# Homework Help: Moment of Inertia Compass Needle

1. Nov 22, 2014

### Crush1986

1. The problem statement, all variables and given/known data
You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is 0.321 Hz. Earth's magnetic field at the location of the compass has a horizontal component of 18.0*10^-6 T. The needle has a magnetic moment of 0.680 mJ/T. What is the needle's rotational inertia bout it's (vertical) axis of rotation?

2. Relevant equations
$K= \frac{1} {2} I \omega^2$
$U= B \mu sin \theta$

3. The attempt at a solution

I set the max kinetic energy to the max potential energy (since there is no damping they should be equal i believe). And I set $\omega$ equal to $2\pi f$.However, I'm getting a solution that is twice as big as the solution when I solve for $I=2k/ \omega^2=(2\mu B)/(2\pi f)^2.$

I was thinking that maybe the omega I'm working with is the average, and that I want the max? Since omega changes like a sinusoidal function maybe I should multiply my omega by $\sqrt 2$? Like how you find the RMS values of sinusoidal functions? This would then make my omega equal to $2*\sqrt 2*\pi*f$. I would then get the correct answer.

Is all that completely silly though and I'm just doing the problem flat wrong? A solution manual that I have solved the problem by equating the period to $2*\pi \sqrt(I/ \mu * B)$.

Any help would be greatly appreciated. This problem is driving me nuts because my method to me just makes so much sense, but apparently my first shot was wrong, and I'm not sure if I've corrected it properly or if the whole thing is just junk :(.

Last edited: Nov 22, 2014
2. Nov 22, 2014

### haruspex

I think you are confusing ω as a rate of change of angle with ω as denoting an oscillation frequency. Consider this simple SHM solution for the variation of some angle: $\theta = A \sin(\omega t)$. Differentiating, $\dot \theta = A\omega \cos(\omega t)$. Writing that as $\omega = A\omega \cos(\omega t)$ would be to confuse two quite separate quantities.

3. Nov 22, 2014

### Crush1986

Sooo... It can't be done the way I wanted to do it at all? I can't find the angular speed somehow and solve the problem like this? I mean is there anyway to get the angular speed from the info given or no?

Last edited: Nov 22, 2014
4. Nov 22, 2014

### haruspex

This much is valid
Just be careful how you express those items. I would start with a generic SHM equation, θ=Asin(ωt), and differentiate as necessary, being careful not confuse $\dot \theta$ with $\omega$.

5. Nov 22, 2014

### Crush1986

Oh ok. I must of typed something off a bit in wording. Sorry.

So is my logic right though? The frequency is .312 Hz so the average angular speed is about 1.9 rad/sec, I could multiply it by the square root of two to get the max angular velocity? It seems to work but I don't inow if I'm just lucky. I don't know if I have enough info to make an equation with a trig function since I don't know the amplitude.

Or is all my talk just faulty? This problem is delving up stuff from a past class and I may just be all off, haha.

Last edited: Nov 22, 2014
6. Nov 23, 2014

### haruspex

No, I mean be careful how you express the PE and KE algebraically.
Seems to me this is wrong:
The right hand side looks like the expression for torque, not energy.
After correcting that, try the following:
- assume a simple SHM behaviour
- write out the algebraic expressions for the PE and KE deduced from that formula
- determine the max value of each and equate them
You will need to make an approximation using the fact that theta is small.

7. Nov 23, 2014

### Crush1986

Ohhhh yah that is supposed to be a cos because it's a dot product.

I'm not sure about doing a small angle approximation though. Think I'll just have to make more sense of the solution in the manual and try to think like that. I guess this way wasn't as nice as I had originally thought haha...

8. Nov 23, 2014

### haruspex

No, it's a cross product. See http://en.wikipedia.org/wiki/Magnetic_moment#Definition. But the result is torque, not energy.
Suppose there's a torque T that succeeds in turning an object through a small angle d$\theta$. How much work is done?

[However, I note that one valid form for the energy is in terms of 1 - cos(theta). Can you provide a link for your equation?]

9. Nov 23, 2014

### Crush1986

I though $U = - \mu B cos \theta$ the dot product of mu and B?

Work done by torque through a very small angle? Is that torque times the change in angle? Assuming the change in angle is small enough I guess to consider the torque constant.

10. Nov 23, 2014

### haruspex

I can't find such an equation on the net, which is why I asked for a link. It's almost right, as you will see...
Yes. So if the torque is $B \mu \sin(\theta)$, what is the work done?

11. Nov 23, 2014

### Crush1986

Using a small angle approximation isn't it $B \mu \theta$ ? I'm just not seeing how I'm closer to an answer though if it's right.

12. Nov 23, 2014

### haruspex

The torque can be approximated to that, yes, but what is the work done for a small change in theta? Use what you wrote here:

13. Nov 23, 2014

### Crush1986

Oh I meant to write the work. You're right though I think I left out an extra theta?
$W= B \mu \theta^2$ ?

14. Nov 23, 2014

### haruspex

Almost. What's $\int \theta d\theta$?

15. Nov 23, 2014

### Crush1986

Oh dang it... My typing tonight at this late hour...

$W= \frac {1} {2} B \mu \theta^2$

16. Nov 23, 2014

### haruspex

Good.
So if theta obeys $\theta = A \sin(\omega t)$, what is the max of W? And what is the max of the KE?

17. Nov 23, 2014

### Crush1986

Well the max theta will ever be is A. Would both be equal to $1/2 B \mu A^2$? I felt before I could reach an answer like this. I wasn't sure how I'd get past the Amplitude though, I don't know what it is.

18. Nov 23, 2014

### haruspex

Yes, but you can compute the max KE another way. Differentiate the SHM equation to get the equation for $\dot \theta$. The A's will cancel eventually.

19. Nov 23, 2014

### Crush1986

Wow I just got it! Thx so much for your help and patience. I'm not usually this slow. No sleep, night shift work tonight... Not good for physics work haha. Cheers!