Question

A bucket, full of water is revolved in a vertical circle of radius $2m$. What should be the maximum time-period of revolution so that the water does not fall out of the bucket?

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Answer 1

Answer: (A)

The minimum velocity which the bucket should have to complete the full circle is $v\ge \sqrt{5gR}$.
The water contained in the revolving bucket experiences a centrifugal force which is always

equal and opposite to the centripetal force. The minimum velocity required to complete the circle with radius $R$ and gravity $g$ is $v\ge \sqrt{5gR}$.
Also $v=R\omega$ where, $\omega$ is angular velocity
$\therefore R\omega \ge \sqrt{5gR}$
$\Rightarrow \omega \ge \sqrt{\frac{5g}{R}}$
Also $\omega =\frac{2\pi }{T}$ where, $T$ is time period
$\therefore T\le 2\pi \sqrt{\frac{R}{5g}}$
Given, $R=2m,g=10m/s^2$
$\therefore T_{\max }=2\pi \sqrt{\frac{2}{5\times 10}}$
$=\frac{2\pi }{5}=1.256s\approx 1s$