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 $Mv\frac{L}{4}=I\omega =[\frac{M{L}^2}{12}+M(\frac{L}{4}{)}^2]\omega$
or $Mv\frac{L}{4}=\frac{7}{48}M{L}^2\omega$
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,
$K=[\frac{M{L}^2}{12}+M{x}^2]\omega$
Hence, $\frac{dK}{dt}=2M\omega x\frac{dx}{dt}$ $(\omega =constant)$
$\Rightarrow \tau =2M\omega x\frac{dx}{dt}\Rightarrow Mgxcos\theta =2M\omega x\frac{dx}{dt}$
$\Rightarrow dx=\left(\frac{g}{2\omega }\right)cos\omega tdt$ $(\because \theta =\omega t)$
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
${\int }_{L/4}^{L/2}dx=\frac{g}{2\omega }{\int }_0^{\pi /2\omega }(cos\omega t)dt$
$[x{]}_{L/4}^{L/2}=\frac{g}{2{\omega }^2}[sin\omega t{]}_0^{\pi /2\omega }$
$\Rightarrow (\frac{L}{2}-\frac{L}{4})=\frac{g}{2{\omega }^2}[sin\frac{\pi }{2}-sin0]$
$\frac{L}{4}=\frac{g}{2{\omega }^2}$ or $\omega =\sqrt{\frac{2g}{L}}$
Substituting in Eq. (1), we get $\sqrt{\frac{2g}{L}}=\frac{12}{7}\frac{v}{L}$
or $v=\frac{7}{12}\sqrt{2gl}=\frac{7}{12}\sqrt{2\times 10\times 1.8}$
$\therefore v=3.5m/s$