# Torsion spring problem

by freshcoast
Tags: spring, torsion
 P: 166 1. Problem statement. 2. Relative formulas http://en.m.wikipedia.org/wiki/Torsion_spring 3. Attempt. Part a) Since I am given the arc length S and radius, I use the equation s = r(theta) and I solve for theta, with r being the length of the rod and the addition of the radius of the ball. Now I need to find torque, which is given since I know the radius and the force. Now applying the equation t = k(theta) , I just solve for k. Part b) For this part I need to find the center of mass of the solid sphere (2/5mr^2) and I added distance from the axis of rotation by parallel axis theorem (ML^2) Now I can find the period by using the equation 2(pi) * sqrt(I/k) since I have I and K. Part c) I'm thinking the work done by bending the spring would just be the equation 1/2k(theta^2)
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P: 9,839
 Quote by freshcoast Part b) For this part I need to find the center of mass of the solid sphere (2/5mr^2) and I added distance from the axis of rotation by parallel axis theorem (ML^2)
Be careful here. When the sphere has moved through an arc of θ, it will have rotated about its centre by rather more. This is because the rod is not a rigid rod sprung at its base.
P: 166
 Quote by haruspex Be careful here. When the sphere has moved through an arc of θ, it will have rotated about its centre by rather more. This is because the rod is not a rigid rod sprung at its base.
Hmm, I am having trouble understanding this. The Moment of inertia of a solid sphere is (2/5 * MR^2) and you're saying that since the rod it is connected to is not a rigid rod, it would have rotated about its center more? I don't see what changes in my attempt, is the parallel axis theorem still valid? I think the ML^2 is for solid cylinders, is that the value that need's to be modified?

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