Chapter 10, Problem 28PQ

(a)

To determine

To show that the velocity of center of mass is given by ${\overrightarrow{v}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{v}}_j}$.

Answer to Problem 28PQ

It is shown that the velocity of center of mass is given by ${\overrightarrow{v}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{v}}_j}$.

Explanation of Solution

Write the expression to find the position of the center of mass.

    ${\overrightarrow{r}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}$

Here, ${\overrightarrow{r}}_{\text{CM}}$ is the position of the center of mass, $M$ is the total mass of the system, $m_j$ is the mass of $j^{th}$ particle, and ${\overrightarrow{r}}_j$ is the position vector of $j^{th}$ particle.

Write the expression for the velocity of the center of mass.

    ${\overrightarrow{v}}_{\text{CM}}=\frac{d{\overrightarrow{r}}_{\text{CM}}}{dt}$

Conclusion:

Substitute $\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}$ for ${\overrightarrow{r}}_{\text{CM}}$ in the above equation.

    $\begin{array}{c}{\overrightarrow{v}}_{\text{CM}}=\frac{d\left(\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}\right)}{dt}\\ =\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{v}}_j}\end{array}$

Therefore, it is shown that the velocity of center of mass is given by ${\overrightarrow{v}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{v}}_j}$.

(b)

To determine

The expression for acceleration of center of mass and to show that $M{\overrightarrow{a}}_{\text{CM}}={\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}={\displaystyle \sum _{j=1}^n{\overrightarrow{F}}_j}$.

Answer to Problem 28PQ

The expression for acceleration of center of mass is ${\overrightarrow{a}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}$ and it is shown that $M{\overrightarrow{a}}_{\text{CM}}={\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}={\displaystyle \sum _{j=1}^n{\overrightarrow{F}}_j}$.

Explanation of Solution

Write the expression to find the position of the center of mass.

    ${\overrightarrow{r}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}$

Here, ${\overrightarrow{r}}_{\text{CM}}$ is the position of the center of mass.

Write the expression for the acceleration of the center of mass.

    ${\overrightarrow{a}}_{\text{CM}}=\frac{d^2{\overrightarrow{r}}_{\text{CM}}}{d{t}^2}$

Conclusion:

Substitute $\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}$ for ${\overrightarrow{r}}_{\text{CM}}$ in the above equation.

    $\begin{array}{c}{\overrightarrow{a}}_{\text{CM}}=\frac{d^2\left(\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{r}}_j}\right)}{dt}\\ =\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}\end{array}$

Multiply the above equation with $M$.

    $\begin{array}{c}M{\overrightarrow{a}}_{\text{CM}}=M\left(\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}\right)\\ ={\displaystyle \sum _{j=1}^n{\overrightarrow{F}}_j}\end{array}$

Therefore, the expression for acceleration of center of mass is ${\overrightarrow{a}}_{\text{CM}}=\frac{1}{M}{\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}$ and it is shown that $M{\overrightarrow{a}}_{\text{CM}}={\displaystyle \sum _{j=1}^n{m}_j{\overrightarrow{a}}_j}={\displaystyle \sum _{j=1}^n{\overrightarrow{F}}_j}$.

(c)

To determine

To show that the total force acting on a body is the sum of external forces only by using Newton’s third law.

Answer to Problem 28PQ

It is shown that the total force acting on a body is the sum of external forces only by Newton’s third law.

Explanation of Solution

Write the expression to find the total force acting on a system of two particles.

    ${\displaystyle \sum _{j=1}^2{\overrightarrow{F}}_j}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$                                                                                                          (I)

Write the expression to find the total force acting on the first particle.

    ${\overrightarrow{F}}_1={\overrightarrow{F}}_{1\text{ext}}+{\overrightarrow{F}}_{\left(2\text{ on }1\right)}$                                                                                                    (II)

Write the expression to find the total force acting on the second particle.

    ${\overrightarrow{F}}_2={\overrightarrow{F}}_{2\text{ext}}+{\overrightarrow{F}}_{\left(\text{1 on 2}\right)}$                                                                                                   (III)

Newton’s third law states that the internal force exerted by one particle on the other is equal in magnitude but opposite in direction.

Write the expression for the relationship between internal forces using Newton’s third law.

    ${\overrightarrow{F}}_{\left(2\text{ on }1\right)}=-{\overrightarrow{F}}_{\left(\text{1 on 2}\right)}$                                                                                                     (IV)

Conclusion:

Substitute equation (II), (III) and (IV) in equation (I) and simplify it.

    $\begin{array}{c}{\displaystyle \sum _{j=1}^2{\overrightarrow{F}}_j}={\overrightarrow{F}}_{1\text{ext}}+{\overrightarrow{F}}_{\left(2\text{ on }1\right)}+{\overrightarrow{F}}_{2\text{ext}}+{\overrightarrow{F}}_{\left(\text{1 on 2}\right)}\\ ={\overrightarrow{F}}_{1\text{ext}}-{\overrightarrow{F}}_{\left(\text{1 on 2}\right)}+{\overrightarrow{F}}_{2\text{ext}}+{\overrightarrow{F}}_{\left(\text{1 on 2}\right)}\\ ={\overrightarrow{F}}_{1\text{ext}}+{\overrightarrow{F}}_{2\text{ext}}\\ ={\displaystyle \sum {\overrightarrow{F}}_{\text{ext}}}\end{array}$

Therefore, it is shown that the total force acting on a body is the sum of external forces only by Newton’s third law.