Question

A homogeneous rod AB of length L= 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 (figure ).
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 $\omega$.
(a) Determine the angular velocity $\omega$ in terms of v and L
(b) If the insect reaches the end B when the rod has turned through an angle of 90$^{\circ}$, determine v.

Answer 1

In this problem we will write K for the angular momentum because L has been used for length of the rod.
(a) Angular momentum of the system (rod + insect) about the centre of the rod O will remain conserved just before collision and after collision i.e. $K_i=K_f$
or $Mv\frac{L}{4}=I\omega=\bigg[\frac{ML^2}{12}+M\bigg(\frac{L}{4}\bigg)^2\bigg]\omega$
or $Mv\frac{L}{4}=\frac{7}{48}ML^2\omega$
i.e. $\omega=\frac{12}{7}\frac{v}{L}$...(i)
(b) Due to the torque of weight of insect about O, angular momentum of the system will not remain conserved (although angular velocity $\omega$ is constant). As the insect moves towards B, moment of inertia of the system increases, hence, the angular momentum of the system will increase.
Let at time $t_1$ the insect be at a distance x from O and by then the rod has rotated through an angle $\theta$. Then, angular momentum at that moment,
$K=\bigg[\frac{ML^2}{12}+Mx^2\bigg]\omega$
Hence, $\frac{dK}{dt}=2M\omega x\frac{dx}{dt}$ $(\omega=constant)$
$\Rightarrow \tau=2M\omega x\frac{dx}{dt}\Rightarrow Mgx cos \theta=2M \omega x\frac{dx}{dt}$
$\Rightarrow dx=\bigg(\frac{g}{2\omega}\bigg)cos \omega t dt$ $(\because\theta=\omega t)$
At time $t = 0, x = L/4$ and at time $t = T/4$ or $\pi/2\omega, x = L / 2.$
Substituting these limits, we get
$\int^{L/2}_{L/4}dx=\frac{g}{2\omega}\int^{\pi/2\omega}_0(cos \omega t)dt$
$[x]^{L/2}_{L/4}=\frac{g}{2\omega^2}[sin \omega t]^{\pi/2\omega}_0$
$\Rightarrow \bigg(\frac{L}{2}-\frac{L}{4}\bigg)=\frac{g}{2\omega^2}\bigg[sin\frac{\pi}{2}-sin 0\bigg]$
$\frac{L}{4}=\frac{g}{2\omega^2}$ or $\omega=\sqrt{\frac{2g}{L}}$
Substituting in Eq. (1), we get $\sqrt{\frac{2g}{L}}=\frac{12}{7}\frac{v}{L}$
or $v=\frac{7}{12}\sqrt{2gl}=\frac{7}{12}\sqrt{2\times10\times1.8}$
$\therefore v=3.5m/s$