# Ridiculously complex mechanics problem

1. Jan 21, 2006

### DaMastaofFisix

Hey Everyone, have another doosie for eveyone. I've been crackin' away at it for some time, but to no avail. Here it is:

A semicircular of mass m and radius R is resting on a rough surface. the center of mass is given by 2R/pi, and that's about it.

a) Calculate the moment of inertia of the semicircular ring about its center of mass
b) The ring is displaced by a small angle; find the period T (no slip). Ignore all terms higher than first order theita
c) The surface is now frictionless. Given a small displacement, as well as ignoring second order and beyond theitas, find the Period T (with slip)

Okay, so for part a I thought it would be easiest if I calculated the Moment of inertia about the center of what would be the whole ring, followed by the use of the parralel axis theorem. Problem is, or at least I think it is, is that my integration for the moment yielded mR^2, which doesn't jive well with me. Am I doing something wrong. In the integral, I took out the R^2 because the distance was the same for all mass elements, so the integral became (after some substitution) the integral of dTheita from 0 to pi. Am I missing something.

As for the other two, I know the Period of a physical pendelum is given by the equation T=2pi*sqrrt(I/mgh), but once again I feel that that isn't the whole story. Can someone ;ead me in the right direction? Thanks a bunch

2. Jan 22, 2006

### siddharth

Let us start with the first part of the question. Your approach is very correct. If you find the moment of Inertia of the semi-circular ring about the geometric center, you can use the parallel axis theorem to find the moment of Inertia about the center of mass. I get mr^2 for the Moment of Inertia about the geometric center as well.

For the second part, when you displace the ring by a small angle $\theta$, the ring will oscillate about the point of contact with the surface.
So what you have to figure out is
(i) What provides the necessary restoring torque for the ring to oscillate?
(ii) What is the moment of Inertia of the ring about the point of contact with the surface?
Hint: You can use the parallel axis theorem again to figure this out.

Once you do this, you will be able to calculate the restoring torque as a function of $\theta$ (For small angles, approx $\sin \theta$ as $\theta$) and this will give you the time period of small oscillations.

Last edited: Jan 22, 2006
3. Jan 26, 2006

### DaMastaofFisix

Ah thank you very much. Though I understand everything that you explained to me, I still don't see how I can translate that into mathematical equations. I think that the restoringtorque has to be the frictional force, because its the only other force besides the weight, which I'm pretty sure produces no torque. So now the issue is how does that restoring force behave (in other words, how does it vary with the angle) and other considerations that are essential.

If y'all have any insightful clues to lead me in the right direction, that would be greatly appreciated. And by the way, I recalculated the moment of inertia about the center of mass and using the first step that I used earlier did indeed yield the correct result. Thanks for the check and anyone brave enough to help a physics student in distress

4. Jan 31, 2006

### DaMastaofFisix

OKay I know his isn't the most interseting problem but I'm still trying to work it out, weeks after I started looking at it and I still can't figure out how to determine the restoring torque as a function of theita. Is it simply the frictional force times the tiny angular displacement ( and ah yes,the Radius of the center of mass)? I'm frikkin out here and I need someone to guide me in the right direction. Thanks for any help.

5. Feb 11, 2006

### siddharth

Anyway, When the ring is displaced slightly by an angle $$\theta$$, it will oscillate about the point of contact with the surface. What does this mean? It means that, if you draw the Free Body diagram for the ring, you will notice that the torque due to friction about the point of contact is 0. So, the restoring torque has to be provided by weight.
To calculate it mathematically, you know that the COM of a semi-circle lies at $$\frac{2R}{\pi}$$ from the center. Now the angle between the line joining the center of the semi-circle to the Center of Mass, and the line joining the center of the semi-circle to the point of contact (which is directly below the center of the semi-circle) is the angle by which the ring was displaced, which is $$\theta$$.