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Question

(A) The moment of inertia of a rectangular block about an axis passing through its center of mass and perpendicular to the face with sides $a$ and $b$

Vedclass · Accepted Answer

$\frac{M}{3}(a^2 + b^2)$

Vedclass · Answer

(A) The moment of inertia of a rectangular block about an axis passing through its center of mass and perpendicular to the face with sides $a$ and $b$ is given by $I_{cm} = \frac{M}{12}(a^2 + b^2)$.According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the center of mass to the edge axis.The distance $d$ from the center of the rectangular face to the corner (edge) is $d = \sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2} = \frac{\sqrt{a^2 + b^2}}{2}$.Thus,$d^2 = \frac{a^2 + b^2}{4}$.Substituting these into the parallel axis theorem:$I = \frac{M}{12}(a^2 + b^2) + M(\frac{a^2 + b^2}{4})$$I = M(a^2 + b^2) [\frac{1}{12} + \frac{1}{4}] = M(a^2 + b^2) [\frac{1+3}{12}] = M(a^2 + b^2) [\frac{4}{12}] = \frac{M}{3}(a^2 + b^2)$.

The figure shows a uniform solid block of mass $M$ and edge lengths $a, b,$ and $c$. Its moment of inertia $(M.I.)$ about an axis passing through one edge and perpendicular to the large face of the block (as shown) is:

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