Answer
The y-coordinate of the slab's center of mass is $~~8.3~cm$
Work Step by Step
Note that $\frac{d_1}{2} = \frac{11.0~cm}{2} = 5.5~cm$
By symmetry, the y-coordinate of the center of mass of the iron section is $5.5~cm$
By symmetry, the y-coordinate of the center of mass of the aluminum section is $16.5~cm$
We can write an expression for the mass $m_a$ of the aluminum section:
$m_a = \rho_a~V$
We can express the density of iron in terms of the density of aluminum:
$\frac{\rho_i}{\rho_a} = \frac{7.85~g/cm^3}{2.70~g/cm^3}$
$\rho_i = 2.9~\rho_a$
We can write an expression for the mass $m_i$ of the iron section:
$m_i = \rho_i~V$
$m_i = 2.9~\rho_a~V$
$m_i = 2.9~m_a$
We can find the y-coordinate of the center of mass of the entire slab:
$y_{com} = \frac{5.5~m_i+16.5~m_a}{m_i+m_a}$
$y_{com} = \frac{5.5~(2.9~m_a)+16.5~m_a}{2.9~m_a+m_a}$
$y_{com} = \frac{32.45~m_a}{3.9~m_a}$
$y_{com} = 8.3~cm$
The y-coordinate of the slab's center of mass is $~~8.3~cm$.
