Chapter 5, Problem 35P

(a)

To determine

To Evaluate: The initial location of center of mass for spheres A and B attached to the ends of a rod of negligible mass (Fig. 5-23). The spheres are initially at rest on a frictionless, horizontal surface. A horizontal string is attached to B at point P and is under constant tension of $6.00N$ .

  

  $m_A=2.00\text{ Kg}$

  $m_B=1.00\text{ Kg}$

Fig. 5-23

Answer to Problem 35P

The initial location of center of mass of the two spheres is $0.333\text{ m}$

Explanation of Solution

Given:

The mass of spheres A and B are given as $m_A=2.00\text{ Kg}$ and $m_B=1.00\text{ Kg}$ respectively and the center to center distance between the spheres is given as $1.00\text{ m}$ . A constant tensional force of $T=6.00N$ acts on sphere B at point P.

Formula used:

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

  $x_{cm}=\frac{(x_A{m}_A+x_B{m}_B)}{(m_A+m_B)}$

Where $m_A\text{ and }{m}_B$ are the masses of spheres A and B respectively while $x_A\text{ and }{x}_B$ are the positions of the spheres A and B respectively.

Calculation:

The x-coordinate for the center of mass of spheres is calculated as follows:

  $x_{cm}=\frac{(x_A{m}_A+x_B{m}_B)}{(m_A+m_B)}$

Substituting $m_A=2.00\text{ Kg, }{x}_A=0,m_B=1.00\text{ Kg and }{x}_B=1.00\text{ cm}$

  $\begin{array}{l}{x}_{cm}=\frac{(0\times 2.00\text{ Kg)+(1}\text{.00 m}\times \text{1}\text{.00 Kg})}{(1.00\text{ Kg + 2}\text{.00 Kg})}\\ \text{ = 0}\text{.333 m}\end{array}$

Position of center of mass of the two spheres is thus derived as $0.333\text{ m}$ .

Conclusion:

Thus, the initial location of center of mass of the two spheres is calculated as $0.333\text{ m}$

(b)

To determine

To Evaluate: The distance by which the center of mass will move in $2.00\text{ s}$ , for spheres A and B attached to the ends of a rod of negligible mass (Fig. 5-23). The spheres are initially at rest on a frictionless, horizontal surface. A horizontal string is attached to B at point P and is under constant tension of $6.00N$ .

  

  $m_A=2.00\text{ Kg}$

  $m_B=1.00\text{ Kg}$

Fig. 5-23

Answer to Problem 35P

The position of center of mass after $2.00s$ will be $4.0m$ in the direction of tensional force of $6.00N$ .

Explanation of Solution

Given:

The mass of spheres A and B are given as $m_A=2.00\text{ Kg}$ and $m_B=1.00\text{ Kg}$ respectively and the center to center distance between the spheres is given as $1.00\text{ m}$ . A constant tensional force of $T=6.00N$ acts on sphere B at point P.

Formula used: It is known that $T=F\times r\times Sin\theta$ where

F is the force, r is the distance between the point where the force is applied and the axis of rotation and $\theta$ is the angle through which the force is applied.

In this case the Cos component of $\theta$ has been considered, as the force component along the x-axis needed to be determined.

Calculation:

The net torque that acts on the system is calculated as follows:

  ${\displaystyle \sum \tau =g{m}_A}(1.00\text{ m)+}g{m}_B(0)$

Substituting $TCos{30}^{\circ }(r)$ for ${\displaystyle \sum \tau }$ and r, the mean distance between the masses is calculated as:

  $\begin{array}{l}TCos{30}^{\circ }=g{m}_A(1.00m)+g{m}_B(0)\\ r\text{ =}\frac{g{m}_A(1.00m)}{TCos{30}^{\circ }}\end{array}$

Substituting ${\displaystyle \sum \tau =TCos{30}^{\circ }},g=9.8m/s^2\text{ and }T=6.00N$ , and the mean distance $r$ is calculated as:

  $\begin{array}{l}r=\frac{(9.8m/s^2)(2.00Kg\text{)(1}\text{.00 }m)}{(6.00N)Cos{30}^{\circ }}\\ \text{ = 3}\text{.77 }m\end{array}$

This figure could be rounded off to $4.0m$ .

Thus, the center of mass gets shifted to $4.0m$ in the direction of the tensional force of $6.00N$ after $2.00s$ .

Conclusion:

Thus, the position of center of mass is shifted by $4.0m$ after $2.00s$ .