Question 17.SP.15: A soccer ball tester consists of a 15-kg slender rod AB with......

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 A, and the torsional spring. In stage 2, your system is rod AB, foot A, 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~kg )(0.9~m )^2=1.0125~kg \cdot m ^2

ANALYSIS:

Rod AB Moves Down. Apply the principle of conservation of energy

T_1+V_{g_1}+V_{e_1}=T_2+V_{g_2}+V_{e_2}        (1)

Position 1. The system starts from rest, so T_1 = 0. Using the datum defined in Fig. 1, you know V_{g_1} = 0, and because the spring is unstretched at position 2, you find

V_{e_1}=\frac{1}{2} k_t \theta^2=\frac{1}{2}(910~N \cdot m )\left(\frac{\pi}{2}\right)^2=1123~J

Position 2. The elastic potential energy is V_{e_2} = 0, and the gravitational potential energy is

= −75.93 J

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\left(\frac{l}{2}\right) \text { 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 = 2.4705ω² − 75.93 + 0

Solving for the angular velocity, you find ω = 22.03 rad/s. Knowing ω, you can calculate the velocities v_{G} = 9.912 m/s and v_{A} = 19.824 m/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

+ ↺ moments about B:

m_A v_A l+m_{A B} v_G \frac{l}{2}+\bar{I}_{A B} \omega+0=m_A v^{\prime}_{A} l+m_{A B} v^{\prime}_{G} \frac{l}{2}+\bar{I}_{A B} \omega^{\prime}+m_S v^{\prime}_{S} l        (2)

The equation for the coefficient of restitution is

v^{\prime}_{S}-v^{\prime}_{A}=e\left(v_A-0\right)       (3)

where v^{\prime}_{S} = 30 m/s . From kinematics, you know v^{\prime}_{A}=\omega^{\prime} l \text { and } v^{\prime}_{G}=\omega^{\prime}(l / 2). Using these kinematic equations and Eqs. (2) and (3), you can solve for the unknown quantities

v^{\prime}_{A}=17.61~m / s \quad v^{\prime}_{G}=8.81~m / s \quad \omega^{\prime}=19.57~rad / 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^{\prime}_{G}+m_A v^{\prime}_{A}+m_S v^{\prime}_{S}       (4)

+ ↑y components:                           0+R_x \Delta t=0       (5)

Solving these equations, you find R_x \Delta t = −5.53 N and R_y \Delta t = 0.

RΔt = 5.53 N ← ◂

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.