Professor Anderson

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>> So r1 will be how far coin one is from the center, r2 will be how far coin two is from the center. When I think about this problem I am thinking about it from the top view here, but it might be good to also have a side view. So, the side view would look like this. We would have -- that turntable becomes a disc and then we have particle number one and particle number two sitting on top of the turntable. I now need to think about the free body diagram and let's use the side view to do the free body diagram. So, for coin number one, what forces are acting on coin number one? What force do we always draw first? And if you guys want to turn off your mics so you can talk to each other again that's totally fine. If your mics are off nobody's going to hear you. You can talk all you want on that side of the glass. Okay? The only mic is here on my face. So, which force should I draw first? Which one do we always draw first? Gravity. m1g going down. What other forces are acting on coin number one? Normal force. n1. What else? Any other forces acting on it? Remember, in this side view it's going around like this. So, there's got to be some centripetal force that's keeping it moving in that circle. You're exactly right. What is that centripetal force in this case? What's keeping the coin on the turntable? In other words, if this force was gone it wouldn't stay on the turntable anymore. Just shout it out. Shout out anything you like. Wrong. Wrong. Wrong. No, I'm just kidding. Yes, static friction. f sub s. And since we're worried about particle number one let's make it f sub s1. Okay? That's it for coin number one. That's all the forces that are acting on coin number one. Let's look at coin number two. Coin number two of course has gravity. mg down. It has normal force up, n2, and it has static friction. f sub s 2. Picture free body diagram, we can now go to Newton second laws. So, Newton's second. We have vertical components to worry about and we have radial components to worry about. So, let's write down the radial components first. We have sum of the forces in the radial direction equals mv squared over r. We only have one force in the radial direction. It's fs1 so fs1 equals m1 v1 squared over r1. fs2 for this particle will equal m2 v2 squared over r2. What about the vertical forces? The vertical forces, we've got sum of the forces in the y direction equals the mass times the acceleration in the y direction. And we know there's no acceleration in the y direction because the coin's going to stay on this horizontal surface. It's not going to jump up off or fall through. And so, we have n1 minus m1g is equal to zero. Acceleration is zero in the vertical direction. And for the second one we have m2 minus m2g is equal to zero. All right, we're almost there. Right? We've got a bunch of stuff now, right. Have these four equations to relate everything. But we need one more equation which is of course our static friction. Static friction says f sub s is less than or equal to mu s times the normal force. So, when the thing is not rotating there's no static friction at play at all. When it starts spinning faster and faster, static friction starts to ramp up until it hits the equal sign. That's the maximum it can be and that's when the coins then fly off the table. So, what is f sub s1? It's less than or equal to mu s times n1. f sub s2 is less than or equal to mu s times n2. Okay, these are all the relationships that you need for this problem. This is everything that you need and now you can solve it for whatever you want. But let's go back to our original question and let's see if there's maybe a more straightforward way to get to the answer that we were looking for. That coin two flies off first. And to do that let's go back to this idea of centripetal acceleration. Centripetal acceleration's right there right? It is v squared over r. [ Writing on Board ] But we don't know exactly what v is for coin one or two and we have some idea what r is. How we going to relate these two? Right? v is certainly different for coin one and two. r is certainly different. But we don't know which one is going to win. So, we need to rewrite this and the way we rewrite this is through our good old angular velocity. [ Writing on Board ] V is equal to omega times r. If I take this relationship and I put it into here, what do I get? I get omega r quantity squared divided by r. And so, I just get omega squared times r. So, So, centripetal acceleration you can write as omega squared times r. Whichever one has the bigger centripetal acceleration, that's the one that's going to fly off first. Because it's going to hit its maximum of static friction first and it's going to lose out and that coin is going to fly off first. So, which one has bigger centripetal acceleration? Well we know the answer. It must be two, but how do we see that? Omega is radians per second. So, if coin one goes around once in a second, what is omega for coin one? I'm asking you guys in front of me. What is omega for coin one if it went around in one second? 2 pi radians per second. Right? The angle is a full circle which is 2 pi. And so, coin one would be 2pi radians per second for its omega. What about coin two? The same thing, right? Coin two also went around once. And so, these things have the same omega. All the particles on there have the exact same omega, but of course we have different r and since r2 is bigger than r1 coin number two has the bigger centripetal acceleration and so coin two flies off first. [ Writing on Board ]

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>> So r1 will be how far coin one is from the center, r2 will be how far coin two is from the center. When I think about this problem I am thinking about it from the top view here, but it might be good to also have a side view. So, the side view would look like this. We would have -- that turntable becomes a disc and then we have particle number one and particle number two sitting on top of the turntable. I now need to think about the free body diagram and let's use the side view to do the free body diagram. So, for coin number one, what forces are acting on coin number one? What force do we always draw first? And if you guys want to turn off your mics so you can talk to each other again that's totally fine. If your mics are off nobody's going to hear you. You can talk all you want on that side of the glass. Okay? The only mic is here on my face. So, which force should I draw first? Which one do we always draw first? Gravity. m1g going down. What other forces are acting on coin number one? Normal force. n1. What else? Any other forces acting on it? Remember, in this side view it's going around like this. So, there's got to be some centripetal force that's keeping it moving in that circle. You're exactly right. What is that centripetal force in this case? What's keeping the coin on the turntable? In other words, if this force was gone it wouldn't stay on the turntable anymore. Just shout it out. Shout out anything you like. Wrong. Wrong. Wrong. No, I'm just kidding. Yes, static friction. f sub s. And since we're worried about particle number one let's make it f sub s1. Okay? That's it for coin number one. That's all the forces that are acting on coin number one. Let's look at coin number two. Coin number two of course has gravity. mg down. It has normal force up, n2, and it has static friction. f sub s 2. Picture free body diagram, we can now go to Newton second laws. So, Newton's second. We have vertical components to worry about and we have radial components to worry about. So, let's write down the radial components first. We have sum of the forces in the radial direction equals mv squared over r. We only have one force in the radial direction. It's fs1 so fs1 equals m1 v1 squared over r1. fs2 for this particle will equal m2 v2 squared over r2. What about the vertical forces? The vertical forces, we've got sum of the forces in the y direction equals the mass times the acceleration in the y direction. And we know there's no acceleration in the y direction because the coin's going to stay on this horizontal surface. It's not going to jump up off or fall through. And so, we have n1 minus m1g is equal to zero. Acceleration is zero in the vertical direction. And for the second one we have m2 minus m2g is equal to zero. All right, we're almost there. Right? We've got a bunch of stuff now, right. Have these four equations to relate everything. But we need one more equation which is of course our static friction. Static friction says f sub s is less than or equal to mu s times the normal force. So, when the thing is not rotating there's no static friction at play at all. When it starts spinning faster and faster, static friction starts to ramp up until it hits the equal sign. That's the maximum it can be and that's when the coins then fly off the table. So, what is f sub s1? It's less than or equal to mu s times n1. f sub s2 is less than or equal to mu s times n2. Okay, these are all the relationships that you need for this problem. This is everything that you need and now you can solve it for whatever you want. But let's go back to our original question and let's see if there's maybe a more straightforward way to get to the answer that we were looking for. That coin two flies off first. And to do that let's go back to this idea of centripetal acceleration. Centripetal acceleration's right there right? It is v squared over r. [ Writing on Board ] But we don't know exactly what v is for coin one or two and we have some idea what r is. How we going to relate these two? Right? v is certainly different for coin one and two. r is certainly different. But we don't know which one is going to win. So, we need to rewrite this and the way we rewrite this is through our good old angular velocity. [ Writing on Board ] V is equal to omega times r. If I take this relationship and I put it into here, what do I get? I get omega r quantity squared divided by r. And so, I just get omega squared times r. So, So, centripetal acceleration you can write as omega squared times r. Whichever one has the bigger centripetal acceleration, that's the one that's going to fly off first. Because it's going to hit its maximum of static friction first and it's going to lose out and that coin is going to fly off first. So, which one has bigger centripetal acceleration? Well we know the answer. It must be two, but how do we see that? Omega is radians per second. So, if coin one goes around once in a second, what is omega for coin one? I'm asking you guys in front of me. What is omega for coin one if it went around in one second? 2 pi radians per second. Right? The angle is a full circle which is 2 pi. And so, coin one would be 2pi radians per second for its omega. What about coin two? The same thing, right? Coin two also went around once. And so, these things have the same omega. All the particles on there have the exact same omega, but of course we have different r and since r2 is bigger than r1 coin number two has the bigger centripetal acceleration and so coin two flies off first. [ Writing on Board ]