Two astronauts (figure), each having a mass of 76.0 kg, are connected by a rope

Question

Two astronauts (figure), each having a mass of 76.0 kg, are connected by a

rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.10 m/s.

Two astronauts (figure), each having a mass of 76.0 kg, are connected by a d = 9.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.10 m/s. Treating the astronauts as particles, calculate the magnitude of the angular momentum of the two-astronaut system. Calculate the rotational energy of the system. You know the speed and mass of each astronaut. How do you calculate the kinetic energy? Does it matter that the motion is circular? kJ By pulling on the rope, one astronaut shortens the distance between them to 5.00 m. What is the new angular momentum of the system? There are no external torques applied here. What quantity is conserved in that case? kg · m2/s What are the astronauts' new speeds? What is the new rotational energy of the system? How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope?

Explanation / Answer

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

So L = 2*M*v*r = 2*88.5kg*5.20m/s*5.0 = 4602kg-m^2/s

b) K = 1/2*I*?^2 = 1/2*(2*M*r^2)*v^2/r^2 = M*v^2 = 85.0kg*(5.20m/s)^2 = 2393J

c) I remains constant since there is no outside torque to the system L = 4602kg-m^2/s

e) Now I = 2*M*r^2 = 2*88.5*2.5^2 = 1106kg-m^2 previously I = 2*88.6*5^2 = 4425

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

So K = 1/2*I*?^2 = 1/2*1106*(4*5.2/2.5)^2 = 3.83x10^4J

f) the work done is the change in K = 3.38x10^4 - 2393 = 3.59x10^4J

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