Two astronauts, each having a mass of 76.0 kg, are connected by a 11.0 m rope of

Question

Two astronauts, each having a mass of 76.0 kg, are connected by a 11.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.60m/s.

(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum.


(b) Calculate the rotational energy of the system.


(c) By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. What is the new angular momentum of the system?


(d) What are the astronauts' new speeds?


(e) What is the new rotational energy of the system?


(f) How much work does the astronaut do in shortening the rope?

Explanation / Answer

a)

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

So L = 2*M*v*r = 2*76kg*4.60m/s*5.5 = 3845.6kg-m^2/s

b)

K = 1/2*I*^2 = 1/2*(2*M*r^2)*v^2/r^2 = M*v^2 = 85.0kg*(4.60m/s)^2 = 1798.6J

c)

I remains constant since there is no outside torque to the system L = 3845.6kg-m^2/s

e)

Now I = 2*M*r^2 = 2*76*2.75^2 = 1134.3kg-m^2 previously I = 2*75*5.5^2 = 4537.5

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

So K = 1/2*I*^2 = 1/2*1134.3*(4*4.60/2.75)^2 = 2.53x10^4J

f) the work done is the change in K = 2.53x10^4 - 1798.6 = 2.35x10^4J

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