Question

In Figure a $\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{g}$ bullet is fired into a $\mathbf{0.50}\mathbf{}\mathbf{kg}$block attached to the end of a $\mathbf{0.60}\mathbf{}\mathbf{m}$ non uniform rod of mass $\mathbf{0.50}\mathbf{}\mathbf{kg}\mathbf{.}$The block–rod–bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about that axis at Ais $\mathbf{0}\mathbf{.060}\mathbf{}\mathbf{kg}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}\mathbf{.}$ Treat the block as a particle.

(a) What then is the rotational inertia of the block–rod–bullet system about point A?

(b) If the angular speed of the system about Ajust after impact$\mathbf{is}\mathbf{}\mathbf{4}\mathbf{.5}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}\mathbf{,}$ what is the bullet’s speed just before impact?

Short answer

1. Rotational inertia of block-rod-bullet system is$\mathrm{I}=0.241\text{kg}.{\text{m}}^2.$
2. Bullet’s speed just before impact is $\mathrm{v}=1800\text{m/s.}$

Step-by-step solution

Step 1: Given

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, ${\mathrm{I}}_0=0.0{\text{6kg.m}}^{\text{2}}$

Determining the concept

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.

 Determining the rotational inertia of block-rod-bullet system

(a)

Rotational inertia of block-rod-bullet system is,

$\begin{array}{l}\mathrm{I}={\mathrm{I}}_0+{\mathrm{mr}}^2\\ \mathrm{I}=0.06+0.50\times {0.6}^2=0.241{\text{\hspace{0.17em}\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

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

Determining the bullet’s speed just before impact

(b)

According to law of conservation of angular momentum:

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

$\begin{array}{c}{\mathrm{L}}_1={\mathrm{L}}_2\\ \mathrm{mvr}=\mathrm{I\omega }\\ 0.001\times \mathrm{v}\times 0.6=0.24\times 4.5\\ \mathrm{v}=1800\text{m/s}\end{array}$

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.