Question

In the shown in the figure a mass $m$ slides down the frictionless surface from height $h$ and mass $M$. After a collision, mass $m$ sticks to the rod. The rod is free to rotate in a vertical plane about a fixed axis through $O$. Find the maximum angular deflected of the rod from its initial position.

Answer 1

Just before collision, velocity of the mass $m$ is along the
horizontal and is equal to
$v_o=\sqrt{2gh}$
From the $FBD$ during the collision, we note that net toque of $FBD$ during collision impulses about $O=0$
$\therefore$ The angular momentum of the system about $O$ is conversed.
If $L_1$ and $L_2$ are the angular momentum of the system
just before and just after the collision, then
$L_1=m{v}_{0}L$ and $L_2=l\omega =(\frac{{ML}^2}{3}+{mL}^2)\omega$
By law of conservation of angular momentum.
$(\frac{M}{3}+m)L^2\omega =m{v}_{0}L$
$\Rightarrow \omega =\frac{m{v}_0}{(\frac{M}{3}+m)L}$
Let the rod deflects through an angle $\theta$
Initial energy of the system $=\frac{1}{2}{\omega }^2$, where $I=(\frac{{ML}^2}{3}+{mL}^2)$
Gain in potential energy of the system
$=mgL[1-\cos \theta ]+Mg\frac{L}{2}[1-\cos \theta ]=(m+\frac{M}{3})gL(1-\cos \theta )$
$\because$ For law of conservation of energy,
$\frac{1}{2}(m+\frac{M}{3})gL(1-\cos \theta )$
$\Rightarrow \frac{1}{2}=(\frac{{ML}^2}{3}+{mL}^2)\times \frac{m^2{v}_0^2}{{(\frac{M}{3}+m)}^2{L}^2}(m+\frac{M}{3})gL(1-\cos \theta )$
$\Rightarrow \frac{1}{2}\frac{m^2{v}_0^2}{[\frac{M}{3}+m]}=(m+\frac{M}{3})gL(1-\cos \theta )$
$\therefore \cos \theta =1-\frac{1}{2}\frac{m^2{v}_0^2}{[\frac{M}{3}+m][\frac{M}{2}+m]gL}$
$\Rightarrow \theta ={\cos }^{-1}\left[\frac{mgh}{\{\left(m/3\right)+m\}\{\left(m/2\right)+m\}gL}\right]$.