Question

Two astronauts, each having a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s. Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the system and (b) the rotational energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. (c) What is the new angular momentum of the system? (d) What are the astronauts’ new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut do in shortening the rope?

Answer 1

Answer:

a) 3750 kgm²/s

b) 1875 J

c) 3750 kgm²/s

d) 10 m/s

e) 7500 J

f) 5625 J

Explanation:

a)

d = distance between the two astronauts = 10 m

m = mass of each astronaut = 75 kg

v = speed of each astronaut = = 5 m/s

r = distance of each astronaut from center of mass = (0.5) d = (0.5) (10) = 5 m

$L_{s}$ = Angular momentum of the system

Angular momentum of the system is given as

$L_{si}$ = 2 m v r

$L_{si}$ = 2 (75) (5) (5)

$L_{si}$ = 3750 kgm²/s

b)

Rotational energy of the system is given as

$R_{si}$ = 2 (0.5) m v²

$R_{si}$ = 2 (0.5) (75) (5)² = 1875 J

c)

$L_{sf}$ = Angular momentum of the system after pulling

Using conservation of angular momentum,

$L_{sf}$ = $L_{si}$

$L_{sf}$ = 3750 kgm²/s

d)

r' = new distance of each astronaut from center of mass = (0.5) r = (0.5) (5) = 2.5 m

v' = new speed of each astronaut

Angular momentum of the system after pulling is given as

$L_{sf}$ = 2 m v' r'

3750 = 2 (75) (2.5) v'

v' = 10 m/s

e)

Rotational energy of the system after pulling the rope is given as

$R_{sf}$ = 2 (0.5) m v'²

$R_{sf}$ = 2 (0.5) (75) (10)² = 7500 J

f)

W = work done by astronaut in pulling the rope

Work done is given as

W = $R_{sf}$ - $R_{si}$

W = 7500 - 1875

W = 5625 J