Question

Look at the drawing given in the figure,which has been drawn with ink of uniform line thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6m$. The coordinates of centres of the different parts are: outer circle ($0,0$), left inner circle ($-a,a$), right inner circle ($a,a$), vertical line ($0,0$), and horizontal line $(0,-a)$. The $y-$ coordinate of centre of mass of the ink in this drawing is:

• A. $\frac{a}{10}$
• B. $\frac{a}{8}$
• C. $\frac{a}{12}$
• D. $\frac{a}{3}$

Answer 1

The correct option is A $\frac{a}{10}$

Since the drawing is drawn with uniform line thickness, hence the system is considered to be uniform and $\text{COM}$ of each component can be assumed at its geometric centre itself.
The whole drawing is composed of $5$ individual components, whose masses and coordinates of $\text{COM}$ are given below:

Inner left circle $(x_1,y_1)=(-a,a),m_1=m$
Inner right circle $(x_2,y_2)=(a,a),m_2=m$
Horizontal line $(x_3,y_3)=(0,-a),m_3=m$
Vertical line $(x_4,y_4)=(0,0),m_4=m$
Outer circle $(x_5,y_5)=(0,0),m_5=6m$

The whole system is equavalent to each component placed at their respective $\text{COM}$ :
$\therefore y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4+m_5{y}_5}{m_1+m_2+m_3+m_4+m_5}$
$\Rightarrow y_{\text{CM}}=\frac{(m\times a)+(m\times a)+(m\times (-a\left)\right)+(m\times 0)+(6m\times 0)}{m+m+m+m+6m}$
$\therefore y_{\text{CM}}=\frac{ma}{10m}=\frac{a}{10}$

Hence, $y_{\text{CM}}$ represent the $y-$coordinate of $\text{COM}$ of the whole system or we can say, the $\text{COM}$ of the ink used in the drawing.