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 =m g $
and $T \sin \theta =m r \omega^{2}$
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 $T=2 \pi \sqrt{\frac{r}{g \tan \theta}}$
$=2 \pi \sqrt{\frac{l \sin \theta}{g \tan \theta}}$
$=2 \pi \sqrt{\frac{l \cos \theta}{g}}$