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  • The rod of length L and mass M is hinged at point O. A small bullet of mass m hits the rod as shown in the figure. The bullet gets embedded in the rod. Find the angular velocity of the system just

The rod of length L and mass M is hinged at point O. A small bullet of mass m hits the rod as shown in the figure. The bullet gets embedded in the rod. Find the angular velocity of the system just after impact.

Option: 1

\frac{2.4 \mathrm{mu}}{[\mathrm{M}+1.92 \mathrm{m}] \mathrm{L}}


Option: 2

\frac{2.4 \mathrm{mu}}{[\mathrm{M}+44 \mathrm{m}] \mathrm{L}}


Option: 3

\frac{2.4 \mathrm{mu}}{\mathrm{M} {L}}


Option: 4

\frac{34 \mathrm{mu}}{[\mathrm{M}+1.92 \mathrm{m}] \mathrm{L}}


Answers (1)

best_answer

Before the collision, the rod is at rest and the bullet is moving with velocity u at a distance 0.8 L as shown in the figure


Therefore angular momentum of the system before impact is,
L_{1}=\frac{1}{3} M L^{2}(0)+m u(0.8 L)=0.8 \mathrm{muL}$.....(i)

Bullet after hitting gets embedded into the rod.
Let $\omega$  be the angular velocity of the system after impact.
The angular momentum of system after impact is,
\begin{aligned} L_{2} &=\left[\frac{1}{3} M L^{2}+m(0.8 L)^{2}\right] \omega =\left[\frac{1}{3} M+0.64 m\right] L^{2} \omega \end{aligned}$...(2)
Using the law of conservation of angular momentum
$$ \begin{aligned} 0.8 \mathrm{muL} &=\left[\frac{1}{3} \mathrm{M}+0.64 \mathrm{m}\right] \mathrm{L}^{2} \omega \\ \Rightarrow \quad \omega &=\frac{0.8 \mathrm{muL}}{\left[\frac{1}{3} \mathrm{M}+0.64 \mathrm{m}\right] \mathrm{L}^{2}} =\frac{2.4 \mathrm{mu}}{[\mathrm{M}+1.92 \mathrm{m}] \mathrm{L}} \end{aligned} $$

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rishi.raj

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