# Moment of inertia of a pulley

1. Nov 20, 2012

### Lotus93

1. The problem statement, all variables and given/known data
A 1.3 kg block is tied to a string that is wrapped around the rim of a pulley of radius 7.2cm. The mass of the pulley is 0.31kg. The block is released from rest. If the velocity of the block is 3.0m/s after it falls 0.5m, what is the moment of inertia for the pulley?[/b]

2. Relevant equations

I = (1/2mv^2)/(angular velocity)^2
angular velocity = v/R

3. The attempt at a solution

I= [(1/2)(1.3)(3.0)^2]/[(3.0/7.2)^2]
I= 3.8025/0.1736
I= 21.90

I know this answer doesn't make sense, but I'm not sure what I'm missing. This was a test question I got wrong, so I'm not concerned about the answer, I just want to make sure I understand how to go about solving it to prepare for my final. Any help or guidance is greatly appreciated.

2. Nov 20, 2012

### Staff: Mentor

Looks like you were attempting to equate the KE of the falling mass with the KE of the pulley. No obvious reason to think that is true.

Instead, use conservation of total mechanical energy. Total mechanical energy = rotational KE of pulley + translational KE of falling mass + gravitational PE.

3. Nov 20, 2012

### Simon Bridge

Welcome to PF;
What makes you think the answer makes no sense?

Did you try isolating the pulley and the mass and using free-body diagrams?
(Or cons. of energy...)

4. Nov 24, 2012

### Lotus93

I spoke with one of my classmates, and this is what we worked out...

v^2 = vo^2 + 2ay
3^2 = 2(a)(0.5)
a = 9m/s

Net Torque = I * alpha
a = alpha * r
Trsin90 = I (a/r)
T(0,072)= I (9/0.072)

F = ma
T - mg = -ma
T = mg - ma
T = 1.3(9.8) - 1.3(9)
T = 1.04
(1.04)(0,072)= I (9/0.072)
0.07488 = 125I

I = 5.99e-4 kg m^2

Can anyone verify, or if we made an error point out where we went wrong?

5. Nov 24, 2012

### Staff: Mentor

Looks good to me. (Just for fun, try solving it using conservation of energy.)

6. Nov 24, 2012

### Simon Bridge

So... did you figure out what you were missing?