# Homework Help: Angular velocity and moment of inertia

1. Apr 7, 2009

### kikidoll

Could you let me know if my answers are correct?

1. The problem statement, all variables and given/known data
A light rope is wrapped several times around a large wheel with a radius of 0.400m. The wheel rotates in frictionless bearings about a stationary horizontal axis. The free end of the rope is tied to a suitcase with a mass of 0.15kg. The suitcase is released from rest at a height of 4.00m above the ground. The suitcase has a speed of 3.50 m/s when it reaches the ground.
Calculate:
A) the angular velocity of the wheel when it reaches the ground and
B) the moment of inertia of the wheel

2. Relevant equations
A) angular velocity (w) = speed/radius
B) mgh = 1/2 mv2 + 1/2Iw2

3. The attempt at a solution

After using these equations I found angular velocity to be 8.75 rad/s and the moment of inertia to be 12.9 kg*m2.

Correct or incorrect? Thanks for you help.

2. Apr 7, 2009

### LowlyPion

With ω = v/r then 8.75 looks ok.

But your I looks a little large.

m*g*h is total energy to start with dividing by ω2 where ω2 looks a lot bigger than m*g*h doesn't look right.

3. Apr 7, 2009

### kikidoll

Thanks for the reply. I also though I was rather large, but with my calculations that's what I ended up getting... strange.

4. Apr 7, 2009

### LowlyPion

I think you've done something wrong.

Maybe show your numbers and how you got there.

5. Apr 7, 2009

### kikidoll

A) w = v/r = (3.50)/(0.400) = 8.75 rad/s

B) Ki + Ui = Kf + Uf
Ki and Uf = 0

Ui = Kf
mgh = 1/2 mv2 + 1/2Iw2
(15)(9.8)(4) = 1/2 (15)(3.5)2 + 1/2 (I) (8.75)2
588 = 91.9 + 38.3(I)
496.1 = 38.3(I)
I = 12.9 kgm2

6. Apr 7, 2009