Question

A uniform rod of mass m and length is hinged at point A, which is at a distance $\frac{1}{4}$ from the centre of mass of the rod. The rod is free to rotate in a vertical plane about A. A particle moving horizontally strikes at the end P of the rod perpendicularly. The mass of the particle is m/2 and its speed is $v_0$ , just before collision. If particle stops just after collision, if the angular velocity of the rod just after collision in $\frac{18v_0}{xl}$ Find x. ___

Answer 1

Angular momentum about point A is conserved i.e.
$L_i=L_f$
$\frac{m}{2}{v}_0\frac{3l}{4}=I_A\omega$
$\Rightarrow \frac{3}{8}m{v}_{0}l=\left[\frac{7}{48}m{l}^2\right]\omega |1_A=\frac{m{l}^2}{12}+m{\left(\frac{1}{4}\right)}^2$
$\Rightarrow \omega =\frac{18}{7}\frac{v_0}{1}$