Pendulum Problems - Online Tutor, Practice Problems & Exam Prep

A pendulum consists of a mass at the end of a string that swings back and forth. To find its maximum speed, apply the conservation of energy principle. The kinetic energy at the lowest point equals the gravitational potential energy at the highest point. The pendulum equation, l_y - y_a = l_y cos(θ), helps determine height changes. Ultimately, the maximum speed is calculated as v_b = √(2 × g × y_a), yielding a result of 2.8 m/s.

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 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.

Calculating Tension in a Pendulum with Energy Conservation

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 v_max 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 K_a + U_a + Work_{non-conservative} = K_b + U_b. 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 v_max².

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 v_max is. So this is pretty straightforward. You just move the one-half over to the other side and your v_max is just going to be the square root of 2g*y_a. 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 v_max 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.