Answer
The change in the angular speed is $~~146~rad/s$
Work Step by Step
We can find the rotational inertia:
$I = \frac{1}{2}(0.120~kg)[(0.0160~m)^2+(0.0450~m)^2]+\frac{1}{2}(0.240~kg)[(0.0900~m)^2+(0.1400~m)^2]$
$I = 0.00346~kg~m^2$
We can find the angular acceleration:
$\tau = I~\alpha$
$\alpha = \frac{\tau}{I}$
$\alpha = \frac{R~F}{I}$
$\alpha = \frac{(0.1400~m)(12.0~N)}{0.00346~kg~m^2}$
$\alpha = 485.55~rad/s^2$
We can find the change in the angular speed:
$\omega = \omega_0+\alpha~t$
$\omega = 0+(485.55~rad/s^2)(0.300~s)$
$\omega = 146~rad/s$
The change in the angular speed is $~~146~rad/s$
