Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Anderson Video - Rotational Motion for Merry Go Round

Professor Anderson
Was this helpful?
>> Hello class. Professor Anderson here. Let's talk about rotational motion and then we'll look at a specific example of a merry go round. Let's think about some of the parameters here for rotational motion. So if I start with my normal Cartesian coordinate system X and Y and now I have an object that is moving around in a circle. How can I define that object's position? Well, I can, of course, give it an X and Y, but I can also give it an r and a theta. And these are our polar coordinates. And anytime you're dealing with circular motion, it makes sense to give it r and theta, particularly because r doesn't change as a function of time. Okay. So what are some of the characteristics of this? Well, if I go one full rotation how far have I gone in distance? Anybody remember the circumference of a circle? >> (student speaking) 2 Pi r. >> 2 Pi r, right. Circumference of a circle is 2 Pi r. All right. We remember that. Okay. So, if I do one full rotation and I do it in one period, T, so a measure in time, then what can I say about the speed of that object? Well, speed is just distance over time and so we can say it's S over T, which is 2 Pi r over T. Okay. And the units work out, right. r is meters, T is seconds. That's meters per second. Everything else is unitless. Okay. Let's talk about the merry go round problem and see if we can make some sense of this. So our object is now going to be you sitting on a merry go round. And you are going around on this merry go round and we tell you the radius of the merry go round and we tell you the period of the merry go round. And we need to figure out what our speed is. Okay. And lets try so different numbers than we used earlier. Let's say that the merry go round goes around, what's a typical period of a merry go round? Trying to remember at Disneyland with my kids. All right, maybe 10 seconds go once around the merry go round. Okay. Ten seconds to go around once and a merry go round can be pretty big, right. Certainly the ones at Disneyland are rather large and probably have a diameter of maybe 10 meters. Okay, this the diameter. If that's the diameter then r is, of course, five meters. That's the radius. And now let's calculate the following. Let's figure out what our initial speed is. Okay. Well we just showed you the formula for that. That's not too bad. The initial speed is just how far you go and how long it takes. The distance that we go once around is 2 Pi r. How long it takes is the period T. And so we get 2 Pi times r we said was 5 meters. T we just said was 10 seconds and we chose those numbers rather nicely, right. We've got 10 pi over 10 and so that is in fact pi. And the units are meters per second. So how fast are you going? In our case you're going 3.14 meters per second. But let's add a little complication. Let's say that the merry go round is going to slow down to a stop. And let's see if we can figure out how far it's going to rotate as it slows to a stop. Okay. We've got a merry go round that is initially spinning and is spinning at a speed of 3.14 meters per second. The person on the edge of the merry go round is going around that fast. And let's ask the question, how far will it rotate if it comes to a stop in 50 seconds. How many rotations is it going to through. All right. I need your help on this one. Anybody have an idea of how we can proceed here? Yeah, Doug in the back there. >> (student speaking) Use one the kinematic equations. >> Okay. Let's go back to our kinematic equations. That sounds like a great idea. All right. Kinematic equations we always wrote in X and Y. Let's write one of them as X final equals X initial plus Vx initial times t plus 1/2 axt squared. But, of course, we're in rotational motion here and so we don't really want to use X anymore. We should probably use S final equals S initial plus V initial times t plus 1/2 a t squared. And this is going to be the arc length, how far around have we gone? What is our distance around the edge that we've got? All right. What do we know here? Well, we can measure from where we started. And so we say that's zero. V initial, we already have that. We also know t time, but we don't know a. And we don't know this number right here. All we know is that this thing is slowing to a stop which means it does have some acceleration, right. There's something that's slowing down this merry go round, but we don't know exactly what that acceleration is. So, this equation looks like it's not going to do it for us. Is there another kinematic equation that we can use? Somebody raise your hand. If you came to the lecture, you'd know the answer. Yes. Laura. >> (student speaking) For S final equals S initial plus 1/2 and then times S initial plus S final times t. >> I'm not sure that's what you what said, but that's what you meant to say, right. >> (student speaking) Yeah. >> Okay. So where did this come from? Well, you have a kinematic equation that looks like this. And so we just transfer this into our arc line S and the idea is that if you are moving at constant acceleration or constant deceleration, same thing, then your movement is just the average of your initial velocity and your final velocity. It's your average velocity, okay. And so that's all we're saying here is that how far is this thing going to move. Well, it's the average velocity. And we know all these things now. We start at zero. We have a Vi which we just solved for. We end at 0. The final speed is 0 because we've come to rest. And so now we can calculate Sf. I think we have 1/2 times Vi, which we just found was strangely pi, meters per second. And t we said was 50 seconds. Okay. So 1/2 times pi times 50, well, what's that? Half of 50 is 25, 25 times pi, that's a little bit more than 75. We'll say maybe 78. Somebody punch that into your calculator and tell me what you get. And this is in units of meters. That's how far we've gone around the edge of the merry go round. Okay. So how many rotations is that? Well, it is how far you've gone divided by the distance of one rotation. And now look what, something exciting happens. All right. Because we've had those interesting numbers Sf is 25 pi. Down here we have 2 pi times the radius. What did we say the radius was? >> (student speaking) Five. Five, 5 meters. Okay. So this becomes 25 over 10 which is 2-1/2. Two and a half revolutions and then you come to a stop. All right. Any questions about that? Okay. So I'll rely on these kinematic equations. We had X. We had Y. We're just writing them again with some new variables. Okay? All right. If that's not clear definitely come see me in office hours. Cheers.