Question 17.SP.9: Two solid spheres with a radius 3 in. and weighing 2 lb each...
Two solid spheres with a radius 3 in. and weighing 2 lb each are mounted at A and B on the horizontal rod A′B′ that rotates freely about a vertical axis with a counterclockwise angular velocity of 6 rad/s. The spheres are held in position by a cord, which is suddenly cut. The centroidal moment of inertia of the rod and pivot is \bar{I}_R=0.25 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{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′.
STRATEGY: You can first use the principle of impulse and momentum to find the angular velocity of the rod and then use the definition of kinetic energy to determine the change in energy.
MODELING: Choose the two solid spheres and the horizontal rod as your system and model these as rigid bodies. The impulse–momentum diagram for this system is shown in Fig. 1.
ANALYSIS:
a. Principle of Impulse and Momentum. Apply the principle of impulse and momentum for this system between time t_1 (when the spheres are at r_1) and t_2 (when the spheres are at r_1)
\text { Syst Momenta }{ }_1 + \text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }{ }_2
The external forces consist of the weights and the reaction at the pivot, which have no moment about the y axis. Noting that the rod is undergoing centroidal rotation and \bar{v}_A=\bar{v}_B=\bar{r} \omega, you can equate moments about the y axis as
\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\end{aligned} (1)
This states that the angular momentum of the system about the y axis is conserved. You can now compute
\begin{aligned}&\bar{I}_S=\frac{2}{5} m_S a^2=\frac{2}{5}\left(2 \mathrm{lb} / 32.2 \mathrm{ft} / \mathrm{s}^2\right)\left(\frac{3}{12} \mathrm{ft}\right)^2=0.00155 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\\&m_S \bar{r}_1^2=(2 / 32.2)\left(\frac{5}{12}\right)^2=0.0108 \quad m_S \bar{r}_2^2=(2 / 32.2)\left(\frac{25}{12}\right)^2=0.2696\end{aligned}
Substituting these values, along with \bar{I}_R=0.25 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2 and \omega_1=6 \mathrm{rad} / \mathrm{s}, into Eq. (1) gives
0.275(6 \mathrm{rad} / \mathrm{s})=0.792 \omega_2 \quad \omega_2=2.08 \mathrm{rad} / \mathrm{s}\Lsh
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
Using the numerical values found here, you have
\begin{gathered}T_1=\frac{1}{2}(0.275)(6)^2=4.95 \mathrm{ft} \cdot \mathrm{lb} \quad T_2=\frac{1}{2}(0.792)(2.08)^2=1.713 \mathrm{ft} \cdot \mathrm{lb} \\\Delta T=T_2 – T_1=1.71 – 4.95 \quad \Delta T=-3.24 \mathrm{ft} \cdot \mathrm{lb}\end{gathered}
REFLECT and THINK: As expected, when the spheres move outward, the angular velocity of the system decreases. This is similar to an ice skater who throws her arms outward to reduce her angular speed.