Conversation of energy to determine angular velocity

[deltaOmega]

Homework Statement

[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

Homework 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

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.

 

[SammyS]

Homework Statement

[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

Homework Equations

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

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 kg)(9.8 N/kg)(4 m)
PE = 117.6 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 J = \frac{1}{2} m_{bucket} v^2 + \frac{1}{2} I_{spool} \omega^2
117.6 J = \frac{1}{2} m_{bucket} (r \omega)^2 + \frac{1}{2} (\frac{1}{2} m_{spool} r^2) \omega^2
117.6 J = \frac{1}{2} (3.00 kg)(0.600 m)^2 \omega^2 + \frac{1}{2} ( \frac{1}{2} (5.00 kg)(0.600 m)^2) \omega^2
117.6 J = 0.54 \omega^2 + 0.162 \omega^2
117.6 J = 0.702 \omega^2
167.52 = \omega^2
12.9 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.

Hello deltaOmega. Welcome to PF !

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

 

[deltaOmega]

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

 

[SammyS]

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

So that's OK ?

What is (1/2)(1/2)(5)(0.6)² ?

It's not 0.162

 

[SteamKing]

I think you mean conservation of energy, not conversation.

 

[deltaOmega]

So that's OK ?

What is (1/2)(1/2)(5)(0.6)² ?

It's not 0.162

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.