Question

A mass $m$ is moving at speed $v$ perpendicular to a rod of length $d$ and mass $M=6m$ which pivots around a frictionless axle running through its centre. It strikes and sticks to the end of the rod. The moment of inertia of the rod about its centre is $\frac{M{d}^2}{12}$. Then the angular speed of the system just after the collision is :

• A. $2v/d$
• B. $2v/\left(3d\right)$
• C. $v/d$
• D. $3v/\left(2d\right)$

Answer 1

The correct option is A $2v/d$

Since there is no torque on the system, angular momentum will be conserved.
$L_i=mv\frac{d}{2}$
$L_f=I\omega =(\frac{M{d}^2}{12}+m(\frac{d}{2}{)}^2)\omega =(\frac{6m{d}^2}{12}+m\left(\frac{d}{2}{)}^2\right)\omega =\frac{3m{d}^2}{4}\omega$
$L_i=L_f$
$\Rightarrow mv\frac{d}{2}=\frac{3m{d}^2}{4}\omega$
$\Rightarrow \omega =\frac{2v}{3d}$