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## Homework Statement

A perfectly solid marble of radius R rolls without slipping at the bottom of a spherical bowl of a radius 6R. The marble rolls back and forth in the vertical plane executing simple harmonic motion close to the lowest point. How long does it take the marble to go down one side and up the other from its maximum height to its maximum height?

a) Cannot be determined without knowing amplitude

b) ##\pi \sqrt{\frac{7R}{g}}##

c) ##\pi \sqrt{\frac{7R}{2g}}##

d) ##\pi \sqrt{\frac{7R}{5g}}##

e) ##\pi \sqrt{\frac{5R}{7g}}##

## Homework Equations

##\tau_{net} = I\alpha##

##I = 2/5 mR^2## (for sphere)

##F = ma##

##\tau = rFsin\theta##

## The Attempt at a Solution

I got answer choice d, but the key says b is correct.

I set up ##\tau_{net} = I\alpha = -fR## where ##f## is the static friction force. I believe gravity does not contribute to the torque, since it acts at the center-of-mass; neither does the normal force since it is always antiparallel to the radius vector from the center of the marble to the point of application. That only leaves the static friction force.

Then, I know that ##F = ma = -mgsin\theta + f \approx -mg\theta + f## tangent to the bowl, using the sine approximation for small theta.

Plugging back into torque net, I get that ##(2/5 mR^2)\alpha = -(ma + mg\theta)R##. Divide by an ##mR## and simplify to get ##7/5 R\alpha = -g\theta##. This means that ##\omega = \sqrt{\frac{5g}{7R}}##. Finally I realize that the desired time is half the period and ##T = 2\pi / \omega## so the desired time is ##\pi \sqrt{\frac{7R}{5g}}##, or at least that's what I got.

Where did I go wrong?