Two astronauts, each having a mass of 72.0 kg, are connected by a d = 11.0-m rop

Question

Two astronauts, each having a mass of 72.0 kg, are connected by a d = 11.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.90m/s.

(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum of the two-astronaut system.
kg · m2/s

(b) Calculate the rotational energy of the system.
kJ

(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?
kg · m2/s

(d) What are the astronauts' new speeds?
m/s

(e) What is the new rotational energy of the system?
kJ

(f) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope?
kJ

Explanation / Answer

here,
mass of astronauts, m = 72 kg
distance between them , r = 5.5/2 = 5.5 m

Part A:
Angular momentum, L = I*w(I is moment of inertia, w is angular speer = v/r)
L = 2*m*r^2*v/r
L = 2*m*r*v
L = 2*72*5.5*4.90
L = 3880.8 kg.m^2/s

Part B:
Energy, K = 1/2*I*^2
K.E = 1/2*(2*M*r^2)*v^2/r^2
K.E = M*v^2
K.E = 72*4.90^2
K.E = 1728.72 J

Part C:
Moment of inertia remains constant, since there is no outside torque to the system
L = 3880.8 kg.m^2/s

Part D:
L = 2*m*r*v

solving for new speed, v = L/2*m*r
v = 3880.8/(2*72*2.5)
v = 10.8 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.