Question 17.SP.11: A square package of side a and mass m moves down a conveyor ......

Principle of Impulse and Momentum. Since the impact between the package and the support is perfectly plastic, the package rotates about B during the impact. We apply the principle of impulse and momentum to the package and note that the only impulsive force external to the package is the impulsive reaction at B.

$\text { Syst Momenta }_1+\text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }_2$

$+\uparrow \text { moments about } B: \quad\left(m \bar{v}_1\right)\left(\frac{1}{2} a\right)+0=\left(m \bar{v}_2\right)\left(\frac{1}{2} \sqrt{2} a\right)+\bar{I} \omega_2$       (1)
Since the package rotates about B, we have $\bar{v}_2=(G B) \omega_2=\frac{1}{2} \sqrt{2} a \omega_2$. We substitute this expression, together with $\bar{I}=\frac{1}{6} m a^2$, into Eq. (1):

$\left(m \bar{v}_1\right)\left(\frac{1}{2} a\right)=m\left(\frac{1}{2} \sqrt{2} a\omega_2\right)\left(\frac{1}{2} \sqrt{2} a\right)+\frac{1}{6} m a^2 \omega_2 \quad\bar{v}_1=\frac{4}{3} a \omega_2$               (2)

Principle of Conservation of Energy. We apply the principle of conservation of energy between position 2 and position 3.
Position 2. $V_2=W h_2$. Recalling that $\bar{v}_2=\frac{1}{2} \sqrt{2} a \omega_2$, we write

$T_2=\frac{1}{2} m \bar{v}_2^2+\frac{1}{2} \bar{I} \omega_2^2=\frac{1}{2} m\left(\frac{1}{2}\sqrt{2} a \omega_2\right)^2+\frac{1}{2}\left(\frac{1}{6} m a^2\right) \omega_2^2=\frac{1}{3}m a^2 \omega_2^2$

Position 3. Since the package must reach conveyor belt C, it must pass through position 3 where G is directly above B. Also, since we wish to determine the smallest velocity for which the package will reach this position, we choose $\bar{v}_3=\omega_3=0 \text {. Therefore }  T_3=0 \text { and } V_3=W h_3$.
Conservation of Energy

$\begin{aligned}T_2+V_2 & =T_3+V_3 \\\frac{1}{3} m a^2 \omega_2^2+W h_2 & =0+W h_3 \\\omega_2^2=\frac{3 W}{m a^2}\left(h_3-h_2\right) & =\frac{3 \text{g}}{a^2}\left(h_3-h_2\right)&(3)\end{aligned}$         (3)

Substituting the computed values of $h_2 \text { and } h_3$ into Eq. (3), we obtain

$\begin{array}{rc}bar{v}_1=\frac{4}{3} a \omega_2=\frac{4}{3} a \sqrt{0.285 \text{g} / a} & \bar{v}_1=0.712 \sqrt{\text{g} a}\end{array}$