---
title: "Linear and angular momentum of particle system"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:system-of-particles-linear-angular-momentum"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/system-of-particles-linear-angular-momentum
site: BCT Engineering
---

# Linear and angular momentum of particle system

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2069 Asar, 3+3+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/system-of-particles-linear-angular-momentum
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Shown in figure below is a system of particles at time t. The following data apply at this instant: V1 = 7 m/s, m1 = 0.5 kg; V2 = 6 m/s, m2 = 1.5 kg; V3 = 5 m/s, m3 = 1 kg; V4 = 1.5 m/s, m4 = 0.5 kg. Determine (a) the total linear momentum of the system, (b) the angular momentum of the system about the origin, (c) the angular momentum of the system about point a. [Figure: box with dimensions 2.5 m, 4 m, 3 m, with particles m1 to m4 at corners moving with velocities V1 to V4 along the box edges/diagonals, axes x, y, z.]

## Short answer

Linear momentum is Σ m v = 10.855i - 0.282j + 3.000k kg·m/s. Angular momentum about origin is Σ r×m v = 3.10i - 1.94j - 46.0k kg·m²/s. About point a, H_a = -9.75i - 27.0j - 1.875k kg·m²/s.

## Answer

#### 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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the total linear momentum of the system of particles found?
   A. Add the magnitudes m|V| of all particles
   B. Vector sum of m V for all particles
   C. Total mass times the largest velocity
   D. Sum of m V r
   Answer: B. L = sum of m V as vectors, with components added separately: (10.855, -0.282, 3.000) kg.m/s. (Key point: Total linear momentum)

2. How is the angular momentum of the system about O found?
   A. Sum of m V
   B. Sum of r x m V over all particles
   C. Total mass times the position of the centroid
   D. Sum of r . m V
   Answer: B. H_O = sum of r x m V. A particle at the origin (m4) contributes zero. (Key point: Angular momentum about origin)

3. How is H_a about point a obtained from H_O?
   A. H_a = H_O + r_a x L
   B. H_a = H_O - r_a x L, equivalent to using (r - r_a) for each particle
   C. H_a = H_O
   D. H_a = L x H_O
   Answer: B. Using relative positions r - r_a gives H_a = H_O - r_a x L = (-9.75, -27.0, -1.875) kg.m^2/s. (Key point: Shifting the moment point)

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