--- title: "Three particles: H_O, mass centre, H_G" subject: "Applied Mechanics (Dynamics)" chapter: 3 key: "semester_3/applied_mechanics_dynamics:three-particles-angular-momentum-vectors" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/three-particles-angular-momentum-vectors site: BCT Engineering --- # Three particles: H_O, mass centre, H_G Applied Mechanics (Dynamics) · Chapter 3: System of particles - Asked in: 2080 Bhadra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/three-particles-angular-momentum-vectors - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3 ## Question A system consists of three particles A, B and C with masses $m_A = 1$ kg, $m_B = 2$ kg and $m_C = 3$ kg and that the velocities of the particles expressed in m/s are respectively, $\vec v_A = 3\hat i - 2\hat j + 4\hat k$, $\vec v_B = 4\hat i + 3\hat j$, and $\vec v_C = 2\hat i + 5\hat j - 3\hat k$. Determine: (i) The angular momentum $\vec H_O$ of the system about O. (ii) The position vector $\vec r$ of the mass centre G of the system. (iii) The angular momentum $\vec H_G$ of the system about G. [Figure: 3D axes x, y, z with origin O; particles A, B, C located using the dimensions 1 m, 3 m, 4 m, 3 m, 2 m, 1 m and 1.5 m shown on the figure; the figure is only partly legible.] ## Short answer H_O = Σ r_i × m_i v_i = (-10, 35.5, 7.5) kg m²/s. Mass centre position: r̄ = (1.917, 1.667, 1.5) m. Then H_G = H_O – r̄ × m v̄ = (26.83, 0.417, -0.583) kg m²/s. ## Answer **Positions (figure only partly legible):** the following position vectors, in metres, are assumed from the dimensions marked on the figure: $\vec r_A=1\hat i+3\hat j+4\hat k$, $\vec r_B=3\hat i+2\hat j+1\hat k$, $\vec r_C=1.5\hat i+1\hat j+1\hat k$. Replace them by the figure values if they differ; the method is unchanged. Masses 1, 2, 3 kg and the given velocities. #### Momenta $m_i\vec v_i$ (kg m/s) $$ \vec p_A=3\hat i-2\hat j+4\hat k,\quad \vec p_B=8\hat i+6\hat j,\quad \vec p_C=6\hat i+15\hat j-9\hat k $$ #### (i) Angular momentum about O $$ \vec H_O=\sum\vec r_i\times m_i\vec v_i $$ $$ \begin{aligned} \vec r_A\times\vec p_A&=(3(4)-4(-2))\hat i-(1(4)-4(3))\hat j+(1(-2)-3(3))\hat k=20\hat i+8\hat j-11\hat k\\ \vec r_B\times\vec p_B&=(2(0)-1(6))\hat i-(3(0)-1(8))\hat j+(3(6)-2(8))\hat k=-6\hat i+8\hat j+2\hat k\\ \vec r_C\times\vec p_C&=(1(-9)-1(15))\hat i-(1.5(-9)-1(6))\hat j+(1.5(15)-1(6))\hat k=-24\hat i+19.5\hat j+16.5\hat k \end{aligned} $$ $$ \vec H_O=-10\hat i+35.5\hat j+7.5\hat k\ \ \text{kg m}^2/\text{s} $$ #### (ii) Mass centre G $$ \vec{\bar r}=\frac{\sum m_i\vec r_i}{\sum m_i}=\frac{1(1,3,4)+2(3,2,1)+3(1.5,1,1)}{6}=\frac{(11.5,\ 10,\ 9)}{6} $$ $$ \vec{\bar r}=1.917\hat i+1.667\hat j+1.500\hat k\ \text{m} $$ #### (iii) Angular momentum about G Mean velocity: $\vec{\bar v}=\dfrac{\sum m_i\vec v_i}{6}=\dfrac{(17,\ 19,\ -5)}{6}=2.833\hat i+3.167\hat j-0.833\hat k$ m/s. Total momentum $m\vec{\bar v}=17\hat i+19\hat j-5\hat k$. $$ \vec H_G=\vec H_O-\vec{\bar r}\times m\vec{\bar v} $$ $$ \vec{\bar r}\times m\vec{\bar v}=-36.83\hat i+35.08\hat j+8.08\hat k\quad\Rightarrow\quad\vec H_G=26.83\hat i+0.417\hat j-0.583\hat k\ \ \text{kg m}^2/\text{s} $$ (Checked by computing $\sum(\vec r_i-\vec{\bar r})\times m_i(\vec v_i-\vec{\bar v})$ directly, giving the same result.) **Answer: $\vec H_O=(-10,\ 35.5,\ 7.5)$; $\vec{\bar r}=(1.917,\ 1.667,\ 1.5)$ m; $\vec H_G=(26.83,\ 0.417,\ -0.583)$ kg m$^2$/s (for the assumed positions).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives H_O of the system? A. Sum of m_i v_i B. Sum of r_i times m_i v_i C. Sum of m_i r_i / m D. Sum of r_i . m_i v_i Answer: B. H_O is the vector sum of the moments of momenta: sum r_i x m_i v_i. (Key point: Formula for H_O) 2. How is the mass-centre position found? A. Average of r_i without masses B. Sum of m_i r_i divided by sum of m_i C. Sum of m_i v_i divided by m D. Sum of r_i x m_i Answer: B. r_bar = sum m_i r_i / sum m_i = (11.5, 10, 9)/6 = (1.917, 1.667, 1.5) m. (Key point: Mass centre) 3. How is H_G computed from H_O? A. H_G = H_O + r_bar x m v_bar B. H_G = H_O C. H_G = H_O - r_bar x m v_bar D. H_G = H_O / 6 Answer: C. Since H_O = r_bar x m v_bar + H_G, H_G = H_O - r_bar x m v_bar. (Key point: H_G from H_O) 4. What is the total momentum m v_bar of the system? A. (9, 6, 1) kg m/s B. (17, 19, -5) kg m/s C. (2.83, 3.17, -0.83) kg m/s D. (6, 6, 6) kg m/s Answer: B. Sum of m_i v_i = (3+8+6, -2+6+15, 4+0-9) = (17, 19, -5) kg m/s. 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