Question

Two masses $m$ and M are connected by a light string that passes through a smooth hole O at the centre of a table. Mass $m$ lies on the table and M hangs vertically. $m$is moved round in a horizontal circle with O as the centre. If $l$ is the length of the string from O to $m$ then the frequency with which $m$ should revolve so that M remains stationary is

• A. $\frac{1}{2\pi}\sqrt{\frac{Mg}{m\mspace{6mu} l}}$
• B. $\frac{1}{\pi}\mspace{6mu}\sqrt{\frac{Mg}{m\mspace{6mu} l}}$
• C. $\frac{1}{2\pi}\sqrt{\frac{m\mspace{6mu} l}{Mg}}$
• D. $\frac{1}{\pi}\mspace{6mu}\sqrt{\frac{m\mspace{6mu} l}{Mg}}$

Answer 1

Answer: (A)

$\frac{1}{2\pi}\sqrt{\frac{Mg}{m\mspace{6mu} l}}$

Answer 2

‘m’ Mass performs uniform circular motion on the table. Let n is the frequency of revolution then centrifugal force $= m4\pi^{2}n^{2}l$

For equilibrium this force will be equal to weight Mg

$m4\pi^{2}n^{2}l = Mg$

$\therefore n = \frac{1}{2\pi}\sqrt{\frac{Mg}{ml}}$