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## Homework Statement

Two men with masses 70 kg and 120 kg rotate at 1 rpm on a frictionless surface and are attached by a 15 m rope.

If they pull the rope so that only 10 m is between them when they rotate, how long does it take to make 1 revolution?

## Homework Equations

Angular Momentum and Inertia

## The Attempt at a Solution

So I found the center of masses for when the distances are 15 m apart and 10 m apart. I assume that angular momentum is conserved since there's nothing that really seems to affect it when I read it.

So L = L

Then Inertia = sum mr^2

Hence (m1r1^2 + m2r2^2)w1 = (m1r3^2 + m2r4^2)w2 where r1, r2, are distances from the center of mass for the 15 m, and r3 and r4 are the distances from the center of mass for the 10 m distance.

I1w1 = I2w2; I'm kind of confused by most of the radian conversions and such.

For what I got I1 = 9947.4kgm^2 and I2 = 4420.8kgm^2

Putting this together I got 9947(2pi rad/60s) = 4421(w2) then w2 = 2 .3rad/s. I assume I did something wrong, also I can't figure how I would convert it to 1 revolution, would I divide it by 2pi*r?

I did try it however it should be 40 seconds given as the answer in my book which I did not get

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