Does Rolling a Sphere in a Cylindrical Trough Exhibit Simple Harmonic Motion?

[tsw99]

SHM on a cylindrical trough (solved?)

Homework Statement

A solid sphere of radius R rolls without slipping in a cylindrical trough of radius 5R. Show for small displacement, the sphere executes SHM with period $$T=2\pi \sqrt{\frac{28R}{5g}}$$

[PLAIN]http://img299.imageshack.us/img299/8026/shmt.jpg

Homework Equations

The Attempt at a Solution

since it is rolling without slipping on the cylindrical trough, so $$\frac{ds}{dt}=4R\frac{d\theta}{dt}$$
By 2nd law, the net force in tangential direction is $$mgsin\theta$$
consider translational motion, $$-mgsin\theta=m\frac{d^{2}s}{dt^{2}}$$

Edit:
I see the problem of the torque equation, the CM must always have zero angular velocity,
so the torque equation for the point touching with the trough, $$-mgsin\theta (R)=(\frac{2}{5}MR^{2}+MR^{2})\alpha$$
where $$\alpha$$ is the angular acceleration of that point $$\alpha=\frac{a}{R}$$ where a is the acceleration of CM, which then leads to $$-\frac{gsin\theta}{R}=4\ddot{\theta}$$
So I have $$-mgRsin\theta=\frac{28MR^{2}\ddot{\theta}}{5}$$ which leads to the correct expression?

Are there any mistakes I made? thanks

 

[anathema]

Hello,

Great job on solving the problem! Your approach seems correct and your final expression for the period T is also correct.

However, there is a small mistake in your solution. The net torque equation that you used (-mgsin\theta (R)=(\frac{2}{5}MR^{2}+MR^{2})\alpha) is actually for a solid sphere rolling without slipping on a flat surface. In this problem, the sphere is rolling without slipping in a cylindrical trough, so the torque equation will be slightly different.

The net torque equation for this problem would be:

-Tsin\theta R=(\frac{2}{5}MR^{2}+MR^{2})\alpha

Where T is the tension in the string and \alpha is the angular acceleration of the sphere.

Substituting this into your expression, we get:

-\frac{TRsin\theta}{MR}=\frac{28R\ddot{\theta}}{5}

Solving for T, we get:

T=\frac{28MR}{5sin\theta}\ddot{\theta}

Now, we know that for small displacements, \theta is very small, so we can approximate sin\theta \approx \theta.

Substituting this into our expression for T, we get:

T=\frac{28MR}{5\theta}\ddot{\theta}

Now, we also know that for small displacements, the angular acceleration \ddot{\theta} is directly proportional to the displacement \theta, so we can write:

\ddot{\theta}=-\omega^{2}\theta

Where \omega is the angular frequency.

Substituting this into our expression for T, we get:

T=\frac{28MR}{5\theta}(-\omega^{2}\theta)

Simplifying this, we get:

T=2\pi \sqrt{\frac{28R}{5g}}

Which is the correct expression for the period T.

So, in conclusion, your approach was correct, but there was a small mistake in your torque equation. Keep up the good work!