Tether rotation device in space problem

AI Thread Summary
The discussion revolves around a spaceborne energy storage device with two equal masses connected by a tether, rotating about their center of mass. When the tether is reeled in, the radius decreases, causing an increase in angular velocity due to the conservation of angular momentum. The kinetic energy formula shows that while the product of angular velocity and radius remains constant, the kinetic energy actually increases as the angular velocity doubles. The key point is that the new kinetic energy becomes 2E when the angular velocity reaches 2ω. The confusion arises from understanding how the energy storage mechanism operates during this process.
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 . 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:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top