Answer
$0.791\ kg.m^2$
Work Step by Step
Mass of a small ball $m=1.30\ kg$
Length of the rod $L=0.780\ m$
Angular frequency $\omega = 5010\ \frac{rev}{min}$
Also,
1 revolution = $2\pi $ radians
1 minute = $60$ seconds
We convert given angular frequency in rad/sec:
$\omega = 5010\times\frac{2\pi\ rad}{60\ s} = 524.64\ rad/s$
Rotation inertia of the system about the axis of rotation is:
$I = mL^2$
$I=(1.30\ kg)(0.780\ m)^2$
$I=0.791\ kg.m^2$
