AP Physics 1 - Unit 5 - Live Kahoot! Multiple-Choice with Flipping Physics

flipics. — Ah, good morning everybody. Welcome to my unit 5 AP Physics 1 live cahoot multiplechoice quiz unit five. Here we are. This is our fifth time doing this. I'm pretty excited to have you all here. Obviously, sign up for the Cahoot using the pin that's right there on the screen um and in the chat. Lovely. Okay, so here we go. Um, is that the right first page? Yes. So, please sign up for the Cahoot. Unit 6 will be on Tuesday, March 17 at 7 p. m. Eastern time. So, we're doing unit 5 now, but we'll be doing unit 6 uh on Tuesday, March 17th. Please be aware that section one of the AP physics exams have uh 40 multiple choice questions and you have 80 minutes to do those. That means it's two minutes per multiple choice question. So today we are also going to be doing 2 minutes per multiple choice question. All right, here we go. Um, say hello to our moderator, Miss Bon. Uh, she is uh an AP physics teacher and project lead the way teacher at Addison Trail High School just outside of Chicago. So she's in our um in our chat helping you out if you have any questions, tech support, things like that. Um, I will also try to answer questions, but there's generally more that I can handle all by myself. So, thank you so much, Miss Bonk. All right, I do have to talk about my ultimate exam slayer, flipping physics, and my ultimate review pack. So, my ultimate exam slayer is available um specifically designed for helping you understand the exam itself for AP Physics 1. It has a bunch of videos and t and things about understanding the exam. Uh, different types of free response questions. different multiple choice categories that I put together um and has solutions to every free response question for AP physics one that's been released that is applicable also has two practice exams that um you can take so and then of course I have my flipping physics website and uh the ultimate review packet which is more designed to help you study and learn about the topics and it is the exam itself uh but that also has one practice uh final exam in it or one at practice AP physics one exam minute in it. Um, so as you can hear, I do apologize. Um, I am sick, but I'm okay. Um, I'm actually on the tail end. I'm feeling better. And I know I don't sound it, but I am definitely feeling better. But I do apologize. You're going to hear me hack and cough. Um, maybe you can even hear me breathing. It's a very raspy, but whatever. I'm here. Okay. So, here we go. I feel like we're pretty close. Um, maybe I'll wait a few more moments, get a little water, then I'll get started. Okay. All right. I am going to get started and people can join as they show up because, you know, it is after seven. All right. Oh, congratulations. We'll do that for real quick. Carve O. Best of luck on your exam tomorrow. I'm glad that or your test this worked out well for you. It doesn't work out for everybody. Okay, here we go. The timing I try I do my best. So, make it so the timing is best for as most people as possible, but you never know. Um, actually, before I click start, I want to do this. So, I need to change Yeah. Question one. Okay. So before I qu click start, I am going to I need to change the setup of the screen here. So give me a moment. All right. So this guy goes here and this guy goes here. So let's see. We have Yep. We have that and we have that. Perfect. All right. We are going to get started. Welcome to the flipping physics unit five AP Physics 1 exam. In three, two, one. We have our first question. And I will make it bigger. Whoops. Here we go. Oh, shoot. There we go. Much bigger.

All right, look at that. We have 17 10 people who answered choice C, which does look like it is the correct answer choice. All right, so let's take a look at why that is. All right, here we go. This has got to be page width per. Okay. So, question one looks like this. All right. So, here we go. Talking about question one. So, at point G, the bob is has reached its maximum height. So, when it's at its maximum height, um the bob is at rest for at that brief moment when it's at point G. So, the bob is at rest. So, that means that the tangential velocity of the bob is equal to zero. That means the centrial acceleration which equals the tangential velocity squar divided by the radius is equal to 0^2 over the radius or zero. So we know the centrial acceleration is equal to zero. When the sum of the forces in the ind direction that's equal to the mass time centrial acceleration. We can substitute in zero in for the centrial acceleration. So the net force in the in direction is equal to zero. So there is no net force in the in or out direction. So that means that the only net force is along the tangental direction. So the correct answer is either two or three. So we know now so far that the net force is either in direction two or three. It has to be tangent to the motion. Eventually the bob will be at point F, right? It's going to go down here and it's at rest here at this brief moment at point G. That means the bob must accelerate towards point F. Therefore the net force must be towards F to cause that acceleration. So there must be a net force in the tangential and left direction. So the correct answer is direction three which is answer choice C. So it is useful to understand the free body diagram. So we're going to walk our way through the free body diagram here. So class anytime I put the word class on there and a question mark I ask a question for you. Class, what forces are in the free body diagram and what are their directions? So you're answering in the chat. What forces are in the free body diagram and what are their So, I got I I'm just going to put this one up here. So, we have gravity and tension. So, first off, you should never use the word gravity by itself on an AP physics exam. Is it the force of gravity? Is the gravitational field? Is it the gravitation the acceleration due to gravity? Right? You should never use that word just gravity by itself. And you didn't mention the direction. So, be careful of that. So, we've got uh we've got force of gravity, which is straight down, and force of tension to the upper left. That's fun. Okay. So, let's take a look. Um, we've got in our free body diagram, we've got the force of tension and the force of gravity. The force of gravity is straight down and the force of tension is in the direction of the string and it is therefore inward because if this is the circle that is going to be the end direction. So the force of gravity is down and the force of tension is inward.

So we're going to resolve the force of gravity into its components. we have this is the force of gravity is straight down. which acts outward and is tangent to the motion. So when we redraw the free body diagram we get it looks like this. We still have the force of tension and instead of having the force of gravity we have the force of gravity which is tangent to the circle and the force of gravity which is outward. those two components. When you sum the forces in the ind direction, we know the force of tangent minus the force of or sorry force of tension gravity in the tangental direction is going to be equal to zero because we already determined the net force in the ind direction is equal to zero. Therefore, the magnitude of the force of tension and component of the force of gravity which is um that's a typo. I apologize. This should say out right there. So the force of gravity which is out and the force of tension are equal in magnitude. I apologize for that. I did not notice that. Uh so the net force on the bob at point G is the force of gravity tangential. This guy right here. That should be force of gravity out. Dang it. All right. We are done with question one. Uh Let's see how everybody is doing. All right, soaring barracuda is in the lead. Moving on to question two. Here we go. All right. Interesting. So, we have 20 people answered choice actually. Okay. Choice C is correct and the least number of people answered choice C. My favorite. I love it. That's great. We're gonna talk about why that is. But before we do, I want to take a moment to talk about a comment that Tina Koko made. Uh, give me a moment. I want to talk about this. Okay. So, let's go back. So, here we go. So, the comment, Tina, that you Tenny, I'm sorry, Tenny, uh, Tenny Vokco, uh, is that the block this is like a block on an incline where the force of tension is the force normal. And I hear that I understand that you would think that because the free body diagram looks very similar, but this dotted line is very important to recognize that this is an object that is not moving in a straight line. This is an object that is moving in a circle. If this were a block on an incline, the net force in the um the net force at this point would be equal to zero because the acceleration in all directions. But because it is actually an object moving along a circle, it um like a moment after this it is going to be moving and it will have an acceleration in the ind direction. So you have to be really careful to recognize it is moving in a circle. So

it's very similar but that's just a subtle distinction that you need to understand. Okay. So uh let's get rid of that for a moment. Okay. So now let's talk about question two. Here we go. Okay. Question two. So in the free body diagram we have the force normal and the force of gravity. The force normal is up or outward. Again, we have the dotted line uh dotted circle here indicating it's moving in a circle. And we have the force of gravity which is down or inward. So again, force of gravity down, force normal up. I do want to point out that the forces are not to scale. We don't know enough from the information in the problem to know at this point how those two forces compare. So they're just force normal and force of gravity, not to scale. So the knowns in this question are that the radius is 5 m and that the tangential speed is 10 m/s. And we're solving for the ratio of the force normal to the force of gravity to the magnitudes of those two. So when we sum the forces in the ind direction, we have the force of gravity minus the force normal. This is very important to recognize. The force of gravity looks like it's down, but whenever you're moving in a circle, you have to be really careful to recognize that in is positive and out is negative. So, because the force of gravity is in, it is positive. Because the force normal is out, it is negative. And the net force is always equal to mass times acceleration in whatever direction you're talking about. When we sum the force in the in direction, that's the centrial acceleration. Centrial acceleration equals tangential velocity square radius. We can substitute that in. We can then solve for the force normal. Force normal equals the force of gravity minus this quantity mass time the tangent of velocity square divided by the radius. We can substitute in the radius because we have solved for or we are given the radius. So we can also substitute in mass time the acceleration to gravity for the force of gravity. And then our goal is to solve for the ratio of the force normal to the force of gravity to the magnitudes of those two. So we can substitute that equation into here and we get Everybody brought mass to the party. — Everybody brought mass. — We can be equitable. We can take mass from everyone and we get we can so the mass cancels out. We can substitute in our numbers and we get the number -1 for the ratio of these two. And you might think that the correct answer is negative one and therefore d it is not. Okay, let's talk about why. So mathematically we got -1. And how can that not be the correct answer? Please realize the magnitude of a vector cannot be negative. Therefore the ratio of the magnitudes of two vectors cannot be negative. So remember to always consider what the numbers you're using actually mean. So let's just look at the force normal. So force normal equals mass time acceleration of gravity minus mass time velocity squared divided by radius. We can substitute in all the numbers and we get 10 * the mass minus 20 * the mass or -10 * the mass. So we do not know the mass of the car. But we do know that according to everything we know in the problem the force normal is negative. This is not possible for the force normal to be negative. So what happened at a speed of 10 m/s the car will not stay on a hill with a radius of 5 m. The faster a car goes over a hill the smaller the normal force. there is some critical speed at which that for force normal will reach zero and at any speed above that critical speed the tires of the car will leave the road. So in this example the tires are no longer on the road and the force normal is zero. Therefore the ratio of the force normal to the force of gravity is zero. The correct answer is C. Sorry. So, is this a fair question for the AP Physics 1 exam? I really I would be very surprised if they put something like this on the AP exam. Is does this question help you understand physics? Absolutely. Right. This is a great question. Would I put this on exam for my students? I absolutely would put this on an exam for my students, but I get that it's probably not fair for the AP physics one exam, but again, it's a really good one for understanding physics. All right, so let's go over a common mistake students make. So, a very common mistake is to forget that the in direction is positive and that

out is negative. When you do that, you switch force normal to positive and force of gravity to negative, right? because force normal is up, force of gravity's down. And when you do that, you get positive three as your answer and you get incorrect answer choice A. So just be aware of that. Be very careful of making that mistake. Remember, in is positive, out is negative. All right, moving on to question three. Actually, before we move on to question three, let's see how people did after question two. Oh, excited gator. Sweet. All right, let's move on to question three. That looks good. Make it a little bigger. All right, look at that. Looks like a lot of people got it right. I That's great. So, choice D looks like is the correct answer. Uh, all right. Let's see why choice D for question three is the correct answer. Let's make that a little bit bigger. That's very tiny. Can I move it here? Come on. There it is. Uh okay, question three. So class, what is Newton's second law for rotational motion? Again, this is a question for you. Please put it in the comments. All right, I'm waiting. What is Newton's second law for rotational motion class? All right. Good. Good. All right. We'll just put this one on here. I love this. Okay. Torque equals I multiplied by angular A. Okay, so we're going to assume that's torque equals rotational inertia time angular acceleration. So just take a moment and think about what I find troubling about that. What is missing from that is very important. Right? Just consider that for a moment. And this is the thing that most often gets forgotten by students and becomes a big issue. And that is we have here that the net torque on an object is equal to the moment inertia multiplied by its angular acceleration. So that net is a huge important piece, right? Um I'm not I also they're vectors the but I guess the net is really the most important piece at this point. So realize net torque equals rotational inertia times angular acceleration where both torque and angular acceleration are vectors. Beautiful. Okay. The net torque is the same in both

cases in this question, right? So a larger rotational inertia will result in a smaller angular acceleration. So when the meter stick is held in the middle and rotated, the axis rotation passes through the middle of the meter stick. And when the meter stick is held at the end and rotated, the axis of rotation passes pause passes through the end of the meter stick. Right? So more mass is farther from the end than the middle of the meter stick. So the rotational inertia for the middle of the meter stick is less than the rotational inertia at the end of the meter stick. And with the same net torque, that means the angular acceleration when rotating about the middle of the meter stick is going to be greater than the angular acceleration rotating about the end of the meter stick. The correct answer is D. Great. Moving on to question four. Before we do that, yeah, I don't want to do that. Okay. So, let's move on to question. Oh, actually, let's see how everybody did. Focus glider doing well. All right. Uh, let's see. There we go. Perfect. Next. All right, awesome. Choice C is correct and a lot of people got that. That's wonderful. Let's walk our way through question four. All right, here we go. Class, what is the condition for rotational equilibrium? Again, question for class. All right, we have a lot of people answer answering. So we got this one right here. Net torque equals zero. Beautiful. That is the condition for rotational equilibrium. So net torque is equal to zero. So external forces acting on the cube beam system in the force diagram are the upward force from the pivot which acts at the location of the pivot and we have uh this location will be the axis of rotation when we sum the torqus. So this force will cause zero torque on the system. There's a downward force of gravity on the beam acting at the center of mass of the beam and a uniform beam is directly in its middle. The downward force of gravity acting on the cube will be at the location the center of mass of the cube. So for the net torque to be equal to zero the two torqus from the force of gravity from the two forces of gravity have to be in opposite directions. So therefore the pivot must be between the beam's center of mass and the cube. So the correct answer has to be C. Beautiful. So I do want to take a moment

to look at the force diagram. Again this just goes beyond the question itself but it is important to walk our way through it. So we have our axis of rotation. So realize the relative magnitudes of the force vectors are unknown. So I'm not including the relative magnitudes. It's just looking at the force diagram itself. So when we look at it in terms of an equation, we have the net torque from the force of gravity of the beam minus grav I'm sorry not the net the torque from the force of gravity of the beam minus cube. So realize that the torque from the force of gravity of the beam would cause the uh the beam to rotate counterclockwise and that's positive. The torque from the force of gravity of the cube would cause it to rotate in the clockwise direction which is negative. That's why it's this is positive. This is negative. That's equal to zero because it's in rotational equilibrium. Therefore the two are equal to one another. So that's just important to recognize. Now I do have students that ask me, but what about the normal forces between the cube and the beam? Right? This is the force diagram of all the forces acting on the cube and beam system. But it's important to realize that the upward normal force from the beam on the cube and the downward normal force from the cube on the beam are both internal to the cube beam system and therefore they do not cause an external torque on the system. So they're not going to be in our force diagram of the external forces acting on the system. Just important thing to recognize. Beautiful. Moving on to question five. But of course before we do. All right. Focus glider. Still sorcera. All right. Barracuda. All right. It fits. Beautiful. Question five. Mhm. All right. Interesting. So, we have more people answered choice B, which is incorrect, rather than choice D, which is correct. We will get to that in a moment. Uh, before we do, I do want to address a question I got from Tiny Coco. So, this goes back to question four. So, let's take a look at question four real quick. Uh, oh, wait, got to go here. Okay, so the question was, oh, we don't know. Okay, sorry. So the question is uh we don't know how do we know it will balance the beam because we don't know the mass of the cube. So the answer is that we simply know that the pivot has to be somewhere between these two forces do two forces of gravity. Uh we don't know exactly where it's going to be and where it would be specifically is going to be depend on the relative masses of the two objects. We know the pivot needs to be somewhere between the two forces. We don't know where though, right? And so the question doesn't ask exactly where. It just identifies that it needs to be somewhere between those two um the

location of the cube and center of mass of the beam. Okay, good. All right, let's go back to we're now looking at question five. Okay, question five. So in question five from 1 to 4 seconds the graph shows the net torque acting on the tire is zero the whole time. Again we have Newton's second law in rotational form. The net torque is equal to the rotational inertia times angular acceleration where both torque and angular acceleration are vectors. If the net torque is zero the angular acceleration is also zero. So from 1 to 4 seconds the net torque is equal to zero. Therefore, the angular acceleration is also equal to zero. Right? We just walked our way through that. So, the correct answer is D. From the information given the problem, we do not actually know if the tire is not rotating or is rotating with a constant angular velocity. It doesn't matter for the question, but it's interesting to notice that we don't actually know. Now, answer choice C could be correct if the time we were looking at were from 0 to 5 seconds, but it is not. So please realize that. So always read carefully. Moving on to question six. Actually before we do of course. Okay. Focused glider. Is anybody going to be able to take down focused glider? We will see. Moving on to question All right, answer choice C. Very large percentage of us got it right. That's awesome. Good to know. All right, let's take a look at question six. Why that is correct? Okay, in question six. So more mass is concentrated farther from the axis of rotation in case two. Therefore, the correct answer is either A or C. More mass concentrated far farther from the axis of rotation increases rotational inertia. Therefore, the rotational inertia in case 2 is larger than one. So the correct answer is C in terms of the parallel axis theorem. Right? So let's look at it in terms of mathematically. So the um case one is through the center. So that is the rotational inertia about the center of mass. In case two, we can use the parallel axis theorem. On the equation sheet, they use a little um what is this quote? I don't even know what that is. Um dash I can't come up with the word. So they use this for the rotational inertia about an axis which is not through the center of mass of the object. So it equals the rotational inertia about the center of mass plus the mass of the object times d the distance from the axis of rotation to the new center of mass. So rotational inertia 2 is equal to rotational inertia 1 plus mass time r the radius of squared. Therefore the

rotational inertia of two is greater than rotational inertia of one. So please remember the parallel axis theorem. You will need to do that sir could be prime. Yeah, prime. That makes sense. Okay. Uh, moving on. Yes. Okay. Moving on to question seven. Yeah, words sometimes. That's fine. Uh, let's see how people are doing. Huh. Focus glider still in the lead. Good luck there. All right, moving on to question seven. All right, look at that. A lot of people answered choice A. Before we move on, I do want to highlight a comment. Dang it, I didn't read the whole question. Official Sophie Jen. Um, I will tell you that the number one piece of advice you will hear from most physics teachers is that you need to read the full question. So, please slow down and Thank you official Sophie Jen for doing that for me. Okay. Uh, question seven. Let's look at question seven. class. What is the relationship between tangential velocity and angular velocity? Please, Okay. All right. I'm going to do this. It's going to bother you, I'm sure. Oliver Hood, I'm gonna do this. So, your equation, if I read it correctly, is volume equals radius times angular velocity. So, I just I'm just doing that because uh capital V technically means volume and lowercase V means velocity. I know you meant velocity, but just be careful of that. So, velocity equals radius times angular velocity is the relationship between tangential velocity and angular velocity. Beautiful. All right, let's Okay, now tangential velocity is proportional to angular velocity. The two graphs will have the same shape. However, their amplitudes will be different. But if you look carefully at all the graphs, none of the graphs have any sort of y-axis scale. So you can't compare their y-axis scales. So graph A has the same shape as the angular velocity graph. So graph A is the correct answer. Great. Moving on to question eight. Before we do, of course, let's see how things are. Oh, focus glider aquatic ocelot. Soaring barracuda. Okay. Oh, soaring barracuda.

You guys are close. All right. Uh, moving on to question eight. All right. Look, a lot of people got choice A. That's correct. Perfect. I do want I'm going to highlight another comment and really I saw it was rotational. I was thinking of something else that wasn't the question, right? Like again I cannot stress this enough. Read the full question. You are going to see questions and think that it's something that you've seen before but it won't be. So please, you've got to read the full questions. It's just reality. Okay. Uh moving on to question eight. All right, here we go. Question eight. We have our knowns. We know the radius for one is twice the radius for two, where they both have the same angular velocity, which is constant. So, I'm just going to label that omega. And we're looking for the ratio of the two centrial accelerations. We know centrial acceleration 1 equals radius 1* angular velocity 1. So, it's just just because we know it's constant. So, we could substitute in two times the radius for uh two in for radius 1 and we get that for centrial acceleration for one. Centrial acceleration for two is radius 2 times angular velocity for two. So squared. So we substituted everything and you can see radius 2 times angular velocity squared is both in the numerator and denominator cancels out and we get two. The correct answer is a. Now there are those of you who want to use tangential velocity to solve this. And if you wanted to solve it this way, again, you have the same known, but it it's just going to look a little bit different because you start with tangential velocity squar divided by radius. You again substitute in all the information. You substitute all the information and you get the same answer. The difference is that it takes a few more steps. So, um I don't know why, but some of you really always want to use tangential velocity. But if like if the question is giving you angular velocity, you should use the equations that have angular velocity in them. If the question is giving you tangential velocity or asking you to solve for tangential velocity, you should use the equations that have tangential velocity in them. It's just a good habit to get into. All right, moving on to question nine. Oh, actually, I'm sorry, I forgot. Okay. Soaring barracuda just overtook aquatic ocelot. Oh, and oh, it's gotten a lot closer. Focus glider. Okay. Uh, moving on to question nine. Here we go.

All right, choice B. We got 24 people got it correct. Lovely. All right, let's take a look at solving question nine. All right, the force diagram is already shown and the axis of rotation is already identified plus the positive direction of torque direction of counterclockwise is identified. So we can sum the torqus, right? The torque from torqus from one and three are positive because they would cause the beam to uh rotate in the counterclockwise direction which is positive. Um and the torque two is negative because it would cause the beam to rotate in the clockwise direction which is uh negative. And then I'm using phi in my figures in my equations for torque because theta is already in the figure. So I don't want to have both theta and uh extra thetas lying around. So I'm using phi instead of theta. Uh so let's substitute in what we know. So we know the length for force one is l over2 the distance there from uh from where the axis of rotation is to where the force is applied. And we know force one is 50 newtons. We know um this guy for uh the distance r for distance for force two for torque 2 is all the way from here to here which is going to be l over2 plus l over 4 or 3 l over4. Again we can substitute in 100. And let's talk actually let's talk through the rest of this. I think I have more stuff on here. So um let's take a look at the angles. Right. So this is theta 2 but theta 2 is not the angle for torque caused by force 2. The angle for the torque caused by 42 is 52. Right? So it is the angle between this direction and this direction. So please be aware that phi 2 is the angle for torque from force two and fi3. Actually let's make this a little bit bigger. Um and 53 is the angle for the torque from force 3. Notice it is not theta 3. So phi 2 is going to be 180° minus theta 2 or 180 - 60 or 120°. And phi3 is going to be 90° this whole 90° minus theta 3 or 90 - 30° which is 60°. And L over2 I already talked my way through that. So I just do want to talk about this for a second because I know do know this confuses people a little bit. So the direction of R2 is this direction here to the right and the angle between direction of force 2 is 52. Note it is not R2 is not to the left. Right? It does not go this direction. So 60° theta 2 is the angle for between to the left and force two. But the angle for uh the torque from force 2 is the angle between to the right the direction of R2 and the direction of force two. And the same is

true for the torque due to force three. Okay. And we can substitute 2 meters in for L into the equation. And then we can just plug in the numbers and we get 36 Newton meters. So the correct answer is B. Wonderful. Uh class, what are some common student mistakes? Give them to me. Common student mistakes. Okay. I'm sorry. I got to highlight that one. pretty funny. Okay, sure. Let's talk about this. Focusing too much on 67. Okay, so I'll take this as a more general statement. So, focusing on things that are just distractions because there will be things that distract you. Like, I mean, who knows? Maybe the number 67 will appear on the exam. I sure hope it doesn't, but if it does, you got to let it go. But I could see that being an issue for some people. Using the wrong direction for forces. Now, this is an interesting one. I think what you mean is using the wrong do for the torqus, right? So, because it's not really the forces because you can see the forces there, but I think you mean the wrong direction for the forces. And that's a very important. And of course, not reading the full problem. Thank you. Of course, that's going to happen. uh picking one of the two answers that are close to each other, which is your multiple choice strategy, please. Uh as someone who has made a whole bunch of multiple choice tests, I know what your strategies are and um I make sure to include things that will point you to the wrong answer using your own strategies. So please don't try to have strategies. Just try to do the physics. Please okay let's talk a little bit more. Okay. So some common mistakes that I thought of is leaving all the torqus is positive which I think is what somebody meant by um using the wrong direction for the forces. So just so you know this result results in incorrect answer choice D. Using the angles given in the figure rather than determining what the angles in the torque equations actually are. This is one that I see students do a lot. This results in incorrect answer choice A. And of course, if you make both of these mistakes, you get incorrect answer choice C. All right, moving on to question 10. But before we do, oh, focus glider. I'm so sorry. Soaring barracuda, congratulations. Good luck, aquatic oelot. Ocelot, I don't even know. Okay. Uh, moving on to question 10. It all fits. Question 10. No, you can't quite read the L. Oh, well, I'm sure you guessed it. Sorry.

All right. Oh, this one was a tough one. Okay, we did have the most people answered. Let's see. The highest number of people answer one question was choice A, but let's take a look at what happened. Okay. Uh, all right, question 10. Here we go. Let's make the question a little bit bigger. I think that's going to be important. There it is. All right, question 10. First, recognize the rotational inertia for the beam given in the problem is not about an axis through an end of the beam. Right? This 112th mass time length squar is through the center of mass, but the axis of rotation is about the end. So, we need to use class. What do to determine the rotational lure of a beam about an axis through an end of the beam? I advertised it earlier in the coat. All right, I'll give it to you. That's right. The parallel axis theorem. We have a couple people answered it. Thank you for that. All right, so we need to use the parallel axis theorem. So the parallel axis theorem, we've decided that's a prime. I prime the rotational inertia prime equals the rotational inertia about the center of mass plus the mass of the object times the distance from the center of mass of the object to the new axis of rotation the square of that. So we have the rotational inertia about the end uh or I'm sorry about the center of mass. It's 112 mass* length squ plus then we have the mass capital m and the distance between the center of mass of the beam and the end of the beam is half the beam. And so then we could go through and do the math and we could determine that the rotational inertia about the end of the beam is 1/3 mass time length squared. Now we can draw the force diagram of the forces acting on the beam. So only two forces act on the beam after the upward force is removed. We have the force of gravity down which is on the beam which is at the beam's center of mass and we have the upward force at the hinge right at that axis of rotation. So that force on the hinge acts right at the axis of rotation. So that force causes zero torque on the beam. So now we can sum the torqus on the beam about the axis of rotation with the counterclockwise direction as positive. Right? We have everything we need in order to do that. So the net torque, the only force that causes a torque about this axis of rotation is that torque caused by the force of gravity. So that's equal to the rotational inertia times the angular acceleration. In this particular case, because we're summing the torqus about that axis of rotation, which is the end of the beam, we need the rotational inertia about the end of the beam. So we can then substitute in our equations for torque and rotational inertia. And then we can substitute in our variables for the rotational inertia about the end. And we have the distance from the axis of rotation to where the force is applied the length over two. We have the force of gravity which is mass times the acceleration to gravity and the sign of 90 degrees because the angle between the direction of r for the torque and for the force of gravity and the force of gravity is 90 degrees. The mass of the beam cancels out and the sign of 90° is one and one of the lengths of the beam L cancels out. So we are left with three 3 * the acceleration due to gravity or gravitational field strength divided by 2 * the length of the beam. The correct answer is A. Beautiful. Now I do want to point out that the angular acceleration we determined is only the initial angular acceleration. It specifically identifies that in the problem. It's important to understand why this is the initial angular acceleration. As soon as the beam rotates a little bit, the torque decreases. Therefore, the angular acceleration decreases. So let's talk about why that happens. The reason the torque decreases is because shortly after the upward force was removed, the beam has moved down to this angle right here. And the angle between R and the force of gravity is no longer 90°. In this particular case, it's greater than 90°. So the r is still L over2, the half the beam. However, that angle is no longer 90°. And the sign of theta is at a maximum when theta is equal to 90°. Therefore in this equation everything else remains the same. Therefore the angular acceleration

decreases because that sign of the angle is going to decrease. So the angular acceleration again realize all we did was figure out that initial uh angular acceleration and the angular acceleration is going to change once it moves. Great. And we are done. Before we complete everything, let's take a look. See how we're doing here. See who won. Good luck. Focus glider came in third gator. Interesting. Who took number one? Victorious Wabe. Congratulations. Victorious Wallaby came out of nowhere. I'm impressed. Victorious Wabe. Okay, so my plan is for this video to remain live on YouTube. Uh the this multiple choice is posted in my ultimate exam slayer. And in fact, there are 17 more questions in my quiz like for this for the ultimate exam slayer. So every one of these um cahoots is actually uh part of is the first 10 multiple choice questions that are in my ultimate exam slayer. Again, unit six multiple choice quiz. Uh, the cahoot will be on Tuesday, March 17 at 7 p. m. Eastern time. Thank you very much for learning with me today. I enjoyed learning with you. You all have a beautiful night. I'm going to try to get a good night's sleep and maybe I'll feel better tomorrow. Who knows? I've been saying that for like five days. Okay. Thanks. Bye. Flip physics.