Mechanics ch 6 (6.1-6.4) Course revision questions

Added: 2026-05-18

[00:00:01]Hello guys, how are you today? I hope everyone is doing um well. So today I'll be explaining the chapter 6 kinematics of rotational motion. I'm going to be doing the course revision questions.

[00:00:12]This course revision questions, it's the AP one. It's the newly added AP1. You have to see mechanics. It will be written AP. Please click on it. And um this is the one you're supposed to see.

[00:00:24]Okay. Then you go to chapter six. And this is the one that I'm going to be doing. I'm going to see which questions like I don't think I would be doing all the questions. I'll try my best of course. But anyways, let's get started with the video. We're dealing with angular kinematic quantities. Okay, question number one. It's a grid. Okay.

[00:00:40]The linear velocity of the rim points of uniformly rotating disc is 3 m/s. Okay.

[00:00:46]So, first of all, the linear velocity we know first of all v is equal to omega * r where omega is what? Omega. It is the angular velocity and r is the radius. Of course, they said the linear velocity of the rim points. This is our circle and a rim point would be a point over here. So, when we're talking about the distance r, it would be the radius. Okay, it's equal to what? It's equal to 3 m/ second. And the point located 10 cm closer. So now it would be v is equal to omega * r - 10 cm. But please change everything to SI units 10 cm you divide by 100 you will get 0.1 m. So r minus 0.1 it is equal to what? It's equal to 2 m/ second. Okay they want the angular velocity of the disc you should know one thing at the same time okay any point all throughout the disc or the circle or the sphere okay the uh angular velocity it's going to be the same. So it's same at any point. Okay. So what could this mean?

[00:01:52]This could mean that I could um uh redirect I could change this equation to make omega the subject. So omega would be equal to what? It would be equal to 2 / r minus 0.1. And omega would also be equal to what? From this equation, it would be equal to 3 / r. Now we said omega is the same at any point no matter which point it's going to stay the same.

[00:02:14]So 2 / r minus 0.1 would be equal to what? It would be equal to 3 / r. And then of course you need to do what? You need to do the cross multiplication. So I got uh 2 r is equal to 3r - 0.3. I move this to the other side. - r is equal to -0.3. So then the r is going to be equal to what? R is going to be equal to 0.3 m. Now is this what they're looking for? No. They're looking for the angular velocity of the disc. So I would substitute the 0.3 in any equation. For example, I'd substitute it here. So, omega is equal to 3 / 0.3, which is how much? Which is 10. And what's the unit for the angular velocity? It would be rad per second. Okay, that's it for question one in grid. We'll move on now to question two. What does it say?

[00:03:00]The rotational motion. In one second, let me just move this. The rotational motion of a wheel is governed by this expression over here. This is the angular position. Okay. Find the instantaneous angular velocity at time equal to second and the direction of the rotation. As you can see, this is like the um course practice questions. Okay, so when they say instantaneous angular velocity, I explained this before.

[00:03:22]Instantaneous means I need to derive it.

[00:03:24]If they said average, I would substitute the values in the position. I'd find delta theta and I do it over delta uh time in order to find what? In order to find the average instant uh sorry, the average angular velocity. But when they want instantaneous, we should derive this. So what we're going to do is we're going to derive this. So omega would turn out to be what? It would turn out to be 3 t ^ 2 + 2. Okay. Then what do I do? I substitute the value of 2 over here. So 3 * 2 ^ 2 + 2. This would give me an answer of 14 rad per second. And now the direction, the direction always depends on omega. Okay. So because it is positive, that means it's rotating in the anticlockwise or the counterclockwise direction. Okay, that's question two done. We'll move on now to question number three. What does it say?

[00:04:10]The rotational motion of a wheel is governed by the expression theta= 1 + 2t + t cubed. Okay, we want the instantaneous angular acceleration.

[00:04:19]Again, instantaneous angular acceleration. I would need to derive from my angular position. Not once, once I would get the angular velocity, but twice to get the angular acceleration.

[00:04:29]If they wanted the average acceleration, okay, if they wanted the average angular acceleration, I would need to substitute my two different times in uh omega and I'd get delta omega and I divide it by the change in time. But they want instantaneous. So I'd need to derive not once but two times. And the direction of rotation it's only determined by what is determined by um the angular velocity which is omega. So let's um derive it.

[00:04:54]this would end up becoming 3t ^2 + 2 and then it would end up becoming 6 t right so then what do I substitute I substitute the time 2 seconds so 6 * 2 this would give me an answer of what of 12 rad per second squared now uh of course since omega is positive you can get it from the previous question or you can substitute 3 * 2 ^ 2 + 2 I got + 14 since it's positive it's rotating in which direction it's rotating in the um in the uh anticlockwise direction. Okay, so that is it for question number three.

[00:05:28]We'll move on now to question number four. What does it say? Uh very simply, an object is undergoing circular motion.

[00:05:35]It covers a distance of 2.5 m when the angular displacement is this. So this is what the distance covered. The distance is denoted by S. Angular displacement is denoted by theta. The angular acceleration is 0.5 rad/s squared.

[00:05:49]What's the radius of the circular path covered by the object and its tangential acceleration? Okay, first thing they want the radius of the circular path. We should know that the displacement is equal to the radius multiplied by the angular displacement. So it would be 2.5 is equal to radius time 1.25, right? So the radius would be equal to 2.5 over 1.25, which is what? Which is 2 uh m.

[00:06:14]This is the value of our radius. Now the tangential acceleration you should know a t it's equal to what? It's equal to um the radius multiplied by the angular acceleration. Our radius it's 2 multiplied by the angular acceleration which is 0.5. So I got it to be what? I got it to be 1 m/s squared. This is the tangential acceleration. Okay, that's it for question number four. We'll move on now to question number five. What does it say? A wheel has an angular position that varies with time as follows. How many revolutions does the wheel make in the first five seconds? So, they want what they want, the number of revolutions. Remember that the number of revolutions, it's equal to what? It's equal to the angular displacement divided by what divided by 2 pi. Why?

[00:06:59]Because 2 pi is one revolution. Okay, remember in a circle one revolution is how much? It's 2 pi. Okay, so this is why we divide by 2 pi. So let's find delta theta or theta does not need to be necessarily delta theta just theta. So I'm going to put five. I'll substitute it here to get the value of theta. Uh so then theta it's going to be equal to 6 * what's the time? 5^ 2 6 * 5^ 2. Where's my calculator? One sec. So 6 * 5^ 2 it gave me an answer of 150. Don't forget the unit is rad because it's theta angular displacement. So rad over what divided by 2 pi uh sorry the theta which is 150 divided by 2 pi I'm going to put it on my calculator 150 / 2 pi I got an answer to be what I got it to be approximately to one decimal place um 23.9 revolutions. So uh what's the time or sorry what how many revolutions does the wheel make in the first 5 seconds? It makes uh 23.9 revolutions. Okay, that's it for question five. We move on to question number six. What does it say? Um okay.

[00:08:13]A wheel is turning with a constant angular speed of three rad per second.

[00:08:18]So this is my angular speed omega. What is the time taken by the wheel to complete one full revolution? This is a very important um question. Right? First of all, one full revolution. The delta theta or the angular displacement of one full revolution, it's how much? It's 2 pi. One full revolution, one full entire circle. It's what? It's 2 pi. They want the time taken by the wheel to complete one full revolution. You should know that delta theta over delta time, it's equal to what? It's equal to the angular speed. So if I want the time, okay, I would have to do that delta time, it's equal to delta theta over the angular speed. So what's delta theta? Delta theta? We know that it's 2 pi. It would be divided by the angular speed which is 3 rad per second. Of course, keep everything in the proper units. So, I'll put it on my calculator. 2 pi / 3. I got an answer to be to one decimal place approximately 2.1 seconds. Okay, that is it for uh question number six. We'll move on to question number seven. What does it say?

[00:09:23]The first wheel at a local amusement park takes 40 seconds. This is the time to complete one revolution. So, what's theta theta in one revolution? It is 2 pi because again one full circle takes uh 2 pi. Okay. What is the average angular speed? So they want the average angular speed um of the wheel and the average angular velocity. What's the difference between speed and velocity speed it would be the absolute value. You only take the magnitude and the velocity um it would be with the direction.

[00:09:56]Another way to say it would be that the angular speed it's equal to what? It's equal to the change delta theta. But here theta is the angular distance over time. And here angular velocity it's delta theta but it's angular displacement over time. Right? Now here this takes a little bit of um thinking about what we're dealing with. Okay. We have a first wheel. Let's say it starts here. It does one full revolution. The distance covered would be the full entire circle which is how much? Which is 2 pi. So delta theta here it's equal to 2 pi over the time which is 40 seconds. But the displacement he starts from this position and where does he end up? He ends up in this position. Did he change position? Is there a change in the position? No. So delta theta the change in displacement is what? It is zero. So automatically this is going to be what? This is going to be zero. So uh the angular displacement it is 0 route per second whereas sorry the angular velocity is 0 route per second and the angular speed is going to be 2 pi over 40. Let me put it on my calculator.

[00:10:56]And I got it to be approximately 0.157 rad per second. Okay, that's it for question number seven. We'll move on to question number eight. By the way, look what I noticed. Questions marked top are intended exclusively for AP students. So I would not be doing the top question.

[00:11:13]These are for the AP students. Um if there is one that I feel might come, I would do it. But uh otherwise you could like try by yourself or you could ask your teacher. Okay, we'll move on anyways to question number eight. What does it say? A car is moving along a horizontal road. Its tires are all rotating with an angular velocity of this. So this is the angular velocity.

[00:11:33]The driver accelerates uniformly for radulation of 4.5 um seconds. This is the time. Okay. Uh as a result each tire reaches an angular velocity of this much or sorry it reaches an angular velocity. So this is angular velocity initial. This is angular velocity final. What is the average angular acceleration during d4.5 seconds? This is as if it were a linear question. We have velocity final velocity initial time. They want the average angular acceleration. So it would be change in average sorry change in um angular velocity divided by the change in time. So it would be what? It would be 58 minus 53.5 divided by 4.5 and you would get uh wait let's see 58 - 53.5 that's 4.5. So you would get 4.5 over 4.5 which is what which is 1 rad/ second squared. So what is the angular acceleration? It is 1 rad/s squared.

[00:12:33]That's it for 6.1. Now we'll move on to 6.2. What does it say in 6.2? A wheel at rest. What does this mean? This means angular velocity initial is equal to zero starts rotating with an angular acceleration of this. So this is the angular acceleration after 0.5 seconds.

[00:12:47]This is the time the magnitude of the wheel's acceleration is this. Now now please pay attention. It's in cm/s squared. So this is the net acceleration. Okay. Now net acceleration or acceleration total. Okay. It's equal to radical tangential acceleration squared plus um centrip petal acceleration squared. Okay, they gave us now angular velocity final to be one rod per second squared. Okay, they want us to find the radius of the wheel. First of all, they gave me the net acceleration. Okay, so let me find the tangential acceleration. Tangential acceleration you should know it's what?

[00:13:29]It's the angular acceleration multiplied by the radius. So the tangential acceleration it would be 2 * the radius.

[00:13:36]Now centrip petal acceleration it's equal to the angular velocity squar multiplied by the radius. So it would be what? It would be the centrip petal acceleration it would be 1 2 * what time the radius? Okay. So here I got 2 r here I got 1 r. So now a total it's equal to what? It's equal to radical 2 r^ 2 + 1 r 2 and then a total it's what? It's 13.6 cm/s squared. Okay. Uh so then I'm going to square both sides. What's 13.6 2. So this would be 18 or 184.96 is equal to 4^2 + 1 4 r 2 + 1 r 2. So 5 r 2. I will divide this by 5. So divided by 5 I will get 36.992 is equal to r^ 2 and then I do radical for both sides I would get that r it's equal to what it's equal to 6.08 so it's approximately 6.1 cm and that's it for question n question 10. What does it say in question 10? A girl is playing on a merry go round of a diameter 3 m.

[00:14:49]What does this mean? This means the radius is half of that which is 1.5 m.

[00:14:53]at some instant. Okay, the magnitude of her angular acceleration is this much.

[00:14:57]So this is her angular acceleration and that of her centrip petal acceleration is 3 m/s squared. So this is the centrip petal acceleration. Okay, they're saying if the girl is at the edge of the merrygoround. So look, she's at the edge. So when she's at the edge, she has her centry petal acceleration and her tangential acceleration. When she's not at the edge, she does not have a tangential acceleration, but she is at the edge and she has her tangential acceleration. They want what the magnitude and direction of her net acceleration at that instant. So uh first thing that we need to find is the tangential acceleration. Tangential acceleration it is what? It is the angular acceleration multiplied by the radius. So it's going to be 050 * 1.5 which is what 0.75 m/s squared. Uh where's the five? Here's the five. So now they want the net acceleration a net or a total. Okay, it's equal to radical tangential acceleration squar plus centrip petal acceleration squared. So it's going to be radical uh 3^ 2 + 0.75 squared. You put it on your calculator.

[00:16:01]I got an approximate answer of 3.1 m/s squared. Now this is an important uh part. Remember we said what we said that she is on a merry ground, right? She is here. This is her. She has a centrip petal acceleration and a tangential acceleration. When we are doing radical tangential acceleration squared plus centry petal acceleration squared we're going to get the resultant. They want us to find the what the angle of the resultant but we need to use what we need to use the values that were given here. So uh I could do this like I can copy it here. I can shift it to this side and I see that it's a triangle right and this is the adjacent side and this is the opposite side and this is the sentry petal. Uh oh but actually yeah this is the petal this is the tangential but they don't want this angle here they don't want the angle there they want this angle over here they want the angle between the resultant and the tangential acceleration. Okay. So you could always take it like this if you want and you could memorize this equation that theta which is the um direction okay it's going to be equal to tangent inverse always always it's going to be the centrip petal acceleration over the tangential acceleration. So you put it on your calculator. So tangent inverse 3 over 0.75 you put it on your calculator and you will get an approximate answer of 76°.

[00:17:29]Okay, that is it for question number 10.

[00:17:32]I'll move on now to question number 11.

[00:17:34]What does it say? Uh very simply, a particle is moving in uniform circular motion. So, U C M with an angular displacement given by this equation. The radius of the circular path is 1 meter.

[00:17:47]They want the particles angular and linear speed. Okay. Uh so in order to find the angular speed, angular speed we will denote it omega.

[00:17:58]Okay. We will need to take the derivative of this. So let us derive it over here. We would get three rad per second. So what is the angular speed?

[00:18:06]The angular speed is three rod per second. Now what's the linear speed?

[00:18:09]Linear speed v is equal to omega multiplied by the radius. So it's going to be 3 rod per second time the radius which is 1 m. So it would give me 3 m/s.

[00:18:18]This is the linear speed. Now they said determine the motion's frequency. You should know that one rotation which is 2 pi multipi multiplied by the frequency it would give me what it would give me the angular um velocity. Okay. So I have omega omega is 3 rad/s.

[00:18:36]Uh I need to find f. So f would be equal to omega over 2 pi. So it would be 3/48 htz. So the frequency is 0.48 hertz.

[00:18:53]Okay. Now part C, what does it say? Find the normal acceleration. When they say normal acceleration, this also means the centrip petal acceleration. Why? Because it's normal to the surface and it's pointing towards where towards the center. Okay. Uh so the centrip petal acceleration or the normal acceleration it's equal to what? It's equal to um omega^ 2 * r. Right? So it's going to be 3^ 2 * the radius which is 1. So uh so I got it equal to 9. Now in the marking scheme what they did to get an like accurate answer kind of an accurate answer uh was that they substituted omega for what they substituted did for 2 pi f. So it would be 2 pi f 2 * r. So 4<unk>^ 2 * 0.48 2 * 1. So 4 *<unk> 2 * 0.48 48^ 2 * 1 I got 9.1 uh m/s squared. So this is what they did. This also works and this works. Um so yeah that's it for question number 11. We'll now move on to question number 12. What does it say? Question number 12. Um at time equals 0 seconds a wheel rotating with an angular velocity of this much starts slowing down uniformly until it stops. That means omega v final is equal to zero. If the angular acceleration is -2/s squared, how many degrees will the wheel turn in the first 2 seconds? And what is the wheels angular velocity in the first 2 seconds?

[00:20:28]Okay, so when they're saying how many degrees will the wheel turn, that means they're asking for delta theta. And delta theta that's 1 /2 * angular acceleration * time^ 2 plus v initial or omega initial * d time t. Okay, so it's going to be 1 /2 * the angular acceleration which is -2 * the time which is 2 ^ 2 plus omega initial which was 5 * 2. So you put it on your calculator 1 / 2 * - 2 * 2^ 2 + 5 * 2 and you should get 6 as an answer. So it's 6 rad. Okay. Now they want what they want the wheels angular velocity.

[00:21:08]Okay.

[00:21:10]uh we can find it without the time. We could apply this equation. Do you know that 2 * angular um acceleration time delta is equal to velocity angular velocity final squar minus angular velocity initial squared. So now you substitute all your values. So 2 * -2 * 6 rad is equal to this that we need to find it - 5^ 2 so - 25.

[00:21:38]So 2 * -2 that's -4 * 6 that's -4. So -4 is equal to um - 255. Move this to the other side I got 1 right. So 1 is equal to omega v final squared. You do radical on both sides. So you got you will get that the wheels angular velocity finally it is what it is one rad per second.

[00:22:01]Okay we move on now to question number 13. What does it say? Below are the graphs um of the angular displacement versus time and the angular velocity versus time of an uh object which is an uacm.

[00:22:15]So look as you can see this is the angular um this is sorry a uh angular displacement versus time graph and it's accelerating and this is angular velocity versus time graph and it's what it's also accelerating. Express the angular displacement and angular angular velocity as a function of the time t.

[00:22:35]Now because it's uacm you should know that delta theta okay it's equal to 1 /2 * angular acceleration time the time squared plus uh omega initial uh time the time okay so I will change it because they want it as a function of time theta time like theta time this is the function of displacement as a time it's 1 /2 times the angular acceleration uh how do you find the angular acceleration Very simply, I can use this graph over here. I can find the slope of any point on this graph. For example, here this is 0 1, right? And let's see anything else that I could use. Um, for example, I'm just trying to find the most accurate one. For example, here this is 4 4.

[00:23:23]So, I think the marking scheme is wrong for this one. So, I got what I got the angular acceleration to be 3 over 4. And what is omega initial? Omega initial it is what it is given as 1. So this would be 1 /2 * 34 * t ^ 2 plus omega initial which is what which is 1 * t ^ 2. So uh what is 1 /2 * 3 over 4? It is 3 over 8.

[00:23:46]So 38 t ^2 + t sorry without the t^2 here plus t. Okay. So this is 4 theta.

[00:23:54]Now they want the angular velocity as a function of time. angular velocity final is equal to because it's UACM it's angular velocity initial plus angular acceleration time the time so it would be 1 + 3 over 4 t and there you go that is the solution uh for question 13 we'll move on now to question 14 okay I think I will finish this chapter and then I will post this video then I'll take a break and I'll continue 6.5 until 6.9 Okay. So, uh question number 14.

[00:24:28]Initially, okay, a grinding wheel has an angular velocity of this much. So, this is the angular velocity initial. The wheel accelerates. This is angular acceleration for 2 seconds. Only the time is 2 seconds. What's the angular velocity of the wheel after 2 seconds?

[00:24:42]Um I believe this would be UACM uniformly accelerated circular. So, I would use this equation. Angular velocity final is equal to angular velocity initial plus angular acceleration time d time. So it would be 24 plus 30 * 2. So it would be what? It would be 84 rad per second. So this is the angular velocity after 2 seconds.

[00:25:09]Now through what angle did the wheel turn during the 2 seconds? So they want the angle. When they say the angle, you directly think of theta and the delta theta because it's UACM. Delta theta is 1 / 2 uh * angular acceleration t ^ 2 + omega initial t. Okay, so it's 1 /2 * 30 * 2 ^ 2 + 24 * 2. Put it on your calculator. What will you get?

[00:25:38]Time 2 + 24 * 2. You will get an answer of 108.

[00:25:42]Okay, so this is the angle that the wheel turned during the first two seconds. Now part C, the wheel starts decelerating at a constant rate until it finally stops. That means omega um final yeah angular velocity final is equal to zero turning through uh 432. So this is delta setta. They want the duration of the decelerating phase. So they want the time of how much it decelerated. So, I'm just going to erase this so I have space to solve the others. For this one, let's see what they need. They want the time.

[00:26:17]I have delta theta. I have um velocity final, angular velocity final, and I have angular velocity initial over here.

[00:26:24]So, I could apply this one. Delta theta is equal to angular velocity initial plus angular velocity final over 2 * the time. So then the time would be equal to 2 * delta theta over angular velocity initial plus angular velocity final.

[00:26:40]Right? I just moved it. Okay. So the time would be 2 * 432 / 24 + uh 0. So 2 * 432 / 24 + 0. This would give me what?

[00:26:54]This would give me um or wait no excuse me. This is not um this is not what we're starting with.

[00:27:01]Excuse me. uh the angular velocity initial we're taking it after the two seconds when it reaches the two seconds.

[00:27:08]So what did we get the angular velocity of the wheel at 2 seconds? We got it to be 84 rad. Excuse me that's my mistake.

[00:27:14]So 84 rad per second. Now this is V initial because look the wheel then starts decelerating. So after 2 seconds so when I get a V initial it's what?

[00:27:24]It's 84. So here I'd substitute it with 84. So this would be 2 * 432 over 84. uh I got an approximate answer of 10.3 seconds for part I. Now for part double I, what was the wheels angular acceleration? Um here we could use um many different types of equations but because it's during a certain phase uh so I would say that the angular acceleration sorry for moving it the angular acceleration it's equal to what it's equal to omega final minus omega initial over delta t. So it would be what it would be uh 0 - 84 over 10.3 and I got it to be - 8.16 rad/ second squared. That is it for question number 14. We now move on to question number 15. What does it say? When a DVD player, okay, is turned off. A disc inside it initially rotating counterclockwise.

[00:28:23]Very important. at this much revolutions per minute stops rotating. This is assumed to decelerate uniformly that means UACM right at a rate of this much.

[00:28:35]So this is the angular acceleration when the excuse me when the DVD player is off a red dot is marked on the disc as shown below. What is the magnitude of the disc's angular velocity 3 seconds after the DVD player was turned off? Okay, because it's UACM, we need to think of the equations that we're going to use.

[00:28:53]they want angular velocity final. So we need angular velocity initial and we need the angular acceleration multiplied by the time the angular uh velocity initial. Let's see what I have. I have the number of revolutions per minute.

[00:29:10]Right? Now what are the units for angular velocity? They are rad per second. Now I have 4.60 * 10 3 and it is in what? It's in revolutions per minute.

[00:29:20]First thing that I need to do is I need to get rid of the revolutions. We know that one revolution is 2 pi rad, right?

[00:29:26]It's 2 pi. And we need to get rid of the minute. We know that one minute it is what it is um 60 seconds. Now, how did I know how to put them? First of all, you see where your unit is and then you place the other one in the opposite direction in order to be able to cancel them, right? Then you just need to know the conversion factors. Okay? Uh so then it's going to be what? 4.60 * 10 3 * 2 pi all of it over 60 seconds. So I got uh 460 over 3 pi. Keep it exactly as it is. Keep the exact answer. Okay. So now um angular velocity final is equal to 460 over sorry 460 pi over 3 plus angular acceleration which is -50 * the time what's the time? It is 3 uh seconds. So 460<unk>i over 3 plus -50 * 3. You put it on the calculator and I got an approximate answer of 300. and 32.

[00:30:23]Um, uh, it depends on how you do it.

[00:30:25]They got 331. I got 332 rad per second.

[00:30:29]Okay. Now, part B, what does it say in part B? What is the angular displacement of the red dot 3 seconds after the DVD player was turned off? So, again, because this is ACM, we're dealing with delta theta. Delta theta. That's equal to 1 /2 angular acceleration t^ 2 plus omega initial t. So delta theta is equal to 1 / 2 * -50 * 3^ 2 + uh we got this to be 460<unk> over 3 * what * 3 seconds. So 1 /2 * -50 * 3^ 2 + 4<unk> over 3 * 3. You put it on the calculator and I got approximately two to three significant figures. 1.22 22 * 10 to the um * 10 the power of 3. Again they are mistaken in the answer key. Okay. Uh now part C they said what is the time needed for the disc to stop? What does this mean the disc to stop? That means that the angular velocity final is equal to zero. So this would mean that 0 is equal to the angular velocity initial. What did we find it to be? We found it to be 460<unk> / 3 plus the angular acceleration time the time. What was the angular acceleration or deceleration? We found it to be 50. So you'll move this to the other side. Uh - 460<unk> / 3 is equal to - 50t. Right? Minus min - cancel it. So t it's equal to 460 pi over 3 divided by 50. So I got the time to be approximately 9.63 seconds. Okay. So time is approximately 9.63 seconds. Now I'm getting different answers from them because I'm using the exact values. Okay. Now how many revolutions would the red dot make before it stops? We have now the time.

[00:32:19]Okay. So we need the revolutions. First of all we need to find the displacement which is delta theta. Delta theta is equal to 1 /2* the angular acceleration which is - 50 * 9 sorry time the time which is 9.63 2 + 460<unk> / 3 * 9.63.

[00:32:36]So now we'll get delta theta. So 1 /2 * -50 * 9.63^ 2 + 460<unk> over 3 * 9.63.

[00:32:47]So I got what? I got uh 2 to the correct significant figures of course uh 2.32 * 10 3 rad. Now this is not over yet. This is delta theta. They want what? They want the uh number of revolutions. I would take this and I will divide it by 2 pi. Why? because one revolution is 2 pi. So 2.32 * 10 d3 divided by 2 pi on my calculator I got uh 369 revolutions. Okay. So that is it for part D question 15. Now we'll move on to question 16 and then I'll stop this video here. What does it say? It says a carousel in an amusement park starts rotating from rest. That means angular velocity initial is equal to zero. With a constant angular acceleration at time equals 0 until time 1 equ= 1 minute.

[00:33:37]During this time interval the carousel completes five turns. If it completes five turns then it completes five revolutions. So what is delta theta equal to delta theta? It's equal to what? It's equal to uh the number of revolutions times one revolution which is 2 pi. So it's going to be what? It's going to be 10 pi. Okay. A pink horse and a yellow horse at sorry are at 2 mters and 3 meters away from the center of the carousel respectively. So let's kind of draw it. We have our carousel, right?

[00:34:06]And um we have a pink horse and we have a yellow horse, right? The yellow horse, what's the distance from the center?

[00:34:16]It's 3 m. And the pink horse, what's the distance from the center? It's 2 m.

[00:34:21]Okay, now they're asking us what's the angular displacement of the pink horse between time 0 and time 1 if it has described an arc length of 4 m.

[00:34:32]So the arc length over here it's what?

[00:34:34]It's basically the displacement which is s or like the the distance moved which is s right now we know that s it's equal to the angular displacement time the radius this you should even know it from math that the arc length is equal to theta * r r theta right you should know from math so it's going to be equal to what it's going to be equal to theta what did we find the theta to be we found that um oh you know they are looking for the angular displacement they're looking for theta so theta it would be what it would be s over the radius okay so it would E4 over the radius which is the pink radius which is two. So it would be what? It would be two rad. So theta for the pink horse it's what? It's two rad. Okay. Now what are the values of the angular acceleration and the tangential acceleration of the yellow horse during the 1 minute interval. So they want the angular acceleration and the tangential acceleration during the 1 minute interval. So look what they have given us. They have given us the 1 minute interval. So this is the time. Time equals 60 seconds. Right? Um let's see.

[00:35:36]We have delta theta.

[00:35:41]Yeah, we have delta theta. We have it to be 10 pi.

[00:35:45]Um we have the radius of the yellow horse which is three.

[00:35:50]We have omega initial, right? So using this information, we would be able to find omega final. How can we find it?

[00:35:59]Using the time delta theta and omega initial. Uh very simply uh you should know this equation. Delta theta is equal to omega initial plus omega final over 2 * the time. So uh 10 pi is equal to omega initial. What's omega initial?

[00:36:15]It's 0 + omega final over 2 * the time which is 60 seconds.

[00:36:21]So then omega final it would be equal to 10 pi * 2 over 60 seconds. So 10 pi * 2 over 60 seconds um I got omega final to be what to be 1.05 um rad per second. Okay. Now what can I do with this information? I would be able to find the angular acceleration.

[00:36:47]Angular acceleration it's equal to what?

[00:36:48]It's equal to um let's see it's equal to the change in the angular speed divided by time. So it would be what? It would be 1.05 minus 0 over the time which is 1 minute which is 60 seconds. So you put it on the calculator and you'll get 1.75 * 10 -2 rad/ squared. So now we have the angular acceleration. So now we found first of all the angular acceleration. Now they want the tangential acceleration.

[00:37:15]Tangential acceleration it's equal to what? It is equal to the radius multiplied by the angular acceleration.

[00:37:21]So the radius of the yellow horse it's 3 * the angular acceleration which is 1.75 * 10 the minus2. So I got it to be oh there's no space. I will just write it here. I got it to be 5.25 * 10 the -2 m/s squared. Okay. So that is it for part B.

[00:37:38]Let me clear it. So now we can answer part um C. What is the speed of the yellow horse at time one? So they want the speed of the yellow horse at a certain time. Now the speed, the linear speed, it's equal to the angular velocity multiplied by the radius. So what was our angular velocity? We found it to be 1.05 time the radius of the yellow horse, which is 3 m. So it's going to be 3.15 m/ second. Is it equal to that of the pink horse? Explain without performing any calculations. Um so honestly it does not based on the equation that we have. Um we know that the angular velocity is the same all throughout like the circle or the disc.

[00:38:23]It does not matter where you are or even like in the carousel. It's always the same the angular velocity but the radius changes. So here the pink horse is 2 m away the yellow horse is 3 m away. So because um it depends on the radius of the circle the linear velocity. Okay. Uh they are at different distances from the center. So their speeds are for sure going to be different. Now let's move on to part D. What is the magnitude of the centry pedal acceleration of the yellow horse at time 1? Centry pedal acceler acceleration. So AC it's equal to omega^ 2 * the radius. So 1.05^ 2 * the radius which is how much? Which is 3 because we're dealing with a yellow horse. So 1.05^ 2 * 3. I got an answer to be 3.31 m/s squared. Now finally, last but not least for part E, determine the characteristics of the net acceleration of the yellow horse. When they say the characteristics, that means they want the magnitude and the direction. So let me just clear it out and zoom in a bit.

[00:39:22]So to find the magnitude, which is the net acceleration. So a net or a total, it's the tangential acceleration plus the centry petal acceleration squared.

[00:39:31]Okay. So um let's see, it would be a radical. What was the tangential acceleration? We got it previously. It was 5.25 * 10 -2^ 2 + 3.31^ 2 uh put it on your calculator and I got again 3.31 m/s squared. So this is the net acceleration. Now they said characteristics. So they also need the direction and remember what was the equation that theta is equal to tangent inverse the centrip petal acceleration divided by the tangential acceleration.

[00:40:03]So it would be tan inverse uh 3.31 over 5.25 * 10 the minus uh 2. You would get an approximate answer of 89.1°.

[00:40:15]And that's it for uh 6.4 question 60 part E. So thank you guys for watching this video. I'll see you in the next one.