Question

A meter stick is pivoted about its centre. A piece of wax of mass $20\text{g}$ travelling horizontally and perpendicular to it at $10\text{m/s}$ strikes and adheres to one end of the stick so that the stick starts to rotate in a horizontal circle. Given the moment of inertia of the stick and wax about the pivot is $0.02{\text{kg-m}}^2$, the initial angular velocity of the stick is (in $\text{rad/s}$)

Answer 1

Given,
Mass of wax $m=20\text{g}$,
For metre stick, length $l=1\text{m}$
Speed of wax $v=10\text{m/s}$
MOI of stick about pivot end $I=0.02{\text{kg-m}}^2$


As we can see, since there is no net external force, the linear momentum of the system remains constant. Also there is no net external torque on the system so the angular momentum of the system about the any axis will also remain constant.
As we know,
$L=I\omega$
$mv{r}_{\perp }=I\omega$
$0.02\times 10\times \frac{1}{2}=0.02\times \omega$ $[\because r_{\perp }=\frac{l}{2}]$
$\therefore \omega =5\text{rad/s}$