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, $\left.k=20 \,Nm ^{-1}, a_{0}=10 \,ms ^{-2}, m=1 \,kg , L=1 \,m \right)$

Answer 1

In equilibrium, $m g\left(\frac{L}{2}\right)=k y L$
$\therefore y=\frac{m g}{2 k}$ (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,
$\tau=-\left[m a_{0} \frac{L}{2} \sin \theta+m y\left(\frac{L}{2} \cos \theta\right)-k(y-L \theta) L \cos \theta\right]$
Using Eq. (i), $\tau=-\left[\frac{m a_{0} L}{2} \theta+k L^{2} \theta\right](\because \theta$ is small $)$
$\tau=-\left[\frac{m a_{0} L}{2}+k L^{2}\right] \theta$
$\therefore I \alpha=-\left[\frac{m a_{0} L}{2}+k L\right] L \theta$
$\alpha=-\frac{\left[\frac{m a_{0}}{2}+k L\right]_{\theta}}{m L^{2} / 3} \theta=-\left[\frac{3 a_{0}}{2 L}+\frac{3 k}{m}\right] \theta$
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}} $
$\therefore D=5$