Answer
(a) $\omega=\sqrt{\frac{11}{(0.25)(4.5)}}=2.2 \frac{rad}{s}$
(b) $\omega$ will increase if the length of the rope is shortened.
Work Step by Step
(a) We know that
$\omega=\sqrt{\frac{F}{mr}}$
We plug in the known values to obtain:
$\omega=\sqrt{\frac{11}{(0.25)(4.5)}}=2.2 \frac{rad}{s}$
(b) We also know that angular velocity $\omega $ and r are inversely proportional, hence $\omega$ will increase if the length of the rope is shortened.
