Question

From an uniform circular sheet of radius $a$ units, a circular portion of diameter $a$ units has been removed as shown in the figure. Find, the coordinates of centre of mass of the remaining part.

• A. $(-\frac{a}{6},0)$
  
• B. $(-\frac{a}{3},0)$
  
• C. $(-\frac{a}{4},0)$
  
• D. $(-\frac{a}{6},\frac{a}{2})$
  

Answer 1

The correct option is A $(-\frac{a}{6},0)$



Area of original circular sheet,
$A_1=\pi a^2$
Let its mass be $M_1$
For original circular lamina centre of mass will be at geometrical centre i.e at $O$
$c_1\Rightarrow (x_1,y_1)=(0,0)$
Centre of mass of the removed part is,
$c_2\Rightarrow (x_2,y_2)=(\frac{a}{2},0)$
Area $A_2=\frac{\pi a^2}{4}$
Let the removed part as negative mass $M_2$, applying the formula for $\text{COM}$ of remaining part by considering mass $M_1$ and $M_2$ placed at their respective $\text{COM}$.
$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$\Rightarrow x_{\text{CM}}=\frac{(\pi a^2\times 0)-(\frac{\pi a^2}{4}\times \frac{a}{2})}{\pi a^2-\frac{\pi a^2}{4}}$
$x_{\text{CM}}=-\frac{a}{6}$
$\Rightarrow$ Since, the given body is symmetrical about $x-$ axis, hence its $\text{COM}$ will lie along $x-$ axis only.
$\therefore (x_{\text{CM}},y_{\text{CM}})=(-\frac{a}{6},0)$