Question

A thin rod of mass $0.9\text{kg}$ and length $1\text{m}$ is suspended at rest from one end so that it can freely oscillates in the vertical plane. A particle of move $80\text{m/s}$ hits the rod at its bottom most point and stickes to it (see figure). The angular speed (in $\text{rad/s}$) of the rod immeditely after the collision will be____

Answer 1

Using principal of conservation of angular momentum, we have

${\overrightarrow{L}}_i={\overrightarrow{L}}_f\Rightarrow mvL=I\omega$
$\Rightarrow mvL=(\frac{M{L}^2}{3}+m{L}^2)\omega$
$\Rightarrow 0.1\times 80\times 1=(\frac{0.9\times 1^2}{3}+0.1\times 1^2)\omega$
$\Rightarrow 8=(\frac{3}{10}+\frac{1}{10})\omega$
$\Rightarrow \omega =20\text{rad/sec}$