Finding the Moment of Inertia from a Point to a Ring to a Disk to a Sphere.
If you want to find the moment of inertia for a rigid object (like a disk or a sphere), you should first ask yourself two questions:
- What is the moment of inertia?
- Why am I doing this?
Here are the answers to those questions. For the moment of inertia, here is my short explanation (longer video here — but still fairly short). Imagine that I have two masses on connected together with a massless rod. Like this.
The kinetic energy for this system would be:
Although mass 1 is moving faster than mass 2 (since it has a larger circular radius), the two masses have the same angular velocity (ω). I can write the linear velocity in terms of the angular velocity.
With that substitution, I get the following expression for the total kinetic energy.
Both terms have a 1/2 and an ω², factoring that out, I get:
Now imagine that I have more than two masses — that are all rotating with the same angular velocity. This would be a rigid object. Those mr² terms can be grouped together and this is what we call the moment of inertia (I). For a bunch of masses, it is defined as:
So the kinetic energy can be:
That’s the moment of inertia. BUT WAIT! This is only the moment of inertia for a rigid object that is rotating about a fixed axis. If you take an object and toss it into the air, it’s much more complicated. I’m just giving you a small warning.
Now for the second question. Why do we need the moment of inertia? Well, imagine a sphere rolling down an incline. You could look at the sphere is a whole bunch of individual points with some common angular velocity — but different linear velocities, or you could treat the sphere as a single rigid object. If you want to treat it as a rigid object, you can find the rotational kinetic energy but you are going to need to know an expression for the moment of inertia.
Moment of Inertia for a Ring (Hoop)
Let’s start with a ring. This ring is rotating about an axis through its center. It has a mass of M and a radius R. It’s not a bunch of point masses — instead it’s a continuous object. However, let’s just break this into a bunch of small masses (labeled dm).
If I break this ring into 10 pieces, then each piece will have a mass of M/10. The mass will be even smaller as I break it into more pieces. But no matter how many pieces you have, the sum of masses will still be M.
The key here is to break this into infinitely small pieces and add them up. This means the summation becomes an integral. So, for this ring I have the following.
I didn’t put limits of integration, but I’m integrating over the whole ring. For this integration, the distance from each piece to the axis of rotation is R — they are all the same distance from the axis. Since R is a constant, I can pull it out of the integral.
Now I just have the integral of dm — which is just the total mass, M. That means my total moment of inertia for the ring will be:
That’s it. That’s the moment of inertia for a ring. Of course, this assumes that the thickness of the ring is small compared to the radius — otherwise the assumption that all the mass pieces are the same distance from the axis is not longer true.
Moment of Inertia for a Disk
Now let’s do a disk. This is basically a uniform density cylinder — think of it is a short cylinder, but it actually doesn’t matter. The only important thing is that it rotates about an axis through the center of the disk.
It’s possible to find the moment of inertia of the disk by again breaking it into small pieces just like the ring. However, in that case you would need a double integral to add up pieces in both the x and y-directions (or the r and θ). But I don’t want to do that. Instead, let’s use the moment of inertia for a ring to figure this out (that’s why I did the ring first).
Suppose I break the disks into a bunch of rings? It would look like this.
Here you can see one thin ring. This ring has a radius r and a mass of dm. The ring also has some thickness — but it’s thin, so I will call that dr. Since this is a ring, I know its moment of inertia — but it’s just a small part of the total moment of inertia (I will call it dI).
Oh, I could just integrate both sides and get the moment of inertia — not so fast. On the right side I have the variable r but an integration variable of dm. That’s a problem. I need to get this dm in terms of r. I can do this with the area and mass of the big disk. If the density of the disk is constant, then the area density of the ring and disk should be the same. But what is the area of the ring?
Imagine I take that ring and cut it, then straighten it out. I would be like a super skinny rectangle. Here is a diagram.
Yes, it’s not exactly a rectangle since the two long side are slightly different lengths. However, if dr is super small (it is), this area works fine. Now I can set the mass-to-area ratio for the ring and the full disk equal to each other.
Now, I can solve this for dm.
This gives the following integral for the moment of inertia.
Boom. That’s it.
Moment of Inertia for a Sphere
It’s time to level up to the final boss. This is a solid sphere with a uniform density, a mass M and a radius R. You might have already guessed it — but I am going to find the moment of inertia for this sphere (about an axis through its center) by breaking it into disks. Here is a side-view.
This disk-slice of the sphere gets smaller as it gets farther along the x-axis. The radius of the disk is y and the distance from the origin is x. It’s a disk, so I know its contribution to the total moment of inertia will be:
But I have two problems. I need to get rid of both dm and y. I want to add up these disks along the x-axis, so my integral should be from x = -R to x = R. Let’s take care of the y first. If you look at the outline of the sphere, it’s a circle with a radius of R. I know the following equation for a circle, and I can solve for y².
That means, my differential element of the moment of inertia will now be:
What about the mass? In this case, the disk-slice has the same density (volume density) as the whole sphere. What is the volume of the disk? Well, it has a radius of y and a height of dx. Putting this together, I get:
Solving for dm.
Now putting this all together, I get the following integral for the moment of inertia.
I’m going to complete the square so that I get three terms in the integral.
Now I can integrate these three terms to get the following.
OK, let’s evaluate at the limits. Hold onto your butts.
See. That wasn’t so bad. OK, I didn’t know it would work out — but it did.