Two astronauts (Fig. P11.51), each having a mass of 78.0 kg, are connected by a
ID: 1401623 • 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
here,
mass of each astronauts , m = 78 kg
length of rod , L = 10.5 m
r = 5.25 m
speed , v = 4.8 m/s
(a)
the magnitude of the angular momentum , P = m * v * r
P = 78 * 4.8 * 5.25
P = 1965.6 kg.m^2/s
the magnitude of the angular momentum 1965.6 kg·m2/s
(b)
the rotational energy of the system , E = 2 * 0.5 * m * v^2
E = 78 * 4.8^2
E = 1797.12 J
the rotational energy of the system is 1797.12 J
(c)
new distance , L' = 5 m
r' = 2.5 m
as the momentum remains conserved
the new angular momentum is 1965.6 kg.m^2/s
(d)
let the new speed be v'
the new angular momentum = 2 * m * v' * r'
1965.6 = 2 * 78 * v' * 2.5
v' = 5.04 m/s
the astronauts' new speeds are 5.04 s