Chapter 7, Problem 87P

To determine

The location of the rod in x and y coordinates of the center of mass from the origin.

Answer to Problem 87P

The location of the rod in x and y coordinates of the center of mass from the origin is $\left(5\text{cm, 6}\text{.7}\text{cm}\right)$.

Explanation of Solution

Write the expression for location of the center of mass of the rod in x direction.

  $x_{CM}=\frac{\frac{m}{3}{x}_1+\frac{m}{3}{x}_2+\frac{m}{3}{x}_3}{\frac{m}{3}+\frac{m}{3}+\frac{m}{3}}$        (I)

Here, $x_{CM}$ is the location of the center of mass of the rod in x direction, $m$ is the mass of the rod, and $x_1,x_2,\text{ and }{x}_3$ is the position of the rod along x axis.

Rewrite the above expression.

  $\begin{array}{c}{x}_{CM}=\frac{m\left(\frac{1}{3}{x}_1+\frac{1}{3}{x}_2+\frac{1}{3}{x}_3\right)}{m\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\\ =\frac{\left(\frac{1}{3}{x}_1+\frac{1}{3}{x}_2+\frac{1}{3}{x}_3\right)}{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\end{array}$        (II)

Write the expression for location of the center of mass of the rod in y direction.

  $y_{CM}=\frac{\frac{m}{3}{y}_1+\frac{m}{3}{y}_2+\frac{m}{3}{y}_3}{\frac{m}{3}+\frac{m}{3}+\frac{m}{3}}$        (III)

Here, $y_{CM}$ is the location of the center of mass of the rod in y direction, $m$ is the mass of the rod, and $y_1,y_2,\text{ and }{y}_3$ is the position of the rod along y axis.

Rewrite the above expression.

  $\begin{array}{c}{y}_{CM}=\frac{m\left(\frac{1}{3}{y}_1+\frac{1}{3}{y}_2+\frac{1}{3}{y}_3\right)}{m\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\\ =\frac{\left(\frac{1}{3}{y}_1+\frac{1}{3}{y}_2+\frac{1}{3}{y}_3\right)}{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\end{array}$        (IV)

Conclusion:

Substitute $0\text{cm}+0\text{cm}/2$ for $x_1$, $0\text{cm}+10.0\text{cm}/2$ for $x_2$, and $10.0\text{cm}+10.0\text{cm}/2$ for $x_3$ in the equation (II) to find $x_{CM}$.

  $\begin{array}{c}{x}_{CM}=\frac{\left(\frac{1}{3}\left(\frac{0\text{cm}+0\text{cm}}{2}\right)+\frac{1}{3}\left(\frac{0\text{cm}+10\text{cm}}{2}\right)+\frac{1}{3}\left(\frac{10\text{cm}+10\text{cm}}{2}\right)\right)}{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\\ =\frac{\frac{0}{3}+\frac{5\text{cm}}{3}+\frac{10\text{cm}}{3}}{1}\\ =5\text{cm}\end{array}$

Substitute $0\text{cm}+10\text{cm}/2$ for $y_1$, $10.0\text{cm}+10.0\text{cm}/2$ for $y_2$, and $0\text{cm}+10.0\text{cm}/2$ for $y_3$ in the equation (IV) to find $y_{CM}$.

  $\begin{array}{c}{x}_{CM}=\frac{\left(\frac{1}{3}\left(\frac{0\text{cm}+10\text{cm}}{2}\right)+\frac{1}{3}\left(\frac{10\text{cm}+10\text{cm}}{2}\right)+\frac{1}{3}\left(\frac{0\text{cm}+10\text{cm}}{2}\right)\right)}{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)}\\ =\frac{\frac{5\text{cm}}{3}+\frac{10\text{cm}}{3}+\frac{5\text{cm}}{3}}{1}\\ =6.7\text{cm}\end{array}$

Thus, the location of the rod in x and y coordinates of the center of mass from the origin is $\left(5\text{cm, 6}\text{.7}\text{cm}\right)$.