Two astronauts, each having a mass of 76.0 kg, are connected by a d = 11.0-m rop

Question

Two astronauts, each having a mass of 76.0 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 5.10 m/s. Treating the astronauts as particles, calculate the magnitude of the angular momentum of the two-astronaut system. Calculate the rotational energy of the system. 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? What are the astronauts' new speeds? What is the new rotational energy of the system? ) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope?

Explanation / Answer

Here ,

for the position of centre of mass from

x = d/2 = 11/2

x = 5.5 m

v = 5.1 m/s

m = 76 Kg

a) magnitude of angular momentum = 2 * m * x * v

magnitude of angular momentum = 2 * 76 * 5.5 * 5.1

magnitude of angular momentum = 4263.6 Kg.m^2

b)

rotational energy of system = 2 * 0.5 ( m * v^2)

rotational energy of system = 76 * 5.1^2

rotational energy of system = 1976.8 J

c)

as there is no external torque acting on the system

the angular momentum will be constant

new angular momentum = 4263.6 Kg.m^2

d)

let the new speed is v

2 * m * (dnew/2) * v = angular momentum

2 * 76 * (5/2) * v = 4263.6

v = 11.22 m/s

the astronaut's new speed is 11.22 m/s

e)

new rotational kinetic energy = 2 * 0.5 * m * v^2

new rotational kinetic energy = 76 * 11.22^2

new rotational kinetic energy = 9567.5 J

the rotational kinetic energy is 9567.5 J

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