Question

Find the position of center of mass of the uniform lamina shown in figure.

• A. $(-\frac{a}{2},0)$
• B. $(-\frac{a}{3},0)$
• C. $(-\frac{a}{6},0)$
• D. $(-\frac{a}{9},0)$

Answer 1

Answer: (C)

$(-\frac{a}{6},0)$

Answer is C.
$A_1$ = area of complete circle = $\pi a^2$
$A_2$ = area of small circle = $\pi {\left(\frac{a}{2}\right)}^2=\frac{\pi a^2}{4}$
$(x_1,y_1)$ = coordinates of center of mass of large circle = (0, 0)
$(x_2,y_2)$ = coordinates of center of mass of small circle = $\left(\frac{a}{2},0\right)$
Using $x_{COM}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
we get $x_{COM}=\frac{-\frac{\pi a^2}{4}\left(\frac{a}{2}\right)}{\pi a^2-\frac{{\pi }^2}{4}}=\frac{-\left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)}a=-\frac{a}{6}$
and get $x_{COM}=0$ as $y_1$ and $y_1$ both are zero.
Therefore, coordinates of COM of the lamina shown in figure are $(-\frac{a}{6},0)$.