physics
posted by Anonymous on .
Two astronaut, as shown in the figure, each having a mass of 62.0 kg, are connected by a 12.00 m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.00 m/s. Treating the astronauts as particles, calculate (b) the rotational energy of the system.

The rotational energy of each astronaut is
1/2 * I * omega^2
where I is the moment of inertia, and omega is the angular speed.
The moment inertia of the astronaut is m*r^2
r = 12.0 / 2 = 6 m
v = 5 m/s = omega*r
The total rotational energy is
1/2 * I * omega^2 + 1/2 * I * omega^2 = I * omega^2