>> Hello, class, Professor Anderson here. Let's talk again about curling. I know we've all been watching it during the Olympics, just can't pull away from the TV. It's like, oh what's going to happen on this next shot? Let's talk about the stone itself sliding along, and let's see how long it would slide distance wise if we assume some initial parameters, okay. So let's assume the following, let's say that it is going to slide along at, I don't know, 1 meter per second, okay, something like that. That might be even a little fast for 1 of those curling stones. But let's take that as our number. And let's say that our friction force is going to be 0.5 newtons. And we, of course, have a mass of the stone which is about 20 kilograms. I think that's about right, somewhere in that nature. And let's ask the following question, how far will it slide? Let's say it doesn't bump into anything, it just slides along on the ice. How far is it going to go? Well, let's draw a picture. Here's our ice, we have our stone heading along at VXI, eventually it's going to come to rest over here somewhere, VX final is equal to 0. And we want to figure out how far that is. So let's start our tape measure right there and our tape measure right there and let's figure out X final. All right, what should I do next? Ally, what should be my next approach? >> Find a kinematic equation that we apply to? >> Okay, which one do you like? >> The VX final equals VX initial. >> That one? That's a good one, right? >> Yeah. >> Okay. Let's see what we know in here and what we don't know. We know VX final, that's 0. We know VX initial, that is 1 meter per second. We don't really know AX yet and we don't know T yet. So not totally sure that equation is going to help us just yet, but let's leave it up there for a second. What other kinematic equation might we look at? Yeah, Megan? >> X final equal X initial, that one. >> Okay, keep going. >> V initial. >> Okay. >> Plus 1/2 AT squared. >> Okay. Anything missing there, guys? >> T. >> T? >> Yeah that's the T initial. >> Right here, right? You could just tell that T's got to be there just from units, right? Meters per second, I've got to multiply by seconds to get meters. All right, let's look at that one. That seems like it might be good, right? This is what we want, we don't know it yet, but that's what we want. That one looks like 0, VX initial we certainly know that. T, we do not know yet. AX we don't really know that yet either. So seems like we need a little more information. And I'm looking at this AX right here. It'd be really nice if we knew something about AX. How can we get AX, Stacey, what do you think? >> I honestly have no idea. >> Okay. Excellent, that's what we're here for. What sort of laws have we just been talking about recently in physics? >> Newton's laws. >> Newton's laws. One of those laws help us get A? >> Yes. >> Yeah, probably, right. Let's take a look at that. Okay so Newton's Second Law is sum of the forces equals the mass times the acceleration. This is, of course, a vector equation and so we have X components. And we have Y components. And let's see if we can figure out what those forces are. So, like we said before, when you have an object, draw it again as a dot. This is our free body diagram and let's identify the forces. Gravity, down. Normal force, up. What about friction, Stacey, which way should friction go? >> The opposite way that [inaudible]. >> Yeah, the thing that's moving to the right, then friction is to the left and is trying to slow it down. Okay, so let's rewrite these equations. Sum of the forces in the X direction. What are the forces in the X direction? There's only 1, and it's in the negative X direction. And so that has to be equal to the mass times acceleration in the X direction. All right, what about the forces in the Y direction. N is going up, that's a positive. MG is going down, that's a negative. That's equal to the mass times the acceleration in the Y direction. But if this thing is moving along horizontally, not going up or down, then that's 0. This whole thing is 0 and we just get N is equal to MG. Okay? That doesn't really help us in this particular problem of interest, yet, because we were given a number for the frictional force F. So let's look at this equation right here, the X equation. And look what happens. A sub X is equal to negative F over M. Aha, and I have those numbers, right? I told you what the force was, I told you what the mass was and now we know this number right here. So this is no longer an unknown. This one we, in fact, do know. All right, but we still have T that we don't know and we still have XF that we don't know. So there's a couple ways to do this, right? One is I can solve this first equation for T. And I can plug it in to the second equation and solve that equation for XF. But there is another kinematic equation that already did that. Anybody remember what the kinematic equation is? Megan, do you remember what that kinematic equation is? >> Bits and pieces. Isn't it the one with the velocity final squared? >> Yep, absolutely. Something about? >> Velocity final squared equals velocity initial squared. Something with X final minus X initial. >> Something, something is like that. >> Yeah. >> X final minus X initial, okay, good. What's that something that we want to put in there? Yeah, Ben? >> It's 2 times acceleration. >> Two times acceleration. Ah, whoops, there goes gravity. Two times the acceleration in the X direction. Where did this equation come from? It came from what we just said, solve the first one for T, stick it into the second one. You can manipulate all that stuff to get equations that look like this. As we're going to see later on, this comes from the conservation of energy as well. And now, look, I know this. That's 0. I know that, that was given. I know that, we just solved for that. X final is what we don't know. X initial, we know that. And so we can solve this equation for X final. Let's just write it out. Zero equals VXI squared plus 2 times A sub X which is negative F over M. All of that times X final. X initial is, of course, 0. And now I can solve this equation for X final. This is a negative quantity. I can put it over on the other side. I get 2F over M times X final equals VXI squared. And now I can solve this for XF. XF equals M VXI squared all over 2 times the frictional force F. And now we can check the numbers with the ones that I gave you. So somebody pull out a calculator and let's try some numbers. All right, I think we had 20 kilograms for M. Our VXI was 1 meter per second. We had a 2 and then we had 0.5 newtons for our frictional force so you don't really need a calculator for that. Right? One squared is 1, 20 over 1 is 20 in the limit of small 1. All right, and what is our units? Meters. So how far is that thing going to slide? Twenty meters. I think our initial numbers were pretty good, right? If you think about the curling, I don't even know what they call it, is it a court, is it a lane, it's like an alley? >> Yeah. >> What do they call it? >> A lane. >> A lane? >> Yeah. >> The curling lane is about 20 meters long. Yeah, that sounds pretty reasonable. Okay, so all those initial assumptions were probably pretty close. Questions about that? Let me ask you a slight follow up question. Let's say I push that stone with twice the initial velocity, how far would it go? Delal, what do you think? I had pushed it with 1 meter per second, but now I'm going to push it with 2 meters per second. How far is it going to go? >> It's going to be the double. >> Okay, what do you guys think? Double or more? Probably a little more, right, because we've got a square up there. So it's going to be, in fact, a factor of and that means not 20 meters, but 80 meters. Okay? And so now you can see why curling is kind of hard, right, because that initial velocity has to be just right. If you do it too hard, it's just going to fly out. Fly out of the lane. Okay, any questions on that approach? Everybody okay with that? All right, good, if you're having some trouble, definitely come see me in office hours. Cheers.
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7. Friction, Inclines, Systems
Kinetic Friction
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