Question

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 $\frac{L}{3}$ and $\frac{L}{6}$ 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 :

• A. $\frac{v}{5L}$
• B. $\frac{6v}{5L}$
• C. $\frac{3v}{5L}$
• D. $\frac{v}{6L}$

Answer 1

Answer: (B)

Centre of mass after the collision will be at $O$. Therefore, $2m\frac{L}{6}=m\frac{L}{3}$
Now, using conservation of angular momentum, we get $2m\frac{L}{6}\times v+m\frac{L}{3}\times 2v=I\omega$
Therefore, $2m\frac{L}{6}\times v+m\frac{L}{3}\times 2v=\left[8m\frac{L^2}{12}+2m{\left(\frac{L}{6}\right)}^2+m{\left(\frac{L}{3}\right)}^2\right]\omega$
$\Rightarrow \left(\frac{L}{6}+\frac{1}{3}\right)2mv=\left[8m\frac{L^2}{12}+\frac{2m{L}^2}{36}+\frac{m{L}^2}{9}\right]\omega$
$\Rightarrow \frac{3L}{6}\times 2mV=\frac{5}{6}m{L}^2\omega$
$\Rightarrow Lmv=\frac{5}{6}m{L}^2\omega$
$\Rightarrow \omega =\frac{6Lmv}{5m{L}^2}$
$\Rightarrow \omega =\frac{6}{5}\frac{v}{L}$