A sick of length $\mathit{L}$and mass $\mathit{M}$hangs from a low-friction axle (Figure 11.90). A bullet of mass $\mathit{m}$travelling at a high speedstrikes $\mathit{v}$near the bottom of the stick and quickly buries itself in the stick.

(a) During the brief impact, is the linear momentum of the stick + bullet system constant? Explain why or why not. Include in your explanation a sketch of how the stick shifts on the axle during the impact. (b) During the brief impact, around what point does the angular momentum of the stick + bullet system remain constant? (c) Just after the impact, what is the angular speed $\mathit{\omega }$of the stick (with the bullet embedded in it) ? (Note that the center of mass of the stick has a speed $\mathit{\omega }\mathit{L}\mathbf{/}\mathbf{2}\mathbf{.}$The moment of inertia of a uniform rod about its center of mass is$\frac{\mathbf{1}}{\mathbf{12}}\mathit{M}{\mathit{L}}^{\mathbf{2}}\mathbf{.}$(d) Calculate the change in kinetic energy from just before to just after the impact. Where has this energy gone? (e) The stick (with the bullet embedded in it) swings through a maximum angle${\mathit{\theta }}_{\mathbf{max}}$after the impact, then swing back. Calculate ${\mathit{\theta }}_{\mathbf{max}}$.

Angular speed is the rate at which the central angle of a spinning body varies over time. We will become acquainted with angular speed in this post.

Initial angular momentum of the bullet with respect to axis of rotation $\left(O\right)$is,

$$\begin{gathered} L_i=m\left(L\right)v \\ =mvL \end{gathered}$$

Moment of inertia of the system is,

$$\begin{gathered} I=I_{stick}+I_{bullet} \\ =\frac{1}{12}\left(M{L}^2\right)+\left(m{L}^2\right) \\ =L^2\left(\frac{M}{12}+m\right) \end{gathered}$$

(a) If there is no external force acting on the system, its overall momentum is always conserved. Because the stick and bullet are part of a system, their momentum is preserved.

The following diagram shows a stick of length $L,$mass $M$is suspended from the point $O$. A bullet of mass $m$travelling at high speed $v$strikes the bottom of the stick and after collision both are combined to move with a maximum angular displacement is ${\theta }_{\max }$.
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(b) The angular momentum of the stick and bullet system around the axis of rotation remains constant, as indicated in the diagram.

(c) There are no internal forces working on the system. The law of conservation of angular momentum ensures that the initial and final angular momentum stay constant.

$$\begin{gathered} L_i+L_f \\ mLv=I\omega \\ mLv=L^2\left(\frac{M}{12}+m\right)\omega \\ \omega =\left(\frac{\left(mv\right)}{\left(\frac{M}{12}+m\right)L}\right) \end{gathered}$$

Therefore, the angular speed is $\omega =\left(\frac{\left(mv\right)}{\left(\frac{M}{12}+m\right)L}\right)$.

(d) The kinetic energy of the bullet after collision is calculated as follows,

$$\begin{gathered} K_f=\frac{1}{2}I\omega \\ =\left(\frac{1}{2}\right)L^2\left(\frac{M}{12}+m\right){\left(\frac{\left(mv\right)}{\left(\frac{M}{12}+m\right)L}\right)}^2 \\ =\left(\frac{1}{2}\right)\left(\frac{m^2{v}^2}{\left(\frac{M}{12}+m\right)}\right) \end{gathered}$$

Change in kinetic energy before and after impact is calculated as follows,

Therefore,

$$\begin{gathered} \Delta KE=\frac{1}{2}m{v}^2\left(\frac{\frac{M}{12}}{\left(\frac{M}{12}+m\right)}\right) \\ =\frac{1}{2}\frac{Mm}{(M+12m)}{v}^2 \end{gathered}$$

Here negative sign indicates loss of energy that can be appears as heat.

(e)After collision both stick and the bullet rises to a height $h.$From the triangle $OCB.$

$$\begin{gathered} \cos {\theta }_{\max }=\frac{L-h}{L} \\ \cos {\theta }_{\max }=\left(1-\frac{h}{L}\right) \\ {\theta }_{\max }={\cos }^{-1}\left(1-\frac{h}{L}\right) \end{gathered}$$

The stick swings through a maximum angle${\theta }_{\max }={\cos }^{-1}\left(1-\frac{h}{L}\right)$.