Q.
A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is
A) $\sqrt{\frac{mgR}{M}}$
B) $\sqrt{\frac{mgR}{m}}$
C) $\sqrt{\frac{mg}{MR}}$
D) $\sqrt{\frac{mR}{Mg}}$
Explanation
(b) To keep the mass M steady, let T is the tension in the string joining the two. Then for particle $m,$
$T = \frac{{mv}^{2}}{R}\ldots(i)$
For mass $M,$
$T = Mg\ldots(ii)$
From Eqs. (i) and (ii)
$\frac{{mv}^{2}}{R} = Mg \Longrightarrow v = \sqrt{\frac{MgR}{m}}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.