Three particles: H_O, mass centre, H_G

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).