One application of impact dynamics is the computation of loads and velocities of cargo that is airdropped from a height or from a low altitude parachute extraction system (LAPES). Damage can occur to the cargo if the terminal velocity at impact with the ground is too high or if the container rolls.

If a rectangular 2   m (height) × 3   m (length) × 1   m (width) cargo container weighing 98.1N drops at a rearward angle of 45^○ at a velocity of 20 m/s, what will its angular velocity be at impact? Will it have a translational velocity? Assume that it does not bounce and that the center of mass is the center of the box.

The mass of the box is 10kg. The moment of inertia about the center of the box is I_C = (m(h² + l²))/(12) = (10)/(12)(4 + 9) = 10.83 kg-m². Using the parallel axis theorem, the moment of inertia at the contact point , H, is I_H = 10.83 + 10[1 + 1.5²] = 43.33 kg-m²

The moment of momentum is H_H = I_Hω = ΔhmV where Δh is the height that the midpoint will drop. Δh = 1.5sin(45^○) + 1.0sin(45^○) − 1.0 = 0.768. So that 43.33ω = (0.768)(10)(20) = 153.55 or ω = 3.54 rad/s.

The velocity at the center of the box at the moment of impact may found using V_C⃗ = ω⃗ × r⃗. The angular velocity vector is about the third axis, so ω⃗ = 3.54k̂. A vector, r⃗, needs to be found that connects the point of rotation (the contact point, H) to the center point, C. r⃗ =  − (1.5cos(45^○) − 1.0cos(45^○)î + 1.5sin(45^○) + 1.0sin(45^○)ĵ =  − 0.35î + 1.77ĵ. Therefore, V_C =  − 0.35î × 3.54k̂ + 1.77ĵ × 3.54k̂ =  − 6.27î − 1.24ĵ (m/s).