AP Physics 1 - Unit 6 - Live Kahoot! Multiple-Choice with Flipping Physics

Flip [music and singing] Physics. — Good [clears throat] morning everybody. Welcome to my unit 6 AP Physics 1 live cahoot. It's a multiplechoice quiz. Please sign up for the Cahoot when you get the chance to see that. Um and here we go. I will get started. Uh, all right. Here we go. Um, again, please sign up for the Cahoot. Uh, unit 7 will be on Tuesday, April 7 at 700 p. m. Eastern time. So, that is roughly actually that is three weeks from today. So, get excited for that. All right. So, section one of the AP physics exams have 40 multiple choice questions in 80 minutes. That means two minutes per multiplechoice question. [snorts] So realize we are going to be doing exactly that. We're going to do two multiple choice two minutes per multiple choice question today. Uh just to give you a good feel for how this is all going to work. Um somebody tell me that the volume's working. There we go. I got somebody who's logged in. Give me some sort of response that tells me everything's working fine. All right. Awesome. Okay. So, please say hello to our moderator, Miss Bonk. She is in the chat helping us all uh with any tech questions that you have or maybe some physics questions depending on how things go. Uh Miss Bon is an AP physics and project lead the way teacher at Addison Trail High School outside Chicago. I do need to talk about my act ultimate exam slayer flipping physics and my ultimate review packet. This is right here. Uh the 10 questions we're going to go through today are the first 10 questions in my ultimate exam slayer multiple choice quiz for unit six. There are 16 more questions. I'm going to in that ultimate exam slayer, there's all sorts of fun things in that ultimate exam slayer. Uh and be aware that everything I'm going to go through today, for example, on any uh question going through the solutions, those are all also in the ultimate exam slayer. So, you get to see all the solutions and everything that I would go through uh if I were talking to you about how to make sure you understand how to solve any of the questions. Flipping physics, of course, I have all of my videos for you and my ultimate review packet, which uh just goes through and helps you review and understand the material before the AP exam. Awesome. So uh before we begin question one I think we are going to get started. All right here we go. I'll click on actually no before I click start I want to do this. So I need to excuse me a moment. I need to switch up where the um different yeah where the different um screens are. Give me just a second. So, I need to go. Let's do this. Escape out of here. Yeah, there it is. See, I need to remember to do that. All right. [sighs] Okay. So, I got to go here and quit out of this. Perfect. Ah, nope. You don't want to see that. All right. Um, here we go. Bring the cahoot over to there. Take this and play and put this over there. Perfect. Okay, so we are ready for question one. I keep doing that. All right, [snorts] perfect. So, we are ready. So, I'm going to click start. Best of luck to everyone. All right. So, 3 2 1. You're going to have two minutes to solve this question.

All right, there we go. Look, did pretty well. Looks like the majority of you got it right. Let's walk through why that is going to be correct. Okay, so question one. In question one, let's make this a little bit bigger. There we go. In question one, we the knowns are that the rotational inertia of object P is equal to capital I. The rotational inertia of capital Q or of object Q is 1/2 times capital I. And the angular velocity angular speed, excuse me, of object P is omega and angular velocity angular speed of um object Q is 2 times omega. So the objects are only rotating. So they only have rotational kinetic energy. Therefore, we could solve for the rotational kinetic energy. It's 1/2 times the rotational inertia times the angular speed squared. For object P, that just works out to be 1/2* uh capital I * omega^ 2. For object Q, when we substitute inh when we substitute in the uh information for the rotational inertia, angular speed of object Q, we end up with I capital I* omega^ 2. So it just works out to be when we um the kinetic energy of rotational kinetic energy of object P is 1/2 times rotational inertia um times angular velocity squared. So that is equal to 12 times the kinetic energy of Q. So you end up with 2 * the kinetic energy rotational kinetic energy of P is equal to the kinetic energy of Q which is correct answer B. Beautiful. All right. Now let's move on to question two. Before we do let's see how people are doing. All right, looks like Amusing Badger is first and Royal Manatee is next. All right, here we go. Uh, moving on to the next question. Question two looks like that. I'll make it a little bit bigger in just a moment.

All right, looks like we have fewer people getting it right this time. That's okay. This was a more difficult question. We have, in fact, more people answered choice D than choice C. So, let's see why choice C is the correct answer. Here we go. Question three. Again, I'll make it a little bit bigger. I think that's going to be important for understanding this. All right, question two. So, please notice there are no yaxis scales, right? There's no numbers over here, right? So we do not need any numbers over there in order to solve this problem. We're only finding which one has the largest rotational kinetic energy. Not we're not finding the value of that largest rotational kinetic energy. We're just figuring out which one has the largest. Okay. So we need to assign values to determine the relative magnitudes. So we're going to use the horizontal lines. These horizontal lines right here with the lowest one being the base value. So we're going to use all these horizontal lines. And again this lowest one is the pa the base value. So angular velocity at zero omega kn we're going to identify as omega. We're going to assign that right there. This first horizontal line is having a value of omega. So at angular velocity at 2 seconds is all the way up here. And if you look that is because it's the third line is going to be the 3 * omega times that first one. And then the angular velocity at 3 seconds is all the way up here. And this is the fourth line. So it's going to be equal to four times the omega. And the angular velocity at 6 seconds is all the way up here. It is 7 time omega. It is 7 times that initial angular velocity at 0 seconds. We can do the same thing with rotational inertia. But you'll notice the rotational inertia at 6 seconds is equal to this lowest value of the rotational inertia of just I. So the rotational inertia at 3 seconds and then is four times the rotational inertia I. And the rotational inertia at 2 seconds is 5 I. And rotational inertia at 0 is 7 I. Again, seconds is 7 times as large as the rotational inertia at 6 seconds. So those are our values that we're going to use for this. So now we can determine the relative rotational kinetic energy values in terms of capital I and omega. So rotational kinetic energy again 1/2 rotational inertia times angular velocity squared. So the rotational the kinetic energy rotational kinetic energy kn at 0 seconds is 1 half times I kn omega kn. But we can substitute in then seven times i for that initial angular or that initial rotational inertia and we can substitute in omega for that initial angular velocity. So we end up with 3. 5 times rotational inertia times angular speed squared. So you'll notice we can do this for every one of these values. The kinetic rotational kinetic energy at two when we substitute in the values gives us 22 * i * omega^ 2. The kinetic energy at 3 when we plug in the numbers we get 32 * i * omega^ 2 and the kinetic energy at 6 when we substitute in the numbers gives us 24. 5 * i * omega^ 2. So notice that the largest rotational kinetic energy is at 3 seconds. So that is the correct answer. The correct answer is C. So I do want to point out that questions one and two are the same concept, but there's the information is presented differently and so it makes it so you solve the two questions very differently. So notice like um I'll okay I'll answer that in a second. So realize like you can have different information uh you can have like the same concept presented with different information and so you can solve it in a different way. So realize the question was is there any reason why it happens at where they intersect. The truth is that the maximum kinetic energy is actually somewhere around here. I just gave the largest uh we just asked which one was the largest of the ones that I provided uh 0 2 3 and six. But the maximum kinetic energy happens right around here. You can actually go through and graph them and figure that out. But uh from that's more than I would expect in a multiple choice question. All right. So let's move on. Let's see how everybody's doing. Let's see. those. Oh, lustrous salamander. Nice. Well done. All right, moving on to question three. All right, that should do it. And next

All right, the correct answer is B and we have more. Most people got that. So, let's see why the correct answer is B. All right. Question three. That probably works. Okay. So we are determining two things for the rod clay system. Does the kinetic energy remain the same or does it decrease? And does angle momentum remain the same or decrease? Anytime you're given one of these data tables as the options for the answers to the multiple choice question, it's important to just take a moment and realize what you're looking for. Right? You can see we have decreases and remains the same and remains the same. And so same is over here. So really you're just looking at does the kinetic energy remain the same or decrease and does angular momentum remain the same or does it decrease. The clay sticks to the rod which is a perfectly inelastic collision and you should rec recognize that kinetic energy decreases during all inelastic collisions. Therefore the correct answer is either B or D because those are where kinetic energy decreases. So now we just need to discuss the angular momentum. So let's define the positive X and positive Y directions. It should be pretty clear. We're going to make positive x to the right, positive y, we'll call that up or yeah, up on the screen. Uh, just want to make sure we're all on the same page for that. Okay. So, now class, what condition? Just realize anytime I say class, I'm referring to everybody here and I want an answer in the uh comments. Want you to answer in the comments. Class, what condition must be true for a system for the angular momentum of a system to remain constant? Looking for answers in the comments. uh for the angular momentum of the system remain constant. Okay, interesting. So this is actually not true. alpha equals zero. The rotational inertia equals zero. Not actually true. There we go. This is the correct answer. There needs to be no net. Well, we need to add the word net in there, but no net external torque. So, that's very important, that word net. Just so you know, no net external torque. So, let's talk our way through this. So angular momentum remains the same for a system if uh if the net external torque on acting on the system equals zero. So that's what we're trying to determine. Does the net external torque on the system equal zero. So we need to identify the forces acting on the system. So in our we have now drawn the force diagram of all the forces acting on the system. We have the force from

the clay on the rod in the positive y direction. So this is We have the force from the rod on the clay in the negative y direction. So Realize these are a Newton's third law pair. Right? So they are going to be equal in magnitude but opposite in direction. The force from the pivot on the rod is in the positive y direction. And this prevents the left end of the rod from moving in the negative y direction when the clay st when the clay strikes the rod. So if the pivot were not there, this left end of the rod would move this direction in the negative y direction during the collision. Right? This collides here. This would go down in the negative y direction except this pivot right here prevents that from happening. That's why the force from the pivot on the rod acts in the positive y direction. So we need to identify the torqus on the rod clay system about an axis normal to the screen through the pivot. So this is the pivot right here. Axis of rotation is normal to the screen out of the screen. So the torque from the force from the P from the uh pivot on the rod is uh if using the torque equation torque equals uh r F sin theta where r is the distance from the axis of rotation to where the force acts times the force acting and times the sign of the angle between the force and r. So the force acts at the axis of rotation. So r value for the force from the pivot on the rod is zero. So therefore the torque is zero from the torque due the force from the pivot on the rod. So then we have the torque from the force of clay on the rod and the rod on the clay. Now they have the same r value all the way from here from the axis of rotation to where the force is applied which is the length of the rod. They have the same angle of 90°. This angle for is 90° this way and it's 90 degrees this way. So they're going to have the same angle there and the same force magnitude because they're in Newton's third law force pair. So they are opposite in direction. Therefore the torque from the force from the clay on the rod is counterclockwise goes this way. And the torque from the force from the rod on the clay is clockwise. So this torque from the force of the rod on the clay would cause the rod to rotate clockwise. So please realize both torqus cancel one another out. So the angular momentum remains the same if the net external torque equals zero. So on the rod clay system equals zero. So the angular momentum of the rod clay system remains constant. So remember from before the kinetic energy of the rod system decreases. Therefore the correct answer is B. All right. Good. So, I do want to address this statement. Hold up. Right here. This is really important to talk about. Okay. I really don't know how I would think and remember all of this in the middle of an AP exam. So, this is where practice is important because believe it or not, this one right here is pretty standard. like this type of question about this event is something you'll probably see or there's a good chance you'll see on the AP exam and if you've done this enough you can look at it and actually already know the answer because you've already worked through all of this in previous problems. So it's an important thing to remember. It's good to keep a balance though because a lot of times they give you something that looks very similar to something you've done before but is a little bit different. So you have to be careful of that. But realize this really comes down to practice. Uh, and I will point out what Kira pointed, what Kira said, which I think is also very important. So if you don't know where to start, you can go back to cons conservation principles. Really, that's very often a good way to do it. Um, just a good place to start. So two different answers, two different physics teachers. There you go. All right, let's go back to here. Let's see how everybody did on question three. Oh, we have a lot of changing up this time. I love it. Royal manity, congratulations in being in first place. All right, here we go. Um, question four fits on the screen. I'll keep going.

All right. Okay. Interesting. We had most people answer choice C, choice B, but choice A was the correct answer. So, here we go. Let's figure out why choice A is the correct answer. Uh actually before we look at the why it is I just want to take a moment to identify that this is one of those questions where I ju I actually just talked about how they'll give you something you've seen before but make it slightly different. So it is actually very common to look at this question uh from going down with all the objects going down the incline. But it's rare to see it going up the incline. So that's what we've done here is we've taken something that you've seen before and made it slightly different. So just be aware of that. All right, question four. So before the incline, both objects have the same linear center of mass speed. Therefore, they both have the same translational kinetic energy. Before the incline, the ball is rotating, but the cube is not rotating. That means that the ball has rotational kinetic energy, but the cube does not have rotational kinetic energy. So that means that before the incline the ball has more total kinetic energy than the cube. Again So when we define the system as the object and the earth so we're going to solve this problem here uh using conservation of mechanical energy. We can define the initial point as at the base of the incline. Define the final point at the highest location each object reaches. Define the horizontal zero line at the initial center of mass of the object. And the that means the initial gravitational potential energy of each object is each system excuse me is zero. So the total initial mechanical energy of each system is the total kinetic energy of each system. So notice that the ball earth system has more total initial mechanical energy than the cube earth system. So no work done is not by done by non-conservative forces and no external forces doing work on the system. So the mechanical energy remains the same. So at the final point both systems are at rest and above the horizontal zero line and both systems only have final gravitational potential energy. So because the ball or system had more initial mechanical energy, will have more final mechanical energy and there more therefore more final gravitational potential energy. So the ball travels farther up the incline. So the correct answer is C. The ball travels farther because it has more total kinetic energy than the cube. The correct answer is A. Beautiful. All right. Uh moving on to question five. But before we do, of course, let's see how everybody's doing. I believe that is we've had four different leaders over four questions. I think I'm right there. I'm impressed. I am impressed. I'm doing a good job with my questions then. All right. Next, we have question five. Good luck.

— [snorts] — All right, look at that choice B. That's what the most people answered and that is the correct answer. All right, let's figure out why. Uh, question five, we have all those words. So, choice A, the force of gravity between the Earth and a satellite is not zero. So class, what is the only way to make the force of gravity between the earth and a satellite equal to zero? Again class, this is for all of you. Please answer in the com in the comments. What is the only way to make the force of gravity between the Earth and the satellite equal to zero? There it is. The two objects need to be an infinite distance away from one another. That is the only way. All right, I'll put this one up there. So, uh, no mass question mark. Yeah, but the objects are going to have mass. I'm sorry. So, yeah, we're not talking relativity here. Okay. Uh, okay. So, here we go. Um so recall the equation for the force of gravity universal gravitational um uh force here is big g the gravitational constant times the mass of the earth satellite divided by the distance between earth and the satellite squared. So the only way to make this force of gravity is to have equal to zero is to have r equal to infinity make this infinitely large. So realize that makes choice A is incorrect. Okay. Choice B. The satellite is in freef fall in the gravitational field of the earth is a true statement. While moving forward fast enough to miss the Earth as it falls is also a true statement. So choice B is correct. Choice C. The net force on the satellite equals the inward force of gravity acting on the satellite from the Earth. So the satellite does experience a net force. Therefore, answer choice C is incorrect. Answer choice D, the satellite's forward motion creates an outward force is just not a true statement. So, choice D is incorrect. All right, moving on to question six. Oh, wait. Before we do, of course, see how everybody's doing. Oh, eager dong in the lead. All right, plenty of space there. [snorts] Good luck everybody.

All right, choice D is the correct answer. We got seven people answered choice D. All right, let's figure out why choice D is the correct answer. I think I can leave it that big. All right, question six class. Where do we find the net work on a net torque versus angular position graph? Please again answer in the comments. Where do we find the net work on a net torque versus All right, I love this answer. I'm gonna highlight this. Oh, good. There we go. So we have this which is I love the brevity but unfortunately like I just I like I love clarity over brevity. So this is the better it's the area under the curve. I would even like it better if there were quotes on it. We'll get there under on under. But here we go. So oh I love that. Okay I do want to put this one up here. I love it. If you're fancy the integral. Yes, if you do calculus, it's the integral, but this is not a calculus based class. So there you go. All right. Here we go. So the equation work equals torque times change in theta. Net work equals net angular position or displacement. So the area under quotes a net torque versus angular position curve is going to be the net work. So those quotes are important. Above the horizontal axis is positive. Below negative. That is why I always put area in quotes. Technically it's called the signed area. But there's no reason you need to know that. So remember area above the horizontal axis is positive. Area below the horizontal axis is negative. So when we look at this graph and we separate it into three areas, I've called them area one, area two, area three. They are all triangles and they are all of equal magnitude. So, and area one and area two are positive and area three is negative. So, because area two and area three have the same um amount whereas area three is positive, area I'm sorry, area two is positive and area three is negative. When we add those two together, they add up to zero. Which means the total area is just the area of one. So we can figure out the network is just the area of triangle one. So 1 1/2 base times height. The base is pi / 2 radians. The height is 5 Newton meters. So we get 5 pi over4 which is 3. 927 or 3. 9 Newton m. The correct answer is D. So a

common student mistake here is to forget about the negative area below the horizontal axis. I see it all the time. And that gives you the incorrect answer of 3 * 5< unk> over 4. Right? Three of these areas. So that gives you 11. 781 or roughly 12 Newton meters, which is incorrect answer choice B. Great. All right, let's see how everybody's doing. Eager Duke way in the lead. Interesting. It'll be interesting to see if you lose. you lose. I hope not. All right, here we go. Uh, moving on to question [snorts] seven. A lot of words. Read well. — [snorts] — All right, look at that. Choice C. I'm impressed. Uh, all right. Before we talk through choice C, I do want to answer this question. I thought it was a good question. Uh, get there. So, the question was, uh, going back here. So the question was radians are unitless which is why the work done is still Newton meters. So uh correct. So realize that because uh we have Newton meters over radians for the units here because radians do not have any units. The units just drop out and it just ends up being Newton meters. Good point. Thank you for that. All right. Okay moving on to question seven. Here we go. All those words. Yes, the story. Okay. The net external torque on the student dumbbell system about a vertical axis of rotation through center of mass of student dumbbell system is equal to zero. So the angular momentum of the student dumbbell system remains constant. A the angular momentum of a rigid object equals rotational inertia times angular velocity. So the general equation for rotational inertia of a system is just rotational inertia equals the mass of the object times the distance from that the object is from the axis of rotation squared the sum of all the particles in the system. So when you when the student brings in the two dumbbells and their arms this decreases the average r which decreases the rotational inertia of the student dumbbell system and because angular momentum is constant the angular velocity must increase. So the correct answer is C. It's pretty a standard question. um likely to see something about that on the AP exam. All right, the top three are staying the same. Interesting. Here we go. Moving on to question eight.

Awesome. Answer choice D. Great. All right, let's talk through why answer choice D is correct for choice A. So, uh, so one of the things that you need to memorize and know how to derive about objects rolling without slipping down inclines is this entire thing that I'm about to walk my way through because this is definitely something you're going to see or again almost on the AP exam. So, the larger the factor in front of M R squ in the rotational inertia equation, the larger the percentage of the initial gravitational potential energy that needs to be converted to rotational kinetic energy. The less percentage remaining of the fi is the for the final kinetic energy translational. Let me just hit again. The less percentage remaining of the final kinetic energy translational the object will have. In other words, the larger the factor in front of mr squared means the smaller linear velocity at the bottom. You do not need to memorize rotational inertia equations for objects. That is not something you need to do for this exam. class. Which object in this problem has the largest percentage of its mass the farthest from its center of mass? Again, class answer in the comments. It is the hollow cylinder. The hollow cylinder has all of its mass located a distance r from its center of mass. So, it's going to have the largest rotational inertia. Okay. So, the for the three objects in this problem, the solid hollow cylinder has its mass farthest from its center of mass. So the solid sphere has its mass closest to a center of mass and the solid cinder cylinder is in between the other two. So in terms of the equation I = XMR^2 the X for the solid sphere is less than cylinder which is less than the hollow cylinder. So the speed of this for the center of mass of the solid sphere is going to be greater than the speed for the solid mass uh the center of mass of the solid cylinder then which is going to be greater than the speed for the center of mass of the hollow cylinder. The hollow cylinder is going to have the least and so the correct answer is D. So I do want to walk through it just in terms of rotational energy equations. You don't won't have to know this. If you have to know these numbers and I think this answers your question, Cass lucky. Um, if you need to know I people use

whatever letter they want to use. I used X, they can use K. If you need to actually know the number, they will give it to you on the exam. Right? A lot of times you don't need to know the number. You just have to know how they relate to one another. But I'm walking through with the numbers here just because it it's helpful to do that. So this is what the numbers look like for the solid sphere. It's 2-5ths * m^2 for a solid cinder is 1/2* m^2. For a hollow cylinder, it's mass time radius squ. So the x for the solid sphere two- fifths I just walked through that. So then two fths is less than 1/2 which is less than one. So again we end up with the same answer. Correct answer is still d. Again you do not need to have those equations memorized. And just so you know I have several videos about this entire concept. Of course I do, but I don't know. Felt like mentioning it here. All right, moving on to question nine. Let's see how everybody's doing. Oh, okay. We had a slight shakeup in the first three. We'll see if that continues. [gasps] All right, moving on to question nine. All right. Answer choice D is the correct answer. Let's figure out why that is for question nine. All right. Question nine. The knowns are that the r for the satellite is equal to the radius of the earth plus the altitude of the satellite. So the radius for one is capital r plus capital r or 2 r. The radius for two is capital r + 2 r which is equal to 3r. It's important to remember that altitude of the satellite. So the free body diagram, free body diagram of the forces acting on the satellite is just the force of gravity from the earth on the satellite which acts in the positive in direction. So net force in the ind direction equals force of gravity times mass times centrial acceleration. could substitute in universal gravitational force there and that's equal so the gravitational constant times the mass of the satellite earth all divided by the distance between the satellite and the earth that quantity that squared which is equal to the mass of satellite times the centrial acceleration class what do we do — Yes, there's totally a typo. I will fix that. I apologize for next time. Yeah. So, satellite 2 is a distance of two R above the surface of the Earth. I apologize. Did not see that. So class, what do we do now? We're right here. I want to know what do we do now?

We have this equation. What do we do? We're not going to plug in the racing satellite yet. Nope, not yet. Really? I'm a little bit sad right now. That's what I have to say. All right, fine. I will tell you, but still I'm sad to have to tell you this. Are you ready? — Everybody brought mass. — Everybody brought the mass of the satellite to the party. We can be equitable. We can take the mass of the satellite from everyone. So we end up with the centrial acceleration equals big g times the mass of the earth divided by the distance between the satellite and the earth that quantity squared. We can substitute in our values for uh satellite one. satellite 2 and we can divide one by the other. we get 1/4 over 1 nth and you get 9 over4. All right, beautiful. And that is of course answer D. All right, nice uh cast lucky. That's a lovely series of comments there. All right, uh moving on. Oh, actually, no. We got to see how people are doing. Oh, look at that. We had a little shakeup. I think Eager Dong's going to have it, though. It's going to come down to who gets to be in second place. All right, question 10. Here we go. — [snorts] — Oo, rough. Okay, choice C is the correct answer. Let's figure out why. All right, here we go. Yeah, so the most common student. Yeah, that's fine. Okay, so choice question 10 here. So the knowns, we know the mass of the pulley is capital m. cube is 2 * capital m. The radius of the pulley is capital R. We know the rotational inertia of the pulley equals 1/2* the mass of the pulley* the radius pulley squar which substituting in our values from the

problem gives us 1/2* capital M * capital R 2. So we're looking for the rotation the angular acceleration of the pulley in this problem. So I'm actually going to walk through two different ways to solve this problem. one is using the equations for the both forms of Newton's second law rotational and linear and we're going to use conservation of mechanical energy. So we're going to walk through both of those and I'm going to tell you the reason I'm going to walk through both of these solutions is because on the AP exams sometimes they say when on a free response question for example that you have to solve it using X or Y. So it's impos it's possible they could give you this to you as a free response question and say you must use rotational and linear form of Newton second law to solve this question or they could say you have to use conservation mechanical energy. So just be aware of that. So I'm going to walk through both. So for the net torque solution we need a force diagram of all the forces acting on here. So we have in here we have all the forces acting on the poly cube system. Okay. So you can see we have the force from the axle pushing up. gravity on the pulley pulling down. Uh we're defining clockwise as a positive torque direction. Note this is different than the way it's normally defined, but it makes sense for this particular problem to define it that way. So I have defined it and I've indicated that in the diagram. We have the force of tension uh on either sides of this rope here. One acting on the pulley, cube, and the force of gravity acting down on the cube. So we're going to sum the torqus about the pulley only just the pulley. So the net torque acting on the pulley is just going to be the torque from the force of tension acting the pulley which is going to equal the rot rotational inertia times angular acceleration. So the reason we uh we cannot include the cube in this question for into the sum of the torqus here is because there the cube has no angular acceleration. So the only force acting on the pulley which causes a torque is the force of tension. These other two forces do not cause torqus. So we have r for the force of tension. We have the force of tension and the sign of the angle for the force of tension. So r for the force of tension is the distance from here to here which is capital r the radius of the pulley times the force of tension the sign of 90°. So the angle between here and here between this direction and this direction is 90°. S of 90° is 1. So we end up with capital R times the force of tension is equal to the rotational inertia times the angular acceleration. So substituting in the equation for rotational inertia, we can solve for the force of tension. Uh and then we can solve for Oh no, I'm sorry. Everybody brought R to the party. I don't think I added this the this music. I totally apologize. That's so sad. Um so we solve for the angular acceleration which we're solving for here. Uh so it's equal to 2 * the force of tension divided by capital M * capital R. But unfortunately the force of tension is an unknown. So we can sum the forces on the cube in the positive direction to solve for that force of tension. So we're going to sum the forces in the cube in the positive direction. So notice the positive direction is because we've defined clockwise as positive down is positive on the cube. So that's ends up being the force of gravity which is positive minus the force of tension which is equal to mass times the acceleration in the positive direction. So that's why down is positive on the cube. So we end up with the force of tension equals the force of gravity minus the mass times the acceleration in the positive direction. We can substitute a mass of the cube times the acceleration of gravity for the force of gravity. And we can substitute in the mass of the cube for this mass right here because we sum the forces on the cube. We can then sub substitute in the mass of the cube which is 2 * capital m. And we end up with for the force of tension 2 * capital m equals the quantity gravitational field strength minus the tangential acceleration. So realize the acceleration in the positive direction is the same as the tangential acceleration for a point on the rim of the pulley. That's why we can substitute in tangential acceleration for this positive acceleration because the tangential acceleration is or the acceleration of the cube is the same as the tangential acceleration. It's important to say on the rim of the pulley. So we know tangential acceleration equals radius time angular acceleration. So we can substitute that in for tangential acceleration. And we can now substitute back into the equation for angular acceleration. This equation we got for the force of tension. You can see we substituted in right there. Uh you can see mass cancels out. It's on the top and the bottom. So we end up with this equation right here which you can we can rearrange. Multiply both sides by this radius. We can combine these right um I'm sorry, move this over to this side and we get 5 * the radius times the angular acceleration equals 4 * gravitational field strength. So we

end up with the angular acceleration= 4g / 5 * r. That's correct answer C. But as advertised that was just our first solution. We're going to walk through one again. Our second solution is that the mass uh the mechanical energy of the poly cube earth system remains constant. We're going to define the system as the poly cube and the earth and the initial point when the cube is released from rest and the final point after the cube has fallen a distance capital h. So the horizontal we're going to put the horizontal zero line at the cube center of mass at the final point and then we're going to define again the positive direction as clockwise. So I've uh summarized everything right there. Here we go. Our known these are all the same known we had before but we've added a couple of known. We have the displacement in the y direction of the cube we've identified as capital h. This is not a value that we know but we're just define it as capital h. And we know the initial velocity of the cube is equal to zero. Again, we're solving for the angular acceleration of the pulley. So realize the displacement in the y direction for the cube is positive h because again clockwise is positive. The cube is going in the positive direction. So we have conservation mechanical energy. The initial mechanical energy of the system is the gravitational potential energy of the cube. And the final mechanical energy of the system is the kinetic energy final of the cube which is going to be translational kinetic energy and the rotational kinetic energy final of the pulley. Now I do recognize that the pulley has gravitational potential energy but it has it both initial and final and it cancels out so it doesn't really matter as far as the equation is concerned. So we can now substitute in equations. We have the mass of the cube times the gravitational field strength time the initial height of the cube. And we have the translational kinetic energy and as I said rotational kinetic energy equations. So we can substitute in the equation for the mass of the cube. We can substitute the I'm sorry the value in the value for the initial height of the cube which is capital H. The mass of the cube again two times the capital M. We can substitute in the rotational inertia of the pulley. Now we can um again Oh my I'm sorry. I told I'm so sorry. We could have sung the song again. That's so sad. I'll have to do that. All right. So, everybody brought mass to the party. So, we can take mass from everyone. We end up with this equation right here. You can see 1/2* 2 is 1. 1/2* 1/2 is 1/4. So on and so forth. Now, the final velocity of the cube is the same as the tangential velocity of the point on the rim of the pulley. This is very similar to what we did before, only with the velocity instead of acceleration. So the final velocity of the cube is equal to tangential velocity. So that's equal to the radius of the pulley times the angular velocity of the pulley or capital r* pulley. So then we can square both sides. And if you look this right here is the same as what we have right here and here. So we can substitute in the radius squar times the final angular velocity of the pulley that squared as well. We can substitute that in right there. So and then when you on the right hand side when you add these two together you get 5/4s times the radius squared times the angular velocity of the pulley final squared and we can solve for the angular velocity final of the pulley squared. Now we have used conservation mechanical energy to solve for to get determine an equation for the final angular velocity of the pulley. The question was what is the final angular or what is the angular acceleration of the pulley. So realize the net torque is constant which means the angular acceleration we can use the u fishy m equations to uniformly angularly accelerated motion equations. We're going to use this equation. The final angular velocity of pulley squared equals the initial angular velocity of the pulley squar plus 2 * the angular acceleration of the pulley times the angular displacement of the pulley. Realize the arc length moved by the rim of the pulley is the same as the linear distance moved by the cube. In other words, the arc length is equal to r times displacement. So the displacement in the y direction of the cube is equal to the radius of the pulley times the angular displacement of the pulley or capital h equals capital r times the angular displacement of the pulley. So notice we can solve for the angular displacement of the pulley. Don't worry cast lucky I'm going to talk about it. Um so the displacement uh the angular displacement of the pulley is equal to capital h over r which going back to our equation we can then uh substitute that into our uniformly angular acceleration motion equation. We can substitute in the final angular velocity of the pulley squared in here. We know the initial angular velocity pulley squared is uh I'm sorry is zero and that is squared. So then here we can do it. I apologize but we can do it here.

— Everybody brought masses. — Okay, so everybody brought capital H one over capital R and two to the party. So we just end up with the angular acceleration equals 4g over 5 r. Correct answer is again c. So, is this more than what you should expect to see on the AP Physics One exam for a multiple choice question? Uh, I sure hope so. I can't imagine they would put something this long on there. I thought it would be really helpful to go through because it's very helpful for your understanding of the physics. This is definitely something you could see for a free response question. Um, but yeah. Yeah, it was definitely too long for a multiple choice question, but hey, I do those things. So, we are done. Of course, we do need to go check out the final results in the cahoot. So, here we go. I think we know who took number one. We'll see who took number two. Oh, Royal Manatee, Lucky Hawk. Nice. And of course, eager doo. All right. So, just so you know, my plan is for this video to remain live on YouTube. And this multiple choice player uh multiple choice quiz is posted in my mult ultimate exam slayer. As I mentioned before, there are six more questions in the quiz in my Ultimate Exam Slayer. I do have to say while you're gearing up for the AP exam in less than two months, it is definitely really helpful to get my ultimate exam slayer and ultimate review packet. Please realize in one and a half fortnights, the unit 7 multiple choice quiz uh three weeks from today, Tuesday, April 7 at 7 p. m. Um, okay. So really, the problem here is it makes me feel like a No, I'm sorry. I just It makes me feel weird to have somebody give me money and ask me to sing. It's just weird. Sorry. No, just not going to do it. I will say thank you very much for learning with me today. I enjoyed learning with you. Have a beautiful night everybody. Bye. You can hear it here though.