Question 17.SP.12: A uniformly loaded square crate is falling freely with a vel......
A uniformly loaded square crate is falling freely with a velocity v_{0} when cable AB suddenly becomes taut. Assuming that the impact is perfectly plastic, determine the angular velocity of the crate and the velocity of its mass center immediately after the cable becomes taut.
STRATEGY: Because impact occurs, use the principle of impulse and momentum.
MODELING: Choose the crate as your system and model it as a rigid body. The impulse–momentum diagram for this system is shown in Fig. 1. The mass moment of inertia of the plate about G is \bar{I}=\frac{1}{6} m a^2 .
ANALYSIS:
Principle of Impulse and Momentum. Applying the impulse–momentum principle in the x direction and taking moments about A gives
+ ↺ moments about A: m v_0 a \frac{\sqrt{2}}{2}+0=\bar{I} \omega+m \bar{v}_x \frac{a}{2}-m \bar{v}_y \frac{a}{2} (1)
\stackrel{+}{\searrow} x components: \quad m v_0 \frac{\sqrt{2}}{2}+0=m \bar{v}_x (2)
There are three unknowns in these two equations: \omega, \bar{v}_x,\text { and } \bar{v}_y. For additional equations, you can use kinematics. Because you are told the impact is perfectly
plastic, point A has a velocity perpendicular to the rope (Fig. 2). Therefore, you can relate the acceleration of A to that of G, as
\overline{ v }= v _G= v _A+ v _{G / A}
= [v_A ⦪ 45°] + \left[a \frac{\sqrt{2}}{2} \omega \downarrow\right]
Equating components in the x and y directions, you find
\stackrel{+}{\searrow} x components: \bar{v}_x=v_A+a \frac{\sqrt{2}}{2} \omega \frac{\sqrt{2}}{2}=v_A+\frac{a \omega}{2} (3)
\stackrel{+}{\nearrow} y components: \bar{v}_y=-a \frac{\sqrt{2}}{2} \omega \frac{\sqrt{2}}{2}=-\frac{a \omega}{2} (4)
You now have four equations and four unknowns. Solving these gives
\omega=\frac{3 \sqrt{2}}{5} \frac{v_0}{a} \qquad \bar{v}_x=\frac{\sqrt{2}}{2} v_0 \qquad \bar{v}_y=-\frac{3 \sqrt{2}}{10} v_0 \qquad v_A=\frac{\sqrt{2}}{5} v_0
So
\omega =0.849 \frac{v_0}{d} \circlearrowleft ◂
Resolving the velocity of the center of mass into a magnitude and direction using Fig. 3 gives you
\overline{ v } = 0.825 v_0 ⦪ 76.0° ◂
REFLECT and THINK: If the impact had not been plastic, point A would have rebounded and the rope would have become slack. To solve the problem in this case, you would have needed to use the equation for the coefficient of restitution.
