The moment of inertia of a circular disc of r | vedclass

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Question

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the axis passing through the center of mass is given by

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$6 \,kg \,m^{2}$

Vedclass · Answer

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the axis passing through the center of mass is given by:$I = I_{cm} + Md^{2}$Here,$I_{cm} = 2 \,kg \,m^{2}$ is the moment of inertia about the axis passing through the center of mass.$M = 1 \,kg$ is the mass of the disc.$d = R = 2 \,m$ is the distance between the two parallel axes (which is equal to the radius of the disc).Substituting these values into the formula:$I = 2 + (1)(2)^{2}$$I = 2 + (1)(4)$$I = 2 + 4 = 6 \,kg \,m^{2}$Therefore,the moment of inertia about the axis passing through the edge of the disc is $6 \,kg \,m^{2}$.

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