## Chapter 8

## Q. 8.3

## Q. 8.3

**WHERE IS THE CENTER OF MASS?**

**GOAL** Find the center of mass of a system of objects.

**PROBLEM** (**a**) Three objects are located in a coordinate system as shown in Figure 8.10 a. Find the center of mass. (**b**) How does the answer change if the object on the left is displaced upward by 1.00 \mathrm{~m} and the object on the right is displaced downward by 0.500 \mathrm{~m} (Fig. 8.10b)? Treat the objects as point particles.

**STRATEGY** The y-coordinate and z-coordinate of the center of mass in part (**a**) are both zero because all the objects are on the x axis. We can find the x coordinate of the center of mass using Equation 8.3a.

x_{\mathrm{cg}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3} + \cdots }{m_{1}+m_{2}+m_{3} + \cdots } =\frac{\sum m_{i} x_{i}}{\sum m_{i}} [8.3a]

Part (**b**) requires Equation 8.3b.

y_{\mathrm{cg}}=\frac{\sum m_{i} y_{i}}{\sum m_{i}} [8.3b]

## Step-by-Step

## Verified Solution

(**a**) Find the center of mass of the system in Figure 8.10a.

Apply Equation 8.3 a to the system of three objects:

(**1**) x_{\mathrm{cg}}=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}

Compute the numerator of Equation (1):

\begin{aligned}\sum m_{i} x_{i} &=m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3} \\&=(5.00 \mathrm{~kg})(-0.500 \mathrm{~m})+(2.00 \mathrm{~kg})(0 \mathrm{~m})+(4.00 \mathrm{~kg})(1.00 \mathrm{~m}) \\&=1.50 \mathrm{~kg} \cdot \mathrm{m}\end{aligned}

Substitute the denominator, \Sigma m_{i}=11.0 \mathrm{~kg}, and the numerator into Equation (1).

x_{\mathrm{cg}}=\frac{1.50 \mathrm{~kg} \cdot \mathrm{m}}{11.0 \mathrm{~kg}}=0.136 \mathrm{~m}

(**b**) How does the answer change if the positions of the objects are changed as in Figure 8.10b?

Because the x-coordinates have not been changed, the x-coordinate of the center of mass is also unchanged:

x_{\mathrm{cg}}=0.136 \mathrm{~m}

Write Equation 8.3b:

y_{\mathrm{cg}}=\frac{\sum m_{i} y_{i}}{\sum m_{i}}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}

Substitute values:

\begin{aligned}y_{\mathrm{cg}} &=\frac{(5.00 \mathrm{~kg})(1.00 \mathrm{~m})+(2.00 \mathrm{~kg})(0 \mathrm{~m})+(4.00 \mathrm{~kg})(-0.500 \mathrm{~m})}{5.00 \mathrm{~kg}+2.00 \mathrm{~kg}+4.00 \mathrm{~kg}} \\y_{\mathrm{cg}} &=0.273 \mathrm{~m}\end{aligned}

**REMARKS** Notice that translating objects in the y-direction doesn’t change the x-coordinate of the center of mass. The three components of the center of mass are each independent of the other two cordinates.