Two astronauts of equal mass 70 kg are holding on to opposite ends of a straight
ID: 2202425 • Letter: T
Question
Two astronauts of equal mass 70 kg are holding on to opposite ends of a straight rope in space. The length of the rope between them is 20 meters. They are rotating about an axis that passes through the center of mass midway between the two astronauts with angular velocity 1 radian/sec. One of the astronauts pulls on the rope, decreasing the distance separating them to 20/2 meters. (In the following, take the rope to be massless.)
a) What is the new angular velocity of the two astronauts?
b)How much work did the astronaut who pulled on the rope do?
Explanation / Answer
L = I* = I*v/r= where I = 2*(M*r^2)
So L = 2*M*v*r = 2*70kg*5.20m/s*10 = 7280kg-m^2/s
K = 1/2*I*^2 = 1/2*(2*M*r^2)*v^2/r^2 = M*v^2 = 70.0kg*(5.20m/s)^2 = 1892.8J
the work done is the change in K = 3.38x10^4 - 1892.8 = 3.19x10^4J