Tether rotation device in space problem

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 3K views
nhmllr
Messages
183
Reaction score
1

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.
 
Physics news on Phys.org
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.
 
gneill said:
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.
 
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 :smile: