The moment of inertia of a uniform circular d | vedclass

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(B) According to the parallel axis theorem,the moment of inertia $I$ about any axis is given by $I = I_{CM} + Md^2$,where $I_{CM}$ is the moment of in

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$B$

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(B) According to the parallel axis theorem,the moment of inertia $I$ about any axis is given by $I = I_{CM} + Md^2$,where $I_{CM}$ is the moment of inertia about the center of mass,$M$ is the mass of the disc,and $d$ is the perpendicular distance of the axis from the center of mass.Since $I_{CM}$ and $M$ are constant for the disc,the moment of inertia $I$ is directly proportional to the square of the distance $d$ from the center of mass $(A)$.Comparing the distances of points $A, B, C,$ and $D$ from the center of mass $A$:- For point $A$,$d = 0$.- For point $B$,$d = R$ (radius of the disc).- For point $C$,$d - For point $D$,$d Since point $B$ is at the maximum distance $(d = R)$ from the center of mass,the moment of inertia is maximum about the axis passing through point $B$.

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through which of the following points?

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