Chapter 13.4, Problem 13.184P

A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible mass. Knowing that the arm is released from the horizontal position and that the coefficient of restitution between the foot and the 1-lb ball is 0.8, determine the exit velocity of the ball (a) if the ball is stationary, (b) if the ball is struck when it is rolling toward the foot with a velocity of 10 ft/s.

Fig. P13.184

To determine

Find the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is stationary.

Answer to Problem 13.184P

The exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is stationary is $\underset{\_}{26.65\text{ft/s}\left(30°\right)}$.

Explanation of Solution

Given information:

The weight of the stimulated foot $\left(W_A\right)$ is $5\text{lb}$.

The weight of the ball $\left(W_B\right)$ is $1\text{lb}$.

The length of the pendulum arm (l) is $6\text{ft}$.

The coefficient of restitution between the foot (e) is 0.8.

The angle $\left(\theta \right)$ is $60°$.

The acceleration of gravity (g) is $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Calculate the mass of the stimulated foot $\left(m_A\right)$ using the formula:

$\begin{array}{c}{W}_A=m_{A}g\\ m_A=\frac{W_A}{g}\end{array}$

Substitute $5\text{lb}$ for $W_A$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_A=\frac{5}{32.2}\\ =0.1553\text{slugs}\end{array}$

Calculate the mass of the ball $\left(m_B\right)$ using the formula:

$\begin{array}{c}{W}_B=m_{B}g\\ m_B=\frac{W_B}{g}\end{array}$

Substitute $\text{1}\text{lb}$ for $W_B$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_B=\frac{1}{32.2}\\ =0.0311\text{slugs}\end{array}$

Calculate the speed of the foot $\left(v_A\right)$ before impact by considering the Newton’s law of motion:

$v_A^2-u_A^2=2gl$

Here, $u_A$ is the initial speed of the foot which is zero as it rest initially.

Substitute 0 for $u_A$.

$\begin{array}{l}{v}_A^2-\left(0\right)=2gl\\ v_A=\sqrt{2gl}\end{array}$

Substitute $32.2{\text{ft/s}}^{\text{2}}$ for g and $6\text{ft}$ for l.

$\begin{array}{c}{v}_A=\sqrt{2\times 32.2\times 6}\\ =19.66\text{ft/s}\end{array}$

Show the impulse momentum diagram for the ball and foot as Figure (1).

The expression for the velocity of the ball ${\left(v_B\right)}_t$ in tangent direction as follows:

${\left(v_B\right)}_t=-v_B\cos \theta$

The expression for the conservation of momentum of the ball in tangent t direction as follows:

${\left({v^{\prime }}_B\right)}_t={\left(v_B\right)}_t$

Substitute $-v_B\cos \theta$ for ${\left(v_B\right)}_t$.

${\left({v^{\prime }}_B\right)}_t=-v_B\cos \theta$

Substitute $60°$ for $\theta$.

${\left({v^{\prime }}_B\right)}_t=-v_B\cos 60°$ (1)

The expression for the speed of the ball ${\left({v^{\prime }}_B\right)}_x$ in x direction as follows:

${\left({v^{\prime }}_B\right)}_x={\left({v^{\prime }}_B\right)}_n\sin \theta -{\left({v^{\prime }}_B\right)}_t\cos \theta$

The expression for the conservation of momentum of the system in x direction as follows:

$m_A{\left(v_A\right)}_x+m_B{\left(v_B\right)}_x=m_A{\left({v^{\prime }}_A\right)}_x+m_B{\left({v^{\prime }}_B\right)}_x$

Substitute ${\left({v^{\prime }}_B\right)}_n\sin \theta -{\left({v^{\prime }}_B\right)}_t\cos \theta$ for ${\left({v^{\prime }}_B\right)}_x$, $v_A$ for ${\left(v_A\right)}_x$, $v_B$ for ${\left(v_B\right)}_x$, and ${v^{\prime }}_A$ for ${\left({v^{\prime }}_A\right)}_x$.

$m_A\left(v_A\right)+m_B\left(v_B\right)=m_A\left({v^{\prime }}_A\right)+m_B\left[{\left({v^{\prime }}_B\right)}_n\sin \theta -{\left({v^{\prime }}_B\right)}_t\cos \theta \right]$ (2)

The expression for the velocity of foot ${\left({v^{\prime }}_A\right)}_n$ after impact in normal direction as follows:

${\left({v^{\prime }}_A\right)}_n=\left({v^{\prime }}_A\right)\sin \theta$

The expression for the velocity of foot ${\left(v_A\right)}_n$ before impact in normal direction as follows:

${\left(v_A\right)}_n=\left(v_A\right)\sin \theta$

The expression for the velocity of ball ${\left(v_B\right)}_n$ before impact in normal direction as follows:

${\left(v_B\right)}_n=\left(v_B\right)\sin \theta$

The expression for the coefficient of restitution for both ball and foot in normal direction as follows:

${\left({v^{\prime }}_B\right)}_n-{\left({v^{\prime }}_A\right)}_n=e\left[{\left(v_A\right)}_n-{\left(v_B\right)}_n\right]$

Substitute $\left({v^{\prime }}_A\right)\sin \theta$ for ${\left({v^{\prime }}_A\right)}_n$, $\left(v_A\right)\sin \theta$ for ${\left(v_A\right)}_n$, and $\left(v_B\right)\sin \theta$ for ${\left(v_B\right)}_n$.

${\left({v^{\prime }}_B\right)}_n-\left({v^{\prime }}_A\right)\sin \theta =e\left[\left(v_A\right)\sin \theta -\left(v_B\right)\sin \theta \right]$

Substitute $60°$ for $\theta$.

${\left({v^{\prime }}_B\right)}_n-\left({v^{\prime }}_A\right)\sin 60°=e\left[\left(v_A\right)\sin 60°-\left(v_B\right)\sin 60°\right]$ (3)

Solve the equations (2) and (3).

${\left({v^{\prime }}_B\right)}_n=\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}$

Calculate the magnitude of the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is stationary using the relation:

${v^{\prime }}_B=\sqrt{{\left({v^{\prime }}_B\right)}_t^2+{\left({v^{\prime }}_B\right)}_n^2}$

Substitute $\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}$ for ${\left({v^{\prime }}_B\right)}_n$ and $-v_B\cos 60°$ for ${\left({v^{\prime }}_B\right)}_t$.

${v^{\prime }}_B=\sqrt{{\left(-v_B\cos 60°\right)}^2+{\left[\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}\right]}^2}$

The velocity of the ball $\left(v_B\right)$ is zero when the ball is in stationary.

Substitute 0 for $v_B$, $0.1553\text{slugs}$ for $m_A$, $0.0311\text{slugs}$ for $m_B$, $19.66\text{ft/s}$ for $v_A$, and 0.8 for e.

$\begin{array}{c}{v^{\prime }}_B={\left\{\begin{array}{l}{\left(-0\times \cos 60°\right)}^2+\\ {\left[\frac{\left[0.1553\times 19.66\left(1+0.8\right)-\left(0.8\times 0.1553\times 0\right)+\left(0.0311\times 0\right)\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{0.1553+0.0311{\sin }^{2}60°}\right]}^2\end{array}\right\}}^{1/2}\\ =26.65\text{ft/s}\end{array}$

Calculate the direction of the velocity of the ball $\left(\alpha \right)$ using the relation:

$\alpha =\left(90°-\theta \right)+{\tan }^{-1}\frac{{\left({v^{\prime }}_B\right)}_t}{{\left({v^{\prime }}_B\right)}_n}$

Substitute $60°$ for $\theta$, $\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}$ for ${\left({v^{\prime }}_B\right)}_n$ and $-v_B\cos 60°$ for ${\left({v^{\prime }}_B\right)}_t$.

$\alpha =\left(90°-60°\right)+{\tan }^{-1}\frac{-v_B\cos 60°}{\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}}$

Substitute 0 for $v_B$, $0.1553\text{slugs}$ for $m_A$, $0.0311\text{slugs}$ for $m_B$, $19.66\text{ft/s}$ for $v_A$, and 0.8 for e.

$\begin{array}{c}\alpha =\left(90°-60°\right)+{\tan }^{-1}\frac{0}{26.65}\\ =30°+0\\ =30°\end{array}$

Therefore, the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is stationary is $\underset{\_}{26.65\text{ft/s}\left(30°\right)}$.

To determine

Find the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is struck when it is rolling toward the foot with velocity of $10\text{ft/s}$.

Answer to Problem 13.184P

The exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is struck is $\underset{\_}{\text{31}\text{.93ft/s}\left(39°\right)}$.

Explanation of Solution

Given information:

The weight of the stimulated foot $\left(W_A\right)$ is $5\text{lb}$.

The weight of the ball $\left(W_B\right)$ is $1\text{lb}$.

The length of the pendulum arm (l) is $6\text{ft}$.

The coefficient of restitution between the foot (e) is 0.8.

The angle $\left(\theta \right)$ is $60°$.

The velocity $\left(v_B\right)$ is $10\text{ft/s}$.

Calculation:

Calculate the magnitude of the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is struck using the relation:

${v^{\prime }}_B=\sqrt{{\left({v^{\prime }}_B\right)}_t^2+{\left({v^{\prime }}_B\right)}_n^2}$

Substitute $\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}$ for ${\left({v^{\prime }}_B\right)}_n$ and $-v_B\cos 60°$ for ${\left({v^{\prime }}_B\right)}_t$.

${v^{\prime }}_B=\sqrt{{\left(-v_B\cos 60°\right)}^2+{\left[\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}\right]}^2}$

Substitute $10\text{ft/s}$ for $v_B$, $0.1553\text{slugs}$ for $m_A$, $0.0311\text{slugs}$ for $m_B$, $19.66\text{ft/s}$ for $v_A$, and 0.8 for e.

$\begin{array}{c}{v^{\prime }}_B={\left\{\begin{array}{l}{\left(10\times \cos 60°\right)}^2+\\ {\left[\frac{\left[0.1553\times 19.66\left(1+0.8\right)-\left(0.8\times 0.1553\times -10\right)+\left(0.0311\times -10\right)\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{0.1553+0.0311{\sin }^{2}60°}\right]}^2\end{array}\right\}}^{1/2}\\ =\sqrt{{\left(5\right)}^2+{\left(31.54\right)}^2}\\ =31.93\text{ft/s}\end{array}$

Calculate the direction of the velocity of the ball $\left(\alpha \right)$ using the relation:

$\alpha =\left(90°-\theta \right)+{\tan }^{-1}\frac{{\left({v^{\prime }}_B\right)}_t}{{\left({v^{\prime }}_B\right)}_n}$

Substitute $60°$ for $\theta$, $\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}$ for ${\left({v^{\prime }}_B\right)}_n$ and $-v_B\cos 60°$ for ${\left({v^{\prime }}_B\right)}_t$.

$\alpha =\left(90°-60°\right)+{\tan }^{-1}\frac{-v_B\cos 60°}{\frac{\left[m_A{v}_A\left(1+e\right)-e{m}_A{v}_B+m_B{v}_B\left(1-{\cos }^{2}60°\right)\right]\sin 60°}{m_A+m_B{\sin }^{2}60°}}$

Substitute 0 for $v_B$, $0.1553\text{slugs}$ for $m_A$, $0.0311\text{slugs}$ for $m_B$, $19.66\text{ft/s}$ for $v_A$, and 0.8 for e.

$\begin{array}{c}\alpha =\left(90°-60°\right)+{\tan }^{-1}\frac{5}{31.54}\\ =30°+9°\\ =39°\end{array}$

Therefore, the exit velocity of the ball $\left({v^{\prime }}_B\right)$ if the ball is struck is $\underset{\_}{\text{31}\text{.93ft/s}\left(39°\right)}$.