The Physics Philes, lesson 48: Physicists Make Weird Things for Science
In which I draw a weird thing, do math, and make promises I intend to keep.
OK nerds! Class is over, which means it’s time to get down and dirty with some physics.
Last time this post had something substantive in it, I discussed the kinetic energy of a rigid rotating body and I introduced a new concept called the moment of inertia. Today we’ll do a couple of example problems and talk about how this all relates to gravitational potential energy. Sounds cool? Then let’s roll.
Let’s do a quick example question on how to find the moment of inertia. Let’s pretend that we’ve built this weird thing consisting of three heavy disks and three lightweight struts. The disks weigh 0.30 kg, 0.10 kg, and 0.20 kg, and the strut lengths are 0.50 m, 0.30 m, and 0.40 m.
I know that none of this makes sense, so here’s a diagram of all the relevant information.
Is that any better? I’m not sure, but let’s just go with it. You might be able to kind of make out that there is an axis in the middle of disk A. The other disks swing around disk A. It’s like disk A is the sun and disks B and C are planets.
Does any of that make sense? Good. Then let’s find the moment of inertia of this body about that axis.
Because the struts are lightweight, we’ll consider them to be massless rods and the disks to be massive particles. By doing so we can use the definition of moment of inertia to find…ugh…the moment of inertia, I guess.
Remember the definition of moment of inertia? Just add together the product of the masses of the particles and the square of the particle’s distance from the axis of rotation. Since we’re considering each individual disk to be an individual particle, and since we know the length of the struts, this will be easy. It’s just a matter of plugging in the values.
Voila! You’ll notice that we just used the mass values from disks B and C and the distances from A to B and a to C. This is because disk A lies on the axis, so its distance to the axis is zero. That means that it would add nothing to the moment of inertia. You know, multiplying by zero and all. But that’s how it’s done.
We can do the same thing if we pretend that our weird contraption could swing around an axis going though disks B and C. In that situation, since B and C both lie on the axis, only disk A contributes to the moment of inertia. So our equation looks like this:
Cool, right? Let’s go back to the first part of this question. We know the moment of inertia for this contraption we’ve made when it swings about the axis in the middle of disk A. If we know the angular speed, we can figure out the kinetic energy.
If you’ll remember, the equation we learned a couple of weeks ago for kinetic energy is the product of the moment of inertia and angular speed in radians, all divided by two. So what is the kinetic energy of our contraption when it has the angular speed of 4.0 rad/s? Let’s find out by plugging in some values.
And there you have it! Pretty easy, right?
This problem, however, just deals with kinetic energy. What do we do when we have to a body that is subject to gravitational potential energy? Let’s pretend for a moment that we have a pulley that’s lifting something. If the acceleration due to gravity is the same at all points on this body, then we can do the math as if all the mass is concentrated at the body’s center of mass. If we take the y-axis to be vertically upward, then the gravitational potential energy turns out to be:
In this equation, y stands for the center of mass, and this equation holds true for all extended bodies, not just rigid bodies.
OK, well, that’s all I have for you today. Next week I’ll probably try to explain something called the parallel-axis theorem. What is it? I don’t know yet. You’ll just have to come back next week.
Featured image credit: ninahale