Question 17.SP.8: Two solid spheres of radius 3 in., weighing 2 lb each, are m......
Two solid spheres of radius 3 in., weighing 2 lb each, are mounted at A and B on the horizontal rod A′B′, which rotates freely about the vertical with a counterclockwise angular velocity of 6 rad/s. The spheres are held in position by a cord which is suddenly cut. Knowing that the centroidal moment of inertia of the rod and pivot is \bar{I}_R=0.25 \text{ lb} \cdot \text{ft} \cdot s ^2, determine (a) the angular velocity of the rod after the spheres have moved to positions A′ and B′, (b) the energy lost due to the plastic impact of the spheres and the stops at A′ and B′.
a. Principle of Impulse and Momentum. In order to determine the final angular velocity of the rod, we will express that the initial momenta of the various parts of the system and the impulses of the external forces are together equipollent to the final momenta of the system.
\text { Syst Momenta }_1+\text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }_2
Observing that the external forces consist of the weights and the reaction at the pivot, which have no moment about the y axis, and noting that \bar{v}_A=\bar{v}_B=\bar{r} \omega, we equate moments about the y axis:
\begin{aligned}2\left(m_S \bar{r}_1 \omega_1\right) \bar{r}_1+2 \bar{I}_S \omega_1+\bar{I}_R \omega_1&=2\left(m_S \bar{r}_2 \omega_2\right) \bar{r}_2+2 \bar{I}_S \omega_2+\bar{I}_R \omega_2 \\\left(2 m_S \bar{r}_1^2+2 \bar{I}_S+\bar{I}_R\right) \omega_1 & =\left(2 m_S \bar{r}_2^2+2\bar{I}_S+\bar{I}_R\right) \omega_2&(1)\end{aligned}
which expresses that the angular momentum of the system about the y axis is conserved. We now compute
\begin{gathered}\bar{I}_S=\frac{2}{5} m_{ S } a^2=\frac{2}{5}\left(2\text{ lb}/ 32.2 \text{ ft} / s ^2\right)\left(\frac{3}{12}\text{ ft} \right)^2=0.00155\text{ lb} \cdot \text{ft} \cdot s ^2 \\m_{ S } \bar{r}_1^2=(2 / 32.2)\left(\frac{5}{12}\right)^2=0.0108 \quad\quad\quad m_{ S } \bar{r}_2^2=(2 / 32.2)\left(\frac{25}{12}\right)^2=0.2696\end{gathered}
Substituting these values, and \bar{I}_R=0.25 \text { and } \omega_1=6 rad/s into (1):
0.275(6\text{ rad}/ s )=0.792 \omega_2 \quad\quad \omega_2=2.08 \text{ rad} / s \uparrow
b. Energy Lost. The kinetic energy of the system at any instant is
T=2\left(\frac{1}{2} m_S \bar{v}^2+\frac{1}{2} \bar{I}_S \omega^2\right)+\frac{1}{2}\bar{I}_R \omega^2=\frac{1}{2}\left(2 m_S \bar{r}^2+2 \bar{I}_S+\bar{I}_R\right) \omega^2
Recalling the numerical values found above, we have
\begin{gathered}T_1=\frac{1}{2}(0.275)(6)^2=4.95\text{ ft}\cdot \text{lb} \quad\quad T_2=\frac{1}{2}(0.792)(2.08)^2=1.713 \text{ ft} \cdot \text{lb} \\\Delta T=T_2-T_1=1.71-4.95 \quad\quad \Delta T=-3.24 \text{ ft} \cdot \text{lb}\end{gathered}
