Finding the angular frequency of SHM of a rolling sphere

[Nemode]

Homework Statement

A uniform sphere of radius a is placed at the lowest point of a fixed thin
hemispherical bowl of radius b > a. The sphere is the slightly displaced and
released with zero initial velocity such that it rolls without slipping on the inner
surface of the bowl. By conservation of energy, or otherwise, show that the
sphere executes simple harmonic motion about the lowest point with angular
frequency.

sqrt(5g/(7(b-a)))

Homework Equations

KE = 1/2 mv^2

Angular KE = 1/2 I \theta\dot{}^{2}

dE/dt = 0

PE = mg ((b-a) - (b-a)cos(\theta))
= mg(b-a)(1 - cos(\theta))

I of sphere = 2/5 m (b-a)^{2}

The Attempt at a Solution

KE = 1/2 m (b-a)^{2} \theta\dot{}^{2}

Total E = \theta\dot{}^{2} (b-a)^{2}(1/2 + 1/5) + mg(b-a)(1 - cos(\theta))

dE/dt = 0 = \theta\ddot{}\theta\dot{} (b-a)^{2} * 7/5 + \theta g(b-a)


I'm stuck with how to deal with having both theta dot and double dot in the first term, if I ignore the theta dot it works and I get the answer stated abov, but obviously something's gone wrong somewhere or else I'm not dealing with the thetas correctly!

Thanks

 

[ehild]

The angle theta must be very small so you can replace cos theta with 1-theta^2/2.

You can consider theta as the coordinate and theta dot =omega as velocity. Look at the form of both the PE and KE. Like in case of SHM, PE is proportional to the square of coordinate, and KE is proportional to the square of velocity. They have the form PE=1/2*D (theta)^2, and KE=1/2 M (omega)^2. Find the equivalent mass M and the equivalent direction force (spring force) D.

ehild