Pendulum Problems

Hey guys, So, for this video, I'm going to introduce you to a type of problem. I like to call the pendulum problem. So, let's get started, first of all, what's a pendulum? It's really just a mass that's at the end of a string. And when you release it, it's allowed to swing back and forth in an arc shape. So what I'm gonna show you in this video is that because objects travel in curved paths along these pendulums, then what we solve them is just by using energy conservation. That's what we've solved a lot of our problems so far. All right, so let's go ahead and get started. We're also gonna need a special equation just for pendulums later on, which I call the pendulum equation. We'll talk about that in just a bit here. So, let's get started with our problem. So, we have an unknown mass that's at the end of a two m massless rope. So I'm just gonna call this m. And I know the distance of this rope is too. So what happens is uh in order to solve this problem, I'm gonna go ahead and draw the diagram list all my variables. Right? So I'm gonna pull this block up to a point where it makes 37 degrees with the vertical here. So this is 37 degrees with respect to the y axis, and then we're gonna release it. We want to calculate what the pendulums maximum speed here is So what's gonna happen? There's actually three points of interest that are gonna happen here. Point is the point where I'm releasing the mass point. This is the part where it swings down and reaches its lowest point. And then if there's no energy loss, and basically, it's just going to swing back up to an initial 37°,, it's gonna make that angle again, and it's just going to keep going back and forth forever. So what I want to do is I want to figure out where the maximum speed is. Where do you think that happens? Well, hopefully you guys realize that when it swings lower, it's going to gain some speed. And so at the lowest point here it's gonna be traveling the fastest. This is V. B. And this is really my target variable. So let's move on to the second step. How do we solve that? We might have to write an energy equation for this. So, I'm gonna write a conservation of energy equation in the interval from A. To B. So from A. To B. Here, my conservation of energy looks like K initial plus you. Initial plus. Work done by non conservative equals K. Final plus you. Final. All right. So let's go ahead and eliminate and expand each one of the terms here. So, is there any kinetic energy initial? We're just releasing this pendulums. We're just releasing this past with initial speed of zero. So there's no kinetic energy. What about gravitational potential? Where at some height above the floor here? So for now we'll just say that there's some gravitational potential. What about work done by non conservative forces? Well, there's no work done by you and there's no friction and air resistance. So there's no work done by non conservative. What about KB? Okay, Final. Well, if you guys realize that that's actually what we're interested in, that's the velocity final here. So it's definitely going to be some kinetic energy now. What about the potential energy. Final? So, we haven't yet reached the floor yet. We're basically still at this lowest points, but it's still some height above the floor. Well, it turns out these problems you'll realize is that we actually aren't given the heights of any of these distances or any of these points here. In fact, usually in pendulum problems, the distance between the pendulum and the floor is going to be unknown. So how do we actually solve for this? We don't have any of the distances. Well, it turns out we can actually use a trick that we've seen in gravitational potential energy, which is we can set the point where y equals zero to be at the lowest point of our problem. Remember, we can set this to B. Y equals zero. We can just set the zero point to be any arbitrary point, meaning you pick it. So we're gonna pick, instead of the floor being where y equals zero. We're actually gonna pick the lowest point of our pendulum to be where Y equals zero here. What that means is that the potential energy here at zero at B is equal to zero And all we're interested in. Remember all it's important in these problems is the change in the height from initial to final. So basically this distance right here, which I'm going to call Y. A. Is actually what I'm interested in here. All right. So that means that our potential energy here A is just gonna be MG times Y. A equals R. Kinetic energy at B is one half M V B squared. So, if I want to solve for this VB here, what I can do is I can cancel out the masses, which is good because I was never given mass in the problem and I can move the one half over to the other side and then just take the square root. And what you'll end up with is V. B. Is equal to the square root of two G. Times Y. A. So I'm almost ready to plug this everything in. But the problem is is that I actually don't know what that Y. A. Is. I know it's important. Remember it's only the change in the height that's important. But I actually need to know what that number is before I consult for this. So how do I do that? Well that's actually where the pendulum equation is going to come in. So I'm gonna go down here for a second and we're gonna just very quickly derive this equation for you. The pendulum equation relates the three important variables of a pendulum, the length, the data with respect to the vertical and the heights. Which is why which is going to be our Y. Variable. All right. So basically we have is we have a pendulum that has length of L. And even when it goes all the way down to the lowest point, it still has a length of el, what we're interested in is we're interested in what is the y distance? Right? How do I calculate what this why variable is? Remember That's what we're looking for in our problem. And so do this. We're actually gonna look at actually this piece of this triangle here. We're gonna look at this side of the triangle that we make when we sort of draw these horizontal lines between our initial and final heights, where most things in physics are actually gonna break down into triangles here. So, what we're gonna do is we're gonna write an expression for this piece of the triangle. Well, hopefully you guys realize that if this entire piece here is L. And this is why then this piece right here is l minus y. And that's actually the first half of the pendulum equation. It's l minus y. Now we just need to write another expression for it. How do we do that? We can actually use trig because we know this length here is L. That's the hypothesis of this triangle and this angle here, which is theta. Why is the angle that were usually given in problems? So this side right here, the one that you'll see here is actually the adjacent side. So what you'll see is that this piece right here is actually equal to L. Times the cosine of theta, and that actually ends up being the right side of our equation. These two things, these two expressions here have to be equal to each other. So the pendulum equation is l minus Y. Is equal to L. A. Times the co sign of fatal. Why? All right. So now we can actually figure out what this Y. A. Is. So I'm gonna go here, go ahead over here and we're gonna solve this by using the pendulum equation. So we have l minus Y. A. Equals L. Co sign of theta. Why we have all these variables here. L. And also tha tha y. So I'm just gonna go ahead and rearrange and start plugging in numbers. Basically I'm going to move this to the other side and these two things are going to switch places. So what end up getting here is the length of two minus two times the co sign of 37 is equal to Y. A. And when you plug everything in which you're gonna get 0.4 m. So this is basically the height here above my lowest 0.0 point four. So that's what I'm gonna plug into this equation. So, when you guys and plug this in, you're gonna get VB is equal to the square root of two times 9.8 times 0.4. And what you'll get is 2.8 m per second. All right. That's the answer. All right. So that's it for this one, guys, you're just gonna use energy conservation plus an extra equation whenever you're not given the heights, um, in problems. All right. So that's what this one guys. Let's move on.

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