Chapter 4.8, Problem 93RP

To determine

The position vector $\left(\overline{r}\right)$, velocity vector $\left(\dot{\overline{r}}\right)$, acceleration vector $\left(\ddot{\overline{r}}\right)$, kinetic energy $\left(T\right)$, angular momentum vector $\left(H_O\right)$, and the time rate of angular momentum vector $\left({\dot{H}}_O\right)$.

Answer to Problem 93RP

The position vector $\left(\overline{r}\right)$ of the system of particles is $\frac{d}{7}(i+4j+6k)$.

The velocity vector $\left(\dot{\overline{r}}\right)$ of the system of particles is $\frac{v}{7}(4i+2j+6k)$.

The acceleration vector $\left(\ddot{\overline{r}}\right)$ of the system of particles is $\left(\frac{F}{7m}\right)k$.

The kinetic energy $\left(T\right)$ of the system of particles is $13m{v}^2$.

The angular momentum vector $\left(H_O\right)$ of the system about the origin is

$mvd[12i+6j+2k]$.

The time rate of angular momentum vector $\left({\dot{H}}_O\right)$ of the system of particles is $-(Fd)j$.

Explanation of Solution

Draw the free body diagram of the system f particles is shown in Figure (1).

Write the expression to determine the position vector $\left(\overline{r}\right)$ of the system of particles.

  $\begin{array}{c}\overline{r}=\frac{{\displaystyle \sum m_i}{r}_i}{{\displaystyle \sum m_i}}\\ =\frac{m_A{r}_A+m_B{r}_B+m_C{r}_C}{m_A+m_B+m_C}\end{array}$        (I)

Here, mass of the particles is $m_i$, position of the articles is $r_i$, masses of the particles in positions are $m_A$, $m_B$, and $m_C$, distance between the particles are $r_A$, $r_B$, and $r_C$.

Write the expression to determine the velocity vector $\left(\dot{\overline{r}}\right)$ of the system of particles.

  $\begin{array}{c}\dot{\overline{r}}=\frac{{\displaystyle \sum m_i}{v}_i}{{\displaystyle \sum m_i}}\\ =\frac{m_A{v}_A+m_B{v}_B+m_C{v}_C}{m_A+m_B+m_C}\end{array}$        (II)

Here, velocity acting on the  particles is $v_i$.and velocity of the particles are $v_A$, $v_B$, and $v_C$.

Write the expression to determine the acceleration vector $\left(\ddot{\overline{r}}\right)$ of the system of particles.

  $\begin{array}{c}\ddot{\overline{r}}=\frac{{\displaystyle \sum F_i}}{{\displaystyle \sum m_i}}\\ =\frac{F_A+F_B+F_C}{m_A+m_B+m_C}\end{array}$        (III)

Here, the external forces acting on the particles is $F_i$, and forces of the particles are $F_A$, $F_B$, and $F_C$.

Write the expression to determine the kinetic energy $\left(T\right)$ of the system of particles.

  $\begin{array}{c}T={\displaystyle \sum \frac{1}{2}{m}_i{v}_i}\\ =\frac{1}{2}\left(m_A{v}_A^2+m_B{v}_B^2+m_C{v}_C^2\right)\end{array}$        (IV)

Write the expression to determine the angular momentum vector $\left(H_O\right)$ of the system about the origin.

  $\begin{array}{c}{H}_O={\displaystyle \sum \left(r_i\times m_i{v}_i\right)}\\ =\left(r_A\times m_A{v}_A+r_B\times m_B{v}_B+r_C\times m_C{v}_C\right)\end{array}$        (V)

Write the expression to determine time rate of angular momentum vector $\left({\dot{H}}_O\right)$ of the system of particles.

  $\begin{array}{c}{\dot{H}}_O={\displaystyle \sum M_O}\\ \begin{array}{l}={\displaystyle \sum r_i}\times F_i\hfill \\ =\left(r_A\times F_A\right)+\left(r_B\times F_B\right)+\left(r_C\times F_c\right)\hfill \end{array}\end{array}$        (VI)

Conclusion:

Susbtitute $4m$ for $m_A$, $2m$ for $m_B$, $m$ for $m_C$, $\left(1.5d\right)i$ for $r_A$, $\left(2d\right)j$ for $r_B$, and $\left(d\right)i$ for $r_C$ in Equation (I).

  $\begin{array}{c}\overline{r}=\frac{[4m\times (1.5d)k]+[2m\times ((2d)j)]+[m\times (d)i]}{4m+2m+m}\\ =\frac{m[6dk+4dj+di]}{7m}\\ =\frac{di+4dj+6dk}{7}\\ =\frac{d}{7}(i+4j+6k)\end{array}$

Thus, position vector $\left(\overline{r}\right)$ of the system of particles is $\frac{d}{7}(i+4j+6k)$.

Susbtitute $m$ for $m_A$, $4m$ for $m_B$, $2m$ for $m_C$, $vi$ for $v_A$, $3vk$ for $v_B$, and $2vj$ for $v_C$ in Equation (II).

  $\begin{array}{c}\dot{\overline{r}}=\frac{[4m\times (v)i]+[2m\times (3v)k]+[m\times (2v)j]}{4m+2m+m}\\ =\frac{m[(4v)i+(6v)k+(2v)j]}{7m}\\ =\frac{(4v)i+(6v)k+(2v)j}{7}\\ =\frac{v}{7}(4i+2j+6k)\end{array}$

Thus, the velocity vector $\left(\dot{\overline{r}}\right)$ of the system of particles is $\frac{v}{7}(4i+2j+6k)$.

Susbtitute $4m$ for $m_A$, $2m$ for $m_B$, $m$ for $m_C$, $0$ for $F_A$, $0$ for $F_B$, and $Fk$ for $F_C$ in Equation (III).

  $\begin{array}{c}\ddot{\overline{r}}=\frac{(F)k}{4m+2m+m}\\ =\left(\frac{F}{7m}\right)k\end{array}$

Thus, the acceleration vector $\left(\ddot{\overline{r}}\right)$ of the system of particles is $\left(\frac{F}{7m}\right)k$.

Susbtitute $4m$ for $m_A$, $2m$ for $m_B$, $m$ for $m_C$, $vi$ for $v_A$, $3vk$ for $v_B$, and $2vj$ for $v_C$ in Equation (IV).

  $\begin{array}{c}T=\frac{1}{2}\left[\left(4m\times v^2\right)+\left(2m\times {(3v)}^2\right)+\left(m\times {(2v)}^2\right)\right]\\ =\frac{1}{2}\left(4m{v}^2+18m{v}^2+4m{v}^2\right)\\ =\frac{1}{2}\left(26m{v}^2\right)\\ =13m{v}^2\end{array}$

Thus, the kinetic energy $\left(T\right)$ of the system of particles is $13m{v}^2$.

Susbtitute $4m$ for $m_A$, $2m$ for $m_B$, $m$ for $m_C$, $vi$ for $v_A$, $3vk$ for $v_B$, $2vj$ for $v_C$, $\left(1.5d\right)i$ for $r_A$, $\left(2d\right)j$ for $r_B$, and $\left(d\right)i$ for $r_C$ in Equation (V).

  $\begin{array}{c}{H}_O=[(1.5d)k+(2d)j+(d)i]\times [(4m\times vi)+(2m\times (3v)k)+(m\times 2v)j]\\ =[(1.5d)k+(2d)j+(d)i]\times m[(4vi)+(6v)k+(2v)j]\\ =6mvdj+12mvdi+2mvdk\\ =mvd[12i+6j+2k]\end{array}$

Thus, the angular momentum vector $\left(H_O\right)$ of the system about the origin is $mvd[12i+6j+2k]$

Susbtitute $4m$ for $m_A$, $2m$ for $m_B$, $m$ for $m_C$, $0$ for $F_A$, $0$ for $F_B$, and $Fk$ for $F_C$, $\left(1.5d\right)i$ for $r_A$, $\left(2d\right)j$ for $r_B$, and $\left(d\right)i$ for $r_C$ in in Equation (VI).

  $\begin{array}{c}{\dot{H}}_O=((1.5d)k\times 0)+[2dj\times 0]+((d)i\times Fk)\\ =0+0+(-Fd)j\\ =-(Fd)j\end{array}$

Thus, the time rate of angular momentum vector $\left({\dot{H}}_O\right)$ of the system of particles is $-(Fd)j$.