# Rotational Energy and Pulley System

1. Oct 24, 2007

1. The problem statement, all variables and given/known data
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.

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!

Last edited: Oct 24, 2007
2. Oct 24, 2007

### Dick

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

3. Oct 24, 2007

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:

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

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?

4. Oct 24, 2007

### 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?

5. Oct 24, 2007

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!

6. Oct 24, 2007

### Dick

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