Question

A uniform bar of length $6a$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in opposite direction in the same horizontal plane with speed $2v$ and $v$, respectively. strike the bar and stick to the bar after the collision. Calculate :
The angular velocity of the centre of mass after collision.

Answer 1

Since no force is applied $\overrightarrow{F_{ext}}=0$ so, linear momentum is conserved

$\Rightarrow -2m\nu +m\left(2\nu \right)+0=(2m+m+8m){\nu }_C\Rightarrow {\nu }_C=0$

Since $Y_{ext}=0$ angular momentum is conserved

$\Rightarrow m_1{\nu }_1{r}_1+m_2{\nu }_2{r}_2=(I_1+I_2+I_3)\omega \Rightarrow 2m{\nu }_a+m\left(2m\nu \right)\left(2a\right)=(2m{a}^2+m(2a{)}^2+\left(\frac{8m\times 36a^2}{12}\right))\omega \Rightarrow 6m{\nu }_a=30m\omega a^2$

$\Rightarrow$Angular velocity$\omega =\frac{v}{5a}$