Question 14.SP.3: A system consists of three particles A, B, and C, with masse......
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. The velocities of the particles expressed in m/s are, respectively,\mathbf{v}_{A}=3\mathbf{i}-2\mathbf{j}+4\mathbf{k},\,\mathbf{v}_{B}=4\mathbf{i}+3\mathbf{j},\,\mathrm{and}\,\mathbf{v}_{C}=2\mathbf{i}+5\mathbf{j}-3\mathbf{k}. Determine (a) the angular momentum H_{O} of the system about O, (b) the position vector \overline{r} of the mass center G of the system, (c) the angular momentum H_{G} of the system about G.
STRATEGY: You have a system of particles, so use the definitions of angular momentum and center of mass.
MODELING: Choose the three particles as your system.
ANALYSIS: The linear momentum of each particle expressed in kg·m/s is
m_{A}\mathbf{v}_{A}=3\mathbf{i}-2\mathbf{j}+4\mathbf{k}
m_{B}\mathbf{v}_{B}=\mathrm{8i}+6\mathbf{j}
m_{C}\mathrm{v}_{C}=6\mathrm{i}+15\mathrm{j}-9\mathrm{k}
The position vectors (in meters) are
\mathbf{r}_{A}=3\mathbf{j}+\mathbf{k}\qquad\mathbf{r}_{B}=3\mathbf{i}+2.5\mathbf{k}\qquad\mathbf{r}_{C}=4\mathbf{i}+2\mathbf{j}+\mathbf{k}
a. Angular Momentum About O. Using the definition of angular momentum about O (in kg·m²/s), you find
\mathbf{H}_{O}=\mathbf{r}_{A}\times(m_{A}\mathbf{v}_{A})+\mathbf{r}_{B}\times(m_{B}\mathbf{v}_{B})+\mathbf{r}_{C}\times(m_{C}\mathbf{v}_{C})
=\left|\begin{array}{ccc}i & j & k \\0 & 3 & 1 \\3 & -2 & 4\end{array}\right|+\left|\begin{array}{ccc}i & j & k \\3 & 0 & 2.5 \\8 & 6 & 0\end{array}\right|+\left|\begin{array}{ccc}i & j & k \\ 4 & 2 & 1 \\ 6 & 15 & -9\end{array}\right|
= ( 14i + 3j − 9k) + ( −15i + 20j + 18k) + ( −33i + 42j + 48k)
= 34i + 65j + 57k
{\bf H}_{O}=-(34\mathrm{~kg\cdot m^{2}/s})i+(65\mathrm{~kg\cdot m^{2}/s})\mathbf{j}+(57\mathrm{~kg\cdot m^{2}/s})\mathbf{k} ◂
b. Mass Center. Using the definition of mass center, you find
(m_{A}+m_{B}+m_{C})\mathbf{\bar{r}}=m_{A}\mathbf{r}_{A}+m_{B}\mathbf{r}_{B}+m_{C}\mathbf{r}_{C}
6{\bar{\mathbf{r}}} = ( 1 )( 3j + k) + ( 2 )( 3i + 2.5k) + ( 3 )( 4i + 2j + k)
\bar{\mathbf{r}} = 3i + 1.5j + 1.5k
\bar{\mathbf{r}} = ( 3.00 m )i + ( 1.500 m )j + ( 1.500 m )k ◂
c. Angular Momentum About G. The angular momentum of the system about G is
\mathrm{H}_{G}=\mathbf{r}_{\mathrm{A}}^{\prime}\times m_{A}\mathbf{v}_{A}+\mathbf{r}_{B}^{\prime}\times m_{B}\mathbf{v}_{B}+\mathbf{r}_{C}^{\prime}\times m_{C}\mathbf{v}_{C}
where \mathbf{r}_{A}^{\prime},\mathbf{r}_{B}^{\prime}{\mathrm{~and~}}\mathbf{r}_{C}^{\prime} are the position vectors from the particles to the center of mass; that is,
\mathbf{r}_{A}^{\prime}=\mathbf{r}_{A}-{\bar{\mathbf{r}}}=-3\mathbf{i}+1.5\mathbf{j}-0.5\mathbf{k}
\mathbf{r}_{B}^{\prime}=\mathbf{r}_{B}-{\bar{\mathbf{r}}}=-1.5{\mathbf{j}}+\mathbf{k}
\mathbf{r}_{C}^{\prime}=\mathbf{r}_{C}-{\bar{\mathbf{r}}}={{\mathbf{i}}}+0.5{{\mathbf{j}}}-0.5\mathbf{k}
Therefore, you can calculate the angular momentum as
\mathbf{H}_{G}=\mathbf{r}_{A}^{\prime}\times(m_{A}\mathbf{v}_{A})+\mathbf{r}_{B}^{\prime}\times(m_{B}\mathbf{v}_{B})+\mathbf{r}_{C}^{\prime}\times(m_{C}\mathbf{v}_{C})
=\left|\begin{array}{ccc}i & j & k \\-3 & 1.5 & -0.5 \\3 & -2 & 4\end{array}\right|+\left|\begin{array}{ccc}i & j & k \\0 & -1.5 & 1 \\8 & 6 & 0\end{array}\right|+\left|\begin{array}{ccc}i & j & k \\1 & 0.5 & -0.5 \\6 & 15 & -9\end{array}\right|
= ( 5i + 10.5j + 1.5k) + ( −6i + 8j + 12k) + ( 3i + 6j + 12k)
= 2i + 24.5j + 25.5k
\mathbf{H}_{G}=(2.00\ { kg}{\cdot}{{m}}^{2}/{{s}})\mathbf{i}+(24.5\ {kg}{\cdot}{{m}}^{2}/{{s}})\mathbf{j}+(25.5\ {{kg}}{\cdot}{{m}}^{2}/{{s}})\mathbf{k} ◂
REFLECT and THINK: You should be able to verify that the answers to this problem satisfy the equations given in Prob. 14.27; that is, \mathbf{H}_{O}={\bar{\mathbf{r}}}\times m{\bar{\mathbf{v}}}+\mathbf{H}_{G}. Because no impulses act on the system, the linear momentum of the overall system is constant; the location of the center of mass of the system, however, changes with time.