Question

A point object of mass $m=1\text{kg}$ moving horizontally with $10\text{m/s}$ hits the lower end of the uniform thin rod of length $2\text{m}$ and sticks to it. The rod is rested on a horizontal, frictionless surface and pivoted at the other end as shown in figure. Find out the angular velocity of the system just after collision.

• A. $15\text{rad/s}$
• B. $3.75\text{rad/s}$
• C. $7.5\text{rad/s}$
• D. Zero

Answer 1

The correct option is B $3.75\text{rad/s}$

As we can see, there is no external torque oo the system (rod + point object).
So the angular momentum of the system will remain same before and after the collision.


Hence,
$L_i=L_f$
$mv{r}_{\perp }=I\omega$
where $r_{\perp }=$ perpendicular distance of particle from the axis of rotation
Hence, $r_{\perp }=l=2\text{m}$
and $I$ is the MOI of the system (rod + point object) about axis of rotation $=\frac{m{l}^2}{3}+m{l}^2=\frac{4}{3}m{l}^2$
$\omega =$ angular velocity of system just after the collision
So, $mvl=\left(\frac{4}{3}m{l}^2\right)\omega$
$\Rightarrow \omega =\frac{3v}{4l}$
$\Rightarrow \omega =\frac{3\times 10}{4\times 2}=3.75\text{rad/s}$