(A) For the mass $M$ to remain stationary,the tension $T$ in the string must balance its weight. Therefore,$T = Mg$.This tension $T$ also provides the

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$\frac{1}{{2\pi }}\sqrt {\frac{{Mg}}{{m\;l}}} $

(A) For the mass $M$ to remain stationary,the tension $T$ in the string must balance its weight. Therefore,$T = Mg$.This tension $T$ also provides the necessary centripetal force for the mass $m$ rotating in a circle of radius $l$ with frequency $n$. The centripetal force is given by $F_c = m \omega^2 l = m(2\pi n)^2 l = m 4\pi^2 n^2 l$.Equating the tension to the centripetal force: $m 4\pi^2 n^2 l = Mg$.Solving for the frequency $n$: $n^2 = \frac{Mg}{4\pi^2 ml}$.Taking the square root on both sides: $n = \frac{1}{2\pi} \sqrt{\frac{Mg}{ml}}$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

Medium

With what frequency should a particle of mass $m$ be rotated so that a body of mass $M$ remains stationary?

A
$\frac{1}{{2\pi }}\sqrt {\frac{{Mg}}{{m\;l}}} $
B
$\frac{1}{\pi }\;\sqrt {\frac{{Mg}}{{m\;l}}} $
C
$\frac{1}{{2\pi }}\sqrt {\frac{{m\;l}}{{Mg}}} $
D
$\frac{1}{\pi }\;\sqrt {\frac{{m\;l}}{{Mg}}} $

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With what frequency should a particle of mass $m$ be rotated so that a body of mass $M$ remains stationary?

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