Two astronauts each having a mass of 75.0 kg are connected by a 10.0m rope of ne
ID: 1293310 • Letter: T
Question
Two astronauts each having a mass of 75.0 kg are connected by a 10.0m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.00m/sec. Treating the astronauts as particles
Calculate
a) the magnitude of the angular momentum
b) The rotational energy of the system. By pulling the rope, the astronauts shorten the distance between them to 5.00m
c) What is the new angular momentum of the system?
d) What are their new speeds?
e) What is the new rotational energy of the system?
f) How much work is done by the astronauts 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*75kg*5m/s*5.0 = 3750kg-m^2/s
b) K = 1/2*I*?^2 = 1/2*(2*M*r^2)*v^2/r^2 = M*v^2 = 75.0kg*(5m/s)^2 = 1875J
c) I remains constant since there is no outside torque to the system L = 3750kg-m^2/s
e) Now I = 2*M*r^2 = 2*75*2.5^2 = 937.5kg-m^2 previously I = 2*75*5^2 = 3750
Since I is reduced by a factor of 4 ? is increased by a factor of 4
So K = 1/2*I*?^2 = 1/2*937.5*(4*5/2.5)^2 = 3x10^4J
f) the work done is the change in K = 3x10^4 - 1875 = 2.81x10^4J