Mechanics question - circular motion

[tjackson3]

Homework Statement

A yo-yo of mass m , on a string of length \ell 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, \ell, \theta, and g.

Diagram:

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

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

Homework Equations

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

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 \ell. Then at some angle \theta, the yo-yo would be at a height of \ell\cos\theta (and have a corresponding potential energy of mg\ell\cos\theta). So the equation becomes

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

After some algebra, the answer comes out to be

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

(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

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

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

Thanks for your help!

 

[gabbagabbahey]

Diagram:

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

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 \ell. Then at some angle \theta, the yo-yo would be at a height of \ell\cos\theta (and have a corresponding potential energy of mg\ell\cos\theta). So the equation becomes

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

Looking at your diagram, when \theta=0, the mass will be at the bottom of the arc and should have zero potential energy (or at least less than it had at the top of the arc)...what does your formula for the potential energy give at \theta=0?

 

[tjackson3]

Then you'd get an angular velocity of zero, which now that you mention it, does not make sense. However, where does the mistake in my reasoning come in?

 

[gabbagabbahey]

Your expression for the potential energy is clearly incorrect...how did you derive it?

 

[tjackson3]

I just derived it from the expression for gravitational potential energy, using \ell\cos\theta as the height.

 

[gabbagabbahey]

\ell\cos\theta is the distance below the pivot...not the height

 

[tjackson3]

Ohhh haha, thank you! :)