Question 17.SP.13: A 2-kg sphere moving horizontally to the right with an initi...
A 2-kg sphere moving horizontally to the right with an initial velocity of 5 m/s strikes the lower end of an 8-kg rigid rod AB. The rod is suspended from a hinge at A and is initially at rest. Knowing that the coefficient of restitution between the rod and the sphere is 0.80, determine the angular velocity of the rod and the velocity of the sphere immediately after the impact.
STRATEGY: Since you have an impact, use the principle of impulse and momentum.
MODELING: Choose the sphere and the rod as your system; model the sphere as a particle and the rod as a rigid body. You also need to use the coefficient of restitution equation. The impulse–momentum diagram for this system is shown in Fig. 1. Note that the only impulsive force external to the system is the impulsive reaction at A.
ANALYSIS:
Principle of Impulse and Momentum.
\text { Syst Momenta }{ }_1 + \text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }{ }_2
+\Lsh \text { moments about } A \text { : }m_s v_s(1.2 \mathrm{~m})=m_s v_s^{\prime}(1.2 \mathrm{~m}) + m_R \bar{v}_R^{\prime}(0.6 \mathrm{~m}) + \bar{I} \omega^{\prime} (1)
In this case, the mass of the sphere is not negligible compared to the rod, so we must include it on the right-hand side of Eq. (1). Since the rod rotates about A, from kinematics, you know \bar{v}_R^{\prime}=\bar{r} \omega^{\prime}=(0.6 \mathrm{~m}) \omega^{\prime}. Also,
\bar{I}=\frac{1}{12} m L^2=\frac{1}{12}(8 \mathrm{~kg})(1.2 \mathrm{~m})^2=0.96 \mathrm{~kg} \cdot \mathrm{m}^2
Substituting these values and the given data into Eq. (1), you obtain
\begin{aligned}(2 \mathrm{~kg})(5 \mathrm{~m} / \mathrm{s})(1.2 \mathrm{~m})=&(2 \mathrm{~kg}) v_s^{\prime}(1.2 \mathrm{~m}) + (8 \mathrm{~kg})(0.6 \mathrm{~m}) \omega^{\prime}(0.6 \mathrm{~m}) \\&+\left(0.96 \mathrm{~kg} \cdot \mathrm{m}^2\right) \omega^{\prime}\end{aligned}
12=2.4 v_s^{\prime} + 3.84 \omega^{\prime} (2)
Coefficient of Restitution. Choosing positive to the right, you have
v_B^{\prime} – v_s^{\prime}=e\left(v_s – v_B\right)
Substituting v_s=5 \mathrm{~m} / \mathrm{s}, v_B=0, and e=0.80 gives
v_B^{\prime}-v_s^{\prime}=0.8(5 \mathrm{~m} / \mathrm{s}-0) (3)
Again noting that the rod rotates about A, you have
v_B^{\prime}=(1.2 \mathrm{~m}) \omega^{\prime} (4)
Solving Eqs. (2) to (4) simultaneously, you obtain
\begin{aligned}\omega^{\prime} &=3.21 \mathrm{rad} / \mathrm{s} & \omega^{\prime}=3.21 \mathrm{rad} / \mathrm{s}\circlearrowleft \\v_s^{\prime} &=-0.143 \mathrm{~m} / \mathrm{s} & \mathbf{v}_s^{\prime}=0.143 \mathrm{~m} / \mathrm{s}\end{aligned}
REFLECT and THINK: The negative value for the velocity of the sphere after impact means that it bounces back to the left. Given the masses of the sphere and the rod, this seems reasonable.