Question 17.SP.15: A soccer ball tester consists of a 15-kg slender rod AB with...
A soccer ball tester consists of a 15-kg slender rod AB with a 1.1-kg simulated foot located at A and a torsional spring located at pin B. The torsional spring has a spring constant of k_t = 910 N·m and is unstretched when AB is vertical. The length of AB is 0.9 m, and you can assume that the foot can be modeled as a point mass. Knowing that the velocity of the 0.45-kg soccer ball is 30 ft/s after impact, determine (a) the coefficient of restitution between the simulated foot and the ball, (b) the impulse at B during the impact.
STRATEGY: This problem can be broken into two distinct stages of motion. In stage 1, the arm moves downward under the influence of gravity and the torsional spring. You can use the conservation of energy for this stage. In stage 2, the foot hits the ball, and you need to use both the principle of impulse and momentum and the coefficient of restitution.
MODELING: Each stage requires a different system. For stage 1, your system is rod AB, foot B, and the torsional spring. In stage 2, your system is rod AB, foot B, and the soccer ball. The appropriate diagrams are drawn in the analysis section. You can model AB as a slender rod, so its mass moment of inertia is
\bar{I}_{A B}=\frac{1}{12} m_{A B} l^2=\frac{1}{12}(15 \mathrm{~kg})(0.9 \mathrm{~m})^2=1.0125 \mathrm{~kg} \cdot \mathrm{m}^2
ANALYSIS:
Rod AB Moves Down. Apply the principle of conservation of energy
T_1 + V_{\mathrm{g}_1} + V_{\mathrm{e}_1}=T_2 + V_{\mathrm{g}_2} + V_{\mathrm{e}_2} (1)
Position 1. The system starts from rest, so T_1=0. Using the datum defined in Fig. 1, you know V_{\mathrm{g}_1}=0, and since the spring is unstretched at position 2, you find
V_{\mathrm{e}_1}=\frac{1}{2} k_t \theta^2=\frac{1}{2}(910 \mathrm{~N} \cdot \mathrm{m})\left(\frac{\pi}{2}\right)^2=1123 \mathrm{~J}
Position 2. The elastic potential energy is V_{\mathrm{e}_2}=0, and the gravitational potential energy is
\begin{aligned}V_{\mathrm{g}_2} &=-m_{A B} g \frac{l}{2} – m_A g l=-(15 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)(0.45 \mathrm{~m}) – (1.1 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)(0.9 \mathrm{~m}) \\&=-75.93 \mathrm{~J}\end{aligned}
The kinetic energy is
T_2=\frac{1}{2} m_A v_A^2 + \frac{1}{2} m_{A B} v_G^2 + \frac{1}{2} \bar{I}_{A B} \omega^2
You can relate the velocity of the foot and the velocity of the center of gravity of the rod to the angular velocity of AB by recognizing that AB is undergoing fixed-axis rotation. Therefore, v_G=\omega \frac{l}{2} and v_A=\omega l . Substituting these into the expression for T_2 and putting in values gives
T_2=\frac{1}{2}\left(m_A l^2 + m_{A B}\left(\frac{l}{2}\right)^2 + \bar{I}_{A B}\right) \omega^2=2.4705 \omega^2
Substituting these energy terms into Eq. (1) gives
0 + 0 + 1123=2.4705 \omega^2 – 75.93 + 0
Solving for the angular velocity, you find \omega=22.03 \mathrm{rad} / \mathrm{s}. Knowing ω, you can calculate the velocities v_G=9.912 \mathrm{~m} / \mathrm{s} and v_A=19.824 \mathrm{~m} / \mathrm{s}.
Foot A Impacts the Soccer Ball. Impulse–momentum diagrams for the impact on the ball are shown in Fig. 2.
Taking moments about B gives you
+\Lsh moments about B:
m_A v_A l + m_{A B} v_G \frac{l}{2} + I_{A B} \omega + 0=m_A v_A^{\prime} l + m_{A B} v_G^{\prime} \frac{l}{2} + \bar{I}_{A B} \omega^{\prime} + m_S v_S^{\prime} l (2)
The equation for the coefficient of restitution is
v_S^{\prime} – v_A^{\prime}=e\left(v_A – 0\right) (3)
where v_S^{\prime}=30 \mathrm{~m} / \mathrm{s}. From kinematics, you know v_A^{\prime}=\omega^{\prime} l and v_G^{\prime}=\omega^{\prime}(l / 2). Using these kinematic equations and Eqs. (2) and (3), you can solve for the unknown quantities
v_A^{\prime}=17.61 \mathrm{~m} / \mathrm{s} \quad v_G^{\prime}=8.81 \mathrm{~m} / \mathrm{s} \quad \omega^{\prime}=19.57 \mathrm{rad} / \mathrm{s} \quad e=0.625
e=0.625
Impulses During Impact. Applying impulse–momentum in the x-and y-directions gives
\stackrel{+}{\rightarrow} x \text {-components: } \quad m_{A B} v_G + m_A v_A + R_x \Delta t=m_{A B} v_G^{\prime} + m_A v_A^{\prime}+m_S v_S^{\prime} (4)
+\uparrow y \text {-components: } \quad 0 + R_y \Delta t=0 (5)
Solving these equations, you find R_x \Delta t=-5.53 \mathrm{~N} and R_y \Delta t=0.
\mathrm{R} \Delta t=5.53 \mathrm{~N} \leftarrow
REFLECT and THINK: This coefficient of restitution seems reasonable. As you decrease the pressure in the ball, you would expect the coefficient of restitution to decrease; therefore, the distance the ball travels will decrease. If you had been asked to determine the reactions at B after the impact, you would need to draw a free-body diagram and kinetic diagram for your system and apply Newton’s second law.
