# Angular Freq. of small oscillations on a wheel/spring.

1. Dec 2, 2005

### nweibley

I've been busy finishing my online physics homework, and I cannot get this problem for the life of me (which is annoying because I just finished the relativity and lorentz transformation assignments). If you are good at physics and think you know how to do it, please post your line of thoughts on the matter. I've tried obvious variations of the kinematic equations of oscillation that I can think of, but no dice.

The question is below:

A wheel of mass M = 4.5 kg and radius R = 0.90 m is free to rotate about its fixed axle. A spring, with spring constant k = 270 N/m, is attached to one of its spokes, a distance r = 0.33 m from the axle, as shown in the figure. What is the angular frequency of small oscillations of this system (in rad/s)?

http://img225.imageshack.us/img225/1908/prob211yw.gif [Broken]

I was told to think of the disc as a torsional oscillator, but I had no luck doing that. If this is a pend problem I assumed (incorrectly) the axis cannot actually be the axle. It would seem to me there would only be two torques... the rotational inertia of the wheel (MR^2) and the spring. How would I calculate the torques cause by the spring though?

I thought I was to use the formula T = 2 pi * sqrt ( I / Mgd) but that obviously does not work in this case, since the wheel doesnt spin from gravity. I cannot figure out how to determine K the torque constant from the spring and substitute Mgd for K. I know angular freq = 2 pi / (T) but none the less, I only have one attempt to get this right, and I know that the way I was doing it was wrong ("practice" inputs).

Any help?

Thanks,
---Nate

Last edited by a moderator: May 2, 2017
2. Dec 2, 2005

### Staff: Mentor

Yes, it's a torsional oscillator. Why in the world would you think that the axis would be anything but the axle?
Rotational inertia is not a torque!
The only thing exerting a torque on the wheel is the spring. You should know how to find the torque based on the force exerted by the spring for a small displacement.

This is not a physical pendulum, so forget about that formula. Instead, compare this to a mass on a spring. I assume that you could find the angular frequency of an oscillating mass on a spring (the usual simple case). (If not, better read up on that problem first.) Now just figure out the analogous equations for rotational SHM, which deals with torques instead of forces. Here are two big hints:

(1) In the "mass on a spring" case, the force formula is F = kx. For this problem, figure out the corresponding torque formula. (Hint: Write the torque as a function of angular displacement. That way you'll find the torsional spring constant.)

(2) In the "mass on a spring" case, the mass is used to find the period of the SHM. For this problem, what's the rotational analog to mass?

Once you've figured out these two hints, you can translate the answer from the "mass on a spring" to work with this problem.

3. Dec 4, 2005

### nweibley

Thanks, that makes perfect sense.... I completed the problem and everything now makes sense.

I was just taking the wrong approach to the problem.

Thanks for the help!
-----Nate

4. May 4, 2009

### Eagle Eyes

Hello, i have no idea how to do this problem. Can someone PLEASE PLEASE explain me how to do this problem step by step.

I really need to know how to do this.