The moment of inertia of a triangular plate A | vedclass

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(D) The given triangle $ABC$ is an isosceles right-angled triangle with base $BC = a$ and height $h = a/2$. Consider a square of side $a$ and mass $4M

Vedclass · Accepted Answer

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

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(D) The given triangle $ABC$ is an isosceles right-angled triangle with base $BC = a$ and height $h = a/2$. Consider a square of side $a$ and mass $4M$ formed by four such identical triangular plates. The moment of inertia of this square about an axis passing through its center and perpendicular to its plane is $I_{square} = \frac{1}{6}(4M)a^2 = \frac{2}{3}Ma^2$. By the parallel axis theorem,the moment of inertia of the square about an axis passing through one of its vertices (say $A$) and perpendicular to its plane is $I_{vertex} = I_{CM} + (4M)d^2$,where $d$ is the distance from the center to the vertex,$d^2 = (a/2)^2 + (a/2)^2 = a^2/2$. So,$I_{vertex} = \frac{1}{6}(4M)a^2 + (4M)(a^2/2) = \frac{2}{3}Ma^2 + 2Ma^2 = \frac{8}{3}Ma^2$. Since the square consists of four identical triangles,the moment of inertia of one triangle about the same axis passing through $A$ is $I_{triangle} = \frac{1}{4} I_{vertex} = \frac{1}{4} 	imes \frac{8}{3}Ma^2 = \frac{2}{3}Ma^2$. However,checking the standard result for a triangle of base $a$ and height $h$ about an axis through the apex perpendicular to the plane,$I = M(\frac{h^2}{6} + \frac{a^2}{24})$. With $h = a/2$,$I = M(\frac{a^2}{24} + \frac{a^2}{24}) = M(\frac{2a^2}{24}) = \frac{Ma^2}{12}$.

The moment of inertia of a triangular plate $ABC$ of mass $M$ and side $BC = a,$ about an axis passing through $A$ and perpendicular to the plane of the plate is :-

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