A thin uniform bar of length L and mass 8m lies on a smooth horizontal table . Two point masses m and 2m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively . The masses stick to the bar after collision at a distance $\frac{\mathrm{L}}{3}$and $\frac{\mathrm{L}}{6}$  respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

 

detailed solution

Correct option is C

Here, the two masses crash with the rod and become locked together; as a result, the masses enter the system and the imposed external torque is therefore required to be zero. Therefore, the system's angular momentum must be conserved. We are aware that a body's centre of mass is determined by,

$R=\frac{\sum m_i{r}_i}{\sum m_i}$

After the collision, the system's centre of mass will be located around point O.

$\mathrm{R}=\frac{8\mathrm{m}\times 0+2\mathrm{m}\times {\displaystyle \frac{\mathrm{L}}{6}}-\mathrm{m}\times {\displaystyle \frac{\mathrm{L}}{3}}}{8\mathrm{m}+2\mathrm{m}+\mathrm{m}}=0$
We also know that a body's angular momentum around a point is determined by,

$\mathrm{L}=\mathrm{mvr}=\mathrm{I\omega }$

Now, the system's angular momentum before impact is,

${\mathrm{L}}_{\mathrm{i}}=\mathrm{m}\times 2\mathrm{v}\times \frac{\mathrm{L}}{3}+2\mathrm{m}\times \mathrm{v}\times \frac{\mathrm{L}}{6}$

${\mathrm{L}}_{\mathrm{i}}=\frac{2\mathrm{mvL}}{3}+\frac{\mathrm{mvL}}{3}=\mathrm{mvL}$

Let's assume that the system's angular speed upon collision is,
Thus, the system's angular momentum around O after the impact is,

${\mathrm{L}}_{\mathrm{f}}=({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{I}}_{\mathrm{m}}+{\mathrm{I}}_{2\mathrm{m}})\mathrm{\omega }$

Now, a rod's moment of inertia around its centre of mass is given by,

$\mathrm{I}=\frac{{\mathrm{ML}}^2}{12}$

${\mathrm{I}}_{\mathrm{rod}}=\frac{8{\mathrm{mL}}^2}{12}$

The mass m's moment of inertia is $I_m$$=\mathrm{m}{\left(\frac{\mathrm{L}}{3}\right)}^2=\frac{{\mathrm{mL}}^2}{9}$

The mass 2m moment of inertia is  ${\mathrm{I}}_{2\mathrm{m}}=2\mathrm{m}{\left(\frac{\mathrm{L}}{6}\right)}^2=\frac{{\mathrm{mL}}^2}{18}$

thus we get,

${\mathrm{L}}_{\mathrm{f}}=({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{I}}_{\mathrm{m}}+{\mathrm{I}}_{2\mathrm{m}})\mathrm{\omega }$

${\mathrm{L}}_{\mathrm{f}}=\left(\frac{8{\mathrm{mL}}^2}{12}+\frac{{\mathrm{mL}}^2}{9}+\frac{{\mathrm{mL}}^2}{18}\right)\mathrm{\omega }=\frac{5}{6}{\mathrm{mL}}^2\mathrm{\omega }$

$\mathrm{Thus}, {\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}$

$\mathrm{mvL}=\frac{5}{6}{\mathrm{mL}}^2\mathrm{\omega }$

$\mathrm{\omega }=\frac{6\mathrm{v}}{5\mathrm{L}}$