Chapter 14.1, Problem 14.14P

For the system of particles of Prob. 14.13, determine (a) the position vector $\bar{r}$ of the mass center G of the system, (b) the linear momentum $m\bar{v}$ of the system, (c) the angular momentum H_G of the system about G. Also verify that the answers to this problem and to Prob. 14.13 satisfy the equation given in Prob. 14.27.

Fig. P14.13

To determine

Find the position vector of the mass center of the system.

Answer to Problem 14.14P

The position vector of the mass center of the system is $\underset{\_}{\left(1.867\text{m}\right)\text{i}+\left(1.533\text{m}\right)\text{j}+\left(0.667\text{m}\right)\text{k}}$.

Explanation of Solution

Given information:

The mass of the particles A is $3\text{kg}$.

The mass of the particles B is $\text{2}\text{kg}$.

The mass of the particles C is $4\text{kg}$.

The position vector is $\overline{\text{r}}$.

The mass center is G.

Calculation:

Find the position vectors from point O to each satellite in meters.

Refer to figure P14.13 in the textbook.

Express the position vector point A as follows:

${\text{r}}_A=3\text{j}$

Express the position vector point B as follows:

${\text{r}}_B=1.2\text{i+2}\text{.4j+3k}$

Express the position vector point C as follows:

${\text{r}}_C=3.6\text{i}$

Determine the mass center G of the system using the relation:

$\begin{array}{c}\overline{\text{r}}\text{=}\frac{{\displaystyle \sum _{i=1}^n{m}_i{\text{r}}_i}}{{\displaystyle \sum _{i=1}^n{m}_i}}\\ =\frac{m_A{\text{r}}_A+m_B{\text{r}}_B+m_C{\text{r}}_C}{m_A+m_B+m_C}\end{array}$ . (1)

Here, $\left(m_A,m_B,m_C\right)$ is mass of A, B, and C and $\left({\text{r}}_A,{\text{r}}_B,{\text{r}}_C\right)$ is position vector.

Substitute $3\text{j}$ for ${\text{r}}_A$, $1.2\text{i+2}\text{.4j+3k}$ for ${\text{r}}_B$, $\text{3}\text{.6i}$ for $r_c$, and $3\text{kg}$ for $m_A$, $\text{2}\text{kg}$ for $m_{B,}$, and $4\text{kg}$ for $m_C$ in Equation (1).

$\begin{array}{c}\overline{\text{r}}=\frac{3\times 3\text{j}+\left(2\right)\left(1.2\text{i}+\text{2}\text{.4j}+\text{3k}\right)+4\left(3.6\text{i}\right)}{9}\\ =\frac{6\text{j}+2.4\text{i}+\text{4}\text{.8j}+\text{6k}+\text{14}\text{.4i}}{9}\\ =1.86667\text{i}+\text{1}\text{.53333j}+\text{0}\text{.6667k}\end{array}$

Thus, the position vector of the mass center of the system is $\underset{\_}{\left(1.867\text{m}\right)\text{i+}\left(1.533\text{m}\right)\text{j+}\left(0.667\text{m}\right)\text{k}}$.

To determine

Find the linear momentum of the system.

Answer to Problem 14.14P

The linear momentum of the system is $\underset{\_}{\left(12.00\text{kg}\cdot \text{m}/\text{s}\right)\text{i}+\left(28.0\text{kg}\cdot \text{m}/\text{s}\right)\text{j}+\left(14.00\text{kg}\cdot \text{m}/\text{s}\right)\text{k}}$.

Explanation of Solution

Calculation:

Express the linear momentum of particle A as follows:

$\begin{array}{c}{m}_A{\text{v}}_A=3\left(4\text{i}+\text{2j+2k}\right)\\ =12.0\text{i}+6\text{j+6k}\end{array}$

Express the linear momentum of particle B as follows:

$\begin{array}{c}{m}_B{\text{v}}_B=2\left(\text{4i}+3\text{j}\right)\\ =8\text{i}+6\text{j}\end{array}$

Express the linear momentum of particle C as follows:

$\begin{array}{c}{m}_C{\text{v}}_C=4\left(-\text{2i}+4\text{j+2k}\right)\\ =-8.0\text{i}+16.0\text{j+8}\text{.0k}\end{array}$

Find the linear momentum of the system using the relation:

$m\overline{\text{v}}\text{=}{m}_A{\text{v}}_A+m_B{\text{v}}_B+m_C{\text{v}}_C$ (2)

Substitute $12.0\text{i}+6\text{j+6k}$ for $m_A{\text{v}}_A$, $8\text{i}+6\text{j}$ for $m_B{\text{v}}_B$, and $-8.0\text{i}+16.0\text{j+8}\text{.0k}$ for $m_C{\text{v}}_C$ in Equation (2).

$\begin{array}{c}m\overline{\text{v}}\text{=}12.0\text{i}+6\text{j+6k}+8\text{i}+6\text{j}+\left(-8.0\text{i}+16.0\text{j+8}\text{.0k}\right)\\ =12.0\text{i}+28\text{j}+14\text{k}\\ =\left(12.00\text{kg}\cdot \text{m}/\text{s}\right)\text{i+}\left(28.0\text{kg}\cdot \text{m}/\text{s}\right)\text{j+}\left(14.00\text{kg}\cdot \text{m}/\text{s}\right)\text{k}\end{array}$

Thus, the linear momentum of the system is $\underset{\_}{\left(12.00\text{kg}\cdot \text{m}/\text{s}\right)\text{i+}\left(28.0\text{kg}\cdot \text{m}/\text{s}\right)\text{j+}\left(14.00\text{kg}\cdot \text{m}/\text{s}\right)\text{k}}$.

To determine

Find the angular momentum of the system about G and also verify this problem and to problem 14.13 satisfy the Equation given in problem 14.27.

Answer to Problem 14.14P

The angular momentum of the system about G is $\underset{\_}{-\left(2.80\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{i+}\left(13.33\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{j}-\left(24.30\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}}$.

Explanation of Solution

Calculation:

Find the position vector from the particles ${{\text{r}}^{\prime }}_A$ to the center of mass using the relation:

${{\text{r}}^{\prime }}_A={\text{r}}_A-\overline{r}$ (3)

Here, ${\text{r}}_A$ is position vector at point A and $\overline{r}$ is the mass center.

Substitute $3\text{j}$ for ${\text{r}}_A$, and $1.86667\text{i+1}\text{.53333j+0}\text{.6667k}$ for $\overline{r}$ in Equation (3).

$\begin{array}{c}{{\text{r}}^{\prime }}_A=3\text{j}-\left(1.86667\text{i+1}\text{.53333j+0}\text{.6667k}\right)\\ =1.86667\text{i+1}\text{.4667j+0}\text{.6667k}\end{array}$

Find the position vector from the particles ${{\text{r}}^{\prime }}_B$ to the center of mass using the relation:

${{\text{r}}^{\prime }}_B={\text{r}}_B-\overline{r}$ (4)

Here, ${\text{r}}_B$ is position vector at point B.

Substitute $1.2\text{i+2}\text{.4j+3k}$ for ${\text{r}}_B$ and $1.86667\text{i+1}\text{.53333j+0}\text{.6667k}$ for $\overline{r}$ in Equation (4).

$\begin{array}{c}{{\text{r}}^{\prime }}_B=1.2\text{i+2}\text{.4j+3k}-\left(1.86667\text{i+1}\text{.53333j+0}\text{.6667k}\right)\\ =-\text{0}\text{.6667}-\text{0}\text{.86667j+2}\text{.3333k}\end{array}$

Find the position vector from the particles ${{\text{r}}^{\prime }}_C$ to the center of mass using the relation:

${{\text{r}}^{\prime }}_C={\text{r}}_C-\overline{r}$ (5)

Here, ${\text{r}}_C$ is position vector at point C.

Substitute $\text{3}\text{.6i}$ for ${\text{r}}_C$, and $1.86667\text{i+1}\text{.53333j+0}\text{.6667k}$ for $\overline{r}$ in Equation (5).

$\begin{array}{c}{{\text{r}}^{\prime }}_C=\text{3}\text{.6i}-1.86667\text{i+1}\text{.53333j+0}\text{.6667k}\\ =1.73333\text{i}-\text{1}\text{.53333j}-\text{0}\text{.6667k}\end{array}$

Calculate the angular momentum about point G using the relation:

$H_G={{\text{r}}^{\prime }}_A\times \left(m_A{\text{v}}_A\right)+{{\text{r}}^{\prime }}_B\times \left(m_B{\text{v}}_B\right)+{{\text{r}}^{\prime }}_C\times \left(m_C{\text{v}}_C\right)$ (6)

Here, $\left(m_A,m_B,m_C\right)$ is mass of particles A, B, C, $\left({{\text{r}}^{\prime }}_A,{{\text{r}}^{\prime }}_B,{{\text{r}}^{\prime }}_C\right)$ are positive vector for particles and $\left({\text{v}}_A,{\text{v}}_B,{\text{v}}_C\right)$ is velocity of particles A, B, C.

Substitute $1.86667\text{i+1}\text{.4667j+0}\text{.6667k}$ for ${{\text{r}}^{\prime }}_A$, $1.73333\text{i}-\text{1}\text{.53333j}-\text{0}\text{.6667k}$ for ${{\text{r}}^{\prime }}_C$, $-\text{0}\text{.6667}-\text{0}\text{.86667j+2}\text{.3333k}$ for ${{\text{r}}^{\prime }}_B$, $12.0\text{i}+6\text{j+6k}$ for $m_A{\text{v}}_A$, $8\text{i}+6\text{j}$ for $m_B{\text{v}}_B$, and $-8.0\text{i}+16.0\text{j+8}\text{.0k}$ for $m_C{\text{v}}_C$ in Equation (6).

$\begin{array}{c}{H}_G=\left[\begin{array}{l}1.86667\text{i+1}\text{.4667j+0}\text{.6667k}\times \left(12.0\text{i}+6\text{j+6k}\right)\\ -\text{0}\text{.6667}-\text{0}\text{.86667j+2}\text{.3333k}\times \left(8\text{i}+6\text{j}\right)+\\ 1.73333\text{i}-\text{1}\text{.53333j}-\text{0}{\text{.6667k}}_C\times \left(-8.0\text{i}+16.0\text{j+8}\text{.0k}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -1.86667& 1.46667& -0.6667\\ 12& 6& 6\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -0.66667& 0.86667& 2.3333\\ 8& 6& 0\end{array}\right|+\\ \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1.73333& -1.5333& -0.6667\\ -8& 16& 8\end{array}\right|\end{array}\right]\\ =\left(12.8\text{i}+3.2\text{j}-28.8\text{k}\right)+\left(-14\text{i+18}\text{.6667j-10}\text{.9333k}\right)+\left(-1.6\text{i-8}\text{.5333j+15}\text{.4667k}\right)\\ =-2.8\text{i+13}\text{.3333j}-\text{24}\text{.2667k}\\ \text{=}-\left(2.80\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{i+}\left(13.33\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{j}-\left(24.30\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}\end{array}$

Thus, the angular momentum of the system about G is $\underset{\_}{-\left(2.80\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{i+}\left(13.33\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{j}-\left(24.30\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}}$.

Find the value of $\overline{\text{r}}\times \text{m}\overline{\text{v}}$ as follows:

Substitute $1.86667\text{i+1}\text{.53333j+0}\text{.6667k}$ for $\overline{\text{r}}$ and $\left(12.00\text{kg}\cdot \text{m}/\text{s}\right)\text{i+}\left(28.0\text{kg}\cdot \text{m}/\text{s}\right)\text{j+}\left(14.00\text{kg}\cdot \text{m}/\text{s}\right)\text{k}$ for $\text{m}\overline{\text{v}}$.

$\begin{array}{c}\overline{\text{r}}\times \text{m}\overline{\text{v}}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1.86667& 1.5333& 0.6667\\ 12& 28& 14\end{array}\right|\\ =\text{i}\left(21.4662-18.6676\right)-\text{j}\left(26.1333-8.0000\right)+\text{k}\left(52.266676-18.3996\right)\\ =2.8\text{i}-\text{18}\text{.33j+33}\text{.8667k}\\ \text{=}\left(\text{2}\text{.8}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{i}-\left(\text{18}\text{.133}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{j+}\left(33.8667\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{k}\\ {\text{H}}_G+\overline{\text{r}}\times \text{m}\overline{\text{v}}=\left[\begin{array}{c}-\left(2.80\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{i+}\left(13.33\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{j}-\left(24.30\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k+}\\ \left(\text{2}\text{.8}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{i}-\left(\text{18}\text{.133}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{j+}\left(33.8667\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{k}\end{array}\right]\\ =-\left(4.8\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{j+}\left(9.6\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\right)\text{k}\end{array}$

Show that the $H_O=H_G+\overline{\text{r}}\times m\overline{\text{v}}$.

Express the angular momentum about point O as follows:

$H_O={\text{r}}_A\times \left(m_A{\text{v}}_A\right)+r_B\times \left(m_B{\text{v}}_B\right)+{\text{r}}_C\times \left(m_C{\text{v}}_C\right)$ (7)

Substitute $3\text{j}$ for ${\text{r}}_A$, $1.2\text{i+2}\text{.4j+3k}$ for ${\text{r}}_B$, $\text{3}\text{.6i}$ for $r_c$, $12.0\text{i}+6\text{j+6k}$ for $m_A{\text{v}}_A$, $8\text{i}+6\text{j}$ for $m_B{\text{v}}_B$, and $-8.0\text{i}+16.0\text{j+8}\text{.0k}$ for $m_C{\text{v}}_C$ in Equation (7).

$\begin{array}{c}{H}_O=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 3& 0\\ 12& 6& 6\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1.2& 2.4& 3\\ 8& 6& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 3.6& 0& 0\\ -8& 16& 8\end{array}\right|\\ =\left(18\text{i}-\text{36k}\right)+\left(-18\text{i+24j}-\text{12k}\right)+\left(-\text{28}\text{.8j}+\text{57}\text{.6k}\right)\\ =-\left(4.8\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{j+}\left(9.6\text{kg}\cdot {\text{m}}^2/\text{s}\right)\text{k}\end{array}$

Hence, the $H_O=H_G+\overline{\text{r}}\times m\overline{\text{v}}$ is proved.