Question

A uniform rod of mass M and length is hanging from its central point as shown. A small ball of equal mass M is attached to the lower end. Time period of small oscillation of the rod is

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

Answer 1

The correct option is D $2\pi \sqrt{\frac{2l}{3g}}$

$(\frac{M{l}^2}{4}+\frac{<l^2}{12})\alpha =-Mg\frac{l}{2}\theta$

$\frac{M{l}^2}{3}\alpha =-Mg\frac{l}{2}\theta$

${\omega }^2=\frac{3g}{2l}$

$T=2\pi \sqrt{\frac{2l}{3g}}$