Two astronauts, each having a mass of 76.9 kg, are connected by a 12.8 m rope of
ID: 1964516 • Letter: T
Question
Two astronauts, each having a mass of 76.9 kg, are connected by a 12.8 m rope of neg- ligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.76 m/s.a) Calculate the magnitude of the initial angular momentum of the system by treating the astronauts as particles. Answer in units of kg m2/s
b) Calculate the rotational energy of the system. Answer in units of J
c) By pulling the rope, the astronauts shorten the distance between them to 3.21 m. What is the new angular velocity of the astronauts? Answer in units of rad/s
d) How much work is done by the astronauts in shortening the rope?
Answer in units of J
Explanation / Answer
a) Angular momentum = mvr + mvr
= 2mvr = 2x76.9 x 5.76 x 6.4 ( as cnter of mass will be at the middle of the rope, r= 6.4m)
= 5669.68 kg m2/s
b) Rotational KE = 1/2 I w2 + 1/2 I w2
= Iw2
= mr2 w2 = mv2 = 2551.36 J
c) angular momentum will be conserved
=> 2 m v1r1 = 2 m v2r2
or m v1r1= m w2 r22( as r2 = 3.21/2 = 1.605 m)
w2 =14. 31 rad/s
d) work done = change in KE
= 1/2 (2m w22 x r22 ) - 1/2 (2mv12)
= 38016.41 J