# Angular velcoity of point masses

1. Jul 29, 2010

### mizzy

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

The system of point masses m shown in the above figure is rotating at an angular velocity of 2rev/s. The masses are equal and connected by light flexible spokes that can be lengthened or shortened. What is the new angular velocity if the spokes are shortened from 1m to 0.5m?

2. Relevant equations

3. The attempt at a solution

Not sure how to approach this question. Someone please guide me.

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2. Jul 29, 2010

### Staff: Mentor

I presume this is meant to be a conservation problem. What's conserved, assuming no external forces act?

3. Jul 29, 2010

### mizzy

how do you know that? When do you know that you are dealing with a conservation problem??

4. Jul 29, 2010

### Staff: Mentor

That's just my guess (otherwise I don't see the point of the question). I'd say that the problem is very poorly worded.

5. Jul 29, 2010

### mizzy

From reading my text, I say that angular momentum is conserved.

Using the equation: I1 * omega1 = I2 * omega 2

To solve the equation, we have to solve for omega 2. Is that right?

6. Jul 29, 2010

### Staff: Mentor

Exactly.

7. Jul 29, 2010

### Aeroneer

its like a figure skater bringing in his/her arms. So the ratio between their circumferences (the revolutions) is 1/2. So 2*2 is 4 and so 4 is the new angular velocity.
Its been a while since I've done something like this. let me know if I'm right, and what I did wrong?

8. Jul 29, 2010

### Staff: Mentor

How about you let the OP solve the problem? (And your solution is incorrect.)

9. Jul 29, 2010

### mizzy

Here is my calculation:

I1 * omega1 = I2 * omega 2
therefore omega 2 = I1 * omega 1/ I2
= mr^2 *omega 1/ mr^2
= r^2 * omega 1/ r^2
= 1^2 * 2rev/s/ 0.5^2
= 8 rev/s

so when the spokes are shortened, the velocity increases. am i correct??

10. Jul 29, 2010

### glennpagano

I just finished my first physics course. You know that it is a conservation problem from doing many different problems and realizing the type of problem it is. You just have to put conservation of momentum in your tool belt. No to solve it think of Momentum before the length change will be equal to the length afterward. I believe you would use the angular momentum equation to do this. Using I as mr^2.

11. Jul 29, 2010

### Staff: Mentor

Perfectly correct.

12. Jul 29, 2010

### mizzy

Thanks Doc Al!

13. Jul 29, 2010

### Aeroneer

hahah! oh, did I forget about the moment of inertia?
Thanks for the heads up, Doc.
And yes, you're right, Mizzy...

EDIT: sry for the late posts... kind of distracted at the moment...

Last edited: Jul 29, 2010