Question

A circular plate of uniform thickness has a diameter of $28cm$. A circular portion of diameter $21cm$ is removed from the plate as shown. $O$ is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from $O$ by

• A. $5cm$
• B. $9cm$
• C. $4.5cm$
• D. $5.5cm$

Answer 1

Answer: (C)

$C_1$ is the centre of mass of cut portion and $C_2$ that of remaining portion. We have to find $x_2$.
$x_1=14-10.5=3.5cm$
Mass will be proportional to area. So
mass of the whole disc is
$M=k\pi (14{)}^2$
Mass of cut portion $m_1=k\pi (10.5{)}^2$
Mass of the remaining portion $m_2=$
$M-m_1=k\pi \left(14^2-10.5^2\right)=k\pi (24.5)$
$\times (3.5)$
Now, $m_1{x}_1=m_2{x}_2$
$\Rightarrow x_2=\frac{m_1{x}_1}{m_2}=\frac{k\pi (10.5{)}^2\times 3.5}{k\pi (24.5)\times 3.5}=4.5cm$