One quarter sector is cut from a uniform circ | vedclass

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(A) Let the mass of the full circular disc be $M_{total}$. Since the quarter sector has mass $M$,the mass of the full disc is $M_{total} = 4M$.The mom

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$\frac{1}{2} M R^2$

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(A) Let the mass of the full circular disc be $M_{total}$. Since the quarter sector has mass $M$,the mass of the full disc is $M_{total} = 4M$.The moment of inertia of a full uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is given by $I_{total} = \frac{1}{2} M_{total} R^2$.Substituting $M_{total} = 4M$,we get $I_{total} = \frac{1}{2} (4M) R^2 = 2 M R^2$.By the principle of symmetry,the moment of inertia of any sector of the disc about the same axis is proportional to its mass.Therefore,the moment of inertia $I$ of the quarter sector is $I = \frac{I_{total}}{4} = \frac{2 M R^2}{4} = \frac{1}{2} M R^2$.

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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