## Tether rotation device in space problem

1. The problem statement, all variables and given/known data
A spaceborne energy storage device consists of two equal masses connected by a tether and rotating about their center of mass. Additional energy is stored by reeling in the tether; no external forces are applied. Initially the device has kinetic energy E and rotates at angular velocity ω. Energy is added until the device rotates at angular velocity . What is the new kinetic energy of the device?

(The answer is 2E but I don't see how)

2. Relevant equations
kinetic energy = 1/2*mv^2
momentum = mv = mωr
initial momentum = final momentum

3. The attempt at a solution
I don't see how this "energy storage" works. If I real the tether in, the radius r of the device decreases but the angular velocity ω of the device increases because of the conservation of momentum. The kinetic energy of the device is 1/2*m(ωr)^2, but the quantity ωr does not change. So I don't see how the potential energy of the device can be converted.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Homework Help ω and r both change when the device is reeled in. Call them ω1 and r1. The product of angular velocity and radius is what remains constant, so that ω1*r1 = ω*r.

 Quote by gneill ω and r both change when the device is reeled in. Call them ω1 and r1. The product of angular velocity and radius is what remains constant, so that ω1*r1 = ω*r.
Right. If ω1*r1 = ω*r, then the kinetic energy stays the same. I still don't understand what the problem is talking about with the "stored energy," because reeling in the tether doesn't affect the energy.

Recognitions:
Homework Help

## Tether rotation device in space problem

Ah. Sorry, I misspoke. Angular momentum is conserved, so it's Mωr2 that remains constant. Since M is the same in both cases, ωr2 is what you need to worry about. The square makes a difference