91Ó°ÊÓ

1. Mass of the rod,$m=0.\text{5kg}$
2. Length of the rod,$\mathrm{r}=0.6\text{m}$
3. Rotational inertia of the rod, ${\mathrm{I}}_0=0.0{\text{6kg.m}}^{\text{2}}$

Find the total rotational inertia of the system by adding the rotational inertia of rod and bullet. Then, using the law of conservation of angular momentum, find speed of the bullet just before the impact.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

$\begin{array}{l}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{0}}\mathbf{+}{\mathbf{mr}}^{\mathbf{2}}\\ \mathbf{L}\mathbf{=}\mathbf{mvr}\\ \mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\end{array}$

where, $\mathbf{I}$is final moment of inertia,$\begin{array}{l}{\mathbf{I}}_{\mathbf{0}}\end{array}$is initial moment of inertia, $\begin{array}{l}\mathbf{m}\end{array}$ is mass, $\mathbf{v}$ is velocity, $\mathbf{L}$is angular momentum and$\mathbf{r}$is radius.

(a)

Rotational inertia of block-rod-bullet system is,

$\begin{array}{l}\mathrm{I}={\mathrm{I}}_0+{\mathrm{mr}}^2\\ \mathrm{I}=0.06+0.50×{0.6}^2=0.241{\text{ â¶Ä‰k²µ.³¾}}^{\text{2}}\end{array}$

Hence, rotational inertia of block-rod-bullet system is $\mathrm{I}=0.241\text{kg}.{\text{m}}^2.$

(b)

According to law of conservation of angular momentum:

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{c}{\mathrm{L}}_1={\mathrm{L}}_2\\ \mathrm{mvr}=\mathrm{\pm õÓ\neg }\\ 0.001×\mathrm{v}×0.6=0.24×4.5\\ \mathrm{v}=1800\text{m/s}\end{array}$

Hence,bullet’s speed just before impact is$\mathrm{v}=1800\text{m/s.}$

Therefore, the total rotational inertia of system of block rod and bullet can be found.

Then, using law conservation of angular momentum, the unknown particle’s speed can be calculated.