A uniform circular disc of radius a is taken. | vedclass

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Question

Let the circular disc of radius a be made up of the circular section of radius $b$ and remainder. Further let the line of symmetry joining the centres

Vedclass · Accepted Answer

$\frac{{ - c{b^2}}}{{\left( {{a^2} - {b^2}} \right)}}$

Vedclass · Answer

Let the circular disc of radius a be made up of the circular section of radius $b$ and remainder. Further let the line of symmetry joining the centres $\mathrm{O}$ and $\mathrm{O}_{1}$ be the $\mathrm{x}$ $-$axis with $\mathrm{O}$ as origin. The centre of mass of the disc of radius a will be given by$:$

$X_{C M}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} \ldots(i)$

while $Y_{\mathrm{CM}}$ and $Z_{\mathrm{CM}}$ will be zero (as for all points on $x-$ axis, $y$ and $z=0$ )

If $\sigma$ be the density of material of disc.

$\mathrm{m}_{1}=\pi \mathrm{b}^{2} \sigma \quad 	ext { and } \mathrm{x}_{1}=\mathrm{c}$

$\mathrm{m}_{2}=\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}ight) \sigma \quad 	ext { and } \mathrm{x}_{2}=?$

$M=\left(m_{1}+m_{2}ight)=\pi a^{2} \sigma$ and $X_{C M}=0$

From equation $(i),$

${0=\frac{\pi \mathrm{b}^{2} \sigma(\mathrm{c})+\pi\left(\mathrm{a}^{2}-\mathrm{b}^{2}ight) \sigma \mathrm{x}_{2}}{\pi \mathrm{a}^{2} \sigma}}$

${\mathrm{x}_{2}=\frac{-\mathrm{cb}^{2}}{\left(\mathrm{a}^{2}-\mathrm{b}^{2}ight)}}$

i.e. the centre of mass of the remainder (say $\mathrm{O}_{2}$ )

is at a distance $c b^{2} /\left(a^{2}-b^{2}ight)$ to the left of $O$ on the line joining the centres $\mathrm{O}$ and $\mathrm{O}_{1}$

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