Two astronauts (Fig. P11.51), each having a mass of 78.0 kg, are connected by a
ID: 1401356 • Letter: T
Question
Two astronauts (Fig. P11.51), each having a mass of 78.0 kg, are connected by a 10.5 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 4.80 m/s.
(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum.
kg·m2/s
(b) Calculate the rotational energy of the system.
J
(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?
J
(f) How much work does the astronaut do in shortening the rope?
kJ
Explanation / Answer
a) for the astronauts, r1 = r2 = 10.5/2 = 5.25 m and angular momentum L = mvr1 + mvr2 = 2 m v r
ie., L = 2x78 x4.8 x5.25 = 3931.2 kg m^2 sec
b) rotational kinetic energy
Er = L^2 / 2 I = 3931.2x3931.2 / 2x78x4.80x4.80 = 4299.75 J
c) new angular momentum
L = 3931.2 Kg m^2 sec ( by conservation of angular momentum)
d) new speed of each astronaut
v = L / m r = 3931.2 / 78 x 2.5 = 20.16 m/sec
e) new rotational kinetic energy = L^2 / 2I = 3931.2x3931.2 / 2 x78 x 2.5x2.5 = 15850.59J
f) work done W = Er2 - Er1 = 15850.59 - 4299.75 = 11550.84 J.