Question

A uniform rod of mass m and length $L$ is hinged at one of its end with the ceiling and another end of the rod is attached with a thread which is attached with the horizontal ceiling at point $P$ . If one end of the rod is slightly displaced horizontally and perpendicular to the rod and released. If the time period of small oscillation is $2\pi \sqrt{\frac{2lsin\theta }{xg}}$ . Find $x.$

Answer 1

Answer: 3

Consider a small angular displacement $\beta$
Torque about the rotational axis
$\tau =-\left(mg\frac{l}{2}sin\theta \right)\left(\beta \right)$
$\tau =I\alpha$
$1=m\frac{{\left(lsin\theta \right)}^2}{3}=m\frac{l^2{\left(sin\right)}^2\theta }{3}$
$-\left(mg\frac{l}{2}sin\theta \right)\left(\beta \right)=\frac{ml{\left(sin\right)}^2\theta }{3}\alpha$
$\alpha =-\left(\frac{3g}{2lsin\theta }\right)\beta$
$\therefore T=2\pi \sqrt{\frac{2lsin\theta }{3g}}$