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?
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