Question 17.SP.10: 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.
Principle of Impulse and Momentum. We consider the rod and sphere as a single system and express that the initial momenta of the rod and sphere and the impulses of the external forces are together equipollent to the final momenta of the system. We note that the only impulsive force external to the system is the impulsive reaction at A.
\text { Syst Momenta }_1+\text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }_2
Since the rod rotates about A, we have
\bar{v}_R^{\prime}=\bar{r} \omega^{\prime}=(0.6 m ) \omega^{\prime}. Also,
\bar{I}=\frac{1}{12} m L^2=\frac{1}{12}(8\text{ kg})(1.2 m )^2=0.96 \text{ kg} \cdot m ^2
Substituting these values and the given data into Eq. (1), we have
\begin{aligned}(2 \text{ kg} )(5 m / s )(1.2 m )= & (2 \text{ kg} ) v_s^{\prime}(1.2 m )+(8 \text{ kg} )(0.6 m ) \omega^{\prime}(0.6 m ) \\& +\left(0.96 \text{ kg} \cdot m ^2\right) \omega^{\prime} \\& 12=2.4 v_s^{\prime}+3.84 \omega^{\prime}&(2)\end{aligned}
Relative Velocities. Choosing positive to the right, we write
v_B^{\prime}-v_s^{\prime}=e\left(v_s-v_B\right)
Substituting v_s=5 m / s , v_B=0, and e = 0.80, we obtain
v_B^{\prime}-v_s^{\prime}=0.80(5 m / s ) (3)
Again noting that the rod rotates about A, we write
v_B^{\prime}=(1.2 m ) \omega^{\prime} (4)
Solving Eqs. (2) to (4) simultaneously, we obtain
\begin{aligned}\omega^{\prime} & =3.21\text{ rad} / s & \omega^{\prime}=3.21 \text{ rad} / s \uparrow \\v_s^{\prime} & =-0.143 m / s &\text{v} _s^{\prime}=-0.143 m / s \leftarrow\end{aligned}
