Chapter 5, Problem 34P

(a)

To determine

To Evaluate: The position of center of mass at $t=0$ , wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  

Answer to Problem 34P

The position of center of mass of particles is $0$ for the value of $t=0$ .

Explanation of Solution

Given:

The time $t=0$ . The location of the particles and the force acting on them has been shown in the Fig. 5-22.

Formula used:

The x-coordinate of center of mass of particles shown in the figure can be calculated by the following formula:

  $x_{cm}=\frac{(x_1{m}_1+x_2{m}_2)}{(m_1+m_2)}$

Where $m_1\text{ and }{m}_2$ are the masses of particles 1 and 2 respectively while $x_1\text{ and }{x}_2$ are the positions of the particles 1 and 2 respectively.

Similarly the y-coordinates can also be calculated:

  $y_{cm}=\frac{(y_1{m}_1+y_2{m}_2)}{(m_1+m_2)}$

Where $m_1\text{ and }{m}_2$ are the masses and $y_1\text{ and }{y}_2$ are the positions of particles 1 and 2 respectively.

Calculation:

When $t=0$ , the position of center of mass of particles is calculated as:

  $x_{cm}=\frac{(x_1{m}_1+x_2{m}_2)}{(m_1+m_2)}$

Substituting $m_1=1.00\text{ Kg, }{m}_2=2.00\text{Kg, }{x}_1=20.0\text{ cm and }{x}_2=-10.0\text{ cm}$

  $\begin{array}{l}{x}_{cm}=\frac{(20.0\text{ cm}\times 1.00\text{ Kg)+(-10}\text{.0 cm}\times 2.00\text{ Kg)}}{(1.00\text{ Kg+2}\text{.00 Kg)}}\\ \text{ = 0 }cm\end{array}$

Conclusion:

At $t=0$ , the position of center of mass of the particles is $0$ .

(b)

To determine

To Evaluate: The resultant force acting on the system, wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  

Answer to Problem 34P

System’s resultant force is calculated as $\text{14}\text{.14 N}$ .

Explanation of Solution

Given:

The magnitude of the force acting on $1\text{ Kg}$ particle, in the positive direction, along the x-axis is $10.0\text{N}$

  $\therefore F_x=10.0N$

The magnitude of force acting on $2.00\text{ Kg}$ particle, in the positive direction, along the y-axis is $10.0\text{ N}$ .

  $\therefore F_y=10.0N$

Formula used:

Formula for the resultant force $(F)$ is given by:

  $F=\sqrt{F_x^2+F_y^2}$

Where $F_x$ is the force acting in the x direction and $F_y$ is the force acting in the y direction.

Calculation:

The resultant force is being calculated as:

  $\begin{array}{l}F=\sqrt{F_x^2+F_y^2}\\ \text{ =}\sqrt{{(10.0\text{ N)}}^2+{(10.0\text{N)}}^2}\\ \text{ = 14}\text{.14 N}\end{array}$

Conclusion:

Thus, the system’s resultant force $=\text{ 14}\text{.14 N}$

(c)

To determine

To Evaluate: The location of center of mass at $t=5.00\text{ s}$ , wherein the location of two particles, which are initially at rest, is shown in Fig. 5-22, and each particle is subject to a constant force, as indicated.

  

Answer to Problem 34P

At $t=5.0\text{ s}$ the location for center of mass is $(41.67\text{ m, 41}\text{.67 m)}$

Explanation of Solution

Given:

The magnitude of the force acting on $1\text{ Kg}$ particle, in the positive direction, along the x-axis is $10.0\text{N}$

  $\therefore F_x=10.0N$

The magnitude of force acting on $2.00\text{ Kg}$ particle, in the positive direction, along the y-axis is $10.0\text{ N}$ .

  $\therefore F_y=10.0N$

Formula used: It is known that $Force\text{ = }Mass\times Acceleration$

  $x=vt+\frac{1}{2}a{t}^2$

Where $x$ is the displacement, v is the velocity, a is the acceleration and t is the time.

The x-coordinate of center of mass of particles shown in the figure can be calculated by the following formula:

  $x_{cm}=\frac{(x_1{m}_1+x_2{m}_2)}{(m_1+m_2)}$

Where $m_1\text{ and }{m}_2$ are the masses of particles 1 and 2 respectively, while $x_1\text{ and }{x}_2$ are the positions of the particles 1 and 2 respectively.

Similarly the y-coordinates can also be calculated:

  $y_{cm}=\frac{(y_1{m}_1+y_2{m}_2)}{(m_1+m_2)}$

Where $m_1\text{ and }{m}_2$ are the masses and $y_1\text{ and }{y}_2$ are the positions of particles 1 and 2 respectively.

Calculation:

As a constant force is acting on each particle, the acceleration of particle $m_1$ is calculated as follows:

  $\begin{array}{l}F=m_1\times a_1\\ \therefore a_1=\frac{F}{m_1}\end{array}$

Substituting $F=10.0\text{ N and }{m}_1=1.00\text{ Kg}$

  $\begin{array}{l}{a}_1=\frac{10.0\text{ N}}{1.0\text{ Kg}}\\ \text{ = 10}\text{.0 }m/s^2\end{array}$

Position of particle $m_1$ is calculated as follows:

  $x_1=(v_{1}t)+(\frac{1}{2}{a}_1{t}^2)$

When the particle is at rest the velocity $v_1=0$ and

  $x_1=(\frac{1}{2}{a}_1{t}^2)$

Substituting $a_1=10.0m/s^2\text{ and }t=5.00\text{ s}$

  $\begin{array}{l}{x}_1=(\frac{1}{2}\times 10.0m/s^2)\times {(}^{5.00}\\ \text{ = 125 m}\end{array}$

Similarly the acceleration of $m_2$ is calculated as follows:

  $\begin{array}{l}F=m_2\times a_2\\ \therefore a_2=\frac{F}{m_2}\end{array}$

Substituting $F=10.0\text{ N and }{m}_2=2.00\text{ Kg}$

  $\begin{array}{l}{a}_2=\frac{10.0\text{ N}}{2.0\text{ Kg}}\\ \text{ = 5}\text{.0 }m/s^2\end{array}$

And position of $m_2$ is calculated as follows:

  $y_1=(\frac{1}{2}{a}_2{t}^2)$

Substituting values of $a_2=5.00m/s^2\text{ and }t=5.00\text{ s}$

  $\begin{array}{l}{y}_1=(\frac{1}{2}\times 5.0m/s^2){(}^{5.0}\\ \text{ = 62}\text{.5 m}\end{array}$

At $t=5.0\text{ s}$ , the x-coordinate of center of mass of two particles is calculated as follows:

  $x_{cm}=\frac{(x_1{m}_1+x_2{m}_2)}{(m_1+m_2)}$

Substituting $m_1=1.00\text{ Kg, }{x}_1=125m,m_2=2.00\text{ Kg and }{x}_2=0$

  $\begin{array}{l}{x}_{cm}=\frac{(125\text{ m }\times \text{ 1}\text{.00 Kg})+(0\times \text{ 2}\text{.00 Kg)}}{(1.00\text{ Kg + 2}\text{.00 Kg})}\\ \text{ = 41}\text{.67 m}\end{array}$

Similarly the y-coordinate of center of system’s mass is calculated at $t=5.0\text{ s}$

  $y_{cm}=\frac{(y_1{m}_1+y_2{m}_2)}{(m_1+m_2)}$

Substituting $m_1=1.00{\text{ Kg, y}}_1=0,m_2=2.00{\text{ Kg and y}}_2=62.5\text{ m}$

  $\begin{array}{l}{y}_{cm}=\frac{(0\times 1.00\text{ Kg})+(62.5\text{ m}\times \text{2}\text{.00 Kg)}}{(1.00\text{ Kg + 2}\text{.00 Kg})}\\ \text{ = 41}\text{.67 m}\end{array}$

The $x$ and $y$ coordinates of the center of mass for $t=5.0\text{ s}$ is $(41.67\text{ m, 41}\text{.67 m)}$

Conclusion:

Thus, the location of center of mass at $t=5.0\text{ s}$ is $(41.67\text{ m, 41}\text{.67 m)}$ .