Moment of inertia of an irregularly shaped object

[UrbanXrisis]

This problem describes one experimental method of determining the moment of inertia of an irregularly shaped objectt such as the payload for a satelite. http://home.earthlink.net/~suburban-xrisis/clip2.jpg shows a mass m suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. When the mass is released from rest, it descends through a distance h, acquiring a speed v. Show that the moment of inertia I of the equptment (including the turntable) is mr^2(2gh/v^2-1).

here's what I tried, by using this fromula I found in the book:
v=(\frac{2mgh}{m+\frac{I}{r^2}})^{1/2}

v^2(m+\frac{I}{r^2})=2mgh

v^2m+\frac{v^2I}{r^2}=2mgh

\frac{v^2I}{r^2}=2mgh-v^2m

v^2I=2ghmr^2-v^2mr^2

I=\frac{mr^2(2gh-v^2)}{v^2}

I'm not sure how to get the right answer.

 

[Nylex]

Divide each of the terms in the bracket by v^2.

 

[UrbanXrisis]

I=\frac{mr^2(2gh-v^2)}{v^2}

I=\frac{mr^2}{v^2}*(\frac{2gh}{v^2}-1)

I=\frac{2ghmr^2}{v^4}-\frac{mr^2}{v^2}

I=\frac{mr^22gh-mr^2v^2}{v^4}

I=mr^2\frac{2gh-v^2}{v^4}

I=mr^2\frac{2gh}{v^4}-\frac{1}{2}

This does not equal
mr^2\frac{2gh}{v^2}-1

what did I do wrong?

 

[mooshasta]

I=\frac{mr^2(2gh-v^2)}{v^2}

I=mr^2 * \frac{2gh-v^2}{v^2}

\frac{2gh-v^2}{v^2} = \frac{2gh}{v^2} - 1

I=mr^2(\frac{2gh}{v^2} - 1)