Question

A uniform stick of length $l$ is hinged so as to rotate about a horizontal axis perpendicular to the stick, at a distance ${}^{\prime }{x}^{\prime }$ from the centre. The value of $x$, for which the time period is minimum will be

• A. $\frac{l}{2}$
• B. $\frac{l}{4}$
• C. $\frac{l}{2\sqrt{3}}$
• D. $\frac{l}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{l}{2\sqrt{3}}$

$T=2\pi \sqrt{\frac{I}{mdg}}$
Where $I$ is the moment of inertia about the point of suspension,
$d$ is the distance of centre of mass from the point of suspension.
m is the mass.
$I=\frac{m{l}^2}{12}+m{x}^2$
$T=2\pi \sqrt{\frac{\frac{m{l}^2}{12}+m{x}^2}{mgx}}$
$T^2=4{\pi }^2\frac{\frac{m{l}^2}{12}+m{x}^2}{mgx}$

For time period to be minimum $\frac{dT}{dx}=0$ or $\frac{d{T}^2}{dx}=0$
$4{\pi }^2\left(\right(\frac{l^2}{12g}\frac{-1}{x^2})+\frac{1}{g})=0$
$x=\frac{l}{\sqrt{12}}$