# Rotational Inertia

1. Oct 28, 2007

### ace214

A 15.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 9.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless.

(a) Determine the tension in the rope.
58.46681 N
(b) Determine the moment of inertia of the wheel.
kg·m2
(c) Determine the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
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The tension is correct. I'm stuck on the inertia. I'll use 't' for torque and 'T' for the tension.

t=Iar => I = t/ar = Fr/ar = T/a but this doesn't work. I also tried using the force of gravity and the sigma force in the above equation and none of them are correct.

Due tomorrow morning at 8:30 AM EST...... :-(

Last edited: Oct 28, 2007
2. Oct 28, 2007

### saket

I guess, the formula is t = Ia/r ... ain't it?

3. Oct 28, 2007

### ace214

........ Crap

Yeah that did it. Thanks.

So for anyone searching the formula becomes
I = Fr^2/a

Last edited: Oct 28, 2007