Chapter 11, Problem 11.51AP

A projectile of mass m moves to the right with a speed ν_i (Fig. P11.51a). The projectile strikes and sticks to the end of a stationary rod of mass M, length d. pivoted about a frictionless axle perpendicular to the page through O (Fig. PH.51b). We wish to find the fractional change of kinetic energy in the system due to the collision, (a) What is the appropriate analysis model to describe the projectile and the rod? (b) What is the angular momentum of the system before the collision about an axis through O? (c) What is the moment of inertia of the system about an axis through O after the projectile sticks to the rod? (d) If the angular speed of the system after the collision is ω. what is the angular momentum of the system after the collision? (e) Find the angular speed to after the collision in terms of the given quantities. (f) What is the kinetic energy of the system before the collision? (g) What is the kinetic energy of the system after the collision? (h) Determine the fractional change of kinetic energy due to the collision.

To determine

The appropriate analysis model to describe the projectile and the rod.

Answer to Problem 11.51AP

The appropriate analysis model to describe the projectile and the rod is isolated system (angular momentum).

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$

From the conservation of the law of angular momentum the total angular momentum of the system remains constant when the net external torque acting on the system about the given axis is $0$.

It is clear from the given explanation and the figure that there is no external force that acts on the rod or the ball and so the angular momentum of the isolated system will remain conserved and so the model for the angular moment of the isolated system will be appropriate model for analysis of projectile and rod.

Conclusion:

Therefore, the appropriate analysis model to describe the projectile and the rod is isolated system (angular momentum).

To determine

The angular momentum of the system before an collision about an axis through $O$.

Answer to Problem 11.51AP

The angular momentum of the system before an collision about an axis through $O$ is $\frac{m{v}_{\text{i}}d}{2}$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$

The distance from the origin to the ball is,

$r=\frac{d}{2}$

Here,

$r$ is the distance from origin to the ball.

The formula to calculate the moment of inertia of the ball about the point $O$ is,

$I_{\text{b}}=m{r}^2$

Here,

$I_{\text{b}}$ is the moment of inertia of the ball.

Substitute $\frac{d}{2}$ for $r$ in the above equation.

$\begin{array}{c}{I}_{\text{b}}=m{\left(\frac{d}{2}\right)}^2\\ =\frac{m{d}^2}{4}\end{array}$

The expression for the initial angular moment of the ball is,

$L_{\text{i}}=I_{\text{b}}{\omega }_{\text{i}}$ (1)

Here,

${\omega }_{\text{i}}$ is the initial angular speed.

$L_{\text{i}}$ is the initial angular moment.

The relation between the angular speed and the linear speed is,

${\omega }_{\text{i}}=\frac{v_{\text{i}}}{r}$

Substitute $\frac{d}{2}$ for $r$ in the above equation.

$\begin{array}{c}{\omega }_{\text{i}}=\frac{v_{\text{i}}}{\left(\frac{d}{2}\right)}\\ =\frac{2v_{\text{i}}}{d}\end{array}$

Substitute $\frac{2v_{\text{i}}}{d}$ for ${\omega }_{\text{i}}$ and $\frac{m{d}^2}{4}$ for $I_{\text{b}}$ in the equation (1) for the initial angular momentum.

$\begin{array}{c}{L}_{\text{i}}=\left(\frac{m{d}^2}{4}\right)\left(\frac{2v_{\text{i}}}{d}\right)\\ =m{v}_{\text{i}}\left(\frac{d}{2}\right)\\ =\frac{m{v}_{\text{i}}d}{2}\end{array}$

Conclusion:

Therefore, the angular momentum of the system before an collision about an axis through $O$ is $\frac{m{v}_{\text{i}}d}{2}$.

To determine

The momentum of inertia of the system about an axis through $O$ after the projectile sticks to the rod.

Answer to Problem 11.51AP

The momentum of inertia of the system about an axis through $O$ after the projectile sticks to the rod is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$

From part (b) the moment of inertia of the ball is $\frac{m{d}^2}{4}$.

The formula to calculate the moment of inertia of the rod about the point $O$ is,

$I_{\text{r}}=\frac{1}{12}M{d}^2$

The moment of inertia of the system when projectile sticks to the rod is,

$I=I_{\text{r}}+I_{\text{b}}$

Here,

$I$ is the moment of inertia after collision.

Substitute $\frac{m{d}^2}{4}$ for $I_{\text{b}}$ and $\frac{1}{12}M{d}^2$ for $I_{\text{r}}$ in the above equation.

$\begin{array}{c}I=\frac{m{d}^2}{4}+\frac{1}{12}M{d}^2\\ =\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\end{array}$

Conclusion:

Therefore, the momentum of inertia of the system about an axis through $O$ after the projectile sticks to the rod is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$.

To determine

The angular momentum of the system after collision.

Answer to Problem 11.51AP

The angular momentum of the system after collision is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\omega$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$ and the angular speed of the system after collision is $\omega$.

The expression for the final angular moment of the ball is,

$L_{\text{f}}=I\omega$

Here,

$L_{\text{f}}$ is the final angular momentum.

From part (c) momentum of inertia of the system about an axis through $O$ after the projectile sticks to the rod is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$.

Substitute $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$ for $I$ in the above equation for the final angular momentum.

$\begin{array}{c}{L}_{\text{i}}=\left(\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\right)\omega \\ =\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\omega \end{array}$

Conclusion:

Therefore, the angular momentum of the system after collision is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\omega$.

To determine

The angular speed $\omega$ after collision in terms of the given quantities.

Answer to Problem 11.51AP

The angular speed $\omega$ after collision in terms of the given quantities is $\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$

From the law of conservation of angular momentum the expression for the angular momentum is,

$L_{\text{f}}=L_{\text{i}}$

Substitute $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$ for $L_{\text{f}}$ and $\frac{m{v}_{\text{i}}d}{2}$ for $L_{\text{i}}$ in the above equation.

$\begin{array}{c}\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2\omega =\frac{m{v}_{\text{i}}d}{2}\\ \omega =\left(\frac{6m{v}_{\text{i}}}{Md+3md}\right)\\ =\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}\end{array}$

Conclusion:

Therefore, the angular speed $\omega$ after collision in terms of the given quantities is $\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}$.

To determine

The kinetic energy of the system before collision.

Answer to Problem 11.51AP

The kinetic energy of the system before collision is $\frac{1}{2}m{v}_{\text{i}}{}^2$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$ and the angular speed of the system after collision is $\omega$.

The formula to calculate the kinetic energy of the system is,

$K{E}_{\text{i}}=\frac{1}{2}m{v}^2$

Here,

$K{E}_{\text{i}}$ is the translational kinetic energy of the system.

$v$ is the velocity of the object.

Substitute $v_{\text{i}}$ for $v$ in the above equation to calculate the initial kinetic energy of the system.

$E_{\text{k}}=\frac{1}{2}m{v}_{\text{i}}{}^2$

Conclusion:

Therefore, the kinetic energy of the system before collision is $\frac{1}{2}m{v}_{\text{i}}{}^2$.

To determine

The kinetic energy of the system after collision.

Answer to Problem 11.51AP

The kinetic energy of the system after collision is $\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$ and the angular speed of the system after collision is $\omega$.

The expression for the final kinetic energy of the system is,

$K{E}_{\text{f}}=\frac{1}{2}I{\omega }^2$

Here,

$K{E}_{\text{f}}$ is the final kinetic energy of the system.

From part (c) momentum of inertia of the system about an axis through $O$ after the projectile sticks to the rod is $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$.

And from part (e) the angular speed $\omega$ after collision in terms of the given quantities is $\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}$.

Substitute $\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2$ for $I$ and $\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}$ for $\omega$ in the above equation for the final angular momentum.

$\begin{array}{c}{L}_{\text{i}}=\frac{1}{2}\left(\frac{1}{4}m+\frac{1}{12}M\right)d^2{\left(\frac{6m{v}_{\text{i}}}{\left(M+3m\right)d}\right)}^2\\ =\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}\end{array}$

Conclusion:

Therefore, the kinetic energy of the system after collision is $\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}$.

To determine

The fractional change of kinetic energy due to collision.

Answer to Problem 11.51AP

The fractional change in the kinetic energy of the system is $-\frac{M}{\left(M+3m\right)}$.

Explanation of Solution

Given info: The mass of the projectile is $m$, the initial speed of the projectile is $v_{\text{i}}$, the mass of the rod is $M$, length of the rod is $d$ and the angular speed of the system after collision is $\omega$.

The expression to calculate the change in the kinetic energy of the system is,

$\Delta KE=K{E}_{\text{f}}-K{E}_{\text{i}}$

Here,

$\Delta KE$ is the change in the kinetic energy of the system.

From part (g) kinetic energy of the system after collision is $\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}$ and from part (f) kinetic energy of the system before collision is $\frac{1}{2}m{v}_{\text{i}}{}^2$.

Substitute $\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}$ for $K{E}_{\text{f}}$ and $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K{E}_{\text{i}}$ in the above equation for the final angular momentum.

$\begin{array}{c}\Delta KE=\frac{3m^2{v}_{\text{i}}^2}{2\left(M+3m\right)}-\frac{1}{2}m{v}_{\text{i}}{}^2\\ =-\left[\frac{mM}{2\left(M+3m\right)}\right]v_{\text{i}}^2\end{array}$

Thus, the change in the kinetic energy is $-\left[\frac{mM}{2\left(M+3m\right)}\right]v_{\text{i}}^2$.

The formula to calculate the fractional change in the kinetic energy is,

$\text{Fractional}\text{change}\text{in}\text{kinetic}\text{energy}=\frac{\Delta KE}{K{E}_{\text{i}}}$

Substitute $-\left[\frac{mM}{2\left(M+3m\right)}\right]v_{\text{i}}^2$ for $\Delta KE$ and $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K{E}_{\text{i}}$ in the above equation.

$\begin{array}{c}\text{Fractional}\text{change}\text{in}\text{kinetic}\text{energy}=\frac{-\left[\frac{mM}{2\left(M+3m\right)}\right]v_{\text{i}}^2}{\frac{1}{2}m{v}_{\text{i}}{}^2}\\ =-\frac{M}{\left(M+3m\right)}\end{array}$

Conclusion:

Therefore, the fractional change in the kinetic energy of the system is $-\frac{M}{\left(M+3m\right)}$.