":"25:30.334 ","End":"25:33.379","Text":"Now we know that our potential energy is always going to be equal"},{"Start":"25:33.379 ","End":"25:37.619","Text":"to our mass times gravity times our height."},{"Start":"25:37.810 ","End":"25:40.175","Text":"When we\u0027re speaking about heights,"},{"Start":"25:40.175 ","End":"25:42.934","Text":"when we\u0027re dealing with a rigid body,"},{"Start":"25:42.934 ","End":"25:44.314","Text":"a large body like this,"},{"Start":"25:44.314 ","End":"25:47.704","Text":"then we\u0027re speaking about the height of our center of mass."},{"Start":"25:47.704 ","End":"25:50.239","Text":"Our center of mass has been marked over here,"},{"Start":"25:50.239 ","End":"25:55.309","Text":"whereas that is where this dot is and we\u0027re trying to find what this height is,"},{"Start":"25:55.309 ","End":"25:58.579","Text":"this distance between this point and the flow."},{"Start":"25:58.579 ","End":"26:01.760","Text":"We have a few options of where we can mark our 0."},{"Start":"26:01.760 ","End":"26:07.009","Text":"We can say that our 0 or h is equal to 0 is on this surface over here."},{"Start":"26:07.009 ","End":"26:09.679","Text":"Or we can say that our h is equal to 0 is over here,"},{"Start":"26:09.679 ","End":"26:15.469","Text":"which might make it a little bit easier to calculate. We\u0027ll take a look."},{"Start":"26:15.469 ","End":"26:17.464","Text":"Before we start with that,"},{"Start":"26:17.464 ","End":"26:22.700","Text":"something that\u0027s very important to notice is that our distance from this point over here,"},{"Start":"26:22.700 ","End":"26:25.070","Text":"so also known as our origin,"},{"Start":"26:25.070 ","End":"26:27.260","Text":"the distance from our origin until"},{"Start":"26:27.260 ","End":"26:30.050","Text":"our surface over here or the floor, whatever you want to call it,"},{"Start":"26:30.050 ","End":"26:36.364","Text":"is always going to be the length of our radius so here a."},{"Start":"26:36.364 ","End":"26:39.230","Text":"Even if we tilt it right until this,"},{"Start":"26:39.230 ","End":"26:41.315","Text":"the radius is still going to be a,"},{"Start":"26:41.315 ","End":"26:45.500","Text":"so that this point over here is on the floor and"},{"Start":"26:45.500 ","End":"26:49.819","Text":"also if we stand it back up to its original position at the beginning of the question,"},{"Start":"26:49.819 ","End":"26:53.209","Text":"then we\u0027ll see that the radius is still going to"},{"Start":"26:53.209 ","End":"26:56.464","Text":"be a and the distance from our origin to the floor,"},{"Start":"26:56.464 ","End":"26:58.859","Text":"will still be a."},{"Start":"26:59.530 ","End":"27:06.994","Text":"Let\u0027s begin. Let\u0027s start by saying that our h is equal to 0."},{"Start":"27:06.994 ","End":"27:10.909","Text":"Here\u0027s our origin that our line for our h is equal to 0,"},{"Start":"27:10.909 ","End":"27:14.070","Text":"is going to be right in the center."},{"Start":"27:14.070 ","End":"27:16.555","Text":"Imagine that this is a straight line."},{"Start":"27:16.555 ","End":"27:18.984","Text":"This is where h is equal to 0."},{"Start":"27:18.984 ","End":"27:21.879","Text":"Now, of course, we could have said that our h is equal to 0"},{"Start":"27:21.879 ","End":"27:25.270","Text":"is on our surface and then we would have just had to add an a."},{"Start":"27:25.270 ","End":"27:30.259","Text":"Because as we just said before, our radius is a."},{"Start":"27:30.259 ","End":"27:33.875","Text":"The distance between the origin and the floor is always going to be a."},{"Start":"27:33.875 ","End":"27:38.105","Text":"However, it\u0027s important to understand why we\u0027re doing it this way,"},{"Start":"27:38.105 ","End":"27:42.484","Text":"is that anytime we\u0027re dealing with some circular shape,"},{"Start":"27:42.484 ","End":"27:46.339","Text":"a sphere or a half sphere that\u0027s rolling, for instance,"},{"Start":"27:46.339 ","End":"27:52.340","Text":"then the origin or its center of mass gives us a uniform sphere."},{"Start":"27:52.340 ","End":"27:55.985","Text":"Then it\u0027s always going to be moving in a straight line."},{"Start":"27:55.985 ","End":"27:59.570","Text":"Its height is never ever going to change."},{"Start":"27:59.570 ","End":"28:04.799","Text":"It\u0027s always going to be traveling like this because it\u0027s distance,"},{"Start":"28:06.370 ","End":"28:11.119","Text":"sorry, from the ground is always going to be that of the radius."},{"Start":"28:11.119 ","End":"28:15.980","Text":"Because of that, it\u0027s easier to choose the line of that point."},{"Start":"28:15.980 ","End":"28:21.364","Text":"The line going through that point to be our height is equal to 0,"},{"Start":"28:21.364 ","End":"28:23.180","Text":"h is equal to 0 there."},{"Start":"28:23.180 ","End":"28:25.684","Text":"Because we know that that height is never,"},{"Start":"28:25.684 ","End":"28:30.935","Text":"ever changing and it\u0027s a lot easier to work out. Let\u0027s begin."},{"Start":"28:30.935 ","End":"28:36.214","Text":"Our h_CM is going to be this height over here,"},{"Start":"28:36.214 ","End":"28:42.454","Text":"from here to here is going to be our h center of mass."},{"Start":"28:42.454 ","End":"28:49.385","Text":"We know that from here to here is our Z_CM."},{"Start":"28:49.385 ","End":"28:53.479","Text":"Because before we said that we were always trying to"},{"Start":"28:53.479 ","End":"28:59.089","Text":"find our center of mass of the entire body relative to the connecting plane."},{"Start":"28:59.089 ","End":"29:01.534","Text":"From our previous question,"},{"Start":"29:01.534 ","End":"29:08.120","Text":"we found that our z center of mass is going to be 8h^2 minus"},{"Start":"29:08.120 ","End":"29:16.295","Text":"27 divided by 2 a^2 divided by 8h plus 9a."},{"Start":"29:16.295 ","End":"29:21.830","Text":"Now we have to translate our Z_CM in terms of our h_CM."},{"Start":"29:21.830 ","End":"29:25.003","Text":"If we can see this angle is our Theta,"},{"Start":"29:25.003 ","End":"29:31.444","Text":"our Z_CM is our hypotenuse and our h_CM is our adjacent angle."},{"Start":"29:31.444 ","End":"29:35.524","Text":"What trigonometric identity or equation"},{"Start":"29:35.524 ","End":"29:41.075","Text":"links our hypotenuse and adjacent angles? That\u0027s cosine."},{"Start":"29:41.075 ","End":"29:47.509","Text":"As we know, cosine of our angle here Theta is going to"},{"Start":"29:47.509 ","End":"29:53.765","Text":"be equal to our adjacent angle divided by our hypotenuse."},{"Start":"29:53.765 ","End":"29:58.699","Text":"Here our adjacent angle is going to be our h center of"},{"Start":"29:58.699 ","End":"30:04.415","Text":"mass and our hypotenuse is our z center of mass."},{"Start":"30:04.415 ","End":"30:08.660","Text":"Therefore, we can isolate out our h_CM,"},{"Start":"30:08.660 ","End":"30:11.390","Text":"because that\u0027s what we need over here and we\u0027ll get"},{"Start":"30:11.390 ","End":"30:15.109","Text":"that our h center of mass is going to be equal to"},{"Start":"30:15.109 ","End":"30:22.610","Text":"our z center of mass multiplied by our cosine of Theta."},{"Start":"30:22.610 ","End":"30:25.849","Text":"Then of course, the mass over here is our total mass"},{"Start":"30:25.849 ","End":"30:28.669","Text":"because we\u0027re speaking about the whole system,"},{"Start":"30:28.669 ","End":"30:30.949","Text":"both the cone and the half sphere."},{"Start":"30:30.949 ","End":"30:34.805","Text":"Now we can plug everything in to our equation."},{"Start":"30:34.805 ","End":"30:38.209","Text":"Again, that our potential energy as a function of"},{"Start":"30:38.209 ","End":"30:41.780","Text":"Theta is going to be equal to our total mass,"},{"Start":"30:41.780 ","End":"30:45.515","Text":"which is going to be m_1 plus our"},{"Start":"30:45.515 ","End":"30:50.255","Text":"m_2 and we worked out what our m_1 and m_2 were in the previous question."},{"Start":"30:50.255 ","End":"30:55.429","Text":"You can go back to that section of the video if you want to completely solve this"},{"Start":"30:55.429 ","End":"31:01.280","Text":"fully multiplied by g and then multiplied by our h center of mass,"},{"Start":"31:01.280 ","End":"31:03.679","Text":"which is going to be our z center of mass."},{"Start":"31:03.679 ","End":"31:05.659","Text":"I won\u0027t write it out because it\u0027s a bit long,"},{"Start":"31:05.659 ","End":"31:11.029","Text":"but it\u0027s this over here and then multiplied by our cosine of Theta."},{"Start":"31:11.029 ","End":"31:16.160","Text":"As we can see, we got our potential energy as a function of Theta over here."},{"Start":"31:16.160 ","End":"31:19.459","Text":"This is our final answer when we\u0027re taking our h is equal"},{"Start":"31:19.459 ","End":"31:22.625","Text":"to 0 to be right here in the origin."},{"Start":"31:22.625 ","End":"31:27.995","Text":"However, if we were going to take our floor as being where h is equal to 0,"},{"Start":"31:27.995 ","End":"31:31.834","Text":"then we would just have to add over here our constant of a."},{"Start":"31:31.834 ","End":"31:35.090","Text":"It\u0027s shifted along. But anyway,"},{"Start":"31:35.090 ","End":"31:38.360","Text":"it doesn\u0027t really matter because we know that our energy is"},{"Start":"31:38.360 ","End":"31:42.455","Text":"only useful to us up until a constant."},{"Start":"31:42.455 ","End":"31:45.095","Text":"Anyway, that constant is irrelevant."},{"Start":"31:45.095 ","End":"31:47.495","Text":"This is our answer for question 3."},{"Start":"31:47.495 ","End":"31:50.734","Text":"This is not the end of the video."},{"Start":"31:50.734 ","End":"31:54.380","Text":"I tricked you by making you think that there were only 3 questions,"},{"Start":"31:54.380 ","End":"31:55.925","Text":"but there\u0027s actually 4."},{"Start":"31:55.925 ","End":"31:58.455","Text":"Let\u0027s take a look at question Number 4."},{"Start":"31:58.455 ","End":"32:01.300","Text":"Question 4 and our last question."},{"Start":"32:01.300 ","End":"32:05.005","Text":"We\u0027re being asked under which geometric circumstances,"},{"Start":"32:05.005 ","End":"32:08.739","Text":"so that means when using what\u0027s given to us in the question."},{"Start":"32:08.739 ","End":"32:12.060","Text":"That\u0027s our h value or a value and our Alpha."},{"Start":"32:12.060 ","End":"32:15.334","Text":"Will the system be the following?"},{"Start":"32:15.334 ","End":"32:18.700","Text":"Our first is a neutral equilibrium."},{"Start":"32:18.700 ","End":"32:21.834","Text":"What does that mean? That means that our potential energy,"},{"Start":"32:21.834 ","End":"32:25.180","Text":"E_p, potential energy is a constant."},{"Start":"32:25.180 ","End":"32:28.435","Text":"We have over here our equation for potential energy,"},{"Start":"32:28.435 ","End":"32:31.355","Text":"and we want this to be something that is constant."},{"Start":"32:31.355 ","End":"32:34.835","Text":"Also, if you don\u0027t know what neutral equilibrium is,"},{"Start":"32:34.835 ","End":"32:39.219","Text":"in basic terms that our energy doesn\u0027t vary."},{"Start":"32:39.219 ","End":"32:45.174","Text":"The potential energy remains the same no matter what you do to the system and"},{"Start":"32:45.174 ","End":"32:51.004","Text":"this is also sometimes referred to as a state of indifference."},{"Start":"32:51.004 ","End":"32:55.045","Text":"That\u0027s the basic principle of neutral equilibrium."},{"Start":"32:55.045 ","End":"32:58.495","Text":"Let\u0027s answer. We\u0027re on 4."},{"Start":"32:58.495 ","End":"33:01.795","Text":"We need our potential energy to be constant."},{"Start":"33:01.795 ","End":"33:06.040","Text":"Let\u0027s take a look. Now, our masses are always going to be constant."},{"Start":"33:06.040 ","End":"33:07.705","Text":"Our gravity is constant,"},{"Start":"33:07.705 ","End":"33:09.610","Text":"our Z_CM is constant,"},{"Start":"33:09.610 ","End":"33:12.505","Text":"and then our cosine of Theta is always changing"},{"Start":"33:12.505 ","End":"33:15.909","Text":"depending on our angle of Theta. It\u0027s not constant."},{"Start":"33:15.909 ","End":"33:18.519","Text":"We can change this our cosine Theta is"},{"Start":"33:18.519 ","End":"33:22.540","Text":"changing because our angle is the only way that we can do this,"},{"Start":"33:22.540 ","End":"33:27.385","Text":"because we can set our masses to be equal to 0 or our gravity to be equal to 0."},{"Start":"33:27.385 ","End":"33:33.685","Text":"The only way that we can get that our potential energy will be equal to 0."},{"Start":"33:33.685 ","End":"33:38.380","Text":"Some constant, if our Z_CM is equal to 0."},{"Start":"33:38.380 ","End":"33:43.615","Text":"Again, neutral equilibrium is when our potential energy is equal to constant."},{"Start":"33:43.615 ","End":"33:45.130","Text":"Now we said that our constants,"},{"Start":"33:45.130 ","End":"33:49.854","Text":"it doesn\u0027t matter what it is when we\u0027re looking at any energy,"},{"Start":"33:49.854 ","End":"33:51.819","Text":"the constant is less important."},{"Start":"33:51.819 ","End":"33:55.075","Text":"We can just say any constants such as 0."},{"Start":"33:55.075 ","End":"34:00.129","Text":"We want our potential energy to be equal to 0."},{"Start":"34:00.129 ","End":"34:03.309","Text":"Now our masses and our gravity can never be equal to"},{"Start":"34:03.309 ","End":"34:06.706","Text":"0 and our cosine Theta is always changing,"},{"Start":"34:06.706 ","End":"34:08.185","Text":"and we can\u0027t do anything about that."},{"Start":"34:08.185 ","End":"34:12.025","Text":"The only thing that we can set to 0 is our Z_CM."},{"Start":"34:12.025 ","End":"34:17.320","Text":"That means that our Z_CM must be equal to 0."},{"Start":"34:17.320 ","End":"34:19.330","Text":"Let\u0027s take a look at our Z_CM."},{"Start":"34:19.330 ","End":"34:21.745","Text":"We have this fraction over here."},{"Start":"34:21.745 ","End":"34:24.684","Text":"Now in order to have something equal to 0,"},{"Start":"34:24.684 ","End":"34:27.789","Text":"we have to say that the numerator is equal to"},{"Start":"34:27.789 ","End":"34:30.705","Text":"0 when we have a fraction and we set it to 0,"},{"Start":"34:30.705 ","End":"34:34.434","Text":"then the easiest way to work it out is that the numerator is equal to 0."},{"Start":"34:34.434 ","End":"34:40.074","Text":"Don\u0027t try and rearrange the denominator to be equal to 0 because any number divided by 0,"},{"Start":"34:40.074 ","End":"34:45.805","Text":"even 0 divided by 0 is something that you\u0027ll learn in calculus lessons."},{"Start":"34:45.805 ","End":"34:48.325","Text":"You can say that it\u0027s equal to 0."},{"Start":"34:48.325 ","End":"34:50.620","Text":"Always stick to the numerator."},{"Start":"34:50.620 ","End":"35:00.850","Text":"That means that we have to have our 8h^2 minus 27 divided by 2 a^2 to equal 0."},{"Start":"35:00.850 ","End":"35:04.929","Text":"Rearranging that we\u0027ll get that 8h^2 is equal to"},{"Start":"35:04.929 ","End":"35:12.024","Text":"27 over 2 a^2 and divide both sides by 8 and square root both sides."},{"Start":"35:12.024 ","End":"35:20.450","Text":"We\u0027ll get that our h is equal to the square root of 27 divided by 16 a^2,"},{"Start":"35:20.450 ","End":"35:30.315","Text":"which is simply equal to the square root of 27 divided by 4 a^2."},{"Start":"35:30.315 ","End":"35:33.290","Text":"This answers our first question."},{"Start":"35:33.290 ","End":"35:41.219","Text":"Because we\u0027re using our geometric values of h and a in order to define this."},{"Start":"35:41.219 ","End":"35:44.010","Text":"Okay, now onto ii."},{"Start":"35:44.010 ","End":"35:50.185","Text":"On ii, now we\u0027re trying to find where the system will be at a stable equilibrium,"},{"Start":"35:50.185 ","End":"35:52.479","Text":"which allows small oscillations."},{"Start":"35:52.479 ","End":"35:55.705","Text":"What does stable equilibrium mean?"},{"Start":"35:55.705 ","End":"35:57.730","Text":"It means that our U,"},{"Start":"35:57.730 ","End":"35:59.065","Text":"our potential energy,"},{"Start":"35:59.065 ","End":"36:01.194","Text":"is at a local minimum."},{"Start":"36:01.194 ","End":"36:04.360","Text":"Right now, just understand it at this later."},{"Start":"36:04.360 ","End":"36:08.275","Text":"Maybe we\u0027ll speak about it a little bit more in terms of intuition."},{"Start":"36:08.275 ","End":"36:13.615","Text":"But right now, stable equilibrium is when our potential energy is at a local minimum."},{"Start":"36:13.615 ","End":"36:15.400","Text":"Here\u0027s specifically potential energy."},{"Start":"36:15.400 ","End":"36:16.494","Text":"It can be something else as well,"},{"Start":"36:16.494 ","End":"36:18.160","Text":"but a local minimum."},{"Start":"36:18.160 ","End":"36:24.385","Text":"How do we check when a system is at a local minimum?"},{"Start":"36:24.385 ","End":"36:27.280","Text":"What do we have to do is we take our equation"},{"Start":"36:27.280 ","End":"36:30.804","Text":"for our potential energy and we take its derivative."},{"Start":"36:30.804 ","End":"36:34.375","Text":"Let\u0027s write this out. I\u0027m using this equation over here."},{"Start":"36:34.375 ","End":"36:36.520","Text":"My potential energy tag,"},{"Start":"36:36.520 ","End":"36:43.195","Text":"so U tag as a function of Theta still is going to be equal to?"},{"Start":"36:43.195 ","End":"36:47.350","Text":"Let\u0027s take the derivative of cosine of Theta is what we\u0027re deriving."},{"Start":"36:47.350 ","End":"36:49.510","Text":"It\u0027s going to be negative sine of Theta."},{"Start":"36:49.510 ","End":"36:56.065","Text":"We\u0027re going to have negative of our m_1 plus m_2 multiplied by g,"},{"Start":"36:56.065 ","End":"36:58.345","Text":"multiplied by Z_CM,"},{"Start":"36:58.345 ","End":"37:02.240","Text":"multiplied by my sine of Theta."},{"Start":"37:02.250 ","End":"37:08.005","Text":"Now what we have to do is we have to set this equal to 0."},{"Start":"37:08.005 ","End":"37:09.520","Text":"You always, always,"},{"Start":"37:09.520 ","End":"37:12.265","Text":"always have to do that. Let\u0027s underline it."},{"Start":"37:12.265 ","End":"37:16.209","Text":"We take our first derivative in order to find a local minimum,"},{"Start":"37:16.209 ","End":"37:18.009","Text":"we take our first derivative."},{"Start":"37:18.009 ","End":"37:19.824","Text":"We take that derivative,"},{"Start":"37:19.824 ","End":"37:22.795","Text":"and then we set that equal to 0."},{"Start":"37:22.795 ","End":"37:25.794","Text":"Now what do we have to do is we have to find"},{"Start":"37:25.794 ","End":"37:28.945","Text":"how we can get this equation over here to equal to 0."},{"Start":"37:28.945 ","End":"37:32.005","Text":"Obviously, that\u0027s through our sine Theta."},{"Start":"37:32.005 ","End":"37:36.115","Text":"Which value for Theta will give sine of Theta is equal to 0?"},{"Start":"37:36.115 ","End":"37:38.830","Text":"If our Theta is equal to 0,"},{"Start":"37:38.830 ","End":"37:41.320","Text":"because sine of 0 is equal to 0."},{"Start":"37:41.320 ","End":"37:43.645","Text":"Of course, we can also use Pi."},{"Start":"37:43.645 ","End":"37:48.189","Text":"However, Pi is less interesting to us because that means that our body is just,"},{"Start":"37:48.189 ","End":"37:50.305","Text":"or our a system is upside down."},{"Start":"37:50.305 ","End":"37:53.935","Text":"We\u0027re not going to look at that 0, nice and easy."},{"Start":"37:53.935 ","End":"37:57.130","Text":"I\u0027m going to underline this."},{"Start":"37:57.130 ","End":"38:04.821","Text":"We found which value our variable has to be equal to in order to have a 0 over here,"},{"Start":"38:04.821 ","End":"38:07.119","Text":"in order for this equation to be equal to 0."},{"Start":"38:07.119 ","End":"38:11.785","Text":"Then what we do is we take the second derivative of this equation."},{"Start":"38:11.785 ","End":"38:14.695","Text":"U tag of Theta."},{"Start":"38:14.695 ","End":"38:17.650","Text":"Now we\u0027re taking the derivative of sine Theta,"},{"Start":"38:17.650 ","End":"38:19.390","Text":"which is just going to be cosine Theta."},{"Start":"38:19.390 ","End":"38:28.449","Text":"It\u0027s going to be equal to negative m_1 plus m_2 multiplied by g multiplied by Z_CM,"},{"Start":"38:28.449 ","End":"38:31.960","Text":"multiplied by cosine of Theta."},{"Start":"38:31.960 ","End":"38:35.709","Text":"Of course, we have to set in this Theta over here."},{"Start":"38:35.709 ","End":"38:39.099","Text":"Cosine of Theta when Theta is equal to 0."},{"Start":"38:39.099 ","End":"38:42.504","Text":"Now what is this equal to?"},{"Start":"38:42.504 ","End":"38:44.905","Text":"When Theta is equal to 0,"},{"Start":"38:44.905 ","End":"38:46.990","Text":"our cosine of Theta is equal to 1,"},{"Start":"38:46.990 ","End":"38:50.514","Text":"which means that we have a negative expression over here."},{"Start":"38:50.514 ","End":"39:00.130","Text":"That\u0027s going to be equal to negative m_1 plus m_2 g multiplied by Z_CM."},{"Start":"39:00.130 ","End":"39:03.774","Text":"Now, in order for us to have a local minimum,"},{"Start":"39:03.774 ","End":"39:09.310","Text":"that means that our second derivative has to be bigger than 0."},{"Start":"39:09.310 ","End":"39:11.815","Text":"This is very important in order to have"},{"Start":"39:11.815 ","End":"39:16.795","Text":"stable equilibrium our potential energy has to be at a local minimum."},{"Start":"39:16.795 ","End":"39:19.134","Text":"In order to know if it\u0027s a local minimum,"},{"Start":"39:19.134 ","End":"39:25.854","Text":"our second derivative of our potential energy has to be bigger than 0."},{"Start":"39:25.854 ","End":"39:30.534","Text":"I know it\u0027s confusing minimum and here it\u0027s bigger than 0. Remember it."},{"Start":"39:30.534 ","End":"39:33.955","Text":"In order for this to happen,"},{"Start":"39:33.955 ","End":"39:37.389","Text":"so my m_1 and my m_2 are constants, nothing here is happening."},{"Start":"39:37.389 ","End":"39:40.209","Text":"My g also, I can\u0027t change."},{"Start":"39:40.209 ","End":"39:44.620","Text":"The only thing that I can change over here according"},{"Start":"39:44.620 ","End":"39:48.775","Text":"to my geometric conditions is my Z_CM."},{"Start":"39:48.775 ","End":"39:52.900","Text":"Because all I can change the value for is my h,"},{"Start":"39:52.900 ","End":"39:56.455","Text":"my a, or my Alpha, that\u0027s my Z_CM."},{"Start":"39:56.455 ","End":"39:59.394","Text":"Now because I have a negative sign over here,"},{"Start":"39:59.394 ","End":"40:05.830","Text":"I have to make sure that my z center of mass is a negative expression so that"},{"Start":"40:05.830 ","End":"40:09.700","Text":"the negative value for this and this negative sign will cancel and"},{"Start":"40:09.700 ","End":"40:14.529","Text":"then this whole expression will be positive, will be bigger than 0."},{"Start":"40:14.529 ","End":"40:22.840","Text":"What I need is that my z center of mass must be then smaller than 0."},{"Start":"40:22.840 ","End":"40:24.759","Text":"It has to be a negative."},{"Start":"40:24.759 ","End":"40:28.645","Text":"Let\u0027s write out our value for our Z_CM."},{"Start":"40:28.645 ","End":"40:30.624","Text":"That means that we have"},{"Start":"40:30.624 ","End":"40:35.335","Text":"8h^2 minus 27 over 2 a^2"},{"Start":"40:35.335 ","End":"40:42.520","Text":"divided by 8h plus 9a has to be smaller than 0."},{"Start":"40:42.520 ","End":"40:46.359","Text":"Again, using our numerator,"},{"Start":"40:46.359 ","End":"40:54.860","Text":"so we\u0027ll get that our 8h^2 has to be smaller than our 27 over 2 a^2."},{"Start":"40:55.980 ","End":"41:02.979","Text":"Then that will result in our h having to"},{"Start":"41:02.979 ","End":"41:10.220","Text":"be smaller than the square root of 27 divided by 4a."},{"Start":"41:10.290 ","End":"41:15.115","Text":"Notice that for first option,"},{"Start":"41:15.115 ","End":"41:17.139","Text":"we had to have for neutral equilibrium,"},{"Start":"41:17.139 ","End":"41:19.810","Text":"we had to have that our h is equal to this value."},{"Start":"41:19.810 ","End":"41:24.415","Text":"For stable equilibrium, we have to have that our h is smaller to that value."},{"Start":"41:24.415 ","End":"41:27.490","Text":"Now let\u0027s go to our third question."},{"Start":"41:27.490 ","End":"41:29.365","Text":"This is the answer to ii."},{"Start":"41:29.365 ","End":"41:34.090","Text":"ii is asking for unstable equilibrium."},{"Start":"41:34.090 ","End":"41:36.714","Text":"We solve for neutral equilibrium."},{"Start":"41:36.714 ","End":"41:39.534","Text":"We had to have our h is equal to this number."},{"Start":"41:39.534 ","End":"41:44.830","Text":"For stable we had our h is smaller and when U as at a local minimum."},{"Start":"41:44.830 ","End":"41:51.025","Text":"Now for unstable, we can say that our unstable equilibrium is when are U,"},{"Start":"41:51.025 ","End":"41:54.025","Text":"our potential energy is at a local maximum."},{"Start":"41:54.025 ","End":"41:57.894","Text":"We can already see that our h is going to be bigger than this value over here."},{"Start":"41:57.894 ","End":"42:03.940","Text":"We\u0027re going to do this in the exact same way. Let\u0027s take a look."},{"Start":"42:03.940 ","End":"42:06.609","Text":"You can already tell that our h is going to be bigger than this value."},{"Start":"42:06.609 ","End":"42:10.060","Text":"However, if you just want to go over the working out again and see how I do it,"},{"Start":"42:10.060 ","End":"42:12.520","Text":"then please carry on watching."},{"Start":"42:12.520 ","End":"42:16.720","Text":"The first thing that we\u0027re going to do is we\u0027re going to take our U,"},{"Start":"42:16.720 ","End":"42:19.434","Text":"and we\u0027re going to find its first derivative."},{"Start":"42:19.434 ","End":"42:23.784","Text":"Just like before, are U tag of Theta is going to be equal to"},{"Start":"42:23.784 ","End":"42:29.249","Text":"negative m_1 plus m_2 multiplied by g,"},{"Start":"42:29.249 ","End":"42:33.630","Text":"multiplied by our Z_CM multiplied by instead of cosine Theta,"},{"Start":"42:33.630 ","End":"42:36.630","Text":"it\u0027s going to be sine Theta because we took the derivative."},{"Start":"42:36.630 ","End":"42:41.115","Text":"Then we say that we set it to 0."},{"Start":"42:41.115 ","End":"42:45.510","Text":"Once again, how can we have that this expression is equal to 0?"},{"Start":"42:45.510 ","End":"42:49.045","Text":"It\u0027s if our sine Theta is equal to 0,"},{"Start":"42:49.045 ","End":"42:54.459","Text":"which means that our Theta has to be equal to 0. Up until here."},{"Start":"42:54.459 ","End":"42:57.654","Text":"I\u0027ll just underline this."},{"Start":"42:57.654 ","End":"43:01.180","Text":"Now, we\u0027re going to take our second derivative."},{"Start":"43:01.180 ","End":"43:03.820","Text":"Again, this is important."},{"Start":"43:03.820 ","End":"43:08.814","Text":"Now, our second derivative is going to be equal to"},{"Start":"43:08.814 ","End":"43:16.375","Text":"negative of m_1 plus m_2 multiplied by g multiplied by Z_CM."},{"Start":"43:16.375 ","End":"43:20.170","Text":"Instead of sine Theta, it\u0027s going to be cosine of Theta,"},{"Start":"43:20.170 ","End":"43:22.554","Text":"where Theta is this over here."},{"Start":"43:22.554 ","End":"43:25.209","Text":"Theta is equal to 0 and"},{"Start":"43:25.209 ","End":"43:30.790","Text":"this expression for an unstable equilibrium our U is at a local maximum."},{"Start":"43:30.790 ","End":"43:33.429","Text":"Now, when it\u0027s a maximum,"},{"Start":"43:33.429 ","End":"43:36.924","Text":"then that means it has to be smaller than 0."},{"Start":"43:36.924 ","End":"43:40.914","Text":"Exactly the opposite before a local minimum was bigger than 0,"},{"Start":"43:40.914 ","End":"43:47.260","Text":"the second derivative, and here local maximum is smaller than 0 at the second derivative."},{"Start":"43:47.260 ","End":"43:49.899","Text":"Now again, we take a look."},{"Start":"43:49.899 ","End":"43:52.119","Text":"When our Theta is equal to 0,"},{"Start":"43:52.119 ","End":"43:54.100","Text":"so our cosine Theta is equal to 1."},{"Start":"43:54.100 ","End":"44:02.365","Text":"That means that our negative m_1 plus m_2 multiplied by g,"},{"Start":"44:02.365 ","End":"44:06.175","Text":"multiplied by our z center of mass."},{"Start":"44:06.175 ","End":"44:09.310","Text":"This has to be smaller than 0,"},{"Start":"44:09.310 ","End":"44:14.470","Text":"which means therefore that my Z_CM has to be bigger than 0."},{"Start":"44:14.470 ","End":"44:17.890","Text":"Because again, I can\u0027t play around with my masses or when my gravity,"},{"Start":"44:17.890 ","End":"44:21.640","Text":"the only thing I can play around with is something in terms of my h,"},{"Start":"44:21.640 ","End":"44:25.435","Text":"a or Alpha, which in this case is solely my Z_CM."},{"Start":"44:25.435 ","End":"44:28.480","Text":"In order for this expression to remain negative,"},{"Start":"44:28.480 ","End":"44:31.285","Text":"I need my Z_CM to be positive."},{"Start":"44:31.285 ","End":"44:36.894","Text":"Therefore, my z center of mass must be bigger than 0."},{"Start":"44:36.894 ","End":"44:44.589","Text":"Then we can go back to what our Z_CM is again taking the numerator."},{"Start":"44:44.589 ","End":"44:46.974","Text":"Let\u0027s write this out."},{"Start":"44:46.974 ","End":"44:55.975","Text":"That means that my 8h^2 minus 27 over 2 a^2 must be bigger than 0."},{"Start":"44:55.975 ","End":"45:00.804","Text":"Then playing around, I will get that my h has to be bigger"},{"Start":"45:00.804 ","End":"45:06.385","Text":"than the square root of 27 divided by 4a."},{"Start":"45:06.385 ","End":"45:13.405","Text":"That\u0027s our answer to our last question and our iii."},{"Start":"45:13.405 ","End":"45:17.845","Text":"Now, let\u0027s speak about states of equilibrium and to give"},{"Start":"45:17.845 ","End":"45:22.270","Text":"a little bit more intuition. Let\u0027s explain this."},{"Start":"45:22.270 ","End":"45:24.790","Text":"Let\u0027s imagine this as a mountain,"},{"Start":"45:24.790 ","End":"45:27.249","Text":"so we have our mountain,"},{"Start":"45:27.249 ","End":"45:31.374","Text":"some valley, and then another mountain over here."},{"Start":"45:31.374 ","End":"45:35.410","Text":"Let\u0027s talk about stable equilibrium."},{"Start":"45:35.410 ","End":"45:42.539","Text":"Stable equilibrium is, let\u0027s say we have a ball and we place a ball over here."},{"Start":"45:42.539 ","End":"45:46.990","Text":"Here will be stable equilibrium and why is that?"},{"Start":"45:46.990 ","End":"45:50.530","Text":"If I try and push my ball out,"},{"Start":"45:50.530 ","End":"45:55.450","Text":"so it\u0027s going to roll back down into this trough over here."},{"Start":"45:55.450 ","End":"45:57.939","Text":"That\u0027s what stable equilibrium means."},{"Start":"45:57.939 ","End":"46:00.820","Text":"If I\u0027m at a point where no matter which"},{"Start":"46:00.820 ","End":"46:04.164","Text":"direction I move it in this direction or this direction,"},{"Start":"46:04.164 ","End":"46:08.770","Text":"it\u0027s always going to fall back into its equilibrium point."},{"Start":"46:08.770 ","End":"46:11.859","Text":"It\u0027s always going to go back to its point of equilibrium."},{"Start":"46:11.859 ","End":"46:14.694","Text":"That\u0027s stable. I wrote it over here."},{"Start":"46:14.694 ","End":"46:20.980","Text":"Stable equilibrium is, when a ball is moved from a P.O.E is point of equilibrium,"},{"Start":"46:20.980 ","End":"46:23.454","Text":"it returns to a point of equilibrium."},{"Start":"46:23.454 ","End":"46:26.064","Text":"It will fall back to its trough."},{"Start":"46:26.064 ","End":"46:29.419","Text":"As we can see, it\u0027s a minimum point."},{"Start":"46:29.670 ","End":"46:34.284","Text":"That\u0027s also something, so min point,"},{"Start":"46:34.284 ","End":"46:37.000","Text":"because it\u0027s a local minimum."},{"Start":"46:37.000 ","End":"46:39.339","Text":"Now let\u0027s talk about unstable equilibrium."},{"Start":"46:39.339 ","End":"46:43.074","Text":"Unstable equilibrium, so U is at a local maximum."},{"Start":"46:43.074 ","End":"46:46.029","Text":"If we put a ball over here,"},{"Start":"46:46.029 ","End":"46:49.645","Text":"so we can see that it\u0027s still at equilibrium."},{"Start":"46:49.645 ","End":"46:53.275","Text":"Let\u0027s just again talk about what equilibrium is."},{"Start":"46:53.275 ","End":"47:02.035","Text":"Equilibrium is when the sum of all of our forces is equal to 0."},{"Start":"47:02.035 ","End":"47:03.594","Text":"At both these points,"},{"Start":"47:03.594 ","End":"47:06.010","Text":"the sum of all of that forces is equal to 0."},{"Start":"47:06.010 ","End":"47:07.749","Text":"They\u0027re both points of equilibrium."},{"Start":"47:07.749 ","End":"47:09.774","Text":"However, what\u0027s the difference over here?"},{"Start":"47:09.774 ","End":"47:14.020","Text":"Here, if I place it right on the point of equilibrium,"},{"Start":"47:14.020 ","End":"47:16.599","Text":"it\u0027s fine, the sum of all of the forces are equal to 0."},{"Start":"47:16.599 ","End":"47:19.644","Text":"However, if I move it slightly to either side,"},{"Start":"47:19.644 ","End":"47:22.045","Text":"the ball is just going to roll down the mountain."},{"Start":"47:22.045 ","End":"47:26.725","Text":"It\u0027s going to move further and further away from its point of equilibrium."},{"Start":"47:26.725 ","End":"47:28.704","Text":"That\u0027s unstable."},{"Start":"47:28.704 ","End":"47:30.010","Text":"Here we saw it stable."},{"Start":"47:30.010 ","End":"47:32.350","Text":"We can move it here, we can move it there and it\u0027s"},{"Start":"47:32.350 ","End":"47:35.050","Text":"just going to fall back to its point of equilibrium."},{"Start":"47:35.050 ","End":"47:40.270","Text":"Here we see it\u0027s at a point of equilibrium and if we move it any which way,"},{"Start":"47:40.270 ","End":"47:43.900","Text":"even slightly away from its exact point,"},{"Start":"47:43.900 ","End":"47:46.629","Text":"it\u0027s just going to roll away from that point of equilibrium."},{"Start":"47:46.629 ","End":"47:48.790","Text":"That is unstable."},{"Start":"47:48.790 ","End":"47:52.315","Text":"Our unstable point of equilibrium is when"},{"Start":"47:52.315 ","End":"47:56.515","Text":"our ball or our object or whatever it is is moved from its point of equilibrium,"},{"Start":"47:56.515 ","End":"47:59.665","Text":"it does not return to the point of equilibrium."},{"Start":"47:59.665 ","End":"48:02.230","Text":"It moves away. Again,"},{"Start":"48:02.230 ","End":"48:06.429","Text":"our equilibrium is defined as when the sum of the forces is equal to 0."},{"Start":"48:06.429 ","End":"48:09.940","Text":"Also, to give you a bit of intuition on our first derivative,"},{"Start":"48:09.940 ","End":"48:15.055","Text":"when we take our first derivative in general of functions and we set it equal to 0."},{"Start":"48:15.055 ","End":"48:17.484","Text":"That means, what is our first derivative?"},{"Start":"48:17.484 ","End":"48:19.465","Text":"It\u0027s like taking the gradient,"},{"Start":"48:19.465 ","End":"48:25.329","Text":"the spline of the slope. How steep it is."},{"Start":"48:25.329 ","End":"48:27.715","Text":"When we set our first derivative to 0,"},{"Start":"48:27.715 ","End":"48:29.455","Text":"that means that our slope,"},{"Start":"48:29.455 ","End":"48:31.359","Text":"its steepness has to be equal to 0,"},{"Start":"48:31.359 ","End":"48:33.144","Text":"which means a flat line."},{"Start":"48:33.144 ","End":"48:36.790","Text":"The flat line occurs at maximum points and"},{"Start":"48:36.790 ","End":"48:40.975","Text":"minimum points and that\u0027s some horse saddle shape,"},{"Start":"48:40.975 ","End":"48:44.470","Text":"something like this, where it\u0027s not a minimum or it\u0027s not a maximum,"},{"Start":"48:44.470 ","End":"48:50.755","Text":"it\u0027s a specific point where at 1 side it\u0027s going down and the other side it\u0027s going up."},{"Start":"48:50.755 ","End":"48:53.380","Text":"That\u0027s why we set it equal to 0."},{"Start":"48:53.380 ","End":"48:56.859","Text":"Then what we want to do is to find the max point."},{"Start":"48:56.859 ","End":"48:58.464","Text":"We take the second derivative."},{"Start":"48:58.464 ","End":"49:01.000","Text":"If our second derivative is smaller than 0,"},{"Start":"49:01.000 ","End":"49:03.414","Text":"then that says that we have a local maximum,"},{"Start":"49:03.414 ","End":"49:05.949","Text":"which means we have unstable equilibrium and if we take"},{"Start":"49:05.949 ","End":"49:10.285","Text":"the second derivative and our point is bigger than 0,"},{"Start":"49:10.285 ","End":"49:14.049","Text":"then that means that we have a local minimum and then we have stable equilibrium."},{"Start":"49:14.049 ","End":"49:17.079","Text":"Then when dealing with neutral equilibrium,"},{"Start":"49:17.079 ","End":"49:21.129","Text":"so that means that we have some flat plane over here."},{"Start":"49:21.129 ","End":"49:25.795","Text":"Then if I take my ball and I put it here,"},{"Start":"49:25.795 ","End":"49:31.450","Text":"so this is my ball and then if I move it this way or this way, it doesn\u0027t matter,"},{"Start":"49:31.450 ","End":"49:34.120","Text":"it still remains on the same height,"},{"Start":"49:34.120 ","End":"49:37.729","Text":"on the same line, and it doesn\u0027t change,"},{"Start":"49:37.729 ","End":"49:39.970","Text":"it\u0027s still at a point of equilibrium."},{"Start":"49:39.970 ","End":"49:42.730","Text":"It\u0027s always constantly at a point of equilibrium,"},{"Start":"49:42.730 ","End":"49:45.400","Text":"no matter which way I try and push it."},{"Start":"49:45.400 ","End":"49:51.400","Text":"Now let\u0027s briefly look at a cone and half sphere and see what this means."},{"Start":"49:51.400 ","End":"49:55.974","Text":"If I take my Z_CM to be bigger than 0,"},{"Start":"49:55.974 ","End":"49:59.245","Text":"when my z center of mass is over here."},{"Start":"49:59.245 ","End":"50:04.009","Text":"Let\u0027s put this in our straight version."},{"Start":"50:04.850 ","End":"50:08.520","Text":"Something like this. Excuse my drawing."},{"Start":"50:08.520 ","End":"50:13.335","Text":"My z center of mass is located above the 0 line. It\u0027s high up."},{"Start":"50:13.335 ","End":"50:15.569","Text":"Here when it\u0027s standing upright,"},{"Start":"50:15.569 ","End":"50:21.550","Text":"the sum of all of the forces is equal to 0. That\u0027s perfect."},{"Start":"50:21.550 ","End":"50:25.015","Text":"That\u0027s when my angle Theta is equal to 0."},{"Start":"50:25.015 ","End":"50:28.944","Text":"Now let\u0027s say that I increase my angle Theta,"},{"Start":"50:28.944 ","End":"50:32.020","Text":"so I go to this position."},{"Start":"50:32.020 ","End":"50:35.484","Text":"Because my z center of mass is above the 0 line,"},{"Start":"50:35.484 ","End":"50:38.755","Text":"so we can see if we draw in the moments."},{"Start":"50:38.755 ","End":"50:42.550","Text":"This will be my mg,"},{"Start":"50:42.550 ","End":"50:46.399","Text":"and this is our radius,"},{"Start":"50:46.800 ","End":"50:54.519","Text":"our r. We can see that our torque or moment is trying to turn,"},{"Start":"50:54.519 ","End":"50:57.984","Text":"it\u0027s pushing our shape in this direction."},{"Start":"50:57.984 ","End":"51:03.085","Text":"Eventually, it\u0027s going to fall over and get stuck that way."},{"Start":"51:03.085 ","End":"51:07.270","Text":"That means that we were at unstable equilibrium,"},{"Start":"51:07.270 ","End":"51:11.005","Text":"because we\u0027re moving away from our equilibrium point."},{"Start":"51:11.005 ","End":"51:13.180","Text":"On the other hand,"},{"Start":"51:13.180 ","End":"51:15.160","Text":"let\u0027s rub all of this out,"},{"Start":"51:15.160 ","End":"51:18.669","Text":"if my Z_CM is over here,"},{"Start":"51:18.669 ","End":"51:20.514","Text":"below the 0 line,"},{"Start":"51:20.514 ","End":"51:23.709","Text":"so it\u0027s at a minimum."},{"Start":"51:23.709 ","End":"51:30.340","Text":"If I again push this over to some Theta,"},{"Start":"51:30.340 ","End":"51:33.310","Text":"let\u0027s draw it like this."},{"Start":"51:33.310 ","End":"51:35.200","Text":"This is my CM."},{"Start":"51:35.200 ","End":"51:38.889","Text":"Here we\u0027ll have our mg pointing down,"},{"Start":"51:38.889 ","End":"51:43.120","Text":"and here is going to be our vector."},{"Start":"51:43.120 ","End":"51:49.555","Text":"Then we can see that our torque is trying to push it back to its point of equilibrium,"},{"Start":"51:49.555 ","End":"51:53.080","Text":"back to this upright position."},{"Start":"51:53.080 ","End":"51:56.830","Text":"That\u0027s going to be a stable equilibrium because our torque is"},{"Start":"51:56.830 ","End":"52:01.315","Text":"moving us back to the stable equilibrium state."},{"Start":"52:01.315 ","End":"52:08.509","Text":"I hope that explained it a little bit better and that is the end of this lesson."}],"ID":9474},{"Watched":false,"Name":"Masses On A Cone","Duration":"17m 2s","ChapterTopicVideoID":9205,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this question,"},{"Start":"00:01.800 ","End":"00:05.460","Text":"we\u0027re being told that we have some mass m_1,"},{"Start":"00:05.460 ","End":"00:11.580","Text":"and it\u0027s inside a cone with 1/2 angle of Alpha,"},{"Start":"00:11.580 ","End":"00:16.215","Text":"which rotates with a constant angular velocity of Omega."},{"Start":"00:16.215 ","End":"00:21.165","Text":"We\u0027re told that the mass can move up and down the cones wall with no friction."},{"Start":"00:21.165 ","End":"00:26.655","Text":"The mass isn\u0027t moving around in the cone,"},{"Start":"00:26.655 ","End":"00:30.160","Text":"but it\u0027s rather moving with the cone,"},{"Start":"00:30.160 ","End":"00:35.145","Text":"it\u0027s almost stuck to the wall and rotating with the cone."},{"Start":"00:35.145 ","End":"00:41.815","Text":"However, it can move up and down the cones wall, and there\u0027s no friction."},{"Start":"00:41.815 ","End":"00:44.525","Text":"What is the radius of rotation r,"},{"Start":"00:44.525 ","End":"00:47.615","Text":"such that mass m will be at equilibrium?"},{"Start":"00:47.615 ","End":"00:51.875","Text":"What does that mean if mass m_1 is at equilibrium?"},{"Start":"00:51.875 ","End":"00:56.205","Text":"It\u0027s that the mass won\u0027t move up or down the cone wall."},{"Start":"00:56.205 ","End":"01:01.505","Text":"What do we want to do is we want to find this radius,"},{"Start":"01:01.505 ","End":"01:04.985","Text":"where our m_1 is going to be located,"},{"Start":"01:04.985 ","End":"01:09.915","Text":"where it\u0027s just stationary and it\u0027s just rotating around with the cone,"},{"Start":"01:09.915 ","End":"01:11.910","Text":"and aside from that, it\u0027s not moving."},{"Start":"01:11.910 ","End":"01:15.110","Text":"How do I start answering this question?"},{"Start":"01:15.110 ","End":"01:16.790","Text":"It could use energy."},{"Start":"01:16.790 ","End":"01:20.165","Text":"However, because I\u0027m using a angular motion,"},{"Start":"01:20.165 ","End":"01:23.975","Text":"a circular motion, it\u0027s going to be very difficult for me to do that."},{"Start":"01:23.975 ","End":"01:27.005","Text":"Instead, what I can use is forces,"},{"Start":"01:27.005 ","End":"01:30.185","Text":"especially when I know that in circular motion,"},{"Start":"01:30.185 ","End":"01:32.660","Text":"I have forces going through the center."},{"Start":"01:32.660 ","End":"01:37.940","Text":"I can say that the sum of all of my forces in the radial direction,"},{"Start":"01:37.940 ","End":"01:42.800","Text":"it\u0027s going to be equal to my mass multiplied by my acceleration in the radial direction,"},{"Start":"01:42.800 ","End":"01:48.660","Text":"which is going to be equal to mv^2 divided by the radius,"},{"Start":"01:48.660 ","End":"01:51.635","Text":"and that because I\u0027m working with Omega,"},{"Start":"01:51.635 ","End":"01:56.180","Text":"is going to equal to m Omega^2 r. Then I\u0027m"},{"Start":"01:56.180 ","End":"02:01.265","Text":"also going to do the sum of my forces in the y-direction."},{"Start":"02:01.265 ","End":"02:05.480","Text":"Of course, because I want my mass m_1 to be"},{"Start":"02:05.480 ","End":"02:09.375","Text":"at equilibrium so that it won\u0027t move up or down the cone wall,"},{"Start":"02:09.375 ","End":"02:11.825","Text":"which means that 0 movement in the y direction,"},{"Start":"02:11.825 ","End":"02:14.500","Text":"which means 0 over here."},{"Start":"02:14.500 ","End":"02:17.805","Text":"Let\u0027s draw our free body diagram."},{"Start":"02:17.805 ","End":"02:22.225","Text":"We\u0027re going to have our m_1g going downwards,"},{"Start":"02:22.225 ","End":"02:25.600","Text":"and then perpendicular to the cone wall,"},{"Start":"02:25.600 ","End":"02:28.405","Text":"we\u0027re going to have our normal force."},{"Start":"02:28.405 ","End":"02:31.295","Text":"Now in the question we said that there\u0027s no friction."},{"Start":"02:31.295 ","End":"02:34.390","Text":"These are our only forces acting."},{"Start":"02:34.390 ","End":"02:39.155","Text":"Now, what we have to do is we have to add in our axes."},{"Start":"02:39.155 ","End":"02:47.660","Text":"We can say that this is our y-axis and that this is our x-axis."},{"Start":"02:47.660 ","End":"02:51.560","Text":"Now, the reason that I\u0027ve chosen it like this and I haven\u0027t chosen that one of"},{"Start":"02:51.560 ","End":"02:55.385","Text":"my axes is going along the diagonal of the cone wall,"},{"Start":"02:55.385 ","End":"02:57.770","Text":"is because usually we choose at least one of"},{"Start":"02:57.770 ","End":"03:02.045","Text":"our axes directions to be in the direction of the acceleration."},{"Start":"03:02.045 ","End":"03:05.350","Text":"We\u0027re using centripetal force."},{"Start":"03:05.350 ","End":"03:09.845","Text":"Angular motion, so we know that our acceleration is going to be in words,"},{"Start":"03:09.845 ","End":"03:14.720","Text":"then our other axes has to be perpendicular to that axis."},{"Start":"03:14.720 ","End":"03:20.880","Text":"Our x is going in the direction of the acceleration and our y is perpendicular to that."},{"Start":"03:21.050 ","End":"03:25.710","Text":"Now, because we have our angle Alpha over here,"},{"Start":"03:25.710 ","End":"03:31.085","Text":"now notice that this isn\u0027t the apex angles now going all across, it\u0027s a half-angle."},{"Start":"03:31.085 ","End":"03:33.455","Text":"If this is our Alpha,"},{"Start":"03:33.455 ","End":"03:39.610","Text":"so then we know that this is also going to be our Alpha, over here."},{"Start":"03:39.610 ","End":"03:42.740","Text":"Now we can write our equation."},{"Start":"03:42.740 ","End":"03:44.370","Text":"In the y-direction,"},{"Start":"03:44.370 ","End":"03:48.805","Text":"so the sum of all of our forces in the y-direction is going to equal to 0."},{"Start":"03:48.805 ","End":"03:53.590","Text":"Then we see that the component for N in the y-direction is equal"},{"Start":"03:53.590 ","End":"03:58.590","Text":"to n multiplied by sine of Alpha,"},{"Start":"03:58.590 ","End":"04:04.290","Text":"and then the component for our m_1g in the y-direction assembly m_1g,"},{"Start":"04:04.290 ","End":"04:06.349","Text":"but it\u0027s in the negative y-direction,"},{"Start":"04:06.349 ","End":"04:09.240","Text":"so it\u0027s going to be negative m_1g."},{"Start":"04:09.860 ","End":"04:14.175","Text":"Now we can take our N out,"},{"Start":"04:14.175 ","End":"04:15.525","Text":"we can isolate it,"},{"Start":"04:15.525 ","End":"04:23.215","Text":"and we\u0027ll get that our N is equal to m_1g divided by sine of Alpha."},{"Start":"04:23.215 ","End":"04:28.790","Text":"Now let\u0027s do the sum of all of our forces in the x-direction,"},{"Start":"04:28.790 ","End":"04:33.810","Text":"and that\u0027s also equal to the sum of all of our forces in the radial direction,"},{"Start":"04:33.810 ","End":"04:38.375","Text":"because the x is in the radial direction in this diagram."},{"Start":"04:38.375 ","End":"04:41.450","Text":"Then we can see the only force that we have going in"},{"Start":"04:41.450 ","End":"04:44.975","Text":"the radial direction is the x-component of our normal force."},{"Start":"04:44.975 ","End":"04:48.270","Text":"It\u0027s going to be N cosine of Alpha."},{"Start":"04:48.270 ","End":"04:51.000","Text":"M_1g has no x component,"},{"Start":"04:51.000 ","End":"04:56.600","Text":"and that is going to be equal to our m Omega^r."},{"Start":"04:56.650 ","End":"04:59.225","Text":"Now we know what our N is,"},{"Start":"04:59.225 ","End":"05:02.320","Text":"so we can substitute in this."},{"Start":"05:02.320 ","End":"05:12.290","Text":"We\u0027ll have that are m_1g divided by sine of Alpha multiplied by our cosine of Alpha,"},{"Start":"05:12.290 ","End":"05:18.855","Text":"that\u0027s our N cosine Alpha is equal to m_1 as well."},{"Start":"05:18.855 ","End":"05:20.940","Text":"This is also m_1."},{"Start":"05:20.940 ","End":"05:25.260","Text":"M_1 Omega squared r. Now,"},{"Start":"05:25.260 ","End":"05:27.645","Text":"we can cross from both sides are m_1,"},{"Start":"05:27.645 ","End":"05:30.860","Text":"and all we have to do is we have to isolate out"},{"Start":"05:30.860 ","End":"05:37.160","Text":"r. We can see that we have over here a cosine Alpha divided by our sine Alpha."},{"Start":"05:37.160 ","End":"05:41.420","Text":"That is the inverse of tan or not the inverse."},{"Start":"05:41.420 ","End":"05:44.098","Text":"It\u0027s 1 divided by our tangents Alpha"},{"Start":"05:44.098 ","End":"05:48.620","Text":"because we know the tangents itself as sine Alpha divided by cosine of Alpha."},{"Start":"05:48.620 ","End":"05:56.460","Text":"What we can do, is we can write g divided by tan of Alpha,"},{"Start":"05:56.460 ","End":"06:02.720","Text":"and then we can further divide it by our Omega^2 and this is going to"},{"Start":"06:02.720 ","End":"06:08.540","Text":"be our r. This is our answer to question number 1."},{"Start":"06:08.540 ","End":"06:11.250","Text":"Let\u0027s move on to question number 2."},{"Start":"06:11.750 ","End":"06:16.650","Text":"Question 2, says that a mass m_2 is placed on m_1."},{"Start":"06:16.650 ","End":"06:19.250","Text":"We\u0027re looking at this diagram over here."},{"Start":"06:19.250 ","End":"06:23.150","Text":"We\u0027re told that the coefficient of friction between the 2 masses is Mu_S,"},{"Start":"06:23.150 ","End":"06:24.680","Text":"so we have static friction."},{"Start":"06:24.680 ","End":"06:27.890","Text":"The rotational velocity of m_1 remains unchanged."},{"Start":"06:27.890 ","End":"06:32.130","Text":"It\u0027s still rotating with Omega with a cone."},{"Start":"06:32.130 ","End":"06:35.200","Text":"M_2 does not slip when it\u0027s on m_1."},{"Start":"06:35.200 ","End":"06:36.410","Text":"Then we\u0027re being asked, well,"},{"Start":"06:36.410 ","End":"06:42.230","Text":"the radius of motion when the system is at equilibrium change, and to explain that."},{"Start":"06:42.230 ","End":"06:44.705","Text":"Basically what they\u0027re asking is,"},{"Start":"06:44.705 ","End":"06:50.785","Text":"how does this value for r change when we add on our mass m_2?"},{"Start":"06:50.785 ","End":"06:53.150","Text":"There are 2 ways to solve this."},{"Start":"06:53.150 ","End":"06:57.155","Text":"Either we could draw our forces just like we did on diagram"},{"Start":"06:57.155 ","End":"07:01.205","Text":"a over here for both mass number 1 and mass number 2,"},{"Start":"07:01.205 ","End":"07:03.080","Text":"and then try and solve it like that."},{"Start":"07:03.080 ","End":"07:05.180","Text":"Or the other way."},{"Start":"07:05.180 ","End":"07:07.520","Text":"This is super easy."},{"Start":"07:07.520 ","End":"07:11.330","Text":"What do we can do is look at our answer for question number 1."},{"Start":"07:11.330 ","End":"07:13.250","Text":"Because we\u0027re being told that there\u0027s"},{"Start":"07:13.250 ","End":"07:17.030","Text":"a static coefficient of friction between the 2 masses,"},{"Start":"07:17.030 ","End":"07:21.064","Text":"that means that they\u0027re not moving, they\u0027re not slipping."},{"Start":"07:21.064 ","End":"07:24.910","Text":"They\u0027re just moving together in the cone."},{"Start":"07:24.910 ","End":"07:28.085","Text":"In that case, we can consider them as 1 body."},{"Start":"07:28.085 ","End":"07:30.470","Text":"Now, if we can consider them as 1 body,"},{"Start":"07:30.470 ","End":"07:33.530","Text":"then we can look at our equation over here,"},{"Start":"07:33.530 ","End":"07:36.425","Text":"and instead of saying that our mass is m_1,"},{"Start":"07:36.425 ","End":"07:39.865","Text":"we\u0027ll say that our total mass is therefore m_1 plus m_2."},{"Start":"07:39.865 ","End":"07:43.610","Text":"However, when we look at our answer from question 1,"},{"Start":"07:43.610 ","End":"07:49.420","Text":"we can see that our equation for finding the radius is independent of the mass."},{"Start":"07:49.420 ","End":"07:53.435","Text":"Now we can say that our radius does not change."},{"Start":"07:53.435 ","End":"07:57.515","Text":"We can add or take away as much mass as we want,"},{"Start":"07:57.515 ","End":"08:00.560","Text":"as long as the coefficient of friction is static between them,"},{"Start":"08:00.560 ","End":"08:02.465","Text":"so we can consider it 1 body,"},{"Start":"08:02.465 ","End":"08:06.670","Text":"our radius is always going to remain the same."},{"Start":"08:06.670 ","End":"08:09.345","Text":"Although we have this answer,"},{"Start":"08:09.345 ","End":"08:13.700","Text":"let\u0027s do it the first way said by drawing in our forces on"},{"Start":"08:13.700 ","End":"08:19.320","Text":"our diagram because it\u0027s also going to help us with our third question anyway."},{"Start":"08:19.370 ","End":"08:23.180","Text":"We\u0027ll carry on, we\u0027ll leave the forces in here for m_1"},{"Start":"08:23.180 ","End":"08:26.120","Text":"and we\u0027ll add in our friction and then our forces for m_2,"},{"Start":"08:26.120 ","End":"08:30.010","Text":"we\u0027ll draw over here, so as not to complicate the diagram."},{"Start":"08:30.010 ","End":"08:37.869","Text":"Even though, I mean that my forces for my m_1 will also be over here."},{"Start":"08:37.869 ","End":"08:43.900","Text":"In our m_1 we\u0027re going to have our frictional force acting as well."},{"Start":"08:43.900 ","End":"08:47.200","Text":"In which direction is it going? It doesn\u0027t really matter."},{"Start":"08:47.200 ","End":"08:53.050","Text":"I can draw it going right or left because I don\u0027t know in which direction it\u0027s acting."},{"Start":"08:53.050 ","End":"08:56.190","Text":"It doesn\u0027t really matter because if I choose, let\u0027s say,"},{"Start":"08:56.190 ","End":"08:58.710","Text":"the left direction and it comes out positive"},{"Start":"08:58.710 ","End":"09:01.560","Text":"then the frictional force is acting in the left direction."},{"Start":"09:01.560 ","End":"09:05.610","Text":"If I choose it in the left direction and then I get a negative sign in front of it and"},{"Start":"09:05.610 ","End":"09:10.255","Text":"the forces then I know that it\u0027s pointing in the right direction."},{"Start":"09:10.255 ","End":"09:14.080","Text":"Let\u0027s say that my frictional force is going here,"},{"Start":"09:14.080 ","End":"09:15.499","Text":"and it\u0027s static friction,"},{"Start":"09:15.499 ","End":"09:17.440","Text":"of course, so f_s."},{"Start":"09:17.440 ","End":"09:22.390","Text":"These are all of my forces now acting on m_1 when I have my m_2 on top."},{"Start":"09:22.390 ","End":"09:24.190","Text":"Now, for my m_2,"},{"Start":"09:24.190 ","End":"09:29.500","Text":"so I\u0027m going to have my m_2g working downwards."},{"Start":"09:29.500 ","End":"09:36.460","Text":"Then in this direction I\u0027m going to have my N but it\u0027s not this N so I\u0027ll call this N_2,"},{"Start":"09:36.460 ","End":"09:38.965","Text":"and of course, I\u0027m going to have my frictional force."},{"Start":"09:38.965 ","End":"09:42.430","Text":"Now, because here I drew that my frictional force is going in"},{"Start":"09:42.430 ","End":"09:48.944","Text":"my left direction as a result of this m_2 sitting on top,"},{"Start":"09:48.944 ","End":"09:52.870","Text":"so that means that the frictional force acting from N_2 on"},{"Start":"09:52.870 ","End":"09:57.316","Text":"my mass number 1 according to Newton\u0027s third law of motion,"},{"Start":"09:57.316 ","End":"09:59.670","Text":"so that means that this frictional force has to"},{"Start":"09:59.670 ","End":"10:02.760","Text":"be the same size but in the opposite direction."},{"Start":"10:02.760 ","End":"10:08.740","Text":"That means that over here my static friction is going to be acting in this direction."},{"Start":"10:08.740 ","End":"10:12.234","Text":"Now, we can write in our equations."},{"Start":"10:12.234 ","End":"10:14.530","Text":"For mass number 1,"},{"Start":"10:14.530 ","End":"10:17.320","Text":"so we\u0027re going to have again,"},{"Start":"10:17.320 ","End":"10:22.870","Text":"so the sum of all of our forces in the y-direction is going to be equal to 0,"},{"Start":"10:22.870 ","End":"10:30.835","Text":"which is going to be equal to our N sine Alpha minus m_1g."},{"Start":"10:30.835 ","End":"10:37.585","Text":"Then we\u0027ll have that our N is equal to m_1g divided by sine of Alpha."},{"Start":"10:37.585 ","End":"10:42.460","Text":"Then we\u0027ll have the sum of all of our forces in the x-axis,"},{"Start":"10:42.460 ","End":"10:45.235","Text":"which is also the radial axis."},{"Start":"10:45.235 ","End":"10:49.015","Text":"Then we\u0027ll have our N cosine of Alpha."},{"Start":"10:49.015 ","End":"10:55.630","Text":"This time we also have negative our f_s and this is going to be equal"},{"Start":"10:55.630 ","End":"11:03.130","Text":"to our m Omega squared r in here, m_1, of course."},{"Start":"11:03.130 ","End":"11:10.600","Text":"This is for our m_1 and this is also m_1."},{"Start":"11:10.600 ","End":"11:13.630","Text":"For m_2 we\u0027re going to have that the sum of all of"},{"Start":"11:13.630 ","End":"11:17.095","Text":"our forces in the y-direction is going to be,"},{"Start":"11:17.095 ","End":"11:19.915","Text":"so we\u0027re going to have N_2 in the positive direction minus"},{"Start":"11:19.915 ","End":"11:24.265","Text":"m_2g in the negative y-direction."},{"Start":"11:24.265 ","End":"11:26.695","Text":"That, of course, is going to be equal to 0"},{"Start":"11:26.695 ","End":"11:29.373","Text":"because there\u0027s no movement in the y-direction."},{"Start":"11:29.373 ","End":"11:31.150","Text":"Then in the x-direction,"},{"Start":"11:31.150 ","End":"11:32.440","Text":"this is also m_2,"},{"Start":"11:32.440 ","End":"11:35.650","Text":"so we\u0027re going to have the sum of all of the forces in our x-direction,"},{"Start":"11:35.650 ","End":"11:38.035","Text":"which is also the radial direction."},{"Start":"11:38.035 ","End":"11:41.980","Text":"The only force that we have is our frictional force,"},{"Start":"11:41.980 ","End":"11:45.835","Text":"so f_s and that is going to be equal to"},{"Start":"11:45.835 ","End":"11:51.520","Text":"m_2 Omega squared r. Of course, my static friction,"},{"Start":"11:51.520 ","End":"11:55.360","Text":"I always have to choose in the direction of the acceleration now because"},{"Start":"11:55.360 ","End":"11:59.230","Text":"we\u0027re working with centripetal force so the direction of"},{"Start":"11:59.230 ","End":"12:03.760","Text":"the acceleration is into my circle or into my cone over"},{"Start":"12:03.760 ","End":"12:08.725","Text":"here so my f_s for M_2 has to be going inside."},{"Start":"12:08.725 ","End":"12:11.980","Text":"I\u0027ve chosen it in the right direction so I have a positive here."},{"Start":"12:11.980 ","End":"12:17.020","Text":"If I would have chosen here my f_s going in the leftwards direction and for my m_1,"},{"Start":"12:17.020 ","End":"12:18.130","Text":"my f_s is going in the rightwards,"},{"Start":"12:18.130 ","End":"12:20.864","Text":"so I just would have here in the end,"},{"Start":"12:20.864 ","End":"12:23.335","Text":"gotten a negative f_s."},{"Start":"12:23.335 ","End":"12:30.160","Text":"Now, all we would do is isolate out our N_2 and substitute that in here with"},{"Start":"12:30.160 ","End":"12:33.445","Text":"our coefficient of friction and then just solve"},{"Start":"12:33.445 ","End":"12:36.775","Text":"all of the algebra and we\u0027ll get to this same answer,"},{"Start":"12:36.775 ","End":"12:40.345","Text":"that our radius is unchanged."},{"Start":"12:40.345 ","End":"12:43.600","Text":"Let\u0027s move on to the next question."},{"Start":"12:43.600 ","End":"12:45.700","Text":"Question number 3 is asking,"},{"Start":"12:45.700 ","End":"12:48.745","Text":"what is the minimum value that the coefficient of friction,"},{"Start":"12:48.745 ","End":"12:53.815","Text":"my Mu_s, can be such that there won\u0027t be slipping between the masses?"},{"Start":"12:53.815 ","End":"12:56.620","Text":"We can say that the top of m_1 is horizontal,"},{"Start":"12:56.620 ","End":"12:58.885","Text":"we\u0027ve always assumed that anyway."},{"Start":"12:58.885 ","End":"13:00.340","Text":"What we\u0027re being asked is,"},{"Start":"13:00.340 ","End":"13:04.270","Text":"what is the minimum value for my Mu_s?"},{"Start":"13:04.270 ","End":"13:06.010","Text":"Anytime I have a question,"},{"Start":"13:06.010 ","End":"13:10.450","Text":"when they\u0027re asking me for something with Mu_s and its minimum value,"},{"Start":"13:10.450 ","End":"13:15.220","Text":"so it\u0027s generally to do with my static friction."},{"Start":"13:15.220 ","End":"13:20.793","Text":"The equation that we\u0027re going to be using is that our static friction, f_s,"},{"Start":"13:20.793 ","End":"13:25.570","Text":"has to be smaller or equal to my coefficient of static friction multiplied by"},{"Start":"13:25.570 ","End":"13:32.515","Text":"N. Remember that our f_s has to be smaller or equal to, not equal to."},{"Start":"13:32.515 ","End":"13:33.970","Text":"Just as a reminder,"},{"Start":"13:33.970 ","End":"13:36.820","Text":"when we\u0027re dealing with kinetic friction, f _k,"},{"Start":"13:36.820 ","End":"13:40.630","Text":"so it\u0027s always equal to Mu_k multiplied by"},{"Start":"13:40.630 ","End":"13:45.745","Text":"N. It doesn\u0027t matter what my motion is relative to the surface,"},{"Start":"13:45.745 ","End":"13:49.810","Text":"this is always going to be my value for my kinetic friction."},{"Start":"13:49.810 ","End":"13:53.525","Text":"But with static friction it\u0027s always smaller or equal to."},{"Start":"13:53.525 ","End":"13:57.270","Text":"O is my value for my static friction is going to be some"},{"Start":"13:57.270 ","End":"14:02.760","Text":"unknown and the only thing I know about it is that it has some maximum value,"},{"Start":"14:02.760 ","End":"14:04.110","Text":"which is going to be this."},{"Start":"14:04.110 ","End":"14:08.265","Text":"But it could also be every single value which is smaller than."},{"Start":"14:08.265 ","End":"14:11.835","Text":"Anytime we have a Mu_s with minimum or maximum,"},{"Start":"14:11.835 ","End":"14:14.615","Text":"then we know that we have to use this equation."},{"Start":"14:14.615 ","End":"14:16.720","Text":"How we\u0027re going to solve this is we\u0027re going to look at"},{"Start":"14:16.720 ","End":"14:21.505","Text":"our f_s value from our equations that we wrote over here."},{"Start":"14:21.505 ","End":"14:24.670","Text":"Then we\u0027re going to say that it has to be smaller or equal"},{"Start":"14:24.670 ","End":"14:28.978","Text":"to Mu_s multiplied by N. We can see,"},{"Start":"14:28.978 ","End":"14:32.215","Text":"over here that our f_s is going to be equal to"},{"Start":"14:32.215 ","End":"14:36.295","Text":"m_2 Omega squared r. Let\u0027s write that down."},{"Start":"14:36.295 ","End":"14:43.810","Text":"We have that our f_s is equal to our m_2 Omega squared r. Now,"},{"Start":"14:43.810 ","End":"14:47.845","Text":"our r we got from our first question."},{"Start":"14:47.845 ","End":"14:52.075","Text":"It\u0027s g divided by Omega squared tan of Alpha."},{"Start":"14:52.075 ","End":"14:54.010","Text":"Let\u0027s rewrite that."},{"Start":"14:54.010 ","End":"15:00.790","Text":"It\u0027s going to be m_2 multiplied by Omega squared and then multiplied by our r,"},{"Start":"15:00.790 ","End":"15:07.240","Text":"which is g divided by Omega squared tan of Alpha."},{"Start":"15:07.240 ","End":"15:13.473","Text":"Then we also know that this has to be because our f_s is smaller than Mu_sN,"},{"Start":"15:13.473 ","End":"15:16.375","Text":"so smaller than Mu_sN."},{"Start":"15:16.375 ","End":"15:21.110","Text":"We can see that this Omega squared and this Omega squared cancel out."},{"Start":"15:21.240 ","End":"15:29.440","Text":"Then we can say that our g divided by tan of Alpha is going to be"},{"Start":"15:29.440 ","End":"15:37.165","Text":"small or equal to our Mu_s multiplied by our N. What N are we speaking about?"},{"Start":"15:37.165 ","End":"15:39.355","Text":"Let\u0027s go back to our diagram."},{"Start":"15:39.355 ","End":"15:42.355","Text":"Because we\u0027re trying to find our frictional force,"},{"Start":"15:42.355 ","End":"15:45.040","Text":"which is acting between our mass,"},{"Start":"15:45.040 ","End":"15:46.924","Text":"m_1 and m_2,"},{"Start":"15:46.924 ","End":"15:51.865","Text":"it means that we want to find the normal force which is dealing with"},{"Start":"15:51.865 ","End":"15:58.105","Text":"the reaction between our masses, m_1 and m_2."},{"Start":"15:58.105 ","End":"16:00.205","Text":"I can see it rubbed out."},{"Start":"16:00.205 ","End":"16:02.845","Text":"We had our normal force over here."},{"Start":"16:02.845 ","End":"16:04.660","Text":"Now, we don\u0027t want this normal force,"},{"Start":"16:04.660 ","End":"16:08.800","Text":"which is between m_1 and the cone because it has no relevance."},{"Start":"16:08.800 ","End":"16:11.490","Text":"Our frictional forces acting between m_1 and m_2,"},{"Start":"16:11.490 ","End":"16:15.980","Text":"so we\u0027re going to take this frictional force over here, N_2."},{"Start":"16:16.260 ","End":"16:19.450","Text":"Let\u0027s go back to our equation."},{"Start":"16:19.450 ","End":"16:21.610","Text":"This is going to be N_2."},{"Start":"16:21.610 ","End":"16:24.520","Text":"Then this is going to be equal to Mu_s."},{"Start":"16:24.520 ","End":"16:26.170","Text":"What is our N_2?"},{"Start":"16:26.170 ","End":"16:28.210","Text":"We can isolate out from here."},{"Start":"16:28.210 ","End":"16:32.770","Text":"Our N_2 is simply going to be equal to our m_2g,"},{"Start":"16:32.770 ","End":"16:36.700","Text":"so m_2g over here."},{"Start":"16:36.700 ","End":"16:40.720","Text":"Now, we can see that our g can cancel out."},{"Start":"16:40.720 ","End":"16:44.110","Text":"I meant to have an m_2 over here as well, I forgot that."},{"Start":"16:44.110 ","End":"16:46.990","Text":"My N_2 will cancel out as well."},{"Start":"16:46.990 ","End":"16:49.479","Text":"Then we can see that we get that our Mu_s,"},{"Start":"16:49.479 ","End":"16:52.150","Text":"our coefficient for static friction,"},{"Start":"16:52.150 ","End":"16:59.410","Text":"has to be bigger or equal to 1 divided by tan of Alpha."},{"Start":"16:59.410 ","End":"17:02.690","Text":"That\u0027s the end of this question."}],"ID":9475},{"Watched":false,"Name":"Balls Hit Rods","Duration":"47m 6s","ChapterTopicVideoID":9206,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.024","Text":"Hello. In this question,"},{"Start":"00:02.024 ","End":"00:04.350","Text":"we have 2 thin and long rods,"},{"Start":"00:04.350 ","End":"00:07.049","Text":"so we\u0027re ignoring that here they\u0027re drawn as cylinders."},{"Start":"00:07.049 ","End":"00:10.364","Text":"They\u0027re rods. They\u0027re both at rest."},{"Start":"00:10.364 ","End":"00:13.470","Text":"Each of mass m and of length L,"},{"Start":"00:13.470 ","End":"00:17.970","Text":"each rod, and they\u0027re joined together at right angles at the origin."},{"Start":"00:17.970 ","End":"00:20.294","Text":"The origin is over here, and they\u0027re here."},{"Start":"00:20.294 ","End":"00:22.460","Text":"2 masses of mass m;"},{"Start":"00:22.460 ","End":"00:25.109","Text":"they both have identical masses,"},{"Start":"00:25.109 ","End":"00:31.605","Text":"move perpendicularly to the rods and collide at the tips of the rod at velocity V_1,"},{"Start":"00:31.605 ","End":"00:34.440","Text":"which is negative v_0 in the x-direction,"},{"Start":"00:34.440 ","End":"00:37.575","Text":"and V_2 is v_0 in the y-direction."},{"Start":"00:37.575 ","End":"00:39.605","Text":"They have the same size."},{"Start":"00:39.605 ","End":"00:41.299","Text":"Each velocity is of the same size."},{"Start":"00:41.299 ","End":"00:43.915","Text":"They\u0027re just going in different directions, as we can see."},{"Start":"00:43.915 ","End":"00:46.639","Text":"Then we\u0027re told at time t=0,"},{"Start":"00:46.639 ","End":"00:50.067","Text":"the masses collide with the rods,"},{"Start":"00:50.067 ","End":"00:51.895","Text":"and they stick to the rods as well."},{"Start":"00:51.895 ","End":"00:53.885","Text":"As they collide, they stick there."},{"Start":"00:53.885 ","End":"00:55.444","Text":"Our first question is,"},{"Start":"00:55.444 ","End":"01:04.290","Text":"what is the position vector of the center of mass r_CM as a function of t at time t=0."},{"Start":"01:04.290 ","End":"01:05.945","Text":"This is our diagram,"},{"Start":"01:05.945 ","End":"01:07.909","Text":"and now I\u0027m just going to copy it out in a way"},{"Start":"01:07.909 ","End":"01:10.835","Text":"that\u0027s a little bit easier for us to work with."},{"Start":"01:10.835 ","End":"01:15.419","Text":"Here\u0027s our diagram redrawn for ease and our question number"},{"Start":"01:15.419 ","End":"01:20.015","Text":"1 was asking what r^CM at t=0, is equal to?"},{"Start":"01:20.015 ","End":"01:22.950","Text":"Sorry, here it\u0027s not a 0 it\u0027s meant to be a question mark."},{"Start":"01:22.950 ","End":"01:25.429","Text":"How do we find our r_CM?"},{"Start":"01:25.429 ","End":"01:29.959","Text":"The first thing we have to do is find what our x center of mass is equal"},{"Start":"01:29.959 ","End":"01:35.035","Text":"to and what our y center of mass is equal to."},{"Start":"01:35.035 ","End":"01:37.440","Text":"Let\u0027s start with our x center of mass."},{"Start":"01:37.440 ","End":"01:41.675","Text":"We know that our equation for x center of mass is equal to the sum"},{"Start":"01:41.675 ","End":"01:47.920","Text":"of m_i multiplied by x_i divided by the sum of m_i."},{"Start":"01:47.920 ","End":"01:51.510","Text":"Let\u0027s take a look. In our diagram, we have 4 bodies."},{"Start":"01:51.510 ","End":"01:53.280","Text":"We have 1 rod, 2 rod,"},{"Start":"01:53.280 ","End":"01:55.860","Text":"a ball and another ball,"},{"Start":"01:55.860 ","End":"01:57.292","Text":"so 4 bodies,"},{"Start":"01:57.292 ","End":"01:59.955","Text":"and we\u0027re being asked at t=0."},{"Start":"01:59.955 ","End":"02:03.400","Text":"+In the question t=0, our balls,"},{"Start":"02:03.400 ","End":"02:07.210","Text":"each one is attached to the end of each rod,"},{"Start":"02:07.210 ","End":"02:08.895","Text":"so it looks like this."},{"Start":"02:08.895 ","End":"02:11.280","Text":"Let\u0027s take a look at what we can do,"},{"Start":"02:11.280 ","End":"02:14.520","Text":"so our balls are spheres over here,"},{"Start":"02:14.520 ","End":"02:18.020","Text":"are considered as point masses of mass m,"},{"Start":"02:18.020 ","End":"02:21.694","Text":"and they are located a distance L away from the origin."},{"Start":"02:21.694 ","End":"02:23.599","Text":"Then we have our rod,"},{"Start":"02:23.599 ","End":"02:27.560","Text":"which is with a uniform mass and so because it\u0027s a large body,"},{"Start":"02:27.560 ","End":"02:29.044","Text":"it\u0027s a rigid body,"},{"Start":"02:29.044 ","End":"02:32.060","Text":"so we have to take its center of mass to be in the middle over"},{"Start":"02:32.060 ","End":"02:35.794","Text":"here as a point mass located at L"},{"Start":"02:35.794 ","End":"02:42.710","Text":"divided by 2 and same over here because this rod is also a rigid body of uniform mass,"},{"Start":"02:42.710 ","End":"02:45.890","Text":"so we\u0027re going to take it as a point mass located"},{"Start":"02:45.890 ","End":"02:49.448","Text":"in the center at L divided by 2, and, of course,"},{"Start":"02:49.448 ","End":"02:53.060","Text":"this is another mass of mass m located at"},{"Start":"02:53.060 ","End":"02:57.920","Text":"distance L. Now let\u0027s substitute everything into the equation."},{"Start":"02:57.920 ","End":"03:00.649","Text":"Now because we\u0027re doing the x center of mass,"},{"Start":"03:00.649 ","End":"03:04.909","Text":"we can see that our x position for this rod and this sphere is equal to 0,"},{"Start":"03:04.909 ","End":"03:09.300","Text":"so we don\u0027t have to add it in and now we just have to add in these details."},{"Start":"03:09.300 ","End":"03:12.859","Text":"We\u0027re going to have our mass of the rod."},{"Start":"03:12.859 ","End":"03:15.290","Text":"They\u0027re all of similar masses of the same mass."},{"Start":"03:15.290 ","End":"03:17.599","Text":"The M, the m are the same thing,"},{"Start":"03:17.599 ","End":"03:23.790","Text":"so this is going to be m multiplied by L divided by 2,"},{"Start":"03:23.790 ","End":"03:25.905","Text":"for the center of mass of the rod,"},{"Start":"03:25.905 ","End":"03:32.210","Text":"and then plus M multiplied by L for the sphere located at the end of the rod."},{"Start":"03:32.210 ","End":"03:35.014","Text":"Then divide it by the total mass of the system,"},{"Start":"03:35.014 ","End":"03:37.730","Text":"which is the mass of the rod plus the mass of"},{"Start":"03:37.730 ","End":"03:40.580","Text":"this rod plus the mass of this sphere plus the mass of this sphere,"},{"Start":"03:40.580 ","End":"03:42.520","Text":"so it\u0027s going to be at 4m."},{"Start":"03:42.520 ","End":"03:46.129","Text":"Then once we rearrange it and play around with the algebra,"},{"Start":"03:46.129 ","End":"03:51.920","Text":"we\u0027re going to get that our x_CM is located 3/8L."},{"Start":"03:51.920 ","End":"03:57.770","Text":"Now, instead of working out the whole same thing for the y_CM, because of symmetry,"},{"Start":"03:57.770 ","End":"03:59.840","Text":"we can see everything is symmetrical,"},{"Start":"03:59.840 ","End":"04:05.220","Text":"so we can say that our y_CM is too located at 3/8L."},{"Start":"04:06.740 ","End":"04:10.640","Text":"Now because we\u0027re asked for our r_CM,"},{"Start":"04:10.640 ","End":"04:18.694","Text":"so we can say therefore that our r_CM at time of t=0,"},{"Start":"04:18.694 ","End":"04:23.269","Text":"is going to be equal to a combination of these 2 coordinates."},{"Start":"04:23.269 ","End":"04:26.914","Text":"Because we can write them as 3/8L"},{"Start":"04:26.914 ","End":"04:31.564","Text":"multiplied by 1 in the x direction and 1 in the y-direction."},{"Start":"04:31.564 ","End":"04:35.485","Text":"Then we can see that we get these exact coordinates."},{"Start":"04:35.485 ","End":"04:37.555","Text":"That\u0027s the end of question number 1."},{"Start":"04:37.555 ","End":"04:39.785","Text":"Now let\u0027s go on to question number 2."},{"Start":"04:39.785 ","End":"04:44.839","Text":"Question number 2 is asking us what is the position vector of the center of mass?"},{"Start":"04:44.839 ","End":"04:47.239","Text":"Again, r_CM is a function of t. However,"},{"Start":"04:47.239 ","End":"04:50.320","Text":"now when t is bigger than 0."},{"Start":"04:50.320 ","End":"04:55.414","Text":"After the point where the spheres stick to the ends of the rods,"},{"Start":"04:55.414 ","End":"04:58.864","Text":"and we\u0027re being asked to find it not relative to the origin,"},{"Start":"04:58.864 ","End":"05:03.995","Text":"but rather relative to the position of the center of mass at t=0."},{"Start":"05:03.995 ","End":"05:07.182","Text":"Relative to the position that we found in question number 1,"},{"Start":"05:07.182 ","End":"05:11.010","Text":"we\u0027re given this equation that our r_CM,"},{"Start":"05:11.010 ","End":"05:17.015","Text":"when t is bigger than 0 minus our r_CM when t is equal to 0 is equal to what?"},{"Start":"05:17.015 ","End":"05:19.785","Text":"Let\u0027s take a look at how we solve this."},{"Start":"05:19.785 ","End":"05:27.965","Text":"We found our center of mass to be somewhere around here where this and this,"},{"Start":"05:27.965 ","End":"05:32.925","Text":"these distances are each 3/8 of L,"},{"Start":"05:32.925 ","End":"05:38.615","Text":"and also over here, 3/8 of L. This is our center of mass."},{"Start":"05:38.615 ","End":"05:41.630","Text":"Now we\u0027re being asked to solve this equation,r_CM,"},{"Start":"05:41.630 ","End":"05:45.665","Text":"when t is bigger than 0 minus our r_CM when t is equal to 0."},{"Start":"05:45.665 ","End":"05:52.979","Text":"Because we went to find r_CM when t is bigger than 0 relative to this point,"},{"Start":"05:52.979 ","End":"05:56.569","Text":"so let\u0027s call this point where we found our center of mass to be when"},{"Start":"05:56.569 ","End":"06:00.820","Text":"t=0 so let\u0027s call this point our origin."},{"Start":"06:00.820 ","End":"06:04.864","Text":"Now we\u0027re going to use our center of mass as our origin."},{"Start":"06:04.864 ","End":"06:07.670","Text":"Now, as the motion continues,"},{"Start":"06:07.670 ","End":"06:13.085","Text":"we\u0027re going to get some curve where our center of mass is going to move."},{"Start":"06:13.085 ","End":"06:16.074","Text":"Then we have to find its position as a function of time"},{"Start":"06:16.074 ","End":"06:19.330","Text":"relative to a new origin over here."},{"Start":"06:19.330 ","End":"06:25.074","Text":"Let\u0027s see how we can begin answering these types of questions,"},{"Start":"06:25.074 ","End":"06:27.895","Text":"so +also this question are questions in general."},{"Start":"06:27.895 ","End":"06:32.000","Text":"The first things that we usually check is conservation of something."},{"Start":"06:32.000 ","End":"06:34.415","Text":"The first thing is conservation of momentum."},{"Start":"06:34.415 ","End":"06:37.735","Text":"Now how do we know if there is conservation of momentum?"},{"Start":"06:37.735 ","End":"06:44.844","Text":"If the sum of all of the external forces are equal to 0,"},{"Start":"06:44.844 ","End":"06:47.754","Text":"then we have conservation of momentum."},{"Start":"06:47.754 ","End":"06:51.745","Text":"The next thing that we check is conservation of energy."},{"Start":"06:51.745 ","End":"06:56.090","Text":"How do we know if there\u0027s conservation of energy if all,"},{"Start":"06:56.090 ","End":"07:00.688","Text":"and this is very important so I\u0027m going to underline all,"},{"Start":"07:00.688 ","End":"07:03.535","Text":"the forces are conservation."},{"Start":"07:03.535 ","End":"07:06.435","Text":"Now, why did I highlight my all,"},{"Start":"07:06.435 ","End":"07:08.974","Text":"because over here and conservation of momentum,"},{"Start":"07:08.974 ","End":"07:10.909","Text":"we only need that the sum of"},{"Start":"07:10.909 ","End":"07:17.375","Text":"all the external forces must be equal to 0, only external forces."},{"Start":"07:17.375 ","End":"07:18.590","Text":"When we\u0027re dealing, however,"},{"Start":"07:18.590 ","End":"07:19.820","Text":"with conservation of energy,"},{"Start":"07:19.820 ","End":"07:22.369","Text":"we need all the forces to be conservation,"},{"Start":"07:22.369 ","End":"07:26.160","Text":"so it\u0027s a different thing and you have to differentiate between the 2."},{"Start":"07:26.160 ","End":"07:31.260","Text":"The next thing that we check is if we have conservation of angular momentum."},{"Start":"07:31.260 ","End":"07:32.978","Text":"Here we had linear momentum,"},{"Start":"07:32.978 ","End":"07:34.539","Text":"here its angular momentum."},{"Start":"07:34.539 ","End":"07:36.395","Text":"When does this happen?"},{"Start":"07:36.395 ","End":"07:42.850","Text":"If the sum of all of the external torques, and again,"},{"Start":"07:42.850 ","End":"07:46.999","Text":"external torques, so if that is equal to 0,"},{"Start":"07:46.999 ","End":"07:50.585","Text":"then we have conservation of angular momentum."},{"Start":"07:50.585 ","End":"07:54.500","Text":"Now how do I check if the sum of all of my external forces is equal to 0,"},{"Start":"07:54.500 ","End":"07:58.460","Text":"so what I have to do is check all of my external forces,"},{"Start":"07:58.460 ","End":"08:01.370","Text":"which they are, then I have to work out the moment"},{"Start":"08:01.370 ","End":"08:04.400","Text":"or the torque of each of those forces,"},{"Start":"08:04.400 ","End":"08:07.174","Text":"and then I have to see if they are equal to 0."},{"Start":"08:07.174 ","End":"08:10.295","Text":"If they do, then I have conservation of angular momentum."},{"Start":"08:10.295 ","End":"08:13.400","Text":"Now, there\u0027s also no connection between this and this,"},{"Start":"08:13.400 ","End":"08:16.880","Text":"conservation of momentum and conservation of angular momentum."},{"Start":"08:16.880 ","End":"08:20.659","Text":"We can have a case where we have conservation of momentum but not of"},{"Start":"08:20.659 ","End":"08:25.433","Text":"angular momentum or we have angular momentum and not of momentum."},{"Start":"08:25.433 ","End":"08:28.185","Text":"Now let\u0027s begin with solving this question."},{"Start":"08:28.185 ","End":"08:30.124","Text":"Let\u0027s take a look at our system."},{"Start":"08:30.124 ","End":"08:35.540","Text":"Now, when we have that our masses over here are spheres are attached to our rod,"},{"Start":"08:35.540 ","End":"08:37.450","Text":"which is the case that we\u0027re looking at now,"},{"Start":"08:37.450 ","End":"08:40.519","Text":"so that means that we don\u0027t have any external bodies."},{"Start":"08:40.519 ","End":"08:42.575","Text":"Now, if we don\u0027t have any external bodies,"},{"Start":"08:42.575 ","End":"08:45.340","Text":"that means that we have no external forces."},{"Start":"08:45.340 ","End":"08:47.434","Text":"If we don\u0027t have any external forces,"},{"Start":"08:47.434 ","End":"08:51.930","Text":"then that means that the sum of all of the external forces is equal to 0,"},{"Start":"08:51.930 ","End":"08:54.335","Text":"so we have conservation of momentum."},{"Start":"08:54.335 ","End":"08:56.599","Text":"Now what about conservation of energy?"},{"Start":"08:56.599 ","End":"09:01.130","Text":"We don\u0027t have conservation of energy because in the question we were told that"},{"Start":"09:01.130 ","End":"09:05.764","Text":"our masses are spheres collide with the rods and are attached to them,"},{"Start":"09:05.764 ","End":"09:08.509","Text":"which means that we have plastic collision."},{"Start":"09:08.509 ","End":"09:13.149","Text":"Plastic collision means that there is no conservation of energy."},{"Start":"09:13.149 ","End":"09:15.694","Text":"This can have a tick. This can have an X,"},{"Start":"09:15.694 ","End":"09:18.500","Text":"and then conservation of angular momentum."},{"Start":"09:18.500 ","End":"09:21.185","Text":"Because we don\u0027t have any external forces,"},{"Start":"09:21.185 ","End":"09:23.869","Text":"that means that the external torque of"},{"Start":"09:23.869 ","End":"09:28.099","Text":"each force is going to also be equal to 0 because the forces are equal to 0,"},{"Start":"09:28.099 ","End":"09:30.909","Text":"which means that this also happens."},{"Start":"09:30.909 ","End":"09:34.165","Text":"We also have conservation of angular momentum."},{"Start":"09:34.165 ","End":"09:37.595","Text":"I\u0027ve written over here the way that we"},{"Start":"09:37.595 ","End":"09:42.319","Text":"calculated which conservation we have in this type of question."},{"Start":"09:42.319 ","End":"09:43.966","Text":"If you want, you can read this."},{"Start":"09:43.966 ","End":"09:46.055","Text":"You can copy this out for your notes."},{"Start":"09:46.055 ","End":"09:47.354","Text":"Because a lot of the time,"},{"Start":"09:47.354 ","End":"09:50.870","Text":"we\u0027ll get questions where similar cases are at hand,"},{"Start":"09:50.870 ","End":"09:52.924","Text":"such as no external forces,"},{"Start":"09:52.924 ","End":"09:58.900","Text":"and then we also have no external torques and such as also plastic collision."},{"Start":"09:58.900 ","End":"10:00.930","Text":"Let\u0027s begin answering."},{"Start":"10:00.930 ","End":"10:02.779","Text":"Now that we know that we have conservation of"},{"Start":"10:02.779 ","End":"10:05.344","Text":"momentum and conservation of angular momentum,"},{"Start":"10:05.344 ","End":"10:09.775","Text":"we can begin by writing our equations for motion and then to solve."},{"Start":"10:09.775 ","End":"10:15.544","Text":"Let\u0027s start with conservation of momentum and let\u0027s write our equations for that."},{"Start":"10:15.544 ","End":"10:17.405","Text":"Of course, when dealing with momentum,"},{"Start":"10:17.405 ","End":"10:20.930","Text":"we have to split it up into it\u0027s different components,"},{"Start":"10:20.930 ","End":"10:25.840","Text":"which means the momentum in the x-direction and the momentum in the y-direction."},{"Start":"10:25.840 ","End":"10:31.625","Text":"Let\u0027s write our momentum in x we have before the collision."},{"Start":"10:31.625 ","End":"10:36.049","Text":"Before the collision, we have this mass over"},{"Start":"10:36.049 ","End":"10:41.120","Text":"here moving in the negative x-direction with a velocity of v_0,"},{"Start":"10:41.120 ","End":"10:44.330","Text":"so let\u0027s write that down, and then, of course,"},{"Start":"10:44.330 ","End":"10:45.964","Text":"our rods are stationary,"},{"Start":"10:45.964 ","End":"10:48.060","Text":"so they have a velocity of 0,"},{"Start":"10:48.060 ","End":"10:51.110","Text":"so we don\u0027t have to add them into the equation."},{"Start":"10:51.110 ","End":"10:54.050","Text":"This rod down here with this mass over here,"},{"Start":"10:54.050 ","End":"10:59.994","Text":"it only has a velocity component in the y-direction,"},{"Start":"10:59.994 ","End":"11:03.659","Text":"so we don\u0027t write it in the momentum for the x."},{"Start":"11:03.659 ","End":"11:09.665","Text":"Let\u0027s write, so we\u0027re going to have the mass of the sphere multiplied by its velocity,"},{"Start":"11:09.665 ","End":"11:13.498","Text":"which is going in the negative x-direction."},{"Start":"11:13.498 ","End":"11:15.865","Text":"That is our momentum before."},{"Start":"11:15.865 ","End":"11:21.790","Text":"Then momentum after is going to be the mass of the whole system."},{"Start":"11:21.790 ","End":"11:29.005","Text":"We know that that is 4m multiplied by the velocity which the whole system is moving in."},{"Start":"11:29.005 ","End":"11:34.885","Text":"Let\u0027s call that u. U is of course also in the x-direction."},{"Start":"11:34.885 ","End":"11:36.925","Text":"Now we can isolate out our u,"},{"Start":"11:36.925 ","End":"11:44.755","Text":"so will get that our u in the x-direction is equal to negative 1/4 of v_0."},{"Start":"11:44.755 ","End":"11:48.205","Text":"Just to explain what I meant here by my 4m,"},{"Start":"11:48.205 ","End":"11:50.710","Text":"I didn\u0027t mean the mass of the whole system."},{"Start":"11:50.710 ","End":"11:54.084","Text":"It\u0027s only the mass of the whole system after collision."},{"Start":"11:54.084 ","End":"11:58.510","Text":"Because now we can consider it as 1 body with"},{"Start":"11:58.510 ","End":"12:04.480","Text":"these 4 masses of m. This was the momentum in the x-direction before the collision,"},{"Start":"12:04.480 ","End":"12:08.259","Text":"and this is the momentum in the x-direction after the collision."},{"Start":"12:08.259 ","End":"12:12.790","Text":"Now, let\u0027s speak a little bit about our u in the x-direction."},{"Start":"12:12.790 ","End":"12:16.225","Text":"When we\u0027re working out our equation for momentum,"},{"Start":"12:16.225 ","End":"12:21.955","Text":"so it doesn\u0027t take into account all the different components of the system."},{"Start":"12:21.955 ","End":"12:26.440","Text":"It takes into account the center of mass."},{"Start":"12:26.440 ","End":"12:33.445","Text":"That means that here we worked out our center of mass multiplied by the velocity."},{"Start":"12:33.445 ","End":"12:35.695","Text":"Specifically here we had a point mass,"},{"Start":"12:35.695 ","End":"12:37.669","Text":"now after the collision,"},{"Start":"12:37.669 ","End":"12:41.890","Text":"we\u0027re dealing with the mass of the entire system multiplied by this velocity."},{"Start":"12:41.890 ","End":"12:43.525","Text":"Now what is this velocity?"},{"Start":"12:43.525 ","End":"12:47.410","Text":"Because our momentum doesn\u0027t take into account every little tiny component,"},{"Start":"12:47.410 ","End":"12:50.770","Text":"but rather looks at it as if it\u0027s a point mass."},{"Start":"12:50.770 ","End":"12:54.595","Text":"This is therefore the center of mass."},{"Start":"12:54.595 ","End":"12:56.994","Text":"It\u0027s like we\u0027re dealing with a rigid body"},{"Start":"12:56.994 ","End":"12:59.455","Text":"and we find the center of mass of the rigid body."},{"Start":"12:59.455 ","End":"13:04.690","Text":"This is the velocity of the center of mass."},{"Start":"13:04.690 ","End":"13:07.570","Text":"It\u0027s like considering the whole rigid body as"},{"Start":"13:07.570 ","End":"13:11.275","Text":"just some point-mass with this velocity over here."},{"Start":"13:11.275 ","End":"13:14.829","Text":"This is in fact the velocity in the x-direction of the center of"},{"Start":"13:14.829 ","End":"13:18.625","Text":"mass of the entire system after the collision."},{"Start":"13:18.625 ","End":"13:24.415","Text":"This U_xcm is in fact the velocity in the x-direction of this point over here."},{"Start":"13:24.415 ","End":"13:27.129","Text":"Our center of mass, every single other points in"},{"Start":"13:27.129 ","End":"13:31.900","Text":"the system moves at a different velocity and we\u0027re not going to go into that now."},{"Start":"13:31.900 ","End":"13:34.765","Text":"Now let\u0027s speak about the y-direction."},{"Start":"13:34.765 ","End":"13:37.345","Text":"We\u0027re working out on momentum in the y-direction."},{"Start":"13:37.345 ","End":"13:39.085","Text":"Again before the collision,"},{"Start":"13:39.085 ","End":"13:43.525","Text":"we\u0027re going to have the mass of this sphere colliding with the rod."},{"Start":"13:43.525 ","End":"13:50.350","Text":"It\u0027s going to be mass multiplied by v_0 and it\u0027s going in the positive y-direction."},{"Start":"13:50.350 ","End":"13:53.335","Text":"It\u0027s just mv_0, that\u0027s before the collision."},{"Start":"13:53.335 ","End":"13:56.620","Text":"Of course, this rod in this mass over here,"},{"Start":"13:56.620 ","End":"13:58.779","Text":"I have no y component."},{"Start":"13:58.779 ","End":"14:00.400","Text":"The y component is equal to 0,"},{"Start":"14:00.400 ","End":"14:03.295","Text":"so we don\u0027t have to add this into the equation."},{"Start":"14:03.295 ","End":"14:05.245","Text":"Of course, this rod over here,"},{"Start":"14:05.245 ","End":"14:06.760","Text":"although it has a y component,"},{"Start":"14:06.760 ","End":"14:10.015","Text":"it has 0 velocity, it\u0027s stationary."},{"Start":"14:10.015 ","End":"14:13.330","Text":"It also doesn\u0027t have momentum before the collision."},{"Start":"14:13.330 ","End":"14:15.564","Text":"Then after the collision, again,"},{"Start":"14:15.564 ","End":"14:18.939","Text":"we\u0027re taking the momentum of the entire system,"},{"Start":"14:18.939 ","End":"14:21.385","Text":"which is now a rigid body,"},{"Start":"14:21.385 ","End":"14:24.835","Text":"it\u0027s weight or its mass is 4m."},{"Start":"14:24.835 ","End":"14:30.310","Text":"Its velocity is u in the y-direction though this time."},{"Start":"14:30.310 ","End":"14:34.900","Text":"Of course, they\u0027re both of the center of mass as previously explained."},{"Start":"14:34.900 ","End":"14:37.780","Text":"Now I can isolate out my u."},{"Start":"14:37.780 ","End":"14:45.565","Text":"It\u0027s going to be u in my y-direction of the center of mass is going to be 1/4v_0."},{"Start":"14:45.565 ","End":"14:48.984","Text":"What\u0027s important to notice over here is that"},{"Start":"14:48.984 ","End":"14:53.095","Text":"our center of mass moves at a constant velocity."},{"Start":"14:53.095 ","End":"14:54.759","Text":"It\u0027s a constant velocity."},{"Start":"14:54.759 ","End":"14:56.605","Text":"Now, let\u0027s explain why."},{"Start":"14:56.605 ","End":"15:02.800","Text":"Our center of mass moves only due to external forces."},{"Start":"15:02.800 ","End":"15:07.839","Text":"Our center of mass moves only according to external forces."},{"Start":"15:07.839 ","End":"15:11.379","Text":"Now, because we said that we have 0 external forces,"},{"Start":"15:11.379 ","End":"15:14.079","Text":"okay, that\u0027s how we got to the idea of conservation of momentum."},{"Start":"15:14.079 ","End":"15:15.729","Text":"We had no external forces,"},{"Start":"15:15.729 ","End":"15:18.039","Text":"which means that the sum of the external forces was equal to 0,"},{"Start":"15:18.039 ","End":"15:20.065","Text":"which means conservation of momentum."},{"Start":"15:20.065 ","End":"15:21.970","Text":"Because there were no external forces,"},{"Start":"15:21.970 ","End":"15:25.180","Text":"then we know that our center of mass is therefore going to"},{"Start":"15:25.180 ","End":"15:28.750","Text":"move at either a constant velocity or velocity 0,"},{"Start":"15:28.750 ","End":"15:31.525","Text":"i.e will remain stationary."},{"Start":"15:31.525 ","End":"15:36.040","Text":"We can see that this is our velocity after collision,"},{"Start":"15:36.040 ","End":"15:37.600","Text":"and we can see that it\u0027s constant."},{"Start":"15:37.600 ","End":"15:41.229","Text":"We can see that our velocity after collision is"},{"Start":"15:41.229 ","End":"15:44.799","Text":"moving constantly because we have no external forces."},{"Start":"15:44.799 ","End":"15:46.615","Text":"Now, also before the collision,"},{"Start":"15:46.615 ","End":"15:51.129","Text":"we had no external forces and during the collision we have no external forces."},{"Start":"15:51.129 ","End":"15:55.179","Text":"Which means that the velocity was always constant both before,"},{"Start":"15:55.179 ","End":"15:56.995","Text":"during and after collision,"},{"Start":"15:56.995 ","End":"16:02.290","Text":"and our velocity was the same also before and also after the collision."},{"Start":"16:02.290 ","End":"16:06.849","Text":"Right now, we have our velocity of the center of mass."},{"Start":"16:06.849 ","End":"16:11.185","Text":"But what we\u0027re being asked is our I_cm our position."},{"Start":"16:11.185 ","End":"16:15.024","Text":"All we have to do, how do we get from velocity to position?"},{"Start":"16:15.024 ","End":"16:17.530","Text":"We simply use integration."},{"Start":"16:17.530 ","End":"16:20.395","Text":"Let\u0027s see how we do that."},{"Start":"16:20.395 ","End":"16:25.585","Text":"Instead of our u in the x-direction of the center of mass."},{"Start":"16:25.585 ","End":"16:31.285","Text":"We\u0027ll have our x in the x-direction of the center of mass."},{"Start":"16:31.285 ","End":"16:32.665","Text":"Let\u0027s just make it clear."},{"Start":"16:32.665 ","End":"16:36.295","Text":"We\u0027ll have our x in the center of mass, its position,"},{"Start":"16:36.295 ","End":"16:38.334","Text":"and of course, as a function of time,"},{"Start":"16:38.334 ","End":"16:41.380","Text":"is simply going to be this integrated."},{"Start":"16:41.380 ","End":"16:46.779","Text":"It\u0027s going to be equal to our u in the x-direction of the center of mass multiplied"},{"Start":"16:46.779 ","End":"16:53.060","Text":"by t. That\u0027s simply going to be negative 1/4v_0t."},{"Start":"16:53.340 ","End":"16:57.490","Text":"Then similarly, for our possession for a y center of mass,"},{"Start":"16:57.490 ","End":"16:58.879","Text":"again as a function of t,"},{"Start":"16:58.879 ","End":"17:03.655","Text":"so it\u0027s simply going to be our u in the y-direction of the center of mass,"},{"Start":"17:03.655 ","End":"17:06.385","Text":"this velocity integrated as well."},{"Start":"17:06.385 ","End":"17:08.650","Text":"It\u0027s going to be multiplied by t,"},{"Start":"17:08.650 ","End":"17:12.290","Text":"which is going to be 1/4v_0t."},{"Start":"17:12.690 ","End":"17:17.215","Text":"Now we have our positions for x and y_cm."},{"Start":"17:17.215 ","End":"17:22.945","Text":"That means that we have our I_cm as a function of time when t is bigger than 0."},{"Start":"17:22.945 ","End":"17:25.884","Text":"Now when we look at the position,"},{"Start":"17:25.884 ","End":"17:29.095","Text":"look at what we have when we go back to our diagram."},{"Start":"17:29.095 ","End":"17:35.600","Text":"We can see that our center of mass is simply moving in the diagonal in this direction."},{"Start":"17:35.640 ","End":"17:39.055","Text":"A straight line diagonal."},{"Start":"17:39.055 ","End":"17:43.164","Text":"Now let\u0027s write out our equation in a neater way for"},{"Start":"17:43.164 ","End":"17:47.425","Text":"I center of mass when t is bigger than 0."},{"Start":"17:47.425 ","End":"17:53.090","Text":"That\u0027s going to be, so we have 1/4 of v_0t."},{"Start":"17:53.220 ","End":"18:01.690","Text":"Then we have that going in the negative x-direction and the positive y-direction."},{"Start":"18:01.690 ","End":"18:04.750","Text":"Now this is our final answer,"},{"Start":"18:04.750 ","End":"18:07.239","Text":"because notice what we had to do was we wanted to"},{"Start":"18:07.239 ","End":"18:12.955","Text":"find our position when our t was bigger than 0."},{"Start":"18:12.955 ","End":"18:19.750","Text":"This negative over here represents relative to I_cm at t is equal to 0,"},{"Start":"18:19.750 ","End":"18:21.535","Text":"relative to this over here."},{"Start":"18:21.535 ","End":"18:25.510","Text":"We did that section when we moved our origin on"},{"Start":"18:25.510 ","End":"18:29.830","Text":"the diagram to our center of mass when t is equal to 0."},{"Start":"18:29.830 ","End":"18:35.425","Text":"When we did that, all our calculations afterwards became relative to that point."},{"Start":"18:35.425 ","End":"18:41.379","Text":"Now we have our final position for our center of mass at t is bigger than 0."},{"Start":"18:41.379 ","End":"18:43.645","Text":"Let\u0027s go on to question number 3."},{"Start":"18:43.645 ","End":"18:45.775","Text":"Question number 3 is asking us,"},{"Start":"18:45.775 ","End":"18:47.709","Text":"what is the angular velocity?"},{"Start":"18:47.709 ","End":"18:51.370","Text":"We have our Omega as a function of t of the system in"},{"Start":"18:51.370 ","End":"18:56.860","Text":"circular motion relative to the center of mass calculated in Question 2."},{"Start":"18:56.860 ","End":"19:00.415","Text":"That\u0027s the center of mass that we worked out over here."},{"Start":"19:00.415 ","End":"19:02.769","Text":"We\u0027re trying to find our Omega as a function of"},{"Start":"19:02.769 ","End":"19:07.225","Text":"time relative to our center of mass calculated over here."},{"Start":"19:07.225 ","End":"19:09.174","Text":"Relative to this point over here."},{"Start":"19:09.174 ","End":"19:13.196","Text":"Now, because we\u0027re trying to find our angular velocity,"},{"Start":"19:13.196 ","End":"19:16.224","Text":"so we know that we\u0027re going to have some circular motion."},{"Start":"19:16.224 ","End":"19:19.930","Text":"That means something to do with our angular momentum."},{"Start":"19:19.930 ","End":"19:23.319","Text":"Now we already know that we have conservation of angular momentum,"},{"Start":"19:23.319 ","End":"19:24.895","Text":"as we spoke about before,"},{"Start":"19:24.895 ","End":"19:27.864","Text":"because we saw that we had no external forces,"},{"Start":"19:27.864 ","End":"19:30.835","Text":"which meant that the sum of the external forces was equal to 0."},{"Start":"19:30.835 ","End":"19:36.280","Text":"Then if you try and work out your torque force being that is equal to 0,"},{"Start":"19:36.280 ","End":"19:40.614","Text":"then we see that the sum of our external torques is equal to 0,"},{"Start":"19:40.614 ","End":"19:44.380","Text":"which means that we have conservation of angular momentum."},{"Start":"19:44.380 ","End":"19:47.065","Text":"We have conservation of angular momentum."},{"Start":"19:47.065 ","End":"19:49.000","Text":"Let\u0027s write out our equation for"},{"Start":"19:49.000 ","End":"19:53.050","Text":"angular momentum before the collision and after the collision."},{"Start":"19:53.050 ","End":"19:56.935","Text":"Our L_i is our angular momentum before,"},{"Start":"19:56.935 ","End":"19:58.670","Text":"and that\u0027s equal to."},{"Start":"19:58.670 ","End":"20:00.450","Text":"What do we have to do is,"},{"Start":"20:00.450 ","End":"20:06.309","Text":"we have to work out the angular momentum of each component or each body in the system."},{"Start":"20:06.309 ","End":"20:12.535","Text":"Now, of course, before the collision our 2 rods over here are stationary."},{"Start":"20:12.535 ","End":"20:18.010","Text":"They\u0027re not moving, which means that their angular momentum is going to be equal to 0."},{"Start":"20:18.010 ","End":"20:22.540","Text":"Now we can work out the angular momentum of each 1 of these spheres."},{"Start":"20:22.540 ","End":"20:27.880","Text":"That will be our equation for angular momentum before."},{"Start":"20:27.880 ","End":"20:33.100","Text":"Let\u0027s take a look at this fear for this 1 over here."},{"Start":"20:33.100 ","End":"20:36.309","Text":"The first thing that we have to do is we have to find"},{"Start":"20:36.309 ","End":"20:41.560","Text":"its shortest distance from this point over here,"},{"Start":"20:41.560 ","End":"20:44.245","Text":"from the origin, from my center of mass."},{"Start":"20:44.245 ","End":"20:47.785","Text":"That\u0027s our r effective, if you remember."},{"Start":"20:47.785 ","End":"20:53.124","Text":"We can see that it moves along this line over here,"},{"Start":"20:53.124 ","End":"20:59.425","Text":"which means that it\u0027s r effective is going to be from this point until over here."},{"Start":"20:59.425 ","End":"21:01.779","Text":"This is it\u0027s r effective."},{"Start":"21:01.779 ","End":"21:04.885","Text":"Now, let\u0027s take a look at what this is equal to."},{"Start":"21:04.885 ","End":"21:07.270","Text":"We know that this full length,"},{"Start":"21:07.270 ","End":"21:10.809","Text":"because it\u0027s located at the tip of our rod,"},{"Start":"21:10.809 ","End":"21:13.735","Text":"which is of length L. It\u0027s going to be L,"},{"Start":"21:13.735 ","End":"21:15.969","Text":"the full length minus this over here,"},{"Start":"21:15.969 ","End":"21:18.230","Text":"3 divided by 8L."},{"Start":"21:18.360 ","End":"21:26.455","Text":"Then we\u0027re going to get that our r effective is simply going to be 5 divided by 8L."},{"Start":"21:26.455 ","End":"21:30.505","Text":"That\u0027s how our effective 5 divided by 8L."},{"Start":"21:30.505 ","End":"21:35.739","Text":"Then we have to multiply it by its momentum,"},{"Start":"21:35.739 ","End":"21:39.875","Text":"which is its mass multiplied by the velocity."},{"Start":"21:39.875 ","End":"21:43.770","Text":"Now I\u0027m not rising in my negative sign over here because if you remember,"},{"Start":"21:43.770 ","End":"21:48.750","Text":"the velocity of this sphere was going at v_0 in the negative x-direction."},{"Start":"21:48.750 ","End":"21:50.369","Text":"Now, why am I not doing that?"},{"Start":"21:50.369 ","End":"21:56.770","Text":"Because I\u0027m going to say that the positive direction for rotation is anticlockwise."},{"Start":"21:56.770 ","End":"21:59.890","Text":"Now, this is generally what is accepted."},{"Start":"21:59.890 ","End":"22:04.690","Text":"You can of course define it being clockwise direction. It doesn\u0027t really matter."},{"Start":"22:04.690 ","End":"22:09.430","Text":"But generally, the positive rotational direction is going to be anticlockwise."},{"Start":"22:09.430 ","End":"22:13.465","Text":"If we do that, we can see intuitively"},{"Start":"22:13.465 ","End":"22:17.844","Text":"that this sphere is pushing the whole system in this direction,"},{"Start":"22:17.844 ","End":"22:19.540","Text":"which is the anticlockwise direction,"},{"Start":"22:19.540 ","End":"22:21.820","Text":"which is the positive direction."},{"Start":"22:21.820 ","End":"22:26.769","Text":"But also another way that you can do is if you want to do the cross-product"},{"Start":"22:26.769 ","End":"22:32.434","Text":"between the r effective and your momentum,"},{"Start":"22:32.434 ","End":"22:33.755","Text":"your linear momentum,"},{"Start":"22:33.755 ","End":"22:34.999","Text":"so when you do the cross-product,"},{"Start":"22:34.999 ","End":"22:39.874","Text":"you\u0027ll see that it\u0027s also going in the negative anticlockwise direction,"},{"Start":"22:39.874 ","End":"22:42.754","Text":"sorry, which is the positive rotational direction."},{"Start":"22:42.754 ","End":"22:47.220","Text":"Now let\u0027s work out the angular momentum of this sphere over here."},{"Start":"22:47.220 ","End":"22:52.145","Text":"Again, we can draw its line is over here,"},{"Start":"22:52.145 ","End":"22:53.524","Text":"which means that it hits here,"},{"Start":"22:53.524 ","End":"22:55.295","Text":"which means that it\u0027s r effective,"},{"Start":"22:55.295 ","End":"22:58.055","Text":"is going to be this over here."},{"Start":"22:58.055 ","End":"23:03.380","Text":"Now notice that it\u0027s r effective is also going to be L minus 3/8."},{"Start":"23:03.380 ","End":"23:08.190","Text":"It\u0027s our effective is also going to be the same as this,"},{"Start":"23:08.190 ","End":"23:10.420","Text":"and then its mass is of course the same,"},{"Start":"23:10.420 ","End":"23:12.969","Text":"and it\u0027s velocity is also the same."},{"Start":"23:12.969 ","End":"23:17.160","Text":"The angular momentum before is of the 2 spheres, which is identical,"},{"Start":"23:17.160 ","End":"23:21.545","Text":"so all I have to do is multiply here by 2."},{"Start":"23:21.545 ","End":"23:25.440","Text":"Of course, it\u0027s also going in the positive direction because we can see via"},{"Start":"23:25.440 ","End":"23:30.705","Text":"the cross-product or just intuitively that it\u0027s going in the anticlockwise direction."},{"Start":"23:30.705 ","End":"23:35.765","Text":"What we want to do is we want to find our angular momentum after,"},{"Start":"23:35.765 ","End":"23:38.440","Text":"L_final after the collision."},{"Start":"23:38.440 ","End":"23:46.135","Text":"We have to work out the angular momentum of this whole system as 1 large rigid body."},{"Start":"23:46.135 ","End":"23:48.724","Text":"We have an equation for this and I\u0027m"},{"Start":"23:48.724 ","End":"23:51.550","Text":"going to remind you of it because it\u0027s very important."},{"Start":"23:51.550 ","End":"23:53.150","Text":"This is the equation."},{"Start":"23:53.150 ","End":"23:54.789","Text":"It\u0027s very important."},{"Start":"23:54.789 ","End":"23:56.060","Text":"That\u0027s why it\u0027s in red."},{"Start":"23:56.060 ","End":"24:01.794","Text":"We have that the angular momentum of a large rigid body is equal to our L_cm tag."},{"Start":"24:01.794 ","End":"24:03.660","Text":"Soon we\u0027ll speak a bit more about this,"},{"Start":"24:03.660 ","End":"24:07.359","Text":"but this is the angular momentum relative to the center of"},{"Start":"24:07.359 ","End":"24:12.055","Text":"mass plus our Mr_cm cross our V_cm."},{"Start":"24:12.055 ","End":"24:18.950","Text":"This entire expression is actually the angular momentum of the center of mass itself."},{"Start":"24:18.950 ","End":"24:21.680","Text":"Let\u0027s take a look at what this means"},{"Start":"24:21.680 ","End":"24:24.290","Text":"here and you\u0027ll understand this equation a little bit more."},{"Start":"24:24.290 ","End":"24:27.005","Text":"Let\u0027s first start with this over here."},{"Start":"24:27.005 ","End":"24:31.835","Text":"The angular momentum of the point which is our center of mass."},{"Start":"24:31.835 ","End":"24:36.852","Text":"We\u0027re looking at the entire rigid body as a point mass,"},{"Start":"24:36.852 ","End":"24:41.140","Text":"and then we\u0027re working its angular momentum out."},{"Start":"24:41.140 ","End":"24:44.574","Text":"We have to work out the angular momentum of a point mass,"},{"Start":"24:44.574 ","End":"24:51.436","Text":"and we\u0027re not taking into consideration at all the size or the shape of the object."},{"Start":"24:51.436 ","End":"24:52.975","Text":"Now let\u0027s take a look."},{"Start":"24:52.975 ","End":"24:56.495","Text":"We can see that our center of mass as we saw before these gray dots,"},{"Start":"24:56.495 ","End":"25:02.125","Text":"is moving in some diagonal line away from our origin."},{"Start":"25:02.125 ","End":"25:04.535","Text":"Let\u0027s see how we work this section out."},{"Start":"25:04.535 ","End":"25:11.854","Text":"We can look at our center of mass a split second after our collision,"},{"Start":"25:11.854 ","End":"25:13.470","Text":"so right over here."},{"Start":"25:13.470 ","End":"25:19.685","Text":"Then we can say that our r therefore is going to be equal to 0."},{"Start":"25:19.685 ","End":"25:22.060","Text":"Then this whole thing is equal to 0."},{"Start":"25:22.060 ","End":"25:23.704","Text":"We can either look at it like that."},{"Start":"25:23.704 ","End":"25:27.580","Text":"Alternatively, we can look at it in this way."},{"Start":"25:27.580 ","End":"25:29.994","Text":"We can say that,"},{"Start":"25:29.994 ","End":"25:32.304","Text":"let\u0027s look at our center of mass when it\u0027s"},{"Start":"25:32.304 ","End":"25:36.875","Text":"some place over here a few seconds after the collision."},{"Start":"25:36.875 ","End":"25:40.254","Text":"But then we can say that"},{"Start":"25:40.254 ","End":"25:48.025","Text":"this r_cm vector is in the same direction as our V_cm vector."},{"Start":"25:48.025 ","End":"25:50.934","Text":"They\u0027re both parallel. In that case,"},{"Start":"25:50.934 ","End":"25:57.210","Text":"the cross-product between 2 parallel vectors is also going to be equal to 0."},{"Start":"25:57.210 ","End":"26:00.215","Text":"Either way, this expression will be equal to 0."},{"Start":"26:00.215 ","End":"26:05.174","Text":"Now it\u0027s important to notice that only in this specific case over here,"},{"Start":"26:05.174 ","End":"26:08.705","Text":"this expression over here is equal to 0."},{"Start":"26:08.705 ","End":"26:12.560","Text":"Not always will it be equal to 0 such as even in this example."},{"Start":"26:12.560 ","End":"26:16.314","Text":"If we would have taken our origin to be at the different points,"},{"Start":"26:16.314 ","End":"26:23.265","Text":"then our L of the cm would not be equal to 0."},{"Start":"26:23.265 ","End":"26:27.360","Text":"Let\u0027s talk about this L_cm tag."},{"Start":"26:27.360 ","End":"26:31.689","Text":"This is the angular momentum of the whole body,"},{"Start":"26:31.689 ","End":"26:32.970","Text":"whichever shape it is,"},{"Start":"26:32.970 ","End":"26:34.550","Text":"of this giant rigid body,"},{"Start":"26:34.550 ","End":"26:38.115","Text":"moving relative to this center of mass,"},{"Start":"26:38.115 ","End":"26:40.930","Text":"so moving around this center of mass."},{"Start":"26:40.930 ","End":"26:42.520","Text":"What this in fact is,"},{"Start":"26:42.520 ","End":"26:46.129","Text":"is our internal angular momentum."},{"Start":"26:46.129 ","End":"26:49.584","Text":"How all the particles or the bodies are moving"},{"Start":"26:49.584 ","End":"26:53.300","Text":"within the system around the center of mass."},{"Start":"26:53.300 ","End":"26:56.890","Text":"This expression here was not taking into account anything and"},{"Start":"26:56.890 ","End":"27:01.250","Text":"just the center of mass as a point mass and how it\u0027s moving and it\u0027s angular momentum,"},{"Start":"27:01.250 ","End":"27:04.824","Text":"and this over here is the internal angular momentum."},{"Start":"27:04.824 ","End":"27:08.554","Text":"Everything moving around our point over here,"},{"Start":"27:08.554 ","End":"27:10.270","Text":"which is our center of mass."},{"Start":"27:10.270 ","End":"27:18.419","Text":"We can look at it as if some person is standing over here at our center of mass,"},{"Start":"27:18.419 ","End":"27:20.185","Text":"which is our axis of rotation."},{"Start":"27:20.185 ","End":"27:25.970","Text":"He\u0027s moving with it, and then he sees the whole body moving around him."},{"Start":"27:26.240 ","End":"27:33.115","Text":"That means that our L_final is simply going to be our L_cm tag."},{"Start":"27:33.115 ","End":"27:35.740","Text":"What is this going to be equal to?"},{"Start":"27:35.740 ","End":"27:40.075","Text":"If we go back, it\u0027s going to be equal to our I, our moment of inertia,"},{"Start":"27:40.075 ","End":"27:44.064","Text":"multiplied by Omega, where our Omega is our unknown,"},{"Start":"27:44.064 ","End":"27:45.775","Text":"which we\u0027re trying to find."},{"Start":"27:45.775 ","End":"27:50.560","Text":"All we have to do is we have to find out what our moment of inertia of this body is,"},{"Start":"27:50.560 ","End":"27:53.300","Text":"and then we simply rearrange to find our Omega."},{"Start":"27:53.300 ","End":"27:54.930","Text":"This is the challenge."},{"Start":"27:54.930 ","End":"27:56.304","Text":"What is our moment of inertia?"},{"Start":"27:56.304 ","End":"27:59.630","Text":"As we can see, our body is pretty complex."},{"Start":"27:59.630 ","End":"28:05.035","Text":"Our moment of inertia is going to be made up of many moments of inertia."},{"Start":"28:05.035 ","End":"28:07.855","Text":"We can say that this is our I total,"},{"Start":"28:07.855 ","End":"28:10.100","Text":"the total moment of inertia of the body."},{"Start":"28:10.100 ","End":"28:11.694","Text":"How am I going to do this?"},{"Start":"28:11.694 ","End":"28:16.240","Text":"I\u0027m going to use the trick about the additivity of I."},{"Start":"28:18.170 ","End":"28:21.065","Text":"Notice we\u0027re not adding up all the masses,"},{"Start":"28:21.065 ","End":"28:23.170","Text":"we\u0027re adding up the moments of inertia."},{"Start":"28:23.170 ","End":"28:32.279","Text":"Our total I is going to be I_1 plus I_2 plus I_3 plus I_4,"},{"Start":"28:32.279 ","End":"28:36.020","Text":"the moments of inertia of each body in the system."},{"Start":"28:36.020 ","End":"28:39.640","Text":"Obviously our I_1 and I_2 of the 2 rods will be the same,"},{"Start":"28:39.640 ","End":"28:43.415","Text":"and our I_3 and I_4 of the 2 spheres will also be the same,"},{"Start":"28:43.415 ","End":"28:46.975","Text":"but let\u0027s go through this in an organized fashion."},{"Start":"28:46.975 ","End":"28:50.780","Text":"What are we going to do is we\u0027re going to work out the moment of inertia of each body,"},{"Start":"28:50.780 ","End":"28:54.890","Text":"and of course, it\u0027s going to be about its axis of rotation."},{"Start":"28:54.890 ","End":"28:59.555","Text":"Now our axis of rotation is over here at the center of mass where our origin is."},{"Start":"28:59.555 ","End":"29:01.160","Text":"This is our axis rotation,"},{"Start":"29:01.160 ","End":"29:05.765","Text":"so every moment of inertia is going to be about this point."},{"Start":"29:05.765 ","End":"29:09.590","Text":"Let\u0027s be reminded for our equation for our moment of inertia."},{"Start":"29:09.590 ","End":"29:13.730","Text":"Our I is given by the sum of each mass"},{"Start":"29:13.730 ","End":"29:18.740","Text":"multiplied by its distance squared from the axis of rotation."},{"Start":"29:18.740 ","End":"29:21.899","Text":"This is not the radius of the body,"},{"Start":"29:21.899 ","End":"29:27.680","Text":"but rather distance from axis of rotation."},{"Start":"29:27.680 ","End":"29:30.660","Text":"Let\u0027s begin from our spheres."},{"Start":"29:30.660 ","End":"29:34.509","Text":"Because we aren\u0027t given radiuses for our spheres in the questions,"},{"Start":"29:34.509 ","End":"29:37.305","Text":"we can consider them as a point mass."},{"Start":"29:37.305 ","End":"29:40.630","Text":"Let\u0027s write our moment of inertia for our sphere."},{"Start":"29:40.630 ","End":"29:46.655","Text":"Our I_3 is going to be m multiplied by r^2,"},{"Start":"29:46.655 ","End":"29:49.334","Text":"where again, it\u0027s the distance from the axis."},{"Start":"29:49.334 ","End":"29:52.800","Text":"This is also the moment of inertia for a point mass."},{"Start":"29:52.800 ","End":"29:55.294","Text":"We have to work out what our r is."},{"Start":"29:55.294 ","End":"29:58.180","Text":"Of course this is going to be our r,"},{"Start":"29:58.180 ","End":"30:00.360","Text":"this pink arrow over here,"},{"Start":"30:00.360 ","End":"30:02.470","Text":"and because this is a point mass,"},{"Start":"30:02.470 ","End":"30:06.060","Text":"we don\u0027t have to take into consideration the size of the sphere,"},{"Start":"30:06.060 ","End":"30:08.675","Text":"or of course, the width of the rod."},{"Start":"30:08.675 ","End":"30:10.254","Text":"As far as we\u0027re concerned,"},{"Start":"30:10.254 ","End":"30:14.469","Text":"our sphere is located at exactly the tip,"},{"Start":"30:14.469 ","End":"30:18.335","Text":"which is at L_0."},{"Start":"30:18.335 ","End":"30:20.615","Text":"Now, we\u0027re going to find the distance from the axis of rotation,"},{"Start":"30:20.615 ","End":"30:23.400","Text":"which is our r. We can see this is the horizontal,"},{"Start":"30:23.400 ","End":"30:25.630","Text":"we\u0027re going to use Pythagoras."},{"Start":"30:25.630 ","End":"30:31.315","Text":"We\u0027ll get that our r^2 is equal to this height squared."},{"Start":"30:31.315 ","End":"30:41.494","Text":"That is going to be (3/8L)^2 multiplied by this distance over here squared."},{"Start":"30:41.494 ","End":"30:45.605","Text":"This distance over here squared is going to be L minus our"},{"Start":"30:45.605 ","End":"30:50.465","Text":"3 divided by 8L because the whole length is L minus this section over here,"},{"Start":"30:50.465 ","End":"30:52.730","Text":"which is 3 divided by 8L."},{"Start":"30:52.730 ","End":"30:58.070","Text":"Plus, and that\u0027s going to be 5 divided by 8L,"},{"Start":"30:58.070 ","End":"31:00.520","Text":"simple mathematics, squared."},{"Start":"31:00.520 ","End":"31:10.302","Text":"Then what we\u0027ll get is that our r^2 is equal to 17 divided by 32L^2."},{"Start":"31:10.302 ","End":"31:14.170","Text":"Then we substitute that into here."},{"Start":"31:14.170 ","End":"31:18.954","Text":"I get that our I_3 is equal to 17 divided by"},{"Start":"31:18.954 ","End":"31:26.470","Text":"32 mL^2 and this is of course equal also to I_4,"},{"Start":"31:26.470 ","End":"31:29.185","Text":"the moment of inertia of this sphere over here,"},{"Start":"31:29.185 ","End":"31:33.235","Text":"because they\u0027re symmetrical about our origin."},{"Start":"31:33.235 ","End":"31:35.155","Text":"Now let\u0027s take a look at the rods,"},{"Start":"31:35.155 ","End":"31:37.615","Text":"so let\u0027s first look at this rod over here,"},{"Start":"31:37.615 ","End":"31:40.270","Text":"so let\u0027s call that I_1."},{"Start":"31:40.270 ","End":"31:44.155","Text":"Now the rods are going to be slightly more complicated but not much."},{"Start":"31:44.155 ","End":"31:47.590","Text":"We want to calculate the I of"},{"Start":"31:47.590 ","End":"31:51.790","Text":"the rod and of course we have to take into account our center of mass."},{"Start":"31:51.790 ","End":"31:58.059","Text":"The I about the center of mass of a rod is simply according to a known equation,"},{"Start":"31:58.059 ","End":"32:01.135","Text":"is mL^2 divided by 12."},{"Start":"32:01.135 ","End":"32:03.969","Text":"Now what we want is our moment of inertia of"},{"Start":"32:03.969 ","End":"32:07.240","Text":"the rod however not about its center of mass,"},{"Start":"32:07.240 ","End":"32:10.690","Text":"but around our axis of rotation which is over here."},{"Start":"32:10.690 ","End":"32:12.584","Text":"Now let\u0027s write this again."},{"Start":"32:12.584 ","End":"32:15.519","Text":"We\u0027re going to have I_1 and tag because"},{"Start":"32:15.519 ","End":"32:19.300","Text":"I\u0027m showing that it\u0027s around a different axis of rotation,"},{"Start":"32:19.300 ","End":"32:24.775","Text":"which is further away, and now we have to take into account our Steiner equation."},{"Start":"32:24.775 ","End":"32:31.960","Text":"That\u0027s going to be equal to our I around the center of mass plus our Steiner,"},{"Start":"32:31.960 ","End":"32:36.460","Text":"which is equal to m multiplied by d^2,"},{"Start":"32:36.460 ","End":"32:41.425","Text":"and our d is the distance between the axis squared."},{"Start":"32:41.425 ","End":"32:44.935","Text":"We already know what our I_1cm is. It\u0027s this over here."},{"Start":"32:44.935 ","End":"32:51.055","Text":"Our mass is just the mass of 1 of the rods and now we just have to find our d^2."},{"Start":"32:51.055 ","End":"32:55.989","Text":"Our d^2 is going to be the distance from this point from our center of"},{"Start":"32:55.989 ","End":"33:01.090","Text":"mass of the rod over here until our axis of rotation."},{"Start":"33:01.090 ","End":"33:07.015","Text":"This green arrow over here is our d and we want to work out its length."},{"Start":"33:07.015 ","End":"33:10.569","Text":"This whole length, until half of the rod,"},{"Start":"33:10.569 ","End":"33:12.115","Text":"is L divided by 2,"},{"Start":"33:12.115 ","End":"33:13.705","Text":"because we\u0027re at the center of the rod."},{"Start":"33:13.705 ","End":"33:16.510","Text":"Then we just have to minus this length over here,"},{"Start":"33:16.510 ","End":"33:19.690","Text":"which is 3 divided by 8L."},{"Start":"33:19.690 ","End":"33:22.409","Text":"Let\u0027s work this out."},{"Start":"33:22.409 ","End":"33:28.165","Text":"We will have L divided by 2 minus 3 divided by 8L,"},{"Start":"33:28.165 ","End":"33:33.025","Text":"and this is going to be equal to L divided by 8."},{"Start":"33:33.025 ","End":"33:36.220","Text":"Then we have to take this height,"},{"Start":"33:36.220 ","End":"33:39.910","Text":"so we know that this height is 3/8L and then we\u0027ll"},{"Start":"33:39.910 ","End":"33:43.690","Text":"get that our d^2 is going to be our Pythagoras."},{"Start":"33:43.690 ","End":"33:47.502","Text":"It\u0027s going to be L divided by 8^2,"},{"Start":"33:47.502 ","End":"33:50.965","Text":"so that\u0027s this distance over here,"},{"Start":"33:50.965 ","End":"33:56.015","Text":"plus our 3 divided by 8L^2."},{"Start":"33:56.015 ","End":"33:59.469","Text":"Now I\u0027m going to save time by not working this all out and I\u0027m just going to"},{"Start":"33:59.469 ","End":"34:04.075","Text":"substitute in over here the final answer."},{"Start":"34:04.075 ","End":"34:06.790","Text":"Taking into account our I_1cm,"},{"Start":"34:06.790 ","End":"34:11.155","Text":"which is this over here, and our m multiplied by this our d^2,"},{"Start":"34:11.155 ","End":"34:18.535","Text":"so we\u0027re going to get 23 divided by 96mL^2."},{"Start":"34:18.535 ","End":"34:21.580","Text":"Of course, because we know that our I_1 is equal to"},{"Start":"34:21.580 ","End":"34:25.435","Text":"our I_2 because of the symmetry between the rods,"},{"Start":"34:25.435 ","End":"34:28.347","Text":"this is also equal to I_2."},{"Start":"34:28.347 ","End":"34:31.959","Text":"Now let\u0027s substitute in our equations for I_1,"},{"Start":"34:31.959 ","End":"34:36.055","Text":"2, 3, and 4 in order to find out what our I total is."},{"Start":"34:36.055 ","End":"34:42.100","Text":"We\u0027ll get that our I total is equal to 2 times this,"},{"Start":"34:42.100 ","End":"34:51.609","Text":"so I\u0027ll have 2 times 23 divided by 96mL^2 plus 2 times our I_3,"},{"Start":"34:51.609 ","End":"34:53.230","Text":"so 2 times this,"},{"Start":"34:53.230 ","End":"34:59.830","Text":"plus 2 times 17 divided by 32mL^2."},{"Start":"34:59.830 ","End":"35:02.349","Text":"Then we will get"},{"Start":"35:02.349 ","End":"35:10.750","Text":"37 divided by 24mL^2."},{"Start":"35:10.750 ","End":"35:14.410","Text":"This is our I total and now all we have to do is we have to substitute"},{"Start":"35:14.410 ","End":"35:21.080","Text":"in our I total into this equation over here."},{"Start":"35:22.650 ","End":"35:26.560","Text":"Now we have conservation of angular momentum,"},{"Start":"35:26.560 ","End":"35:30.175","Text":"so that means that our L_i is going to equal our L_f."},{"Start":"35:30.175 ","End":"35:36.865","Text":"Our L_i is going to be 5 divided by 4LmV_0,"},{"Start":"35:36.865 ","End":"35:39.295","Text":"and this is going to be equal to our L_f,"},{"Start":"35:39.295 ","End":"35:40.750","Text":"which is our I total,"},{"Start":"35:40.750 ","End":"35:47.184","Text":"so 37 divided by 24mL^2 multiplied by"},{"Start":"35:47.184 ","End":"35:50.140","Text":"our unknown which is our Omega as a function of"},{"Start":"35:50.140 ","End":"35:53.887","Text":"t. Now all we have to do is isolate out our Omega."},{"Start":"35:53.887 ","End":"35:59.514","Text":"We can cancel out this L with this L and our m\u0027s also cancel out."},{"Start":"35:59.514 ","End":"36:06.054","Text":"Then we just have to isolate out our Omega and we\u0027ll get that it is equal to 30"},{"Start":"36:06.054 ","End":"36:13.105","Text":"divided by 37V_0 divided by L. This is our final answer,"},{"Start":"36:13.105 ","End":"36:14.349","Text":"this is what we were looking for,"},{"Start":"36:14.349 ","End":"36:16.434","Text":"and we can see that again,"},{"Start":"36:16.434 ","End":"36:19.840","Text":"our Omega is equal to a constant."},{"Start":"36:19.840 ","End":"36:22.135","Text":"We have constant angular velocity."},{"Start":"36:22.135 ","End":"36:26.234","Text":"Let\u0027s take a look if this is something which is a logical."},{"Start":"36:26.234 ","End":"36:31.020","Text":"Our whole system is rotating at a constant angular velocity about this point,"},{"Start":"36:31.020 ","End":"36:32.384","Text":"about the center of mass."},{"Start":"36:32.384 ","End":"36:35.669","Text":"Now, the reason that we could do this equation is"},{"Start":"36:35.669 ","End":"36:39.670","Text":"because we were using this idea over here,"},{"Start":"36:39.670 ","End":"36:46.525","Text":"that the sum of all external torque is equal to 0."},{"Start":"36:46.525 ","End":"36:48.235","Text":"Now another equation for it,"},{"Start":"36:48.235 ","End":"36:57.890","Text":"the sum of all of the external torques is equal to I multiplied by d Omega by dt."},{"Start":"36:58.050 ","End":"37:01.974","Text":"In order for this to equal 0,"},{"Start":"37:01.974 ","End":"37:03.939","Text":"which is what we wanted over here,"},{"Start":"37:03.939 ","End":"37:07.458","Text":"that means that our d Omega by dt must be equal 0."},{"Start":"37:07.458 ","End":"37:12.159","Text":"Our differentiation of our Omega has to be equal to"},{"Start":"37:12.159 ","End":"37:17.500","Text":"0 and that only happens if our Omega is a constant."},{"Start":"37:17.500 ","End":"37:20.229","Text":"Because when you differentiate a constant,"},{"Start":"37:20.229 ","End":"37:22.495","Text":"then you get 0."},{"Start":"37:22.495 ","End":"37:25.045","Text":"We got that this is logical,"},{"Start":"37:25.045 ","End":"37:30.205","Text":"our answer over here for question number 3 is logical."},{"Start":"37:30.205 ","End":"37:34.359","Text":"Now onto our fourth and final question and we\u0027re being"},{"Start":"37:34.359 ","End":"37:38.605","Text":"told to find the position vector r(t) of the origin,"},{"Start":"37:38.605 ","End":"37:40.300","Text":"so this point over here,"},{"Start":"37:40.300 ","End":"37:42.820","Text":"relative to the origin\u0027s position at t=0."},{"Start":"37:42.820 ","End":"37:48.669","Text":"As we saw, our origin at t=0 is over here."},{"Start":"37:48.669 ","End":"37:54.625","Text":"And then we have to find the function of its position as a function of time."},{"Start":"37:54.625 ","End":"38:00.280","Text":"In this question we\u0027re trying to find the position of our origin,"},{"Start":"38:00.280 ","End":"38:02.170","Text":"this point over here,"},{"Start":"38:02.170 ","End":"38:04.525","Text":"as a function of time."},{"Start":"38:04.525 ","End":"38:10.299","Text":"This equation is going to be made up of 2 parts."},{"Start":"38:10.299 ","End":"38:13.704","Text":"It\u0027s going to be the position of the origin"},{"Start":"38:13.704 ","End":"38:18.055","Text":"relative to our center of mass in circular motion,"},{"Start":"38:18.055 ","End":"38:24.535","Text":"because this point is spinning around with the rest of the body around our center of mass"},{"Start":"38:24.535 ","End":"38:27.789","Text":"and the second piece of motion that is going into"},{"Start":"38:27.789 ","End":"38:32.245","Text":"here is that our center of mass is also moving in this diagonal."},{"Start":"38:32.245 ","End":"38:37.855","Text":"This point aside from rotating is also going to be moving in the diagonal."},{"Start":"38:37.855 ","End":"38:39.834","Text":"An important fact to remember,"},{"Start":"38:39.834 ","End":"38:42.385","Text":"and something which will allow us to solve this,"},{"Start":"38:42.385 ","End":"38:47.650","Text":"is to know that every single particle inside a rigid body,"},{"Start":"38:47.650 ","End":"38:51.415","Text":"when the rigid body is moving in circular motion,"},{"Start":"38:51.415 ","End":"38:55.584","Text":"every single point in that rigid body is moving with"},{"Start":"38:55.584 ","End":"39:00.409","Text":"circular motion with the same angular velocity."},{"Start":"39:00.409 ","End":"39:03.010","Text":"This is our new arbitrary position."},{"Start":"39:03.010 ","End":"39:07.645","Text":"We can see our center of mass was located here at the origin,"},{"Start":"39:07.645 ","End":"39:10.630","Text":"and these gray dots represent the motion"},{"Start":"39:10.630 ","End":"39:14.005","Text":"of the center of mass in this diagonal straight line,"},{"Start":"39:14.005 ","End":"39:16.374","Text":"because it\u0027s our axis of rotation."},{"Start":"39:16.374 ","End":"39:18.685","Text":"Then we can see over here,"},{"Start":"39:18.685 ","End":"39:24.700","Text":"we have this point which is our origin where 2 rods met at origin."},{"Start":"39:24.700 ","End":"39:27.715","Text":"What we want to do is we want to find"},{"Start":"39:27.715 ","End":"39:31.359","Text":"this position vector from this origin over here in blue,"},{"Start":"39:31.359 ","End":"39:38.410","Text":"the original origin, until it\u0027s position over here during motion."},{"Start":"39:38.410 ","End":"39:40.240","Text":"Let\u0027s draw the vector."},{"Start":"39:40.240 ","End":"39:46.219","Text":"This is our vector for the position vector of our origin and now,"},{"Start":"39:46.219 ","End":"39:49.614","Text":"we want to try and describe the motion."},{"Start":"39:49.614 ","End":"39:52.660","Text":"Now it\u0027s going to be very difficult to do this,"},{"Start":"39:52.660 ","End":"39:57.055","Text":"because it\u0027s moving both diagonally and it\u0027s rotating."},{"Start":"39:57.055 ","End":"40:00.249","Text":"The way that we\u0027re going to describe this vector is as"},{"Start":"40:00.249 ","End":"40:04.984","Text":"a product or the sum of a few other vectors."},{"Start":"40:04.984 ","End":"40:08.294","Text":"The first vector that we\u0027re going to use is this,"},{"Start":"40:08.294 ","End":"40:12.150","Text":"from the origin until our center of mass and we\u0027re"},{"Start":"40:12.150 ","End":"40:16.140","Text":"going to call that vector our r center of mass."},{"Start":"40:16.140 ","End":"40:18.334","Text":"The position of the center of mass."},{"Start":"40:18.334 ","End":"40:22.809","Text":"Then the next vector that we\u0027re going to define is a vector that goes from"},{"Start":"40:22.809 ","End":"40:27.565","Text":"the center of mass until our origin which is moving,"},{"Start":"40:27.565 ","End":"40:31.524","Text":"and that\u0027s going to be called r_o tag."},{"Start":"40:31.524 ","End":"40:34.719","Text":"Why are we defining all of these vectors?"},{"Start":"40:34.719 ","End":"40:38.815","Text":"We\u0027re going to use the idea of relativistic motion,"},{"Start":"40:38.815 ","End":"40:41.724","Text":"so not relativistic as in the speed of light,"},{"Start":"40:41.724 ","End":"40:44.800","Text":"but motion relative to other points,"},{"Start":"40:44.800 ","End":"40:49.765","Text":"so we\u0027re going to find 1 vector relative to other positions."},{"Start":"40:49.765 ","End":"40:57.414","Text":"We say that our movement of the origin is equal to the vector position"},{"Start":"40:57.414 ","End":"41:05.019","Text":"of our center of mass plus the movement of our origin relative to our center of mass."},{"Start":"41:05.019 ","End":"41:07.735","Text":"Let\u0027s write that out as an equation."},{"Start":"41:07.735 ","End":"41:11.949","Text":"We have our r_o which is a vector,"},{"Start":"41:11.949 ","End":"41:13.345","Text":"and this is what we\u0027re trying to find,"},{"Start":"41:13.345 ","End":"41:17.380","Text":"so that\u0027s equal to our r center of mass."},{"Start":"41:17.380 ","End":"41:21.385","Text":"This we found in question number 1 so soon we\u0027ll add it,"},{"Start":"41:21.385 ","End":"41:25.750","Text":"plus our r_0 tag."},{"Start":"41:25.750 ","End":"41:29.109","Text":"Let\u0027s find out what our r_0 tag is equal to."},{"Start":"41:29.109 ","End":"41:35.350","Text":"As we can see, our r_0 tag is the movement of our origin or rather,"},{"Start":"41:35.350 ","End":"41:40.554","Text":"where our 2 rods connected when it was connected to the origin."},{"Start":"41:40.554 ","End":"41:45.430","Text":"This point over here, moving relative to our center of mass."},{"Start":"41:45.430 ","End":"41:50.019","Text":"What I need to imagine is that there\u0027s someone standing over here on the center of mass,"},{"Start":"41:50.019 ","End":"41:52.900","Text":"and so he doesn\u0027t see that he\u0027s moving himself."},{"Start":"41:52.900 ","End":"41:55.210","Text":"He\u0027s in the reference frame of the center of"},{"Start":"41:55.210 ","End":"41:58.960","Text":"mass and all he can see is this point over here,"},{"Start":"41:58.960 ","End":"42:01.539","Text":"this is the only point he\u0027s looking at, focusing on,"},{"Start":"42:01.539 ","End":"42:05.409","Text":"this point over here rotating around him."},{"Start":"42:05.409 ","End":"42:08.605","Text":"Now we\u0027re going to use this sentence over here."},{"Start":"42:08.605 ","End":"42:15.460","Text":"Our r_0 tag is always going to be defined as some x movement in"},{"Start":"42:15.460 ","End":"42:22.615","Text":"the x direction plus some movement in our y direction."},{"Start":"42:22.615 ","End":"42:26.575","Text":"Let\u0027s take a look at what our x component is going to be."},{"Start":"42:26.575 ","End":"42:29.199","Text":"It\u0027s always going to be when we\u0027re dealing with circular motion,"},{"Start":"42:29.199 ","End":"42:38.401","Text":"it\u0027s going to be some radius R multiplied by cosine of our Omega t plus some phase Phi."},{"Start":"42:38.401 ","End":"42:40.292","Text":"Our y component,"},{"Start":"42:40.292 ","End":"42:43.060","Text":"so our y tag is going to be the same."},{"Start":"42:43.060 ","End":"42:45.505","Text":"Our radius away from the axis of rotation,"},{"Start":"42:45.505 ","End":"42:52.795","Text":"but this time multiplied by sine of our angular velocity multiplied by t plus our phase."},{"Start":"42:52.795 ","End":"42:54.624","Text":"What is our radius?"},{"Start":"42:54.624 ","End":"42:56.544","Text":"Let\u0027s take a look at what this is."},{"Start":"42:56.544 ","End":"43:01.570","Text":"Our radius is this horizontal line over here in pink."},{"Start":"43:01.570 ","End":"43:04.210","Text":"In order to find this, we have to use Pythagoras."},{"Start":"43:04.210 ","End":"43:07.299","Text":"We\u0027re going to take this side squared plus this side"},{"Start":"43:07.299 ","End":"43:11.139","Text":"squared is going to be our radius squared."},{"Start":"43:11.139 ","End":"43:16.110","Text":"If you remember, each side was of length 3 divided by 8L."},{"Start":"43:16.110 ","End":"43:22.470","Text":"So 3 divided by 8L all of this squared plus,"},{"Start":"43:22.470 ","End":"43:28.555","Text":"because it\u0027s the square section so this height is also 3 divided by 8L,"},{"Start":"43:28.555 ","End":"43:30.999","Text":"and of course, squared."},{"Start":"43:30.999 ","End":"43:38.515","Text":"Then we\u0027ll get that our R^2 is 9 divided by 32L^2."},{"Start":"43:38.515 ","End":"43:40.150","Text":"Now we have our R^2,"},{"Start":"43:40.150 ","End":"43:43.330","Text":"so all we have to do is square root this and we\u0027ll have our R. We"},{"Start":"43:43.330 ","End":"43:46.180","Text":"have our Omega from question number 3 that we worked out,"},{"Start":"43:46.180 ","End":"43:49.840","Text":"and now all we have to do is find out what this Phi is."},{"Start":"43:49.840 ","End":"43:51.564","Text":"Let\u0027s see how we work this out."},{"Start":"43:51.564 ","End":"43:54.830","Text":"If we\u0027re going to look at some circle,"},{"Start":"43:54.830 ","End":"43:57.945","Text":"and this is r origin,"},{"Start":"43:57.945 ","End":"43:59.400","Text":"the center of the circle."},{"Start":"43:59.400 ","End":"44:04.365","Text":"When our body starts somewhere over here at the rightmost section,"},{"Start":"44:04.365 ","End":"44:07.239","Text":"so we can see if its angle is 0,"},{"Start":"44:07.239 ","End":"44:11.245","Text":"then its phase is also going to be equal to 0."},{"Start":"44:11.245 ","End":"44:13.255","Text":"Let\u0027s plug this in."},{"Start":"44:13.255 ","End":"44:15.549","Text":"If our phase is equal to 0,"},{"Start":"44:15.549 ","End":"44:18.475","Text":"then we\u0027ll get over here t is equal to 0,"},{"Start":"44:18.475 ","End":"44:22.344","Text":"that our y value is equal to 0, so that\u0027s perfect,"},{"Start":"44:22.344 ","End":"44:26.260","Text":"and our x value is going to be equal to R,"},{"Start":"44:26.260 ","End":"44:28.299","Text":"so that make sense."},{"Start":"44:28.299 ","End":"44:31.660","Text":"We just saw what happens when our angle is 0."},{"Start":"44:31.660 ","End":"44:36.145","Text":"However, in our starting position if we go back to what it looks like,"},{"Start":"44:36.145 ","End":"44:39.520","Text":"our body is starting somewhere over here."},{"Start":"44:39.520 ","End":"44:41.845","Text":"Looking at our original position,"},{"Start":"44:41.845 ","End":"44:45.460","Text":"this is what our system looks like at t is equal to 0."},{"Start":"44:45.460 ","End":"44:47.890","Text":"We can see that our r_0 tag,"},{"Start":"44:47.890 ","End":"44:49.524","Text":"it\u0027s meant to go up until the origin,"},{"Start":"44:49.524 ","End":"44:52.179","Text":"is this diagonal over here."},{"Start":"44:52.179 ","End":"44:56.560","Text":"That means that our angle is going to be all of this,"},{"Start":"44:56.560 ","End":"44:59.290","Text":"this is our Phi, this is our phase."},{"Start":"44:59.290 ","End":"45:02.620","Text":"Now, how do we find what this angle is?"},{"Start":"45:02.620 ","End":"45:06.865","Text":"Because we know that this point here is our center of mass,"},{"Start":"45:06.865 ","End":"45:10.434","Text":"and we worked out that this length over here"},{"Start":"45:10.434 ","End":"45:14.169","Text":"was 3/8L and this length over here was also 3/8L,"},{"Start":"45:14.169 ","End":"45:17.140","Text":"which means that here we have a square."},{"Start":"45:17.140 ","End":"45:22.165","Text":"A perfect square we\u0027re going from 1 corner diagonally to the bottom,"},{"Start":"45:22.165 ","End":"45:25.465","Text":"the other corner across the entire shape."},{"Start":"45:25.465 ","End":"45:30.445","Text":"In a square, we know that this angle over here is going to be"},{"Start":"45:30.445 ","End":"45:38.950","Text":"45 degrees because it\u0027s the bisection of a 90-degree angle right in the center."},{"Start":"45:38.950 ","End":"45:44.440","Text":"Then we know that this angle over here is 180."},{"Start":"45:44.440 ","End":"45:48.940","Text":"Therefore, we can say that our Phi angle is equal"},{"Start":"45:48.940 ","End":"45:53.365","Text":"to our 180 degrees which is this semicircle over here,"},{"Start":"45:53.365 ","End":"45:59.320","Text":"plus our bisected angle over here which was 45 degrees,"},{"Start":"45:59.320 ","End":"46:05.679","Text":"and then we\u0027re going to get that it is equal to 225 degrees."},{"Start":"46:05.679 ","End":"46:08.110","Text":"Now we have our value for R,"},{"Start":"46:08.110 ","End":"46:09.925","Text":"we just have to square root this."},{"Start":"46:09.925 ","End":"46:11.320","Text":"We have our value for Omega,"},{"Start":"46:11.320 ","End":"46:13.510","Text":"and we have our value for Phi."},{"Start":"46:13.510 ","End":"46:17.155","Text":"We have our x and y values over here,"},{"Start":"46:17.155 ","End":"46:24.370","Text":"and now all we need to do is we have to substitute in what we get for our r_0 tag,"},{"Start":"46:24.370 ","End":"46:28.435","Text":"and what we got for our r_cm in 1 of the previous questions."},{"Start":"46:28.435 ","End":"46:29.859","Text":"Now let\u0027s just convert,"},{"Start":"46:29.859 ","End":"46:34.195","Text":"because it\u0027s better to work in radians in these types of questions."},{"Start":"46:34.195 ","End":"46:42.640","Text":"Our 225 degrees is the same as 5Pi divided by 4 in radians."},{"Start":"46:42.640 ","End":"46:44.590","Text":"Let\u0027s substitute everything in."},{"Start":"46:44.590 ","End":"46:49.030","Text":"What I did was I took the square root for R,"},{"Start":"46:49.030 ","End":"46:50.889","Text":"I wrote down our Omega from"},{"Start":"46:50.889 ","End":"46:56.245","Text":"our previous question and our r_cm also from our previous question,"},{"Start":"46:56.245 ","End":"47:01.840","Text":"and I substituted everything back in into this top equation over here,"},{"Start":"47:01.840 ","End":"47:06.650","Text":"and I got my equation for my r_0 vector."}],"ID":9476},{"Watched":false,"Name":"Float In Harmonic Motion","Duration":"34m 2s","ChapterTopicVideoID":9207,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.879","Text":"Hello, we\u0027re given this question."},{"Start":"00:02.879 ","End":"00:08.699","Text":"Now, let\u0027s not go over that but let\u0027s rather look at our diagram."},{"Start":"00:08.699 ","End":"00:11.895","Text":"We have some sphere of mass, m,"},{"Start":"00:11.895 ","End":"00:16.575","Text":"which is attached to a spring of spring constant, k. Now,"},{"Start":"00:16.575 ","End":"00:21.419","Text":"up until this line over here we have some liquid with density,"},{"Start":"00:21.419 ","End":"00:26.350","Text":"Rho, and our mass floats on the surface."},{"Start":"00:26.350 ","End":"00:34.380","Text":"Now, our y-axis is pointing upwards and our y=0 at the surface of our liquid."},{"Start":"00:34.510 ","End":"00:38.030","Text":"In our first question, we\u0027re being told that our mass has"},{"Start":"00:38.030 ","End":"00:43.099","Text":"an initial velocity of V_0 going upwards and they\u0027re asking us,"},{"Start":"00:43.099 ","End":"00:46.714","Text":"what is the maximum height that the ball will reach?"},{"Start":"00:46.714 ","End":"00:48.545","Text":"This is our question."},{"Start":"00:48.545 ","End":"00:51.635","Text":"Now, as the ball goes up,"},{"Start":"00:51.635 ","End":"00:55.084","Text":"so let\u0027s imagine it over here,"},{"Start":"00:55.084 ","End":"00:58.820","Text":"it\u0027s still attached to the spring."},{"Start":"00:58.820 ","End":"01:05.600","Text":"We know that at its maximum height its velocity is going to be equal to 0."},{"Start":"01:05.600 ","End":"01:13.550","Text":"Here we have a conservation of energy and this height over here,"},{"Start":"01:13.550 ","End":"01:15.380","Text":"this distance traveled,"},{"Start":"01:15.380 ","End":"01:20.445","Text":"is our h and it also equals to Delta y,"},{"Start":"01:20.445 ","End":"01:25.210","Text":"our change in length of the spring because the spring is now stretched."},{"Start":"01:25.210 ","End":"01:28.400","Text":"Let\u0027s write our equations for energy conservation."},{"Start":"01:28.400 ","End":"01:31.565","Text":"First, I\u0027m going to write my initial energy."},{"Start":"01:31.565 ","End":"01:34.220","Text":"Now, my starting velocity is V_0"},{"Start":"01:34.220 ","End":"01:38.010","Text":"so it\u0027s going to have initial kinetic energy of 1/2mV_0^2."},{"Start":"01:39.670 ","End":"01:45.575","Text":"It\u0027s going to have no potential energy because our y here is equal to 0."},{"Start":"01:45.575 ","End":"01:48.229","Text":"Then our final energy,"},{"Start":"01:48.229 ","End":"01:50.300","Text":"so our Ef is going to be."},{"Start":"01:50.300 ","End":"01:53.600","Text":"So first of all, we\u0027re going to have the force from the spring,"},{"Start":"01:53.600 ","End":"02:00.640","Text":"which is equal to 1/2k multiplied by Delta y^2."},{"Start":"02:00.640 ","End":"02:04.909","Text":"Our kinetic energy, as we know,"},{"Start":"02:04.909 ","End":"02:08.120","Text":"at the maximum height our velocity is equal to 0."},{"Start":"02:08.120 ","End":"02:10.474","Text":"That\u0027s how we know we\u0027ve gotten to maximum height."},{"Start":"02:10.474 ","End":"02:14.209","Text":"Our kinetic energy is going to be equal to 0."},{"Start":"02:14.209 ","End":"02:17.089","Text":"Then we have to add in our potential energy,"},{"Start":"02:17.089 ","End":"02:21.064","Text":"which is going to be mg multiplied by our height,"},{"Start":"02:21.064 ","End":"02:23.339","Text":"which is Delta y."},{"Start":"02:23.339 ","End":"02:26.184","Text":"We\u0027re dealing with conservation of energy,"},{"Start":"02:26.184 ","End":"02:29.875","Text":"which means that our Ei is going to be equal to our Ef."},{"Start":"02:29.875 ","End":"02:31.090","Text":"We can rewrite this,"},{"Start":"02:31.090 ","End":"02:36.310","Text":"our Ei is 1/2 mV_0^2 and that is equal to our Ef,"},{"Start":"02:36.310 ","End":"02:44.079","Text":"which is 1/2 k Delta y^2 plus mg Delta y."},{"Start":"02:44.079 ","End":"02:49.134","Text":"Now all we have to do is isolate out Delta y over here."},{"Start":"02:49.134 ","End":"02:54.124","Text":"What we can do is we can move this term over here to this side of the equation."},{"Start":"02:54.124 ","End":"02:57.435","Text":"Then we have a quadratic equation,"},{"Start":"02:57.435 ","End":"03:01.239","Text":"and solving with a quadratic equation formula we\u0027ll get that our h,"},{"Start":"03:01.239 ","End":"03:03.789","Text":"which is also equal to our Delta y,"},{"Start":"03:03.789 ","End":"03:11.044","Text":"is going to be equal to negative mg plus or minus the square root"},{"Start":"03:11.044 ","End":"03:15.670","Text":"of mg^2"},{"Start":"03:15.670 ","End":"03:21.770","Text":"plus kmV_0^2."},{"Start":"03:21.770 ","End":"03:29.809","Text":"Square root of all that and then divide it by k. Now we can see that we have two answers."},{"Start":"03:29.809 ","End":"03:33.664","Text":"We have our answer using our plus and our answer using our negative."},{"Start":"03:33.664 ","End":"03:36.019","Text":"Now, we have to look at our question and see"},{"Start":"03:36.019 ","End":"03:40.019","Text":"which answer is relevant and which is the correct answer."},{"Start":"03:40.130 ","End":"03:44.419","Text":"We know that our Delta y or our height has to be something"},{"Start":"03:44.419 ","End":"03:48.544","Text":"positive because it\u0027s above the y=0 line."},{"Start":"03:48.544 ","End":"03:52.650","Text":"We need some positive value for our Delta y."},{"Start":"03:52.650 ","End":"03:55.710","Text":"If we look at our square root sign and what we have"},{"Start":"03:55.710 ","End":"03:59.344","Text":"inside we can see that the expression inside the square root sign"},{"Start":"03:59.344 ","End":"04:01.820","Text":"is bigger than mg. Because we have"},{"Start":"04:01.820 ","End":"04:06.140","Text":"mg^2 plus something else squared and then the square root of that,"},{"Start":"04:06.140 ","End":"04:09.079","Text":"which is going to be at least equal to mg."},{"Start":"04:09.079 ","End":"04:17.735","Text":"So if I choose the plus sign then I\u0027ll get some positive answer over here."},{"Start":"04:17.735 ","End":"04:22.309","Text":"If I choose the negative I\u0027ll get a negative answer and that makes no sense."},{"Start":"04:22.309 ","End":"04:30.239","Text":"I\u0027m just going to erase the negative sign over here and put a plus."},{"Start":"04:30.290 ","End":"04:33.389","Text":"This is our answer for Question Number 1"},{"Start":"04:33.389 ","End":"04:37.135","Text":"and now I\u0027m going to move on to Question Number 2."},{"Start":"04:37.135 ","End":"04:41.045","Text":"In Question Number 2, we\u0027re told that our ball,"},{"Start":"04:41.045 ","End":"04:42.814","Text":"from being at its maximum value,"},{"Start":"04:42.814 ","End":"04:45.214","Text":"has now fallen back down."},{"Start":"04:45.214 ","End":"04:49.384","Text":"We can not look at this arrow and it has,"},{"Start":"04:49.384 ","End":"04:52.414","Text":"therefore, some downwards velocity."},{"Start":"04:52.414 ","End":"04:55.624","Text":"Now, because we already saw that we have conservation of energy,"},{"Start":"04:55.624 ","End":"04:57.570","Text":"that means that it\u0027s downward velocity,"},{"Start":"04:57.570 ","End":"05:00.710","Text":"its velocity now is going to be the same velocity with"},{"Start":"05:00.710 ","End":"05:04.610","Text":"which it was going when it was moving upwards."},{"Start":"05:04.610 ","End":"05:07.400","Text":"Its downwards velocity is V_0."},{"Start":"05:07.400 ","End":"05:12.080","Text":"Then we\u0027re told that our ball enters into the liquid."},{"Start":"05:12.080 ","End":"05:16.624","Text":"Now we have to analyze its motion within this liquid."},{"Start":"05:16.624 ","End":"05:21.799","Text":"Now, in this question we\u0027re being asked to consider our ball as some point-mass."},{"Start":"05:21.799 ","End":"05:25.340","Text":"As soon as one section of the ball has touched the water then"},{"Start":"05:25.340 ","End":"05:28.675","Text":"we can consider the whole entire ball to be under water."},{"Start":"05:28.675 ","End":"05:31.130","Text":"Because otherwise the question gets a little bit"},{"Start":"05:31.130 ","End":"05:34.385","Text":"more complicated if we have to also take into account"},{"Start":"05:34.385 ","End":"05:43.730","Text":"the motion of the ball as it itself enters the water in layers or stages."},{"Start":"05:43.730 ","End":"05:45.994","Text":"When the ball is in the water,"},{"Start":"05:45.994 ","End":"05:49.370","Text":"we\u0027re told that there are two forces working on it."},{"Start":"05:49.370 ","End":"05:51.650","Text":"Let\u0027s take a look at what they are."},{"Start":"05:51.650 ","End":"05:53.330","Text":"We have our fv,"},{"Start":"05:53.330 ","End":"05:57.820","Text":"which is our Stokes force,"},{"Start":"05:57.820 ","End":"06:08.745","Text":"and is equal to negative 6 Pi Eta Ry dot."},{"Start":"06:08.745 ","End":"06:13.235","Text":"Now, we can see that all of this is a constant."},{"Start":"06:13.235 ","End":"06:15.995","Text":"Our Eta is some constant value,"},{"Start":"06:15.995 ","End":"06:17.705","Text":"our radius is the radius of the ball,"},{"Start":"06:17.705 ","End":"06:19.150","Text":"which is unchanging,"},{"Start":"06:19.150 ","End":"06:23.869","Text":"and then our y dot is obviously the velocity in the y-direction."},{"Start":"06:23.869 ","End":"06:28.160","Text":"Instead of writing it like this let\u0027s shorten it and"},{"Start":"06:28.160 ","End":"06:32.374","Text":"say that it\u0027s equal to negative and then instead of 6 Pi Eta R,"},{"Start":"06:32.374 ","End":"06:37.919","Text":"we\u0027ll write Lambda and then our y dot over here."},{"Start":"06:38.680 ","End":"06:41.705","Text":"Something we can notice from our fv,"},{"Start":"06:41.705 ","End":"06:44.030","Text":"which is our Stokes force,"},{"Start":"06:44.030 ","End":"06:48.609","Text":"is that it\u0027s acting in the opposite direction to our velocity."},{"Start":"06:48.609 ","End":"06:51.469","Text":"If our ball is let\u0027s say inside the water,"},{"Start":"06:51.469 ","End":"06:53.381","Text":"so let\u0027s draw it over here."},{"Start":"06:53.381 ","End":"06:54.839","Text":"Here\u0027s our line,"},{"Start":"06:54.839 ","End":"06:56.359","Text":"so our ball is inside the water."},{"Start":"06:56.359 ","End":"07:00.229","Text":"If our ball is moving in the downwards direction then"},{"Start":"07:00.229 ","End":"07:05.009","Text":"our fv is in the negative direction to its velocity."},{"Start":"07:05.009 ","End":"07:09.259","Text":"Our fv will be in this direction and vice versa."},{"Start":"07:09.259 ","End":"07:12.185","Text":"If our ball was moving up in this direction"},{"Start":"07:12.185 ","End":"07:15.559","Text":"then our fv will be going in the opposite direction,"},{"Start":"07:15.559 ","End":"07:18.369","Text":"in the downwards direction."},{"Start":"07:18.369 ","End":"07:20.139","Text":"That\u0027s Number 1,"},{"Start":"07:20.139 ","End":"07:25.704","Text":"so we can see it\u0027s acting in the negative direction to the velocity and aside from that,"},{"Start":"07:25.704 ","End":"07:28.225","Text":"because it\u0027s acting in the negative direction of the velocity,"},{"Start":"07:28.225 ","End":"07:31.735","Text":"we can see that it\u0027s constantly changing direction."},{"Start":"07:31.735 ","End":"07:34.494","Text":"One time the force will be pointing downwards"},{"Start":"07:34.494 ","End":"07:37.315","Text":"and the next time it will be pointing upwards."},{"Start":"07:37.315 ","End":"07:41.334","Text":"It\u0027s a little bit difficult to substitute this into a force equation,"},{"Start":"07:41.334 ","End":"07:44.080","Text":"but we\u0027re going to speak about it in a second."},{"Start":"07:44.080 ","End":"07:48.625","Text":"Now a little bit of information on what the Stokes force is,"},{"Start":"07:48.625 ","End":"07:52.045","Text":"so it\u0027s just our friction from dealing with water."},{"Start":"07:52.045 ","End":"07:54.219","Text":"If you can imagine yourself putting"},{"Start":"07:54.219 ","End":"07:57.115","Text":"your hand into a body of water and moving your hand around,"},{"Start":"07:57.115 ","End":"08:02.005","Text":"you\u0027ll feel some force which is pushing against your hand,"},{"Start":"08:02.005 ","End":"08:03.445","Text":"so that\u0027s just this."},{"Start":"08:03.445 ","End":"08:06.670","Text":"Now, we also have it aside from in water or liquid here,"},{"Start":"08:06.670 ","End":"08:09.544","Text":"we can also have it in air."},{"Start":"08:09.544 ","End":"08:12.430","Text":"If we have it in the air then Lambda over here,"},{"Start":"08:12.430 ","End":"08:14.184","Text":"our constant over here,"},{"Start":"08:14.184 ","End":"08:18.925","Text":"will just represent a different set of constant over here."},{"Start":"08:18.925 ","End":"08:21.985","Text":"This is the first force that we have,"},{"Start":"08:21.985 ","End":"08:25.910","Text":"and then the second force that we have is buoyancy."},{"Start":"08:26.130 ","End":"08:30.295","Text":"We can see our F_b, b representing buoyancy."},{"Start":"08:30.295 ","End":"08:33.295","Text":"This is our buoyancy force and it equals Rho,"},{"Start":"08:33.295 ","End":"08:37.870","Text":"which is the density of our liquid here."},{"Start":"08:37.870 ","End":"08:40.360","Text":"V, this is our volume,"},{"Start":"08:40.360 ","End":"08:43.300","Text":"so the volume of our sphere."},{"Start":"08:43.300 ","End":"08:46.149","Text":"Not to be confused with our velocity."},{"Start":"08:46.149 ","End":"08:50.630","Text":"This is a capital V and g our gravitational force."},{"Start":"08:51.090 ","End":"08:54.610","Text":"Again, our Rho is the density of our fluid,"},{"Start":"08:54.610 ","End":"08:58.555","Text":"our V is the volume of our shape,"},{"Start":"08:58.555 ","End":"09:00.468","Text":"here of our sphere,"},{"Start":"09:00.468 ","End":"09:03.145","Text":"and our g is gravitational force."},{"Start":"09:03.145 ","End":"09:05.470","Text":"What we can see is that it\u0027s positive,"},{"Start":"09:05.470 ","End":"09:08.815","Text":"so it\u0027s always acting in the upwards direction."},{"Start":"09:08.815 ","End":"09:13.179","Text":"It\u0027s always trying to push whatever you put in upwards and"},{"Start":"09:13.179 ","End":"09:18.175","Text":"another thing is that it\u0027s a constant."},{"Start":"09:18.175 ","End":"09:20.380","Text":"Our density of the fluid,"},{"Start":"09:20.380 ","End":"09:21.879","Text":"if we have a uniform fluid,"},{"Start":"09:21.879 ","End":"09:23.514","Text":"is a constant,"},{"Start":"09:23.514 ","End":"09:28.674","Text":"our volume of the shape that\u0027s coming in is also a constant and our gravity is also."},{"Start":"09:28.674 ","End":"09:31.464","Text":"Now, the way that you can look at this is that it\u0027s"},{"Start":"09:31.464 ","End":"09:36.490","Text":"almost the opposite of our mg. Just like"},{"Start":"09:36.490 ","End":"09:38.410","Text":"your mg is pulling the pull"},{"Start":"09:38.410 ","End":"09:44.501","Text":"downwards with a force of mg pointing in the downwards direction."},{"Start":"09:44.501 ","End":"09:47.260","Text":"So this you can look at as something like the opposite."},{"Start":"09:47.260 ","End":"09:53.725","Text":"It\u0027s some force which is always pushing the shape upwards with a force of Rho V_g."},{"Start":"09:53.725 ","End":"09:57.339","Text":"What it does is it just reduces the force of"},{"Start":"09:57.339 ","End":"10:05.725","Text":"our mg. Now let\u0027s take a look at what Question Number 2 is asking us."},{"Start":"10:05.725 ","End":"10:11.125","Text":"It is, what is the mass\u0027 equation of motion when traveling through the liquid?"},{"Start":"10:11.125 ","End":"10:14.994","Text":"Assume that from the moment that the mass comes into contact with the liquid,"},{"Start":"10:14.994 ","End":"10:17.170","Text":"the whole sphere is submerged."},{"Start":"10:17.170 ","End":"10:19.930","Text":"Assume also that the face of the surface of"},{"Start":"10:19.930 ","End":"10:24.009","Text":"the liquid is unchanged by the entering sphere."},{"Start":"10:24.009 ","End":"10:29.810","Text":"Hint for how to answer this for ease, use variable substitution."},{"Start":"10:30.330 ","End":"10:35.320","Text":"Question 2 is asking us for our equation of motion."},{"Start":"10:35.320 ","End":"10:40.705","Text":"Again, we\u0027re taking it from when the ball is inside the liquid,"},{"Start":"10:40.705 ","End":"10:42.235","Text":"so I\u0027ll rub out all this."},{"Start":"10:42.235 ","End":"10:44.770","Text":"When the ball is inside the liquid and we\u0027re trying to"},{"Start":"10:44.770 ","End":"10:47.770","Text":"find its equation of motion there when we are taking into"},{"Start":"10:47.770 ","End":"10:51.819","Text":"account that it\u0027s a point mass that just enters the liquid and we don\u0027t have to work out"},{"Start":"10:51.819 ","End":"10:56.769","Text":"the stages of the whole sphere entering step-by-step,"},{"Start":"10:56.769 ","End":"10:59.410","Text":"different volume of sphere inside the liquid."},{"Start":"10:59.410 ","End":"11:02.364","Text":"We don\u0027t have to do that."},{"Start":"11:02.364 ","End":"11:08.604","Text":"The surface of the liquid is unchanged from the entering sphere. Let\u0027s take a look."},{"Start":"11:08.604 ","End":"11:10.180","Text":"We\u0027re trying to find this equation of motion."},{"Start":"11:10.180 ","End":"11:11.649","Text":"Now, what is an equation of motion?"},{"Start":"11:11.649 ","End":"11:16.120","Text":"An equation of motion is something that involves our position,"},{"Start":"11:16.120 ","End":"11:17.904","Text":"so here we\u0027re changing in the y direction."},{"Start":"11:17.904 ","End":"11:21.310","Text":"So our position is going to be some variable of y,"},{"Start":"11:21.310 ","End":"11:24.835","Text":"our velocity and acceleration."},{"Start":"11:24.835 ","End":"11:30.115","Text":"Some equation which links 2 or more of these variables."},{"Start":"11:30.115 ","End":"11:34.435","Text":"For instance, an equation would be for offspring,"},{"Start":"11:34.435 ","End":"11:38.080","Text":"so negative ky. Because again,"},{"Start":"11:38.080 ","End":"11:39.670","Text":"we\u0027re working on the y-axis,"},{"Start":"11:39.670 ","End":"11:43.720","Text":"is equal to mass times our acceleration."},{"Start":"11:43.720 ","End":"11:45.550","Text":"Here we have our position,"},{"Start":"11:45.550 ","End":"11:47.861","Text":"and here we have our acceleration."},{"Start":"11:47.861 ","End":"11:54.705","Text":"Of course, our acceleration is equal to the second derivative of our y of our position."},{"Start":"11:54.705 ","End":"11:58.745","Text":"Then we have a differential equation which we can solve."},{"Start":"11:58.745 ","End":"12:03.520","Text":"Anytime we have some equation linking our position or"},{"Start":"12:03.520 ","End":"12:08.050","Text":"velocity or acceleration and also possibly t,"},{"Start":"12:08.050 ","End":"12:12.610","Text":"our time variable, then we can call that an equation of motion."},{"Start":"12:12.610 ","End":"12:16.869","Text":"The general way that we get to this type of equation, an equation of motion,"},{"Start":"12:16.869 ","End":"12:20.335","Text":"is by drawing a free-fall diagram of all of our forces,"},{"Start":"12:20.335 ","End":"12:25.345","Text":"writing an equation for the sum of all of our forces and then playing around with it."},{"Start":"12:25.345 ","End":"12:29.240","Text":"Let\u0027s see how we do this in our example over here."},{"Start":"12:29.550 ","End":"12:34.569","Text":"Let\u0027s look over here where our sphere is inside the liquid."},{"Start":"12:34.569 ","End":"12:36.175","Text":"The forces that we have acting are,"},{"Start":"12:36.175 ","End":"12:39.550","Text":"we have our mg pointing downwards."},{"Start":"12:39.550 ","End":"12:42.189","Text":"Then we said that we have our buoyancy force,"},{"Start":"12:42.189 ","End":"12:44.320","Text":"which is always pointing upwards."},{"Start":"12:44.320 ","End":"12:50.065","Text":"That\u0027s our F_b and now we need to speak about our Stokes."},{"Start":"12:50.065 ","End":"12:53.289","Text":"As we said, our Stokes is constantly changing"},{"Start":"12:53.289 ","End":"12:56.935","Text":"direction depending on the direction of the velocity."},{"Start":"12:56.935 ","End":"13:02.229","Text":"What we generally do is we say that it\u0027s pointing upwards in"},{"Start":"13:02.229 ","End":"13:07.615","Text":"the positive direction and when it\u0027s pointing in the positive direction,"},{"Start":"13:07.615 ","End":"13:10.390","Text":"then as we write out our equation,"},{"Start":"13:10.390 ","End":"13:13.959","Text":"the sign will sort itself out."},{"Start":"13:13.959 ","End":"13:16.660","Text":"Now, the last force that we have,"},{"Start":"13:16.660 ","End":"13:19.179","Text":"because we have a spring in the equation,"},{"Start":"13:19.179 ","End":"13:22.225","Text":"we also have our spring force."},{"Start":"13:22.225 ","End":"13:26.860","Text":"That\u0027s going to be negative kx."},{"Start":"13:26.860 ","End":"13:32.110","Text":"Again, our spring force is equal to negative kx and so remember there\u0027s a negative,"},{"Start":"13:32.110 ","End":"13:35.050","Text":"sorry, here it will be y because we\u0027re working on the y-axis."},{"Start":"13:35.050 ","End":"13:39.010","Text":"Again, it\u0027s going to be pointing upwards in the positive y-direction."},{"Start":"13:39.010 ","End":"13:43.780","Text":"Then it will sort itself out as we write our equation. We\u0027ll see."},{"Start":"13:43.780 ","End":"13:46.030","Text":"Whenever you have some force,"},{"Start":"13:46.030 ","End":"13:50.358","Text":"which is changing direction depending on the movement."},{"Start":"13:50.358 ","End":"13:53.515","Text":"So such as your Stokes force or your spring force,"},{"Start":"13:53.515 ","End":"13:57.160","Text":"you always have the arrow representing the direction of"},{"Start":"13:57.160 ","End":"14:01.915","Text":"the force pointing in the positive direction of your axis."},{"Start":"14:01.915 ","End":"14:05.620","Text":"Now, what we\u0027re going to do is we\u0027re going to write the sum of all of"},{"Start":"14:05.620 ","End":"14:11.395","Text":"our forces and only in the y direction because that\u0027s all that interests us right now."},{"Start":"14:11.395 ","End":"14:13.990","Text":"Let\u0027s look at our negative ky,"},{"Start":"14:13.990 ","End":"14:16.225","Text":"so the force from my spring."},{"Start":"14:16.225 ","End":"14:21.055","Text":"We said it\u0027s negative ky pointing in this upwards positive direction."},{"Start":"14:21.055 ","End":"14:22.989","Text":"Let\u0027s see if this makes sense."},{"Start":"14:22.989 ","End":"14:25.935","Text":"If our spring has been shortened."},{"Start":"14:25.935 ","End":"14:29.590","Text":"So we\u0027re right now below our line of y=0,"},{"Start":"14:29.590 ","End":"14:32.934","Text":"then we can see that negative k multiplied by"},{"Start":"14:32.934 ","End":"14:39.625","Text":"some negative value for our y is going to be positive k multiplied by y."},{"Start":"14:39.625 ","End":"14:41.465","Text":"It\u0027s going to be a pushing force,"},{"Start":"14:41.465 ","End":"14:43.700","Text":"which is true if the spring is compressed,"},{"Start":"14:43.700 ","End":"14:47.150","Text":"it\u0027s going to be pushing our ball backup."},{"Start":"14:47.150 ","End":"14:52.040","Text":"Alternatively, if our ball is above our line of y=0,"},{"Start":"14:52.040 ","End":"14:54.695","Text":"so we have some positive value for y,"},{"Start":"14:54.695 ","End":"14:57.879","Text":"then we\u0027re going to have negative k multiplied by our y."},{"Start":"14:57.879 ","End":"15:00.319","Text":"So we\u0027ll have a negative expression over here,"},{"Start":"15:00.319 ","End":"15:05.554","Text":"which means that when our ball is above our y=0,"},{"Start":"15:05.554 ","End":"15:11.895","Text":"the spring is going to try and pull it back to the point where the spring is slack."},{"Start":"15:11.895 ","End":"15:14.960","Text":"That makes sense. When it\u0027s a positive,"},{"Start":"15:14.960 ","End":"15:18.440","Text":"it\u0027s a pushing force and when this is a negative, it\u0027s a pulling force."},{"Start":"15:18.440 ","End":"15:19.950","Text":"That makes total sense and,"},{"Start":"15:19.950 ","End":"15:21.634","Text":"of course, at y=0,"},{"Start":"15:21.634 ","End":"15:27.490","Text":"then our spring force is equal to 0 because it\u0027s slack. That\u0027s perfect."},{"Start":"15:27.490 ","End":"15:35.219","Text":"We\u0027ll have negative ky then we have our F_b in the positive y direction,"},{"Start":"15:35.219 ","End":"15:38.520","Text":"so plus buoyancy force."},{"Start":"15:38.520 ","End":"15:43.095","Text":"Then we\u0027re going to have our mg going in the negative y-direction."},{"Start":"15:43.095 ","End":"15:50.550","Text":"It\u0027s going to be negative mg. Now let\u0027s look at our fv, our stokes force."},{"Start":"15:50.550 ","End":"15:57.030","Text":"We can write it as negative Lambda y dot."},{"Start":"15:57.030 ","End":"15:59.805","Text":"Here similarly, with our negative,"},{"Start":"15:59.805 ","End":"16:03.825","Text":"it\u0027s going to work just like our spring force works."},{"Start":"16:03.825 ","End":"16:07.380","Text":"It\u0027s going to just work itself out."},{"Start":"16:07.380 ","End":"16:12.960","Text":"Just like now in our velocity is pointing in the downwards direction,"},{"Start":"16:12.960 ","End":"16:17.280","Text":"we can see that our stokes force is going to be pushing upwards."},{"Start":"16:17.280 ","End":"16:20.040","Text":"It\u0027s going to be negative to the downwards direction."},{"Start":"16:20.040 ","End":"16:21.945","Text":"Negative, negative and then positive."},{"Start":"16:21.945 ","End":"16:27.850","Text":"It will just work itself out depending on the direction of our velocity."},{"Start":"16:27.950 ","End":"16:30.240","Text":"That\u0027s all of our forces."},{"Start":"16:30.240 ","End":"16:34.950","Text":"This is, of course, meant to be equal to our mass multiplied by acceleration."},{"Start":"16:34.950 ","End":"16:37.943","Text":"I\u0027ll write acceleration as y double dot,"},{"Start":"16:37.943 ","End":"16:40.840","Text":"the second derivative of our y."},{"Start":"16:40.990 ","End":"16:44.825","Text":"Now we\u0027re left with this differential equation."},{"Start":"16:44.825 ","End":"16:49.605","Text":"This is in fact, our equation of motion because we have our position,"},{"Start":"16:49.605 ","End":"16:52.515","Text":"we have our velocity, and we have our acceleration."},{"Start":"16:52.515 ","End":"16:58.020","Text":"Now, let\u0027s just tidy it up a little bit. What does that mean?"},{"Start":"16:58.020 ","End":"17:03.630","Text":"Now before I go into how we\u0027re going to tidy this up a little bit, a quick thing,"},{"Start":"17:03.630 ","End":"17:07.965","Text":"usually you would have remembered that your force for your spring would be"},{"Start":"17:07.965 ","End":"17:12.855","Text":"negative k multiplied by y minus y0 or x minus x0."},{"Start":"17:12.855 ","End":"17:16.289","Text":"Now the reason I didn\u0027t have to do that by adding"},{"Start":"17:16.289 ","End":"17:21.720","Text":"my y0 is because my y0 is where my spring is slack."},{"Start":"17:21.720 ","End":"17:26.110","Text":"Now specifically here my spring is slack at y is equal to 0."},{"Start":"17:26.150 ","End":"17:29.400","Text":"Having this where this is equal to 0,"},{"Start":"17:29.400 ","End":"17:31.379","Text":"if I open up my brackets,"},{"Start":"17:31.379 ","End":"17:35.170","Text":"I\u0027ll just be left with negative ky like here."},{"Start":"17:35.360 ","End":"17:39.089","Text":"Back to tidying up and sorting out this equation."},{"Start":"17:39.089 ","End":"17:40.815","Text":"Now, because we have a spring,"},{"Start":"17:40.815 ","End":"17:42.944","Text":"it\u0027s already a hint at that,"},{"Start":"17:42.944 ","End":"17:45.345","Text":"we\u0027re going to be dealing with harmonic motion."},{"Start":"17:45.345 ","End":"17:48.270","Text":"We have our negative ky here, harmonic motion,"},{"Start":"17:48.270 ","End":"17:51.960","Text":"and then we also have our negative lambda y dot,"},{"Start":"17:51.960 ","End":"17:53.460","Text":"which is our stokes."},{"Start":"17:53.460 ","End":"17:59.505","Text":"We said that this was the friction that is in different fluids, different liquids."},{"Start":"17:59.505 ","End":"18:02.340","Text":"We can already see that we\u0027re going to have something to"},{"Start":"18:02.340 ","End":"18:05.985","Text":"do with harmonic motion and damping."},{"Start":"18:05.985 ","End":"18:08.805","Text":"Because we have 2 forces which are changing"},{"Start":"18:08.805 ","End":"18:12.360","Text":"direction depending on where we are in the motion,"},{"Start":"18:12.360 ","End":"18:16.035","Text":"and so we\u0027re going to have harmonic motion with damping."},{"Start":"18:16.035 ","End":"18:19.440","Text":"We\u0027ll notice that our differential equation over"},{"Start":"18:19.440 ","End":"18:24.015","Text":"here is of the same type of differential equation as this."},{"Start":"18:24.015 ","End":"18:27.524","Text":"We have our y double dot, x double dot,"},{"Start":"18:27.524 ","End":"18:30.960","Text":"some constant multiplied by x"},{"Start":"18:30.960 ","End":"18:34.725","Text":"dot plus another constant squared multiplied by x is equal to 0."},{"Start":"18:34.725 ","End":"18:39.989","Text":"All we have to do is rearrange this equation in order to get it into this format,"},{"Start":"18:39.989 ","End":"18:45.000","Text":"and then from there we can solve or we know how to solve this differential equation."},{"Start":"18:45.000 ","End":"18:48.044","Text":"Also, this differential equation should be written in"},{"Start":"18:48.044 ","End":"18:53.040","Text":"your notes that you can bring in with you to the exam."},{"Start":"18:53.040 ","End":"18:55.484","Text":"Let\u0027s see how to sort this out."},{"Start":"18:55.484 ","End":"18:59.415","Text":"Now the first thing we can see is that our variable with the double dots,"},{"Start":"18:59.415 ","End":"19:04.770","Text":"our second derivative has to have a coefficient of 1 but here,"},{"Start":"19:04.770 ","End":"19:10.269","Text":"we see that we have a coefficient of m. Let\u0027s begin by sorting everything out."},{"Start":"19:10.310 ","End":"19:15.090","Text":"I\u0027m going to do a trick which I did in the unit dealing with harmonic motion."},{"Start":"19:15.090 ","End":"19:23.130","Text":"That is to look at this entire expression up until my y dot,"},{"Start":"19:23.130 ","End":"19:26.444","Text":"and to take my k out as a common factor."},{"Start":"19:26.444 ","End":"19:31.139","Text":"I\u0027ll have negative k and then I\u0027ll open my brackets."},{"Start":"19:31.139 ","End":"19:37.320","Text":"I\u0027ll have a y and then I\u0027ll have negative my F_b plus"},{"Start":"19:37.320 ","End":"19:44.565","Text":"my mg divided by k. Close the brackets."},{"Start":"19:44.565 ","End":"19:52.834","Text":"Then I\u0027ll carry on with my negative lambda y dot is equal to my my double dot."},{"Start":"19:52.834 ","End":"19:54.320","Text":"Now if you open the brackets,"},{"Start":"19:54.320 ","End":"19:57.940","Text":"you\u0027ll see that you\u0027ll get this exact expression over here."},{"Start":"19:57.940 ","End":"20:06.345","Text":"Now what I\u0027m going to do is I\u0027m going to call this expression over here my y0."},{"Start":"20:06.345 ","End":"20:08.670","Text":"Now, what is our y0?"},{"Start":"20:08.670 ","End":"20:11.800","Text":"It\u0027s a point of equilibrium."},{"Start":"20:14.210 ","End":"20:17.204","Text":"A point of equilibrium. What does that mean?"},{"Start":"20:17.204 ","End":"20:21.015","Text":"It means that the sum of all of the forces at that point,"},{"Start":"20:21.015 ","End":"20:22.845","Text":"are equal to 0."},{"Start":"20:22.845 ","End":"20:24.975","Text":"At our y is equal to 0,"},{"Start":"20:24.975 ","End":"20:27.119","Text":"that\u0027s not our point of equilibrium."},{"Start":"20:27.119 ","End":"20:30.810","Text":"Our point of equilibrium is some distance below."},{"Start":"20:30.810 ","End":"20:33.465","Text":"Somewhere over here is our y0."},{"Start":"20:33.465 ","End":"20:38.010","Text":"That means that if we place our sphere over here,"},{"Start":"20:38.010 ","End":"20:40.319","Text":"then it will not move."},{"Start":"20:40.319 ","End":"20:41.969","Text":"Its forces are balanced,"},{"Start":"20:41.969 ","End":"20:45.405","Text":"which means that it will not move up or down."},{"Start":"20:45.405 ","End":"20:51.030","Text":"Now what I want to do is I want to get this equation over here into this format,"},{"Start":"20:51.030 ","End":"20:54.105","Text":"the format of the equation in my blue."},{"Start":"20:54.105 ","End":"20:57.165","Text":"What I can see over here is that my y0,"},{"Start":"20:57.165 ","End":"21:01.150","Text":"my points of equilibrium is a constant."},{"Start":"21:01.340 ","End":"21:04.860","Text":"What I\u0027m going to do is I\u0027m going to move"},{"Start":"21:04.860 ","End":"21:09.330","Text":"my point of the origin to be located at this constant."},{"Start":"21:09.330 ","End":"21:13.275","Text":"I\u0027m going to shift my origin over here and then I can get rid of this."},{"Start":"21:13.275 ","End":"21:14.639","Text":"How am I going to do that?"},{"Start":"21:14.639 ","End":"21:18.824","Text":"I\u0027m going to do that via variable substitution."},{"Start":"21:18.824 ","End":"21:21.839","Text":"I\u0027m going to name some variable, let\u0027s say Z,"},{"Start":"21:21.839 ","End":"21:26.910","Text":"and I\u0027m going to say that that is equal to what is inside here."},{"Start":"21:26.910 ","End":"21:32.280","Text":"My y minus my F_b plus mg"},{"Start":"21:32.280 ","End":"21:38.310","Text":"divided by k. Now I have my Z variable,"},{"Start":"21:38.310 ","End":"21:41.535","Text":"but what about my y dot and my y double dot?"},{"Start":"21:41.535 ","End":"21:46.283","Text":"We can see that if I take the derivative of my Z, so Z dot."},{"Start":"21:46.283 ","End":"21:48.494","Text":"Taking the derivative of this side,"},{"Start":"21:48.494 ","End":"21:52.739","Text":"we already saw that my F_b plus mg divided by k is a constant,"},{"Start":"21:52.739 ","End":"21:54.840","Text":"so the derivative, it will disappear,"},{"Start":"21:54.840 ","End":"21:57.510","Text":"and we\u0027ll be left with y dot."},{"Start":"21:57.510 ","End":"21:59.370","Text":"Also with my y double dot,"},{"Start":"21:59.370 ","End":"22:03.015","Text":"if I take the second derivative of my Z double dots,"},{"Start":"22:03.015 ","End":"22:05.489","Text":"I\u0027m going to take the second derivative of my y."},{"Start":"22:05.489 ","End":"22:08.580","Text":"So I\u0027ll have also is equal to y double dot."},{"Start":"22:08.580 ","End":"22:12.584","Text":"Now I can substitute all of this into this equation."},{"Start":"22:12.584 ","End":"22:14.625","Text":"I\u0027m going to have negative k,"},{"Start":"22:14.625 ","End":"22:17.430","Text":"then all of this inside the brackets is equal to Z,"},{"Start":"22:17.430 ","End":"22:20.835","Text":"and then negative Lambda, my y dot,"},{"Start":"22:20.835 ","End":"22:23.489","Text":"I can see is equal to my Z dot,"},{"Start":"22:23.489 ","End":"22:27.090","Text":"and then all of that is equal to my my double dot,"},{"Start":"22:27.090 ","End":"22:31.935","Text":"which is equal to mZ double dot."},{"Start":"22:31.935 ","End":"22:37.410","Text":"What we can see is that we\u0027re getting pretty close to this equation in the blue."},{"Start":"22:37.410 ","End":"22:41.009","Text":"Now all I want to do is I want to rearrange this equation"},{"Start":"22:41.009 ","End":"22:45.179","Text":"so that everything is on the same side of the equation."},{"Start":"22:45.179 ","End":"22:47.939","Text":"I\u0027ll have mZ double dot,"},{"Start":"22:47.939 ","End":"22:58.020","Text":"and I\u0027ll add to both sides my kZ and my Lambda Z. I\u0027ll have plus kZ plus Lambda Z,"},{"Start":"22:58.020 ","End":"23:02.370","Text":"dot is going to be equal to 0."},{"Start":"23:02.370 ","End":"23:06.190","Text":"Now, I want my z dots in order,"},{"Start":"23:06.190 ","End":"23:08.080","Text":"so I have my z double dot, then my z dot,"},{"Start":"23:08.080 ","End":"23:10.944","Text":"and then my z so I\u0027ll just rearrange this."},{"Start":"23:10.944 ","End":"23:12.864","Text":"Now, as I said before,"},{"Start":"23:12.864 ","End":"23:18.280","Text":"I need my coefficient of my variable double-dot to be 1."},{"Start":"23:18.280 ","End":"23:22.555","Text":"Here I have to divide by m to get rid of this coefficient."},{"Start":"23:22.555 ","End":"23:33.174","Text":"I\u0027ll have my z double dot plus my Lambda divided by m z dot plus Omega k divided by mz,"},{"Start":"23:33.174 ","End":"23:34.999","Text":"and that will equal to 0."},{"Start":"23:34.999 ","End":"23:39.475","Text":"So I divided everything by m. Now,"},{"Start":"23:39.475 ","End":"23:42.565","Text":"we\u0027ve really got into an equation which is identical to this."},{"Start":"23:42.565 ","End":"23:47.575","Text":"We can say that my Lambda divided by m is equal to"},{"Start":"23:47.575 ","End":"23:53.184","Text":"this Gamma over here and then my k divided by m is equal to my coefficient over here,"},{"Start":"23:53.184 ","End":"23:56.965","Text":"which is my Omega 0 squared."},{"Start":"23:56.965 ","End":"24:00.265","Text":"What I have over here for my Gamma is"},{"Start":"24:00.265 ","End":"24:04.450","Text":"some coefficient to do with my damping force and over here,"},{"Start":"24:04.450 ","End":"24:06.340","Text":"my k divided by m is"},{"Start":"24:06.340 ","End":"24:13.570","Text":"my oscillating frequency of the system itself, so it\u0027s self-frequency."},{"Start":"24:13.570 ","End":"24:17.215","Text":"This is our answer to Question Number 2."},{"Start":"24:17.215 ","End":"24:19.194","Text":"We needed an equation of motion,"},{"Start":"24:19.194 ","End":"24:20.469","Text":"and here we have it,"},{"Start":"24:20.469 ","End":"24:23.169","Text":"and we\u0027ve used variable substitution,"},{"Start":"24:23.169 ","End":"24:25.900","Text":"which is what they asked from us in the question."},{"Start":"24:25.900 ","End":"24:28.690","Text":"Now we\u0027re going on to Question Number 3."},{"Start":"24:28.690 ","End":"24:30.700","Text":"Assuming weak damping, what is"},{"Start":"24:30.700 ","End":"24:34.735","Text":"the general solution for the equation of motion inside the liquid?"},{"Start":"24:34.735 ","End":"24:36.759","Text":"What we\u0027re going to do is we\u0027re going to solve"},{"Start":"24:36.759 ","End":"24:39.925","Text":"the differential equation that we got from Question Number 2."},{"Start":"24:39.925 ","End":"24:41.530","Text":"Then we\u0027re also being asked,"},{"Start":"24:41.530 ","End":"24:46.270","Text":"what are the initial conditions for the motion and that our final answers must be"},{"Start":"24:46.270 ","End":"24:51.290","Text":"presented in terms of the variable that was used before variable substitution was used."},{"Start":"24:51.290 ","End":"24:53.170","Text":"Instead of using our variables z,"},{"Start":"24:53.170 ","End":"24:57.444","Text":"we have to work or back to our variable being in terms of y."},{"Start":"24:57.444 ","End":"25:00.055","Text":"Hint, when solving a differential equation,"},{"Start":"25:00.055 ","End":"25:02.630","Text":"use the equation sheets."},{"Start":"25:03.030 ","End":"25:07.644","Text":"Question Number 3; there are actually 3 parts to the question."},{"Start":"25:07.644 ","End":"25:11.229","Text":"One, we\u0027re going to find the general solution to our equation of motion,"},{"Start":"25:11.229 ","End":"25:13.689","Text":"so that means solving this differential equation."},{"Start":"25:13.689 ","End":"25:18.370","Text":"The second part is to write that solution in terms of our original variable,"},{"Start":"25:18.370 ","End":"25:21.295","Text":"which was y, not in terms of z."},{"Start":"25:21.295 ","End":"25:27.730","Text":"The third part is asking us what are the initial conditions of the system?"},{"Start":"25:27.730 ","End":"25:30.280","Text":"Now in the question they hinted for solving for"},{"Start":"25:30.280 ","End":"25:34.120","Text":"the general solution to look in our equation notes,"},{"Start":"25:34.120 ","End":"25:36.440","Text":"so let\u0027s take a look at that."},{"Start":"25:36.720 ","End":"25:38.830","Text":"In our equation sheets,"},{"Start":"25:38.830 ","End":"25:42.939","Text":"will see that for a differential equation of this type,"},{"Start":"25:42.939 ","End":"25:45.685","Text":"its general solution is simply this."},{"Start":"25:45.685 ","End":"25:47.980","Text":"Now, of course, here we\u0027re working in terms of x,"},{"Start":"25:47.980 ","End":"25:50.920","Text":"and in fact we have to write it in terms of z."},{"Start":"25:50.920 ","End":"25:54.520","Text":"So this is our general solution where our Gamma over"},{"Start":"25:54.520 ","End":"25:59.619","Text":"here we defined over here as our Lambda divided by m and our Omega 0"},{"Start":"25:59.619 ","End":"26:05.469","Text":"squared we defined as this our k divided by m. Here we"},{"Start":"26:05.469 ","End":"26:07.659","Text":"have an Omega so notice it\u0027s not an Omega"},{"Start":"26:07.659 ","End":"26:11.701","Text":"0 and our Omega is given to us here in this equation."},{"Start":"26:11.701 ","End":"26:15.350","Text":"Here, our Omega 0 squared is incorporated."},{"Start":"26:16.080 ","End":"26:20.365","Text":"This is our general solution to our equation of motion,"},{"Start":"26:20.365 ","End":"26:22.974","Text":"so that\u0027s it, nothing to it."},{"Start":"26:22.974 ","End":"26:25.344","Text":"All I did was from this equation,"},{"Start":"26:25.344 ","End":"26:28.044","Text":"I just instead of having it as x as a function of t,"},{"Start":"26:28.044 ","End":"26:31.420","Text":"I wrote it as z as a function of t. Of course,"},{"Start":"26:31.420 ","End":"26:32.619","Text":"we know what our Gamma is,"},{"Start":"26:32.619 ","End":"26:36.475","Text":"what our Omega and Omega 0 is equal to."},{"Start":"26:36.475 ","End":"26:39.745","Text":"Now on to the second part of Question Number 3."},{"Start":"26:39.745 ","End":"26:43.750","Text":"Write this in terms of our original variable, which was y."},{"Start":"26:43.750 ","End":"26:45.595","Text":"Now as we remember over here,"},{"Start":"26:45.595 ","End":"26:50.215","Text":"we said that our z was y minus y0 over here,"},{"Start":"26:50.215 ","End":"26:52.960","Text":"so let\u0027s write this out."},{"Start":"26:52.960 ","End":"27:02.920","Text":"We said that our z as a function of t was equal to our y as a function of t minus our y0."},{"Start":"27:02.920 ","End":"27:06.519","Text":"Now all I have to do is isolate out my y as a function of"},{"Start":"27:06.519 ","End":"27:09.594","Text":"t so I\u0027ll get that my y as a function of t"},{"Start":"27:09.594 ","End":"27:17.570","Text":"is equal to my z as a function of t plus my y0."},{"Start":"27:18.540 ","End":"27:22.540","Text":"Now all I have to do in order to write out"},{"Start":"27:22.540 ","End":"27:27.024","Text":"my general solution in terms of my original variable is just to write this."},{"Start":"27:27.024 ","End":"27:31.704","Text":"So I have that my y as a function of t is equal to my z as a function of t,"},{"Start":"27:31.704 ","End":"27:38.860","Text":"which is my Ae negative Gamma divided by 2 multiplied by t,"},{"Start":"27:38.860 ","End":"27:44.514","Text":"multiplied by my cosine of my Omega t plus Phi and then"},{"Start":"27:44.514 ","End":"27:50.785","Text":"plus my y0 where my y0 is this over here."},{"Start":"27:50.785 ","End":"27:57.235","Text":"Now, of course, we can substitute in all our unknowns over here all of these,"},{"Start":"27:57.235 ","End":"28:00.100","Text":"so my Lambda, my Omega,"},{"Start":"28:00.100 ","End":"28:03.205","Text":"my y0, everything can go in."},{"Start":"28:03.205 ","End":"28:06.265","Text":"Now the only two things that I don\u0027t know yet,"},{"Start":"28:06.265 ","End":"28:11.620","Text":"what my A is, my amplitude is and what my Phi, my phase is."},{"Start":"28:11.620 ","End":"28:16.630","Text":"Now, our third and final part of this question is,"},{"Start":"28:16.630 ","End":"28:19.449","Text":"what are the initial conditions?"},{"Start":"28:19.449 ","End":"28:22.825","Text":"So let\u0027s go back to our diagram."},{"Start":"28:22.825 ","End":"28:26.724","Text":"We can see here that our initial velocity,"},{"Start":"28:26.724 ","End":"28:30.669","Text":"because our ball is going inside the liquid,"},{"Start":"28:30.669 ","End":"28:35.330","Text":"into the fluid, so we know that its velocity is v_0."},{"Start":"28:36.780 ","End":"28:40.602","Text":"We said our initial velocity."},{"Start":"28:40.602 ","End":"28:44.095","Text":"So that\u0027s y dots at t = 0,"},{"Start":"28:44.095 ","End":"28:46.177","Text":"is going to be equal to our V_0."},{"Start":"28:46.177 ","End":"28:48.490","Text":"Remember, let\u0027s take a look again,"},{"Start":"28:48.490 ","End":"28:51.669","Text":"that our arrow was pointing in the downwards direction,"},{"Start":"28:51.669 ","End":"28:53.604","Text":"which is the negative y-direction,"},{"Start":"28:53.604 ","End":"28:55.420","Text":"so it\u0027s going to be negative V_0."},{"Start":"28:55.420 ","End":"28:57.639","Text":"Then in initial conditions,"},{"Start":"28:57.639 ","End":"29:00.490","Text":"we\u0027re asking also for the initial velocity,"},{"Start":"29:00.490 ","End":"29:02.545","Text":"but also the initial position,"},{"Start":"29:02.545 ","End":"29:05.245","Text":"so that\u0027s our y at t = 0,"},{"Start":"29:05.245 ","End":"29:06.640","Text":"so let\u0027s take a look."},{"Start":"29:06.640 ","End":"29:12.040","Text":"We\u0027re looking at rights when the ball hits the surface of the fluid."},{"Start":"29:12.040 ","End":"29:16.765","Text":"The surface of the fluid is located at y = 0."},{"Start":"29:16.765 ","End":"29:19.480","Text":"Notice it\u0027s not at z = 0,"},{"Start":"29:19.480 ","End":"29:21.980","Text":"but y = 0."},{"Start":"29:22.740 ","End":"29:25.839","Text":"That answered what our initial conditions,"},{"Start":"29:25.839 ","End":"29:29.140","Text":"so we have our initial velocity and our initial position."},{"Start":"29:29.140 ","End":"29:32.049","Text":"Now in order to find what my A and my Phi is,"},{"Start":"29:32.049 ","End":"29:35.935","Text":"I\u0027m going to need to use my initial conditions."},{"Start":"29:35.935 ","End":"29:41.080","Text":"In order to fight, I would substitute into y at t = 0,"},{"Start":"29:41.080 ","End":"29:47.409","Text":"so I would say that this is equal to 0 and set my t here and here, t equals 0."},{"Start":"29:47.409 ","End":"29:49.539","Text":"Then for my velocity,"},{"Start":"29:49.539 ","End":"29:54.430","Text":"I would take the derivative of this equation because I need"},{"Start":"29:54.430 ","End":"30:01.330","Text":"my y dot and then again set my t = 0 and set the entire equation equal to negative V_0."},{"Start":"30:01.330 ","End":"30:03.745","Text":"Now I\u0027m not going to solve this because"},{"Start":"30:03.745 ","End":"30:06.985","Text":"the question didn\u0027t ask us to find what our A and what our Phi was."},{"Start":"30:06.985 ","End":"30:08.469","Text":"However, if it did ask,"},{"Start":"30:08.469 ","End":"30:11.919","Text":"that is what we would have to do to find these 2 unknowns,"},{"Start":"30:11.919 ","End":"30:14.710","Text":"we have to use our initial conditions."},{"Start":"30:14.710 ","End":"30:17.590","Text":"Now we\u0027re going on to Question Number 4."},{"Start":"30:17.590 ","End":"30:19.749","Text":"Once the mass has entered the liquid,"},{"Start":"30:19.749 ","End":"30:23.643","Text":"how long will it take the mass to return to the surface?,"},{"Start":"30:23.643 ","End":"30:26.800","Text":"The state described at the start of 2."},{"Start":"30:26.800 ","End":"30:32.500","Text":"Here in 2 a sphere is traveling through the liquid,"},{"Start":"30:32.500 ","End":"30:38.440","Text":"and now we want it to return to the top of the liquid."},{"Start":"30:38.440 ","End":"30:40.210","Text":"Looking at our diagram,"},{"Start":"30:40.210 ","End":"30:45.744","Text":"we\u0027re being asked to find when our sphere is going to return to this line over here;"},{"Start":"30:45.744 ","End":"30:48.025","Text":"of y is equal to 0."},{"Start":"30:48.025 ","End":"30:51.264","Text":"We\u0027ve already got our equation of motion."},{"Start":"30:51.264 ","End":"30:56.380","Text":"We have our general solution where we had that our y of t was equal to"},{"Start":"30:56.380 ","End":"31:02.869","Text":"our Ae to the power of what we found over here,"},{"Start":"31:03.090 ","End":"31:06.695","Text":"so that\u0027s that equation."},{"Start":"31:06.695 ","End":"31:09.490","Text":"Now what we want to do is we want to"},{"Start":"31:09.490 ","End":"31:13.420","Text":"see when a sphere returns to the line of y is equal to 0."},{"Start":"31:13.420 ","End":"31:16.810","Text":"That means that we\u0027ll set our function for our y"},{"Start":"31:16.810 ","End":"31:20.515","Text":"is a function of t = 0 because we want our y equals 0."},{"Start":"31:20.515 ","End":"31:23.800","Text":"Then what we should get is two answers,"},{"Start":"31:23.800 ","End":"31:28.149","Text":"so we\u0027ll get 1 answer where a mass went out of"},{"Start":"31:28.149 ","End":"31:33.059","Text":"the water or starting when it was at the water surface or the liquid surface."},{"Start":"31:33.059 ","End":"31:36.190","Text":"Then our second answer will be after the mass has gone"},{"Start":"31:36.190 ","End":"31:40.090","Text":"down into the liquid and has resurfaced,"},{"Start":"31:40.090 ","End":"31:42.560","Text":"so let\u0027s see how we do that."},{"Start":"31:42.660 ","End":"31:49.375","Text":"What we\u0027re trying to do is we\u0027re setting y at some t = 0,"},{"Start":"31:49.375 ","End":"31:51.234","Text":"we\u0027re going to get 2 answers,"},{"Start":"31:51.234 ","End":"31:54.380","Text":"and 1 of them is the correct answer."},{"Start":"31:54.740 ","End":"31:58.154","Text":"What we have to do is we have to use"},{"Start":"31:58.154 ","End":"32:03.884","Text":"our initial conditions over here in order to find what our A and our Phi is equal to."},{"Start":"32:03.884 ","End":"32:06.180","Text":"Otherwise, we can\u0027t answer this question,"},{"Start":"32:06.180 ","End":"32:07.605","Text":"so if you want to do this,"},{"Start":"32:07.605 ","End":"32:09.540","Text":"you can do the algebra."},{"Start":"32:09.540 ","End":"32:10.874","Text":"However, to save time,"},{"Start":"32:10.874 ","End":"32:13.930","Text":"I\u0027ll just write down what the A and Phi come out to be,"},{"Start":"32:13.930 ","End":"32:16.580","Text":"then you can check your answer at home."},{"Start":"32:16.620 ","End":"32:22.764","Text":"Our A, our amplitude is going to be this really complicated expression over here,"},{"Start":"32:22.764 ","End":"32:25.930","Text":"so we have our g multiplied by m minus Rho,"},{"Start":"32:25.930 ","End":"32:27.940","Text":"the density of the fluid multiplied by v,"},{"Start":"32:27.940 ","End":"32:30.654","Text":"the volume of the sphere,"},{"Start":"32:30.654 ","End":"32:32.740","Text":"divided by k, the spring constant,"},{"Start":"32:32.740 ","End":"32:37.165","Text":"multiplied by the square root of 1 plus our Gamma,"},{"Start":"32:37.165 ","End":"32:41.545","Text":"which we defined over here."},{"Start":"32:41.545 ","End":"32:46.555","Text":"That\u0027s our Gamma divided by r_2 Omega,"},{"Start":"32:46.555 ","End":"32:49.569","Text":"which was also defined to us over here."},{"Start":"32:49.569 ","End":"32:53.815","Text":"This is what our Omega is, plus kV_0,"},{"Start":"32:53.815 ","End":"33:00.654","Text":"where V_0 is our velocity divided by Omega g multiplied by m minus Rho,"},{"Start":"33:00.654 ","End":"33:04.210","Text":"and V volume again squared."},{"Start":"33:04.210 ","End":"33:10.260","Text":"Then our Phi, so I phase is going to be equal to negative the inverse of tan,"},{"Start":"33:10.260 ","End":"33:14.355","Text":"so negative arctan of our Gamma,"},{"Start":"33:14.355 ","End":"33:22.089","Text":"again defined to us divided by 2 Omega plus kV_0 velocity divided by again,"},{"Start":"33:22.089 ","End":"33:29.000","Text":"our Omega g multiplied by m minus Rho V for volume again."},{"Start":"33:29.520 ","End":"33:33.280","Text":"Now it\u0027s going to be some long expression,"},{"Start":"33:33.280 ","End":"33:37.119","Text":"but all we have to do is we have to substitute in our A and our Phi"},{"Start":"33:37.119 ","End":"33:42.220","Text":"into this equation over here."},{"Start":"33:42.220 ","End":"33:45.549","Text":"Then we have to set this whole equation to equal 0."},{"Start":"33:45.549 ","End":"33:51.580","Text":"Of course, you can substitute in your Gamma and your y0 and your Omega and everything."},{"Start":"33:51.580 ","End":"33:53.631","Text":"Then you set it equal to 0."},{"Start":"33:53.631 ","End":"33:59.980","Text":"Then you see what t you get and that will be your final answer."},{"Start":"33:59.980 ","End":"34:02.989","Text":"That\u0027s the end of this lesson."}],"ID":9477},{"Watched":false,"Name":"Tower Of Bricks","Duration":"11m 49s","ChapterTopicVideoID":9208,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"Hello. In this question,"},{"Start":"00:01.680 ","End":"00:05.550","Text":"Charlie is trying to build a tower from 5 identical blocks,"},{"Start":"00:05.550 ","End":"00:07.980","Text":"each with sides of length a."},{"Start":"00:07.980 ","End":"00:10.680","Text":"What is the maximum distance that he can place"},{"Start":"00:10.680 ","End":"00:14.415","Text":"the uppermost block such that the tower will not fall?"},{"Start":"00:14.415 ","End":"00:17.130","Text":"Measure the distance between the left side of"},{"Start":"00:17.130 ","End":"00:20.040","Text":"the bottom block to the left side of the top block."},{"Start":"00:20.040 ","End":"00:23.055","Text":"The left side of the button block to the left side of the top block."},{"Start":"00:23.055 ","End":"00:25.050","Text":"This is our x max, hint,"},{"Start":"00:25.050 ","End":"00:28.140","Text":"begin calculating from the top block."},{"Start":"00:28.140 ","End":"00:32.355","Text":"Each mass its bottom and it\u0027s top of length a,"},{"Start":"00:32.355 ","End":"00:35.985","Text":"and they all have the same mass m. Now,"},{"Start":"00:35.985 ","End":"00:37.440","Text":"in the hint they mentioned,"},{"Start":"00:37.440 ","End":"00:39.300","Text":"begin calculating from the top block."},{"Start":"00:39.300 ","End":"00:42.255","Text":"What we\u0027re going to do is we\u0027re going to calculate"},{"Start":"00:42.255 ","End":"00:46.660","Text":"every single block relative to the block under it."},{"Start":"00:46.660 ","End":"00:48.815","Text":"Let\u0027s look at our top block."},{"Start":"00:48.815 ","End":"00:52.385","Text":"Now, in order for this block not to fall over,"},{"Start":"00:52.385 ","End":"00:57.540","Text":"I need my center of mass to be somewhere over here."},{"Start":"00:57.540 ","End":"01:04.420","Text":"What is this place if this is my right side of the block?"},{"Start":"01:04.420 ","End":"01:08.314","Text":"I need my center of mass to be anywhere before"},{"Start":"01:08.314 ","End":"01:13.355","Text":"this dotted line because once the center of mass is over this dotted line,"},{"Start":"01:13.355 ","End":"01:15.580","Text":"my block will just tip over."},{"Start":"01:15.580 ","End":"01:18.980","Text":"Now let\u0027s call this distance over here."},{"Start":"01:18.980 ","End":"01:21.200","Text":"Let\u0027s call that X1."},{"Start":"01:21.200 ","End":"01:25.850","Text":"The distance between the left side of the lower block,"},{"Start":"01:25.850 ","End":"01:30.130","Text":"to the left side of the block on top of it."},{"Start":"01:30.130 ","End":"01:32.140","Text":"If this is X1,"},{"Start":"01:32.140 ","End":"01:35.315","Text":"and if this point over here is our center of mass,"},{"Start":"01:35.315 ","End":"01:39.065","Text":"so because our block is of uniform mass,"},{"Start":"01:39.065 ","End":"01:40.489","Text":"of uniform density,"},{"Start":"01:40.489 ","End":"01:46.690","Text":"we know that our center of mass is located halfway into the block."},{"Start":"01:46.690 ","End":"01:49.805","Text":"It\u0027s going to be at a divided by 2."},{"Start":"01:49.805 ","End":"01:53.585","Text":"If so, we know that our X1,"},{"Start":"01:53.585 ","End":"01:55.280","Text":"this distance over here,"},{"Start":"01:55.280 ","End":"01:59.220","Text":"plus our a divided by 2,"},{"Start":"01:59.220 ","End":"02:01.295","Text":"so halfway into the box,"},{"Start":"02:01.295 ","End":"02:08.940","Text":"has to be smaller or equal to our a. X1 plus a divided by"},{"Start":"02:08.940 ","End":"02:11.990","Text":"2 is the distance from the left side of"},{"Start":"02:11.990 ","End":"02:17.445","Text":"my lower block until this blue dotted line over here."},{"Start":"02:17.445 ","End":"02:21.240","Text":"If I have that my X1 plus my a divided by 2."},{"Start":"02:21.240 ","End":"02:24.500","Text":"If that whole distance is smaller or equal to a,"},{"Start":"02:24.500 ","End":"02:30.215","Text":"then that means that my center of mass will be on the left side of this blue line,"},{"Start":"02:30.215 ","End":"02:32.735","Text":"which means that my block won\u0027t topple over."},{"Start":"02:32.735 ","End":"02:38.900","Text":"Therefore, my maximum X1 that I can have,"},{"Start":"02:38.900 ","End":"02:44.055","Text":"is going to be equal to a minus a divided by 2."},{"Start":"02:44.055 ","End":"02:45.910","Text":"It\u0027s going to be a divided by 2."},{"Start":"02:45.910 ","End":"02:48.650","Text":"We also could have said this intuitively by saying"},{"Start":"02:48.650 ","End":"02:52.070","Text":"that if my X1 is equal to a divided by 2,"},{"Start":"02:52.070 ","End":"02:55.730","Text":"then that means that a divided by 2 plus a divided by 2 will"},{"Start":"02:55.730 ","End":"03:00.290","Text":"mean that my center of mass is right on this blue dotted line."},{"Start":"03:00.290 ","End":"03:05.195","Text":"That\u0027s our X1. Now, let\u0027s go on to the next block down."},{"Start":"03:05.195 ","End":"03:10.540","Text":"Then I\u0027m going to label this distance over here, X2."},{"Start":"03:10.540 ","End":"03:14.915","Text":"Now what I have to do is I have to take the center of mass"},{"Start":"03:14.915 ","End":"03:19.005","Text":"of the top 2 blocks over here,"},{"Start":"03:19.005 ","End":"03:24.080","Text":"the center of mass of these 2 together and I have to make sure that my center of"},{"Start":"03:24.080 ","End":"03:29.435","Text":"mass doesn\u0027t go over this blue dotted line over here."},{"Start":"03:29.435 ","End":"03:32.030","Text":"For the same reason that our center of mass"},{"Start":"03:32.030 ","End":"03:35.140","Text":"couldn\u0027t go over this blue dotted line over here."},{"Start":"03:35.140 ","End":"03:41.505","Text":"Let\u0027s find the center of mass of blocks 1 and 2."},{"Start":"03:41.505 ","End":"03:48.540","Text":"Our X center of mass of blocks 1 and 2 is going to be equal too."},{"Start":"03:48.540 ","End":"03:52.190","Text":"Now we\u0027re measuring from this side over here."},{"Start":"03:52.190 ","End":"03:53.990","Text":"For block number 1,"},{"Start":"03:53.990 ","End":"03:59.610","Text":"our center of mass is going to be at X2,"},{"Start":"03:59.610 ","End":"04:05.945","Text":"which is this distance, plus our X1,"},{"Start":"04:05.945 ","End":"04:14.820","Text":"which is that distance plus our a divided by 2, that distance."},{"Start":"04:14.820 ","End":"04:18.780","Text":"We\u0027re going a divided by 2 plus X1 plus X2 and all of"},{"Start":"04:18.780 ","End":"04:24.800","Text":"this multiplied by m and then plus our center of mass of block Number 2."},{"Start":"04:24.800 ","End":"04:30.035","Text":"We\u0027re going to go X2 plus our a divided by 2."},{"Start":"04:30.035 ","End":"04:35.930","Text":"Because our center of mass is located in the center of the block at a divided by 2,"},{"Start":"04:35.930 ","End":"04:41.240","Text":"so plus a divided by 2 and again multiplied by m and then all of this is going"},{"Start":"04:41.240 ","End":"04:47.200","Text":"to be divided by the total mass of these 2 blocks, which is 2m."},{"Start":"04:47.200 ","End":"04:53.075","Text":"Now our center of mass of these 2 blocks has to be smaller than"},{"Start":"04:53.075 ","End":"04:59.420","Text":"or equal to a because this total length over here is a,"},{"Start":"04:59.420 ","End":"05:02.425","Text":"because I\u0027m measuring from over here."},{"Start":"05:02.425 ","End":"05:05.615","Text":"Now all we have to do is we have to rearrange."},{"Start":"05:05.615 ","End":"05:08.945","Text":"First of all, these m\u0027s can cross out."},{"Start":"05:08.945 ","End":"05:12.755","Text":"Then what we\u0027ll get once we open up these brackets,"},{"Start":"05:12.755 ","End":"05:19.850","Text":"is we\u0027ll get that our X1 plus 2X2 plus 2 multiplied"},{"Start":"05:19.850 ","End":"05:27.510","Text":"by a divided by 2 has to be smaller or equal to our 2a."},{"Start":"05:27.510 ","End":"05:31.340","Text":"In that case, if we move this over to the other side,"},{"Start":"05:31.340 ","End":"05:39.235","Text":"we\u0027ll have that our X1 plus our 2X2 has to be smaller or equal to a."},{"Start":"05:39.235 ","End":"05:43.020","Text":"We know what our X1 max is, it\u0027s a/2."},{"Start":"05:43.020 ","End":"05:45.050","Text":"In that case,"},{"Start":"05:45.050 ","End":"05:46.460","Text":"we can substitute that in."},{"Start":"05:46.460 ","End":"05:50.780","Text":"A divided by 2 plus 2X2,"},{"Start":"05:50.780 ","End":"05:54.320","Text":"we\u0027re going to say is equal to a because we want to"},{"Start":"05:54.320 ","End":"05:57.585","Text":"find our maximum value that our X2 can be,"},{"Start":"05:57.585 ","End":"06:00.970","Text":"because we want to find the maximum distance."},{"Start":"06:01.160 ","End":"06:07.025","Text":"In that case, once we move our a/2 over here and divide it by 2,"},{"Start":"06:07.025 ","End":"06:09.905","Text":"to isolate out our X2 max,"},{"Start":"06:09.905 ","End":"06:17.680","Text":"we\u0027ll get that our maximum distance that our X2 can be is equal to a divided by 4."},{"Start":"06:17.690 ","End":"06:24.920","Text":"Now we have to do this exact same thing over and over for the rest of the blocks."},{"Start":"06:24.920 ","End":"06:29.150","Text":"Let\u0027s just highlight over here our X1 and our X2."},{"Start":"06:29.150 ","End":"06:33.885","Text":"Now we\u0027re dealing with also our block Number 3."},{"Start":"06:33.885 ","End":"06:40.415","Text":"Let\u0027s say that this length over here is our X3,"},{"Start":"06:40.415 ","End":"06:45.540","Text":"and we\u0027re measuring from this side over here now."},{"Start":"06:45.650 ","End":"06:50.935","Text":"Let\u0027s take a look our center of mass is over here again at a divided by 2."},{"Start":"06:50.935 ","End":"06:54.470","Text":"Now we have to find our Xcm of masses 1,"},{"Start":"06:54.470 ","End":"06:56.775","Text":"2, and 3."},{"Start":"06:56.775 ","End":"07:03.785","Text":"That\u0027s going to be equal to our distance of X1 plus X2"},{"Start":"07:03.785 ","End":"07:11.910","Text":"plus X3 plus a divided by 2 multiplied by m, so that\u0027s this."},{"Start":"07:11.910 ","End":"07:14.975","Text":"Then we\u0027re going to do plus the center of mass."},{"Start":"07:14.975 ","End":"07:19.489","Text":"It\u0027s going to be X2 plus"},{"Start":"07:19.489 ","End":"07:25.800","Text":"X3 plus a divided by 2 multiplied by m,"},{"Start":"07:25.800 ","End":"07:31.395","Text":"and then plus our, oh sorry,"},{"Start":"07:31.395 ","End":"07:36.045","Text":"X3 plus a divided by 2 multiplied by"},{"Start":"07:36.045 ","End":"07:38.915","Text":"m. That\u0027s going to be"},{"Start":"07:38.915 ","End":"07:43.160","Text":"this distance over here from the center of mass until this side over here."},{"Start":"07:43.160 ","End":"07:46.775","Text":"Then of course, all of this is going to be divided by"},{"Start":"07:46.775 ","End":"07:51.415","Text":"the total mass of the system, which is 3m."},{"Start":"07:51.415 ","End":"07:57.035","Text":"Just like before, it has to be smaller or equal to a,"},{"Start":"07:57.035 ","End":"07:59.855","Text":"otherwise, the tower will topple over."},{"Start":"07:59.855 ","End":"08:05.050","Text":"Then we can divide everything by m, they cancel out."},{"Start":"08:09.260 ","End":"08:11.990","Text":"Once we open up all of the brackets,"},{"Start":"08:11.990 ","End":"08:17.540","Text":"we\u0027ll get that we have 3 multiplied by X3 plus 2 multiplied by"},{"Start":"08:17.540 ","End":"08:24.350","Text":"X2 plus X1 plus 3 multiplied by a divided by 2."},{"Start":"08:24.350 ","End":"08:31.350","Text":"This has to be smaller than we multiply both sides by 3, so 3a."},{"Start":"08:31.690 ","End":"08:36.460","Text":"Then we\u0027re going to have our 3X3 plus,"},{"Start":"08:36.460 ","End":"08:39.230","Text":"now we\u0027re going to substitute in what our X2 is."},{"Start":"08:39.230 ","End":"08:41.630","Text":"We\u0027re going to have 2 multiplied by our X2,"},{"Start":"08:41.630 ","End":"08:45.485","Text":"which is a divided by 4 plus our X1,"},{"Start":"08:45.485 ","End":"08:52.755","Text":"which we saw was a divided by 2 and then plus our 3a divided by 2."},{"Start":"08:52.755 ","End":"08:58.910","Text":"This, now because I want to find my maximum value for my X3,"},{"Start":"08:58.910 ","End":"09:02.420","Text":"so I\u0027m going to say that it\u0027s equal to 3a."},{"Start":"09:02.420 ","End":"09:05.390","Text":"Now when we solve this,"},{"Start":"09:05.390 ","End":"09:06.860","Text":"we\u0027re going to get that"},{"Start":"09:06.860 ","End":"09:12.425","Text":"our 3X3 max is"},{"Start":"09:12.425 ","End":"09:17.615","Text":"simply going to be equal to a divided by 6."},{"Start":"09:17.615 ","End":"09:20.825","Text":"Just do a bit of algebra to rearrange this."},{"Start":"09:20.825 ","End":"09:24.395","Text":"Then we have to do the exact same thing,"},{"Start":"09:24.395 ","End":"09:28.680","Text":"however, for our X4 over here."},{"Start":"09:28.680 ","End":"09:33.170","Text":"Now we can already see the type of pattern that we\u0027re getting."},{"Start":"09:33.170 ","End":"09:35.270","Text":"Let\u0027s write this out."},{"Start":"09:35.270 ","End":"09:37.085","Text":"For our X4,"},{"Start":"09:37.085 ","End":"09:45.250","Text":"we\u0027re going to have 4X4 plus 3X3 plus 2X2,"},{"Start":"09:45.250 ","End":"09:52.110","Text":"plus X1 plus 4a divided by 2,"},{"Start":"09:52.110 ","End":"09:55.830","Text":"and this is going to be smaller or equal to 4a,"},{"Start":"09:56.060 ","End":"10:00.160","Text":"according to our pattern over here."},{"Start":"10:00.730 ","End":"10:06.605","Text":"Then in order to find our X4 max,"},{"Start":"10:06.605 ","End":"10:09.560","Text":"the maximum value for this distance over here."},{"Start":"10:09.560 ","End":"10:16.160","Text":"We\u0027re going to substitute what do we have for our X3 which is over here,"},{"Start":"10:16.160 ","End":"10:18.050","Text":"sorry, without a 3 over here."},{"Start":"10:18.050 ","End":"10:22.340","Text":"For our X3 and then what we have for X2, which we have over here,"},{"Start":"10:22.340 ","End":"10:25.430","Text":"and what we have for our X1,"},{"Start":"10:25.430 ","End":"10:29.145","Text":"which is over here."},{"Start":"10:29.145 ","End":"10:31.250","Text":"Once we substitute in all of"},{"Start":"10:31.250 ","End":"10:34.340","Text":"the values and set because we want to find our maximum value,"},{"Start":"10:34.340 ","End":"10:36.215","Text":"so we set it equal to 4a,"},{"Start":"10:36.215 ","End":"10:44.730","Text":"we\u0027ll get that our X4 max is going to be equal to a divided by 8."},{"Start":"10:45.220 ","End":"10:50.315","Text":"Now in order for us to find our X max,"},{"Start":"10:50.315 ","End":"10:57.180","Text":"our total X max in the question that we had over here."},{"Start":"10:57.180 ","End":"11:06.880","Text":"It\u0027s going to be equal to our X1 max plus our X2 max plus our X3 max plus our X4 max."},{"Start":"11:07.550 ","End":"11:15.135","Text":"That is going to be equal to a divided by 2 plus a"},{"Start":"11:15.135 ","End":"11:22.705","Text":"divided by 4 plus a divided by 6 plus a divided by 8."},{"Start":"11:22.705 ","End":"11:25.460","Text":"Then that\u0027s going to be equal to,"},{"Start":"11:25.460 ","End":"11:28.085","Text":"once we add all of this up,"},{"Start":"11:28.085 ","End":"11:32.130","Text":"25 divided by 24a."},{"Start":"11:34.130 ","End":"11:36.545","Text":"This is our final answer."},{"Start":"11:36.545 ","End":"11:40.295","Text":"This is the maximum distance that"},{"Start":"11:40.295 ","End":"11:47.210","Text":"our uppermost block can be placed such that the tower will not fall."},{"Start":"11:47.210 ","End":"11:50.700","Text":"Okay, that\u0027s the end of this question."}],"ID":9478},{"Watched":false,"Name":"Fly On Disk","Duration":"45m 23s","ChapterTopicVideoID":9209,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this question,"},{"Start":"00:01.920 ","End":"00:05.969","Text":"we\u0027re being told that we have a flat disk of mass M and a radius R,"},{"Start":"00:05.969 ","End":"00:12.270","Text":"which is rotating at an initial angular velocity of Omega_0 about its center,"},{"Start":"00:12.270 ","End":"00:15.539","Text":"which is lying stationary on a frictionless table."},{"Start":"00:15.539 ","End":"00:18.660","Text":"The disk isn\u0027t attached to the table,"},{"Start":"00:18.660 ","End":"00:20.655","Text":"is just lying on a frictionless table."},{"Start":"00:20.655 ","End":"00:23.429","Text":"On the table under the center of the disc,"},{"Start":"00:23.429 ","End":"00:25.350","Text":"a green dot is drawn,"},{"Start":"00:25.350 ","End":"00:27.629","Text":"called a capital O."},{"Start":"00:27.629 ","End":"00:29.504","Text":"That\u0027s what it\u0027s called from now on."},{"Start":"00:29.504 ","End":"00:31.110","Text":"At the center of the disk,"},{"Start":"00:31.110 ","End":"00:34.604","Text":"a green fly of mass, small m sleeps."},{"Start":"00:34.604 ","End":"00:36.254","Text":"This is our green fly,"},{"Start":"00:36.254 ","End":"00:42.270","Text":"a green radial line that this green line over here is drawn on the disk."},{"Start":"00:42.270 ","End":"00:45.359","Text":"Question number 1, at t is equal to 0,"},{"Start":"00:45.359 ","End":"00:48.790","Text":"the fly awakes and begins walking on the radial line."},{"Start":"00:48.790 ","End":"00:53.209","Text":"Find the position of point O so our green dot on the table,"},{"Start":"00:53.209 ","End":"00:56.779","Text":"which is on the table relative to the fly as a function of"},{"Start":"00:56.779 ","End":"01:01.235","Text":"the distance h between the fly and the center of the disk."},{"Start":"01:01.235 ","End":"01:04.385","Text":"The y-axis is perpendicular to the x-axis,"},{"Start":"01:04.385 ","End":"01:08.855","Text":"and the x-axis is in the direction of the center of the disc."},{"Start":"01:08.855 ","End":"01:14.370","Text":"The y-axis is on the plane of the disk."},{"Start":"01:15.230 ","End":"01:18.130","Text":"Our disk is moving, however,"},{"Start":"01:18.130 ","End":"01:22.579","Text":"we\u0027re going to be working with our axes that are moving"},{"Start":"01:22.579 ","End":"01:28.440","Text":"with our fly relative to the flight in the same frame of reference as the fly."},{"Start":"01:28.460 ","End":"01:31.999","Text":"Our frame of reference is rotating with"},{"Start":"01:31.999 ","End":"01:36.745","Text":"the disk because the fly is on the disk and it is rotating."},{"Start":"01:36.745 ","End":"01:39.995","Text":"If our axis is rotating with the disc,"},{"Start":"01:39.995 ","End":"01:42.709","Text":"that means that from our point of reference,"},{"Start":"01:42.709 ","End":"01:45.125","Text":"which is the fly,"},{"Start":"01:45.125 ","End":"01:48.260","Text":"we can see our rotation."},{"Start":"01:48.260 ","End":"01:50.419","Text":"Potentially later on when we\u0027re doing"},{"Start":"01:50.419 ","End":"01:53.870","Text":"our force equations will need to take into account our rotation."},{"Start":"01:53.870 ","End":"01:57.475","Text":"But currently, it\u0027s not relevant."},{"Start":"01:57.475 ","End":"02:00.549","Text":"This is our starting diagram,"},{"Start":"02:00.549 ","End":"02:02.319","Text":"this is starting position."},{"Start":"02:02.319 ","End":"02:07.670","Text":"Let\u0027s see what happens once the fly has moved forward a little bit."},{"Start":"02:07.670 ","End":"02:12.369","Text":"What we have over here is that our fly moved"},{"Start":"02:12.369 ","End":"02:15.490","Text":"a little bit so we can see that because our fly is"},{"Start":"02:15.490 ","End":"02:19.169","Text":"moved and because we don\u0027t have any friction between the disk and the table."},{"Start":"02:19.169 ","End":"02:22.194","Text":"As the fly moves in the right direction,"},{"Start":"02:22.194 ","End":"02:25.855","Text":"our disk is going to slide slightly leftwards."},{"Start":"02:25.855 ","End":"02:29.380","Text":"Our green point, which is drawn on the table called O,"},{"Start":"02:29.380 ","End":"02:30.700","Text":"we can see it\u0027s not moving,"},{"Start":"02:30.700 ","End":"02:31.884","Text":"it\u0027s still on the table."},{"Start":"02:31.884 ","End":"02:37.650","Text":"Now we can see the disk\u0027s position relative to our green point O,"},{"Start":"02:37.650 ","End":"02:40.419","Text":"and how there\u0027s been a shift."},{"Start":"02:40.670 ","End":"02:43.910","Text":"The fly has moved forward some distance,"},{"Start":"02:43.910 ","End":"02:49.079","Text":"and the center of the disk has also moved some distance to the side."},{"Start":"02:50.690 ","End":"02:55.880","Text":"This green dot has got nothing to do with that disk and fly system."},{"Start":"02:55.880 ","End":"03:00.784","Text":"It\u0027s drawn on the table. It\u0027s like our disk is see-through and we can see this dot."},{"Start":"03:00.784 ","End":"03:05.660","Text":"We can see that the center of the disk is going"},{"Start":"03:05.660 ","End":"03:10.624","Text":"to shift and we can say it\u0027s going to move to the opposite direction."},{"Start":"03:10.624 ","End":"03:17.210","Text":"We can say that because of conservation of momentum or from the idea of center of mass,"},{"Start":"03:17.210 ","End":"03:18.949","Text":"that if the 2 bodies are moving,"},{"Start":"03:18.949 ","End":"03:22.554","Text":"then our center of mass has to remain in the center."},{"Start":"03:22.554 ","End":"03:24.469","Text":"If 1 body moves,"},{"Start":"03:24.469 ","End":"03:27.080","Text":"then that means that the second body has to move in order"},{"Start":"03:27.080 ","End":"03:30.774","Text":"to maintain the balance of where the center of mass is."},{"Start":"03:30.774 ","End":"03:33.509","Text":"Now let\u0027s draw in our axes."},{"Start":"03:33.509 ","End":"03:38.090","Text":"We\u0027re told that our x-axis is in the direction of the center of the disk."},{"Start":"03:38.090 ","End":"03:42.010","Text":"That means going in this direction."},{"Start":"03:42.010 ","End":"03:46.910","Text":"This is our x and then we\u0027re being told that our y-axis is"},{"Start":"03:46.910 ","End":"03:53.510","Text":"perpendicular to the x-axis on the plane of the disk so going in this direction."},{"Start":"03:53.510 ","End":"03:56.660","Text":"A distance h, which we\u0027re trying to"},{"Start":"03:56.660 ","End":"04:01.249","Text":"measure is the distance between the fly and the center of the disk."},{"Start":"04:01.249 ","End":"04:05.640","Text":"Over here, we have our center of the disk."},{"Start":"04:05.640 ","End":"04:11.780","Text":"Our h is going to be this whole distance over here."},{"Start":"04:11.780 ","End":"04:16.799","Text":"From my center of the disk until our fly."},{"Start":"04:17.110 ","End":"04:23.825","Text":"Now what we want to do is we want to find the position of our green dot O."},{"Start":"04:23.825 ","End":"04:30.745","Text":"Let\u0027s call this distance our x_o."},{"Start":"04:30.745 ","End":"04:35.245","Text":"This is what interests us and it\u0027s relative to our flight."},{"Start":"04:35.245 ","End":"04:41.989","Text":"As we can see, the positive x-direction is in the direction of our center."},{"Start":"04:41.989 ","End":"04:47.219","Text":"That means that our x_o is meant to be some positive number."},{"Start":"04:47.390 ","End":"04:51.310","Text":"Let\u0027s begin by answering our question number 1."},{"Start":"04:51.310 ","End":"04:57.385","Text":"Now we know that our x center of mass at our time of t is equal to 0."},{"Start":"04:57.385 ","End":"05:01.340","Text":"Right at the start is going to be at x_o."},{"Start":"05:01.340 ","End":"05:05.259","Text":"Because our origin is over here where the fly is and I fly"},{"Start":"05:05.259 ","End":"05:09.564","Text":"is a distance x_o away from our point O,"},{"Start":"05:09.564 ","End":"05:11.810","Text":"from the green dot."},{"Start":"05:12.260 ","End":"05:17.110","Text":"But our x_m, we can write a whole equation for our location,"},{"Start":"05:17.110 ","End":"05:18.369","Text":"for our x center of mass."},{"Start":"05:18.369 ","End":"05:21.160","Text":"We\u0027re going to write the mass of the fly,"},{"Start":"05:21.160 ","End":"05:24.820","Text":"which is small m multiplied by its distance from the origin."},{"Start":"05:24.820 ","End":"05:27.864","Text":"Now because it\u0027s at the origin, it\u0027s multiplied by 0."},{"Start":"05:27.864 ","End":"05:35.479","Text":"Then plus the mass of our disk multiplied by its distance from the origin,"},{"Start":"05:35.479 ","End":"05:40.759","Text":"which is going to be h. We\u0027re using the center of mass of the disk,"},{"Start":"05:40.759 ","End":"05:43.369","Text":"which has a distance h from the origin."},{"Start":"05:43.369 ","End":"05:47.209","Text":"Then all of this is going to be divided by the total mass of the system,"},{"Start":"05:47.209 ","End":"05:50.585","Text":"which is m plus m. Now,"},{"Start":"05:50.585 ","End":"05:54.109","Text":"the center of mass remains at the exact same point."},{"Start":"05:54.109 ","End":"05:59.599","Text":"This is actually also equal to our x center of mass at"},{"Start":"05:59.599 ","End":"06:07.010","Text":"some general time t. I say that it remains at the exact same point the whole time."},{"Start":"06:07.010 ","End":"06:08.660","Text":"Our center of mass,"},{"Start":"06:08.660 ","End":"06:16.119","Text":"because the sum of all of the external forces is equal to 0."},{"Start":"06:16.119 ","End":"06:21.469","Text":"Also, because our initial velocity for the center of mass,"},{"Start":"06:21.469 ","End":"06:24.065","Text":"so v center of mass at t is equal to 0,"},{"Start":"06:24.065 ","End":"06:26.644","Text":"is also equal to 0."},{"Start":"06:26.644 ","End":"06:35.244","Text":"We know that our point for the center of mass moves only due to our external forces."},{"Start":"06:35.244 ","End":"06:38.150","Text":"If there aren\u0027t external forces,"},{"Start":"06:38.150 ","End":"06:45.450","Text":"then we can know that the acceleration of the center of mass is equal to 0."},{"Start":"06:45.800 ","End":"06:50.349","Text":"The fact that our acceleration of the center of mass is equal to 0,"},{"Start":"06:50.349 ","End":"06:53.740","Text":"coupled with our initial velocity being equal to 0,"},{"Start":"06:53.740 ","End":"06:57.540","Text":"so we can say that our x center of mass,"},{"Start":"06:57.540 ","End":"07:01.555","Text":"so the position of our center of mass is a constant."},{"Start":"07:01.555 ","End":"07:05.239","Text":"It remains at the exact same point."},{"Start":"07:05.580 ","End":"07:08.469","Text":"I know that right at the start,"},{"Start":"07:08.469 ","End":"07:10.929","Text":"my x center of mass at t is equal to 0,"},{"Start":"07:10.929 ","End":"07:13.045","Text":"is located at x_o."},{"Start":"07:13.045 ","End":"07:16.044","Text":"Which means that throughout all the movement,"},{"Start":"07:16.044 ","End":"07:17.785","Text":"no matter what t I have,"},{"Start":"07:17.785 ","End":"07:20.875","Text":"my center of mass is still going to be at this point,"},{"Start":"07:20.875 ","End":"07:23.970","Text":"x_o at O over here."},{"Start":"07:23.970 ","End":"07:30.974","Text":"Now we can say that my x_o is equal to this."},{"Start":"07:30.974 ","End":"07:40.245","Text":"I\u0027m going to have Mh divided by m plus M. This is our answer to question number 1."},{"Start":"07:40.245 ","End":"07:42.374","Text":"Let\u0027s move on to question number 2."},{"Start":"07:42.374 ","End":"07:46.419","Text":"Question number 2 is asking us to find"},{"Start":"07:46.419 ","End":"07:51.504","Text":"the angular velocity of the disc when the fly is at its edge."},{"Start":"07:51.504 ","End":"07:54.954","Text":"Now the way that we\u0027re going to answer this is to find"},{"Start":"07:54.954 ","End":"07:58.089","Text":"our angular velocity as a function of"},{"Start":"07:58.089 ","End":"08:00.579","Text":"this distance h. As"},{"Start":"08:00.579 ","End":"08:05.395","Text":"the fly moves further and further away from the center and towards the edge."},{"Start":"08:05.395 ","End":"08:09.520","Text":"Then we\u0027re going to use this fact and then we\u0027re going to"},{"Start":"08:09.520 ","End":"08:13.899","Text":"find our angular velocity when the fly is at edge."},{"Start":"08:13.899 ","End":"08:16.600","Text":"First, we\u0027re going to find the angular velocity as a function of"},{"Start":"08:16.600 ","End":"08:20.184","Text":"h. How are we going to solve this?"},{"Start":"08:20.184 ","End":"08:23.309","Text":"Now because we saw in question number"},{"Start":"08:23.309 ","End":"08:27.704","Text":"1 that the sum of our external forces was equal to 0."},{"Start":"08:27.704 ","End":"08:33.030","Text":"If the sum of our external forces is equal to 0,"},{"Start":"08:33.030 ","End":"08:43.075","Text":"that means that the sum of external torques will also be equal to 0."},{"Start":"08:43.075 ","End":"08:46.150","Text":"If the sum of our external torques is equal to 0,"},{"Start":"08:46.150 ","End":"08:51.370","Text":"so that means we have conservation of angular momentum."},{"Start":"08:51.370 ","End":"08:58.989","Text":"Angular momentum is denoted by L. The way we\u0027re going to write out our equation for"},{"Start":"08:58.989 ","End":"09:06.970","Text":"our angular momentum is by using an axis which is working around our point O."},{"Start":"09:06.970 ","End":"09:13.120","Text":"Because we know that our point O is not rotating."},{"Start":"09:13.120 ","End":"09:17.320","Text":"That\u0027s an easy way or an easy place to"},{"Start":"09:17.320 ","End":"09:21.880","Text":"place our axis where our origin is at our green point O."},{"Start":"09:21.880 ","End":"09:26.575","Text":"Our axis at O are these pink arrows over here."},{"Start":"09:26.575 ","End":"09:30.805","Text":"Let\u0027s write our initial angular momentum."},{"Start":"09:30.805 ","End":"09:32.830","Text":"Let\u0027s see what this is equal to,"},{"Start":"09:32.830 ","End":"09:35.559","Text":"so let\u0027s draw our diagram again."},{"Start":"09:35.559 ","End":"09:39.325","Text":"Here\u0027s my disc and I had my center of mass,"},{"Start":"09:39.325 ","End":"09:40.750","Text":"or the center of my disc,"},{"Start":"09:40.750 ","End":"09:44.905","Text":"which is also located on top of my green point O on the table"},{"Start":"09:44.905 ","End":"09:50.514","Text":"and my fly was a point-mass on that."},{"Start":"09:50.514 ","End":"09:56.515","Text":"Then my entire disc rotated at an angular velocity of Omega_0."},{"Start":"09:56.515 ","End":"10:01.510","Text":"In order to find my angular momentum in this situation,"},{"Start":"10:01.510 ","End":"10:08.605","Text":"so all I have to do is write out my angular momentum for my disc."},{"Start":"10:08.605 ","End":"10:13.780","Text":"This is just the moment of inertia of the disc, so id."},{"Start":"10:13.780 ","End":"10:17.499","Text":"We don\u0027t have to take the angular momentum of"},{"Start":"10:17.499 ","End":"10:22.315","Text":"our fly because our fly is a point mass located at the center of the disc,"},{"Start":"10:22.315 ","End":"10:24.970","Text":"where we said our origin is."},{"Start":"10:24.970 ","End":"10:29.419","Text":"Its moment of inertia is going to be equal to 0."},{"Start":"10:29.550 ","End":"10:34.899","Text":"Any point mass which is located on our axis of rotation,"},{"Start":"10:34.899 ","End":"10:39.430","Text":"so on our point of rotation has no angular momentum."},{"Start":"10:39.430 ","End":"10:44.619","Text":"Now let\u0027s just rewrite this where we fill in our moment of inertia of a disc."},{"Start":"10:44.619 ","End":"10:49.959","Text":"As we know, it\u0027s simply going to be half multiplied by the mass of the disc which is M,"},{"Start":"10:49.959 ","End":"10:56.994","Text":"multiplied by its radius squared and then multiply it over here by r Omega_0."},{"Start":"10:56.994 ","End":"11:01.525","Text":"This is our initial angular momentum and now let\u0027s write"},{"Start":"11:01.525 ","End":"11:06.490","Text":"our final angular momentum or better yet,"},{"Start":"11:06.490 ","End":"11:09.115","Text":"let\u0027s write it as a function of our h."},{"Start":"11:09.115 ","End":"11:13.285","Text":"The distance that the fly is from the center of the disc."},{"Start":"11:13.285 ","End":"11:16.060","Text":"My origin is still over here,"},{"Start":"11:16.060 ","End":"11:19.585","Text":"my pink axis on my green dot O."},{"Start":"11:19.585 ","End":"11:23.620","Text":"Now, I have to work out because I have also"},{"Start":"11:23.620 ","End":"11:28.225","Text":"my disc rotating around this point and also my fly rotating around this point."},{"Start":"11:28.225 ","End":"11:37.224","Text":"My L_f is going to be my angular momentum of my fly,"},{"Start":"11:37.224 ","End":"11:42.070","Text":"plus my angular momentum of my disc."},{"Start":"11:42.070 ","End":"11:47.679","Text":"My angular momentum of the fly is going to be that of a point-mass,"},{"Start":"11:47.679 ","End":"11:52.660","Text":"which is rotating about the axis of rotation at some distance."},{"Start":"11:52.660 ","End":"11:58.059","Text":"We can see that it\u0027s going to be a distance of x_o away from our origin."},{"Start":"11:58.059 ","End":"12:00.130","Text":"It\u0027s going to be its mass."},{"Start":"12:00.130 ","End":"12:06.070","Text":"The mass of the fly is m multiplied by its distance from the origin squared."},{"Start":"12:06.070 ","End":"12:11.090","Text":"Here, where our r is equal to our x_o."},{"Start":"12:11.720 ","End":"12:15.855","Text":"Now for the angular momentum of our disc,"},{"Start":"12:15.855 ","End":"12:19.260","Text":"so we know that the angular momentum of our disc is going to"},{"Start":"12:19.260 ","End":"12:22.650","Text":"be equal to 1/2 multiplied by the mass of the disc,"},{"Start":"12:22.650 ","End":"12:26.869","Text":"which is M, multiplied by its radius squared."},{"Start":"12:26.869 ","End":"12:31.135","Text":"But we can see that our axis of rotation for the disc,"},{"Start":"12:31.135 ","End":"12:33.235","Text":"its center is right over here,"},{"Start":"12:33.235 ","End":"12:37.705","Text":"this blue point, but our origin is some distance away."},{"Start":"12:37.705 ","End":"12:40.760","Text":"We\u0027re going to have to use our Steiner."},{"Start":"12:40.760 ","End":"12:42.719","Text":"How do we do that?"},{"Start":"12:42.719 ","End":"12:46.574","Text":"We add in the total mass of the disc, which is M,"},{"Start":"12:46.574 ","End":"12:52.075","Text":"multiplied by the distance between our 2 axis of rotation."},{"Start":"12:52.075 ","End":"12:57.700","Text":"The distance between this blue dot and our green dot over here. What\u0027s that going to be?"},{"Start":"12:57.700 ","End":"13:01.360","Text":"We can see that the distance from our blue dot until the fly is"},{"Start":"13:01.360 ","End":"13:07.190","Text":"h and from our fly until our origin is x_o."},{"Start":"13:07.890 ","End":"13:18.530","Text":"This distance over here is simply going to be h minus our X_o and all of that squared."},{"Start":"13:19.560 ","End":"13:23.380","Text":"Now let\u0027s substitute in all of our values."},{"Start":"13:23.380 ","End":"13:27.580","Text":"We\u0027re going to have our m multiplied by r squared,"},{"Start":"13:27.580 ","End":"13:29.499","Text":"which is X_o squared,"},{"Start":"13:29.499 ","End":"13:39.190","Text":"so multiplied by our Mh divided by m plus m^2."},{"Start":"13:39.190 ","End":"13:43.810","Text":"Then we\u0027re going to have plus our 1/2 M R^2,"},{"Start":"13:43.810 ","End":"13:50.290","Text":"plus our m multiplied by our h minus our x_o,"},{"Start":"13:50.290 ","End":"13:56.649","Text":"which is Mh divided by m plus M squared."},{"Start":"13:56.649 ","End":"14:02.275","Text":"Then let\u0027s do a little bit more algebra to make this look a little bit nicer."},{"Start":"14:02.275 ","End":"14:04.344","Text":"We\u0027re going to have m,"},{"Start":"14:04.344 ","End":"14:13.690","Text":"M^2 h^2 divided by m plus M^2 plus"},{"Start":"14:13.690 ","End":"14:21.490","Text":"M R^2 divided by 2 plus M multiplied by"},{"Start":"14:21.490 ","End":"14:32.150","Text":"m^2 h^2 divided by m plus m squared."},{"Start":"14:32.190 ","End":"14:38.769","Text":"Then of course, I forgot to multiply everything by our angular velocity,"},{"Start":"14:38.769 ","End":"14:44.485","Text":"which changes as a function of h. Multiplied by our Omega as a function of h,"},{"Start":"14:44.485 ","End":"14:48.609","Text":"so that means also here we\u0027re multiplying by our angular velocity as"},{"Start":"14:48.609 ","End":"14:52.930","Text":"a function of h and also over here,"},{"Start":"14:52.930 ","End":"14:57.264","Text":"Omega as a function of h. Now,"},{"Start":"14:57.264 ","End":"14:59.770","Text":"because of conservation of angular momentum,"},{"Start":"14:59.770 ","End":"15:04.659","Text":"we can say that our L_f is equal to our L_i."},{"Start":"15:04.659 ","End":"15:11.364","Text":"Now, notice that our L_f as a function of h is equal to this and we\u0027re being asked,"},{"Start":"15:11.364 ","End":"15:14.425","Text":"when our fly gets to the edge."},{"Start":"15:14.425 ","End":"15:19.060","Text":"When our h is equal to the radius of the circle,"},{"Start":"15:19.060 ","End":"15:20.904","Text":"so it\u0027s right at the edge."},{"Start":"15:20.904 ","End":"15:24.940","Text":"L_f, when h is equal to r,"},{"Start":"15:24.940 ","End":"15:27.995","Text":"so that is going to be equal to,"},{"Start":"15:27.995 ","End":"15:29.210","Text":"so everywhere I see an h,"},{"Start":"15:29.210 ","End":"15:39.640","Text":"I substitute it in r. I\u0027ll have m M^2 R^2 divided by m plus M^2"},{"Start":"15:39.640 ","End":"15:44.380","Text":"plus M R^2 divided by"},{"Start":"15:44.380 ","End":"15:49.306","Text":"2 plus M m^2 R^2"},{"Start":"15:49.306 ","End":"15:55.169","Text":"divided by M plus m squared."},{"Start":"15:55.169 ","End":"16:01.310","Text":"All of this multiplied by my angular velocity when my h is equal to R,"},{"Start":"16:01.310 ","End":"16:04.159","Text":"and this is equal to my L_i,"},{"Start":"16:04.159 ","End":"16:10.820","Text":"which is going to be my 1/2 M R^2 Omega_0."},{"Start":"16:10.820 ","End":"16:19.075","Text":"Now what we can do is we can cancel out our capital M\u0027s,"},{"Start":"16:19.075 ","End":"16:23.110","Text":"divide both sides by 1 capital M. This will cross out,"},{"Start":"16:23.110 ","End":"16:24.399","Text":"this will cross out,"},{"Start":"16:24.399 ","End":"16:25.780","Text":"this will cross out,"},{"Start":"16:25.780 ","End":"16:27.475","Text":"and this will cross out."},{"Start":"16:27.475 ","End":"16:31.345","Text":"We can divide both sides by R^2."},{"Start":"16:31.345 ","End":"16:37.345","Text":"R^2, R^2, R^2, R^2."},{"Start":"16:37.345 ","End":"16:40.525","Text":"Now what I want to isolate out is my Omega"},{"Start":"16:40.525 ","End":"16:47.545","Text":"R. All I have to do is I have to find a common denominator and do some algebra."},{"Start":"16:47.545 ","End":"16:49.180","Text":"I\u0027m not going to waste any time."},{"Start":"16:49.180 ","End":"16:51.865","Text":"You can do this on your own and check your answer."},{"Start":"16:51.865 ","End":"16:54.670","Text":"But we\u0027ll get that our Omega,"},{"Start":"16:54.670 ","End":"16:56.650","Text":"so our angular velocity,"},{"Start":"16:56.650 ","End":"17:02.530","Text":"when our fly is at the edge at a distance R away from the center is going to"},{"Start":"17:02.530 ","End":"17:08.680","Text":"be equal to M plus m squared,"},{"Start":"17:08.680 ","End":"17:15.565","Text":"multiplied by Omega_0 and all of this is going to be divided by"},{"Start":"17:15.565 ","End":"17:21.660","Text":"3m^2 plus"},{"Start":"17:21.660 ","End":"17:28.030","Text":"4mM plus m^2."},{"Start":"17:28.030 ","End":"17:29.830","Text":"Here, we\u0027ve answered question number 2."},{"Start":"17:29.830 ","End":"17:32.755","Text":"Let\u0027s move on to question number 3."},{"Start":"17:32.755 ","End":"17:39.655","Text":"Question number 3 is asking us to check our answer for 2 when in the first instance,"},{"Start":"17:39.655 ","End":"17:45.550","Text":"the mass of the fly is significantly much smaller than the mass of the disk."},{"Start":"17:45.550 ","End":"17:47.845","Text":"Then the second case, the opposite,"},{"Start":"17:47.845 ","End":"17:53.830","Text":"so when the mass of the fly is significantly bigger than the mass of the disk."},{"Start":"17:53.830 ","End":"17:57.715","Text":"Let\u0027s go back to our answer to question 2."},{"Start":"17:57.715 ","End":"18:03.879","Text":"We\u0027ll start with our case where our small m is significantly smaller than our"},{"Start":"18:03.879 ","End":"18:10.869","Text":"large M. The fly mass is significantly smaller than the mass of the disk."},{"Start":"18:10.869 ","End":"18:16.615","Text":"Now, if we look at our equation for our Omega R,"},{"Start":"18:16.615 ","End":"18:19.930","Text":"so we can see if our small m is very small,"},{"Start":"18:19.930 ","End":"18:23.005","Text":"so very small number squared is really,"},{"Start":"18:23.005 ","End":"18:26.566","Text":"really tiny and we don\u0027t have to take it into account."},{"Start":"18:26.566 ","End":"18:29.890","Text":"When we have our m plus some number,"},{"Start":"18:29.890 ","End":"18:31.480","Text":"plus a tiny, tiny number,"},{"Start":"18:31.480 ","End":"18:35.995","Text":"so we can just say that it\u0027s M and then squared multiplied by our Omega_0,"},{"Start":"18:35.995 ","End":"18:40.600","Text":"and then divided by 3 times a tiny number squared is tiny,"},{"Start":"18:40.600 ","End":"18:42.129","Text":"so we don\u0027t have to look at that."},{"Start":"18:42.129 ","End":"18:48.729","Text":"Then we have a 4mM because a small number multiplied by big number,"},{"Start":"18:48.729 ","End":"18:53.300","Text":"we still don\u0027t really know what to do with that plus our M^2."},{"Start":"18:53.940 ","End":"18:57.954","Text":"Now what we can do in order to try and simplify this,"},{"Start":"18:57.954 ","End":"19:02.485","Text":"in order to understand what this 4 small m multiplied by big M is,"},{"Start":"19:02.485 ","End":"19:03.820","Text":"so we can rewrite this,"},{"Start":"19:03.820 ","End":"19:08.019","Text":"so M^2 multiplied by Omega_0 divided by,"},{"Start":"19:08.019 ","End":"19:12.055","Text":"and let\u0027s take a common factor of capital M. Capital M,"},{"Start":"19:12.055 ","End":"19:15.054","Text":"and then we have 4m plus capital"},{"Start":"19:15.054 ","End":"19:20.035","Text":"M. Just like here where we had a capital M plus a tiny number,"},{"Start":"19:20.035 ","End":"19:22.630","Text":"so we didn\u0027t have to look or take into"},{"Start":"19:22.630 ","End":"19:26.379","Text":"consideration this tiny number and we just took capital M,"},{"Start":"19:26.379 ","End":"19:30.324","Text":"so just like here, we have our capital M and 4 times a tiny number,"},{"Start":"19:30.324 ","End":"19:32.335","Text":"which is still a very small number."},{"Start":"19:32.335 ","End":"19:35.905","Text":"Again, we don\u0027t have to take it into account. We can cross it out."},{"Start":"19:35.905 ","End":"19:45.069","Text":"Then we\u0027ll get that our M^2 Omega_0 divided by M^2 and then these 2 cancel out and we"},{"Start":"19:45.069 ","End":"19:49.600","Text":"get that when the mass of the fly is significantly smaller"},{"Start":"19:49.600 ","End":"19:54.414","Text":"than the mass of the disk that our Omega when the fly is at the edge of the disk,"},{"Start":"19:54.414 ","End":"19:58.190","Text":"is equal to our initial Omega."},{"Start":"19:58.530 ","End":"20:01.000","Text":"That actually makes sense."},{"Start":"20:01.000 ","End":"20:07.810","Text":"If the fly is so light when we\u0027re comparing it to the disk,"},{"Start":"20:07.810 ","End":"20:11.304","Text":"so we can see that as it goes to the edge of the disk,"},{"Start":"20:11.304 ","End":"20:15.400","Text":"because it\u0027s so light as applying barely any force or whatever,"},{"Start":"20:15.400 ","End":"20:17.544","Text":"so the disk is barely going to move."},{"Start":"20:17.544 ","End":"20:20.349","Text":"Nothing is really going to change because it\u0027s so insignificant"},{"Start":"20:20.349 ","End":"20:24.774","Text":"and our angular velocity will just remain the same as well."},{"Start":"20:24.774 ","End":"20:28.075","Text":"Now let\u0027s check the other case,"},{"Start":"20:28.075 ","End":"20:35.360","Text":"so where the mass of our fly was significantly bigger than the mass of our disk."},{"Start":"20:35.420 ","End":"20:38.910","Text":"Now we\u0027ll do the same thing that we did with"},{"Start":"20:38.910 ","End":"20:42.119","Text":"our small m when we took it out of the calculation,"},{"Start":"20:42.119 ","End":"20:45.420","Text":"except now with our capital M. We\u0027ll get"},{"Start":"20:45.420 ","End":"20:49.825","Text":"that our Omega has a function of h when h is equal to R,"},{"Start":"20:49.825 ","End":"20:53.440","Text":"so when our fly is at the edge of the disk is going to equal 2."},{"Start":"20:53.440 ","End":"20:58.509","Text":"Now our small m is actually the big number or the normal size number,"},{"Start":"20:58.509 ","End":"21:00.805","Text":"and our capital M is the teeny tiny number."},{"Start":"21:00.805 ","End":"21:10.180","Text":"I\u0027ll be left with small m^2 multiplied by our Omega_0 divided by 3m^2."},{"Start":"21:10.180 ","End":"21:12.490","Text":"Again, like we saw before,"},{"Start":"21:12.490 ","End":"21:16.780","Text":"if we have this tiny number multiplied by our normal number,"},{"Start":"21:16.780 ","End":"21:20.950","Text":"so we can just cancel out this term as well and of course this term as well,"},{"Start":"21:20.950 ","End":"21:22.630","Text":"because it\u0027s a tiny number squared,"},{"Start":"21:22.630 ","End":"21:25.645","Text":"so it\u0027s just going to be 3m^2."},{"Start":"21:25.645 ","End":"21:30.490","Text":"Then we can cancel out our m^2 and we\u0027ll get that our Omega when"},{"Start":"21:30.490 ","End":"21:36.280","Text":"the fly is at the edge is equal to 1/3 of our initial angular velocity."},{"Start":"21:36.280 ","End":"21:38.154","Text":"Let\u0027s take a look now,"},{"Start":"21:38.154 ","End":"21:41.325","Text":"if this answer is logical"},{"Start":"21:41.325 ","End":"21:45.585","Text":"because it\u0027s a little bit harder to understand than this one over here."},{"Start":"21:45.585 ","End":"21:48.555","Text":"If we look at our disk,"},{"Start":"21:48.555 ","End":"21:54.200","Text":"this is meant to be a perfect circle and this is our center of mass,"},{"Start":"21:54.200 ","End":"21:57.250","Text":"at the beginning when the fly is right on top of it."},{"Start":"21:57.250 ","End":"22:00.249","Text":"But then as the fly moves along,"},{"Start":"22:00.249 ","End":"22:04.105","Text":"so because its mass is so much bigger than the mass of the disk,"},{"Start":"22:04.105 ","End":"22:10.015","Text":"then the center of mass moves along with the fly as well, more or less."},{"Start":"22:10.015 ","End":"22:11.829","Text":"Instead of the center of mass being here now,"},{"Start":"22:11.829 ","End":"22:14.785","Text":"it will be probably more or less over here."},{"Start":"22:14.785 ","End":"22:20.485","Text":"Now I\u0027m reminding you that our disk is rotating about the center of mass."},{"Start":"22:20.485 ","End":"22:24.219","Text":"Now, the way to look at it is to imagine that my fly"},{"Start":"22:24.219 ","End":"22:27.670","Text":"has like in the question finally gotten to the edge,"},{"Start":"22:27.670 ","End":"22:31.014","Text":"which means that the center of mass is going to be somewhere over here."},{"Start":"22:31.014 ","End":"22:36.445","Text":"Now, a way that we can imagine this is if we have our perfectly circular disk,"},{"Start":"22:36.445 ","End":"22:43.285","Text":"and we have some nail put into the right at the edge over here of the disk."},{"Start":"22:43.285 ","End":"22:49.120","Text":"Then we can see that the disk is going to rotate about this point more or less."},{"Start":"22:49.120 ","End":"22:51.115","Text":"Then in that case,"},{"Start":"22:51.115 ","End":"22:54.760","Text":"I can look at this as a question where at the beginning,"},{"Start":"22:54.760 ","End":"23:00.130","Text":"the entire disk was rotating about its center,"},{"Start":"23:00.130 ","End":"23:05.859","Text":"the center of the disk and then suddenly we stuck a nail in right at"},{"Start":"23:05.859 ","End":"23:11.875","Text":"the edge of the desk and suddenly the disk is rotating about a point on its edge."},{"Start":"23:11.875 ","End":"23:18.040","Text":"Then we can use our idea of conservation of momentum or angular momentum."},{"Start":"23:18.040 ","End":"23:20.784","Text":"Through conservation of angular momentum,"},{"Start":"23:20.784 ","End":"23:26.230","Text":"I can say that my initial angular momentum is that of a disk rotating about its center."},{"Start":"23:26.230 ","End":"23:32.500","Text":"That\u0027s equal to 1/2 M R^2 multiplied by our Omega_0."},{"Start":"23:32.500 ","End":"23:34.840","Text":"Because of conservation of angular momentum,"},{"Start":"23:34.840 ","End":"23:38.094","Text":"it has to be equal to our final angular momentum,"},{"Start":"23:38.094 ","End":"23:42.205","Text":"which is equal to a disk rotating on its side."},{"Start":"23:42.205 ","End":"23:49.959","Text":"That\u0027s going to equal to 1/2 M R^2 plus our [inaudible],so it\u0027s going to"},{"Start":"23:49.959 ","End":"23:58.435","Text":"be M R^2 and this is multiplied by its angular velocity then."},{"Start":"23:58.435 ","End":"24:02.950","Text":"Now we don\u0027t have to add in the angular momentum of the fly,"},{"Start":"24:02.950 ","End":"24:06.850","Text":"only of the disk because we said that our axes of rotation,"},{"Start":"24:06.850 ","End":"24:08.364","Text":"which is our center of mass,"},{"Start":"24:08.364 ","End":"24:10.075","Text":"is moving with the fly."},{"Start":"24:10.075 ","End":"24:14.109","Text":"We can consider our fly as being on the axes of rotation,"},{"Start":"24:14.109 ","End":"24:17.590","Text":"in which case it has no angular momentum"},{"Start":"24:17.590 ","End":"24:20.755","Text":"because the fly is significantly heavier than the disk,"},{"Start":"24:20.755 ","End":"24:25.450","Text":"the center of mass is going to be pretty much on the same point as the fly."},{"Start":"24:25.450 ","End":"24:28.239","Text":"Throughout the motion in this type of case,"},{"Start":"24:28.239 ","End":"24:31.194","Text":"where our small m is significantly bigger than our capital M,"},{"Start":"24:31.194 ","End":"24:32.845","Text":"we\u0027ll get this equation."},{"Start":"24:32.845 ","End":"24:40.359","Text":"This is our angular momentum of the disk when it\u0027s rotating about its center."},{"Start":"24:40.359 ","End":"24:45.025","Text":"This is the angular momentum of the disk rotating about its edge."},{"Start":"24:45.025 ","End":"24:51.330","Text":"What we have in our brackets over here is going to be equal to 3 divided"},{"Start":"24:51.330 ","End":"24:58.380","Text":"by 2M R^2 and then 1/2 and our 1/2 over here can cancel out,"},{"Start":"24:58.380 ","End":"25:03.400","Text":"our M R^2 and our M R^2 over here can cancel out."},{"Start":"25:03.400 ","End":"25:07.230","Text":"Then when we isolate out our Omega R,"},{"Start":"25:07.230 ","End":"25:13.660","Text":"we\u0027ll really get that our Omega R is equal to 1/3 of our Omega_0."},{"Start":"25:13.660 ","End":"25:16.510","Text":"That\u0027s the answer that we got."},{"Start":"25:16.510 ","End":"25:19.615","Text":"Here, we finished our question 3."},{"Start":"25:19.615 ","End":"25:20.949","Text":"These are our answers."},{"Start":"25:20.949 ","End":"25:24.095","Text":"Now let\u0027s go on to our question 4."},{"Start":"25:24.095 ","End":"25:28.290","Text":"Question number 4 is asking if the fly moves with"},{"Start":"25:28.290 ","End":"25:31.889","Text":"a constant velocity of v_0 relative to the disc,"},{"Start":"25:31.889 ","End":"25:35.160","Text":"what will be the frictional force between the fly and the disk"},{"Start":"25:35.160 ","End":"25:39.620","Text":"a moment before the fly reaches the disk\u0027s edge?"},{"Start":"25:39.620 ","End":"25:44.320","Text":"Question 4 is telling us that our fly is moving at"},{"Start":"25:44.320 ","End":"25:50.260","Text":"some constant velocity of v_0 on this line."},{"Start":"25:50.260 ","End":"25:52.524","Text":"What we\u0027re being asked is,"},{"Start":"25:52.524 ","End":"25:54.820","Text":"what is our frictional force,"},{"Start":"25:54.820 ","End":"26:01.267","Text":"so static frictional force a moment before our fly gets to the edge of the disk?"},{"Start":"26:01.267 ","End":"26:03.760","Text":"That\u0027s where our h,"},{"Start":"26:03.760 ","End":"26:06.924","Text":"our distance between the center of the disk and our fly"},{"Start":"26:06.924 ","End":"26:11.395","Text":"is almost R. It\u0027s almost the radius of the disc,"},{"Start":"26:11.395 ","End":"26:14.600","Text":"which represents that the fly is at the edge."},{"Start":"26:14.790 ","End":"26:19.150","Text":"Let\u0027s write the equation for the sum of all of our forces."},{"Start":"26:19.150 ","End":"26:24.280","Text":"The sum of our forces on both the x and the y axis,"},{"Start":"26:24.280 ","End":"26:25.629","Text":"so on the entire x,"},{"Start":"26:25.629 ","End":"26:30.880","Text":"y plane is simply equal to this frictional force."},{"Start":"26:30.880 ","End":"26:33.890","Text":"This static friction."},{"Start":"26:33.890 ","End":"26:35.604","Text":"Now, of course,"},{"Start":"26:35.604 ","End":"26:38.200","Text":"on the axis which is perpendicular to the disk,"},{"Start":"26:38.200 ","End":"26:42.415","Text":"we have our mg force pointing downwards and our N force,"},{"Start":"26:42.415 ","End":"26:44.170","Text":"our normal force pointing upwards."},{"Start":"26:44.170 ","End":"26:46.734","Text":"Now they\u0027re not really relevant to me because they\u0027re not"},{"Start":"26:46.734 ","End":"26:50.274","Text":"on the plane that I\u0027m trying to work out."},{"Start":"26:50.274 ","End":"26:53.935","Text":"Also, either way, they both cancel each other out."},{"Start":"26:53.935 ","End":"27:01.090","Text":"All that\u0027s interesting to me is the force acting on the fly on my x, y plane."},{"Start":"27:01.090 ","End":"27:06.219","Text":"The only force acting there is my static friction, and of course,"},{"Start":"27:06.219 ","End":"27:11.009","Text":"this is equal to my mass multiplied by my acceleration,"},{"Start":"27:11.009 ","End":"27:14.309","Text":"where my acceleration is of the fly."},{"Start":"27:14.309 ","End":"27:16.949","Text":"Now, in order to solve this,"},{"Start":"27:16.949 ","End":"27:21.905","Text":"we\u0027re going to be working in our frame of reference of the lab."},{"Start":"27:21.905 ","End":"27:26.754","Text":"Relative to the lab and not using a rotating frame of reference."},{"Start":"27:26.754 ","End":"27:28.464","Text":"Now, why are we doing this?"},{"Start":"27:28.464 ","End":"27:32.844","Text":"A rotating frame of reference is very difficult for the following reason."},{"Start":"27:32.844 ","End":"27:36.969","Text":"We\u0027re used to working with a rotating frame of"},{"Start":"27:36.969 ","End":"27:42.235","Text":"reference when our angular velocity is a constant."},{"Start":"27:42.235 ","End":"27:44.889","Text":"Now, in our question at hand,"},{"Start":"27:44.889 ","End":"27:48.610","Text":"our angular velocity is constantly changing."},{"Start":"27:48.610 ","End":"27:52.149","Text":"We saw that our equation for our angular velocity,"},{"Start":"27:52.149 ","End":"27:53.965","Text":"we can write it in terms of h,"},{"Start":"27:53.965 ","End":"27:55.675","Text":"in terms of this distance."},{"Start":"27:55.675 ","End":"27:57.564","Text":"As the fly advances,"},{"Start":"27:57.564 ","End":"27:59.949","Text":"our angular velocity is changing."},{"Start":"27:59.949 ","End":"28:02.635","Text":"It\u0027s going to be equal to something else."},{"Start":"28:02.635 ","End":"28:07.360","Text":"Here we saw that our angular velocity is dependent on h. Now,"},{"Start":"28:07.360 ","End":"28:12.939","Text":"if I really want to see the movement both of the center of my disk and of the fly,"},{"Start":"28:12.939 ","End":"28:17.830","Text":"where the fly is walking on the green line across the radius,"},{"Start":"28:17.830 ","End":"28:22.225","Text":"then I have to have a rotating frame of reference when my Omega,"},{"Start":"28:22.225 ","End":"28:26.500","Text":"my angular velocity is not constant."},{"Start":"28:26.500 ","End":"28:28.569","Text":"In a case like that,"},{"Start":"28:28.569 ","End":"28:32.995","Text":"when my angular velocity is not constant,"},{"Start":"28:32.995 ","End":"28:37.250","Text":"then I have to add an imaginary force."},{"Start":"28:37.860 ","End":"28:44.259","Text":"Or instead of imaginary forces, fictitious forces."},{"Start":"28:44.259 ","End":"28:50.155","Text":"It will be equal to negative m Alpha r,"},{"Start":"28:50.155 ","End":"28:57.160","Text":"where my Alpha is the change in my rotation of the system."},{"Start":"28:57.160 ","End":"29:01.870","Text":"Now, we don\u0027t really know how to use this and we don\u0027t really want to use this,"},{"Start":"29:01.870 ","End":"29:03.430","Text":"especially at this stage,"},{"Start":"29:03.430 ","End":"29:07.839","Text":"so that means that I can\u0027t use a rotational frame if I want to"},{"Start":"29:07.839 ","End":"29:13.555","Text":"see how my fly is moving just on the x-axis."},{"Start":"29:13.555 ","End":"29:16.870","Text":"If I am going to use a rotational frame of"},{"Start":"29:16.870 ","End":"29:19.915","Text":"reference when my angular velocity is constant,"},{"Start":"29:19.915 ","End":"29:23.410","Text":"so when my fly will reach the edge,"},{"Start":"29:23.410 ","End":"29:27.730","Text":"so we\u0027ll see that it\u0027s the angular velocity of my fly at"},{"Start":"29:27.730 ","End":"29:31.959","Text":"the edge was not the angular velocity of the fly a moment before the edge,"},{"Start":"29:31.959 ","End":"29:36.200","Text":"which means that it\u0027s going to have some angular acceleration."},{"Start":"29:36.480 ","End":"29:42.219","Text":"If we use a rotational frame of reference when my Omega is not constant,"},{"Start":"29:42.219 ","End":"29:45.250","Text":"then we\u0027re going to have to use this fictitious force,"},{"Start":"29:45.250 ","End":"29:48.020","Text":"which we don\u0027t yet know how to use."},{"Start":"29:48.300 ","End":"29:52.615","Text":"That\u0027s why we\u0027re not using our rotational frame of reference,"},{"Start":"29:52.615 ","End":"29:59.240","Text":"but what we will do is we\u0027re going to use our frame of reference of the lab."},{"Start":"29:59.670 ","End":"30:03.565","Text":"We\u0027re using our lab frame of reference."},{"Start":"30:03.565 ","End":"30:06.565","Text":"Now, let\u0027s write out our acceleration."},{"Start":"30:06.565 ","End":"30:09.770","Text":"Now, because I\u0027m working in circular motion,"},{"Start":"30:10.620 ","End":"30:12.910","Text":"our motion is in a circle,"},{"Start":"30:12.910 ","End":"30:15.355","Text":"so I want to use polar coordinates."},{"Start":"30:15.355 ","End":"30:18.579","Text":"My acceleration in polar coordinates is equal to r double dot"},{"Start":"30:18.579 ","End":"30:26.484","Text":"minus Theta dot r in the r direction,"},{"Start":"30:26.484 ","End":"30:31.135","Text":"plus 2r dot Theta dot"},{"Start":"30:31.135 ","End":"30:39.110","Text":"plus r Theta double dot in the Theta direction."},{"Start":"30:39.300 ","End":"30:42.669","Text":"Notice with our Theta double dot over here,"},{"Start":"30:42.669 ","End":"30:45.519","Text":"that even in the labs frame of reference,"},{"Start":"30:45.519 ","End":"30:48.460","Text":"we\u0027re going to have angular acceleration."},{"Start":"30:48.460 ","End":"30:52.280","Text":"Soon we\u0027re going to see how to calculate that."},{"Start":"30:53.010 ","End":"30:58.840","Text":"Let\u0027s start calculating the different components for this equation."},{"Start":"30:58.840 ","End":"31:02.650","Text":"We\u0027re going to have our r. Now, what is our r?"},{"Start":"31:02.650 ","End":"31:07.090","Text":"It\u0027s going to be the distance the fly is from our origin."},{"Start":"31:07.090 ","End":"31:10.960","Text":"Now, notice our origin is still this green point over here."},{"Start":"31:10.960 ","End":"31:18.440","Text":"We saw that the distance our fly is from the origin is always going to be x_0."},{"Start":"31:18.740 ","End":"31:22.994","Text":"This was also equal to our x center of mass,"},{"Start":"31:22.994 ","End":"31:25.140","Text":"which we worked out in question number 1,"},{"Start":"31:25.140 ","End":"31:33.894","Text":"I think for capital M multiplied by h divided by m plus M. Now,"},{"Start":"31:33.894 ","End":"31:35.710","Text":"this is just our size."},{"Start":"31:35.710 ","End":"31:36.955","Text":"It\u0027s not a vector."},{"Start":"31:36.955 ","End":"31:39.265","Text":"I\u0027m not looking at the direction."},{"Start":"31:39.265 ","End":"31:43.990","Text":"This is our r, so now if I want my r dot,"},{"Start":"31:43.990 ","End":"31:48.760","Text":"my first derivative, then it\u0027s going to be equal to the derivative of this."},{"Start":"31:48.760 ","End":"31:54.115","Text":"My variable is my h. My h is my variable."},{"Start":"31:54.115 ","End":"31:56.935","Text":"I just take the derivative of my h,"},{"Start":"31:56.935 ","End":"32:01.694","Text":"so then I\u0027ll be left with my M divided by"},{"Start":"32:01.694 ","End":"32:09.050","Text":"my m plus M and then multiplied by my h dot."},{"Start":"32:09.050 ","End":"32:14.229","Text":"To remind you, my h is the distance between the center"},{"Start":"32:14.229 ","End":"32:18.655","Text":"of the disk and my fly and it\u0027s changing at a constant rate,"},{"Start":"32:18.655 ","End":"32:22.900","Text":"which is dependent on my v_0 over here."},{"Start":"32:22.900 ","End":"32:26.725","Text":"Therefore it\u0027s moving at a constant velocity of v_0."},{"Start":"32:26.725 ","End":"32:31.330","Text":"I can just say that this instead of my h dot I can put in my v_0."},{"Start":"32:31.330 ","End":"32:38.090","Text":"It\u0027s going to be M divided by m plus M multiplied by v_0."},{"Start":"32:38.700 ","End":"32:44.005","Text":"Then what is our r double dot going to be equal to?"},{"Start":"32:44.005 ","End":"32:50.560","Text":"We know that it\u0027s going to be M divided by m plus M multiplied by h double dot."},{"Start":"32:50.560 ","End":"32:55.390","Text":"Now because we saw that our h double dot is a constant velocity of v_0,"},{"Start":"32:55.390 ","End":"32:58.660","Text":"so our h double dot which is the acceleration,"},{"Start":"32:58.660 ","End":"33:03.710","Text":"is going to be equal to 0. r double dot is equal to 0."},{"Start":"33:04.680 ","End":"33:07.494","Text":"Now I have my h over here."},{"Start":"33:07.494 ","End":"33:11.814","Text":"Now, I don\u0027t want to have my R in terms of my h,"},{"Start":"33:11.814 ","End":"33:14.949","Text":"especially because I\u0027m speaking about what will happen"},{"Start":"33:14.949 ","End":"33:18.670","Text":"when the fly is at the edge of the disk."},{"Start":"33:18.670 ","End":"33:22.810","Text":"I\u0027m speaking about my R when my h,"},{"Start":"33:22.810 ","End":"33:26.440","Text":"this, is equal to my radius of the disk."},{"Start":"33:26.440 ","End":"33:29.050","Text":"I\u0027m just going to substitute in R over here,"},{"Start":"33:29.050 ","End":"33:33.114","Text":"so it\u0027s going to be MR divided by m plus"},{"Start":"33:33.114 ","End":"33:40.000","Text":"M. Now our Theta double is equal to our Omega final."},{"Start":"33:40.000 ","End":"33:42.039","Text":"Our Omega R over here,"},{"Start":"33:42.039 ","End":"33:47.409","Text":"which we found in question number 2 and also over here,"},{"Start":"33:47.409 ","End":"33:51.055","Text":"it\u0027s going to be this, which we have."},{"Start":"33:51.055 ","End":"33:54.054","Text":"Now we have our r double dot which is 0,"},{"Start":"33:54.054 ","End":"33:56.530","Text":"our Theta dot which is our Omega R. Our r,"},{"Start":"33:56.530 ","End":"33:59.650","Text":"we have, our r dot, we have."},{"Start":"33:59.650 ","End":"34:01.180","Text":"Our Theta dot again, we have."},{"Start":"34:01.180 ","End":"34:02.425","Text":"Our R, again we have."},{"Start":"34:02.425 ","End":"34:06.894","Text":"Now all we have to find is what our Theta double dot is equal to."},{"Start":"34:06.894 ","End":"34:12.220","Text":"My Theta double dot is equal to my Omega dot."},{"Start":"34:12.220 ","End":"34:14.275","Text":"We saw that with my r,"},{"Start":"34:14.275 ","End":"34:16.465","Text":"I could take the derivative of my r,"},{"Start":"34:16.465 ","End":"34:19.930","Text":"because I can take the derivative of my h, which is my variable."},{"Start":"34:19.930 ","End":"34:24.564","Text":"Now what I want to do is I want to write out my equation for"},{"Start":"34:24.564 ","End":"34:29.635","Text":"Omega as a function of h and leave it as a function of h,"},{"Start":"34:29.635 ","End":"34:32.965","Text":"so not to substitute in our r over here."},{"Start":"34:32.965 ","End":"34:37.255","Text":"Now let\u0027s see what our Omega is as a function of"},{"Start":"34:37.255 ","End":"34:43.495","Text":"h. Let\u0027s go back 1 second to question number 2."},{"Start":"34:43.495 ","End":"34:46.555","Text":"Here we\u0027re at question number 2,"},{"Start":"34:46.555 ","End":"34:50.605","Text":"where we use the idea of our conservation of angular momentum."},{"Start":"34:50.605 ","End":"34:57.760","Text":"A moment before we substituted in that our h has"},{"Start":"34:57.760 ","End":"35:05.919","Text":"to be equal to R. We can see that we have this equation over here as a function of h,"},{"Start":"35:05.919 ","End":"35:08.439","Text":"and not just at the final moment,"},{"Start":"35:08.439 ","End":"35:11.920","Text":"but throughout the entire motion of the disk."},{"Start":"35:11.920 ","End":"35:15.340","Text":"In order to get my Omega as a function of h,"},{"Start":"35:15.340 ","End":"35:19.330","Text":"I\u0027m going to take this whole expression over here for my L_f as"},{"Start":"35:19.330 ","End":"35:23.920","Text":"a function of h and I\u0027m going to say that it\u0027s equal my L_i,"},{"Start":"35:23.920 ","End":"35:25.300","Text":"which is this over here."},{"Start":"35:25.300 ","End":"35:28.015","Text":"Then I\u0027m just going to leave in my h\u0027s."},{"Start":"35:28.015 ","End":"35:31.319","Text":"I\u0027m not going to substitute in that h is equal to"},{"Start":"35:31.319 ","End":"35:36.030","Text":"R. Then all I\u0027m going to do is going to isolate out this my Omega as"},{"Start":"35:36.030 ","End":"35:42.040","Text":"a function of h. I\u0027m going to get that my Omega as a function of"},{"Start":"35:42.040 ","End":"35:49.840","Text":"h is equal to this divided by all of this giant bracket."},{"Start":"35:49.840 ","End":"35:52.824","Text":"Now I\u0027m going to scroll down and write that out."},{"Start":"35:52.824 ","End":"35:56.245","Text":"This is my Omega as a function of h,"},{"Start":"35:56.245 ","End":"35:59.170","Text":"so I did a little bit of algebra in order to tidy it up"},{"Start":"35:59.170 ","End":"36:02.815","Text":"a little bit and I can further tidy it up as well."},{"Start":"36:02.815 ","End":"36:08.545","Text":"What I can do is I can take out my 1/2 M R^2 as a common factor."},{"Start":"36:08.545 ","End":"36:15.209","Text":"Then what I will be left with is Omega_0 divided by 1"},{"Start":"36:15.209 ","End":"36:21.290","Text":"plus 2m h^2 divided by"},{"Start":"36:21.290 ","End":"36:28.855","Text":"M plus m multiplied by R^2."},{"Start":"36:28.855 ","End":"36:32.125","Text":"Now what I want to find is my Omega dot."},{"Start":"36:32.125 ","End":"36:35.530","Text":"My Omega dot is using the chain rule,"},{"Start":"36:35.530 ","End":"36:45.110","Text":"I\u0027m going to work out my dOmega by dt and then multiply it by my dh by dt."},{"Start":"36:45.110 ","End":"36:51.535","Text":"First, we\u0027re going to take the derivative of this,"},{"Start":"36:51.535 ","End":"36:58.824","Text":"for h. We\u0027re going to get negative Omega_0 divided by"},{"Start":"36:58.824 ","End":"37:04.510","Text":"our 1 plus 2m h^2 divided"},{"Start":"37:04.510 ","End":"37:11.125","Text":"by M plus m multiplied by R^2,"},{"Start":"37:11.125 ","End":"37:16.060","Text":"and then all of this squared."},{"Start":"37:16.060 ","End":"37:20.980","Text":"Then we\u0027re going to multiply it by our d Omega by dt,"},{"Start":"37:20.980 ","End":"37:24.055","Text":"so that\u0027s our inner derivative."},{"Start":"37:24.055 ","End":"37:29.320","Text":"Then we\u0027re going to get 4m h divided"},{"Start":"37:29.320 ","End":"37:35.919","Text":"by M plus m multiplied by our R^2,"},{"Start":"37:35.919 ","End":"37:40.375","Text":"and this is going to be multiplied by our h dot."},{"Start":"37:40.375 ","End":"37:45.740","Text":"We know that our h dot is equal to our V_0."},{"Start":"37:46.410 ","End":"37:49.884","Text":"Now what we want is we want"},{"Start":"37:49.884 ","End":"37:58.130","Text":"our Omega dot when our h is equal to R. When the fly is on the edge of the disk."},{"Start":"37:58.320 ","End":"38:02.350","Text":"What we can do is everywhere we see an h is we"},{"Start":"38:02.350 ","End":"38:06.159","Text":"substitute in our R. Then what we\u0027ll get is something along"},{"Start":"38:06.159 ","End":"38:14.815","Text":"the lines of negative Omega_0 multiplied by 4mR multiplied by V_0."},{"Start":"38:14.815 ","End":"38:20.289","Text":"Then divided by our m plus M,"},{"Start":"38:20.289 ","End":"38:23.650","Text":"and this is going to be multiplied by our R^2."},{"Start":"38:23.650 ","End":"38:26.650","Text":"Then this R and 1 of these Rs can cancel out,"},{"Start":"38:26.650 ","End":"38:31.824","Text":"and all of this is going to be divided by with the common denominator."},{"Start":"38:31.824 ","End":"38:37.225","Text":"It\u0027s going to be M plus"},{"Start":"38:37.225 ","End":"38:43.599","Text":"3m divided by M plus m,"},{"Start":"38:43.599 ","End":"38:46.210","Text":"and this is going to be squared."},{"Start":"38:46.210 ","End":"38:48.789","Text":"Then this is going to be equal to,"},{"Start":"38:48.789 ","End":"38:54.700","Text":"let\u0027s sort this out a little bit."},{"Start":"38:54.700 ","End":"38:56.755","Text":"It\u0027s going to be equal to our negative over here stays,"},{"Start":"38:56.755 ","End":"39:06.745","Text":"and then it\u0027s going to be 4m multiplied by M plus m multiplied by V_0 and Omega_0."},{"Start":"39:06.745 ","End":"39:19.299","Text":"Then all of this is going to be divided by M plus 3m"},{"Start":"39:19.299 ","End":"39:25.704","Text":"squared multiplied by R. This is our equation for"},{"Start":"39:25.704 ","End":"39:29.214","Text":"our Omega dot when h is equal to"},{"Start":"39:29.214 ","End":"39:34.640","Text":"R. I\u0027m going to rub out all of the equations and let\u0027s substitute everything in."},{"Start":"39:35.130 ","End":"39:39.400","Text":"Now going back to our original question,"},{"Start":"39:39.400 ","End":"39:45.309","Text":"we wanted to see what our f_s is equal to when our h is around about equal"},{"Start":"39:45.309 ","End":"39:51.460","Text":"to our R. We came up with the sum of all of the forces is equal to our f_s,"},{"Start":"39:51.460 ","End":"39:53.810","Text":"which is equal to our ma."},{"Start":"39:54.000 ","End":"39:57.385","Text":"We\u0027re trying to find out what our f_s is equal to."},{"Start":"39:57.385 ","End":"40:00.985","Text":"Our a, we said was equal to this expression over here."},{"Start":"40:00.985 ","End":"40:02.769","Text":"Then at r double-dot,"},{"Start":"40:02.769 ","End":"40:06.369","Text":"we\u0027re going to substitute in 0 over here at r Theta dot,"},{"Start":"40:06.369 ","End":"40:10.480","Text":"which is our Omega R, we\u0027ll substitute what we found for question number 2."},{"Start":"40:10.480 ","End":"40:14.200","Text":"Our r will substitute in this over"},{"Start":"40:14.200 ","End":"40:17.950","Text":"here when our h is equal to r. That\u0027s in our r direction."},{"Start":"40:17.950 ","End":"40:19.270","Text":"Then in our Theta direction,"},{"Start":"40:19.270 ","End":"40:21.580","Text":"we\u0027re going to have 2 multiply by r dots,"},{"Start":"40:21.580 ","End":"40:24.610","Text":"which is this, multiplied by our Omega r,"},{"Start":"40:24.610 ","End":"40:26.590","Text":"which we\u0027ve already found in question 2,"},{"Start":"40:26.590 ","End":"40:31.390","Text":"plus our R multiplied by r Theta double dot,"},{"Start":"40:31.390 ","End":"40:37.900","Text":"which is simply this with a dot on top."},{"Start":"40:37.900 ","End":"40:42.010","Text":"Now, we can just substitute all of this in and I\u0027m just"},{"Start":"40:42.010 ","End":"40:45.880","Text":"going to write it out what our f_s is equal to."},{"Start":"40:45.880 ","End":"40:51.160","Text":"You can do this on your own on a piece of paper and see if you get the same answer."},{"Start":"40:51.160 ","End":"40:56.769","Text":"This is going to be the final answer that we get for our frictional force."},{"Start":"40:56.769 ","End":"40:59.739","Text":"Now, it\u0027s filled with a lot of tedious algebra,"},{"Start":"40:59.739 ","End":"41:01.270","Text":"so that\u0027s why I didn\u0027t do it."},{"Start":"41:01.270 ","End":"41:05.809","Text":"I\u0027m also not going to read out the answer. You can take a look at it."},{"Start":"41:05.820 ","End":"41:08.949","Text":"But what we can do in order to check that"},{"Start":"41:08.949 ","End":"41:11.919","Text":"this answer makes sense and is in the right direction?"},{"Start":"41:11.919 ","End":"41:16.254","Text":"We can look back at what we did in our question number 3, where we said,"},{"Start":"41:16.254 ","End":"41:21.513","Text":"what if the mass of the fly is significantly smaller than the mass of the disk?"},{"Start":"41:21.513 ","End":"41:25.840","Text":"Then we saw that our Omega at point R when h was equal"},{"Start":"41:25.840 ","End":"41:31.220","Text":"to R was equal to our initial angular velocity of Omega_0."},{"Start":"41:31.590 ","End":"41:37.270","Text":"If now we apply this to our f_s,"},{"Start":"41:37.270 ","End":"41:43.555","Text":"we\u0027ll see that our f_s is going to equal to negative."},{"Start":"41:43.555 ","End":"41:45.550","Text":"Our m, M,"},{"Start":"41:45.550 ","End":"41:54.782","Text":"and then this m can cancel out so this will be M^4 Omega_0^2 R in the r direction,"},{"Start":"41:54.782 ","End":"41:58.285","Text":"and then divided by our m^2,"},{"Start":"41:58.285 ","End":"42:05.305","Text":"and 4mM here cancel out divided by our M and this is also squared, so M^4."},{"Start":"42:05.305 ","End":"42:07.120","Text":"Then these cancel out,"},{"Start":"42:07.120 ","End":"42:13.660","Text":"and we\u0027ll be left with negative small m Omega_0^2 R in the r direction."},{"Start":"42:13.660 ","End":"42:20.395","Text":"Now if we look and we imagine that we have a disc of center here,"},{"Start":"42:20.395 ","End":"42:26.920","Text":"and we have some object or body moving at a constant velocity V_0 in a straight line,"},{"Start":"42:26.920 ","End":"42:31.405","Text":"and our desk is moving at a constant angular velocity of Omega_0."},{"Start":"42:31.405 ","End":"42:39.070","Text":"Then what we\u0027ll get is a centrifugal force pushing inwards, with this value,"},{"Start":"42:39.070 ","End":"42:42.760","Text":"the equation for a centrifugal force is going to be negative m multiplied"},{"Start":"42:42.760 ","End":"42:47.455","Text":"by Omega_0^2 R. Our Omega_0,"},{"Start":"42:47.455 ","End":"42:50.005","Text":"or angular velocity is a constant."},{"Start":"42:50.005 ","End":"42:55.720","Text":"Here we can also work with a rotating frame of reference,"},{"Start":"42:55.720 ","End":"43:00.550","Text":"because we can see that our angular velocity is constant throughout."},{"Start":"43:00.550 ","End":"43:04.030","Text":"Then we can see that our frictional force is going to"},{"Start":"43:04.030 ","End":"43:07.779","Text":"be equal to negative our centrifugal force."},{"Start":"43:07.779 ","End":"43:10.495","Text":"Our frictional force is going to be negative"},{"Start":"43:10.495 ","End":"43:15.385","Text":"our fictitious force in the radial direction."},{"Start":"43:15.385 ","End":"43:22.570","Text":"Now let\u0027s also apply this rule over to what we have going on in the Theta direction."},{"Start":"43:22.570 ","End":"43:24.865","Text":"We\u0027re going to have plus"},{"Start":"43:24.865 ","End":"43:33.430","Text":"our m M V_0 Omega_0 multiplied by our 2M,"},{"Start":"43:33.430 ","End":"43:39.310","Text":"because this M will cancel out and then divide it by just this our capital M^2."},{"Start":"43:39.310 ","End":"43:44.155","Text":"This will cancel out and then minus our 4m"},{"Start":"43:44.155 ","End":"43:52.495","Text":"divided by capital M^2."},{"Start":"43:52.495 ","End":"43:54.699","Text":"Now looking at these terms,"},{"Start":"43:54.699 ","End":"44:01.795","Text":"we have 2 divided by capital M minus a very small number divided by a very big number."},{"Start":"44:01.795 ","End":"44:06.070","Text":"When we have a very small number divided by a large denominator,"},{"Start":"44:06.070 ","End":"44:10.435","Text":"so we can say that this is equal to 0."},{"Start":"44:10.435 ","End":"44:14.980","Text":"Then I don\u0027t have this expression anymore,"},{"Start":"44:14.980 ","End":"44:18.715","Text":"and then I can cancel out this M with this M,"},{"Start":"44:18.715 ","End":"44:24.009","Text":"and then I can just re-write my Theta-hat section of"},{"Start":"44:24.009 ","End":"44:30.680","Text":"the equation as 2 from here, m V_0 Omega_0."},{"Start":"44:30.840 ","End":"44:37.060","Text":"This is exactly the size of my Coriolis force."},{"Start":"44:37.060 ","End":"44:42.145","Text":"We know that my Coriolis force is always going to be 2 times the mass,"},{"Start":"44:42.145 ","End":"44:44.664","Text":"multiplied by my V_0"},{"Start":"44:44.664 ","End":"44:47.710","Text":"cross with my Omega_0, sorry, there are meant to be in the opposite direction."},{"Start":"44:47.710 ","End":"44:54.985","Text":"When these 2 are perpendicular to 1 another."},{"Start":"44:54.985 ","End":"44:59.215","Text":"This is our Coriolis force and this is exactly what we get."},{"Start":"44:59.215 ","End":"45:03.620","Text":"My Coriolis force is perpendicular."},{"Start":"45:03.990 ","End":"45:08.185","Text":"Now we\u0027ve seen that it makes sense for this."},{"Start":"45:08.185 ","End":"45:12.550","Text":"Now of course, that doesn\u0027t mean that a 100 percent we\u0027re correct,"},{"Start":"45:12.550 ","End":"45:15.710","Text":"but it\u0027s a good start to check yourself."},{"Start":"45:15.900 ","End":"45:18.220","Text":"Here we\u0027ve finished the question."},{"Start":"45:18.220 ","End":"45:19.644","Text":"It\u0027s a bit of a difficult question,"},{"Start":"45:19.644 ","End":"45:21.680","Text":"but it\u0027s a good 1."}],"ID":9479},{"Watched":false,"Name":"Half Sphere In Harmonic Motion","Duration":"29m 2s","ChapterTopicVideoID":9210,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.280","Text":"Hello. In this question we\u0027re being told that we have a half-sphere"},{"Start":"00:04.280 ","End":"00:08.775","Text":"of radius R and mass M which is resting on a surface."},{"Start":"00:08.775 ","End":"00:12.090","Text":"The half-sphere is tilted a small angle"},{"Start":"00:12.090 ","End":"00:15.735","Text":"from its state of equilibrium and is released from rest."},{"Start":"00:15.735 ","End":"00:18.135","Text":"We\u0027re being asked to find the frequency of"},{"Start":"00:18.135 ","End":"00:24.150","Text":"small oscillations if the half-sphere rolls without slipping."},{"Start":"00:24.150 ","End":"00:28.620","Text":"The center of mass of the half-sphere is located a distance of"},{"Start":"00:28.620 ","End":"00:34.045","Text":"d is equal to 3/8R from the center of mass of a full sphere."},{"Start":"00:34.045 ","End":"00:37.550","Text":"Here we can see our drawing of half-sphere."},{"Start":"00:37.550 ","End":"00:40.310","Text":"You can imagine another line going"},{"Start":"00:40.310 ","End":"00:43.015","Text":"over here in order to complete and make it a full sphere,"},{"Start":"00:43.015 ","End":"00:45.090","Text":"this point over here is the origin."},{"Start":"00:45.090 ","End":"00:46.535","Text":"If this was a full sphere,"},{"Start":"00:46.535 ","End":"00:48.635","Text":"this would be the center of the sphere."},{"Start":"00:48.635 ","End":"00:51.980","Text":"This d, this distance and this point over"},{"Start":"00:51.980 ","End":"00:55.985","Text":"here represents the center of mass of the half-sphere."},{"Start":"00:55.985 ","End":"01:00.044","Text":"This dotted line represents our point of equilibrium is,"},{"Start":"01:00.044 ","End":"01:04.420","Text":"and usually and this is the angle that our half-sphere was tilted."},{"Start":"01:04.420 ","End":"01:10.425","Text":"Of course, the radius of the half-sphere over here is R and"},{"Start":"01:10.425 ","End":"01:12.870","Text":"our half-spheres of mass M."},{"Start":"01:12.870 ","End":"01:17.360","Text":"What we\u0027re trying to find is the frequency of small oscillations."},{"Start":"01:17.360 ","End":"01:20.910","Text":"We know that we\u0027re going to be trying to find our harmonic motion"},{"Start":"01:20.910 ","End":"01:24.715","Text":"and equation representing the harmonic motion of this half-sphere."},{"Start":"01:24.715 ","End":"01:30.695","Text":"In order to do that we\u0027re going to use equations for our forces and for our torques."},{"Start":"01:30.695 ","End":"01:36.410","Text":"The hardest thing in this question is to decide on an axis of rotation."},{"Start":"01:36.410 ","End":"01:42.400","Text":"If I choose my axis of rotation to be at my point here O,"},{"Start":"01:42.400 ","End":"01:45.600","Text":"so then I\u0027d have to work out my I_o."},{"Start":"01:45.600 ","End":"01:49.835","Text":"The problem with that is that my point o is actually"},{"Start":"01:49.835 ","End":"01:55.575","Text":"an acceleration and that\u0027s because it\u0027s not stationary as our sphere is rolling,"},{"Start":"01:55.575 ","End":"01:58.835","Text":"so we can see that our point o is going to be moving."},{"Start":"01:58.835 ","End":"02:01.295","Text":"If our point o is accelerating"},{"Start":"02:01.295 ","End":"02:04.180","Text":"that means that if we choose it to be our axis of rotation,"},{"Start":"02:04.180 ","End":"02:07.160","Text":"then our axis of rotation is accelerating."},{"Start":"02:07.160 ","End":"02:08.434","Text":"Now this is a problem,"},{"Start":"02:08.434 ","End":"02:11.285","Text":"we\u0027re not used to working with"},{"Start":"02:11.285 ","End":"02:16.010","Text":"a axis of rotation which is moving let alone an acceleration."},{"Start":"02:16.010 ","End":"02:20.165","Text":"The only time that we have used an axis which is moving"},{"Start":"02:20.165 ","End":"02:24.810","Text":"is when our axis of rotation was located at our center of mass."},{"Start":"02:25.150 ","End":"02:28.535","Text":"We\u0027re not going to choose this point."},{"Start":"02:28.535 ","End":"02:34.650","Text":"The next point that I could choose is for my center of mass."},{"Start":"02:34.970 ","End":"02:39.065","Text":"Also my center of mass has some weird motion,"},{"Start":"02:39.065 ","End":"02:41.525","Text":"so it\u0027s also moving."},{"Start":"02:41.525 ","End":"02:44.480","Text":"It\u0027s going in circular motion and also"},{"Start":"02:44.480 ","End":"02:48.275","Text":"moving and even if I were to be able to deal with that."},{"Start":"02:48.275 ","End":"02:54.750","Text":"Over here I\u0027m going to have a torques due to the normal force and the frictional force,"},{"Start":"02:54.750 ","End":"02:58.030","Text":"and they\u0027ll be very difficult to calculate."},{"Start":"02:58.030 ","End":"03:00.425","Text":"I wrote that for my center of mass."},{"Start":"03:00.425 ","End":"03:04.010","Text":"It has complex movement and the torque of"},{"Start":"03:04.010 ","End":"03:10.280","Text":"the frictional force and our torque for our normal force are difficult to calculate."},{"Start":"03:10.280 ","End":"03:15.560","Text":"Although technically speaking we can work with this, it\u0027s not good."},{"Start":"03:15.560 ","End":"03:19.060","Text":"It\u0027s got to be very difficult, so this also not."},{"Start":"03:19.060 ","End":"03:23.315","Text":"Our last option is to choose this point over here,"},{"Start":"03:23.315 ","End":"03:24.905","Text":"let\u0027s call it A."},{"Start":"03:24.905 ","End":"03:30.230","Text":"This is the exact point of contact between the surface and our half-sphere."},{"Start":"03:30.230 ","End":"03:32.840","Text":"The axis of rotation at this point of"},{"Start":"03:32.840 ","End":"03:37.195","Text":"contact is an axes which is stationary, it doesn\u0027t move."},{"Start":"03:37.195 ","End":"03:41.107","Text":"Even though our point of contact is always going to be different,"},{"Start":"03:41.107 ","End":"03:45.630","Text":"we still consider it as an axis which is constant with time."},{"Start":"03:45.630 ","End":"03:50.165","Text":"A way to think of that is for every short miniscule moment."},{"Start":"03:50.165 ","End":"03:53.420","Text":"We can see that even though our point has changed for that moment"},{"Start":"03:53.420 ","End":"03:57.433","Text":"that our axis that our point A is at the new position,"},{"Start":"03:57.433 ","End":"04:00.715","Text":"so our axes are stationary for that moment."},{"Start":"04:00.715 ","End":"04:02.340","Text":"At our point A,"},{"Start":"04:02.340 ","End":"04:08.830","Text":"our axis is fixed in time for short periods of time."},{"Start":"04:08.830 ","End":"04:12.260","Text":"What we\u0027re going to do is we\u0027re going to choose this axis."},{"Start":"04:12.260 ","End":"04:13.760","Text":"I hear you asking,"},{"Start":"04:13.760 ","End":"04:21.125","Text":"why can I say that my point A is a fixed axis even though the axis is still moving?"},{"Start":"04:21.125 ","End":"04:24.687","Text":"It\u0027s going to be located at a different point as a function of time,"},{"Start":"04:24.687 ","End":"04:27.740","Text":"then to say that my axis if it was at o it would"},{"Start":"04:27.740 ","End":"04:31.330","Text":"be an accelerating axis because my Point O is accelerating."},{"Start":"04:31.330 ","End":"04:33.845","Text":"How come this difference?"},{"Start":"04:33.845 ","End":"04:35.690","Text":"Isn\u0027t it just the same?"},{"Start":"04:35.690 ","End":"04:42.050","Text":"The velocity or the acceleration of an axis is decided upon"},{"Start":"04:42.050 ","End":"04:45.020","Text":"by the point located on the axis of"},{"Start":"04:45.020 ","End":"04:49.130","Text":"a rotation and finding its velocity or its acceleration."},{"Start":"04:49.130 ","End":"04:52.850","Text":"If we were to put an axis at our origin we"},{"Start":"04:52.850 ","End":"04:56.660","Text":"can see that our point on the origin which is our o point."},{"Start":"04:56.660 ","End":"05:00.845","Text":"We can see that it has a velocity and it has an acceleration,"},{"Start":"05:00.845 ","End":"05:03.350","Text":"and so therefore because our o is moving,"},{"Start":"05:03.350 ","End":"05:06.855","Text":"so would be the axis at our Point O."},{"Start":"05:06.855 ","End":"05:11.975","Text":"Let\u0027s write this over here."},{"Start":"05:11.975 ","End":"05:17.420","Text":"Because in the question, we\u0027re being told that our half-sphere rolls without slipping,"},{"Start":"05:17.420 ","End":"05:21.760","Text":"so because we have rolling without slipping;"},{"Start":"05:21.760 ","End":"05:29.090","Text":"RWS, then that means that the point of contact is stationary."},{"Start":"05:29.090 ","End":"05:34.835","Text":"In other words, the velocity at point A=0"},{"Start":"05:34.835 ","End":"05:38.045","Text":"because there is rolling without slipping so"},{"Start":"05:38.045 ","End":"05:41.120","Text":"that means that even though our point is moving along."},{"Start":"05:41.120 ","End":"05:46.965","Text":"Every single short miniscule moment are point that is in contact"},{"Start":"05:46.965 ","End":"05:49.940","Text":"between our sphere and our surface is going"},{"Start":"05:49.940 ","End":"05:53.141","Text":"to be stationary because we\u0027re rolling without slipping."},{"Start":"05:53.141 ","End":"05:55.610","Text":"That means that the point on our axis of"},{"Start":"05:55.610 ","End":"05:59.210","Text":"rotation which is located at our point of contact,"},{"Start":"05:59.210 ","End":"06:04.010","Text":"is going to be stationary and that means that we have a stationary axis."},{"Start":"06:04.010 ","End":"06:05.675","Text":"I changed the wording over here."},{"Start":"06:05.675 ","End":"06:11.685","Text":"At my point A I have a momentarily fixed axis,"},{"Start":"06:11.685 ","End":"06:16.660","Text":"and that\u0027s because we can see that the point on our axis;"},{"Start":"06:16.660 ","End":"06:19.870","Text":"so the point on our axis of rotation which is this point A."},{"Start":"06:19.870 ","End":"06:23.510","Text":"Even though it\u0027s changing because we have rolling without slipping,"},{"Start":"06:23.510 ","End":"06:27.860","Text":"the velocity at our point A is going to be equal to 0 which means that"},{"Start":"06:27.860 ","End":"06:30.830","Text":"the velocity of our axis is also going to be equal"},{"Start":"06:30.830 ","End":"06:34.340","Text":"to 0 which means that we have a stationary axis."},{"Start":"06:34.340 ","End":"06:39.740","Text":"In that case I can just work with this axis as per usual and I don\u0027t have to worry about"},{"Start":"06:39.740 ","End":"06:45.535","Text":"fictitious forces and axis of rotation which are accelerating."},{"Start":"06:45.535 ","End":"06:52.610","Text":"In general, when dealing with these types of questions where we have that one,"},{"Start":"06:52.610 ","End":"06:57.310","Text":"our center of mass is not in the center of the shape."},{"Start":"06:57.310 ","End":"07:03.155","Text":"Like here in our half-sphere or if we have a half-sphere with some cone on top,"},{"Start":"07:03.155 ","End":"07:06.875","Text":"or a full sphere where our center of mass is somewhere here and not in the center,"},{"Start":"07:06.875 ","End":"07:12.495","Text":"anything like that and number 2 that we have rolling without"},{"Start":"07:12.495 ","End":"07:18.600","Text":"slipping and number 3 we have a point of contact."},{"Start":"07:18.600 ","End":"07:21.325","Text":"Then when we have all these three,"},{"Start":"07:21.325 ","End":"07:24.770","Text":"then all the time you\u0027re going to choose your axis of"},{"Start":"07:24.770 ","End":"07:29.746","Text":"rotation to be at your point of contact always."},{"Start":"07:29.746 ","End":"07:33.955","Text":"This can be in your test in many different forms as you can see,"},{"Start":"07:33.955 ","End":"07:37.000","Text":"and this makes it a lot easier."},{"Start":"07:37.000 ","End":"07:39.535","Text":"I hope that that was understandable."},{"Start":"07:39.535 ","End":"07:46.525","Text":"Now let\u0027s go to solving the question using our Point A is our axes of rotation."},{"Start":"07:46.525 ","End":"07:50.890","Text":"We chose our Point A to be our axis of rotation."},{"Start":"07:50.890 ","End":"07:56.440","Text":"That\u0027s also going to be our origin and we can take these red lines to be our 2 axes."},{"Start":"07:56.440 ","End":"08:01.660","Text":"The next thing that we\u0027re going to do is we\u0027re going to find our sum of all of"},{"Start":"08:01.660 ","End":"08:08.215","Text":"our torques that are acting on the shape relative to this Point A over here,"},{"Start":"08:08.215 ","End":"08:12.190","Text":"and we\u0027re going to say that they are equal to I Alpha."},{"Start":"08:12.190 ","End":"08:19.330","Text":"Well, our Alpha is our angular acceleration of the entire body about this Point A."},{"Start":"08:19.330 ","End":"08:23.560","Text":"The forces that we have acting is we have from our center of mass,"},{"Start":"08:23.560 ","End":"08:25.300","Text":"we have our mg,"},{"Start":"08:25.300 ","End":"08:29.665","Text":"and then we also have over here our normal force,"},{"Start":"08:29.665 ","End":"08:34.045","Text":"and going from over here we have our static friction."},{"Start":"08:34.045 ","End":"08:37.030","Text":"We don\u0027t really know in which direction our static friction is going,"},{"Start":"08:37.030 ","End":"08:38.200","Text":"but it doesn\u0027t really matter."},{"Start":"08:38.200 ","End":"08:42.175","Text":"The torques for our normal and our f_s,"},{"Start":"08:42.175 ","End":"08:46.270","Text":"for a normal force and our frictional force are equal to 0."},{"Start":"08:46.270 ","End":"08:49.840","Text":"Why? Because if you remember the equation for the torque and"},{"Start":"08:49.840 ","End":"08:55.090","Text":"involves the position that the force is acting from relative to our origin."},{"Start":"08:55.090 ","End":"08:57.190","Text":"Now because both are normal force and"},{"Start":"08:57.190 ","End":"09:00.640","Text":"our frictional force are acting from the point of the origin,"},{"Start":"09:00.640 ","End":"09:05.395","Text":"that means that there are vector which is"},{"Start":"09:05.395 ","End":"09:11.380","Text":"their distance relative to the origin for both the normal force and our frictional force."},{"Start":"09:11.380 ","End":"09:14.980","Text":"They\u0027re going to be equal to 0."},{"Start":"09:14.980 ","End":"09:19.780","Text":"Which means that our torques for these two forces will be equal to 0."},{"Start":"09:19.780 ","End":"09:25.615","Text":"In that case, let\u0027s about our normal force and a frictional force,"},{"Start":"09:25.615 ","End":"09:28.195","Text":"and let\u0027s just leave our mg,"},{"Start":"09:28.195 ","End":"09:30.595","Text":"what we have to work out a torque for."},{"Start":"09:30.595 ","End":"09:36.970","Text":"Now when I\u0027m trying to work out my moment or my torque from my mg,"},{"Start":"09:36.970 ","End":"09:46.045","Text":"I have to use the equation of r cross F. Now my r is going to be this vector over here."},{"Start":"09:46.045 ","End":"09:50.650","Text":"This is my r, it\u0027s the vector going from"},{"Start":"09:50.650 ","End":"09:55.360","Text":"our origin until the point from which my force is acting."},{"Start":"09:55.360 ","End":"09:59.680","Text":"My mg is acting from my center of mass."},{"Start":"09:59.680 ","End":"10:03.025","Text":"My r is going from the origin until my center of mass."},{"Start":"10:03.025 ","End":"10:07.570","Text":"Now it\u0027s going to be easier to work with our r_effective."},{"Start":"10:07.570 ","End":"10:12.670","Text":"Our r_effective is going to be this vector over here."},{"Start":"10:12.670 ","End":"10:20.750","Text":"It is the components of our r vector which is perpendicular to our mg force."},{"Start":"10:20.880 ","End":"10:27.220","Text":"Then we can calculate our torque for our mg to be equal"},{"Start":"10:27.220 ","End":"10:33.280","Text":"to a force which is mg multiplied by our r_effective."},{"Start":"10:33.280 ","End":"10:35.200","Text":"What does that equal to?"},{"Start":"10:35.200 ","End":"10:38.365","Text":"That is equal to mg and our r_effective,"},{"Start":"10:38.365 ","End":"10:41.050","Text":"if this is our angle Theta,"},{"Start":"10:41.050 ","End":"10:45.235","Text":"then we can see that this angle over here is also."},{"Start":"10:45.235 ","End":"10:51.175","Text":"Then we can say that it\u0027s multiplied by sine of Theta multiplied of course,"},{"Start":"10:51.175 ","End":"10:54.640","Text":"by d. Now we have the size of"},{"Start":"10:54.640 ","End":"10:59.170","Text":"out torque but now we have to work out the direction because it\u0027s a vector,"},{"Start":"10:59.170 ","End":"11:01.300","Text":"which means it has size and direction."},{"Start":"11:01.300 ","End":"11:03.760","Text":"Now when we\u0027re dealing with harmonic motion,"},{"Start":"11:03.760 ","End":"11:08.590","Text":"we already know that our sign over here is going to be negative."},{"Start":"11:08.590 ","End":"11:11.395","Text":"However, let\u0027s work it out the correct way."},{"Start":"11:11.395 ","End":"11:19.630","Text":"A good way to start is to work with our diagram being positive. What does that mean?"},{"Start":"11:19.630 ","End":"11:22.000","Text":"If here I\u0027ve from my dotted line,"},{"Start":"11:22.000 ","End":"11:24.895","Text":"opened up this angle Theta."},{"Start":"11:24.895 ","End":"11:28.420","Text":"I want this angle Theta to be a positive angle."},{"Start":"11:28.420 ","End":"11:32.110","Text":"In that case, everything to the right of"},{"Start":"11:32.110 ","End":"11:36.115","Text":"this dotted line going in in anticlockwise direction."},{"Start":"11:36.115 ","End":"11:38.260","Text":"Everything here is going to be positive,"},{"Start":"11:38.260 ","End":"11:39.880","Text":"and if I opened my Theta,"},{"Start":"11:39.880 ","End":"11:46.210","Text":"then over here, so then I would have some negative angle over there."},{"Start":"11:46.210 ","End":"11:51.130","Text":"In that case, I can say that my positive direction of"},{"Start":"11:51.130 ","End":"11:56.300","Text":"rotation is going to be this anticlockwise direction."},{"Start":"11:56.300 ","End":"12:01.030","Text":"Now I can look at the torque that my mg has created."},{"Start":"12:01.030 ","End":"12:06.355","Text":"We can see that my mg is trying to close my Theta angle."},{"Start":"12:06.355 ","End":"12:09.175","Text":"It\u0027s trying to move in this direction,"},{"Start":"12:09.175 ","End":"12:10.810","Text":"in my clockwise direction,"},{"Start":"12:10.810 ","End":"12:12.610","Text":"which is my negative direction."},{"Start":"12:12.610 ","End":"12:16.220","Text":"Then I can add a negative sign over here."},{"Start":"12:16.220 ","End":"12:20.760","Text":"Another way to think of it is that we can see that our mg is always"},{"Start":"12:20.760 ","End":"12:25.485","Text":"trying to push our system back to its point of equilibrium."},{"Start":"12:25.485 ","End":"12:28.665","Text":"Which means that our mg is a restorative force."},{"Start":"12:28.665 ","End":"12:33.490","Text":"It\u0027s trying to restore equilibrium to the system."},{"Start":"12:33.490 ","End":"12:35.965","Text":"Every time we have a restorative force,"},{"Start":"12:35.965 ","End":"12:39.385","Text":"the sign is going to be a negative."},{"Start":"12:39.385 ","End":"12:47.205","Text":"This is our torque of mg and that has to be equal to our I multiplied by Alpha."},{"Start":"12:47.205 ","End":"12:53.130","Text":"Now our I has to be relative to this Point A over here,"},{"Start":"12:53.130 ","End":"12:55.895","Text":"because this is our axis of rotation."},{"Start":"12:55.895 ","End":"12:57.700","Text":"How am I going to do this?"},{"Start":"12:57.700 ","End":"13:06.470","Text":"I\u0027m first going to start by writing out my moment of inertia for a half-sphere,"},{"Start":"13:06.570 ","End":"13:12.655","Text":"which is equal to the moment of inertia of a full sphere."},{"Start":"13:12.655 ","End":"13:19.045","Text":"It\u0027s equal to 2/5 multiplied by mR^2."},{"Start":"13:19.045 ","End":"13:20.470","Text":"Let\u0027s just explain this,"},{"Start":"13:20.470 ","End":"13:25.975","Text":"we know that the moment of inertia of a full sphere is going to be equal like so."},{"Start":"13:25.975 ","End":"13:28.735","Text":"Let\u0027s say I have a full sphere"},{"Start":"13:28.735 ","End":"13:33.100","Text":"and the center is right over here at the center of the sphere."},{"Start":"13:33.100 ","End":"13:42.445","Text":"My moment of inertia for my full sphere is going to be 2/5(mR)^2."},{"Start":"13:42.445 ","End":"13:46.520","Text":"Our sphere is made up of 2 halves."},{"Start":"13:46.950 ","End":"13:51.865","Text":"Our moment of inertia of the full sphere is made up of I,"},{"Start":"13:51.865 ","End":"13:56.575","Text":"of I first 1/2 plus I moment of inertia of our second 1/2."},{"Start":"13:56.575 ","End":"14:01.960","Text":"Now we know that our I_1 and I_2 are equal to each other because they are symmetrical."},{"Start":"14:01.960 ","End":"14:08.200","Text":"In that case we can get that are I_1 is going to be equal to our I_2,"},{"Start":"14:08.200 ","End":"14:14.230","Text":"which is equal to 1/2 of our I for our full sphere."},{"Start":"14:14.230 ","End":"14:19.840","Text":"That is going to be equal to 1/2 of m"},{"Start":"14:19.840 ","End":"14:25.255","Text":"over here multiplied by 2/5(R)^2."},{"Start":"14:25.255 ","End":"14:30.730","Text":"Now 1/2m is exactly the mass of 1/2 a sphere."},{"Start":"14:30.730 ","End":"14:37.480","Text":"Then we can see that our I for a half-sphere is simply equal"},{"Start":"14:37.480 ","End":"14:45.415","Text":"to the mass of a half-sphere multiplied by 2/5 times R^2."},{"Start":"14:45.415 ","End":"14:48.800","Text":"That\u0027s exactly what we have over here."},{"Start":"14:48.840 ","End":"14:54.440","Text":"Our m here is the mass of 1/2 of a sphere and not of a full sphere."},{"Start":"14:54.600 ","End":"14:57.445","Text":"That\u0027s an explanation for this equation."},{"Start":"14:57.445 ","End":"15:01.010","Text":"Now I\u0027m going to rub this out and let\u0027s carry on from here."},{"Start":"15:01.380 ","End":"15:06.610","Text":"This moment of inertia that I have is at my Point O,"},{"Start":"15:06.610 ","End":"15:11.440","Text":"on my origin, which was the center of my full sphere."},{"Start":"15:11.440 ","End":"15:19.510","Text":"Now I need to find my moment of inertia relative to this axis of rotation."},{"Start":"15:19.510 ","End":"15:24.250","Text":"I can now do Steiner from this point to this point."},{"Start":"15:24.250 ","End":"15:27.775","Text":"Because neither one of these points is"},{"Start":"15:27.775 ","End":"15:31.975","Text":"my center of mass and I always have to work with my center of mass."},{"Start":"15:31.975 ","End":"15:33.785","Text":"In order to solve this,"},{"Start":"15:33.785 ","End":"15:36.140","Text":"I have to find my I at c.m,"},{"Start":"15:36.140 ","End":"15:40.350","Text":"and then from there to work it out to my Point A."},{"Start":"15:40.830 ","End":"15:47.655","Text":"We\u0027ll get that our I_c.m is equal to our I_o"},{"Start":"15:47.655 ","End":"15:56.455","Text":"minus our mass multiplied by the distance between the axis of rotation squared."},{"Start":"15:56.455 ","End":"16:01.255","Text":"This is a different d, let\u0027s call it d wave, d Tilde."},{"Start":"16:01.255 ","End":"16:04.140","Text":"Notice that there is a minus over here."},{"Start":"16:04.140 ","End":"16:10.005","Text":"Because it usually my I_o would be my I_c.m plus my Steiner."},{"Start":"16:10.005 ","End":"16:13.280","Text":"I\u0027ve rearranged the equation to isolate out my I_c.m,"},{"Start":"16:13.280 ","End":"16:16.125","Text":"and that\u0027s why there\u0027s a minus over here."},{"Start":"16:16.125 ","End":"16:19.685","Text":"Now we can substitute in our values over here."},{"Start":"16:19.685 ","End":"16:26.215","Text":"We have that our I_o is equal to 2/5mR^2"},{"Start":"16:26.215 ","End":"16:35.035","Text":"minus our m. Then what we have over here is d Tilde over here."},{"Start":"16:35.035 ","End":"16:38.620","Text":"Because usually our sine equation really is"},{"Start":"16:38.620 ","End":"16:42.680","Text":"some I for your center of mass"},{"Start":"16:42.680 ","End":"16:47.225","Text":"plus m multiplied by the distance between the axes squared."},{"Start":"16:47.225 ","End":"16:50.420","Text":"Now here the distance is some kind of distance."},{"Start":"16:50.420 ","End":"16:51.620","Text":"It can be anything,"},{"Start":"16:51.620 ","End":"16:57.110","Text":"and here specifically in the question they told us that we have some value for"},{"Start":"16:57.110 ","End":"17:03.350","Text":"our d. This is some a distance that isn\u0027t necessarily relating to this same distance."},{"Start":"17:03.350 ","End":"17:09.385","Text":"This is the arbitrary letter given to represent this in our Steiner equation."},{"Start":"17:09.385 ","End":"17:14.830","Text":"But here our d=3 divided by"},{"Start":"17:14.830 ","End":"17:20.485","Text":"8 multiplied by R and all of this is squared."},{"Start":"17:20.485 ","End":"17:22.325","Text":"Now if we play around with this,"},{"Start":"17:22.325 ","End":"17:26.195","Text":"we\u0027ll get that our answer is equal to 83"},{"Start":"17:26.195 ","End":"17:33.500","Text":"divided by 320 mR^2."},{"Start":"17:33.500 ","End":"17:36.620","Text":"Now we have our I for our center of mass,"},{"Start":"17:36.620 ","End":"17:39.605","Text":"and then we want to get to this position over here."},{"Start":"17:39.605 ","End":"17:44.795","Text":"We\u0027ll say therefore that our I at our Point A over here at the bottom"},{"Start":"17:44.795 ","End":"17:50.590","Text":"is going to be equal to our I_c.m plus our Steiner."},{"Start":"17:50.590 ","End":"17:55.660","Text":"It\u0027s going to be m multiplied by our d Tilde^2."},{"Start":"17:55.660 ","End":"17:57.220","Text":"It\u0027s this over here,"},{"Start":"17:57.220 ","End":"18:00.265","Text":"our Steiner theorem additive."},{"Start":"18:00.265 ","End":"18:02.795","Text":"That means that we\u0027re going to have to add"},{"Start":"18:02.795 ","End":"18:07.850","Text":"our I center of mass plus what is our d over here."},{"Start":"18:07.850 ","End":"18:09.440","Text":"It\u0027s from our axes,"},{"Start":"18:09.440 ","End":"18:11.840","Text":"it\u0027s the distance between our 2 axes."},{"Start":"18:11.840 ","End":"18:17.760","Text":"It\u0027s going to be at plus our m multiplied by our r vector^2."},{"Start":"18:18.770 ","End":"18:21.750","Text":"What is this r vector?"},{"Start":"18:21.750 ","End":"18:25.585","Text":"Let\u0027s take a look. We have over here,"},{"Start":"18:25.585 ","End":"18:27.490","Text":"r vector is equal to."},{"Start":"18:27.490 ","End":"18:33.560","Text":"Let\u0027s see, it\u0027s x component this distance over here, our r_effective,"},{"Start":"18:33.560 ","End":"18:38.760","Text":"it is going to be equal to our d,"},{"Start":"18:38.760 ","End":"18:46.050","Text":"multiplied by sine of Theta in the x-direction."},{"Start":"18:46.060 ","End":"18:49.910","Text":"We worked it out over here, where we have over here,"},{"Start":"18:49.910 ","End":"18:53.810","Text":"and then in our y-direction so I\u0027ll have plus."},{"Start":"18:53.810 ","End":"18:55.220","Text":"Now let\u0027s see what it is."},{"Start":"18:55.220 ","End":"19:04.475","Text":"We know that this total length over here is equal to r, it\u0027s the radius."},{"Start":"19:04.475 ","End":"19:12.725","Text":"Then we know that this length over here until this point is going to be,"},{"Start":"19:12.725 ","End":"19:13.970","Text":"so if this is our d,"},{"Start":"19:13.970 ","End":"19:15.050","Text":"and this is our Theta,"},{"Start":"19:15.050 ","End":"19:17.854","Text":"so we want the adjacent."},{"Start":"19:17.854 ","End":"19:22.690","Text":"That means that this length over here is"},{"Start":"19:22.690 ","End":"19:28.225","Text":"going to be equal to d multiplied by cosine of Theta."},{"Start":"19:28.225 ","End":"19:31.510","Text":"Because our d is our hypotenuse and then we"},{"Start":"19:31.510 ","End":"19:35.870","Text":"have our adjacent angle to our Theta angle over here."},{"Start":"19:36.240 ","End":"19:39.460","Text":"In order to find this length,"},{"Start":"19:39.460 ","End":"19:43.270","Text":"we wanted to find here what\u0027s in gray."},{"Start":"19:43.270 ","End":"19:44.965","Text":"We have to take our R,"},{"Start":"19:44.965 ","End":"19:49.195","Text":"our total length minus this section over here."},{"Start":"19:49.195 ","End":"19:53.245","Text":"Therefore will have in y-direction we have"},{"Start":"19:53.245 ","End":"20:01.085","Text":"R minus d multiplied by cosine of Theta like so."},{"Start":"20:01.085 ","End":"20:04.975","Text":"In our equation we see that we have our r vector^2."},{"Start":"20:04.975 ","End":"20:07.510","Text":"Whenever we see some vector^2,"},{"Start":"20:07.510 ","End":"20:11.020","Text":"it means that we\u0027re trying to find the size of the vector."},{"Start":"20:11.020 ","End":"20:13.450","Text":"Let\u0027s start it from here."},{"Start":"20:13.450 ","End":"20:21.400","Text":"We have our r vector^2 which is equal to the size of our vector^2."},{"Start":"20:21.400 ","End":"20:26.095","Text":"That is going to be equal to d^2"},{"Start":"20:26.095 ","End":"20:30.910","Text":"sine^2 Theta plus"},{"Start":"20:30.910 ","End":"20:37.435","Text":"our R minus d cosine Theta^2."},{"Start":"20:37.435 ","End":"20:40.030","Text":"Notice when we\u0027re squaring our vector quantity,"},{"Start":"20:40.030 ","End":"20:43.270","Text":"it\u0027s just the size of the vector^2."},{"Start":"20:43.270 ","End":"20:46.630","Text":"We can get rid of our x-hat and our y-hat."},{"Start":"20:46.630 ","End":"20:56.785","Text":"Now when we open up this brackets we\u0027ll get d^2 sine^2 Theta plus R^2"},{"Start":"20:56.785 ","End":"21:02.530","Text":"minus 2Rd cosine of"},{"Start":"21:02.530 ","End":"21:10.720","Text":"Theta plus d^2 cosine^2 Theta."},{"Start":"21:10.720 ","End":"21:13.330","Text":"Now we can simplify this even more."},{"Start":"21:13.330 ","End":"21:17.050","Text":"We know that whenever we have sine^2 Theta plus cosine^2 Theta,"},{"Start":"21:17.050 ","End":"21:19.160","Text":"it\u0027s equal to 1."},{"Start":"21:23.880 ","End":"21:26.740","Text":"Here we have our d^2,"},{"Start":"21:26.740 ","End":"21:31.630","Text":"a common factor multiplied by our sine^2 Theta plus our"},{"Start":"21:31.630 ","End":"21:37.350","Text":"cosine^2 Theta and then plus"},{"Start":"21:37.350 ","End":"21:44.335","Text":"our R^2 minus 2Rd cosine of Theta."},{"Start":"21:44.335 ","End":"21:49.940","Text":"Then what\u0027s inside this bracket is going to be equal to 1."},{"Start":"21:50.430 ","End":"21:55.105","Text":"Then that\u0027s simply going to be equal to"},{"Start":"21:55.105 ","End":"22:06.080","Text":"d^2 plus R^2 minus 2Rd cosine of Theta."},{"Start":"22:06.600 ","End":"22:09.550","Text":"Let\u0027s look at the question again."},{"Start":"22:09.550 ","End":"22:13.900","Text":"Now we\u0027re being told that we\u0027re using small oscillations."},{"Start":"22:13.900 ","End":"22:17.845","Text":"Small oscillations mean small angles."},{"Start":"22:17.845 ","End":"22:23.545","Text":"That means that our Theta is going to be significantly smaller than 1."},{"Start":"22:23.545 ","End":"22:34.060","Text":"When that happens, so we can write our approximations as"},{"Start":"22:34.060 ","End":"22:39.760","Text":"our cosine Theta is simply going to be equal to 1"},{"Start":"22:39.760 ","End":"22:47.650","Text":"and our sine Theta is going to be around about equal to Theta."},{"Start":"22:47.650 ","End":"22:50.785","Text":"Now notice over here our cosine Theta,"},{"Start":"22:50.785 ","End":"22:58.165","Text":"were approximating to 1 and not 1 minus Theta^2 divided by 2."},{"Start":"22:58.165 ","End":"23:00.325","Text":"Now why are we doing it like that?"},{"Start":"23:00.325 ","End":"23:03.460","Text":"Because over here we have our equation model a sine"},{"Start":"23:03.460 ","End":"23:06.430","Text":"Theta and when we\u0027re working with small angles,"},{"Start":"23:06.430 ","End":"23:10.405","Text":"we saw that our sine Theta is approximately equal to Theta."},{"Start":"23:10.405 ","End":"23:15.130","Text":"If we would approximate our cosine Theta to equal around"},{"Start":"23:15.130 ","End":"23:19.810","Text":"about 1 minus Theta^2 divided by 2"},{"Start":"23:19.810 ","End":"23:26.500","Text":"then our Theta^2 divided by 2 will be significantly smaller than our Theta over here."},{"Start":"23:26.500 ","End":"23:33.145","Text":"We would get a negligible results and aside from a negligible result,"},{"Start":"23:33.145 ","End":"23:35.275","Text":"generally when we\u0027re working with our forces,"},{"Start":"23:35.275 ","End":"23:38.830","Text":"we don\u0027t want to have our Theta to the power of"},{"Start":"23:38.830 ","End":"23:42.805","Text":"2 or Theta^2 because we don\u0027t know how to work with that enforces."},{"Start":"23:42.805 ","End":"23:46.480","Text":"Only with energy we know how to solve those types of equations."},{"Start":"23:46.480 ","End":"23:52.105","Text":"Now let\u0027s substitute in everything that we have into our equation over here."},{"Start":"23:52.105 ","End":"23:53.725","Text":"What we\u0027re going to have,"},{"Start":"23:53.725 ","End":"23:55.570","Text":"we\u0027ll start with this side of the equation."},{"Start":"23:55.570 ","End":"23:59.460","Text":"We\u0027ll have negative mg multiplied by"},{"Start":"23:59.460 ","End":"24:03.660","Text":"d and then out sine Theta because we\u0027re working with our small angles so"},{"Start":"24:03.660 ","End":"24:07.725","Text":"our sine Theta is going to be equal to Theta and that"},{"Start":"24:07.725 ","End":"24:12.315","Text":"is equal to our I at our point A over here."},{"Start":"24:12.315 ","End":"24:17.255","Text":"Our I at our point A is equal to our I at our center of mass"},{"Start":"24:17.255 ","End":"24:26.080","Text":"plus M multiplied by our r vector^2 which is what we worked out over here."},{"Start":"24:26.080 ","End":"24:29.155","Text":"What was our I at our center of mass?"},{"Start":"24:29.155 ","End":"24:30.580","Text":"It\u0027s this over here."},{"Start":"24:30.580 ","End":"24:33.355","Text":"We\u0027re going to have 83"},{"Start":"24:33.355 ","End":"24:40.780","Text":"divided by 320MR^2,"},{"Start":"24:40.780 ","End":"24:50.320","Text":"that\u0027s our I_cm plus our M multiplied by our I vector^2,"},{"Start":"24:50.320 ","End":"24:52.840","Text":"which is over here,"},{"Start":"24:52.840 ","End":"24:58.405","Text":"our d^2 plus R^2 minus 2dR."},{"Start":"24:58.405 ","End":"25:01.885","Text":"Then here we have our cosine Theta."},{"Start":"25:01.885 ","End":"25:08.335","Text":"Our cosine Theta when dealing with small angles is equal to 1, so simply that."},{"Start":"25:08.335 ","End":"25:12.760","Text":"Then, because we have to multiply our I at our point A by Alpha,"},{"Start":"25:12.760 ","End":"25:20.845","Text":"so we can multiply everything over here by Alpha."},{"Start":"25:20.845 ","End":"25:23.335","Text":"Put in brackets and Alpha."},{"Start":"25:23.335 ","End":"25:25.375","Text":"Then instead of writing Alpha,"},{"Start":"25:25.375 ","End":"25:28.690","Text":"I can write it because I know that I have my equation in terms of"},{"Start":"25:28.690 ","End":"25:33.470","Text":"Theta so I can write it in terms of Theta double dot."},{"Start":"25:33.510 ","End":"25:37.075","Text":"Now let\u0027s simplify this equation."},{"Start":"25:37.075 ","End":"25:44.770","Text":"We\u0027re going to have negative mg and then our value for d if we look back at the question,"},{"Start":"25:44.770 ","End":"25:46.510","Text":"let\u0027s go up a little bit."},{"Start":"25:46.510 ","End":"25:50.630","Text":"Our value for d is 3 divided by 8R."},{"Start":"25:50.670 ","End":"25:57.265","Text":"We can write instead of d we\u0027ll have 3 divided by 8 multiplied by I."},{"Start":"25:57.265 ","End":"26:03.070","Text":"Then this is multiplied by Theta and this is going to be equal to"},{"Start":"26:03.070 ","End":"26:11.875","Text":"so we\u0027ll have our 83 divided by 320 MR^2."},{"Start":"26:11.875 ","End":"26:19.040","Text":"We can actually take our M outside of the brackets because it\u0027s common factor."},{"Start":"26:19.080 ","End":"26:23.260","Text":"R^2 plus our M which is located here."},{"Start":"26:23.260 ","End":"26:28.030","Text":"Then if we look at what we have over here inside our brackets,"},{"Start":"26:28.030 ","End":"26:31.510","Text":"so we\u0027ll see that it\u0027s simply a quadratic equation."},{"Start":"26:31.510 ","End":"26:40.675","Text":"If we simplify the equation we\u0027ll get that this is equal to d minus R^2^2."},{"Start":"26:40.675 ","End":"26:44.140","Text":"Then we can put this over here."},{"Start":"26:44.140 ","End":"26:46.945","Text":"Rd minus I so what\u0027s our d?"},{"Start":"26:46.945 ","End":"26:53.380","Text":"It\u0027s 3/8R so it\u0027s going to be plus our 3/8R^2."},{"Start":"26:53.380 ","End":"26:54.955","Text":"Let\u0027s just write it over here,"},{"Start":"26:54.955 ","End":"27:02.620","Text":"3/8(r) minus R and all of this squared."},{"Start":"27:02.620 ","End":"27:10.405","Text":"That\u0027s going to be equal to 5 divided by 8 R^2."},{"Start":"27:10.405 ","End":"27:15.610","Text":"Now we can substitute that in here so 5 divided by 8^2 is going to"},{"Start":"27:15.610 ","End":"27:21.295","Text":"be 25 divided by 64 multiplied by R^2."},{"Start":"27:21.295 ","End":"27:28.240","Text":"Then all of this multiplied by our Alpha or our Theta double-dot."},{"Start":"27:28.240 ","End":"27:31.217","Text":"Now we\u0027re looking at this equation so,"},{"Start":"27:31.217 ","End":"27:34.510","Text":"we can see that we can divide both sides by our M."},{"Start":"27:34.510 ","End":"27:38.335","Text":"Our m and our M are the same so divide both sides by"},{"Start":"27:38.335 ","End":"27:46.225","Text":"m. We can also divide both sides by our R. Now let\u0027s do the calculation."},{"Start":"27:46.225 ","End":"27:50.215","Text":"Then we\u0027ll get that we have negative 3 divided by"},{"Start":"27:50.215 ","End":"27:55.765","Text":"8 Theta is equal 2 and then once we add these two together,"},{"Start":"27:55.765 ","End":"27:57.940","Text":"we have to find the common denominator."},{"Start":"27:57.940 ","End":"28:08.485","Text":"I have that this is equal to 208 divided by 320 multiplied by R Theta double dot."},{"Start":"28:08.485 ","End":"28:14.905","Text":"Then all we have to do in order to find our frequency or Omega_0."},{"Start":"28:14.905 ","End":"28:22.285","Text":"We know that it\u0027s going to be the square root of this divided by this so 3"},{"Start":"28:22.285 ","End":"28:25.540","Text":"divided by 8 get the coefficient of our Theta"},{"Start":"28:25.540 ","End":"28:29.980","Text":"divided by the coefficient of our Theta double dot."},{"Start":"28:29.980 ","End":"28:36.310","Text":"When I\u0027m doing a fraction divided by another fraction so I just multiply by its inverse."},{"Start":"28:36.310 ","End":"28:45.655","Text":"Multiplied by 320 divided by 208 R. Then we can simplify this a little bit further,"},{"Start":"28:45.655 ","End":"28:51.655","Text":"and we\u0027ll get it\u0027s the square root of 15 divided by 26"},{"Start":"28:51.655 ","End":"28:59.000","Text":"multiplied by g divided by R. That\u0027s our final answer."},{"Start":"28:59.310 ","End":"29:03.200","Text":"That\u0027s the end of the question."}],"ID":9480},{"Watched":false,"Name":"Energy Lost","Duration":"17m 55s","ChapterTopicVideoID":9211,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Hello. In this question,"},{"Start":"00:03.045 ","End":"00:06.135","Text":"we\u0027re dealing with energy and pulse."},{"Start":"00:06.135 ","End":"00:11.715","Text":"We\u0027re told that a mass m here is placed resting on a conveyor belt."},{"Start":"00:11.715 ","End":"00:14.010","Text":"This is the conveyor belt."},{"Start":"00:14.010 ","End":"00:17.130","Text":"The conveyor belt has a friction of Mu"},{"Start":"00:17.130 ","End":"00:21.960","Text":"and the mass is placed at a time equal to t equals 0."},{"Start":"00:21.960 ","End":"00:24.180","Text":"The conveyor belt is acted on by"},{"Start":"00:24.180 ","End":"00:28.695","Text":"an external force which pulls the bells at a constant velocity u,"},{"Start":"00:28.695 ","End":"00:32.530","Text":"so the belt is being pulled at you."},{"Start":"00:33.710 ","End":"00:39.080","Text":"What is happening is that the mass is placed on the conveyor belt,"},{"Start":"00:39.080 ","End":"00:42.760","Text":"which is already moving at a constant velocity u."},{"Start":"00:42.760 ","End":"00:46.460","Text":"From the moment that I placed the mass on the conveyor belt,"},{"Start":"00:46.460 ","End":"00:49.370","Text":"there\u0027ll be some kind of friction acting on it,"},{"Start":"00:49.370 ","End":"00:54.440","Text":"which will begin to move the mass along with the conveyor belt."},{"Start":"00:54.440 ","End":"00:57.320","Text":"However, the mass will still not be moving at"},{"Start":"00:57.320 ","End":"01:01.210","Text":"the exact velocity of the conveyor belt, it\u0027s not attached."},{"Start":"01:01.210 ","End":"01:05.945","Text":"It will take a while and then the mass will begin to move at the same velocity."},{"Start":"01:05.945 ","End":"01:11.415","Text":"Now, this period is what we call sliding."},{"Start":"01:11.415 ","End":"01:15.480","Text":"We\u0027re going to have sliding with friction."},{"Start":"01:15.480 ","End":"01:18.420","Text":"Let\u0027s answer all of the questions."},{"Start":"01:18.420 ","End":"01:21.303","Text":"Our first question is,"},{"Start":"01:21.303 ","End":"01:25.865","Text":"what is the force acting on the conveyor belt?"},{"Start":"01:25.865 ","End":"01:27.890","Text":"That\u0027s what we\u0027re going to answer first."},{"Start":"01:27.890 ","End":"01:29.705","Text":"Let\u0027s look at a diagram."},{"Start":"01:29.705 ","End":"01:34.295","Text":"Now, I\u0027ve put a point here to symbolize"},{"Start":"01:34.295 ","End":"01:39.695","Text":"this area where the mass is being placed on the conveyor belt."},{"Start":"01:39.695 ","End":"01:44.330","Text":"Once the mass has been placed and the conveyor belt has moved at velocity Mu,"},{"Start":"01:44.330 ","End":"01:47.870","Text":"we can see that this point has moved to here."},{"Start":"01:47.870 ","End":"01:54.560","Text":"However, the mass m has moved significantly less."},{"Start":"01:54.560 ","End":"01:57.710","Text":"Now, we expected this because the mass isn\u0027t yet"},{"Start":"01:57.710 ","End":"02:01.870","Text":"moving at the same velocity as the conveyor belt."},{"Start":"02:01.870 ","End":"02:07.235","Text":"This is exactly the sliding that I was talking about."},{"Start":"02:07.235 ","End":"02:10.565","Text":"If there wasn\u0027t this sliding tendency then"},{"Start":"02:10.565 ","End":"02:14.705","Text":"the mass would be moving at the exact same velocity as the conveyor belt."},{"Start":"02:14.705 ","End":"02:18.725","Text":"Now, as soon as the sliding period of time is over,"},{"Start":"02:18.725 ","End":"02:22.775","Text":"then the mass and the conveyor belt will move at the exact same velocity."},{"Start":"02:22.775 ","End":"02:28.280","Text":"It\u0027s only at the beginning until the frictional forces and other forces balance."},{"Start":"02:28.280 ","End":"02:30.724","Text":"Now, in order to answer,"},{"Start":"02:30.724 ","End":"02:33.170","Text":"what is the force acting on the conveyor belt,"},{"Start":"02:33.170 ","End":"02:34.550","Text":"let\u0027s look at this."},{"Start":"02:34.550 ","End":"02:36.920","Text":"When the mass is placed on the conveyor belt,"},{"Start":"02:36.920 ","End":"02:38.420","Text":"so let\u0027s look over here."},{"Start":"02:38.420 ","End":"02:45.180","Text":"Now, the friction that the mass producers on the conveyor belt is in this direction,"},{"Start":"02:45.180 ","End":"02:48.140","Text":"so the friction that the conveyor belt"},{"Start":"02:48.140 ","End":"02:52.055","Text":"produces on the mass must be in the opposite direction,"},{"Start":"02:52.055 ","End":"02:55.850","Text":"so it\u0027s in the right direction."},{"Start":"02:55.850 ","End":"02:57.815","Text":"Now as a rule,"},{"Start":"02:57.815 ","End":"03:01.565","Text":"if the conveyor belt is moving in rightwards direction,"},{"Start":"03:01.565 ","End":"03:06.340","Text":"then the friction on the mass will also be in a rightwards direction."},{"Start":"03:06.340 ","End":"03:12.230","Text":"Now, if there\u0027s this friction on the conveyor belt,"},{"Start":"03:12.230 ","End":"03:16.370","Text":"and we\u0027re told that it\u0027s moving at a constant velocity of u,"},{"Start":"03:16.370 ","End":"03:18.380","Text":"which means that there\u0027s no acceleration,"},{"Start":"03:18.380 ","End":"03:20.570","Text":"the acceleration equals to 0."},{"Start":"03:20.570 ","End":"03:24.875","Text":"Then there must be another force going in this direction,"},{"Start":"03:24.875 ","End":"03:27.590","Text":"which must be exactly equal and opposite to"},{"Start":"03:27.590 ","End":"03:32.630","Text":"this force in order to keep the velocity constant."},{"Start":"03:32.630 ","End":"03:35.900","Text":"Now this force is an external force,"},{"Start":"03:35.900 ","End":"03:38.585","Text":"so let\u0027s write external,"},{"Start":"03:38.585 ","End":"03:41.795","Text":"and then we can write down the equation,"},{"Start":"03:41.795 ","End":"03:48.380","Text":"so F external equals the kinetic friction,"},{"Start":"03:48.380 ","End":"03:53.455","Text":"which means that it\u0027s equals not smaller than or equal to Mu"},{"Start":"03:53.455 ","End":"03:59.255","Text":"of mg. What happens here is that we have the F external,"},{"Start":"03:59.255 ","End":"04:01.475","Text":"which is acting on the conveyor belt."},{"Start":"04:01.475 ","End":"04:04.385","Text":"After it\u0027s acted on the conveyor belt,"},{"Start":"04:04.385 ","End":"04:07.065","Text":"it also acts on the mass."},{"Start":"04:07.065 ","End":"04:12.005","Text":"Now let\u0027s talk about the acceleration of the mass."},{"Start":"04:12.005 ","End":"04:16.295","Text":"I\u0027m going to call acceleration of the mass a\u0027."},{"Start":"04:16.295 ","End":"04:21.745","Text":"This isn\u0027t the sign to derive, it\u0027s a\u0027."},{"Start":"04:21.745 ","End":"04:24.635","Text":"Now every time I\u0027m going to speak about the forces"},{"Start":"04:24.635 ","End":"04:27.560","Text":"acting on the mass and not on the conveyor belt,"},{"Start":"04:27.560 ","End":"04:30.380","Text":"I\u0027m going to call them something prime."},{"Start":"04:30.380 ","End":"04:33.935","Text":"Whenever I\u0027m talking about forces acting on the conveyor belt,"},{"Start":"04:33.935 ","End":"04:37.620","Text":"such as here, then it will be without a prime."},{"Start":"04:37.620 ","End":"04:39.800","Text":"As we know, F=ma,"},{"Start":"04:39.800 ","End":"04:47.345","Text":"so I can say that a=F over m. Now,"},{"Start":"04:47.345 ","End":"04:48.845","Text":"which F are we talking about?"},{"Start":"04:48.845 ","End":"04:50.705","Text":"We\u0027re talking about the frictional force."},{"Start":"04:50.705 ","End":"04:54.200","Text":"This frictional force is equal to, as we know,"},{"Start":"04:54.200 ","End":"04:57.650","Text":"Mu times the normal. Now what is the normal?"},{"Start":"04:57.650 ","End":"05:01.205","Text":"We have mg going downwards and the normal is"},{"Start":"05:01.205 ","End":"05:06.270","Text":"equal and opposite to the mg. We know that it\u0027s Mumg."},{"Start":"05:07.570 ","End":"05:11.975","Text":"Here, this is equal to the frictional force."},{"Start":"05:11.975 ","End":"05:16.355","Text":"So we have Mumg divided by m,"},{"Start":"05:16.355 ","End":"05:22.225","Text":"these cross out, which means that the acceleration of the mass is equal to Mug."},{"Start":"05:22.225 ","End":"05:27.320","Text":"Now we\u0027re being asked for how long will the mass continue sliding."},{"Start":"05:27.320 ","End":"05:31.925","Text":"Now what does this mean? The mass is placed at rest on the conveyor belt."},{"Start":"05:31.925 ","End":"05:36.095","Text":"We\u0027re being asked how long it will take for the masses velocity"},{"Start":"05:36.095 ","End":"05:41.975","Text":"being 0 over here until it equals the velocity of the conveyor belt,"},{"Start":"05:41.975 ","End":"05:44.090","Text":"which is equal to u."},{"Start":"05:44.090 ","End":"05:46.454","Text":"How long will it slide?"},{"Start":"05:46.454 ","End":"05:48.040","Text":"That\u0027s just what it means."},{"Start":"05:48.040 ","End":"05:52.275","Text":"How long from resting until velocity u,"},{"Start":"05:52.275 ","End":"05:54.460","Text":"velocity of the conveyor belt."},{"Start":"05:54.460 ","End":"05:58.690","Text":"Let\u0027s see how we do this. The first thing that we know is"},{"Start":"05:58.690 ","End":"06:05.245","Text":"that once the mass reaches the velocity of u of the conveyor belt,"},{"Start":"06:05.245 ","End":"06:08.140","Text":"that the forces acting on it will equal"},{"Start":"06:08.140 ","End":"06:11.590","Text":"0 because the mass will be moving at a constant velocity."},{"Start":"06:11.590 ","End":"06:14.200","Text":"The forces acting on it are 0."},{"Start":"06:14.200 ","End":"06:15.730","Text":"Then, first of all,"},{"Start":"06:15.730 ","End":"06:17.260","Text":"there won\u0027t be any kinetic friction,"},{"Start":"06:17.260 ","End":"06:20.905","Text":"if anything, there will be a dynamic friction and also the dynamic friction,"},{"Start":"06:20.905 ","End":"06:23.410","Text":"sorry, not dynamic, static friction,"},{"Start":"06:23.410 ","End":"06:29.480","Text":"static friction will equal 0 because there are no forces acting on the mass."},{"Start":"06:30.020 ","End":"06:35.450","Text":"Now let\u0027s find how the velocity of the mass changes with time,"},{"Start":"06:35.450 ","End":"06:39.815","Text":"so we have v prime because it\u0027s the velocity of the mass,"},{"Start":"06:39.815 ","End":"06:41.960","Text":"which is a function of time,"},{"Start":"06:41.960 ","End":"06:44.450","Text":"because as time moves on,"},{"Start":"06:44.450 ","End":"06:48.515","Text":"the velocity of the mass increases."},{"Start":"06:48.515 ","End":"06:54.020","Text":"We know that the velocity is just acceleration multiplied by time,"},{"Start":"06:54.020 ","End":"07:00.200","Text":"so we have Mug multiplied by time."},{"Start":"07:00.200 ","End":"07:08.590","Text":"Now, if we want to know at what time the mass will reach the velocity u."},{"Start":"07:08.590 ","End":"07:12.045","Text":"Let\u0027s separate the T,"},{"Start":"07:12.045 ","End":"07:13.830","Text":"let\u0027s put it to one side."},{"Start":"07:13.830 ","End":"07:16.475","Text":"I\u0027m going to call it big T because this is"},{"Start":"07:16.475 ","End":"07:21.005","Text":"the specific time at which the velocity will equal u,"},{"Start":"07:21.005 ","End":"07:26.620","Text":"and that just equals u divided by Mug,"},{"Start":"07:26.780 ","End":"07:32.390","Text":"so this is the time that it will take the mass to reach velocity u."},{"Start":"07:32.390 ","End":"07:35.240","Text":"Now, I could have called this T,"},{"Start":"07:35.240 ","End":"07:38.195","Text":"T final or whatever,"},{"Start":"07:38.195 ","End":"07:40.190","Text":"decided here to call it a big T,"},{"Start":"07:40.190 ","End":"07:42.990","Text":"but it\u0027s the same thing. Same meaning."},{"Start":"07:43.030 ","End":"07:47.045","Text":"Now the next question that we\u0027re being asked is,"},{"Start":"07:47.045 ","End":"07:51.125","Text":"what distance will the conveyor travel this time,"},{"Start":"07:51.125 ","End":"07:52.595","Text":"so this is our time."},{"Start":"07:52.595 ","End":"07:55.715","Text":"Now, if you can see here in the original diagram,"},{"Start":"07:55.715 ","End":"08:02.780","Text":"we\u0027ve placed the mass and the mass and the dot are on some imaginary line here."},{"Start":"08:02.780 ","End":"08:04.490","Text":"They start at the same place."},{"Start":"08:04.490 ","End":"08:09.155","Text":"Now, because the conveyor belt is initially"},{"Start":"08:09.155 ","End":"08:14.300","Text":"moving at a faster velocity relative to the mass,"},{"Start":"08:14.300 ","End":"08:17.660","Text":"so the blue dot from here ends up here."},{"Start":"08:17.660 ","End":"08:22.310","Text":"Now, let\u0027s see the distance that the blue dot traveled in the time that"},{"Start":"08:22.310 ","End":"08:27.415","Text":"it took for the velocity of the mass to catch up to that of the conveyor belt."},{"Start":"08:27.415 ","End":"08:30.345","Text":"Let\u0027s write down this equation."},{"Start":"08:30.345 ","End":"08:35.730","Text":"The displacement of the conveyor belt,"},{"Start":"08:35.730 ","End":"08:39.055","Text":"so it\u0027s x without a prime because it\u0027s of the conveyor belt,"},{"Start":"08:39.055 ","End":"08:44.165","Text":"is equal to velocity multiplied by time."},{"Start":"08:44.165 ","End":"08:48.700","Text":"The velocity we know is u multiplied by time."},{"Start":"08:48.700 ","End":"08:51.150","Text":"This is our time, the time that it took."},{"Start":"08:51.150 ","End":"08:58.210","Text":"We have u multiplied by u divided by Mug,"},{"Start":"08:58.210 ","End":"09:02.420","Text":"which equals u^2 over Mug."},{"Start":"09:02.880 ","End":"09:09.370","Text":"This is the distance that the conveyor belt has traveled in that time."},{"Start":"09:09.370 ","End":"09:14.575","Text":"Now, the next question that we\u0027re being asked is,"},{"Start":"09:14.575 ","End":"09:18.545","Text":"what distance the mass moves in this time?"},{"Start":"09:18.545 ","End":"09:21.655","Text":"The mass begins here,"},{"Start":"09:21.655 ","End":"09:30.940","Text":"and then it accelerates until it reaches this velocity."},{"Start":"09:30.940 ","End":"09:33.710","Text":"Let\u0027s see the distance."},{"Start":"09:35.220 ","End":"09:44.650","Text":"The displacement of the mass x\u0027 is equal to from one of the kinematic equations,"},{"Start":"09:44.650 ","End":"09:50.855","Text":"1/2at^2, which equals 1/2,"},{"Start":"09:50.855 ","End":"09:53.280","Text":"and then our a, where is it?"},{"Start":"09:53.280 ","End":"09:59.715","Text":"Is here, Mug multiplied by Mug multiplied by t^2,"},{"Start":"09:59.715 ","End":"10:01.530","Text":"which is this,"},{"Start":"10:01.530 ","End":"10:08.226","Text":"so it\u0027s u divided by Mug^2."},{"Start":"10:08.226 ","End":"10:13.890","Text":"This basically equals, we can cross out certain things,"},{"Start":"10:13.890 ","End":"10:18.930","Text":"U^2 divided by 2Mug."},{"Start":"10:18.930 ","End":"10:22.125","Text":"That\u0027s the displacement of the mass."},{"Start":"10:22.125 ","End":"10:25.215","Text":"Now, the next question we\u0027re being asked is,"},{"Start":"10:25.215 ","End":"10:29.100","Text":"how much work has the system done."},{"Start":"10:29.100 ","End":"10:33.750","Text":"Now, another way of defining"},{"Start":"10:33.750 ","End":"10:40.440","Text":"what work is is how much energy has been changed in the system."},{"Start":"10:40.440 ","End":"10:46.110","Text":"Now, the reason that this isn\u0027t exactly a 100 percent correct to say is because some of"},{"Start":"10:46.110 ","End":"10:49.335","Text":"the energy that this f external puts"},{"Start":"10:49.335 ","End":"10:53.745","Text":"into the system will be lost to heat energy and things like that."},{"Start":"10:53.745 ","End":"10:58.755","Text":"We\u0027ll check that soon but right now what we\u0027re checking"},{"Start":"10:58.755 ","End":"11:07.080","Text":"is the work done by f external on the system."},{"Start":"11:07.080 ","End":"11:13.905","Text":"Now, work is force multiplied by the distance."},{"Start":"11:13.905 ","End":"11:18.825","Text":"Now, the more formal description of what work is"},{"Start":"11:18.825 ","End":"11:24.225","Text":"is the integral of force multiplied by the distance."},{"Start":"11:24.225 ","End":"11:28.860","Text":"However, because here the force is a constant and it\u0027s not changing,"},{"Start":"11:28.860 ","End":"11:31.810","Text":"we can just say it like this."},{"Start":"11:32.210 ","End":"11:41.670","Text":"The energy that the external force has put into the system is going to be the force which"},{"Start":"11:41.670 ","End":"11:51.066","Text":"is Mumg multiplied by the distance that it\u0027s moved,"},{"Start":"11:51.066 ","End":"11:54.030","Text":"so from this dot here to here."},{"Start":"11:54.030 ","End":"11:59.325","Text":"It\u0027s going to be multiplied by u^2 divided by"},{"Start":"11:59.325 ","End":"12:05.745","Text":"Mug and then this and this cross off."},{"Start":"12:05.745 ","End":"12:09.720","Text":"Then we\u0027re left with Mu^2."},{"Start":"12:09.720 ","End":"12:16.575","Text":"Now, this is the energy that the external force put into the system."},{"Start":"12:16.575 ","End":"12:21.230","Text":"Now we\u0027re going to work out the work prime and"},{"Start":"12:21.230 ","End":"12:26.860","Text":"this is the work put in from the friction of the mass."},{"Start":"12:26.860 ","End":"12:33.780","Text":"It\u0027s going to be the same. It\u0027s got to be the force which is equal to Mumg like before"},{"Start":"12:33.780 ","End":"12:41.045","Text":"and multiplied by the distance prime of the mass."},{"Start":"12:41.045 ","End":"12:43.715","Text":"Now, this distance is going to be different."},{"Start":"12:43.715 ","End":"12:47.660","Text":"Why? Because the first distance was over the conveyor belt,"},{"Start":"12:47.660 ","End":"12:51.780","Text":"which was from here till here but now,"},{"Start":"12:51.780 ","End":"12:54.343","Text":"we\u0027re looking at the distance moved by the mass."},{"Start":"12:54.343 ","End":"12:58.260","Text":"The distance that the friction worked on,"},{"Start":"12:58.260 ","End":"13:01.457","Text":"which is from here till here."},{"Start":"13:01.457 ","End":"13:04.215","Text":"Which is a smaller distance."},{"Start":"13:04.215 ","End":"13:12.135","Text":"Let\u0027s do this. We have Mumg multiplied by here,"},{"Start":"13:12.135 ","End":"13:16.240","Text":"u^2 divided by 2Mug."},{"Start":"13:16.280 ","End":"13:20.085","Text":"The Mug crosses off,"},{"Start":"13:20.085 ","End":"13:26.925","Text":"which leaves us with mu^2 divided by 2."},{"Start":"13:26.925 ","End":"13:29.610","Text":"Let\u0027s take a look at what we have here."},{"Start":"13:29.610 ","End":"13:38.280","Text":"Now, we\u0027re being told that f external is putting in energy into the system,"},{"Start":"13:38.280 ","End":"13:39.975","Text":"onto the conveyor belt,"},{"Start":"13:39.975 ","End":"13:43.035","Text":"and it\u0027s putting in mu^2."},{"Start":"13:43.035 ","End":"13:47.580","Text":"Now, the conveyor belt is moving at a constant velocity at u,"},{"Start":"13:47.580 ","End":"13:50.910","Text":"which means that none of the energy is"},{"Start":"13:50.910 ","End":"13:55.365","Text":"acting on the conveyor belt because it\u0027s moving at a constant velocity."},{"Start":"13:55.365 ","End":"13:59.865","Text":"Now we can see that the mass took"},{"Start":"13:59.865 ","End":"14:06.465","Text":"half Mu^2 in order to get from resting to a velocity of u."},{"Start":"14:06.465 ","End":"14:12.375","Text":"What we can tell is the difference between mu^2 and"},{"Start":"14:12.375 ","End":"14:16.080","Text":"half mu^2 is the energy that was lost to"},{"Start":"14:16.080 ","End":"14:21.190","Text":"heat or sound or whatever it may be due to friction."},{"Start":"14:22.190 ","End":"14:25.890","Text":"How would we write the energy loss?"},{"Start":"14:25.890 ","End":"14:28.035","Text":"We write delta E,"},{"Start":"14:28.035 ","End":"14:33.585","Text":"which represents energy loss and it equals energy put in,"},{"Start":"14:33.585 ","End":"14:42.026","Text":"which is here mu^2 minus energy out or useful,"},{"Start":"14:42.026 ","End":"14:43.305","Text":"energy that was used,"},{"Start":"14:43.305 ","End":"14:52.930","Text":"which here is 1/2mu^2 and this equals this minus this is 1/2mu^2."},{"Start":"14:53.600 ","End":"14:58.380","Text":"This is the total energy lost in the system."},{"Start":"14:58.380 ","End":"15:04.515","Text":"Now we can see that we worked out the force acting on the conveyor belt."},{"Start":"15:04.515 ","End":"15:07.875","Text":"We worked out the acceleration of the mass."},{"Start":"15:07.875 ","End":"15:12.540","Text":"We worked out how long the mass will be sliding for."},{"Start":"15:12.540 ","End":"15:19.440","Text":"We worked out how much the conveyor belt would have moved in this time."},{"Start":"15:19.440 ","End":"15:25.607","Text":"We worked out the distance that the mass moved in this time."},{"Start":"15:25.607 ","End":"15:32.490","Text":"We saw how much work the external force put into the system and we saw"},{"Start":"15:32.490 ","End":"15:36.435","Text":"the energy that was used in order to move"},{"Start":"15:36.435 ","End":"15:41.565","Text":"the mass and we worked out how much energy was lost in the system."},{"Start":"15:41.565 ","End":"15:45.510","Text":"Now, let\u0027s look at the change in momentum,"},{"Start":"15:45.510 ","End":"15:48.390","Text":"which is another thing that they could potentially ask."},{"Start":"15:48.390 ","End":"15:54.435","Text":"What they mean by the change in momentum in these sorts of questions is,"},{"Start":"15:54.435 ","End":"16:00.675","Text":"how much momentum did this force transfer to the mass?"},{"Start":"16:00.675 ","End":"16:03.409","Text":"Let\u0027s take a look on the conveyor belt,"},{"Start":"16:03.409 ","End":"16:06.270","Text":"there\u0027s no there\u0027s no momentum."},{"Start":"16:06.270 ","End":"16:10.215","Text":"Why? Because the conveyor belt is moving at a constant velocity,"},{"Start":"16:10.215 ","End":"16:13.850","Text":"which means that the sum of all the forces on the conveyor belt is equal to 0."},{"Start":"16:13.850 ","End":"16:19.860","Text":"Which means that there is no momentum acting but now,"},{"Start":"16:19.860 ","End":"16:25.470","Text":"let\u0027s see the momentum that was transferred onto the mass."},{"Start":"16:25.470 ","End":"16:30.720","Text":"This equals to the force multiplied by the time."},{"Start":"16:30.720 ","End":"16:32.670","Text":"Because the force is constant,"},{"Start":"16:32.670 ","End":"16:35.895","Text":"we don\u0027t have to integrate here."},{"Start":"16:35.895 ","End":"16:38.685","Text":"The force, as before,"},{"Start":"16:38.685 ","End":"16:41.745","Text":"as always here in this question is"},{"Start":"16:41.745 ","End":"16:49.423","Text":"Mumg and the time that we\u0027re talking about is this time."},{"Start":"16:49.423 ","End":"16:53.955","Text":"So multiplied by u divided by Mug."},{"Start":"16:53.955 ","End":"16:58.410","Text":"Then we can cross out Mug and Mug and then we"},{"Start":"16:58.410 ","End":"17:04.695","Text":"get that the momentum is equal to m multiplied by u."},{"Start":"17:04.695 ","End":"17:07.410","Text":"Now, what\u0027s the logic behind this?"},{"Start":"17:07.410 ","End":"17:14.670","Text":"At the beginning, the mass is placed at rest and then it receives momentum in"},{"Start":"17:14.670 ","End":"17:21.450","Text":"order to start moving at the velocity of u and so it equals mu."},{"Start":"17:21.450 ","End":"17:26.430","Text":"What\u0027s good about this question is that it shows us the difference between"},{"Start":"17:26.430 ","End":"17:32.547","Text":"forces and work and momentum between different bodies in a system."},{"Start":"17:32.547 ","End":"17:37.950","Text":"It also shows us and lets us practice how to work out all sorts of different things,"},{"Start":"17:37.950 ","End":"17:43.500","Text":"including the time, distance traveled by the different bodies in the system."},{"Start":"17:43.500 ","End":"17:49.185","Text":"It also shows us how to work out how much work is put into a system"},{"Start":"17:49.185 ","End":"17:56.230","Text":"to how much useful energy is taken out of the system. That\u0027s it."}],"ID":9481},{"Watched":false,"Name":"Skateboarder","Duration":"15m 30s","ChapterTopicVideoID":9212,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.415","Text":"Hello. In this question,"},{"Start":"00:02.415 ","End":"00:07.080","Text":"we\u0027re told that there\u0027s a skater who slides down,"},{"Start":"00:07.080 ","End":"00:09.705","Text":"skates down this over here,"},{"Start":"00:09.705 ","End":"00:12.930","Text":"which is L, an angle of Theta."},{"Start":"00:12.930 ","End":"00:21.675","Text":"He goes through this loop whose radius is R then skates here,"},{"Start":"00:21.675 ","End":"00:22.935","Text":"gets to a ramp,"},{"Start":"00:22.935 ","End":"00:25.905","Text":"which is also at an angle of Theta."},{"Start":"00:25.905 ","End":"00:31.500","Text":"We\u0027re also told that this height is R,"},{"Start":"00:31.500 ","End":"00:33.030","Text":"jumps the distance,"},{"Start":"00:33.030 ","End":"00:35.550","Text":"d, and gets to the other side."},{"Start":"00:35.550 ","End":"00:39.525","Text":"In the first question, we\u0027re being asked,"},{"Start":"00:39.525 ","End":"00:45.100","Text":"what is the minimal height that L can be in order for the skater to complete the loop?"},{"Start":"00:45.100 ","End":"00:49.415","Text":"The first thing we have to look at is to find"},{"Start":"00:49.415 ","End":"00:53.720","Text":"the velocity at which the skater enters the loop."},{"Start":"00:53.720 ","End":"00:58.475","Text":"We\u0027re going to do this through conservation of energy."},{"Start":"00:58.475 ","End":"01:00.875","Text":"When the skater starts off here,"},{"Start":"01:00.875 ","End":"01:04.100","Text":"he has gravitational potential energy."},{"Start":"01:04.100 ","End":"01:05.615","Text":"As he moves down,"},{"Start":"01:05.615 ","End":"01:10.470","Text":"the potential energy turns into kinetic energy."},{"Start":"01:10.470 ","End":"01:16.555","Text":"Then through that, we can find the velocity with which the skater enters the loop."},{"Start":"01:16.555 ","End":"01:18.750","Text":"Let\u0027s see how we do this."},{"Start":"01:18.750 ","End":"01:25.230","Text":"We have the equation for gravitational potential energy, which is mgh."},{"Start":"01:25.300 ","End":"01:27.800","Text":"What is our h?"},{"Start":"01:27.800 ","End":"01:29.775","Text":"Our h, if we notice,"},{"Start":"01:29.775 ","End":"01:33.390","Text":"is going to be here because this is L and this is Theta,"},{"Start":"01:33.390 ","End":"01:39.530","Text":"so it\u0027s going to be L sine of Theta because"},{"Start":"01:39.530 ","End":"01:47.210","Text":"it\u0027s the opposite side to Theta, which equals 1/2mv^2."},{"Start":"01:47.210 ","End":"01:49.685","Text":"We know what m is,"},{"Start":"01:49.685 ","End":"01:51.260","Text":"we know what g is,"},{"Start":"01:51.260 ","End":"01:52.940","Text":"we know what L is,"},{"Start":"01:52.940 ","End":"01:55.610","Text":"what Theta is, and again, what m is."},{"Start":"01:55.610 ","End":"01:58.460","Text":"That leaves that this is our only unknown."},{"Start":"01:58.460 ","End":"02:01.070","Text":"Then we just have to isolate out the V and then we know"},{"Start":"02:01.070 ","End":"02:04.160","Text":"the velocity with which the skater enters the loop."},{"Start":"02:04.160 ","End":"02:06.590","Text":"For every distance, L,"},{"Start":"02:06.590 ","End":"02:09.000","Text":"there\u0027s a certain velocity that comes with it"},{"Start":"02:09.000 ","End":"02:11.960","Text":"because there\u0027s a relationship between L and V. Now we need to"},{"Start":"02:11.960 ","End":"02:18.850","Text":"find what the minimum velocity can be in order for the skater to complete the loop."},{"Start":"02:18.850 ","End":"02:22.340","Text":"Let\u0027s see what forces are acting on the loop."},{"Start":"02:22.340 ","End":"02:25.265","Text":"When the skater gets to this point,"},{"Start":"02:25.265 ","End":"02:27.650","Text":"the pinnacle of the loop,"},{"Start":"02:27.650 ","End":"02:35.228","Text":"he has mg pointing downwards and the normal is also pointing downwards."},{"Start":"02:35.228 ","End":"02:39.890","Text":"The more correct way to say it is that mg and the normal are pointing inwards"},{"Start":"02:39.890 ","End":"02:44.990","Text":"because when we\u0027re speaking about circular bodies such as loops,"},{"Start":"02:44.990 ","End":"02:48.649","Text":"we always speak about the forces pointing inwards or outwards."},{"Start":"02:48.649 ","End":"02:53.330","Text":"Let\u0027s write down the forces acting on the circle."},{"Start":"02:53.330 ","End":"03:01.490","Text":"The sum of all the forces in the radial direction is equal to ma in the radial direction."},{"Start":"03:01.490 ","End":"03:03.775","Text":"As we know, so m is m,"},{"Start":"03:03.775 ","End":"03:11.390","Text":"the acceleration in the radial direction is equal to Omega^2 multiplied by R. Here,"},{"Start":"03:11.390 ","End":"03:16.065","Text":"it\u0027s going to be capital R. I almost forgot,"},{"Start":"03:16.065 ","End":"03:18.140","Text":"there has to be a minus in front."},{"Start":"03:18.140 ","End":"03:19.430","Text":"Why is it a minus?"},{"Start":"03:19.430 ","End":"03:24.320","Text":"Because whenever we\u0027re speaking about the forces acting inwards into a circle,"},{"Start":"03:24.320 ","End":"03:26.420","Text":"then it\u0027s with a minus and when it\u0027s outwards,"},{"Start":"03:26.420 ","End":"03:28.620","Text":"it\u0027s with a plus."},{"Start":"03:28.620 ","End":"03:31.215","Text":"Let\u0027s see what forces are working here."},{"Start":"03:31.215 ","End":"03:34.175","Text":"We have minus, as I previously explained,"},{"Start":"03:34.175 ","End":"03:38.255","Text":"mg minus the normal,"},{"Start":"03:38.255 ","End":"03:41.360","Text":"because mg and the normal are pointing in the same direction here,"},{"Start":"03:41.360 ","End":"03:45.095","Text":"equals minus m,"},{"Start":"03:45.095 ","End":"03:46.775","Text":"I\u0027m going to write the R first,"},{"Start":"03:46.775 ","End":"03:49.595","Text":"you\u0027ll see why in a second, Omega^2."},{"Start":"03:49.595 ","End":"03:52.895","Text":"Because I\u0027m working with velocities,"},{"Start":"03:52.895 ","End":"03:57.005","Text":"I don\u0027t want to be working with this Omega because it will make it too difficult."},{"Start":"03:57.005 ","End":"04:02.790","Text":"I remember that there\u0027s an equation that states that V"},{"Start":"04:02.790 ","End":"04:10.120","Text":"equals Omega R. Then I can just sub this in."},{"Start":"04:10.130 ","End":"04:14.145","Text":"I can say that this equals to"},{"Start":"04:14.145 ","End":"04:22.600","Text":"minus mR multiplied by V^2,"},{"Start":"04:22.600 ","End":"04:28.050","Text":"so it\u0027s V^2 over R^2."},{"Start":"04:28.050 ","End":"04:30.530","Text":"Then this crosses out and this crosses out,"},{"Start":"04:30.530 ","End":"04:38.870","Text":"which gives me a final answer of minus mv^2 over"},{"Start":"04:38.870 ","End":"04:43.250","Text":"R. The condition that"},{"Start":"04:43.250 ","End":"04:49.190","Text":"promises that I\u0027ll go through an entire loop is if the normal is equal to 0,"},{"Start":"04:49.190 ","End":"04:51.605","Text":"because once the normal is equal to 0,"},{"Start":"04:51.605 ","End":"04:54.140","Text":"then that\u0027s an indication,"},{"Start":"04:54.140 ","End":"04:56.885","Text":"that\u0027s the definition, if you will,"},{"Start":"04:56.885 ","End":"05:00.960","Text":"that the skater will complete an entire loop."},{"Start":"05:01.180 ","End":"05:05.150","Text":"Now that we\u0027ve said that the n has to equal 0,"},{"Start":"05:05.150 ","End":"05:08.345","Text":"so I can say that minus mg"},{"Start":"05:08.345 ","End":"05:15.050","Text":"equals minus mv^2 over R. Then I can cross out the m\u0027s from both side and"},{"Start":"05:15.050 ","End":"05:24.480","Text":"the minuses from both sides and then I\u0027ll be left with g equals V^2 over R."},{"Start":"05:24.760 ","End":"05:32.240","Text":"Then all I do is I isolate out the V. We\u0027ll get that V equals"},{"Start":"05:32.240 ","End":"05:40.550","Text":"the square root of g multiplied by R. Then I just substitute that in into this equation,"},{"Start":"05:40.550 ","End":"05:45.765","Text":"so this V will be substituted into here."},{"Start":"05:45.765 ","End":"05:51.140","Text":"Then I just do all the calculations in order to find L and then that will be"},{"Start":"05:51.140 ","End":"05:58.290","Text":"the minimum L that can be in order for the skater to complete the loop."},{"Start":"05:58.430 ","End":"06:00.880","Text":"Question number 2,"},{"Start":"06:00.880 ","End":"06:02.990","Text":"we\u0027re going to put to the side for a moment and"},{"Start":"06:02.990 ","End":"06:06.025","Text":"we\u0027re going to go and answer now question number 3."},{"Start":"06:06.025 ","End":"06:08.085","Text":"In question number 3,"},{"Start":"06:08.085 ","End":"06:11.420","Text":"we\u0027re told that a skater can dismount the skateboard mid-air at"},{"Start":"06:11.420 ","End":"06:15.670","Text":"a horizontal velocity of p relative to the skateboard."},{"Start":"06:15.670 ","End":"06:18.995","Text":"Then we\u0027re being asked how long after the original jump"},{"Start":"06:18.995 ","End":"06:22.565","Text":"must the skater begin the next jump, the original jump,"},{"Start":"06:22.565 ","End":"06:29.581","Text":"as in here he jumps and then the next jump is talking about the jump from the skateboard."},{"Start":"06:29.581 ","End":"06:34.265","Text":"How long after the original jump must the skater begin the next jump,"},{"Start":"06:34.265 ","End":"06:35.825","Text":"the jump off of the skateboard,"},{"Start":"06:35.825 ","End":"06:39.305","Text":"in order to land exactly on the edge of the ditch?"},{"Start":"06:39.305 ","End":"06:41.465","Text":"This is the edge of the ditch."},{"Start":"06:41.465 ","End":"06:47.300","Text":"The skater leaves the loop with his velocity, travels here,"},{"Start":"06:47.300 ","End":"06:52.370","Text":"gets to the ramp of height capital R angle Theta,"},{"Start":"06:52.370 ","End":"06:58.565","Text":"does a jump mid-air can jump off the skateboard with"},{"Start":"06:58.565 ","End":"07:05.135","Text":"a velocity relative to the skateboard of p and to land exactly here?"},{"Start":"07:05.135 ","End":"07:09.065","Text":"Let\u0027s see how we go about answering this question."},{"Start":"07:09.065 ","End":"07:12.715","Text":"Now the first thing that we\u0027re going to talk about is how long"},{"Start":"07:12.715 ","End":"07:16.885","Text":"from the jump the skater will remain in the air?"},{"Start":"07:16.885 ","End":"07:18.310","Text":"How do we do this?"},{"Start":"07:18.310 ","End":"07:20.095","Text":"We\u0027re going to look at it in"},{"Start":"07:20.095 ","End":"07:25.779","Text":"the vertical y-direction because in the horizontal direction,"},{"Start":"07:25.779 ","End":"07:30.160","Text":"the skater potentially can change velocities."},{"Start":"07:30.160 ","End":"07:32.830","Text":"But in the y-direction it\u0027s going to be the same the whole time,"},{"Start":"07:32.830 ","End":"07:36.805","Text":"and then we can check how long the skater is in the air."},{"Start":"07:36.805 ","End":"07:40.660","Text":"How do we do this? It\u0027s simple kinematic equations."},{"Start":"07:40.660 ","End":"07:44.140","Text":"We remember that this equation d,"},{"Start":"07:44.140 ","End":"07:47.980","Text":"the distance in the y-directional right here,"},{"Start":"07:47.980 ","End":"07:57.370","Text":"equals v_0 multiplied by t plus 1/2 at^2."},{"Start":"07:57.370 ","End":"07:59.050","Text":"We all remember this equation."},{"Start":"07:59.050 ","End":"08:01.435","Text":"Now in our case,"},{"Start":"08:01.435 ","End":"08:05.635","Text":"we\u0027re looking for the distance in the y-direction,"},{"Start":"08:05.635 ","End":"08:13.045","Text":"and it\u0027s a function of t equals v_0, which here,"},{"Start":"08:13.045 ","End":"08:15.130","Text":"because we\u0027re only looking at it in the y-direction,"},{"Start":"08:15.130 ","End":"08:18.970","Text":"soon we\u0027ll work out what v_0 is,"},{"Start":"08:18.970 ","End":"08:20.320","Text":"but just for a moment."},{"Start":"08:20.320 ","End":"08:22.240","Text":"The skater leaves the loop,"},{"Start":"08:22.240 ","End":"08:23.905","Text":"gets to here with a v_0,"},{"Start":"08:23.905 ","End":"08:29.610","Text":"moves up this slanted ramp and jumps."},{"Start":"08:29.610 ","End":"08:34.830","Text":"We can see that the velocity is in a slant and we just want it in the y-direction."},{"Start":"08:34.830 ","End":"08:37.320","Text":"We want it in this direction,"},{"Start":"08:37.320 ","End":"08:39.885","Text":"which is the side opposite to the Theta,"},{"Start":"08:39.885 ","End":"08:44.800","Text":"so we know that it\u0027s going to be v_0 sine Theta."},{"Start":"08:44.800 ","End":"08:52.945","Text":"Then because the acceleration working on the skater is gravity,"},{"Start":"08:52.945 ","End":"08:55.375","Text":"and gravity is pulling the skater down,"},{"Start":"08:55.375 ","End":"08:57.130","Text":"instead of it being a plus,"},{"Start":"08:57.130 ","End":"09:02.980","Text":"it\u0027s going to be a minus 1/2gt^2."},{"Start":"09:02.980 ","End":"09:05.590","Text":"Now in this question specifically,"},{"Start":"09:05.590 ","End":"09:08.845","Text":"we\u0027re considering that this height,"},{"Start":"09:08.845 ","End":"09:12.970","Text":"I, is our ground 0,"},{"Start":"09:12.970 ","End":"09:16.915","Text":"and we\u0027re landing at the exact same height just on the other side."},{"Start":"09:16.915 ","End":"09:19.524","Text":"We want to know when this equation,"},{"Start":"09:19.524 ","End":"09:23.635","Text":"because we\u0027re starting from this position,"},{"Start":"09:23.635 ","End":"09:26.793","Text":"and ending in the exact same y-position,"},{"Start":"09:26.793 ","End":"09:29.890","Text":"so we\u0027re going to say that this equals 0 because"},{"Start":"09:29.890 ","End":"09:34.765","Text":"the total displacement after everything is going to be 0."},{"Start":"09:34.765 ","End":"09:40.460","Text":"Now let\u0027s take a look at how we find out what this v_0 is."},{"Start":"09:41.040 ","End":"09:45.445","Text":"As we remember, we\u0027re dealing with conservation of energy."},{"Start":"09:45.445 ","End":"09:49.240","Text":"The energy from when he goes from here to here,"},{"Start":"09:49.240 ","End":"09:50.650","Text":"to here, to here,"},{"Start":"09:50.650 ","End":"09:53.870","Text":"to here is the same."},{"Start":"09:54.150 ","End":"09:58.615","Text":"Here, if we imagine an imaginary dotted line,"},{"Start":"09:58.615 ","End":"10:00.130","Text":"it\u0027s around about here."},{"Start":"10:00.130 ","End":"10:04.105","Text":"This height is going to be R. Let\u0027s see,"},{"Start":"10:04.105 ","End":"10:07.690","Text":"at the beginning, he travels down."},{"Start":"10:07.690 ","End":"10:09.669","Text":"To work out this,"},{"Start":"10:09.669 ","End":"10:11.800","Text":"the energy at this point,"},{"Start":"10:11.800 ","End":"10:17.290","Text":"we\u0027re going to see that the energy is all of this that he traveled down,"},{"Start":"10:17.290 ","End":"10:21.535","Text":"this height minus this R to get to this point,"},{"Start":"10:21.535 ","End":"10:28.555","Text":"and then the energy at this point is going to be equal to the energy at this point."},{"Start":"10:28.555 ","End":"10:36.640","Text":"Let\u0027s do this. We have mgh."},{"Start":"10:36.640 ","End":"10:44.515","Text":"Now our h is going to be L sine of Theta minus R,"},{"Start":"10:44.515 ","End":"10:53.065","Text":"because we want this height."},{"Start":"10:53.065 ","End":"11:00.310","Text":"Minus R, and this equals 1/2mv_0^2."},{"Start":"11:00.310 ","End":"11:03.625","Text":"Then we just have to isolate the v_0,"},{"Start":"11:03.625 ","End":"11:07.945","Text":"and then we can plug it into this equation."},{"Start":"11:07.945 ","End":"11:11.950","Text":"Then with our v_0 plugged into this equation,"},{"Start":"11:11.950 ","End":"11:17.125","Text":"we multiply it by sine Theta in order to get it in the vertical direction."},{"Start":"11:17.125 ","End":"11:21.715","Text":"Then all we have to do is do our calculations to isolate out t,"},{"Start":"11:21.715 ","End":"11:23.605","Text":"and then we find out what t is,"},{"Start":"11:23.605 ","End":"11:31.580","Text":"which is the time that the skater is in the air from takeoff until landing."},{"Start":"11:32.160 ","End":"11:36.460","Text":"Now what we want to find is what this distance is,"},{"Start":"11:36.460 ","End":"11:37.990","Text":"what does d equal?"},{"Start":"11:37.990 ","End":"11:40.780","Text":"Then we\u0027re going to find out all the forces working in"},{"Start":"11:40.780 ","End":"11:46.430","Text":"the x-direction in order to work out the velocity."},{"Start":"11:47.100 ","End":"11:49.720","Text":"Let\u0027s do this in green."},{"Start":"11:49.720 ","End":"11:53.680","Text":"Now it\u0027s important to remember the skater jumps,"},{"Start":"11:53.680 ","End":"11:57.475","Text":"and then somewhere in the air at any moment that he chooses to,"},{"Start":"11:57.475 ","End":"12:00.685","Text":"he can then jump off the skateboard and have"},{"Start":"12:00.685 ","End":"12:06.110","Text":"an added velocity of p. Let\u0027s see how we do this."},{"Start":"12:06.480 ","End":"12:09.640","Text":"The first thing we\u0027re going to do is we\u0027re going to say that"},{"Start":"12:09.640 ","End":"12:18.460","Text":"its placement on the x-axis is equal to velocity multiplied by time."},{"Start":"12:18.460 ","End":"12:23.980","Text":"Now velocity is going to be v_0 for the x-component,"},{"Start":"12:23.980 ","End":"12:26.770","Text":"we\u0027re going to multiply it by cosine of Theta."},{"Start":"12:26.770 ","End":"12:29.740","Text":"Just like here, we multiplied it by sine Theta,"},{"Start":"12:29.740 ","End":"12:33.190","Text":"so it\u0027s the same idea, multiplied by t,"},{"Start":"12:33.190 ","End":"12:35.230","Text":"and we\u0027re going to call this t_1."},{"Start":"12:35.230 ","End":"12:37.030","Text":"Why am I calling this t_1?"},{"Start":"12:37.030 ","End":"12:39.955","Text":"Because we\u0027re working out its position"},{"Start":"12:39.955 ","End":"12:46.404","Text":"when the skaters at his original velocity with the skateboard,"},{"Start":"12:46.404 ","End":"12:50.260","Text":"and then we\u0027re going to work out a different position,"},{"Start":"12:50.260 ","End":"12:54.175","Text":"different t from when the skater jumps off the skateboard,"},{"Start":"12:54.175 ","End":"12:58.810","Text":"and then has an added velocity of p relative to the skateboard."},{"Start":"12:58.810 ","End":"13:03.010","Text":"Now here comes the second section of this equation."},{"Start":"13:03.010 ","End":"13:06.250","Text":"Plus, and now we\u0027re going to put this in brackets,"},{"Start":"13:06.250 ","End":"13:10.975","Text":"plus the velocity with the added p relative to the skateboard."},{"Start":"13:10.975 ","End":"13:16.107","Text":"We have v_0 cosine Theta plus p"},{"Start":"13:16.107 ","End":"13:23.690","Text":"because it\u0027s relative to this velocity multiplied by t_2."},{"Start":"13:24.750 ","End":"13:29.290","Text":"Now we know that all of this has to be equal"},{"Start":"13:29.290 ","End":"13:33.370","Text":"to d. The position can only equal a maximum of d,"},{"Start":"13:33.370 ","End":"13:40.765","Text":"so it equals d. Now I can also say that t_1,"},{"Start":"13:40.765 ","End":"13:44.755","Text":"the time at the original velocity plus t_2,"},{"Start":"13:44.755 ","End":"13:52.446","Text":"the time of the original velocity plus p is equal to t. What t am I speaking about here?"},{"Start":"13:52.446 ","End":"13:54.370","Text":"I\u0027m speaking about this t,"},{"Start":"13:54.370 ","End":"13:59.860","Text":"which is when we worked out the maximum time that the skater will be in the air."},{"Start":"13:59.860 ","End":"14:02.230","Text":"I know it\u0027s a bit confusing working with the t_1,"},{"Start":"14:02.230 ","End":"14:03.775","Text":"t_2, t,"},{"Start":"14:03.775 ","End":"14:06.880","Text":"it\u0027s a bit confusing, but we have to get used to it."},{"Start":"14:06.880 ","End":"14:09.460","Text":"Practice makes perfect."},{"Start":"14:09.460 ","End":"14:12.895","Text":"Now I\u0027ve given you all the information that you need,"},{"Start":"14:12.895 ","End":"14:17.870","Text":"all you have to do is sub in everything into this equation and"},{"Start":"14:17.870 ","End":"14:23.280","Text":"then you can work out the answer to question number 3."},{"Start":"14:23.280 ","End":"14:25.540","Text":"Now in question 4, we\u0027re being asked,"},{"Start":"14:25.540 ","End":"14:28.775","Text":"what is the maximum distance the skater can travel safely?"},{"Start":"14:28.775 ","End":"14:32.059","Text":"Now we\u0027re trying to find what the maximum value d"},{"Start":"14:32.059 ","End":"14:35.225","Text":"can have with which the skater can still cross."},{"Start":"14:35.225 ","End":"14:37.775","Text":"Ignore whatever we did in question 3,"},{"Start":"14:37.775 ","End":"14:39.980","Text":"this is a new question."},{"Start":"14:39.980 ","End":"14:46.445","Text":"We can have that v_0 cosine of Theta plus p. Now why have I done this like this?"},{"Start":"14:46.445 ","End":"14:50.420","Text":"Because if we want the maximum distance that the skater can pass,"},{"Start":"14:50.420 ","End":"14:54.440","Text":"then he\u0027ll have to have maximum speed as early as possible,"},{"Start":"14:54.440 ","End":"14:57.890","Text":"which means that as soon as he does the jump from the ramp,"},{"Start":"14:57.890 ","End":"15:00.225","Text":"he also jumps off the skateboard."},{"Start":"15:00.225 ","End":"15:04.090","Text":"That\u0027s v_0 cosine of Theta plus p,"},{"Start":"15:04.090 ","End":"15:12.710","Text":"which multiplied by t equals d max."},{"Start":"15:13.050 ","End":"15:15.310","Text":"Now, this t, when I\u0027m talking about this t,"},{"Start":"15:15.310 ","End":"15:18.685","Text":"I\u0027m talking about the t that we find over here,"},{"Start":"15:18.685 ","End":"15:20.375","Text":"and then that is it."},{"Start":"15:20.375 ","End":"15:23.480","Text":"That is the maximum value that d can be in order to"},{"Start":"15:23.480 ","End":"15:27.440","Text":"allow the skateboarder to still get over the ditch."},{"Start":"15:27.440 ","End":"15:30.960","Text":"Here we finished the question."}],"ID":9482},{"Watched":false,"Name":"Metronome","Duration":"17m 13s","ChapterTopicVideoID":9213,"CourseChapterTopicPlaylistID":5360,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.520","Text":"Hello. In this question,"},{"Start":"00:02.520 ","End":"00:06.090","Text":"we\u0027re being asked to find the frequency of a metronome."},{"Start":"00:06.090 ","End":"00:09.810","Text":"This right here is a metronome."},{"Start":"00:09.810 ","End":"00:12.495","Text":"Now, for those of you who don\u0027t know what this is,"},{"Start":"00:12.495 ","End":"00:17.205","Text":"it\u0027s what musicians use in order to help them keep the beats."},{"Start":"00:17.205 ","End":"00:18.945","Text":"Now how it works,"},{"Start":"00:18.945 ","End":"00:24.897","Text":"there\u0027s a type of pole going in the middle with a mass attached to the bottom."},{"Start":"00:24.897 ","End":"00:28.725","Text":"The metronome, the stick moves sideways like this,"},{"Start":"00:28.725 ","End":"00:34.590","Text":"and it goes ding to a certain beat that you can set it to."},{"Start":"00:34.590 ","End":"00:36.040","Text":"How do you set the beats?"},{"Start":"00:36.040 ","End":"00:41.356","Text":"Here, there\u0027s another mass placed which you can move up or down."},{"Start":"00:41.356 ","End":"00:43.130","Text":"The higher you move it,"},{"Start":"00:43.130 ","End":"00:45.770","Text":"the slower the beat and the lower you move it,"},{"Start":"00:45.770 ","End":"00:47.060","Text":"the faster the beat,"},{"Start":"00:47.060 ","End":"00:49.385","Text":"the faster the frequency."},{"Start":"00:49.385 ","End":"00:51.805","Text":"Here\u0027s a sketch,"},{"Start":"00:51.805 ","End":"00:54.680","Text":"here\u0027s the mass at the bottom that I was talking about,"},{"Start":"00:54.680 ","End":"00:59.430","Text":"here\u0027s the mass which can be moved up and down."},{"Start":"00:59.510 ","End":"01:02.235","Text":"Let\u0027s go to the question now."},{"Start":"01:02.235 ","End":"01:05.495","Text":"Find that frequency of the metronome in the sketch,"},{"Start":"01:05.495 ","End":"01:10.580","Text":"which changes according to the placement of the mass which glides along the metronome."},{"Start":"01:10.580 ","End":"01:12.710","Text":"That\u0027s this here."},{"Start":"01:12.710 ","End":"01:15.590","Text":"It is given that the axis of the metronome is"},{"Start":"01:15.590 ","End":"01:19.775","Text":"located a 1/4 of the way up. What does this mean?"},{"Start":"01:19.775 ","End":"01:28.515","Text":"If we\u0027re going to say that all of this is length l,"},{"Start":"01:28.515 ","End":"01:34.805","Text":"the axis here is at a 1/4(l) way up."},{"Start":"01:34.805 ","End":"01:38.600","Text":"Now all of the items in here have masses."},{"Start":"01:38.600 ","End":"01:41.675","Text":"Let\u0027s call the movable mass,"},{"Start":"01:41.675 ","End":"01:45.120","Text":"we\u0027ll call it m_1."},{"Start":"01:45.120 ","End":"01:48.675","Text":"Let\u0027s call the bottom mass m_2,"},{"Start":"01:48.675 ","End":"01:54.155","Text":"and then the mass of the entire pole,"},{"Start":"01:54.155 ","End":"01:58.635","Text":"the red line let\u0027s call m_3."},{"Start":"01:58.635 ","End":"02:03.950","Text":"What we\u0027re being asked to do is to find the frequency of the metronome as"},{"Start":"02:03.950 ","End":"02:09.820","Text":"a function of the location of this mass, mass Number 1."},{"Start":"02:09.820 ","End":"02:14.720","Text":"If we label the location of mass Number 1 as x,"},{"Start":"02:14.720 ","End":"02:21.290","Text":"as this, then we\u0027re trying to find Omega as a function of x."},{"Start":"02:21.290 ","End":"02:23.120","Text":"Let\u0027s work this out."},{"Start":"02:23.120 ","End":"02:27.380","Text":"Now of course, the first thing that we have to do is to draw a free body diagram."},{"Start":"02:27.380 ","End":"02:28.715","Text":"Now, in this case,"},{"Start":"02:28.715 ","End":"02:35.575","Text":"it\u0027s more correct to say free body diagram showing the moments rather than the forces."},{"Start":"02:35.575 ","End":"02:37.320","Text":"In the diagram here,"},{"Start":"02:37.320 ","End":"02:44.390","Text":"we can see that if this is where the axes is,"},{"Start":"02:44.390 ","End":"02:50.130","Text":"then we can see that our pole is at an angle of Theta."},{"Start":"02:50.130 ","End":"02:54.250","Text":"Theta is a positive number."},{"Start":"02:54.250 ","End":"02:59.075","Text":"Now we can start writing down all of the equations."},{"Start":"02:59.075 ","End":"03:01.520","Text":"Let\u0027s write the sum of all the moments,"},{"Start":"03:01.520 ","End":"03:03.530","Text":"we\u0027ll write it here."},{"Start":"03:03.530 ","End":"03:10.630","Text":"The sum of all the moments will represent them as Tau is equal to i Alpha,"},{"Start":"03:10.630 ","End":"03:14.810","Text":"which is something that we\u0027ve already learned earlier on in the units."},{"Start":"03:14.810 ","End":"03:18.740","Text":"Now, let\u0027s see what that equals to in our example."},{"Start":"03:18.740 ","End":"03:26.350","Text":"As we said, this section is l divided by 4,"},{"Start":"03:26.570 ","End":"03:29.295","Text":"so it\u0027s the same here."},{"Start":"03:29.295 ","End":"03:33.000","Text":"This is l divided by 4."},{"Start":"03:33.000 ","End":"03:35.915","Text":"Now we want to find the force,"},{"Start":"03:35.915 ","End":"03:40.810","Text":"the x-component is over here of this m_1g."},{"Start":"03:40.810 ","End":"03:48.540","Text":"We can say that it\u0027s m_1g multiplied by, now here the thing."},{"Start":"03:48.540 ","End":"03:55.445","Text":"We said that this entire length over here is equal to x,"},{"Start":"03:55.445 ","End":"04:06.035","Text":"this but we want only from this imaginary line until the axes here, until this dot."},{"Start":"04:06.035 ","End":"04:13.271","Text":"We want it of length x minus l over 4."},{"Start":"04:13.271 ","End":"04:19.020","Text":"Then I\u0027m just going to rub out everything to make it a bit clearer again."},{"Start":"04:19.090 ","End":"04:21.260","Text":"This was Theta."},{"Start":"04:21.260 ","End":"04:26.245","Text":"Now we have to find what the horizontal component is."},{"Start":"04:26.245 ","End":"04:28.555","Text":"If this is Theta,"},{"Start":"04:28.555 ","End":"04:31.040","Text":"then because of alternate angles,"},{"Start":"04:31.040 ","End":"04:32.690","Text":"this is also Theta."},{"Start":"04:32.690 ","End":"04:35.660","Text":"Then if we want to find the horizontal component,"},{"Start":"04:35.660 ","End":"04:38.100","Text":"it\u0027s the side opposite to Theta,"},{"Start":"04:38.100 ","End":"04:42.560","Text":"so we have to multiply it by sine Theta."},{"Start":"04:42.560 ","End":"04:45.350","Text":"That\u0027s where that section for m_1."},{"Start":"04:45.350 ","End":"04:48.080","Text":"Another thing that\u0027s important to note is"},{"Start":"04:48.080 ","End":"04:51.280","Text":"because the force is pointing downwards and at the beginning,"},{"Start":"04:51.280 ","End":"04:56.620","Text":"we said that Theta was a positive number,"},{"Start":"04:56.620 ","End":"05:02.450","Text":"we\u0027re just going to define it as pointing in the clockwise direction."},{"Start":"05:02.450 ","End":"05:09.950","Text":"Then m_1g is pushing the metronome pole downwards and also into a clockwise direction."},{"Start":"05:09.950 ","End":"05:15.170","Text":"Therefore, we\u0027re going to leave this whole expression also as a positive."},{"Start":"05:15.170 ","End":"05:21.175","Text":"Now the next expression that we\u0027re going to look at is for this mass m_2."},{"Start":"05:21.175 ","End":"05:23.780","Text":"Let\u0027s, before we put down assign,"},{"Start":"05:23.780 ","End":"05:28.190","Text":"we\u0027ll talk about in a second whether it\u0027s positive or negative. Let\u0027s write it out."},{"Start":"05:28.190 ","End":"05:33.605","Text":"We\u0027re going to have m_2g multiplied by,"},{"Start":"05:33.605 ","End":"05:35.180","Text":"because it\u0027s at a distance,"},{"Start":"05:35.180 ","End":"05:39.135","Text":"remember this is l over 4 this distance,"},{"Start":"05:39.135 ","End":"05:45.150","Text":"multiplied by l divided by 4, which is the distance."},{"Start":"05:45.150 ","End":"05:48.724","Text":"Then again, because as you\u0027ll notice,"},{"Start":"05:48.724 ","End":"05:53.945","Text":"all of the forces of the masses are pointing vertically downwards,"},{"Start":"05:53.945 ","End":"05:56.453","Text":"which means that they\u0027re all parallel."},{"Start":"05:56.453 ","End":"06:01.025","Text":"So it\u0027s going to be the same again multiplied by sine Theta."},{"Start":"06:01.025 ","End":"06:05.840","Text":"Another way, whether we\u0027re looking at the angle here or the angle here,"},{"Start":"06:05.840 ","End":"06:09.500","Text":"it doesn\u0027t make any difference because we\u0027re using sine,"},{"Start":"06:09.500 ","End":"06:13.895","Text":"because sine has even parity then we know that sine of"},{"Start":"06:13.895 ","End":"06:19.634","Text":"180 or Pi minus Theta is the same as sine of Theta."},{"Start":"06:19.634 ","End":"06:21.410","Text":"Therefore, it doesn\u0027t really matter."},{"Start":"06:21.410 ","End":"06:26.075","Text":"Now whether this expression is plus or minus, let\u0027s take a look."},{"Start":"06:26.075 ","End":"06:27.905","Text":"If we\u0027re looking at this mass,"},{"Start":"06:27.905 ","End":"06:31.864","Text":"it\u0027s trying to push the whole system in this direction,"},{"Start":"06:31.864 ","End":"06:34.085","Text":"in an anticlockwise direction."},{"Start":"06:34.085 ","End":"06:39.260","Text":"Which means because we defined the clockwise direction as positive,"},{"Start":"06:39.260 ","End":"06:41.645","Text":"we\u0027ll put a minus in front of the expression."},{"Start":"06:41.645 ","End":"06:44.755","Text":"Now for the third mass, as we said,"},{"Start":"06:44.755 ","End":"06:50.190","Text":"m_3 represents the mass of the entire pole."},{"Start":"06:50.240 ","End":"06:55.550","Text":"We\u0027re going to look at it as if the mass is acting from the center of mass,"},{"Start":"06:55.550 ","End":"06:57.230","Text":"from the mid point of the pole."},{"Start":"06:57.230 ","End":"06:59.810","Text":"Now the midpoint of the pole is around about here,"},{"Start":"06:59.810 ","End":"07:02.165","Text":"I really should have drawn this arrow here."},{"Start":"07:02.165 ","End":"07:04.590","Text":"Hold on, let me just do that."},{"Start":"07:05.360 ","End":"07:08.000","Text":"It\u0027s acting from the mid point of the pole,"},{"Start":"07:08.000 ","End":"07:09.440","Text":"which is around about here."},{"Start":"07:09.440 ","End":"07:14.680","Text":"Because we\u0027ve already said that this dot here is at l divided by 4,"},{"Start":"07:14.680 ","End":"07:20.990","Text":"so we know that the distance between a 1/4(l) to 1/2l is 1/4(l)."},{"Start":"07:20.990 ","End":"07:25.205","Text":"Then again, without writing yet whether it\u0027s plus or minus,"},{"Start":"07:25.205 ","End":"07:32.730","Text":"we\u0027re going to write m_3g multiplied by l divided by 4,"},{"Start":"07:32.730 ","End":"07:35.325","Text":"and again multiplied by"},{"Start":"07:35.325 ","End":"07:40.580","Text":"sine Theta for the same reason as this was sine Theta and this was sine Theta."},{"Start":"07:40.580 ","End":"07:42.403","Text":"They\u0027re all parallel."},{"Start":"07:42.403 ","End":"07:46.940","Text":"This is also Theta and then whether it\u0027s a plus or minus,"},{"Start":"07:46.940 ","End":"07:50.270","Text":"because it\u0027s on this side of the pole,"},{"Start":"07:50.270 ","End":"07:53.897","Text":"which we\u0027ve already designated as being the positive side."},{"Start":"07:53.897 ","End":"07:56.390","Text":"Because the forces are pushing down"},{"Start":"07:56.390 ","End":"07:59.150","Text":"and pushing the whole pole to move in a clockwise direction,"},{"Start":"07:59.150 ","End":"08:02.590","Text":"we can say that this expression is positive."},{"Start":"08:02.590 ","End":"08:08.207","Text":"This is the sum of moments and all of this is equal to i Alpha."},{"Start":"08:08.207 ","End":"08:13.520","Text":"Let\u0027s just remember that Alpha equals Theta double dot,"},{"Start":"08:13.520 ","End":"08:17.750","Text":"as we already know but we\u0027re going to get to that later in the question."},{"Start":"08:17.750 ","End":"08:21.740","Text":"Now what\u0027s left to do is to find out what this i equals."},{"Start":"08:21.740 ","End":"08:26.255","Text":"Now i changes with relation to x."},{"Start":"08:26.255 ","End":"08:29.680","Text":"As I move m_1 up or down,"},{"Start":"08:29.680 ","End":"08:33.470","Text":"a value for i will also change accordingly."},{"Start":"08:33.470 ","End":"08:39.920","Text":"It\u0027s important to remember that I\u0027m changing i with relation to this axis,"},{"Start":"08:39.920 ","End":"08:41.390","Text":"the axis of rotation."},{"Start":"08:41.390 ","End":"08:43.745","Text":"Now let\u0027s work out what i is."},{"Start":"08:43.745 ","End":"08:46.940","Text":"Let\u0027s write that here. I,"},{"Start":"08:46.940 ","End":"08:55.970","Text":"and I\u0027m going to write this as i total is equal to the i of every single body or mass."},{"Start":"08:55.970 ","End":"08:59.050","Text":"Let\u0027s start with m_1."},{"Start":"08:59.050 ","End":"09:04.440","Text":"It equals mass times the distance squared,"},{"Start":"09:04.440 ","End":"09:06.970","Text":"so we have m_1,"},{"Start":"09:06.970 ","End":"09:10.640","Text":"which is the mass multiplied by the distance."},{"Start":"09:10.640 ","End":"09:15.080","Text":"Which as we said, was x minus l divided"},{"Start":"09:15.080 ","End":"09:21.385","Text":"by 4^2 and then also for m_2 mass times the distance squared."},{"Start":"09:21.385 ","End":"09:25.780","Text":"Then it\u0027s going to be plus m_2,"},{"Start":"09:25.780 ","End":"09:29.960","Text":"which is the mass, multiplied by the distance squared."},{"Start":"09:29.960 ","End":"09:32.375","Text":"Which here is l over 4,"},{"Start":"09:32.375 ","End":"09:36.670","Text":"so it\u0027s l over 4^2."},{"Start":"09:36.670 ","End":"09:40.255","Text":"As for mass Number 3, it\u0027s slightly different."},{"Start":"09:40.255 ","End":"09:49.825","Text":"Mass Number 3 represents the mass of the entire pole or rod and it\u0027s uniform."},{"Start":"09:49.825 ","End":"09:55.795","Text":"If you remember there is a proof that you can look back in previous lessons;"},{"Start":"09:55.795 ","End":"10:02.215","Text":"it\u0027s there, that shows that at the center of mass of the pole."},{"Start":"10:02.215 ","End":"10:06.430","Text":"Then the moment of inertia; the i there,"},{"Start":"10:06.430 ","End":"10:12.730","Text":"is equal to 1/12mL^2 squared."},{"Start":"10:12.730 ","End":"10:15.670","Text":"L representing the length of the pole of course,"},{"Start":"10:15.670 ","End":"10:18.160","Text":"m representing the mass of the pole,"},{"Start":"10:18.160 ","End":"10:20.770","Text":"and 1/12 represents the center of mass."},{"Start":"10:20.770 ","End":"10:27.040","Text":"Let\u0027s write this, 1/12m_3L^2."},{"Start":"10:27.040 ","End":"10:32.572","Text":"However, the center of mass is not the axis of rotation."},{"Start":"10:32.572 ","End":"10:36.946","Text":"So we have to add on an extra section which is this section"},{"Start":"10:36.946 ","End":"10:41.470","Text":"to represent that change which means that we have to add"},{"Start":"10:41.470 ","End":"10:51.085","Text":"on m_3 multiplied by the length which is L divided by 4^2."},{"Start":"10:51.085 ","End":"10:56.350","Text":"This addition is called Steiner."},{"Start":"10:56.350 ","End":"10:59.065","Text":"We\u0027ve also gone through this in previous lessons."},{"Start":"10:59.065 ","End":"11:00.715","Text":"If you don\u0027t remember what this is,"},{"Start":"11:00.715 ","End":"11:06.170","Text":"then go over that lesson again because it explains that there are more in detail."},{"Start":"11:06.390 ","End":"11:15.850","Text":"We\u0027ve found our value for I. I_tot will be subbed into here, into this I."},{"Start":"11:15.850 ","End":"11:19.060","Text":"Let\u0027s see what else we have to find."},{"Start":"11:19.060 ","End":"11:21.985","Text":"Let\u0027s sort out these equations."},{"Start":"11:21.985 ","End":"11:28.585","Text":"We have what I is and we know that Alpha is Theta-dot dot."},{"Start":"11:28.585 ","End":"11:34.377","Text":"I times Alpha is equal to this whole long expression here."},{"Start":"11:34.377 ","End":"11:44.830","Text":"Then we can say that Theta-dot dot or Alpha is equal to this expression divided by I."},{"Start":"11:44.830 ","End":"11:47.720","Text":"Let\u0027s write it out."},{"Start":"11:48.000 ","End":"11:53.680","Text":"Let\u0027s see what I\u0027ve done. I\u0027ve done m_1g x minus L/4."},{"Start":"11:53.680 ","End":"11:58.407","Text":"The sine Theta,"},{"Start":"11:58.407 ","End":"12:03.220","Text":"when we\u0027re dealing with small angles which in harmonic motion we generally are,"},{"Start":"12:03.220 ","End":"12:10.000","Text":"we can say that sine of Theta is approximately equal to Theta."},{"Start":"12:10.000 ","End":"12:12.145","Text":"This is only true for small angles."},{"Start":"12:12.145 ","End":"12:13.990","Text":"Here we\u0027re dealing with small angles."},{"Start":"12:13.990 ","End":"12:16.720","Text":"I can put the whole expression"},{"Start":"12:16.720 ","End":"12:20.605","Text":"multiplied by because everything is being multiplied by sine Theta,"},{"Start":"12:20.605 ","End":"12:29.005","Text":"so multiplied by Theta and then divided by the I."},{"Start":"12:29.005 ","End":"12:32.335","Text":"First I\u0027m going to add a minus to the front over here"},{"Start":"12:32.335 ","End":"12:36.114","Text":"because we want it to match the equation."},{"Start":"12:36.114 ","End":"12:39.265","Text":"Because we put the minus over here then"},{"Start":"12:39.265 ","End":"12:44.545","Text":"this has to become a minus over here, this is a plus."},{"Start":"12:44.545 ","End":"12:48.190","Text":"This becomes a minus as well in order to keep"},{"Start":"12:48.190 ","End":"12:52.675","Text":"the signs and then we have to divide it by I."},{"Start":"12:52.675 ","End":"12:55.639","Text":"This is the final equation,"},{"Start":"13:00.360 ","End":"13:05.080","Text":"and then all of this that I\u0027m circling or squaring in"},{"Start":"13:05.080 ","End":"13:09.355","Text":"red is equal to the frequency squared."},{"Start":"13:09.355 ","End":"13:14.660","Text":"The frequency is just the square root of this whole expression."},{"Start":"13:15.270 ","End":"13:21.040","Text":"Let\u0027s just look at it from a more intuitive perspective."},{"Start":"13:21.040 ","End":"13:27.700","Text":"If we\u0027re looking at the rod over here at m_1 and we decide to move m_1 up,"},{"Start":"13:27.700 ","End":"13:33.625","Text":"this means that we\u0027re increasing the size of x because we\u0027re increasing the distance."},{"Start":"13:33.625 ","End":"13:37.060","Text":"If we look at this equation for frequency squared,"},{"Start":"13:37.060 ","End":"13:45.460","Text":"then we can see that here in the numerator we have an x,"},{"Start":"13:45.460 ","End":"13:48.700","Text":"but in the denominator we have something,"},{"Start":"13:48.700 ","End":"13:51.063","Text":"an expression here with x squared."},{"Start":"13:51.063 ","End":"13:53.875","Text":"Which means that as x increases,"},{"Start":"13:53.875 ","End":"13:58.405","Text":"the denominator will increase more than the numerator will."},{"Start":"13:58.405 ","End":"14:02.410","Text":"Meaning that the frequency squared"},{"Start":"14:02.410 ","End":"14:07.825","Text":"or the frequency as well will decrease which is exactly what we said at the beginning."},{"Start":"14:07.825 ","End":"14:12.055","Text":"As this mass moves up, the frequency decreases."},{"Start":"14:12.055 ","End":"14:16.782","Text":"I don\u0027t know who has experience with a metronome,"},{"Start":"14:16.782 ","End":"14:20.740","Text":"but if you do then you will know that this is the truth."},{"Start":"14:20.740 ","End":"14:26.920","Text":"Another thing to look at is because we said that this is frequency squared,"},{"Start":"14:26.920 ","End":"14:32.185","Text":"so in order to find the frequency we have to find the square root of this."},{"Start":"14:32.185 ","End":"14:40.600","Text":"The problem is that if this expression becomes negative,"},{"Start":"14:40.600 ","End":"14:44.875","Text":"then we cannot find the square root of it."},{"Start":"14:44.875 ","End":"14:46.375","Text":"In which case,"},{"Start":"14:46.375 ","End":"14:49.000","Text":"this is a problem. What does this mean?"},{"Start":"14:49.000 ","End":"14:54.745","Text":"It means that we\u0027ve lifted this mass, mass Number 1,"},{"Start":"14:54.745 ","End":"15:02.004","Text":"too high and therefore the center of mass has moved above the axis of rotation."},{"Start":"15:02.004 ","End":"15:09.415","Text":"Therefore, the whole rod will instead of moving in harmonic motion from side-to-side,"},{"Start":"15:09.415 ","End":"15:14.230","Text":"it will just flip over to the other side and that\u0027s something that we don\u0027t want."},{"Start":"15:14.230 ","End":"15:19.615","Text":"That as a matter of fact is why this mass m Number 2 exists in the system."},{"Start":"15:19.615 ","End":"15:25.254","Text":"It\u0027s in order to keep the center of mass below a certain point,"},{"Start":"15:25.254 ","End":"15:30.565","Text":"and below the certain point is below the axis of rotation."},{"Start":"15:30.565 ","End":"15:34.720","Text":"Another thing that they could ask or that maybe is good to"},{"Start":"15:34.720 ","End":"15:38.170","Text":"know is what\u0027s the maximum height that we"},{"Start":"15:38.170 ","End":"15:46.038","Text":"can lift this mass up until the center of mass becomes above the center of rotation."},{"Start":"15:46.038 ","End":"15:50.725","Text":"Then the rod will flip upside down which is something that we don\u0027t want to do."},{"Start":"15:50.725 ","End":"15:52.750","Text":"Let\u0027s see what this is."},{"Start":"15:52.750 ","End":"15:55.510","Text":"What we would have to write is the maximum height,"},{"Start":"15:55.510 ","End":"15:58.105","Text":"so it\u0027s in the y-direction here."},{"Start":"15:58.105 ","End":"16:06.400","Text":"The center of mass is equal to m_1;"},{"Start":"16:06.400 ","End":"16:07.915","Text":"so this mass,"},{"Start":"16:07.915 ","End":"16:11.914","Text":"multiplied by the distance which is x"},{"Start":"16:11.914 ","End":"16:19.570","Text":"plus m_2 multiplied by its distance."},{"Start":"16:19.570 ","End":"16:23.530","Text":"It\u0027s right at the bottom, so it\u0027s multiplied by 0."},{"Start":"16:23.530 ","End":"16:26.140","Text":"This crosses out."},{"Start":"16:26.140 ","End":"16:33.760","Text":"Plus m_3 which is mass of the rod and we\u0027ll take it from its midpoint multiply it"},{"Start":"16:33.760 ","End":"16:38.650","Text":"by L over 2 divided by m_1"},{"Start":"16:38.650 ","End":"16:44.110","Text":"plus m_2, plus m_3."},{"Start":"16:44.110 ","End":"16:49.570","Text":"This is the center of mass of the height and it always has to be smaller or equal to L"},{"Start":"16:49.570 ","End":"16:56.065","Text":"divided by 4 because L divided by 4 in this instance is the center of rotation."},{"Start":"16:56.065 ","End":"17:01.660","Text":"Another thing that they could have asked is what is the minimum mass"},{"Start":"17:01.660 ","End":"17:07.195","Text":"that m Number 2 could be in order to keep the system moving in harmonic motion?"},{"Start":"17:07.195 ","End":"17:11.995","Text":"Then again you just rearrange this equation and there you go."},{"Start":"17:11.995 ","End":"17:14.600","Text":"That\u0027s the end of the question."}],"ID":9483}],"Thumbnail":null,"ID":5360}]

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