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A bucket containing water is revolved in a vertical circle of radius $r$ .To prevent the water from falling down, the minimum frequency of revolution required is ( $g=$ acceleration due to gravity)
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Verified Answer
The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{g}{r}}$
Let its angular velocity be $\omega$ at all points (uniform motion). At the highest point, weight of the body is balanced centrifugal force, so
$$
\begin{aligned}
& m \omega^2 r=m g \\
& \Rightarrow \omega=\sqrt{\frac{g}{r}}
\end{aligned}
$$
Angular frequency is related to frequency as follows:
$\omega=2 \pi f$
$\therefore f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{r}}$
$$
\begin{aligned}
& m \omega^2 r=m g \\
& \Rightarrow \omega=\sqrt{\frac{g}{r}}
\end{aligned}
$$
Angular frequency is related to frequency as follows:
$\omega=2 \pi f$
$\therefore f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{r}}$
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