Question 9.3: Figure 9-36 shows a slab with dimensions d1 = 11.0 cm, d2 = ......
Figure 9-36 shows a slab with dimensions d_{{1}} = 11.0 cm, d_{2} = 2.80 cm, and d_{{3}} = 13.0 cm. Half the slab consists of aluminum (density = 2.70 g/cm³) and half consists of iron (density = 7.85 g/cm³). What are (a) the x coordinate, (b) the y coordinate, and (c) the z coordinate of the slab’s center of mass?
We use Eq. 9-5 to locate the coordinates.
x_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}x_{i},\;\;\;y_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}y_{i},\;\;\;z_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}z_{i}, (9-5)
(a) By symmetry x_{\mathrm{com}}=-d_{1}/2=-(13\ \mathrm{cm})/2=-\ 6.5\ \mathrm{cm}. The negative value is due to our choice of the origin.
(b) We find y_{\mathrm{com}} as
y_{\mathrm{con}}={\frac{m_{i}y_{\mathrm{com,i}}+m_{a}y_{\mathrm{com,a}}}{m_{i}+m_{a}}}=\frac{\rho_{i}V_{i}y_{\mathrm{com},i}+\rho_{a}V_{a}y_{\mathrm{cm},a}}{\rho_{i}V_{i}+\rho_{a}V_{a}}
=\frac{(11 cm/2)(7.85 g/cm^3)+3(11 cm/2)(2.7 g/cm^3)}{7.85 g/cm^3 + 2.7 g/cm^3}=8.3 cm
(c) Again by symmetry, we have z_{\mathrm{com}}=(2.8~\mathrm{cm})/2=1.4~\mathrm{cm}.