Chapter 14.1, Problem 14.30P

Show that the relation $\sum M_A={\stackrel{\dot{}}{H}}_A^{\text{'}}$ where $H_A^{\text{'}}$ is defined by Eq. (1) of Prob. 14.29 and where $\sum M_A$ represents the sum of the moments about A of the external forces acting on the system of particles, is valid if, and only if, one of the following conditions is satisfied: (a) the frame Axyz¹is itself a newtonian frame of reference, (b) A coincides with the mass center G,

To determine

(a)

$\Sigma M_A={\dot{H}}_A^{\text{'}}$ If the frame $A{x}^{\text{'}}{y}^{\text{'}}{z}^{\text{'}}$ is itself a Newtonian frame of reference.

Explanation of Solution

Given information:

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(v_i-v_A\right)\right]$

Now, differentiate the above equation with respect to time:

${\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left({\dot{r}}_i-{\dot{r}}_A\right)\times m_i\left(v_i-v_A\right)\right]+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left({\dot{v}}_i-{\dot{v}}_A\right)\right]$

$\begin{array}{l}But,\\ {\dot{r}}_i=v_i\\ {\dot{v}}_i=a_i\\ {\dot{r}}_A=v_A\\ {\dot{v}}_A=a_A\end{array}$

Hence,

$\begin{array}{l}{\dot{H}}_A^{\text{'}}=0+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(a_i-a_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times \left(F_i-m_i{a}_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times F\right]-\sum _{i=1}^n\left[m_i\left(r_i-r_A\right)\right]\times a_A\\ {\dot{H}}_A^{\text{'}}={{\rm M}}_A-m\left(r-r_A\right)\times a_A\end{array}$

The term $\Sigma M_A={\dot{H}}_A^{\text{'}}$ is satisfied if the term $m\left(r-r_A\right)\times a_A$ is equals to zero.

Hence, if the frame of A is Newtonian then,

$a_A=0$

Thus, $m\left(r-r_A\right)\times a_A=0$ then, $\Sigma M_A={\dot{H}}_A^{\text{'}}$

Conclusion:

The given condition is satisfied the given equation.

To determine

(b)

$\Sigma M_A={\dot{H}}_A^{\text{'}}$ If A coincides with the mass centre G.

Explanation of Solution

Given information:

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(v_i-v_A\right)\right]$

Now, differentiate the above equation with respect to time:

${\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left({\dot{r}}_i-{\dot{r}}_A\right)\times m_i\left(v_i-v_A\right)\right]+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left({\dot{v}}_i-{\dot{v}}_A\right)\right]$

$\begin{array}{l}But,\\ {\dot{r}}_i=v_i\\ {\dot{v}}_i=a_i\\ {\dot{r}}_A=v_A\\ {\dot{v}}_A=a_A\end{array}$

Hence,

$\begin{array}{l}{\dot{H}}_A^{\text{'}}=0+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(a_i-a_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times \left(F_i-m_i{a}_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times F\right]-\sum _{i=1}^n\left[m_i\left(r_i-r_A\right)\right]\times a_A\\ {\dot{H}}_A^{\text{'}}={{\rm M}}_A-m\left(r-r_A\right)\times a_A\end{array}$

The term $\Sigma M_A={\dot{H}}_A^{\text{'}}$ is satisfied if the term $m\left(r-r_A\right)\times a_A$ is equals to zero.

Hence, if point A coincides with mass centre then,

$r=r_A$

Thus, $m\left(r-r_A\right)\times a_A=0$ then, $\Sigma M_A={\dot{H}}_A^{\text{'}}$

Conclusion:

The given condition is satisfied the given equation.

To determine

(c)

$\Sigma M_A={\dot{H}}_A^{\text{'}}$ the acceleration $a_A$ of A is directed along the line AG.

Explanation of Solution

Given information:

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n{r}_i^{\text{'}}\times m_i{v}_i^{\text{'}}$

$H_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(v_i-v_A\right)\right]$

Now, differentiate the above equation with respect to time:

${\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left({\dot{r}}_i-{\dot{r}}_A\right)\times m_i\left(v_i-v_A\right)\right]+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left({\dot{v}}_i-{\dot{v}}_A\right)\right]$

$\begin{array}{l}But,\\ {\dot{r}}_i=v_i\\ {\dot{v}}_i=a_i\\ {\dot{r}}_A=v_A\\ {\dot{v}}_A=a_A\end{array}$

Hence,

$\begin{array}{l}{\dot{H}}_A^{\text{'}}=0+\sum _{i=1}^n\left[\left(r_i-r_A\right)\times m_i\left(a_i-a_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times \left(F_i-m_i{a}_A\right)\right]\\ {\dot{H}}_A^{\text{'}}=\sum _{i=1}^n\left[\left(r_i-r_A\right)\times F\right]-\sum _{i=1}^n\left[m_i\left(r_i-r_A\right)\right]\times a_A\\ {\dot{H}}_A^{\text{'}}={{\rm M}}_A-m\left(r-r_A\right)\times a_A\end{array}$

The term $\Sigma M_A={\dot{H}}_A^{\text{'}}$ is satisfied if the term $m\left(r-r_A\right)\times a_A$ is equals to zero.

Hence, if the acceleration of A is directed along the line AG then,

$a_{A}isparallelto\left(r-r_A\right)$

Thus, $m\left(r-r_A\right)\times a_A=0$ then, $\Sigma M_A={\dot{H}}_A^{\text{'}}$

Conclusion:

The given condition is satisfied the given equation.