Two astronauts (m = 75 kg) are connected by a 10.0 m rope of negligible mass. Th

Question

Two astronauts (m = 75 kg) are connected by a 10.0 m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.00 m/s. Treating the astronauts as particles, calculate:

The magnitude of the angular momentum

The rotational energy of the system

By pulling the rope, the astronauts shorten the distance between them to 5.00 m.

What is the new angular momentum of the system?

What are their new speeds?

What is the new rotational energy of the system?

How much work is done in shortening the rope?

Explanation / Answer

m =75 kg , v =5 m/s

a) L = I* = I*v/r= where I = 2*(m*r^2)

So L = 2*m*v*r = 2*75*5*5 = 3750 kg-m^2/s

b) K = 1/2*I*^2 = 1/2*(2*M*r^2)*v^2/r^2 = M*v^2 = 75*5*5 = 1875 J

c) I remains constant since there is no outside torque to the system L = 3750 kg-m^2/s

e) From conservation of momentum m1v1r1 = m2v2r2

v2 = v1r1r2 = 5*5/2.5 = 10 m/s

f) Now I = 2*M*r^2 = 2*75*2.5*2.5 = 937.5 kg-m^2 previously I = 2*75*5*5 = 3750

Since I is reduced by a factor of 4 is increased by a factor of 4

So K = 1/2*I*^2 = 0.5*937.5*(4*5/2.5)^2 = 3*10^4 J

g) the work done is the change in K = 3*10^4 - 1875 = 2.8*10^4J

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