# Eneregy Conservation

1. Aug 7, 2008

### kaka2007

A 1.3kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.2cm. The block is released from rest. (a) Assuming the pulley is a uniform disk with a mass of 0.31kg, find the speed of the block after is has fallen through a height of 0.5m.

I can't seem to find the right answer (above). I set this equation below. I tried to use Ki+Ui=Kf+Uf with Ki and Uf being zero

Mass(of ball) x gravity x height = 1/2mv^2 x (1+ I/mr^2) with I = 3/2mr^2 and m= mass of pulley

2. Aug 7, 2008

### Hootenanny

Staff Emeritus
Are you sure that your expression for rotational kinetic energy is correct? What is the general expression for rotational kinetic energy? You also may want to check your moment of inertia for a disk.

3. Aug 7, 2008

### kaka2007

Appreciate the help. I figured something was wrong w/ that side of the equation.

So the general rotational equation is (I think) K = 1/2Iw^2. I assume I should replace w with v/r? Giving me K = 1/2Iv^2/r^2.

The book gives me two moments of inertia for a disk. One is 1/2MR^2 but there is an arrow at the middle; the other (which I used) is 3/2MR^2 which has the arrow at the end of the disk. I figured I use the latter since the string is probably on the side of the pulley?

4. Aug 7, 2008

### Staff: Mentor

No. The "arrow" represents the axis of rotation. (Moment of inertia varies with choice of axis.) In your problem, the axis of rotation is in the center of the pulley.