From a uniform disc of radius R,an equilatera | vedclass

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(B) Let the disc have its center at $(0, R)$ and radius $R$. The origin $(0, 0)$ is at the bottom of the disc.An equilateral triangle of side $a = \sq

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$(0, R)$

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(B) Let the disc have its center at $(0, R)$ and radius $R$. The origin $(0, 0)$ is at the bottom of the disc.An equilateral triangle of side $a = \sqrt{3}R$ is inscribed in the disc. The height of this triangle is $h = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2}(\sqrt{3}R) = 1.5R$.The centroid of the triangle coincides with the center of the disc at $(0, R)$.Since the disc is uniform and the equilateral triangle is symmetric about the $y$-axis and centered at $(0, R)$,the removal of the triangle maintains the symmetry of the remaining mass about the $y$-axis.However,the mass distribution is no longer symmetric about the horizontal line passing through $(0, R)$.Let $M$ be the mass of the disc and $m$ be the mass of the triangle. The center of mass of the remaining part is given by $Y_{cm} = \frac{M Y_{disc} - m Y_{triangle}}{M - m}$.Since $Y_{disc} = R$ and $Y_{triangle} = R$,the center of mass remains at $(0, R)$ only if the triangle is centered at the disc's center. Looking at the figure,the triangle's top vertex is at $(0, 2R)$ and its base is at $y = 0.5R$. The centroid of this triangle is at $y = \frac{2R + 0.5R + 0.5R}{3} = R$. Thus,the triangle is indeed centered at the disc's center $(0, R)$.Therefore,the center of mass of the remaining portion remains at $(0, R)$.

From a uniform disc of radius $R$,an equilateral triangle of side $\sqrt{3}R$ is cut as shown. The new position of the centre of mass is:

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