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## 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]

## 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.