# Angular velocity of a rod

1. Feb 24, 2014

### BrownBoi7

The figure attached shows two masses held together by a thread on a rod that is rotating about its center with angular velocity w. If the thread breaks, the masses will slide out to the ends of the rod. How will that affect the rod's angular velocity. Will it increase, decrease, or remain unchanged? Explain.

My attempt:

w= V/R

Neither of the two variables change, wouldn't it remain unchanged?

Thanks!

#### Attached Files:

• ###### velocity.JPG
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2. Feb 24, 2014

### BvU

How do you know that neither variable changes? What do you consider a conserved quantity, and why ?

3. Feb 24, 2014

### BrownBoi7

Radius remains the same. Velocity will change I guess. It has something to do with angular momentum may be?
Can you guide me here?

4. Feb 25, 2014

### BvU

Radius remains the same ? I read that the masses will slide to the ends of the rod ? What is it that we call radius here ?

5. Feb 25, 2014

### hilbert2

Hint: Rotational energy is conserved, but what happens to the moment of inertia of the system?

The rotational energy is $E=\frac{1}{2}I\omega^{2}$, where $I$ is the moment of inertia and $\omega$ the angular velocity.

EDIT: Just to be sure, let's rather say that angular momentum, $L=I\omega$ is the quantity that is conserved... The answer is same, anyway.

Last edited: Feb 25, 2014
6. Feb 25, 2014

### haruspex

Yes, you must use angular momentum here, not energy - for two reasons.
First, we don't know whether there is friction. Second, we don't know how much KE there is in the radial motion.