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A circular portion of radius $R_2$ has been removed from one edge of a circular disc of radius $R_1$. The correct expression for the centre of mass for the remaining portion of the disc is
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Verified Answer
The correct answer is:
$-\frac{R_2^2}{R_1+R_2}$
Given situation is shown below.
Here we treat removed mass as negative mass. As solid is symmetrical about $X$ - axis, its CM lies along $X$ - axis.
Now
$x_{\mathrm{CM}}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$
$=\frac{d \times \pi R_1^2 \times 0-d \times \pi R_2^2\left(R_1-R_2\right)}{d \times \pi R_1^2-d \times \pi R_2^2}$
Here, $d=$ Area wise mass density
$\Rightarrow x_{\mathrm{CM}}=\frac{-R_2^2\left(R_1-R_2\right)}{R_1^2-R_2^2}$
$=\frac{-R_2^2\left(R_1-R_2\right)}{\left(R_1+R_2\right)\left(R_1-R_2\right)}=\frac{-R_2^2}{R_1+R_2^{\prime}}$
Here we treat removed mass as negative mass. As solid is symmetrical about $X$ - axis, its CM lies along $X$ - axis.
Now
$x_{\mathrm{CM}}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$
$=\frac{d \times \pi R_1^2 \times 0-d \times \pi R_2^2\left(R_1-R_2\right)}{d \times \pi R_1^2-d \times \pi R_2^2}$
Here, $d=$ Area wise mass density
$\Rightarrow x_{\mathrm{CM}}=\frac{-R_2^2\left(R_1-R_2\right)}{R_1^2-R_2^2}$
$=\frac{-R_2^2\left(R_1-R_2\right)}{\left(R_1+R_2\right)\left(R_1-R_2\right)}=\frac{-R_2^2}{R_1+R_2^{\prime}}$
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