- #1
Crush1986
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- 10
Homework Statement
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?[/B]
Homework Equations
[itex]K= \frac{1} {2} I \omega^2[/itex]
[itex]U= B \mu sin \theta[/itex]
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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 [itex]\omega[/itex] equal to [itex] 2\pi f [/itex].However, I'm getting a solution that is twice as big as the solution when I solve for [itex] I=2k/ \omega^2=(2\mu B)/(2\pi f)^2.[/itex]
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 [itex] \sqrt 2[/itex]? Like how you find the RMS values of sinusoidal functions? This would then make my omega equal to [itex]2*\sqrt 2*\pi*f[/itex]. 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 [itex]2*\pi \sqrt(I/ \mu * B)[/itex].
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 :(.
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