# Time for Revolution (Rotating about the Center of Mass)

## 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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Bystander
Homework Helper
Gold Member
1 revolution, would I divide it by 2pi*r?
2π radians is one full circle. Multiplying by r gives you the length of the arc, or circumference.

2π radians is one full circle. Multiplying by r gives you the length of the arc, or circumference.
Okay thanks I'll look into it.
Edit: yeah I actually meant 2pir/w instead. Mistyped

Bystander
Homework Helper
Gold Member
I meant 2rpi/w. However that doesn't yield anything
It yields m/s, which you are not interested in for this problem. Pay attention to units.
What do you mean by checking my arithmetic?
When I suggest that you check your arithmetic, it means you've made an error, and you should go through your work and find it.

haruspex