Two astronauts, each having a mass of 74 kg, are connected by a 8.53 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.08 m/s. Calculate the magnitude of the initial angular momentum of the system by treating the astronauts as particles. Answer in units of kg m2 /s. 017 (part 2 of 4) 10.0 points Calculate the rotational energy of the system. Answer in units of J. 018 (part 3 of 4) 10.0 points By pulling the rope, the astronauts shorten the distance between them to 4.18 m. What is the new angular velocity of the astronauts

a) 2575 kgm²/s

b) 1.23 kJ

c) 0.478 rad/s

Explanation:

Given

Mass of astronauts, m = 74 kg

Length of rope, l = 8.53 m

Speed of orbit, v = 4.08 m/s

L = m1v1.x1(i) + m2v2.x2(i) = 2mv(d/2)

Thus, L = 2.m.v.(d/2)

L = 2 * 74 * 4.08 * (8.53/2)

L = 2 * 74 * 4.08 * 4.265

L = 2575.38 kgm²/s

Rotational Energy of the system

K(i) = 1/2m1v1(i)² + 1/2m2v2(i)²

K(i) = 2(1/2) * 74 * 4.08²

K(i) = 74 * 16.6464

K(i) = 1231.83 J = 1.23 kJ

Angular momentum is conserved, thus, angular velocity, w = v/r

w = 4.08 / 8.53

w = 0.478 rad/s

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )