# Pulley/Rotational Torque Problem

1. Mar 4, 2009

### mastermike707

1. The problem statement, all variables and given/known data
In Fig. 11-42, one block has a mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 81.4 cm in 2.59 s (without the cord slipping on the pulley).

(a) What is the magnitude of the block's acceleration?

(b) What is the tension in the part of the cord that supports the heavier block?

(c) What is the tension in the part of the cord that supports the lighter block?

(d) What is the magnitude of the pulley's angular acceleration?

(e) What is its rotational inertia?

3. The attempt at a solution

I tried doing .814=1/2a*2.59^2 to find a and it is wrong. :(

2. Mar 4, 2009

### Delphi51

It is a rotational motion problem because of the inertia of the pulley.
Use torque = (moment of inertia)*(angular acceleration)

3. Mar 4, 2009

### mastermike707

How do I find the magnitude of the block's acceleration?

I am pretty sure it has something to do with the heavier block falling .814 meters in 2.59 seconds

4. Mar 4, 2009

### Delphi51

Use an accelerated motion formula such as d = Vi*t + .5a*t^2