10. Conservation of Energy

Pendulum Problems

# Calculating Tension in a Pendulum with Energy Conservation

Patrick Ford

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Alright guys, we have a pendulum here uh and were told some information about this. The block that's on the pendulum is two kg. The length of the pendulum. The length of the string here is gonna be three. And what happens is this block is going to be pulled one m above the lowest point and released. So basically what's gonna happen is I'm gonna draw like a sort of exaggerated diagram like this, it's just gonna swing down and then it's gonna go back and forth like this. So we know here that be between the initial height and the bottom point. It drops a distance of one m. And remember that in pendulum problems were not usually going to be given the heights relative to the floor. So we do here is we can say that this sort of point here, the lowest point of our diagram is actually gonna be y equals zero. All right, so let's go ahead and check out our problem here. We're gonna have to calculate the pendulums maximum speed. So basically we're gonna release the block, It follows this path. And then at some point down here it's traveling with the maximum speed. I'm gonna call this V max and then it just basically does this over and over again. Right? So we draw the diagram. Now we're gonna have to go ahead and write an energy conservation equation. So if we take the point of release to be point A the point at the bottom to be point B. And then this part of the point, see where it reaches the initial height again. Really, That's what we're looking for is we have some information about the release about point A. We want to calculate what the speed is here at B. So we're gonna use the interval from A to B to set up our energy conservation equation. So for part A we're gonna set up the energy conservation equation and we're looking for the maximum speed. So we have here is K. A. Plus you eh plus work. Non conservative equals KB plus you be. So now that we've done that we're gonna eliminate and expand the terms. So do we have kinetic energy here at point A. Well at point A you've just released it has no initial kinetic energy because it's initial speed is equal to zero. So there's no kinetic energy here. What about potential energy? We know it's some height above the floor. But really what matters is the height above our zero point? Remember the zero point is at the bottom of the swing and we are some distance this one m above that point. So there is some potential energy. There's no work done by non conservative forces. Nothing done by you Or friction. What about kinetic energy? And be well, actually, we're looking for the maximum speed of B. So there's definitely some kinetic energy. What about potential energy? You might think that there is some because it's still above the floor. But remember that this is actually are zero points. So there's actually no potential energy here because of y. Is equal to zero. Alright. So we're not just gonna go ahead and expand in the terms and then solve. So this is gonna be MG. Y. A. And that's actually what this distance is over here. This is why A. Is equal to one m and then this is equal to one half and then M. V. B. Except I'm gonna call this V. Max squared. Alright. So what I'm gonna do here is I'm just gonna cancel out the masses like this. They'll cancel out and I'm just looking for what V max is. So this is pretty straightforward. You just moved one half over to the other side and your V. Max, it's just gonna be the square roots of two G. Times Y. A. So this is gonna be the square root of two times 9.8 times the initial height. Which is one. If you go out and work this out, you're gonna get is a maximum speed of 4.43 m per second. So it's just straight up energy conservation pointed to point B. And you'll see this is a pretty you know, familiar results here when something drops some distance we get 4.43 m per second. That's what V. Max is equal to. All right? So let's move on to part B. Now. So in part B. Now we want to calculate is the ropes tension at the very bottom of the swing. So we're still looking at this point right here. But what happens is as this rope as this block is swinging, there's gonna be some tension from the rope or the strength. So this is what the rope is gonna look like. And we know that there's gonna be some tension because of some weight. So there's some tension like this and that's really what we're trying to find here. What is this t. So in order to find what the tension is, we're gonna have to look at the forces that are going on in this diagram. But remember that this object is swinging in a circular path. So if we want to look at for a force and the object is swinging in a circular path, we're gonna have to use F. Equals M. A. This is F. Equals M. A. Here. But we're gonna have to use f centripetal equals M. A. So we're just gonna use the centripetal forces equation. Alright so remember that this centripetal acceleration here is actually equal to M. And this is gonna be V. Squared over. R. Alright so the first thing we have to do is we're gonna have to expand out all of the forces to figure all of them out that are acting on this block. We know that there's a tension pointing up. But remember that that because because this block has mass, it also has a gravitational force. So this is gonna be MG. So there's no other forces, right? There's no normal forces, no applied forces, anything like that. It's just tension and gravity. And because these things point in different directions one points up and the other one points down, we're gonna have to pick a direction. So remember that the convention is that forces that points towards the center of the circle are positive. So this tension here is gonna be positive. Anything that points this direction is gonna be positive. So that means when you expand your forces are gonna get tea and then the MG becomes negative right? Because it points downwards and then this is equal to M. And then this is gonna be this is V. But because we're looking at the bottom of the swing, this is actually gonna be v max squared. And then what happens is we need the radius of the circle. The radius of the circle is really just gonna be this distance over here. Which distance do we know that? This applies to? Well this is actually gonna be the length of the string. So really what happens is that are actually just becomes L. So those are an L. Is equal to the same thing. That's the radius of the circle. So now all we have to do is just go ahead and isolate this tea here and you can just bring everything over to the other side. So this is gonna be T. Is equal to and you're gonna have MG plus M. And this is gonna be V. Max squared divided by L. So we can do this, we can actually group together these M. S. Because there are common factors in both the terms. So this is just to become M. And this is going to be become G. Plus and then V. Max squared over L. Right? So you just sort of pulling out the M. Now we can just go ahead and plug and solve. So this is T. And this is gonna be equal to the mass, which is two. And then we're gonna have parentheses 9.8 plus. And then V max remember it's just the 4.43 that we just calculated. So this is 4.43 squared divided by the length of the string, which is the radius of the circle. And that is gonna be three. So if you go ahead and work the sell out, what you're gonna get is 32.67. And that's going to be in Newton's. So that is the tension that's on the that's on this block here at the bottom of the swing. So that's for this one. Guys, let me know if you have any questions

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