Q. A block of mass m at the end of the string is whirled round a vertical circle of radius r. The critical speed of the block at the top of the swing is
A) $\left( \frac{r}{g}\ \right)^{1/2}\ $
B) $\frac{g}{r}$
C) $\frac{m}{rg}$
D) $(rg)^{1/2\ }\ $
Explanation
(d) The tension in the string at any position is
$T = \frac{mv^{2}}{r} + mg\cos\theta$
For critical position
$\theta = 180{^\circ}\ $
$v = v_{c}$
T=0
${Hence\ \ \ v}_{c}\sqrt{rg}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.