Torque with Kinematic Equations - Video Tutorials & Practice Problems

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1

concept

Torque with Kinematic Equations

Video duration:

8m

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Hey, guys. So now that we've seen how to solve some basic torque acceleration questions, we're going to add motion equations into the mix, which will generate a bunch of extra questions problems we can solve. Let's check it out. So you may remember that when we had forced problems, most forced problems were solved using our vehicles in May. You may remember that some of those problems would also involve are three or four equations of motion or cinematics equations. Same thing is gonna happen here with rotation, where some torque problems will require both torque equals Alfa, which is the rotational version of Newton's second law and rotational motion equations. Motion equations Are you am equations Universe? Um, uniformly accelerated motion telematics, equations of motion equation. Whatever you call it, we got 3 to 4 of these guys. Um, so remember, just like how it was with linear motion. The variable that will connect torque equals Alfa to the rotational motion equations, which are these guys here is going to be acceleration. Okay, acceleration. Now, because we're talking about rotation, this means Alfa right. Notice how there's an Alfa here and there's an Alfa here, here and here Okay, so we got Mawr equations and more variables, but it's not harder. Uh, it's just more stuff. So let's check out this example Here, have a solid sphere and I give you the mass and the diameter. Solid sphere is the shape of the sphere. So I know that because of the solid sphere, I'm gonna use the moment of inertia of a solid sphere, which is to over five m r squared. I got the mass. The mass is 200 I have the diameter. Remember, in physics, you're never gonna use the diameter. So as soon as I see diameter, I convert that immediately into radius, which is half of that. So it's 3 m and it spends about an access through its center. This is the regular rotation of a sphere solid sphere, which is around itself. Okay, it says it does this with 180 rpm clockwise. So the rpm is 1 80. What's up with clockwise? Well, clockwise clockwise is going to mean that it is negative. So it's got a rpm of 1 80 now, remember? And as we did in motion problems, rotational motion problems. Whenever you're given rpm vast majority of the time. You're gonna immediately convert that into Omega because most of our equations have omega little W but not rpm in it. Right? So the first thing I'm gonna do here or the next thing I'm gonna do here is convert this into W so w equals or Omega equals two pi F or two pi. Remember, F frequency is our PM over 60. Okay, so this is gonna be two pi negative. 1 80 over 60. This is going to be a three right there. Which means the whole thing will be negative. Six pie radiance per second radiance per second. Cool. So I got that. That's the initial speed. Um, I wanna know how much torque is needed to stop this thing in just 10 seconds. So I'm asking what is a torque to stop it. Right? So that means that this is my initial mega and I wanna have a final omega of zero. And I want to do this in just 10 seconds. So Delta T equals 10. So I hope you'll notice Here you start seeing all these motion variables. And remember the way I solved motion problems is by setting up the curly braces and putting all five motion variables there. So let's do that. Omega initial equals negative. Six pi Omega final equals we wanted to be zero Alfa Delta data in Delta Teeth. Now the tea is 10. Um, and these two guys, we don't have them, okay? And they're also not what we're looking for. But since I saw all these variables, I decided, Hey, let's start sending this up because I know this is coming. But really, what we're looking for is, um, is torque So you might actually have started this question instead of gone here. You might have just written that the sum of all torques equal Scialfa on. That's perfectly fine as well. That's if anything, that's ah Mawr directed way to the answer, right. More targeted, which is fine. There's only one torque here. We're assuming there's one torque. This thing is spinning, and I guess you're applying torque to it, uh, to make it stop so you can assume that there's only one torque. So the sum of all topics will become just the torque that you're looking for. And that is I Alfa. Okay, so if I could have I and I have Alfa. I'm done and that will be my answer. So let's expand I two pi I'm sorry to over five m r squared and Alfa noticed that I have m I have are, But I don't have Alfa. Right? So what you're gonna do is you're going to go over here and try to find Alfa. Okay, so let me plug in these numbers toe were the only thing we're missing is Alfa m is 200. Our is three squared. So as soon as we have Alfa, we can plug it in there. Okay, Now back to the motion equations so the basic ideas get stuck and you go to the other side back to the motion equations. We have three variables, which means we can solve. This is my target and this is my ignored variable. So I'm going to use the Onley equation that does not have a delta theta in it. And the only equation that doesn't have a delta theta in it is the first equation. Okay, so omega final equals Omega initial plus Alfa T and we're looking for Alfa. So Alfa is going to be Omega final minus omega initial divided by t. This, by the way, is the definition of Alfa. It's the change in omega over the changing T. You could have started there as well. That would have worked. Um, this is zero minus negative. Six pie. And the time is 10 seconds. So these cancel end up with six pi over 10. Positive. 65 or 10 which is 1.88 I got a positive, which should make sense even though I'm slowing down. Right. Let's talk about that real quick. My velocity is negative. If I'm slowing down, I'm trying to make my velocity positive. Eso my acceleration should be positive. I'm trying to make my velocity more positive. Another way to think about this that might be even easier is you are You have a negative omega and you're trying to slow down. You have to go in the other direction, the accelerations to go in the other direction. So the acceleration would oppose it because you're trying to slow down. And this is counter clockwise, which would be positive. Okay. Anyway, my acceleration is 1.8 radiance per second squared. Now I can plug this in here, and we are done. So if I multiply all of that, um and then I multiply that by 1.88 I should get, um I get 1354 Newton meter 1354 Newton meter and that's it. That's the final answer. So just to recap real quick, there's basically two parts to this. Um, we were asked for torque so you could have started here. And then you start plugging stuff in and you realize you don't have Alfa, but you have a bunch of motion equations motion variables so you can find Alfa using one of the motion equations. Plug it back in its the classic, um, standard type of physics question where you got stuck with something. We'll look for another variable, plug it in, come back with the value got on dissolved. Alright, so that's it. Hopefully makes sense. Let me know if you have any questions. Let's keep going

2

Problem

Problem

A light, long rope is wrapped around a solid disc, in such a way that pulling the rope causes the disc to spin about a fixed axis perpendicular to itself and through its center. The disc has mass 40 kg, radius 2 m, and is initially at rest, and the rope unwinds without slipping. You pull on the rope with a constant 200 N. Use the rotational version of Newton's Second Law to calculate how fast (in rad/s) the disc be spinning after you pull 50 m of rope.

A

3.2 rad/s

B

15.8 rad/s

C

22.4 rad/s

D

633 rad/s

3

Problem

Problem

A system is made of two small, 3 kg masses attached to the ends of a 5 kg, 4 m long, thin rod, as shown. The system is free to rotate about an axis perpendicular to the rod and through its center. Two forces, both of magnitude F and perpendicular to the rod, are applied as shown below. What must the value of F be to the system from rest to 10 rad/s in exactly 8 complete revolutions?

A

3.8 N

B

6.0 N

C

7.7 N

D

15.4 N

4

Problem

Problem

Two rotating doors, each 6.0 m long, are fixed to the same central axis of rotation, as shown (top view). When you push on one door with a constant 100 N, directed perpendicular from the face of the door and 50 cm from its outer edge, the rotating door system takes 8 s to complete a full revolution from rest. The doors can be modeled as thin rectangles (moments of inertia for thin rectangles, around two different axes, are shown for reference). Calculate the mass of the system.

A

21 kg

B

84 kg

C

104 kg

D

416 kg

5

example

Stopping flywheel with friction

Video duration:

11m

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Hey, guys. So in this example, we have a flywheel. It's spinning, and we're going to press an object against the wheel to cause the stops. Imagine, got something spinning. And if you squeeze an object against it, if you push an object against it, it would stop. And we wanna know how hard do you have to push against, um, the flywheel so that it stops in a particular amount of time? Okay, so it says here flywheels a rotating disc. This means we're going to use the moment of inertia of a disk, which is half M R Square. Samos. A solid cylinder. Andi. It's used to store energy. Suppose this one has a mass of eight times 10 to the fourth kilograms and has a diameter of m, which means we're going to immediately change the diameter. Is the radius off 2.5? Um, and it's set up vertically. So we got this wheel That's vertical, Um, and it's free to spin around a fixed axis, so there's a fixed access their perpendicular to the wheel. So basically the wheel spins like this around a central axis, um, to slow down the fly. Well, I mentioned this. You push this block here, so there's a force that you apply. Um, I want to point out that once you push here, it's going to have a contact. Therefore, there's going to be a normal force back. And this normal force, we have the same magnitude is your force so action reaction If you push with 10 normal pushes back with 10 um, it says that the coefficients of friction between the block and the wheel right here the coefficients of friction Our 0.6 and 0.8. So I'm gonna write them here. Mu static. Remember, when you have two quick fixes friction, the static coefficient is the greater one. So 10.8 and new kinetic is going to be 0.6. Okay, Onda, we want to know what is how hard do you have to push? So I wanna know what is F um it says here that the wheel is gonna come to a complete stop so Omega Final will be zero from on our PM of 300. So rpm in their show is 300 and it's going to do this in a delta t of 30 seconds. Remember? Rpm almost always gets converted into omega. So I'm gonna do that real quick. I'm gonna say, um, Omega initial equals two pi our PM over 60. That's the equation to convert to. And if you multiply this, you get 10 pie. Okay? 10 pie radiance per second. And notice that Now we have. I'm gonna actually move that over here on dso scratch this outs just because I'm trying to list all my motion variables and I'm trying to make the point to you that as of now, we already have three motion variables, which is good news Means we could solve for the others. The other questions that the other variables that are missing here are Al from and Delta Theta. Okay, so all I'm doing is grabbing the information and sort of fixing it up. Eso how hard do you have to push again? This is a force that's going to cause a torque, right? And so we're gonna use the force. Um, the the some of all torques equals I Alfa Equation. But I want to explain to you what's going on here. So the idea is that the wheel spinning this way, let's say within Omega in issue. And when you push, there is now a normal force here. I had that drawn earlier, and what that means is because there's a normal force and there's friction. There's going to be friction acting against motion. So motion is rotating that way. Friction is going to try to stop the wheel. Which means there's going to be a force of friction this way. Okay, so you got a normal this way on. There is a force of friction this way. What that force of friction does. It causes a torque like this torque of friction, which is opposite to my velocity. So it's trying to slow it down. It's trying to get it to stop. Okay, um, in this particular case just because of the way I drew it, um, this is actually negative. And then this would be positive. As long as they're opposite to each other. You're fine. Okay, so there is a torque. There is an acceleration. This is a force problem with angular acceleration with torque. So we're going to start here and we have to find what this f is. Okay, three Onley. So let's expand this equation real quick. The Onley torque acting on. This is a torque due to friction because you're pushing against this thing. So we're gonna right? Torque of friction, theme, moments of inertia, The moment of inertia. It says here I didn't read this part to you, but it says here you may assume the wheels entire mass is concentrated at its outer rim. This means that this is not a thing is actually not a rotating disc. I apologize. Um, this is not a rotating disc. This is going to be a hollow disc. Okay, so it's going to be hollow disk. So instead of half m R Square, let me just meet this, um, instead of half from our square, it's going to be just m R Square. Okay. Where r is the radius. So I'm gonna put this here m r squared and Alfa. We don't have Alfa, but we could find Alfa if we wanted Thio. Okay, let's take this one step further and expand. This, um, torque is force Little. Our sign of data force in this case is friction. So it's little f r is the distance from the axis of rotation to the point where the force happens. This is the distance that we are talking about. Okay, this is my our vector. Um, because they force happens here. And the axes, obviously, in the middle here, the our vector is as long as the radius of the wheel, because friction happens at the edge there. So we're gonna have f big are and then sign of theta. The angle will be 90 degrees. Notice how they make an angle of 90 degrees Because frictions Shrake down as a result of the pushing against it on. By the way, friction is in situations like this, friction is always gonna make an angle of 90 degrees, and it's always gonna happen at the edge. Right? So sign of 90. So that's nice. Um, and M r square the radius. We have that as well. We're gonna be able to plug it in there and then Alfa and we're looking for F Now, don't get confused. We're not looking for a little f. We're looking for Big F. So what we're gonna do is we're gonna keep expanding this equation. Friction can be expanded. Friction is mu normal. So I could rewrite friction as mu normal. Uh, that's what we're gonna do except there's one change which is normal is the same thing as F. So I'm gonna plug in effort here, and I'm gonna write friction as mu F instead of new Normal. It's the same thing. The reason we do that is because now, finally, our equation actually has are variable, right? Until then, you haven't seen your variable around. Um, so that's that. Let's gonna rewrite this here. I'm gonna cancel This are with this are this is just the one. So it's gonna be m f M u f equals m r Alfa. So f is m r Alfa over mu. And we know these numbers m is or most of them eight times 10 to the fourth radius is 2.5. I gotta go find Alfa. And then we do have, um you which do you think we use kinetic or static? And I I hope you're thinking kinetic because the block is rubbing against the disk. So it's gonna be 0.6, which is kinetic, Okay. And we have to go find Alfa, So let's go do that real quick. So we're gonna go over here and look for Alfa, um, to find out for I can use motion equations. I have three knowns. Uh, this is my target. This is my ignore variable, Which means I can use the first equation to find Alfa Omega Final equals Omega initial plus Alfa T for looking for Alfa. So let's move everything out of the way. You might notice that this is the definition of acceleration of angular acceleration, which is change in omega over Children time. You could have started there as well. The final omega is zero because we're looking to stop. And the initial omega is 10 radiance. Um, 10 pi ratings per second. Now, notice that the way we drew it, it's a negative. So let's plug it in as a negative. Um, and in the time the time we're going forwards seconds. Okay. Now, if you do this, if you do this, you end up with an Alfa of 1.5 radiance per second squared. When you highlight this here. And that's what I'm gonna put right here one point a different color. Um, 1.5 right there. Now, if you multiply once you multiply this whole thing, once you multiply this whole thing, I'm sure I got it right. Yep. You end up with 3.5 times. 10 to the fifth, 3.5 times. 10 to the fifth Newton's. Okay, so that's how much force you have to push against the block with, um and push the block against the wheel so that this thing stops in 30 seconds. All right, so that's it for this one. Thes questions fairly popular. It's a little bit tricky, but again, I think the key thing is to realize you have a force causing a torque causing acceleration that this is the starting place. The trickiest part, I think here is to try to make a connection from this initial equation. And then how we're gonna get f. And sometimes you just have to trust that if you keep expanding the equation like we did here and then we expanded the F here, you have to trust that if you keep expanding, the variable will show up. Um, Alright, so that's it for this one. Let me know if you have any questions and let's keep going

6

Problem

Problem

A 1,000 kg disc that has a 5 m outer radius is mounted on a vertical, inner axle 80 kg in mass and 1 m in radius. A motor acts on the axle to speed up or slow down the system. Suppose the motor stops functioning when the system is spinning at 70 rad/s. To bring it to a complete stop, you apply a constant 200 N friction to the surface of the axle. How many revolutions will the system take to stop?

A

5.8

B

2.6×10^{4}

C

1.6×10^{5}

D

2.5×10^{5}

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