Quick Question about Rotational Energy

[MJC8719]

Two astronauts (Fig. P11.51), each having a mass of 77.0 kg, are connected by a 11.0 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.40 m/s.


(a) Treating the astronauts as particles, calculate the magnitude of the angular momentum.
4573.8 kgm2/s
(b) Calculate the rotational energy of the system.

What is the equation for rotational energy of a system...I thought it was Krot = L^2/(2I) which for this problem would have been (4573.8^2)/(2 x 144 x 5.5^2) but this is not correct. I though 2 I would be equal to 2MR as the astronauts are like a hollow disk rotating in the air

 

[MJC8719]

wow..never mind 2x77 is 154 and not 144 lol

thanks for looking

 

[m2003]

Your equation for rotational energy is correct. However, in this case, the astronauts are not rotating as a single solid object, but rather orbiting around their center of mass. Therefore, the moment of inertia (I) should be calculated using the parallel axis theorem, taking into account the distance between the astronauts and their center of mass.

The correct equation for rotational energy in this case would be: Krot = (L^2)/(2I) = (4573.8^2)/(2 x 2 x 77 x 5.5^2) = 179.4 J.

It is important to note that this is the rotational energy of the system as a whole, and not the individual rotational energies of the two astronauts. As they are connected by a rope, their individual rotational energies would be equal but opposite in direction, and would cancel out when considering the system as a whole.