Physics: Physical Pendulums

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A simple pendulum is just a point mass at the end of an ideal string bobbing back and forth. And real pendulum might be some funny shape like this, and we'll call this a physical pendulum. And it sure acts, in many ways, like a pendulum. When I talk about a physical pendulum, I'm still not worried about the forces of friction. I'm just worried about the fact that this has this funny shape and I would still like to be able to predict, for example, what's the period of motion of such a physical pendulum. So how do we work that out? It seems like it could be a hard problem.
Let's draw the picture. So here's my physical pendulum. It's got a pivot point, and here it is. Now what have I drawn here? What I really am concerned about is, remember, what is the net torque about the pivot point? Because the equation - Newton's second law for angular motion, sum of the external torques - is supposed to equal to . And external torques on a physical pendulum, just like for a simple pendulum, are arising from gravitational force. So let's think about this equation and this picture. would be the moment of inertia, not of the object around its center of mass, but around the pivot point. Alpha would be the angular acceleration, so that would be d^2^^^^dt^2. And sum of the external torques - how do you find that for a physical object? It's the sum of the torques about this point. Remember than when you have a complex object, gravity acts as though all of the mass was concentrated at the center of gravity. So what this point here is is the center of the gravity. Well, careful. In a uniform field, like here on Earth, the center of gravity and the center of mass are the same. So that's really what I'm assuming in this picture. This dot here is the center of mass and the center of gravity. And so I'm safe to argue that the net external torque, which is arising from gravity about this point, is the same as though all of the mass of this funny shape was concentrated right there at the center - at the center of gravity - the center of mass.
So what does that look like? At this point, the problem looks exactly like a simple pendulum of length L and mass m. The formula for the torque will now be exactly the same, except I cannot assume that the moment of inertia is just mL^2 like I did for a simple pendulum. So I still get all the same expressions that I had before for the sum of the external torque. And what you find is the following differential equation for theta: d^2^^^^dt^2 is minus mgL. That was what I had before. What I had before, in fact, for a simple pendulum was divided by and I said is mL^2. So I canceled out the m and had .
So now I'm going to leave it in this general form. This is the mass of the physical pendulum, L. Remember it's important, it's not the length of the physical pendulum. It's not from the top to the bottom. L represents the distance from the pivot to the center of mass of the physical pendulum. And is not the moment of inertia of the object around its center of mass. It's the moment of inertia of this object around this pivot point. So I have to remember the parallel axis theorem, which says if you have a pivot point here, and you have some distribution of matter around it and here's the center of mass - so here's my pendulum - the moment of inertia of this system around the pivot point is equal to the moment of inertia around the center of mass plus m times the distance from the center of mass to the pivot point, which in this case is what we're calling L. So the parallel axis theorem here tells me what I should use. It's the moment of inertia of this funny-shaped object around its center of mass plus mL^2.
So this formula has a similar structure to what we used for a simple pendulum, but the details are more complicated. And by the way, this theta here is really sine theta, so I've already made my small angle approximation. It's just as important for a physical pendulum as it is for a simple pendulum.
And what's the period of oscillation of this physical pendulum? Well, this equation looks just like my generic equation for simple harmonic motion. And the period is always 2 times the square root of this constant upside down. I've got d^2^^^^dt^2 equals negative a constant times theta. So here's the formula, 2. I can't really simplify this any more without knowing what's the moment of the inertia of the object. If the object was just a point mass down there, if this really was, after all, a simple pendulum, then the moment of inertia would just be mL^2, and that's what we did before. The m's cancel and I'd have 2. So this reduces to the correct formula for a simple pendulum, and for a physical pendulum, you just have to put in the numbers.
Let's do an example. Supposing that you think about a person walking, and the person has a leg of about 90 cm. So here's this leg. It's pivoting up here at the hip and there's a foot on the bottom. And let me suppose that the leg is approximately 90 cm long and it's pivoted at the top. And so you're walking along and you're keeping your knee locked, so you're doing kind of this funny-looking walk where you're not bending your knees. And I would like to know what's the period of the pendulum. You leg is penduluming underneath you. You kind of lift it up and then it swings back again. And that will tell me about the gait of a walking person. It will tell me, kind of roughly speaking, if you're just walking at a natural pace, kind of let your leg naturally swing forward. It's acting like a physical pendulum.
So let's just work it out. What I need to know is the total moment of inertia of this complex shape around this point, at the end. So first of all, let's make some simplifying assumptions. I just want to get a quick guess here. Let me assume that the leg is primarily a rod. So I'm going to assume that a leg looks a little bit more like so. And it's a rod with a total length, = .9 meters. And the center of mass of this rod is going to be roughly in the center. So capital L, which is, remember, defined to be the distance from the axis to the center of mass. It's not 90 cm. It's half of that. It's .45 meters. So the L in the formula will be .45 meters. The m would be the mass of my leg. And what's the moment of inertia of a rod around an end? Well, you have to either work that one out with a little calculus, or you can just look up - there's a nice formula for the moment of inertia of a rod around its center. And the moment of inertia of a rod around its center is the mass of the rod times the total length of the rod squared. And I'd better not use the symbol, L, because I'm now calling the total length, . L I was using for half of that.
So that's the moment of inertia about the center. What's the moment of inertia of this point over here? Well, it's , center of mass, plus m times this distance squared, which is .
So you take this and you add this and you notice there's a common factor of m. And that common factor of m is going to cancel in the moment of inertia with the m downstairs. So you don't need to know the mass of the leg to work this problem. And if you plug in L is 90 cm and you work this out, I got approximately 1.6 seconds. Now that's the amount of time, remember. The period is all the way out and all the way back. So just to take one step, your leg was behind you and it's walking forward and then you put it down. That's a half a period. So that's about .8 seconds, and it seems about right. When you're walking, just about one step per second. And the answer depends a little bit on the value of L that I used. And when you take a look at the final result, you can see that when you have littler legs, you tend to get a shorter period. So little people tend to walk with a little bit of a quicker gait. Of course, they're not necessarily going faster, because they're not going as far with each step.
The formula that we derived - period of a physical pendulum - is very similar to the formula for a simple pendulum. And what you really need to be getting out of this, more than just the details for this particular physics problem, is that whenever you have a situation of any kind where you have an object which is displaced from equilibrium, and then there are forces or torques that want to bring it back towards equilibrium, so you're going to get some kind of harmonic motion, if you can write down the equation for the displacement and it looks like d^2 something dt^2 is negative a constant times that something, then you really understand what's going on, even in a complicated situation like this. Once we got to this differential equation, it's the differential equation for simple harmonic motion. This is just a constant, and now I know the period and everything, really, that I might want to know about that motion.
This is really the big idea here, and it's certainly nice to be able to apply these ideas to what looked like a rather complicated problem with complicated, funny-shaped physical objects. And still, just a quick and reasonably simple analysis allows us to understand this kind of simple harmonic motion.
Oscillatory Motion
Pendulums
Physical Pendulums Page [2 of 2]