Question

A circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be : 

• A. $\frac{1}{6}a$
• B. $\frac{5}{6}a$
• C. $\frac{2}{3}a$
• D. $\frac{10}{11}a$

Hint:

For objects with parts removed, treat the removed portion as having "negative mass" or "negative area" in the center of mass formula.

Answer 1

The Correct Option is A

Step 1: Let the center of the original disc be the origin $(0,0)$. The hole is cut with its center at $(a/2, 0)$.
Step 2: Area of original disc $A_1 = \pi a^2$. Area of hole $A_2 = \pi (a/2)^2 = \frac{\pi a^2}{4}$.
Step 3: Position of centroid $x_{cm} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2} = \frac{(\pi a^2)(0) - (\frac{\pi a^2}{4})(\frac{a}{2})}{\pi a^2 - \frac{\pi a^2}{4}} = \frac{-\frac{\pi a^3}{8}}{\frac{3\pi a^2}{4}} = -\frac{a}{6}$. The distance from the center $O$ is $\frac{a}{6}$.