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Chapter 8

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]

8.10

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.