Register to reply

Tether rotation device in space problem

by nhmllr
Tags: device, rotation, space, tether
Share this thread:
nhmllr
#1
Jan15-13, 01:18 PM
P: 184
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.
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
gneill
#2
Jan15-13, 02:44 PM
Mentor
P: 11,621
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
#3
Jan15-13, 08:10 PM
P: 184
Quote Quote by gneill View Post
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
#4
Jan15-13, 08:43 PM
Mentor
P: 11,621
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


Register to reply

Related Discussions
Think of a device that can transfer a payload to a tether! General Physics 13