Question

Figure is an overhead view of a thin uniform rod of length $0.600m$ and mass $M$ rotating horizontally at $80.0rad/s$ counter clockwise about an axis through its center. A particle of mass $\frac{M}{3}$ and traveling horizontally at speed $40.0m/s$ hits the rod and sticks and stops the rod. The particle’s path is perpendicular to the rod at the instant of the hit, at a distance d from the rod’s center. At what value of $d$ are rod and particle stationary after the hit?

• A. $0.25m$
• B. $0.18m$
• C. $0.15m$
• D. $0.36m$

Answer 1

The correct option is B $0.18m$

Formula used:$L=mvr,L=I\omega$
Given : $l=6m,\omega =80rad/s,m_{particle}=\frac{M}{3},v=40m/s$
We consider conservation of angular momentum about the centre of the rod
$L_i=L_f$
$-mvd+I\omega =0$
$-\frac{M}{3}\times 40\times d+\left(\frac{M{l}^2}{12}\right)\omega =0$
$\frac{40}{3}d=\left(\frac{{0.6}^2}{12}\right)\times 80$
$d=\frac{9}{50}=0.18m$

Final answer:(d)