A one-quarter sector is cut from a uniform ci | vedclass

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(A) Let the mass of the complete circular disc be $M_{total}$. Since the sector is one-quarter of the disc,its mass $M = \frac{M_{total}}{4}$,which im

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$\frac{1}{2}MR^2$

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(A) Let the mass of the complete circular disc be $M_{total}$. Since the sector is one-quarter of the disc,its mass $M = \frac{M_{total}}{4}$,which implies $M_{total} = 4M$.The moment of inertia $(I)$ of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is given by $I_{total} = \frac{1}{2}M_{total}R^2$.Substituting $M_{total} = 4M$ into the formula,we get:$I_{total} = \frac{1}{2}(4M)R^2 = 2MR^2$.By the principle of symmetry,the moment of inertia of the quarter sector $(I_{sector})$ about the same axis is one-fourth of the moment of inertia of the complete disc:$I_{sector} = \frac{I_{total}}{4} = \frac{2MR^2}{4} = \frac{1}{2}MR^2$.

$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

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