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Anderson Video - Non-Uniform Circular Motion

Professor Anderson
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>> What do we mean by non-uniform? Does it mean it's not a circle? No. It just means that you're not moving around at constant speed anymore. You are accelerating with some tangential acceleration. So A sub T is the tangential acceleration. And if you are accelerating, then we call that non-uniform circular motion. So uniform is when you're just going around at constant speed. Non-uniform is when you're accelerating. So in the problem on the homework, you had a bicycle that was going around a track, bicycle on a circular track. And in that problem, they told you a little bit about how far around they went in some amount of time, and so there is some AT as they go around. But there is also some AC. And we have to worry about both of those things. AT, you can get from the kinematic equations. So for instance, if they start from rest and they go once around the track in some amount of time, you can calculate their acceleration because you know that they went two Pi R in distance and they did that in some amount of time, T. And so you can calculate exactly what the acceleration is. AC, we know what that is. That is V squared over R. Alright. So let's think about the bicyclist on a circular track. Right, here is the circular track, and we have a bicyclist that is racing around. Here's my bicycle -- -- person going around. And we need to figure out something about the net force or the net acceleration on the bicyclist after they make one full rotation. So we need to think a little bit carefully about this. Let's find the magnitude of net force after one revolution. Alright. What do we know about net force? What we know is net force is equal to mass times acceleration. So if we knew the acceleration, we would basically have it. But we got to be careful because this bicycle in this position as it goes around is going to have A sub T, which is tangential to the circle. And it's going to have A sub C, which is along the radius. So what does this A look like? A, the vector, is a vector sum of those two. It's A sub T plus A sub C. And A sub T we said was up. I'm going to add A sub C, which is to the left. And my net vector A is therefore right there. Okay? So it's actually sort of at an angle is the net acceleration A. Alright, so if I go back to the question at hand, find the magnitude of the net force after one revolution, what do we mean by magnitude of net force? What we mean is this. What is the magnitude of F? The magnitude of F, let's put that summation sign back in there so we're consistent with what we wrote earlier. So we have the magnitude of the net force. But that is the same as MA. And M, that's just a number, so that comes out in front. And so I have M times the magnitude of A. There's my A right there. What is the magnitude of A? This is a right triangle, and so the magnitude of that hypotenuse is just AT squared plus AC squared. You already know what AC is. It's just V squared over R. AT you'll have to figure out by solving the kinematic equations. If we want to know the direction of A, we go back to our picture right here, and you can calculate whatever you want now. If you know AC and you know AT, you can calculate whatever angle you want in that triangle. Okay? Alright. Questions about that one? Was that a challenging one on the homework? Are you guys aware that we have homework? You're like what class is this? I stumbled into the wrong room.