Question 17.SP.14: A square package of side a and mass m moves down a conveyor ......
A square package of side a and mass m moves down a conveyor belt A with a constant velocity \overline{ v }_1. At the end of the conveyor belt, the corner of the package strikes a rigid support at B. Assuming that the impact at B is perfectly plastic, derive an expression for the smallest magnitude of the velocity \overline{ v }_1 for which the package will rotate about B and reach conveyor belt C.
STRATEGY: Because you have an impact, use the principle of impulse and momentum for when the package strikes the rigid support at B, and then apply the conservation of energy for the rotation of the package about the support B after the impact.
MODELING: Choose the package to be your system and model it as a rigid body. The impulse–momentum diagram for this system is shown in Fig. 1. Note that the only impulsive force external to the package is the impulsive reaction at B.
ANALYSIS:
Principle of Impulse and Momentum.
\text { Syst Momenta }_1+\text { Syst Ext Imp }_{1 \rightarrow 2}=\text { Syst Momenta }_2
+ ↺ moments about B: \left(m \bar{v}_1\right)\left(\frac{1}{2} a\right)+0=\left(m \bar{v}_1\right)\left(\frac{1}{2} \sqrt{2} a\right)+\bar{I} \omega_2 (1)
Because the package rotates about B, from kinematics you have \bar{v}_2=(G B) \omega_2=\frac{1}{2} \sqrt{2} a \omega_2.
Substitute this expression, together with \bar{I}=\frac{1}{6} m a^2, into Eq. (1) for
\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 \qquad \bar{v}_1=\frac{4}{3} a \omega_2 (2)
Conservation of Energy. Apply the principle of conservation of energy
between position 2 and position 3 (Fig. 2) as
T_2+V_2=T_3+V_3 (3)
You need to determine the energy at these two positions.
Position 2. V_2=W h_2 \text {. } Because \bar{v}_2=\frac{1}{2} \sqrt{2} a \omega_2, you have
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. The package must reach conveyor belt C, so it must pass through position 3 where G is directly above B. Also, because you wish to determine the smallest velocity for which the package will reach this position, choose \bar{v}_3=\omega_3=0. Therefore, T_3=0 \text { and } V_3=W h_3 \text {. }
Substituting these into Eq. (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 g}{a^2}\left(h_3-h_2\right) (4)
Substituting the computed values of h_2 \text { and } h_3 into Eq. (4), you obtain
\omega_2^2=\frac{3 g}{a^2}(0.707 a-0.612 a)=\frac{3 g}{a^2}(0.095 a) \qquad \omega_2=\sqrt{0.285 g / a}
\begin{array}{ll}\bar{v}_1=\frac{4}{3} a \omega_2=\frac{4}{3} a \sqrt{0.285 g / a} & \qquad \qquad \bar{v}_1=0.712 \sqrt{ga}\end{array}◂
REFLECT and THINK: The combination of energy and momentum methods is typical of many design analyses. If you had been interested in determining the reaction at B immediately after the impact or at some other point in the motion, you would have needed to draw a free-body diagram and a kinetic diagram and apply Newton’s second law.
