Two astronauts, each having a mass of 74.0 kg, are connected by a 10.0 m rope of
ID: 1411486 • Letter: T
Question
Two astronauts, each having a mass of 74.0 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 4.50 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum.
(b) Calculate the rotational energy of the system.
(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?
(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?
Explanation / Answer
(a)The centre of mass is middle of 10 m
angular speed of each=w=4.5/5=0.9 rad/sec
moment of inertia of each=I=mR^2=1850 kgm^2
angular momentum =Iw+Iw=3330 kgm^2/s
(b)rotational energy of each=0.5*I*w^2=749.25 J
Rotational energy of system=1498.5 J
(c)now if distance = 5 m
centre of mass =middle of 5 m =2.5 m
new moment of inertia of each =MR^2=462.5 kgm^2
new angular momentum will remain same =3330 kgm^2/s
(d)by conservation of angular momentum
2*462.5*w=3330
w=3.6 rad/sec
new velocity=w*r=9 m/s
(e)new rotational energy=Iw^2=5994 J
(f)work done =change in energy=5994-1498.5=4495.5 J