Question

A rod of mass $m$ and length $L$ is pivoted at point $O$ in a car whose acceleration towards left is $a_0$. The rod is free to rotate in vertical plane. In equilibrium state, the rod remains horizontal when other end is suspended by a spring of force constant $k$. The time period of small oscillations of rod is $T=\frac{2\pi }{D\sqrt{3}}$. The value of $D$ is ____
(Given, $k=20N{m}^{-1},a_0=10m{s}^{-2},m=1kg,L=1m)$

Answer 1

Answer: 5

In equilibrium, $mg\left(\frac{L}{2}\right)=kyL$
$\therefore y=\frac{mg}{2k}$ (initial extension of spring)
During oscillations, consider a situation when the angular displacement of rod is $\theta$ as shown in the figure.

$m{a}_0=$ Pseudo force
In the reference frame of car, restoring torque is,
$\tau =-\left[m{a}_0\frac{L}{2}\sin \theta +my\left(\frac{L}{2}\cos \theta \right)-k(y-L\theta )L\cos \theta \right]$
Using Eq. (i), $\tau =-\left[\frac{m{a}_{0}L}{2}\theta +k{L}^2\theta \right](\because \theta$ is small $)$
$\tau =-\left[\frac{m{a}_{0}L}{2}+k{L}^2\right]\theta$
$\therefore I\alpha =-\left[\frac{m{a}_{0}L}{2}+kL\right]L\theta$
$\alpha =-\frac{{\left[\frac{m{a}_0}{2}+kL\right]}_{\theta }}{m{L}^2/3}\theta =-\left[\frac{3a_0}{2L}+\frac{3k}{m}\right]\theta$
Using the given values, we get $\alpha =-75\theta$
$\therefore \omega =\sqrt{75}=5\sqrt{3}$
$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{5\sqrt{3}}$
$\therefore D=5$