A heavy small sized sphere is suspended by a string of length $l$. The sphere is rotated uniformly in a horizontal circle with the string making an angle $ heta$ with the vertical. The time period of this conical pendulum is

Question

A heavy small sized sphere is suspended by a string of length $l$. The sphere is rotated uniformly in a horizontal circle with the string making an angle $	heta$ with the vertical. The time period of this conical pendulum is

Dr. Abubacker · Accepted Answer

(d) As is clear from figure
[image:4_image125.emf]
$T\sin	heta = \frac{mv^{2}}{r},\ T\cos	heta = mg$
Dividing, we get
$	an	heta = \frac{v^{2}}{rg} = \frac{r}{g}\left( \frac{2\pi}{T} ight)^{2}$
$\frac{2\pi}{T} = \sqrt{\frac{g	an	heta}{r}} = \sqrt{\frac{g	an	heta}{l\sin	heta}} = \sqrt{\frac{g}{l\cos	heta}}$
or $T = 2\pi\sqrt{\frac{l\cos	heta}{g}}$

Q. A heavy small sized sphere is suspended by a string of length $l$. The sphere is rotated uniformly in a horizontal circle with the string making an angle $\theta$ with the vertical. The time period of this conical pendulum is

Explanation

Reviewed by Dr. Abubacker

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