The moment of inertia of a uniform rectangula | vedclass

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Question

(D) For a uniform rectangular plate of mass $M$,length $l$,and width $b$,the moment of inertia about an axis passing through its center and parallel t

Vedclass · Accepted Answer

$ \frac{Mb^2}{12} $

Vedclass · Answer

(D) For a uniform rectangular plate of mass $M$,length $l$,and width $b$,the moment of inertia about an axis passing through its center and parallel to its length $l$ is calculated by treating the plate as a series of rods of length $l$ stacked along the width $b$.The moment of inertia of a thin rod of mass $m$ and length $l$ about its center is $I = \frac{ml^2}{12}$.For the rectangular plate,we consider an elemental strip of width $dy$ at a distance $y$ from the central axis parallel to the length. The mass of this strip is $dm = \frac{M}{b} dy$.The moment of inertia of this strip about the central axis is $dI = (dm) y^2 = \left( \frac{M}{b} dy \right) y^2$.Integrating this from $y = -b/2$ to $y = b/2$:$I = \int_{-b/2}^{b/2} \frac{M}{b} y^2 dy = \frac{M}{b} \left[ \frac{y^3}{3} \right]_{-b/2}^{b/2} = \frac{M}{b} \left( \frac{b^3}{24} - (-\frac{b^3}{24}) \right) = \frac{M}{b} \left( \frac{2b^3}{24} \right) = \frac{Mb^2}{12}$.Thus,the correct option is $(D)$.

The moment of inertia of a uniform rectangular plate about an axis passing through its center and parallel to its length $(l)$ is ($b =$ width of the rectangular plate):

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