Two astronauts, each having a mass of 78.0 kg, are connected by a d = 11.0-m rop
ID: 2107541 • Letter: T
Question
Two astronauts, each having a mass of 78.0 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 5.30 m/s.
Image: http://www.webassign.net/pse/p11-59.gif
(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
a) for the astronauts, r1 = r2 = 9.5/2 = 4.75 m and angular momentum L = mvr1 + mvr2 = 2 m v r
ie., L = 2x78 x5.1 x4.75 = 3779.1 kg m^2 sec
b) rotational kinetic energy Er = L^2 / 2 I = 3779.1x3779.1 / 2x78x5.1x5.1 = 14281596 / 4057.56
= 3520 J (nearly)
c) new angular momentum L = 3779.1 Kg m^2 sec ( by conservation of angular momentum)
d) new speed of each astronaut v = L / m r = 3779.1 / 78 x 2.5 = 19.38 m/sec
e) new rotational kinetic energy = L^2 / 2I = 3779.1x3779.1 / 2 x78 x 2.5x2.5 = 14281596 / 975
= 14648J
f) work done W = Er2 - Er1 = 14648 - 3520 = 11128 J.