# Moment of inertia of an irregularly shaped object

1. Feb 24, 2005

### 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. Picture 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.

Last edited: Feb 24, 2005
2. Feb 24, 2005

### Nylex

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

3. Feb 24, 2005

### 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?

4. Feb 24, 2005

### mooshasta

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

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

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

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