ABC is a triangular plate of uniform thicknes | vedclass

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(C) The moment of inertia $(I)$ of a body depends on the distribution of mass relative to the axis of rotation. The farther the mass is from the axis,

Vedclass · Accepted Answer

$I_{BC} > I_{AB}$

Vedclass · Answer

(C) The moment of inertia $(I)$ of a body depends on the distribution of mass relative to the axis of rotation. The farther the mass is from the axis,the greater the moment of inertia.For a triangular plate,the moment of inertia about an axis is proportional to the square of the distance of the center of mass from that axis.Let the height of the triangle with respect to base $BC$ be $h_A = AB = 4$ units.Let the height of the triangle with respect to base $AB$ be $h_C = BC = 3$ units.Let the height of the triangle with respect to hypotenuse $AC$ be $h_B$. Since the area is $\frac{1}{2} 	imes 3 	imes 4 = 6$,we have $\frac{1}{2} 	imes 5 	imes h_B = 6$,so $h_B = 2.4$ units.The moment of inertia is proportional to the square of the perpendicular distance of the center of mass from the axis. The distance of the centroid from any side is $\frac{1}{3}$ of the altitude to that side.Thus,$I \propto h^2$.$I_{AB} \propto (BC)^2 = 3^2 = 9$$I_{BC} \propto (AB)^2 = 4^2 = 16$$I_{CA} \propto (h_B)^2 = (2.4)^2 = 5.76$Comparing these values,we get $I_{BC} > I_{AB} > I_{CA}$.Therefore,the relation $I_{BC} > I_{AB}$ is correct.

$ABC$ is a triangular plate of uniform thickness. The sides are in the ratio $3:4:5$ as shown in the figure. $I_{AB}, I_{BC}, I_{CA}$ are the moments of inertia of the plate about axes $AB, BC$ and $CA$ respectively. Which one of the following relations is correct?

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