Rotational Energy and Pulley System

[GoldShadow]

1. The problem statement, all variables and given/known data
Problem reads:
The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kgm^2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.
http://img.photobucket.com/albums/v196/goldshadow/YF-09-36.jpg


2. Relevant equations
E=\frac{1}{2}I\omega^{2}+m_{1}gh_{1}+m_{2}gh_{2}

E_{initial}=E_{final}

v=\omega r


3. The attempt at a solution
I said m_{1} was the 4.00 kg block and m_{2} was the 2.00 kg block. Setting the initial and final energies of the system equal, I got:

m_{1}gh_{10}+m_{2}gh_{20}+\frac{1}{2}I\omega^{2}_{ 0}=m_{1}gh_{1}+m_{2}gh_{2}+\frac{1}{2}I\omega^{2}

Since initial angular momentum is zero and so is the height of block 2, and in the final state, height of block 2 is zero, this simplifies to:

m_{1}gh_{10}=m_{2}gh_{2}+\frac{1}{2}I\omega^{2}

Plugging in numbers:
(4.00)(9.8)(5.00)=(2.00)(9.8)(5.00)+\frac{1}{2}(0. 480)\omega^{2}

I got \omega=20.21 rad/s. Then using v=\omega r I just plugged in the radius and the angular velocity I just found to get v=3.23 m/s

This is not the right answer according to the homework website unfortunately... help would be appreciated!

[Dick]

You are forgetting that not all of the kinetic energy is in the pulley. The blocks have some too.

[GoldShadow]

Alright, I added the kinetic energy into the equation (kept the left side the same since kinetic energy is 0 for both blocks when they're at rest) and solved for v:

m_{1}gh_{10}=m_{2}gh_{2}+\frac{1}{2}I\omega^{2}+\f rac{1}{2}m_{1}v^{2}+\frac{1}{2}m_{2}v^{2}

Using v=\omega r and isolating \omega and substituting that into the equation, I got 5.71 m/s.

But it still says this is wrong... how do I go about getting the right answer?

[Dick]

You are doing everything right and your presentation of this problem is super clear, good job. But if I put the numbers into your equation I don't get what you get for v. Are you just punching in numbers wrong?

[GoldShadow]

Ah alright, I got it (v=2.81 m/s)... it turns out I did punch in a wrong number somewhere, I hate it when I do that, heh.

Anyway, thank you for the help Dick, I really appreciate it!

[Dick]

That's what I get as well. Like I say, I wish all posters presented stuff this well.