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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The moment of inertia of a uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2} M_{total} R^2$.For a uniform disc,the mass is proportional to the area. $A$ quarter disc has an area equal to $1/4$ of the full disc. If the mass of the quarter disc is $M$,then the mass of the full disc would be $4M$.The moment of inertia of the quarter disc about the same axis is the sum of the moments of inertia of its constituent particles. Since the axis of rotation passes through the center of the original disc,the distance of every mass element $dm$ of the quarter disc from the axis is the same as it would be in the full disc.Thus,the moment of inertia $I$ of the quarter disc is given by $I = \int r^2 dm$. Since the quarter disc is a part of the full disc,its moment of inertia is simply $1/4$ of the moment of inertia of the full disc of mass $4M$.$I = \frac{1}{4} 	imes (\frac{1}{2} 	imes (4M) 	imes R^2) = \frac{1}{2} M R^2$.

$A$ quarter part is cut from a uniform circular disc of radius $R$. The mass of this cut 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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