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 $v \geq \sqrt{5 g R}$.
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 $v \geq \sqrt{5 g R}$.
Also $v=R \omega$ where, $\omega$ is angular velocity
$\therefore R \omega \geq \sqrt{5 g R} $
$\Rightarrow \omega \geq \sqrt{\frac{5 g}{R}}$
Also $ \omega=\frac{2 \pi}{T}$ where, $T$ is time period
$\therefore T \leq 2 \pi \sqrt{\frac{R}{5 g}}$
Given, $R=2\, m , g=10 \,m / s ^{2}$
$\therefore T_{\max }=2 \pi \sqrt{\frac{2}{5 \times 10}}$
$=\frac{2 \pi}{5}=1.256 s \approx 1 s$