Find the moment of inertia of a plate cut in | vedclass

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(B) Consider a right-angled triangle $ABC$ with mass $M$ and sides $AC = BC = a$. The hypotenuse $AB = a\sqrt{2}$.If we join another identical triangl

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$\frac{Ma^2}{6}$

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(B) Consider a right-angled triangle $ABC$ with mass $M$ and sides $AC = BC = a$. The hypotenuse $AB = a\sqrt{2}$.If we join another identical triangle of mass $M$ along the hypotenuse $AB$,we form a square plate of side $a$ and total mass $2M$.The axis passing through the midpoint $O$ of $AB$ and perpendicular to the plane of the square is the axis passing through the center of the square.The moment of inertia of a square plate of side $L$ and mass $M_{total}$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{6} M_{total} L^2$.Here,$M_{total} = 2M$ and $L = a$.Thus,the moment of inertia of the square plate is $I_{square} = \frac{1}{6} (2M) a^2 = \frac{Ma^2}{3}$.Since the square is composed of two identical triangles,by the principle of superposition,the moment of inertia of the square is the sum of the moments of inertia of the two triangles about the same axis.$I_{square} = I_{triangle1} + I_{triangle2} = 2 I_{triangle}$.Therefore,$I_{triangle} = \frac{1}{2} I_{square} = \frac{1}{2} \left( \frac{Ma^2}{3} \right) = \frac{Ma^2}{6}$.

Find the moment of inertia of a plate cut in the shape of a right-angled triangle of mass $M$ and sides $AC = BC = a$,about an axis perpendicular to the plane of the plate and passing through the midpoint of the side $AB$.

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