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(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2} M_{total} R^2$.Since the disc is uniform,the mass is proportional to the area. $A$ quarter of the disc has an area equal to one-fourth of the total area.Therefore,the mass of the full disc would be $M_{total} = 4M$.The moment of inertia of the full disc about the given axis is $I_{full} = \frac{1}{2} (4M) R^2 = 2 M R^2$.Since the moment of inertia is an additive property for parts of a rigid body,the moment of inertia of the quarter disc $(I_{quarter})$ is one-fourth of the moment of inertia of the full disc about the same axis.$I_{quarter} = \frac{1}{4} I_{full} = \frac{1}{4} (2 M R^2) = \frac{1}{2} M R^2$.

$A$ quarter of a uniform circular disc of radius $R$ is cut off. The mass of the cut-off part is $M$. It rotates about an axis passing through the center of the original disc and perpendicular to the plane of the disc. What will be its moment of inertia about the axis of rotation?

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