Question

A uniform straight rod of mass 'm; kg and length L is hinged at one end. It is free to oscillate in vertical plane. A point of mass (m kg) is attached to it at a distance 'x' from the hinge. The value of x for which time period of oscillations will be minimum is (use: $\sqrt{\frac{7}{3}=1.5}$)

• A. $2L/3$
• B. $L/2$
• C. $L/4$
• D. $2L/7$

Answer 1

The correct option is C $L/4$

For a rigid body oscillating about horizontal axis passing through the point O the time period is given by $T=2\pi \sqrt{\frac{l}{mg\left(lin\right)}}$

cg$=$length of centre of gravity.

For real centre$=l/2$ from ringed point

For mass of x

Centre of mass of system $c=\frac{mx+ml/2}{2m}$

$=\frac{(2l+l/2)}{2}$

Now,

Moment of inertia of system $=\frac{1}{3}m{l}^2+m{x}^2$

$T=2\pi \sqrt{\frac{\frac{1}{3}m{l}^2+m{x}^2}{\left(2mg\right)\left(\frac{x+l/2}{2}\right)}}$

T should be minimum then $\frac{\frac{1}{3}m{l}^2+m{x}^2}{\left(2mg\right)\left(\frac{x+l/2}{2}\right)}$ should be minimum

Let $=\frac{\frac{1}{3}{l}^2+x^2}{(\frac{x}{3}+\frac{l}{4})}$

Should be minimum

$x=\frac{L}{4}$.