>> Hello class, Professor Anderson here. Welcome to another lecture on physics. Let's talk about one of the classic physics problems which is a catapult. Okay. And, this is related to one of your homework problems in which a flower actually launches a spore through this catapult type of action, but let's go back to the, sort of, medieval notion of a catapult which I think is kind of cool. All right. So, what does a catapult look like? Well, usually they put it on wheels so they can drag it around to various places. There's a big arm here that has some sort of bucket on the end, and in that bucket, they've put a big item, like a cannonball which they are going to launch by rotating this catapult through a various angle. And, let's define that angle as theta. So, we're all sort of familiar with this medieval idea of the catapult. Let's see if we can analyze the torque involved here. What is the torque that we need to make this catapult do this? And, let's give some other parameters. Let's say that we're going to cover this angle theta in a time t. Okay, and then we're going to launch this thing at a speed v. And the catapult has a length to it. We'll call it L, that's the length of that arm. Now, we also have to identify some masses involved here, right? So, this is m1. This is m2. The mass of the ball versus the mass of the pole. Okay. How do we do this? Well, we're looking for torque, right? So, we think back to our definition of torque. What is torque? Well, torque is, you know, it's the twisting force. It's how hard are you going to rotate this thing. And, we had some ideas for calculating that. One of them was force times lever arm. Okay, if you're applying a bigger force to a bigger lever arm, there's more torque associated with that. But, in our case, we don't really know what that force is. Right, we know what the lever arm is. That's L, but we really don't know what the force is. Another definition for torque, which maybe we can take advantage of which is this. Torque is equal to pi times alpha. This is the analogy of f equals ma. f equals ma. In rotational kinematics, it becomes torque equals moment of inertia times angular acceleration. And, this, maybe we can figure out, right? Certainly, we know i. i is i of this arm plus i of the cannonball. Okay, we have two things there. We have, not only this pole, the arm itself, but we have the item that we're launching. All right. And, we can figure those out. Right? We know what the moment of inertia of a uniform beam is. It is 1/3 m2 L squared. That's a uniform beam rotated through one end. That's the moment of inertia. And, we also know what the moment of inertia of the cannonball is at the end. It is just m1 times L squared. Okay, so we know i. What about alpha? What is alpha? Alpha is angular acceleration. So, anytime you hear that word acceleration, you probably think of the kinematic equations, and there are kinematic equations in linear case. But, there are also kinematic equations in the rotational case. And, since we have some information here, like angle and time, maybe there is an equation that will help us get to alpha. The one I'm thinking of is this. Theta f equals theta i plus omega initial times t plus 1/2 alpha t squared. All right. Theta, we can call theta f zero arbitrarily, and then our theta final is how far that thing rotated. It also started from rest, right? This whole thing is tied down with a rope. They cut the rope, and then boom it goes. So, omega initial is equal to zero. And, now we know this, right? This is something they would give you in the problem. We know t. That's something they would give you in a problem. So, we can solve this equation for alpha. What do we get? We get alpha equals 2 theta f divided by t squared. And so, now we can tie all this stuff together and calculate the torque. So, torque is equal to i times alpha which is 1/3 m2 L squared plus m1 L squared. All of that times alpha, which we said was 2 theta f over t squared. And, if you know all these numbers, then you can plug it in and calculate the straightening torque, right? What is the torque to cause this thing to rotate? Now, the second part of this question says what's the speed? How fast does the cannonball actually get launched? Let's see if we can figure that out. So, let's ask the question what is the final speed of launch of that cannonball, and maybe one way to think about it is this thing is rotating with some omega final. And, if I can figure out what that omega final is, I can probably relate it to the speed of the launch of the cannonball. So, let's go back to the kinematic equations, and let's think about kinematic equations for this thing right here. Omega final equals omega initial plus alpha times t, and we know omega initial is 0. It started at 0. And so, we get omega final equals alpha times t, but we know what alpha is. We did that in the first part of this problem. It's that. 2 theta f over t squared. And, now we multiply it by t, one of those t's cancels out, and we get 2 theta f divided by t. This is your omega final. What we're looking for is v, and v relates to omega by the radius. How far are you from the axis of rotation? v is equal to omega times r. So, in this case, it's going to be omega final times l, and we get 2 theta f over t, all of that times L. And now, if you have all those numbers, you can plug it in and calculate what the final speed of the cannonball is. Let's make sure that this makes sense, okay? What it says is if I rotate my catapult through a bigger angle, theta, it's going to be going faster at the end. That makes sense, right? If my catapult just went up like this, it would launch it with some pretty slow speed. If it goes up pretty fast, it has high altitude, it's going to launch it with some bigger speed. All right. That makes sense. It also says that if I do that in a shorter amount of time, it's also going to be launched faster. This makes sense when you see those, like, catapult competitions on reality TV. You guys ever seen those? They have those complicated trebuchet catapults, and they launch these giant objects like pianos. And, they launch them like hundreds and hundreds of feet. You know, like 1000 feet or something ridiculous. Okay. They do it in a very short amount of time. The faster this thing pops up, the faster it's going to be going up there. That's because t gets small. It's in the denominator, and so v gets big. It also says that you want a big lever arm, L. You want to be able to have a long pole that you put your cannonball at the end of. If it swings up, everything else is the same. Then, it's going to be going faster at the end. Now, the downside to having a very long lever arm is it requires a lot more torque. Right? Here's the torque. L squared right here. So, the bigger lever arm you use, the harder it is to rotate it at that same angular acceleration. But, if you can, it will be going faster at the end. All right. Questions about that one? You guys going to go out and build a catapult, now? They're really fun. We had to do a catapult experiment in high school physics, and the idea was you have to launch a tennis ball a certain distance and try to hit a target. And, you have to calculate how much you're going to compress your thing, your spring or whatever you're using, and launch it and try to hit that target. And, my buddy and I did this, and we were able to hit the target on our first shot. And, fortunately, our entire catapult destroyed itself after that first shot. So, we didn't have to do it again.