Asmall ball with mass 2.50 kg is mounted on one end of a rod 0.750 m long and of negligible mass. the system rotates in a horizontal circle about the other end of the rod at 4900 rev/min.

1.40625 kg-m^2

Explanation:

Supposing we have to calculate rotational moment of inertia

Given:

Mass of the ball m= 2.50 kg

Length of the rod, L= 0.78 m

The system rotates in a horizontal circle about the other end of the rod

The constant angular velocity of the system, ω= 5010 rev/min

The rotational inertia of system is equal to rotational inertia of the the ball about other end of the rod because the rod is mass-less

=1.40625 kg-m^2

m= mass of the ball and L= length of the ball

that would be because of both gravity and friction. so     c