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 force problems, most force problems were solved using f=ma. You may remember that some of those problems would also involve our 3 or 4 equations of motion or kinematics equations. The same thing is going to happen here with rotation, where some torque problems will require both τ=Iα, which is the rotational version of Newton's second law, and rotational motion equations. Motion equations are UAM equations, uniformly accelerated motion, kinematics equations, motion equations, whatever you call it. We got 3 to 4 of these guys. So remember, just like how it was with linear motion, the variable that will connect τ=Iα 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 α. Right? Notice how there's an α here and there's an α here, here and here. Okay. So we got more equations and more variables, but it's not harder. It's just more stuff. So let's check out this example here. I 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 it's a solid sphere, I am going to use the moment of inertia of a solid sphere, which is 25mr2. I got the mass. The mass is 200. And I have the diameter. Remember, in physics, you're never going to use diameter. So, as soon as I see diameter, I convert that immediately into radius which is half of that. So it's 3 meters. And it spins about an axis through its center. This is the regular rotation of a sphere, a solid sphere which is around itself. Okay. It says it does this with 180 RPM clockwise. So the RPM is 180. What's up with clockwise? Well, clockwise is going to mean that it is negative. So it's got an RPM of 180. Now remember, as we did in motion problems, rotational motion problems, whenever you're given RPM, a vast majority of the time, you're going to 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=ω=2πf or 2π Remember, f frequency is, RPM over 60. K. So this is going to be 2πnegative 18060. This is going to be 3 right there, which means the whole thing will be negative 6π radians per second radians per second. Cool. So I got that. That's the initial speed. I wanna know how much torque is needed to stop this thing in just 10 seconds. So I'm asking what is the torque to stop it. Right. So that means that this is my initial omega and I want to have a final omega of 0. And I want to do this in just 10 seconds. So ∆t=10. So I hope you notice here, you start seeing all these motion variables. And remember, the way I solve motion problems is by setting up the curly braces and putting all 5 motion variables there. So let's do that. Omega initial equals, negative 6π. Omega final equals, we want it to be 0 alpha ∆θ and delta t. Now the t is 10, and these 2 guys, we don't have them. K. And they're also not what we're looking for. But since I saw all these variables, I decided, hey, let's start setting this up because I know this is coming. But really, what we're looking for is, 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 =Iα, and that's perfectly fine as well. That's if anything, that's a more 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 a torque to it, to make it stop. So you can assume that there's only one torque. So the sum of all torques will become just the torque that you're looking for. And that is Iα. Okay. So if I can have I and I have α, I'm done and that will be my answer. So let's expand I=25,mr2 and α. Notice that I have m, I have r, but I don't have alpha. Right? So what you're gonna do is you're going to go over here and try to find alpha. Okay. So let me plug in these numbers. So were the only thing we're missing is alpha. M is 200. R is 3 squared. Alpha so as soon as we have alpha, we can plug it in there. Okay? Now back to the motion equations. So the basic idea is you get stuck and you go to the other side. Back to the motion equations, we have 3 variables, which means we can solve. This is my target, and this is my ignored variable. So I'm going to use the only equation that does not have a ∆θ in it. And the only equation that doesn't have a ∆θ in it is the first equation. Okay. So ωfinal=ωinitial+αt And we're looking for alpha. So alpha is going to be ωfinal-ωinitial/t. This, by the way, is the definition of alpha. It's the change in omega over the change in t. You could have started there as well. That would have worked. This is 0 minus negative 6π and the time is 10 seconds. So these cancel and end up with 6π over 10, positive 6π over 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. So 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 have a negative omega and you're trying to slow down. So you to go in the other direction, the acceleration is to go in the other direction. So the acceleration would oppose it because you're trying to slow down and this is counterclockwise, which would be positive. Okay? Anyway, my acceleration is 1.8 radians per second squared. Now I can plug this in here and we are done. So if I multiply all of that, and then I multiply that by 1 point 88, I should get, 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 2 parts to this. 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 alpha, but you have a bunch of motion equations, motion variables. So you can find alpha using one of the motion equations, plug it back in. Right? So it's the classic, standard type of physics question where you get stuck with something, go look for another variable, plug it in, come back with the value you got, and and solve. Alright? That's it. Hopefully, it makes sense. Let me know if you have any questions and let's keep going.
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Torque with Kinematic Equations
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