Chapter 14.1, Problem 14.14P

For the system of particles of Prob. 14.13, determine (a) the position vector $\stackrel{\to }{r}$ of the mass center G of the system. (b) the linear momentum $m\stackrel{\to }{v}$ 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.

To determine

(a)

The position vector of the mass centre, G of the system.

Answer to Problem 14.14P

$\overline{r}=\left(1.87m\right)i+\left(1.53m\right)j+\left(0.667m\right)k$

Explanation of Solution

Given information:

Mass of A $m_a=3{\rm K}g$

Mass of B $m_b=2{\rm K}g$

Mass of C $m_b=4{\rm K}g$

Velocity of A $V_A=4i+2j+2k$

Velocity of B $V_B=4i+3j$

Velocity of C $V_C=-2i+4j+2k$

We can find the position vectors at centre O with the help of figure at points A, B and C.

Position vector at A, $r_A=3j$

Position vector at B, $r_B=1.2i+2.4j+3k$

Position vector at C, $r_C=3.6i$

The position vector at mass centre G is, $m\overline{r}=\left(m_A+m_B+m_C\right)\overline{r}$

$\begin{array}{l}\left(m_A+m_B+m_C\right)\overline{r}=m_A{r}_A+m_B{r}_B+m_C{r}_C\\ \left(3+2+4\right)\overline{r}=3\left(3j\right)+2\left(1.2i+2.4j+3k\right)+4\left(3.6i\right)\\ 9\overline{r}=9j+2.4i+4.8j+6k+14.4i\\ 9\overline{r}=16.8i+13.8j+6k\\ \overline{r}=1.87i+1.53j+0.667k\end{array}$

Conclusion:

The position vector of the given system is, $\overline{r}=\left(1.87m\right)i+\left(1.53m\right)j+\left(0.667m\right)k$

To determine

(b)

The linear momentum of the system.

Answer to Problem 14.14P

$m\overline{V}=\left(12{\rm K}g.m/s\right)i+\left(28{\rm K}g.m/s\right)j+\left(14{\rm K}g.m/s\right)k$

Explanation of Solution

Given information:

Mass of A $m_a=3{\rm K}g$

Mass of B $m_b=2{\rm K}g$

Mass of C $m_b=4{\rm K}g$

Velocity of A $V_A=4i+2j+2k$

Velocity of B $V_B=4i+3j$

Velocity of C $V_C=-2i+4j+2k$

The linear momentum of each particle,

Linear momentum of particle A, $m\overline{V}=m_A{V}_A=3\left(4i+2j+2k\right)=12i+6j+6k$

Linear momentum of particle B, $m\overline{V}=m_B{V}_B=2\left(4i+3j\right)=8i+6j$

Linear momentum of particle C, $m\overline{V}=m_C{V}_C=4\left(-2i+4j+2k\right)=-8i+16j+8k$

Now, total linear momentum of the system, $m\overline{V}=m_A{V}_A+m_B{V}_B+m_C{V}_C$

$\begin{array}{l}m\overline{V}=12i+6j+6k+8i+6j-8i+16j+8k\\ m\overline{V}=12i+28j+14k\end{array}$

Conclusion:

Linear moment of the system, $m\overline{V}=\left(12{\rm K}g.m/s\right)i+\left(28{\rm K}g.m/s\right)j+\left(14{\rm K}g.m/s\right)k$

To determine

(c)

The angular momentum of system about point ‘G’. And also to show that the answer is verify the equation $H_O=H_G+\overline{r}\times m\overline{V}.$

Answer to Problem 14.14P

$H_G=\left(-2.8{\rm K}g.m^2/s\right)i+\left(13.33{\rm K}g.m^2/s\right)j-\left(24.3{\rm K}g.m^2/s\right)k$

Explanation of Solution

Given information:

Mass of A $m_a=3{\rm K}g$

Mass of B $m_b=2{\rm K}g$

Mass of C $m_b=4{\rm K}g$

Velocity of A $V_A=4i+2j+2k$

Velocity of B $V_B=4i+3j$

Velocity of C $V_C=-2i+4j+2k$

We can find the position vectors at centre O with the help of figure at points A, B and C.

Position vector at A, $r_A=3j$

Position vector at B, $r_B=1.2i+2.4j+3k$

Position vector at C, $r_C=3.6i$

The position vector at mass centre G is, $m\overline{r}=\left(m_A+m_B+m_C\right)\overline{r}$

$\begin{array}{l}\left(m_A+m_B+m_C\right)\overline{r}=m_A{r}_A+m_B{r}_B+m_C{r}_C\\ \left(3+2+4\right)\overline{r}=3\left(3j\right)+2\left(1.2i+2.4j+3k\right)+4\left(3.6i\right)\\ 9\overline{r}=9j+2.4i+4.8j+6k+14.4i\\ 9\overline{r}=16.8i+13.8j+6k\\ \overline{r}=1.87i+1.53j+0.667k\end{array}$

Position vector relative to the mass centre,

$\begin{array}{l}{r}_A^{\text{'}}=r_A-\overline{r}\\ r_A^{\text{'}}=3j-\left(1.87i+1.543j+0.667k\right)\\ r_A^{\text{'}}=-1.87i+1.467j-0.66k\end{array}$

$\begin{array}{l}{r}_B^{\text{'}}=r_B-\overline{r}\\ r_B^{\text{'}}=1.2i+2.4j+3k-\left(1.87i+1.543j+0.667k\right)\\ r_B^{\text{'}}=-0.667i+0.867j+2.33k\end{array}$

$\begin{array}{l}{r}_C^{\text{'}}=r_C-\overline{r}\\ r_C^{\text{'}}=3.6i-\left(1.87i+1.543j+0.667k\right)\\ r_C^{\text{'}}=1.733i-1.543j-0.667k\end{array}$

Now, angular momentum about G,

$H_G=r_A^{\text{'}}\times m_A{V}_A+r_B^{\text{'}}\times m_B{V}_B+r_C^{\text{'}}\times m_C{V}_C$

$\begin{array}{l}{H}_G=\left|\begin{array}{ccc}i& j& k\\ -1.867& 1.467& -0.667\\ 12& 6& 6\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ -0.667& 0.867& 2.33\\ 8& 6& 0\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 1.733& -1.533& -0.667\\ -8& 16& 8\end{array}\right|\\ H_G=\left(12.8i+3.2j-28.8k\right)+\left(-14i+18.667j-10.933k\right)+\left(-1.6i-8.533j+15.467k\right)\\ H_G=12.8i+3.2j-28.8k-14i+18.667j-10.933k-1.6i-8.533j+15.467k\\ H_G=2.8i+13.33j-24.267k\end{array}$

$H_G=\left(-2.8{\rm K}g.m^2/s\right)i+\left(13.33{\rm K}g.m^2/s\right)j-\left(24.3{\rm K}g.m^2/s\right)k$

Now, we show that this answer is verified the given formula:

$\overline{r}\times m\overline{V}$

$\begin{array}{l}\overline{r}\times m\overline{V}=\left|\begin{array}{ccc}i& j& k\\ 1.87& 1.53& 0.667\\ 12& 28& 14\end{array}\right|\\ \overline{r}\times m\overline{V}=\left(2.8i-18.133j+33.86k\right)\end{array}$

Now, the angular momentum about G is added in the above obtained equation:

$H_G+\overline{r}\times m\overline{V}$

$\begin{array}{l}\Rightarrow \left(-2.8{\rm K}g.m^2/s\right)i+\left(13.33{\rm K}g.m^2/s\right)j-\left(24.3{\rm K}g.m^2/s\right)k+\left(2.8i-18.133j+33.86k\right)\\ \Rightarrow -2.8i+13.33j-24.3k+2.8i-18.133j+33.86k\\ \Rightarrow -4.8j+9.6k\end{array}$

Now, Angular moment of all the particles about point O,

$H_O=r_A\times m_A{V}_A+r_B\times m_B{V}_B+r_C\times m_C{V}_C$

$\begin{array}{l}{H}_O=\left|\begin{array}{ccc}i& j& k\\ 0& 3& 0\\ 12& 6& 6\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 1.2& 2.4& 3\\ 8& 6& 0\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 3.6& 0& 0\\ -8& 16& 8\end{array}\right|\\ H_O=\left(18i-36k\right)+\left(-18i+24j-12k\right)+\left(-28.8j+57.6k\right)\\ H_O=18i-36k-18i+24j-12k-28.8j+57.6k\\ H_O=0i-4.8j+9.6k\end{array}$

$H_O=\left(-4.8{\rm K}g.m^2/s\right)j+\left(9.6{\rm K}g.m^2/s\right)k$

Hence the above obtained values verify the equation:

$H_O=H_G+\overline{r}\times m\overline{V}.$