Answer
$ 1.79\times 10^{-2}\ N.m$
Work Step by Step
Mass of a small ball $m=1.30\ kg$
Length of the rod $L=0.780\ m$
Angular frequency $\omega = 5010\ \frac{rev}{min}$
1 revolution = $2\pi $ radians
1 minute = $60$ seconds
We convert given angular frequency in rad/sec:
$\omega = 5010\times\frac{2\pi\ rad}{60\ s} = 524.64\ rad/s$
Also, air drag on the ball is $F =2.30\times 10^{-2}\ N$.
Torque applied on the system is:
$\tau = F r$
$\tau = F L$
$\tau = (2.30\times 10^{-2}\ N)(0.780\ m) =0.0179 $
$\tau = 1.79\times 10^{-2}\ N.m$
