# Conversation of energy to determine angular velocity

1. Jan 14, 2013

### deltaOmega

1. The problem statement, all variables and given/known data
[From Serway College Physics; Ch 8, problem 52] Use conversation of energy to determine the angular speed of the spool shown in the figure after the 3.00 kg bucket has fallen 4.00 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

For variables:
$m_{bucket}$ = 3.00 kg
$m_{spool}$ = 5.00 kg
$r$ = 0.600 m

2. Relevant equations
$I_{spool} = \frac{1}{2} mr^2$
$PE = mgh$
$KE_{r} = \frac{1}{2} m \omega^2$
$PE_i + KE_i = PE_f + KE_f$
$v_t = r \omega$

3. The attempt at a solution
I've seen variants of this question all over the web, including this site, but none of them were helpful to me.

Obviously the only thing driving the bucket down is gravity, which will unwind our spool. So potential energy is:

$PE = m_{bucket} g h$
$PE = (3 \text{kg})(9.8 \text{N/kg})(4 \text{m})$
$PE = 117.6 \text{J}$

Since potential energy is unwinding the spool, then the spool's kinetic rotational energy ($KE_{r} = \frac{1}{2} m \omega^2$) should be equal to the potential energy ($PE_{bucket} = KE_{r}$).

I tried to derive the angular velocity from this with:
$117.6 \text{J} = \frac{1}{2} m_{bucket} v^2 + \frac{1}{2} I_{spool} \omega^2$
$117.6 \text{J} = \frac{1}{2} m_{bucket} (r \omega)^2 + \frac{1}{2} (\frac{1}{2} m_{spool} r^2) \omega^2$
$117.6 \text{J} = \frac{1}{2} (3.00 \text{kg})(0.600 \text{m})^2 \omega^2 + \frac{1}{2} ( \frac{1}{2} (5.00 \text{kg})(0.600 \text{m})^2) \omega^2$
$117.6 \text{J} = 0.54 \omega^2 + 0.162 \omega^2$
$117.6 \text{J} = 0.702 \omega^2$
$167.52 = \omega^2$
$12.9 \text{m/s} = \omega$

However, this cannot be right, as the question is multiple choice and only offers the following answers: 7.12, 8.66, 9.12, 10.9, or 11.8.

Help would be appreciated.

Last edited: Jan 14, 2013
2. Jan 14, 2013

### SammyS

Staff Emeritus
Hello deltaOmega. Welcome to PF !

It seems to me that your moment of inertia for the spool is incorrect.

3. Jan 14, 2013

### deltaOmega

Oops, it should be $\frac{1}{2} m r^2$, as used in the work later.

4. Jan 14, 2013

### SammyS

Staff Emeritus
So that's OK ?

What is (1/2)(1/2)(5)(0.6)2 ?

It's not 0.162

5. Jan 15, 2013

### SteamKing

Staff Emeritus
I think you mean conservation of energy, not conversation.

6. Jan 15, 2013

### deltaOmega

Well, that's embarassing. That ends up being 0.45, which changes the final answer to 10.9 rad/s, corresponding with one of the multiple choice answers. Thanks.