Calculating Final Angular Velocity | Skater and Rotational Motion [Help]

[shinystar]

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

 

[Hootenanny]

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.

 

[shinystar]

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?

 

[Hootenanny]

Your nearly there, but don't forget she's holding one mass in each hand ...

 

[shinystar]

oh ya..

Thanks for your help! =)

 

[Hootenanny]

oh ya..

Thanks for your help! =)

My pleasure