(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A yo-yo of mass m , on a string of length [itex]\ell[/itex] is held horizontally and then released, as shown in the diagram. Assume that there is no friction and that the yo-yo swings freely in a vertical plane. Give your answers in terms of m, [itex]\ell[/itex], [itex]\theta[/itex], and g.

Diagram:

[PLAIN]http://img43.imageshack.us/img43/6735/physprob2.png [Broken]

The part I'm stuck on: What is the angular speed [itex]\omega = \dot{\theta}[/itex] as a function of [itex]\theta[/itex]?

2. Relevant equations

Kinetic energy: [tex](1/2)mv^2 = (1/2)m\omega^2\ell^2[/tex]

3. The attempt at a solution

The way I solved it was using conservation of energy. If we take the bottom of the arc to be zero, the yo-yo was dropped from a height [itex]\ell[/itex]. Then at some angle [itex]\theta[/itex], the yo-yo would be at a height of [itex]\ell\cos\theta[/itex] (and have a corresponding potential energy of [itex]mg\ell\cos\theta[/itex]). So the equation becomes

[tex](1/2)m\omega^2\ell^2 + mg\ell\cos\theta = mg\ell[/tex]

After some algebra, the answer comes out to be

[tex]\omega = \sqrt{ 2g(1-\cos\theta) /\ell}[/tex].

(for some reason, the LaTeX coding is being silly with the square root; however, the entire expression is supposed to be under the square root)

My professor does it differently. In his solution, he uses conservation of energy with the equation

[tex](1/2)m\omega^2\ell^2 = mg\ell\cos\theta[/tex]

and solves from there to get the same thing I got except with just [itex]\cos\theta[/itex] where I have [itex]1-\cos\theta[/itex]. What is wrong with my reasoning here?

Thanks for your help!

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# Homework Help: Mechanics question - circular motion

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