Answer
$11\ rad/s$
Work Step by Step
It is given that:
Number of revolutions $\theta = 2.5\ rev = 2.5 \times 2\pi rad = 15.7 rev$
Height of the platform $h = 10\ m$
Initial velocity $u=0\ m/s$
Using kinematic equation;
$h =ut + \frac{1}{2}at^2$
$10\ m=0+0.5(9.8\ m/s) t^2$
$t^2 =2.04 $
$t = 1.42\ s$
The angular velocity is $\omega = \frac{d\theta}{dt}$:
$\omega =\frac{15.7\ rad}{1.42\ s} $
$\omega = 11\ rad/s$
