Chapter 14.1, Problem 14.11P

A system consists of three identical 19.32-lb particles A, B, and C. The velocities of the particles are, respectively, v_A = v_A j, v_B = v_Bi, and v_C = v_Ck. Knowing that the angular momentum of the system about O expressed in ft · lb · s is H_O = −1.2k, determine (a) the velocities of the particles, (b) the angular momentum of the system about its mass center G.

Fig. P14.11 and P14.12

To determine

Find the velocities of the particles.

Answer to Problem 14.11P

The velocity of particles A is $\underset{\_}{\left(4.00\text{ft}/\text{s}\right)\text{j}}$

The velocity of particles B is $\underset{\_}{\left(1.00\text{ft}/\text{s}\right)\text{i}}$

The velocity of particles C is $\underset{\_}{\left(3.00\text{ft}/\text{s}\right)\text{k}}$

Explanation of Solution

Given information:

The angular momentum about point O is $-1.2\text{k}$

Calculation:

The mass of three particles A, B, and C is equal.

$m_A=m_B=m_C$

Determine the weight of the identical particle.

$m_A=m_B=m_C=\frac{W}{g}$ (1)

Here, W is weight of each particle, $m_B$ is mass of particle B, $m_A$ is mass of particle A, $m_C$ is mass of particle B, and g is acceleration due to gravity.

Substitute $19.32\text{lb}$ for W and $32.2\text{ft}/{\text{s}}^{\text{2}}$ in Equation (1).

$\begin{array}{c}{m}_A=m_B=m_C\\ =\frac{19.32}{32.2}\\ =0.63\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Write the position vectors for the particles based on the given coordinate system:

$\begin{array}{l}{\text{r}}_A=3\text{k}\\ {\text{r}}_B=2\text{i}+2\text{j}+\text{3k}\\ {\text{r}}_C=\text{i}+\text{4j}\end{array}$

Determine the angular momentum of the system about the origin using the Equation.

$H_o=\left({\text{r}}_A\times m_A{\text{v}}_A\right)+\left({\text{r}}_B\times m_B{\text{v}}_B\right)+\left({\text{r}}_C\times m_C{\text{v}}_C\right)$ (2)

Here, ${\text{v}}_A$ is velocity vector of particle A, $m_B$ is mass of particle is B, ${\text{v}}_B$ is the velocity of particle B, ${\text{v}}_C$ is the velocity of particle C.

Substitute $0.6\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $\left(m_A,m_{B,}{m}_C\right)$, $3\text{k}$ for ${\text{r}}_A$, $2\text{i}+2\text{j+3k}$ for ${\text{r}}_B$, $\text{i}+4\text{j}$  for $r_c$, $\left(v_A\text{j}\right)$ for ${\text{v}}_A$, $\left({\text{v}}_B\text{i}\right)$ for ${\text{v}}_B$, $\left({\text{v}}_C\text{k}\right)$ for ${\text{v}}_C$, and $-1.2\text{k}$ for $H_o$ in Equation (2).

$\begin{array}{l}-1.2\text{k}=\left(\text{3k}\right)\times \left(0.6{\text{v}}_A\text{j}\right)\text{+}\left(2\text{i}+2\text{j}+\text{3k}\right)\times \left(0.6{\text{v}}_B\text{i}\right)\text{+}\left(\text{i}+\text{4j}\right)\times \left(0.6{\text{v}}_c\text{k}\right)\\ -1.2\text{k}=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 3\\ 0& 0.6{\text{v}}_A& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& 2& 3\\ 0.6{\text{v}}_B& 0& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1& 4& 0\\ 0& 0& 0.6{\text{v}}_C\end{array}\right|\\ -1.2\text{k}=\left(-1.8\text{i}\right)+\left(1.8{\text{v}}_B\text{j}-\text{1}{\text{.2v}}_B\text{k}\right)+\left(\text{2}{\text{.4v}}_C\text{i}-\text{0}{\text{.6v}}_C\text{j}\right)\\ -1.2\text{k}=\left(-1.8{\text{v}}_A+2.4{\text{v}}_c\right)\text{i}+\left(1.8{\text{v}}_{\text{B}}-0.6{\text{v}}_{\text{c}}\right)\text{j}+\left(-1.2{\text{v}}_{\text{B}}\right)\text{k}\end{array}$

Equating i, j, k components.

$-1.8{\text{v}}_A+2.4{\text{v}}_C=0$ (3)

$-1.8{\text{v}}_B-0.6{\text{v}}_C=0$ (4)

$-1.2{\text{v}}_B=-1.2$ (5)

Find the velocity at point B as follows:

$\begin{array}{l}-1.2{\text{v}}_B=-1.2\\ {\text{v}}_B=\left(1.00\text{ft}/\text{s}\right)\text{i}\end{array}$

Thus, the velocity of particles B is $\underset{\_}{\left(1.00\text{ft}/\text{s}\right)\text{i}}$

Find the velocity at point C as follows:

Substitute $\left(1.00\text{ft}/\text{s}\right)\text{i}$ for ${\text{v}}_B$ in Equation (4).

$\begin{array}{l}-1.8\left(1.00\right)-0.6{\text{v}}_C=0\\ -1.8-0.6{\text{v}}_C=0\\ {\text{v}}_C=\left(3.00\text{ft}/\text{s}\right)\text{k}\end{array}$

Thus, the velocity of particles C is $\underset{\_}{\left(3.00\text{ft}/\text{s}\right)\text{k}}$

Find the velocity at point A as follows:

Substitute $\left(3.00\text{ft}/\text{s}\right)\text{k}$ for ${\text{v}}_C$ in Equation (3).

$\begin{array}{l}-1.8{\text{v}}_A+2.4\left(\left(3.00\right)\right)=0\\ -1.8{\text{v}}_A+2.4\left(\left(3.00\right)\right)=0\\ -1.8{\text{v}}_A=7.2\\ {\text{v}}_A=\left(4.00\text{ft}/\text{s}\right)\text{j}\end{array}$

Thus, the velocity of particles A is $\underset{\_}{\left(4.00\text{ft}/\text{s}\right)\text{j}}$

Determine position vector $\left(\overline{\text{r}}\right)$ of the mass center G of the system using the relation:

$\begin{array}{c}\overline{\text{r}}\text{=}\frac{{\displaystyle \sum _{i=1}^n{m}_i{\text{r}}_i}}{{\displaystyle \sum _{i=1}^n{m}_i}}\\ =\frac{m_A{\text{r}}_A+m_B{\text{r}}_B+m_C{\text{r}}_C}{m_A+m_B+m_C}\end{array}$ . (6)

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 $3\text{k}$ for ${\text{r}}_A$, $2\text{i}+2\text{j}+\text{3k}$ for ${\text{r}}_B$, $\text{i+4j}$ for $r_c$, and $0.6\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $\left(m_A,m_{B,}{m}_C\right)$, in Equation (6).

$\begin{array}{c}\overline{\text{r}}=\frac{0.6\times 3\text{k+}\left(0.6\right)\left(2\text{i}+\text{2j}+\text{3k}\right)+0.6\left(\text{i}+\text{4j}\right)}{1.8}\\ =\text{i}+\text{2j}+\text{2k}\end{array}$

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

${{\text{r}}^{\prime }}_A={\text{r}}_A-\overline{r}$ (7)

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

Substitute $3\text{k}$ for ${\text{r}}_A$, and $\text{i+2j+2k}$ for $\overline{r}$ in Equation (7).

$\begin{array}{c}{{\text{r}}^{\prime }}_A=3\text{k}-\left(\text{i}+\text{2j}+\text{2k}\right)\\ =-\text{i}-\text{2j}+\text{k}\end{array}$

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

${{\text{r}}^{\prime }}_B={\text{r}}_B-\overline{r}$ (8)

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

Substitute $2\text{i}+2\text{j}+\text{3k}$ for ${\text{r}}_B$ and $\text{i}+\text{2j}+\text{2k}$ for $\overline{r}$  in Equation (8).

$\begin{array}{c}{{\text{r}}^{\prime }}_B=2\text{i}+2\text{j}+\text{3k}-\text{i}+\text{2j}+\text{2k}\\ =\text{i}+\text{k}\end{array}$

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

${{\text{r}}^{\prime }}_C={\text{r}}_C-\overline{r}$ (9)

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

Substitute $\text{i}+\text{4j}$ for ${\text{r}}_C$, and $\text{i}+\text{2j}+\text{2k}$ for $\overline{r}$ in Equation (9).

$\begin{array}{c}{{\text{r}}^{\prime }}_C=\text{i}+\text{4j}-\text{i}+\text{2j}+\text{2k}\\ =\text{2j}-\text{2k}\end{array}$

Express the linear momentum of particle A as follows:

$\begin{array}{c}{m}_A{\text{v}}_A=0.6\left(4\text{j}\right)\\ =\left(2.4\text{lb}\cdot \text{s}\right)\text{j}\end{array}$

Express the linear momentum of particle B as follows:

$\begin{array}{c}{m}_B{\text{v}}_B=0.6\left(\text{i}\right)\\ =\left(0.6\text{lb}\cdot \text{s}\right)\text{i}\end{array}$

Express the linear momentum of particle C as follows:

$\begin{array}{c}{m}_C{\text{v}}_C=0.6\left(3\text{k}\right)\\ =\left(1.8\text{lb}\cdot \text{s}\right)\text{k}\end{array}$

To determine

Find the angular momentum $H_G$ of the system.

Answer to Problem 14.11P

The angular momentum $H_G$ of the system is $\underset{\_}{\left(1.2\text{i}+\text{0}\text{.6j}-\text{2}\text{.4k}\right)\text{ft}\cdot \text{lb}\cdot \text{s}}$.

Explanation of Solution

Calculation:

Calculate the angular momentum about point G using the relation:

$H_G={{\text{r}}^{\prime }}_A\times \left(m_A{\text{v}}_A\right)+{{\text{r}}^{\prime }}_B\times \left(m_B{\text{v}}_B\right)+{{\text{r}}^{\prime }}_C\times \left(m_C{\text{v}}_C\right)$ (10)

Here, $\left({{\text{r}}^{\prime }}_A,{{\text{r}}^{\prime }}_B,{{\text{r}}^{\prime }}_C\right)$ are positive vector for particles and $\left(m_A{\text{v}}_A,m_B{\text{v}}_B,m_C{\text{v}}_C\right)$ are linear momentum.

Substitute $-\text{i}-\text{2j+k}$ for ${{\text{r}}^{\prime }}_A$, $\text{i}+\text{k}$ for ${{\text{r}}^{\prime }}_B$, $\text{2j}-\text{2k}$ for ${{\text{r}}^{\prime }}_C$, $\left(2.4\text{lb}\cdot \text{s}\right)\text{j}$ for $m_A{\text{v}}_A$, $\left(0.6\text{lb}\cdot \text{s}\right)\text{i}$ for $m_B{\text{v}}_B$, and $\left(1.8\text{lb}\cdot \text{s}\right)\text{k}$ for $m_C{\text{v}}_C$ in Equation (10).

$\begin{array}{c}{H}_G=-\text{i}-\text{2j}+\text{k}\times \left(2.4\text{lb}\cdot \text{s}\right)\text{j}+\text{i}+\text{k}\times \left(0.6\text{lb}\cdot \text{s}\right)\text{i}+\text{2j}-\text{2k}\times \left(\left(1.8\text{lb}\cdot \text{s}\right)\text{k}\right)\\ =\left(-2.4\text{i}-2.4\text{k}\right)+\left(0.6\text{j}\right)+\left(3.6\text{i}\right)\\ =\left(1.2\text{i}+\text{0}\text{.6j}-\text{2}\text{.4k}\right)\text{ft}\cdot \text{lb}\cdot \text{s}\end{array}$

Thus, the angular momentum $H_G$ of the system is $\underset{\_}{\left(1.2\text{i}+\text{0}\text{.6j}-\text{2}\text{.4k}\right)\text{ft}\cdot \text{lb}\cdot \text{s}}$.