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More Rollercoaster Problems

Patrick Ford
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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 card is trying to go around and we're just dealing with letters here. All we know is that this loop de loop is a radius arm. 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? Are we comparing two points? We're trying to figure out the speed we need at the bottom where I'm going to call this my initial here, so this is my the initial so that I can reach the top, which is gonna be my final. That's two points. So when you use energy conservation, so two points just means I use energy, it's gonna be energy conservation here. So once we draw my diagram, I'm just gonna go ahead and start writing my energy conservation equation. So this is K initial plus you initial plus work done by non conservative equals K. Final. Plus you find out 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 the 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 zero, then there is no gravitational potential energy. There's also no we're 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 sort of tricky part about this problem. We're trying to figure out the minimum speed so that we just barely reached the top. And the other important part is that the car 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 car were going very, very slow at the at the top of the loop, it would basically just fall off the tracks. If the car remains locked, then that means what happens is it can actually go very, very slow as you reach at the top and it won't have to necessarily fall. So what is 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 a here at the top is actually equal to zero. 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 some sort of limit or boundary condition. What happens is if this minimum speed that we calculate or 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 get all the way up here. But then we would basically fall back down like this. If the uh if the speed here at the top weren't equal to zero, let's say the speed here at the top or equal to five, then whatever speed that we calculate down here actually could have been less than what we calculated, right? So that's why we have to set this equal to zero so we just barely reach the top. So now what happens is we're gonna set our one half mv initial squared equal to now is the gravitational potential energy. This is going to be MG. Why final? So we've risen some height like this. This is going to be my wife final so I can go ahead and cancel out the masses in my expression. And again I just want to solve or figure out an expression for this V initial. So I'm gonna move the one half to the other side and I get the initial squared is equal to two G. Now I can figure out an expression for y final. Remember I'm not going to use my final because I'm giving the radius is our If you take a look here, what happens is you have actually risen sort of twice the radius. There's just the diameter. So your wife final is actually equal to our so that's the that's the expression I'm gonna substitute in here. So now all we have to do is just take the square roots so we're gonna take the square root. And this really just becomes four G. R. And that is your expression one last thing that you could have done but you totally could have left us like this. It's just pull the four outside of the square roots and you would have simplified this and it would become two times the square of gr either one of these is fine and you full credit for these answers. All right. So that's it for this one. Guys let me know if you have any questions.