Rotational KE: Calculate Energy of 2 Astronauts | 739.3 J

[Noor123]

URGENT: rotatonal KE

Two astronauts each having a mass of 54 kg are connected by a 10 m rope of negligible mass. They are isolated in space, orbiting about their centre of mass at speeds of 3.7 m/s. Calculate the rotational energy of the two astronauts.

A. 739.3 J
B. 1478.5 J
C. 29.6 J
D. 317.9 J
E. 369.6 J

Which one is it?

I did:

KE= 1/2 (I + I) omega ^2

and it turned out to be A. Is it correct?

 

[hage567]

That's right.

 

[ogb p]

Yes, your calculation is correct. The rotational energy of the two astronauts would be 739.3 J. This result can be obtained by using the formula for rotational kinetic energy, which takes into account the moment of inertia (I) and the angular velocity (omega) of the system. In this case, the moment of inertia would be equal for both astronauts since they have the same mass and are rotating together, and the angular velocity can be calculated using the given speed and radius of rotation. Therefore, the final result would be the same for both astronauts, resulting in a total rotational energy of 739.3 J.