# Rotation angular velocity of a turntable

1. Nov 21, 2014

### lilyE

I've spent at least an hour trying to figure this out, but can't seem to figure out how to solve this.

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

A 170g , 30.0-cm-diameter turntable rotates on frictionless bearings at 66.0rpm . A 25.0g block sits at the center of the turntable. A compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.

What is the turntable's rotation angular velocity when the block reaches the outer edge?

2. Relevant equations
Conservation of momentum: Ii*wi = If*wf

3. The attempt at a solution
I(turntable) = Mr^2 = (.170 kg)(.15 m)^2 = 3.825*10^-3 kg m^2
I(block)o = mr^2 = 0
I(block)f = mr^2 = (.025 kg)(.15 m)^2 = 5.265*10^-4 kg m^2
wi = ((66 rpm)(2pi))/60 = 6.91 rad/s
I(turntable+block)f = 3.825*10^-3 + 5.265*10^-4 = 4.3875*10^-3 kg m^2

(3.825*10^-3)(6.91) = (4.3875*10^-3)wf
0.1826 = (4.3875*10^-3)wf
wf = 6.02 rad/s -- wrong

Initially, I tried using conservation of energy, and ended up with 5.52 rad/s. The program told me "not quite" and suggested a rounding error, so I think 5.52 might be close to the right answer, but I can't seem to figure out why I'm not getting it. Any help would be much appreciated!

Last edited: Nov 21, 2014
2. Nov 21, 2014

### mo_0820

I think you should recheck the equation of $I_{turntable}$.

3. Nov 21, 2014

### lilyE

Would it just be 1/2*(.170)(.15)? That's what I had at first, but I wasn't sure if that was correct or not.

4. Nov 21, 2014

### lilyE

If I use Iturntable = (0.5)(0.17)(0.15)2, I get Iturntable = 1.9125 * 10-3 kg m2. Plugging this in, I get a final answer of 5.42 rad/s. Does this seem right?

5. Nov 21, 2014

### mo_0820

Believe yourself.

6. Nov 21, 2014

### mo_0820

By the way, due to the error generated by calculating, recommend you to calculate at the last pace.
You can get the best result.

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