Hey, guys. Let's go ahead and work this one out together. So we have a big loop de loop of radius r that this cart is trying to go around, and we're just dealing with letters here. All we know is that this loop de loop is of radius r. Now what happens is we're trying to figure out the minimum speed that we need at the very bottom of the loop. So, we're trying to figure out how if we're going around this loop like this, what's the minimum speed that we need so that we reach the top, but just barely. So, what happens here? Are we using just one point or are we comparing 2 points? We're trying to figure out the speed we need at the bottom, or I'm going to call this my initial here, so this is my v initial, so that I can reach the top, which is going to be my final. That's 2 points. So, I'm going to use energy conservation. So 2 points just means I use energy. It's going to be energy conservation here. Once I draw my diagram, I'm just going to go ahead and start writing my energy conservation equation. So this is k initial plus u initial plus work done by non-conservative equals k final plus u final. We have to do this because we're traveling in a curved or circular path. So, let's go ahead and expand out the terms and eliminate. We're looking for some kinetic energy because we're looking for what's the minimum speed, that's going to be v initial and we're again we're just working with letters. Is there any potential energy initial? Well, if you consider the bottom of the loop the place where you have y equals 0, then there is no gravitational potential energy. There's also no work done by non-conservative forces because you're just sitting there watching the cart and there's also no friction. What about k final? Well, that's actually the tricky part about this problem. We're trying to figure out the minimum speed so that we just barely reach the top and the other important part is that the cart has to stay locked to the tracks. So, what does that mean? It means that if the cart weren't locked to the tracks, if the cart were going very, very slow at the top of the loop, it would just basically just fall off the tracks. If the cart remains locked, then that means what happens is it can actually go very, very slow as it reaches the top and it won't have to necessarily fall. So, what does this just barely reach the top actually mean? What it means here is that we're trying to figure out the minimum speed so that the speed here at the top is actually equal to 0. So, that's the important sort of conceptual point they need to realize. So, this kinetic final here actually has to go away, and the reason for that is we're looking for the minimum speed here. So, we're looking for some sort of limit or boundary condition. What happens is if this minimum speed that we calculate were any less than what we calculated, then that means that this that this cart wouldn't actually reach the top, right? If we were traveling any less than this minimum speed, we would get all the way up here, but then we would basically fall back down like this. If this, if the speed here at the top weren't equal to 0, let's say the speed here at the top were equal to 5, then whatever speed that we calculated down here actually could have been less than what we calculated. Right? So, that's why we have to set this equal to 0, so we just barely reach the top. So now what happens is we're going to set our 12mvinitial² equal to and that's the gravitational potential energy. This is going to be mgyfinal. So we've risen some height like this. This is going to be my y final. So, I can go ahead and cancel out the masses in my expression, and again, I would just want to solve or figure out an expression for this v initial. So, I'm going to move the one-half to the other side and I get vinitial2 is equal to 2g. Now I can figure out an expression for y final. Remember, I'm not going to use y final because I'm given the radius is r. If you take a look here, what happens is you've actually risen sort of twice the radius. There's just the diameter. So your y final is actually equal to 2r. So, that's the that's the expression I'm going to substitute in here. So now all you have to do is just take the square roots. So, we're going to take the square root and this really just becomes 4gr, and that is your expression. One last thing that you could have done, but you totally could have left this like this, is just pull the 4 outside of the square root, and you would have simplified this. It would have become 2gr. Either one of these is fine and you give full credit for these answers. Alright. So that's it for this one, guys. Let me know if you have any questions.
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Rollercoaster Problems
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