# Conversation of energy to determine angular velocity

• deltaOmega
In summary, using the conservation of energy, the angular speed of the spool shown in the figure can be determined to be 10.9 rad/s 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.
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.

Last edited:
deltaOmega said:

## 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.

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

deltaOmega said:
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

I think you mean conservation of energy, not conversation.

SammyS said:
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.

## What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. This means that the total energy in a closed system remains constant over time.

## How does conservation of energy relate to determining angular velocity?

In the context of determining angular velocity, the law of conservation of energy states that the total energy of a rotating object remains constant as long as there are no external forces acting on it. This means that the sum of the object's kinetic and potential energy remains the same at all points during its rotation.

## What factors affect the conservation of energy in determining angular velocity?

The main factors that affect the conservation of energy in determining angular velocity are the mass and shape of the rotating object, as well as the applied torque or force. In addition, the presence of any non-conservative forces such as friction can also impact the conservation of energy.

## How is the conservation of energy used to calculate angular velocity?

To calculate angular velocity using the law of conservation of energy, the initial and final positions of the rotating object must be known. The initial energy of the object is then equated to the final energy, taking into account any changes in potential energy and any external work done on the object. This equation can then be solved for the angular velocity.

## What are some real-life applications of conservation of energy in determining angular velocity?

Some real-life applications of conservation of energy in determining angular velocity include analyzing the motion of objects such as spinning tops, bicycles, and satellites. It is also used in engineering and design to ensure the safe and efficient operation of rotating machinery.

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