Chapter 14.1, Problem 14.10P

For the satellite system of Prob. 14.9. assuming that the velocity of the base satellite is zero, determine (a) the position vector r of the mass center G of the system, (b) the linear momentum L of the system, (c) the angular momentum Hg of the system about G. Also, verify that the answers to this problem and to Prob. 14.9 satisfy the equation given in Prob. 14.27.

To determine

(a)

The position vector r of the mass centre G of the system assuming the base satellite velocity is zero

Explanation of Solution

Given information:

${\overrightarrow{v}}_A=\left(4i-2j+2k\right)$

${\overrightarrow{v}}_B=\left(i+4j\right)$

${\overrightarrow{v}}_C=\left(2i+2j+4k\right)$

$m_A=4kg$

$m_B=6kg$

$m_C=8kg$

Reference equation from Problem 14.27

$H_o=\overline{r}\times m\overline{v}+H_G$

Position vectors is written as below

$\begin{array}{l}{\overrightarrow{r}}_A=30j\\ {\overrightarrow{r}}_B=15i+25j+35k\\ {\overrightarrow{r}}_C=40i\end{array}$

System mass centre is calculated as below,

$\overrightarrow{r}=\frac{m_A{\overrightarrow{r}}_A+m_B{\overrightarrow{r}}_B+m_C{\overrightarrow{r}}_C}{m_A+m_B+m_C}$

Substituting the values we get,

$\overrightarrow{r}=\frac{4\left(30j\right)+6\left(15i+25j+35k\right)+8\left(40i\right)}{4+6+8}$

$\overrightarrow{r}=22.78i+15j+11.67k$

Therefore, mass centre position vector is $\overrightarrow{r}=22.78i+15j+11.67k$

Conclusion:

Position vector of mass centre is $\overrightarrow{r}=22.78i+15j+11.67k$

To determine

(b)

The linear momentum L of the system assuming the base satellite velocity is zero

Explanation of Solution

Given information:

${\overrightarrow{v}}_A=\left(4i-2j+2k\right)$

${\overrightarrow{v}}_B=\left(i+4j\right)$

${\overrightarrow{v}}_C=\left(2i+2j+4k\right)$

$m_A=4kg$

$m_B=6kg$

$m_C=8kg$

Reference equation from Problem 14.27

$H_o=\overline{r}\times m\overline{v}+H_G$

Position vectors is written as below

$\begin{array}{l}{\overrightarrow{r}}_A=30j\\ {\overrightarrow{r}}_B=15i+25j+35k\\ {\overrightarrow{r}}_C=40i\end{array}$

Linear momentum is calculated by using the relation,

$\overrightarrow{L}=m_A{\overrightarrow{v}}_A+m_B{\overrightarrow{v}}_B+m_C{\overrightarrow{v}}_C$

Substituting the values we get,

$\overrightarrow{L}=4\left(4i-2j+2k\right)+6\left(i+4j\right)+8\left(2i+2j+4k\right)$

$\overrightarrow{L}=38i+32j+40k$

Therefore, linear momentum of the system is $\overrightarrow{L}=38i+32j+40k$

Conclusion:

Linear momentum of the system is $\overrightarrow{L}=38i+32j+40k$

To determine

(c)

The angular momentum of the system about G assuming the base satellite velocity is zero

Explanation of Solution

Given information:

${\overrightarrow{v}}_A=\left(4i-2j+2k\right)$

${\overrightarrow{v}}_B=\left(i+4j\right)$

${\overrightarrow{v}}_C=\left(2i+2j+4k\right)$

$m_A=4kg$

$m_B=6kg$

$m_C=8kg$

Reference equation from Problem 14.27

$H_o=\overline{r}\times m\overline{v}+H_G$

Position vectors is written as below

$\begin{array}{l}{\overrightarrow{r}}_A=30j\\ {\overrightarrow{r}}_B=15i+25j+35k\\ {\overrightarrow{r}}_C=40i\end{array}$

Angular momentum is calculated by using the relation,

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

Substituting the values we get,

$H_G=\left\{\begin{array}{l}\left(30j-\left(22.78i+15j+11.67k\right)\right)\times 4\left(4i-2j+2k\right)\\ +\left(15i+25j+35k-\left(22.78i+15j+11.67k\right)\right)\times 6\left(i+4j\right)\\ +\left(40i-\left(22.78i+15j+11.67k\right)\right)\times 8\left(2i+2j+4k\right)\end{array}\right\}$

$=\left(\begin{array}{ccc}i& j& k\\ -22.78& 15& -11.67\\ 16& -8& 8\end{array}\right)+\left(\begin{array}{ccc}i& j& k\\ -7.78& 10& 23.33\\ 6& 24& 0\end{array}\right)+\left(\begin{array}{ccc}i& j& k\\ 17.22& -15& -11.67\\ 16& 16& 32\end{array}\right)$

$=\left\{\begin{array}{l}i\left(120-93.36-559.92-480+186.72\right)\\ -j(-182.24+186.72-139.98+551.04+186.72)\\ +k(182.24-240-186.72-60+275.52+240)\end{array}\right\}$

$=\left(-826.56i-602.26j+211.04k\right)kg{m}^2/s$

$\therefore H_G=\left(-826.56i-602.26j+211.04k\right)kg{m}^2/s$

Therefore, angular momentum of the system is $H_G=\left(-826.56i-602.26j+211.04k\right)kg{m}^2/s$

Assume, mass centre G velocity,

$\overrightarrow{v}=v_{x}i+v_{y}j+v_{z}k$

Linear momentum equation can be written as,

$\overrightarrow{L}=\left(m_A+m_B+m_C\right)\overrightarrow{v}$

Substituting the values we get,

$38i+32j+40k=\left(4+6+8\right)\left(v_{x}i+v_{y}j+v_{z}k\right)$

$v_{x}i+v_{y}j+v_{z}k=2.11i+1.78j+2.22k$

Therefore, individual vector components are,

$\begin{array}{l}{v}_x=2.11\\ v_y=1.78\\ v_z=2.22\end{array}$

Angular momentum of the system about origin is

$H_o={\overrightarrow{r}}_A\times m_A{\overrightarrow{v}}_A+{\overrightarrow{r}}_B\times m_B{\overrightarrow{v}}_B+{\overrightarrow{r}}_C\times m_C{\overrightarrow{v}}_C$

$H_o=\left\{\left(3oj\right)\times 4\left(4i-2j+2k\right)+\left(15i+25j+35k\right)\times 6\left(i+4j\right)+\left(40i\right)\times 8\left(2i+2j+4k\right)\right\}$

$=\left(\begin{array}{ccc}i& j& k\\ 0& 30& 0\\ 16& -8& 8\end{array}\right)+\left(\begin{array}{ccc}i& j& k\\ 15& 25& 35\\ 6& 24& 0\end{array}\right)+\left(\begin{array}{ccc}i& j& k\\ 40& 0& 0\\ 16& 16& 32\end{array}\right)$

$H_o=\left(240-840\right)i-j\left(-210+1280\right)+k\left(-480+360-150+640\right)$

$H_o=-600i-1070j+370k$ ................. (Equation A)

Check whether the above relation satisfies with equation in 14.27

${\overrightarrow{H}}_o=\overrightarrow{r}\times \left(m_A+m_B+m_C\right)\overrightarrow{v}+H_G$

${\overrightarrow{H}}_o=\left\{\begin{array}{l}\left(22.78i+15j+11.67k\right)\times \left(4+6+8\right)\left(2.11i+1.78j+2.22k\right)\\ +\left(-826.56i-602.26j+211.04k\right)\end{array}\right\}$

$=\left(\begin{array}{ccc}i& j& k\\ 22.78& 15& 11.67\\ 38& 32& 40\end{array}\right)+\left(-826.56i-602.26j+211.04k\right)$

${\overrightarrow{H}}_o=-600i-1070j+370k$ ................ (Equation B)

From the above, we can conclude, equation A satisfies with equation B

Conclusion:

Angular momentum of the system is $H_G=\left(-826.56i-602.26j+211.04k\right)kg{m}^2/s$