Question

A horizontal rod of length $l=2\text{m}$ and mass $2\text{kg}$ with the balls $\text{A}$ and $\text{B}$ of mass $1\text{kg}$ each at its ends is clamped at the centre $\text{C}$ in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle $P$ of mass $1\text{kg}$ is dropped from a height of $5\text{m}$ on the ball $B$. The particle collides with $B$ and sticks to it. Find the angular velocity of the system just after the collision.

• A. $\frac{10}{3}\text{rad/s}$
• B. $\frac{20}{3}\text{rad/s}$
• C. $\frac{10}{11}\text{rad/s}$
• D. $\frac{30}{11}\text{rad/s}$

Answer 1

The correct option is D $\frac{30}{11}\text{rad/s}$

Given,
Mass of ball $A,B$ and $P=1\text{kg}$
Mass of rod $=M=2\text{kg}$
Length of rod $=l=2\text{m}$

Using work energy theorem for particle $P$:
$mgh+0=\frac{1}{2}m{v}^2+0$
$v=\sqrt{2gh}$ ... (1) is the speed of particle $P$ just before collision :

There is no external torque on the system. Hence the angular momentum of the system remains same.
Hence, $L_i=L_f$


$L_i=mv{r}_{\perp }$
where, $r_{\perp }=\left(\frac{l}{2}\right)$
Hence, $L_i=m\left(\sqrt{2gh}\right)\left(\frac{l}{2}\right)$
($r_{\perp }$ is the perpendicular distance of velocity vector from the axis of rotation)


$L_f=I\omega$
where $I=$ MOI of the system
$I=\frac{M{l}^2}{12}+3m{\left(\frac{l}{2}\right)}^2$
$I=\frac{M{l}^2}{12}+\frac{3m{l}^2}{4}$
$L_f=(\frac{M{l}^2}{12}+\frac{3m{l}^2}{4})\omega$
From $L_i=L_f$
$m\left(\sqrt{2gh}\right)\times \left(\frac{l}{2}\right)=(\frac{M{l}^2}{12}+\frac{3m{l}^2}{4})\omega$
$1\times \sqrt{2\times 10\times 5}\times \left(\frac{2}{2}\right)=(\frac{2\times 2^2}{12}+\frac{3\times 1\times 2^2}{4})\times \omega$
$10=(\frac{2}{3}+3)\omega$
i.e $\omega =\frac{30}{11}\text{rad/s}$