In the figure,one-fourth part of a uniform di | vedclass

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(A) Let the disc lie in the $xy$-plane with its centre at the origin $O(0,0)$. The disc occupies the first quadrant.Due to symmetry about the line $y=

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$\sqrt{2} \frac{4R}{3\pi}$

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(A) Let the disc lie in the $xy$-plane with its centre at the origin $O(0,0)$. The disc occupies the first quadrant.Due to symmetry about the line $y=x$,the centre of mass $(X_{cm}, Y_{cm})$ must lie on this line,so $X_{cm} = Y_{cm}$.For a semi-circular disc of radius $R$,the centre of mass is at a distance $\frac{4R}{3\pi}$ from the diameter along the axis of symmetry.For a quarter disc,the centre of mass coordinates are $X_{cm} = \frac{4R}{3\pi}$ and $Y_{cm} = \frac{4R}{3\pi}$.The distance of the centre of mass from the origin $O$ is given by $d = \sqrt{X_{cm}^2 + Y_{cm}^2}$.$d = \sqrt{\left(\frac{4R}{3\pi}\right)^2 + \left(\frac{4R}{3\pi}\right)^2} = \sqrt{2 \left(\frac{4R}{3\pi}\right)^2} = \sqrt{2} \frac{4R}{3\pi}$.

In the figure,one-fourth part of a uniform disc of radius $R$ is shown. The distance of the centre of mass of this object from the centre $O$ is:

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