Question

A cylinder of radius R = 1m and height H=3m, two thirds filled with water, is rotated about its vertical axis, as shown in figure. Determine the speed of rotation when,
A. The water starts spilling over the rim
B. The center of the base is just exposed

Answer 1

Hint: As the container rotates the water will experience centrifugal force which radially outwards in direction from the axis of rotation. The level of water at boundaries will increase.

Complete step by step answer:
We know that the maximum height of water to be attained just before spilling out is ${{\text{H}}_{\text{max}}}$.
${{H}_{\max }}=3m$
$h=\dfrac{2}{3}H=2m$
$R=1m$
$g=10m/{{s}^{2}}$
The centrifugal force will be balanced by the normal reaction force, therefore-
$\omega =\dfrac{2}{R}\sqrt{g(H-h)}$
$\omega =\dfrac{2}{1}\sqrt{10(3-2)}=\sqrt{40}$
Therefore, the speed of rotation of container is $\sqrt{\text{40}}\text{rad/s}\text{.}$
As this container rotates, the point $P_o$ which is initially at top of water at height h comes to bottom and touches the base at P.

$y-{{y}_{\min }}=\dfrac{{{\omega }^{2}}r}{2g}$
As the point at top touches the base the $y_{min}$ will get zero.
$\omega =\sqrt{\dfrac{2gH}{{{R}^{2}}}}$
$\omega =\sqrt{\dfrac{2.(10).3}{{{1}^{2}}}}=\sqrt{60}rad/s$
Therefore, the speed of rotation of container is$\sqrt{\text{60}}\text{rad/s}\text{.}$

Additional Information:
Centrifugal force is used by the governor to regulate speed of rotation, centrifugal pumps etc. Also, it acts on concepts of the universe like bulging of the earth at equator and flattening at the poles is due to centrifugal force acting on it.

Note: Students should not confuse between centripetal force and centrifugal force. While rotating the water level due to paraboloid shape but, the volume of water remains constant.