Two astronauts (Fig. P11.51), each having a mass of 80.0 kg, are connected by a
ID: 1610100 • Letter: T
Question
Two astronauts (Fig. P11.51), each having a mass of 80.0 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 5.10 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. 2040 Your response differs from the correct answer by more than 10%. Double check your calculations. Kg middot m^2/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 m^2/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? KJExplanation / Answer
angular velocity = linear velocity / radius
angular velocity = 5.1 / 5
angular velocity = 1.02 rad/s
moment of inertia = mass * radius^2
moment of inertia of 1 astronaut = 80 * 5^2
moment of inertia of 2 astronauts = 2 * 80 * 5^2
angular momentum = moment of inertia * angular velocity
angular momentum = 2 * 80 * 5^2 * 1.02
angular momentum = 4080 kg.m^2/sec
rotational energy = 0.5 * moment of inertia * angular velocity^2
rotational energy = 0.5 * 2 * 80 * 5^2 * 1.02^2
rotational energy = 2080.8 J
by conservation of angular momentum
initial momentum = final momentum
so,
new angular momentum = 4080 kg.m^2/sec
4080 = 2 * 80 * 2.5^2 * new angular velocity
new angular velocity = 4.08 rad/sec
new linear speed = 4.08 * 2.5
new linear speed = 10.02 m/s
new rotational energy = 0.5 * 2 * 80 * 2.5^2 * 4.08^2
new rotational energy = 8323.2 J
work done = change in rotational kinetic energy
work done = 8323.2 - 2080.8
work done = 6242.4 J or 6.2424 kJ