Question

An open cylindrical vessel filled with water is rotated together with water at angular speed $1 \, rad \, s^{- 1}$ . At this angular speed, the centre of the bottom is just exposed. If the angular speed is increased to $\sqrt{\frac{3}{2}} \, rad \, s^{- 1}$ , how much area (in $cm^{2}$ ) of the bottom of the vessel is now exposed to surrounding? [Area of the cross-section of cylinder is $27 \, cm^{2}$ ]

Answer 1

Let $\omega _{1}$ be initial angular velocity

$\frac{\omega _{1}^{2} R^{2}}{2 g}=h$
For angular velocity $\omega _{2}$ let radius of exposed surface be x

$\frac{\omega _{2}^{2}}{2 g}\left[R^{2} - x^{2}\right]=h$
Putting the value of $\omega _{1}$ and $\omega _{2}$
$\pi x^{2}=9$
Since $\pi R^{2}=27$