Alright, guys. We have a pendulum here, and we're told some information about this. The block that's on the pendulum is 2 kilograms. The length of the pendulum, the length of the string here is going to be 3 meters. What happens is this block is going to be pulled 1 meter above the lowest point and released. So basically what's going to happen is, I'm going to draw an exaggerated diagram like this. It's going to swing down and then it's going to go back and forth like this. So we know here that between the initial height and the bottom point, it drops a distance of 1 meter. And remember that in pendulum problems, we're not usually going to be given the heights relative to the floor. So what we do here is we could say that this sort of point here, the lowest point of our diagram, is actually going to be y equals 0. Alright? So let's go ahead and check out our problem here. We're going to calculate the pendulum's maximum speed. So basically, we're going to release the block, it follows this path, and then at some point down here, it's traveling with the maximum speed. I'm going to call this vmax and then it just basically does this over and over again. Right? So we're on the diagram, now we're going to have to go ahead and write an energy conservation equation. So if we take the point of release to be point a, the points at the bottom to be point b, and then this part for you to be point c where it reaches the initial height again, really what we're looking for is we have some information about the release, about point a, but we want to calculate what the speed is here at b. So we're going to use the interval from a to b to set up our energy conservation equation.
So for part a, we're going to set up the energy conservation equation, and we're looking for the maximum speed. So what we have here is Ka + Ua + Worknon-conservative = Kb + Ub. So now that we've done that, we're going to eliminate and expand the terms. So, do we have kinetic energy here at point a? Well, at point a, you've just released it. It has no initial kinetic energy because its initial speed is equal to 0. 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 points. Remember, the zero point is at the bottom of the swing and we are some distance, 1 meter above that point, so there is some potential energy. There's no work done by nonconservative forces, nothing done by air or friction. What about the kinetic energy of b? Well, actually we're looking for the maximum speed at 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, this is actually our 0 point. So, there's actually no potential energy here because y is equal to 0. Alright. So, we're now just going to go ahead and expand in the terms and then solve. So this is going to be mgya, and that's actually what this distance is over here. This is ya is equal to 1 meter, and then this is equal to 12mvb2, except I'm going to call this vmax2.
Alright? So what I'm going to do here is I'm just going to cancel out the masses like this, they'll cancel out, and I'm just looking for what vmax is. So this is pretty straightforward. You just move the one-half over to the other side and your vmax is just going to be the square root of 2g*ya. So this is going to be the square root of 2 times 9.8 times the initial height, which is 1. If you go and work this out, what you're going to get is a maximum speed of 4.43 meters per second. So it's just straight-up energy conservation point a to point b, and you'll see this is a pretty familiar result here when something drops some distance. We get 4.43 meters per second. That's what vmax is equal to. Alright. So let's move on to part b now.
So in part b, now we want to calculate is the rope's 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 going to be some tension from the rope or the string. So this is what the rope is going to look like, and we know that there's going to 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 going to 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 for a force and the object is swinging in a circular path, we're going to have to use f equals ma. This is f equals ma here, but we're going to have to use fcentripetal=ma. So we're just going to use the centripetal forces equation. Alright? So remember that this centripetal acceleration here is actually equal to m, and this is going to be ve2L.
The first thing we're going to have to do is we're going to have to expand out all of the forces and figure all of them out that are acting on this block. We know that there's a tension pointing up, but remember that because this block has mass it also has a gravitational force.