A homogeneous rod $AB$ of length $A = 1.8 m$ and mass $M$ is pivoted at the centre $O$ in such a way that it can rotate freely in the vertical plane. The rod is initially in the horizontal position. An insect $S$of the same mass $M$ falls vertically with speed $v$ on the point $C$, midway between the points $O$ and $B$. Immediately after falling, the insect moves towards the end $B$ such that the rod rotates with a constant angular velocity $\omega$. If the insect reaches the end$B$ when the rod has turned through an angle of $90°,$ determine $v.$

In the process, the angular momentum of the system remains constant.

$Mv\times \frac{L}{4}+0=I\omega \dots \dots \left(1\right)$

where $I$ is the moment of inertia of the system which is

$\frac{M{L}^2}{12}+M{\left(\frac{L}{4}\right)}^2=\frac{7M{L}^2}{48}$

Substituting this value in the above equation, we get

$\omega =\frac{12v}{7L}$ $\dots \left(2\right)$

Let the insect is at a distance $x$from the centre of the rod after time$t$, the moment of inertia of the system will be  $=\frac{M{L}^2}{12}+M{x}^2$, and therefore angular momentum at this instant $\left(\frac{M{L}^2}{12}+M{x}^2\right)\omega$

The torque exerted by the weight of the insect at that moment

$\tau =mgx$.

Since we have $\frac{dL}{dt}=\tau$

Therefore, $\frac{d}{dt}\left(\frac{M{L}^2}{12}+M{x}^2\right)\omega =Mgx \dots \left(3\right)$

or $\frac{dx}{dt}=u=\frac{g}{2\omega }\dots \dots \left(4\right)$

The time taken by rod to rotate through $\frac{\mathrm{\pi }}{2}rad=\frac{\mathrm{\pi }/2}{\omega }$ , it is given that the rod rotates with constant angular velocity. The time is taken by the insect to reach the end $B = (L/4)/u$.

It is given that both times are equal, therefore

$\frac{\mathrm{\pi }/2}{\omega }=\frac{L/4}{u}$

or $\frac{\left(\mathrm{\pi }/2\right)}{\omega }=\frac{\left(L/4\right)}{g/2\omega }$

After solving the above equation, we get

$\omega =\sqrt{\left(\mathrm{\pi g}/\mathrm{L}\right)}$

Substituting this value in equation $\left(2\right)$, we get

$v=\frac{7\omega L}{12}=7\sqrt{\frac{\mathrm{\pi g}}{L}}\times \frac{L}{12}$ where $L=1.8m$

$=4.4 m/s$