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

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Question

(A) The moment of inertia of a disc about an axis passing through its center of mass and perpendicular to its plane is given by $I_{cm} = \frac{1}{2}

Vedclass · Accepted Answer

$6 \,kg \,m^2$

Vedclass · Answer

(A) The moment of inertia of a disc about an axis passing through its center of mass and perpendicular to its plane is given by $I_{cm} = \frac{1}{2} MR^2$. Given $M = 1 \,kg$ and $R = 2 \,m$,$I_{cm} = \frac{1}{2} 	imes 1 	imes (2)^2 = 2 \,kg \cdot m^2$. According to the parallel axis theorem,the moment of inertia about an axis parallel to the $XY$ axis and passing through the edge of the disc is $I' = I_{cm} + Md^2$,where $d = R$ is the distance between the two axes. Substituting the values,$I' = 2 + (1 	imes 2^2) = 2 + 4 = 6 \,kg \cdot m^2$.

The moment of inertia of a circular disc of radius $2 \,m$ and mass $1 \,kg$ about an axis $XY$ passing through its center of mass and perpendicular to the plane of the disc is $2 \,kg \,m^2$. The moment of inertia about an axis parallel to the axis $XY$ and passing through the edge of the disc is

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