Linear and angular momentum of particle system

Reading of the figure and assumptions

The particles sit at corners of a box 2.5 m (x) × 4 m (y) × 3 m (z), with the origin O at the lower back-left corner. Positions: m₁ (0, 0, 3), m₂ (0, 4, 0), m₃ (2.5, 4, 0), m₄ (0, 0, 0). Point a is the top front-right corner (2.5, 4, 3). The velocities (assumed from the arrows) are:

• V₁ = 7 m/s along m₁→a, direction (2.5, 4, 0)/4.717, so V₁ = (3.710, 5.936, 0) m/s
• V₂ = 6 m/s along +x: (6, 0, 0)
• V₃ = 5 m/s along the diagonal of the side face x = 2.5 m (from m₃ towards (2.5, 0, 3)), direction (0, −4, 3)/5: V₃ = (0, −4, 3)
• V₄ = 1.5 m/s along +y: (0, 1.5, 0)

(a) Total linear momentum, L = Σ m V

• m₁V₁ = 0.5(3.710, 5.936, 0) = (1.855, 2.968, 0)
• m₂V₂ = 1.5(6, 0, 0) = (9, 0, 0)
• m₃V₃ = 1(0, −4, 3) = (0, −4, 3)
• m₄V₄ = 0.5(0, 1.5, 0) = (0, 0.75, 0)

L = (10.855 i − 0.282 j + 3.000 k) kg·m/s, magnitude 11.27 kg·m/s.

(b) Angular momentum about the origin, H_O = Σ r × m V

• m₁: (0, 0, 3) × (1.855, 2.968, 0) = (−8.904, 5.565, 0)
• m₂: (0, 4, 0) × (9, 0, 0) = (0, 0, −36)
• m₃: (2.5, 4, 0) × (0, −4, 3) = (12, −7.5, −10)
• m₄: r = 0, so zero

H_O = (3.10 i − 1.94 j − 46.0 k) kg·m²/s, magnitude 46.1 kg·m²/s.

(c) Angular momentum about point a, H_a = Σ (r − r_a) × m V

Using r_a = (2.5, 4, 3) and the relation H_a = H_O − r_a × L:

• r_a × L = (2.5, 4, 3) × (10.855, −0.282, 3) = (12.846, 25.065, −44.125) (the components are 4(3) − 3(−0.282) = 12.846, 3(10.855) − 2.5(3) = 25.065, 2.5(−0.282) − 4(10.855) = −44.125)
• H_a = (3.096 − 12.846, −1.935 − 25.065, −46 + 44.125)

H_a = (−9.75 i − 27.0 j − 1.875 k) kg·m²/s, magnitude 28.8 kg·m²/s.

(The figure is a low-resolution scan, so the direction of each velocity was read from the arrows and the assumptions above are used. The method is the same for any other reading.)