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## Homework Statement

Hi,

this is supposed to be an easy question, but for some reason I can't get it to work. The question is:

A weightless rod is hinged at O so that it can rotate without friction in a vertical plane. A mass m is attached to the end of the rod A, which is balanced vertically above O. At time t = 0, the rod moves away from the vertical with negligible initial angular velocity. Prove that the mass first reaches the position under O at t = √(OA/g) ln (1 + √2).

## The Attempt at a Solution

First let OA = r, and let \theta be the angle from the upward vertical. Now, from conservation of energy:

[tex]\frac{1}{2}m r^2 \dot{\theta}^2 = mgr(1-\cos \theta)[/tex]

This is a seperable equation, and after rearranging we get:

[tex]dt = \sqrt{\frac{r}{2g(1-\cos\theta)}} d\theta[/tex]

This is where things start to go wrong. This integral diverges when I try to take it from theta=0. Why? Is it because I'm starting with zero velocity?

Also, the result of the indefinite integral is (using mathematica):

[tex]

t(\theta) = \sqrt{\frac{2r}{g}} \ln \left( \tan\frac{\theta}{4}\right).[/tex]

I'm trying to think what kind of angles I'm supposed to put in there to get the given result, but I can't work it out.

Thanks in advance,

Lior