Two astronauts, each having a mass of 77.0 kg, are connected by a 11.0 m rope of
ID: 1410240 • Letter: T
Question
Two astronauts, each having a mass of 77.0 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 4.50 m/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) angular momentum = m( r x v)
= 2(77 * 5.5 * 4.50) = 3811.5 kg m^2 /s
b) KErot = Iw^2/2
both have same I and w
so total KE = 2(I w^2 /2 ) = I w^2
= ( 77 x 5.5^2 ) ( 4.50 / 5.5)^2 = 1559.25 J
c) Suppose new speed is v but in this process angular momentum will be constant.
as tension is a internal force and due to this torque is zro on system.
2( 77 * 2.5 * vf) = 3811.5
vf = 9.9 m/s
e) KE = (77 x 2.5^2 ) (9.9/ 2.5)^2 = 7546.77 J
f) Work done = change in KE = 7546.77 - 1559.5 = 5987.27 J