Tether rotation device in space problem

[nhmllr]

Homework Statement

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 2ω. What is the new kinetic energy of the device?

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

Homework Equations

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

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.

 

[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.

 

[nhmllr]

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.

 

[gneill]

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

 

[Nickie]

It seems that the concept of this spaceborne energy storage device is based on the conservation of angular momentum. As the tether is reeled in, the radius of the device decreases, causing an increase in angular velocity. This increase in angular velocity results in an increase in kinetic energy, which is represented by the equation 1/2*m(ωr)^2. Since the device initially has kinetic energy E and rotates at angular velocity ω, the equation would be 1/2*m(ωr)^2 = E. When the device rotates at angular velocity 2ω, the new kinetic energy would be 1/2*m(2ωr)^2 = 4E. This shows that the new kinetic energy is indeed 2E, as stated in the problem. The potential energy is not being converted, but rather the kinetic energy is increasing due to the change in angular velocity.