Question

A circular plate of uniform thickness has a diameter of $56cm$. A circular portion of diameter $42cm$ is removed from one edge of the plate. Find $CM$ of the remaining portion.

Answer 1

Let the surface mass density

of the plate be 5. Then,

Initially,

Let X be distance of com of region II

from O

Mass of circular region - I = ${\left(\frac{21}{100}\right)}^2\pi 5$

Mass of region$-2=\pi [\frac{(28{)}^2}{100}-{\left(\frac{21}{100}\right)}^2]5$

COM of system compressing both

masses about the centre of bigger

circular = $\frac{-m_2\frac{\left(X\right)}{100}+m\frac{(28-21)}{100}}{m_2+m_1}$

As com of the system $(M_1+M_2)$ is

O itself,

$\Rightarrow O=-m_2\frac{\left(X\right)}{100}+M_1\frac{7}{100}$

$X=7\left(\frac{M_1}{M_2}\right)$

$x=\frac{7[\frac{21}{100}{]}^2\pi 5}{{}^{\pi [\frac{(28{)}^2}{(100{)}^2}-(\frac{21}{100}{)}^2{]}^5}}$

$X=9cm$

The COM of remaining position from the center of the particle is 9 cm.