Question

A thin metal rod of mass$\mathbf{\text{1.3 kg}}$ and length$\mathbf{\text{0.4 m}}$ is at rest in outer space, near a space station (Figure 11.99). A tiny meteorite with mass $\mathbf{\text{0.06 kg}}$ travelling at a high speed of strikes the rod a distance $\mathbf{\text{0.2 m}}$ from the center and bounces off with speed 60 m/s as shown in the diagram. The magnitudes of initial and final angles to the$\mathit{x}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$ of the small mass’s velocity are ${\mathit{\theta }}_{\mathbf{i}}\mathbf{=}\mathbf{26}\mathbf{°}$ and ${\mathit{\theta }}_{\mathbf{f}}\mathbf{=}\mathbf{82}\mathbf{°}$.(a). Afterward, what is the velocity of the center of the rod? (b) Afterward, what is the angular velocity $\overrightarrow{\mathbf{\omega }}$ of the rod? (c) What is the increase in internal energy of the objects?

Short answer

The velocity of the center of mass of the rod along horizontal is 8.68m/s.

The magnitude of the center of mass of the rod is 8.78 m/s.

The angular velocity of the rod is 130 rad/s.

The increase in internal kinetic energy of the system is $803J$.

Step-by-step solution

Definition of law of conservation of momentum.

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects.

Use the law of conservation of momentum and the formula for the moment of inertia of the rod. The moment of inertia of the rod with respect to its center is as follows:

$I=\frac{1}{21}M{L}^2$

Here, $M$is the mass of the rod and$L$is the length of the rod.

The motion of the meteorite is shown in the following figure:

In the figure, $m$is the mass of the meteorite, $d$is the distance from the center of the rod to the point of application of velocity, $L$is the length of the rod, and$M$is the mass of the rod.

Find the velocity of the center of mass of the rod.

1. Calculate the center of mass of the rod after the collision as follows:

Apply the law of conservation of momentum along the horizontal as follows:

$m{v}_i\cos {\theta }_i=-m{v}_f\cos {\theta }_f+M{V}_{x,cm}$

Here$V_{x,cm}$is the center of mass of the rod along$x-$direction.

Rearrange the equation for the velocity of the center of mass of the rod.

$V_{x,cm}=\frac{m{v}_i\cos {\theta }_i+m{v}_f\cos {\theta }_f}{M}$

Substitute$200m/s$for$v_i,60m/s$for$v_f,0.06kg$for $m,1.3kg$for$M,26°$for ${\theta }_i,$and $82°$for${\theta }_f$ in the equation $V_{x,cm}=\frac{m{v}_i\cos {\theta }_i+m{v}_f\cos {\theta }_f}{M}$and solve for $V_{x,cm}.$

$V_{cm}=\frac{(0.06kg)(200m/s)\cos 26°+(0.06kg)(60m/s)\cos 82°}{1.3kg}$

$=8.68m/s$

Therefore, the velocity of the center of mass of the rod along horizontal is 8.68m/s.

Find the magnitude of the center of mass of the rod.

Apply the law of conservation of momentum along the vertical as follows:

$m{v}_i\sin {\theta }_i=m{v}_f+M{V}_f\sin {\theta }_f+M{V}_{y,cm}$

Here$V_{y,cm}$is the center of mass of the rod along$y-$direction.

Rearrange the equation for the velocity of the center of mass of the rod.

$V_{y,cm}=\frac{m{v}_i\sin {\theta }_i-m{v}_f\sin {\theta }_f}{M}$

Substitute $200m/s$for$v_i,60m/s$ for$v_f,0.06kg$for$m,1.3kg$for$M,26°$for ${\theta }_i,$and$82°$for${\theta }_f$in the equation $V_{y,cm}=\frac{m{v}_i\sin {\theta }_i-m{v}_f\sin {\theta }_f}{M}$and solve for $V_{y,cm}$

$V_{y,cm}=\frac{(0.06kg)(200m/s)\sin 26°-(0.06kg)(60m/s)\sin 82°}{1.3kg}$

$=1.3m/s$

Therefore, the velocity of the center of mass of the rod along vertical is $1.3m/s$.

Therefore, the center of mass of the rod is

The magnitude of center of mass of the rod is as follows:

$\begin{array}{rcl}{V}_{cm}& =& \sqrt{{(8.68m/s)}^2+{(1.3m/s)}^2+{(0m/s)}^2}\\ & =& 8.78m/s\end{array}$

Therefore, the magnitude of the center of mass of the rod is 8.78m/s.

Find the angular velocity of the rod.

b. Apply law of conservation of angular momentum.

$L_i=L_f$

$\left(m{v}_i\cos {\theta }_i\right)d=-\left(m{v}_f\cos {\theta }_f\right)d+I\omega$$\left(m{v}_i\cos {\theta }_i\right)d=-\left(m{v}_f\cos {\theta }_f\right)d+I\omega$

Here$I$is the moment of inertia of the rod and is the angular velocity.

Substitute$\frac{1}{12}M{L}^2$for$\left(m{v}_i\cos {\theta }_i\right)d=-\left(m{v}_f\cos {\theta }_f\right)d+I\omega$in the equation

$m{v}_i\cos {\theta }_{i}d=-md{v}_f\cos {\theta }_f+\left(\frac{1}{12}M{L}^2\right)\omega$$m{v}_i\cos {\theta }_{i}d=-md{v}_f\cos {\theta }_f+\left(\frac{1}{12}M{L}^2\right)\omega$

Rearrange the equation for the angular velocity.

$\omega =\frac{md(v_i\cos {\theta }_i+v_f\cos {\theta }_f)}{\left(\frac{1}{12}M{L}^2\right)}$

Substitute $0.4m$for$L,0.2m$for$d,200m/s$for $v_i,60m/s$for$v_f,0.06kg$for$m,1.3kg$for $M,26°$for ${\theta }_i,$and $82°$for ${\theta }_f.$

$\omega =\frac{(0.06kg)(0.2m)\left(\right(200m/s)\cos 26°+(60m/s)\cos 82°)}{\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)}$

$\omega =\frac{(0.06kg)(0.2m)\left(\right(200m/s)\cos 26°+(60m/s)\cos 82°)}{\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)}$

$=130rad/s$

Therefore, the angular velocity of the rod is$130rad/s.$

Find the internal kinetic energy of the system.

(c) Calculate the kinetic energy is of the system before collision as follows:

Initially the rod is in rest. So, initial kinetic of the rod is zero.

$K_i=\frac{1}{2}m{v_i}^2$

Calculate the kinetic energy is of the system before collision as follows:

$K_f=\frac{1}{2}m{v_f}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}M{V_{cm}}^2$

Calculate the increase in internal energy as follows:

The expression for the change in internal energy is as follows:

$$\begin{gathered} \Delta E=K_i-K_f \\ \Delta E=\frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_f}^2-\frac{1}{2}I{\omega }^2-\frac{1}{2}M{V_{cm}}^2 \end{gathered}$$

Substitute $\frac{1}{12}M{L}^2$for in the equation $\Delta E=\frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_f}^2-\frac{1}{2}I{\omega }^2-\frac{1}{2}M{V_{cm}}^2$and solve for $\Delta E.$

$\Delta E=\frac{1}{2}m({v_i}^2-{v_f}^2)-\frac{1}{2}\left(\frac{1}{12}M{L}^2\right){\omega }^2-\frac{1}{2}M{V_{cm}}^2$

Substitute $0.4m$for$L,200m/s$for$v_1,60m/s$for$v_f,0.06kg$for $m,1.3kg$for $M,8.78m/s$for$V_{cm},$and 130rad/s for$\omega$in the equation

$\Delta E=\frac{1}{2}m({v_i}^2-{v_f}^2)-\frac{1}{2}\left(\frac{1}{12}M{L}^2\right){\omega }^2-\frac{1}{2}M{V_{cm}}^2$and solve for$\Delta E$.

$$\begin{gathered} \Delta E=\left\{\left(\frac{0.06kg}{2}{\left((200m/s\right)}^2-{(60m/s)}^2\right)-\frac{1}{2}\left(\frac{1}{12}(1.3kg){(0.4m)}^2\right)(130rad/s{)}^2 \\ -\left[\frac{1}{2}(1.3kg){\left((8.68m/s\right)}^2+{(1.3m/s)}^2\right] \\ \right\} \end{gathered}$$

$$\begin{gathered} =895.46J \\ \approx 895.5J \end{gathered}$$

Therefore, the increase in internal kinetic energy of the system is $803J$.