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 going to show you in this video is that because objects travel in curved paths along these pendulums, then when we solve them it's just by using energy conservation. That's where we've solved a lot of our problems so far. Alright. So let's go ahead and get started. We're also going to 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. Let's get started with our problem.

So we have an unknown mass that's at the end of a 2 meter massless rope. So I'm just going to call this \( m \) and I know the distance of this rope is 2. So what happens is, in order to solve this problem, I'm going to go ahead and draw the diagram, list all my variables. Right? So I'm going to 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 going to release it. And we want to calculate what the pendulum's maximum speed here is. So what's going to happen? There are actually 3 points of interest that are going to happen here. Point \( a \) is the point where I'm releasing the mass. Point \( b \) is the part where it swings down and reaches its lowest point. And then if there's no energy loss, then it basically is just going to swing back up to an initial 37 degrees. It's going to 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 going to 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? Well, I might have to write an energy equation for this. So I'm going to 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 \( k_{initial} + u_{initial} + work_{done by nonconservative} = k_{final} + u_{final} \). Alright? So let's go ahead and eliminate and expand each one of the terms here. So is there any kinetic energy initial? Well, we're just releasing this pendulum. So we're just releasing this pass with an initial speed of 0, so there's no kinetic energy. What about gravitational potential? We're 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 nonconservative forces? Well, there's no work done by you and there's no friction or air resistance, so there's no work done by nonconservative. What about \( k_{final} \)? Well, hopefully, 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 point, but it's still at some height above the floor. Well, it turns out with these problems, what 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 if 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 \( y = 0 \) to be at the lowest point of our problem. Remember, we can set this to be \( y = 0 \). We can just set the 0 point to be any arbitrary point, meaning you pick it. So we're going to pick instead of the floor being where \( y = 0 \), we're actually going to pick the lowest point of our pendulum to be where \( y = 0 \) here. What that means is that the potential energy here at 0, there at \( b \), is equal to 0. And all we're interested in, remember all that'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 \( u \), is actually what I'm interested in here. Alright? So that means that our potential energy here \( a \) is just going to be \( mg \times y_a = k_b \) and our kinetic energy at \( b \) is \( \frac{1}{2} m v_b^2 \). So if I want to solve for this \( v_b \) 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 \( 2g \times y_a \). So I'm almost ready to just plug this everything in, but the problem is that I actually don't know what that \( u \) is. I know it's important. Remember, it's only the change in height that's important, but I actually need to know what that number is before I can solve for this. So how do I do that? Well, that's actually where the pendulum equation is going to come in. So I'm going to go down here for a second, and we're going to just very quickly derive this equation for you. The pendulum equation relates the 3 important variables of a pendulum. The length, the \( \theta \) with respect to the vertical, and the heights, which is \( y \), which is going to be our \( y \) variable. Alright? So basically, what we have is we have a pendulum that has a length of \( l \), and even when it goes all the way down to the lowest point, it still has a length of \( l \). What we're interested in is we're interested in what is the \( y \) distance. Right? How do I calculate what this \( y \) variable is? Remember, that's what we're looking for in our problem. And so to do this, we're actually going to look at this piece of this triangle here. We're going to look at this side of the triangle that we make when we sort of draw these horizontal lines between our initial and final heights. Most things in physics are actually going to break down into triangles here. So what we're going to do is we're going to 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 \( y \), then this piece right here is \( l - y \). And that's actually the first half of the pendulum equation. It's \( l - y \). 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 hypotenuse of this triangle, and this angle here, which is \( \theta_y \), is the angle that we're 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 \( l \times \cos(\theta) \), 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 \( l - y = l \times \cos(\theta_y) \). Alright. So now we can actually figure out what this \( u \) is. So I want to go here. Go ahead over here, and we're going to solve this by using the pendulum equation. So we have \( l - u = l \cdot \cos(\theta_y) \). We have all these variables here, \( l \) and also \( \theta_y \). So I'm just going to 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 I end up getting here is, the length of 2 minus 2 times the cosine of 37 is equal to \( y_a \), and when you plug everything in, what you're going to get is 0.4 meters. So this is basically the height here above my lowest point, 0.4. So that's what I'm going to plug into this equation. So when you go ahead and plug this in, you're going to get \( v_b \) is equal to the square root of \( 2 \times 9.8 \times 0.4 \), and what you'll get is 2.8 meters per second. Alright. That's the answer. Alright. So that's it for this one, guys. You're just going to use energy conservation plus an extra equation whenever you're not given the heights in problems. Alright. So that's it for this one, guys. Let's move on.