Question

A heavy sphere of mass $m$ is suspended by string of length $l$. The sphere is made to revolve about a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at an 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. $T=2\pi \sqrt{\frac{l\tan \theta }{g}}$

Answer 1

Answer: (B)

Resolving tension $T$ in string into two rectangular components, we get
$T\cos \theta =mg$
and $T\sin \theta =mr{\omega }^2$
So, $\frac{T\sin \theta }{T\cos \theta }=\tan \theta =\frac{r{\omega }^2}{g}$
or $g\tan \theta =r{\omega }^2=r\times 4{\pi }^2/T^2$
or $T=2\pi \sqrt{\frac{r}{g\tan \theta }}$
$=2\pi \sqrt{\frac{l\sin \theta }{g\tan \theta }}$
$=2\pi \sqrt{\frac{l\cos \theta }{g}}$