# Problem with Rotational Inertia

1. Nov 5, 2006

### Lancen

I am having problem with a relatively simple problem that I probably have done before but because my Physics have become so rusty recently I just can't get the right answer. You have a disc that sitting about some point P on the ground - so its upright - and there is a hole cut out on the top of the disc so that the top of the hole touches the top of the disc. The hole has a radius of a/2 and the disc and radius of a. The center of mass of the disc c is a/6 from the center of the disc. And the disc has a uniform mass density of v. What is the rotational inertia of the disc about the point P that it touches on the ground?

So parallel axis theorem right? Ip = Icm + ML^2. My problem is what is Icm? 1/2M*L^2 by any means. But what is L? Usually if the disc' center of mass is about its center then its just the radius but here the center of mass is a/6 from the center of the disc.

2. Nov 5, 2006

### OlderDan

Hint: If the disk were solid, you could find its moment of inertia. By definition the moment of inertia is a sum of the products (dm)r^2 where dm is the mass of a tiny piece of the disk. Since it is just a sum, you could divide the disk up any way you found convenient and find the moments of inertia of each piece and add them together. Think of a solid disk as being made up of two parts. One part is the thing you have, and the second part is the disk that was cut away. The total angular momentum of a solid disk is the sum of those two parts.

3. Nov 5, 2006

### Lancen

I got that far. I know the cut out hole is presented as "negative" mass, but the problem is Ip = Icm + M*h^2 = M*L^2+M*h^2=1/2((pi*a^2*v)-pi*(a/2)^2*v))*L^2+((pi*a^2*v)-pi*(a/2)^2*v)*(5a/6)^2.

Here Icm is the rotational inertia about the center of mass C which is not at the center of the circle, and h is the distance from the Center of mass to the point at which the circle touches the ground P. My problem is whats L?

4. Nov 6, 2006

### OlderDan

There are a couple of ways to approach this. Here is a calculation based on the "full" disk about point P minus the "missing" disk about point P. It does not require you to find the CM of the object or an L related to that CM. It can be done that way, but it involves more steps. This way uses separate Ls for the full disk and the missing disk.

$$I_{FCM} = \frac{1}{2}Ma^2$$

$$I_{FP} = \frac{1}{2}Ma^2 + Ma^2 = \frac{3}{2}Ma^2$$

$$I_{MCM} = \frac{1}{2}\left( {\frac{M}{4}} \right)\left( {\frac{a}{2}} \right)^2 = \frac{1}{{32}}Ma^2$$

$$I_{MP} = \frac{1}{{32}}Ma^2 + \frac{M}{4}\left( {\frac{{3a}}{2}} \right)^2 = \frac{1}{{32}}Ma^2 + \frac{9}{{16}}Ma^2 = \frac{{19}}{{32}}Ma^2$$

$$I_{NP} = I_{FP} - I_{MP} = \frac{3}{2}Ma^2 - \frac{{19}}{{32}}Ma^2 = \left( {\frac{{48}}{{32}} - \frac{{19}}{{32}}} \right)Ma^2 = \left( {\frac{{29}}{{32}}} \right)Ma^2$$

Last edited: Nov 6, 2006
5. Nov 6, 2006

### Lancen

Ah, I see. I have one more problem with the question. If the disc is upright and is free to roll without slipping along the surface on which it rests under gravitational forces, whats the period T for small oscillations about the vertical position? I am having a bit of trouble understanding the problem. Is it saying that under some condition instead of rolling forward it will just rock back and forth about an imaginary line down the middle? And is this a simple or physical pendulum? Physical I think because it is represented as a rigid body. Do I just go straight ahead and try to solve for T=2*pi/w, where w=sqrt(mgh/I)? Where w is the angular velocity and I is the moment of inertia, and solve this using the moment of inertia that was just solved for?

6. Nov 6, 2006

### OlderDan

Now you tell me I avoided calculating the CM in the I calculation, but now we need it. You had quoted it earlier as a/6 below the center of the big disk, and that is correct. Any displacement of the disk from point P being in contact with the floor will raise the CM, so there will be a restorative force that will bring it back to equilibrium. It's not quite a physical pendulum because there is no fixed point of rotation. Still, if we can show that for small oscillations the potential energy is a quadratic function of diplacement, then we will have shown the motion to be harmonic and we can find the period. See if you can write an expression for the potential energy in terms of the angle of the roll from equilibrium. The potential energy can be taken as zero at equilibrium and Mgh(θ) elsewhere; h(θ) is the height of the CM relative to the lowest point.

Last edited: Nov 6, 2006