Question

A small 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.

Answer 1

Answer:

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

$I_{sys}= mL^2= 2.50\times 0.75^2$

=1.40625 kg-m^2

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