Q. A heavy sphere of mass m is suspended by string of length $l.$ The spear is made to revolve above a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at $\angle\theta.$ What is its period of revolution?
A) $T = 2\pi\sqrt{\frac{l}{g}}$
B) $T = 2\pi\sqrt{\frac{l\cos\theta}{g}}$
C) $T = 2\pi\sqrt{\frac{l\sin\theta}{g}}$
D) $T = 2\pi\sqrt{\frac{l\tan\theta}{g}}$
Explanation
(b) Resolve tension $T$ in string into two rectangular components. Then
$T\cos\theta = mg$
And $T\sin\theta = mr\ \omega^{2}$
So $\frac{T\sin\theta}{T\cos\theta} = \tan\theta = \frac{r\omega^{2}}{g}$
Or $g\tan\theta = r\omega^{2} + r4\frac{\pi^{2}}{T^{2}}$
Or $T = 2\pi\sqrt{\frac{r}{g\tan\theta}}\ \ \ \ \ (\because r = l\sin\theta)$
$= 2\pi\sqrt{\frac{l\sin\theta}{g\tan\theta}}$
$= 2\pi\sqrt{\frac{l\cos\theta}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.