Why does angular velocity change if moment of inertia stays constant?

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shinystar
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Homework Statement



A skater extends her arms horizontally, holding a 5kg mass in each hand. She is rotating about a vertical axis with an angular velocity of 1rev/s. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg.m^2, and the distance of the masses from the axis changes from 1 m to 0.1m?


Homework Equations



I tried using the conservation of momentum. I1w1=I2w2. But since the qtn states that moment of inertia is constant, isn't w1=w2?


The Attempt at a Solution



The answer is 3 rev/s but I can't seem to get it.

Hope to find help
Thank you
 
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The moment of inertia of the skater's body remains constant, but don't neglect the overall change in inertia, since she changes the position of the two masses.
 
oh ok..

but I still can't seem to solve it..

I1w1 + inertia of skater's body*w1 = inertia of skater's body*w2 + I2w2

and I = mr^2

so:

5 (1)^2 * 1 + 5*1 = 5*w2 + 5 (0.1)^2 * w2

5.05w2= 10
w2 = 1.98 rev/s

have I made any mistakes again?
 
Your nearly there, but don't forget she's holding one mass in each hand ... :wink:
 
shinystar said:
oh ya..

Thanks for your help! =)

My pleasure :smile: