Angular velcoity of point masses

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Homework Statement



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?

Homework Equations





The Attempt at a Solution



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

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I presume this is meant to be a conservation problem. What's conserved, assuming no external forces act?
 
how do you know that? When do you know that you are dealing with a conservation problem??
 
mizzy said:
how do you know that? When do you know that you are dealing with a conservation problem??
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.
 
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?
 
mizzy said:
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?
Exactly.
 
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?
 
Aeroneer said:
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?
How about you let the OP solve the problem? (And your solution is incorrect.)
 
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??
 
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.
 
mizzy said:
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??
Perfectly correct.
 
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...
 
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