Chapter 14.1, Problem 14.12P

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, and the magnitude of the linear momentum L of the system is 9 lb · s. Knowing that H_G = H_O, where H_G is the angular momentum of the system about its mass center G and H_O is the angular momentum of the system about O, determine (a) the velocities of the particles, (b) the angular momentum of the system about O.

Fig. P14.11 and P14.12

To determine

Find the velocity of particles.

Answer to Problem 14.12P

The velocity of particle A is $\underset{\_}{\left(10.00\text{ft}/\text{s}\right)\text{j}}$.

The velocity of particle B is $\underset{\_}{\left(5.00\text{ft}/\text{s}\right)\text{j}}$.

The velocity of particle C is $\underset{\_}{\left(10.00\text{ft}/\text{s}\right)\text{j}}$.

Explanation of Solution

Given information:

The velocity of particles is as follows:

$\begin{array}{l}{\text{v}}_A=v_A\text{j}\\ {\text{v}}_B=v_B\text{i}\\ {\text{v}}_C=v_C\text{k}\end{array}$

The linear momentum (L) of the system is $9\text{lb}\cdot \text{s}$.

Calculation:

The mass of three particles 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 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:

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

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

$\begin{array}{c}\overline{\text{r}}\text{=}\frac{{\displaystyle \sum _{i=1}^n{m}_i{\text{r}}_i}}{{\displaystyle \sum _{i=1}^n{m}_i}}\\ =\frac{m_A{\text{r}}_A+m_B{\text{r}}_B+m_C{\text{r}}_C}{m_A+m_B+m_C}\end{array}$ . (2)

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 ${\text{r}}_A$, $2\text{i}+2\text{j+3k}$ 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 (2).

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

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

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

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 (3).

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

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

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

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

Substitute $2\text{i}+2\text{j+3k}$ for ${\text{r}}_B$ and $\text{i+2j+2k}$ for $\overline{r}$ in Equation (4).

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

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

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

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

Substitute $\text{i+4j}$ for ${\text{r}}_C$, and $\text{i+2j+2k}$ for $\overline{r}$ in Equation (5).

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

Express the angular momentum about point O as follows:

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

Calculate the angular momentum about point G using the relation:

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

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.

Subtract the angular momentum $H_O-H_G$ as follows:

$\begin{array}{c}{H}_O-H_G=\left(\begin{array}{l}{\text{r}}_A\times \left(m_A{\text{v}}_A\right)+r_B\times \left(m_B{\text{v}}_B\right)+{\text{r}}_C\times \left(m_C{\text{v}}_C\right)\\ -{{\text{r}}^{\prime }}_A\times \left(m_A{\text{v}}_A\right)+{{\text{r}}^{\prime }}_B\times \left(m_B{\text{v}}_B\right)+{{\text{r}}^{\prime }}_C\times \left(m_C{\text{v}}_C\right)\end{array}\right)\\ =\left({\text{r}}_A-{{\text{r}}^{\prime }}_A\right)\times \left(m_A{\text{v}}_A\right)+\left({{\text{r}}^{\prime }}_B\right)\times m_B{\text{v}}_B+{{\text{r}}^{\prime }}_C\times \left(m_C{\text{v}}_C\right)\\ 0=\overline{\text{r}}\times \left(m_A{\text{v}}_A\right)+\overline{\text{r}}\times \left(m_B{\text{v}}_B\right)\overline{\text{r}}\times \left(m_C{\text{v}}_C\right)\\ =\overline{\text{r}}\left(m_A{\text{v}}_A+m_B{\text{v}}_B+m_C{\text{v}}_C\right)\end{array}$

L is parallel to $\overline{\text{r}}$ as follows:

$\begin{array}{l}L=\lambda \overline{\text{r}}\\ \text{L}\cdot \text{L=}{\lambda }^2\overline{\text{r}}\cdot \overline{\text{r}}\\ {\lambda }^2=\frac{\text{L}\cdot \text{L}}{\overline{\text{r}}\cdot \overline{\text{r}}}\end{array}$ (6)

Substitute $9\text{lb}\cdot \text{s}$ for L and 3 for $\overline{\text{r}}$ in Equation (6).

$\begin{array}{c}{\lambda }^2=\frac{9\times 9}{3\times 3}\\ =\frac{81}{9}\\ =3^2\\ \lambda =\pm 3\text{lb}\cdot \text{s}/\text{ft}\end{array}$

Find the velocity of particles using the relation as follows:

$m_A{\text{v}}_A+m_B{\text{v}}_B+m_C{\text{v}}_C=\lambda \overline{\text{r}}$ (7)

Substitute $0.6\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $\left(m_A,m_{B,}{m}_C\right)$, $\pm 3\text{lb}\cdot \text{s}/\text{ft}$ for $\lambda$, $\text{i+2j+2k}$ for $\overline{\text{r}}$, $\left(v_A\text{j}\right)$ for ${\text{v}}_A$, $\left(v_B\text{i}\right)$ for ${\text{v}}_B$, $\left(v_C\text{k}\right)$ for ${\text{v}}_C$ in Equation (7).

$0.6\left(\left({\text{v}}_A\text{j}\right)\right)+0.6\left({\text{v}}_B\text{i}\right)+0.6\left({\text{v}}_C\text{k}\right)=\pm 3\left(\text{i+2j+2k}\right)$

Resolve the components and solve ${\text{v}}_A,{\text{v}}_B,{\text{v}}_C$.

Find the velocity of particle A using the Equation:

$\begin{array}{l}0.6\left(\left({\text{v}}_A\text{j}\right)\right)=6\text{j}\\ {\text{v}}_A=10\left(\text{ft}/\text{s}\right)\text{j}\end{array}$

Thus, the velocity of particle A is $\underset{\_}{\left(10.00\text{ft}/\text{s}\right)\text{j}}$.

Find the velocity of particle B using the Equation:

$\begin{array}{l}0.6\left(\left({\text{v}}_B\text{i}\right)\right)=3\text{i}\\ {\text{v}}_B=5\left(\text{ft}/\text{s}\right)\text{i}\end{array}$

Thus, the velocity of particle B is $\underset{\_}{\left(5.00\text{ft}/\text{s}\right)\text{j}}$.

Find the velocity of particle C using the Equation:

$\begin{array}{l}0.6\left(\left({\text{v}}_C\text{j}\right)\right)=6\text{k}\\ {\text{v}}_C=10\left(\text{ft}/\text{s}\right)\text{k}\end{array}$

Thus, the velocity of particle C is $\underset{\_}{\left(10.00\text{ft}/\text{s}\right)\text{j}}$.

To determine

Find the angular momentum of the system about O.

Answer to Problem 14.12P

The angular momentum of the system about O is $\underset{\_}{\left(6.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{i+}\left(3.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{j}-\left(6.0\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{k}}$

Explanation of Solution

Calculation:

Calculate the angular momentum about point H using the relation:

$H_O={\text{r}}_A\times \left(m_A{\text{v}}_A\right)+{\text{r}}_B\times \left(m_B{\text{v}}_B\right)+{\text{r}}_C\times \left(m_C{\text{v}}_C\right)$ (8)

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

Substitute $3\text{k}$ for ${\text{r}}_A$, $2\text{i}+2\text{j+3k}$ for ${\text{r}}_B$, $\text{i+4j}$ for $r_c$, $0.6\left(\left({\text{v}}_A\text{j}\right)\right)$ for $m_A{\text{v}}_A$, $0.6\left({\text{v}}_B\text{i}\right)$ for $m_B{\text{v}}_B$, and $0.6\left({\text{v}}_C\text{k}\right)$ for $m_C{\text{v}}_C$ in Equation (8).

$\begin{array}{c}{H}_O=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 3\\ 0& 0.6{\text{v}}_A& 8.0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& 2& 3\\ 0.6{\text{v}}_B& 0& 0\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 1& 4& 0\\ 0& 0& 0.6{\text{v}}_C\end{array}\right|\\ =\left(-1.8{\text{v}}_A\text{i}\right)+\left(1.8{\text{v}}_B\text{j}-1.2{\text{v}}_B\text{k}\right)+\left(2.4{\text{v}}_C\text{j}-0.6{\text{v}}_C\text{j}\right)\\ =\left(-1.8{\text{v}}_A+2.4{\text{v}}_C\right)\text{i}+\left(1.8{\text{v}}_B-0.6{\text{v}}_C\right)\text{j}+\left(-1.2{\text{v}}_B\right)\text{k}\\ =6\text{i}+3\text{j}-\text{6k}\\ \text{=}\left(6.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{i+}\left(3.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{j}-\left(6.0\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{k}\end{array}$

Thus, the angular momentum of the system about O is $\underset{\_}{\left(6.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{i+}\left(3.00\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{j}-\left(6.0\text{ft}\cdot \text{lb}\cdot \text{s}\right)\text{k}}$