Our next force example is the conical pendulum. So that is kind of like a regular pendulum, except it doesn't just swing in a plane. It actually goes around, kind of a circle like that, where you have just like a regular pendulum, you can think of a normal direction at an angle theta, but it's sweeping out a circle. We've looked at them before, we used our big conical pendulum here hanging all the way from the ceiling with our bowling ball and we use it to study two dimensional motion. So if we start this sweeping on a circle like a good conical pendulum should, it makes circular motion in a plane. But now, rather than just thinking about the plane, we can do all the vectors and all three dimensions and think about the motion that way. As we're going to confirm, I have to send it at just the right velocity to get a nice circle. There we go. That's a nice circular conical pendulum. So let's see, well, what do we always do in this situation? We apply Newton's second law of motion. So we draw a free body diagram here, here's our mass, my goodness, here's our mass, we have mg down, and the tension force is the other force, tension from the string. And it's always along the direction of the string. So there's the tension vector, and if that's the vertical, it's also in theta. We're going to do this in Cartesian coordinates or related to Cartesian coordinates y and X. So the reason we could treat this as a case of circular motion is because it's not moving in the y. No acceleration in the y, so we'll start there. Sum of the forces in the y equals the mass times acceleration in the y. And in the upward y direction, we have the vertical component of the tension, and that's the cosine, right? So, T cosine theta is the T y component minus mg pulling it straight down. Minus mg and that has to be equal to zero. So, this little part is what you solve for the tension. The tension in the chord is mg over cosine theta. Looks good. Now we need to do the X. So for sum of the forces equals ma in the X, in the X, a in the X. And this is what I would call a radial snapshot. We don't really want to do it in the X because it's going around in a circle when the X component is changing all the time. But at any position, we could lay it out like this, and say this is a snapshot in time. And the X is sort of the same direction as the radial direction because we're going to apply ideas from circular motion. So we'll just call it the X at this moment in time. And we'll say in this moment in time, the X and the radial direction are kind of the same. They're along the same axis. So we'll say the sum of the forces in the radial direction equals the mass times acceleration in the radial direction for this snapshot. So the force in the radial direction is the sine component of the tension, TX. So it is T sine theta and it's pointing in positive X, but it's pointing in a negative radial direction. So since we wrote it as radial, I'm going to write negative T sine theta. Why not? What else do we have in the X direction in terms of forces? Nothing, that's the only one, the only one pulling it in, making it go in a circle, minus T sine theta equals the mass. And now the radial acceleration, well, we learned that that is v squared over r, because this is circular motion in that direction. So v squared over r and it's actually also negative. Because the acceleration we said is centripetal. It points in v squared over r, radial direction is positive out. So we could then say this is equal to negative v squared over r. And you can say, I don't like radial, I want to do X, well, it'd be easy, t sine theta would be positive, v squared over r v positive. I would get rid of these two negatives and make them positive. And then I did it in X. So it's really just arbitrary which one you call it, it doesn't matter. So, now we have two equations, two unknowns. We don't know the tension, and we don't know the velocity. Let's substitute this tension into this equation. Let's make them both positive and mg over cosine theta, mg over cosine theta times sine theta equals mv squared over r, wherever r is the radius of the circle, the circular motion. And let's see, the ms cancel, like that. Oops, no, none of these cancel, sorry [LAUGH]. The ms cancel, that's all. So, then we keep going and we say, this is g sine over cosine is tangent. G tangent theta equals v squared over r. And then I realized we never asked a question. We just started going crazy. So, what we really wanted to think about is the relationship between how fast it's going and how big the angle is. That's kind of what we're looking for here. So, we can see that here we can solve this and so we find that v is equal to the square root of g r tangent theta. So we get this kind of weird relationship between v and the angle. I could plot it for you because I know what tangent looks like. Basically, if this is theta, and if you think about it, it can't really go beyond 90 degrees, 0 is hanging straight down. 90 is the thing going round really hard. So let's plot it to 90 degrees, 0 degrees, v tangent goes to infinity at 90 degrees. So, it looks something like this, tangent does that famously, I'm taking the square root of a tangent. So, some sort of distorted tangent like that. Wait a minute. No, I'm sorry. Doesn't do that, kind of like that. Take a square root, supposed to come straight through here. Anyway, tangent, look it up. But the point is, as we go to larger and larger angle, we have to go faster and faster. And that actually matches what we observed with the conical pendulum. So here it is, going at a fairly small theta or an intermediate angle, and it's going kind of medium fast, in terms of the speed. I can go to a very small angle and then it's kind of barely moving, really slow like that. Or I can go to a really large angle and to achieve the circular motion of the conical pendulum, then I have to go really fast. So we described this before for circular motion, we said it always has the same period as it goes around. And it does. We didn't prove that here. We just derived the relationship between the angle and v. So you can imagine v can't get to- or the angle can't get too high. We got to really big include- be flying around and it would pull the roof down, we don't want to do that, but the basic relationship there, we did confirm.
