Question

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

• A. $5\text{cm}$
• B. $9\text{cm}$
• C. $4.5\text{cm}$
• D. $5.5\text{cm}$

Answer 1

The correct option is C $4.5\text{cm}$

Let us consider this plate as combination of two plates, one larger plate with areal density $\lambda$ and another smaller plate with areal density $-\lambda$ (corresponding to removed portion)


Now consider larger plate is placed at origin, therefore $COM$ of larger plate is at origin. i.e $x_1=0$
Radius of larger plate is $r_2=\frac{28}{2}=14\text{cm}$
Radius of smaller plate is $r_1=\frac{21}{2}=10.5\text{cm}$
COM of smaller plate will at $x_2=r_2-r_1$
$\Rightarrow x_2=\frac{28}{2}-\frac{21}{2}=3.5\text{cm}$ (from origin)
Mass of larger plate = $M_2$ = area $\times$ density
$=\pi (14{)}^2\times \lambda =\lambda \pi \times {14}^2$
Mass of removed plate = $M_1$=$\pi (10.5{)}^2\times (-\lambda )$ $=-\lambda \pi (10.5{)}^2$

Therefore, $x_{COM}$ of combined plate
$=\frac{M_2{x}_2+M_1{x}_1}{M_2+M_1}$
$=\frac{\lambda \pi (14{)}^2\times 0-\lambda \pi (10.5{)}^2\times 3.5}{\lambda \pi (14{)}^2-\lambda \pi (10.5{)}^2}$
$\therefore x_{COM}=-4.5\text{cm}$

Therefore $COM$ shifts from $x=0\text{cm}$ to $x=-4.5\text{cm}$
Hence distance by which position of $COM$ will shift towards left is $4.5\text{cm}$.