Chapter 5, Problem 30P

To determine

The center of mass of the particles which are positioned at the corners of a square with length of sides measuring $30cm$ , as shown in Fig. 5-19.

  

Answer to Problem 30P

Position of center of mass $=14.14cm$ and its x and y coordinate is $10cmand10cm$ .

Explanation of Solution

Given info:

  Mass of particle A $=10.0gm$

  Mass of particle B $=5.00gm$

  Mass of particle C $=20.0gm$

  Mass of particle D $=10.0gm$

Length of side of square $=30.0cm$

Formula used:

The x and y coordinates, of the center of mass of particles of a system in a two dimensional plane is calculated by the following formula:

  $X_{cm}=\frac{x_1{m}_A+x_2{m}_B+x_3{m}_C+x_4{m}_D}{m_A+m_B+m_C+m_D}$

Wherein $x_{1,}{x}_{2,}{x}_3\text{ and }{x}_4$ the linear horizontal are distances of particles from the origin and $m_A,m_B,m_C\text{ and }{m}_D$ are the various masses of particles.

Similarly the y coordinates are given by the following formula:

  $Y_{cm}=\frac{y_1{m}_A+y_2{m}_B+y_3{m}_C+y_4{m}_D}{m_A+m_B+m_C+m_D}$

Wherein $y_1,y_2,y_3\text{ and }{y}_4$ the linear verticals are distances of particles from the origin and $m_A,m_B,m_C\text{ and }{m}_D$ are the various masses of particles.

The position of center of mass of a particle having x and y coordinate is given by the formula:

  $r=\sqrt{X^{2}cm+Y^{2}cm}$

Wherein r is the position of center of mass of the particle, while $X\text{ and }Y$ are its coordinates.

Calculation:

As per figure 5-19, the x and y coordinates of position of mass at A is

  $(x_{1,}{y}_1)=(0,30.0cm)$

Similarly the x and y coordinates for the position of mass at B, C and D respectively are

  $(x_2,y_2)=(30.0cm,30.0cm)$

  $\begin{array}{l}(x_3,y_3)=(0,0)\\ (x_4,y_4)=(30.0cm,0)\end{array}$

Substituting the various values of masses of particles at points A, B, C and D and also its horizontal distances from origin in the equation:

  $\begin{array}{l}{X}_{cm}=\frac{x_1{m}_A+x_2{m}_B+x_3{m}_C+x_4{m}_D}{m_A+m_B+m_C+m_D}\\ \text{ =}\frac{(0\times 10.0g)+(30.0cm\times 5.00g)+(0\times 20.0g)+(30.0cm\times 10.0g)}{(10.0g+5.00g+20.0g+10.0g)}\\ \text{ =10 cm}\end{array}$

Similarly substituting the various values of masses of particles at points A, B, C and D and also its vertical distances from origin in the equation:

  $\begin{array}{l}{Y}_{cm}=\frac{y_1{m}_A+y_2{m}_B+y_3{m}_C+y_4{m}_D}{m_A+m_B+m_C+m_D}\\ \text{ =}\frac{(30.0cm\times 10.0g)+(30.0cm\times 5.00g)+(0\times 20.0g)+(0\times 10.0g)}{(10.0g+5.00g+20.0g+10.0g)}\\ \text{ =10 cm}\end{array}$

The position of center of mass is given by the formula:

  $r=\sqrt{X^{2}cm+Y^{2}cm}$

Substituting $X_{cm}=10cm;\text{ and }{Y}_{cm}=10cm$ in the formula:

  $\begin{array}{l}r=\sqrt{{(10cm)}^2+{(10cm)}^2}\\ r=14.14cm\end{array}$

Conclusion:

The x and y coordinates of the particles are $10cmand10cm$ , of which its position of center of mass is $14.14cm$