The figure shows an isosceles triangular plat | vedclass

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(C) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and it is an isosceles triangle,the height $h$ bisects the base $L$. Thus,th

Vedclass · Accepted Answer

$\frac{ML^2}{24}$

Vedclass · Answer

(C) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and it is an isosceles triangle,the height $h$ bisects the base $L$. Thus,the triangle is formed by two right-angled triangles with base $L/2$ and height $h$. From the geometry,$h = L/2$.Consider a thin strip of thickness $dy$ at a distance $y$ from the apex. The length of this strip $l(y)$ can be found by similar triangles: $l(y)/y = L/h = L/(L/2) = 2$. So,$l(y) = 2y$.The area of the triangle is $A = \frac{1}{2} 	imes L 	imes h = \frac{1}{2} 	imes L 	imes \frac{L}{2} = \frac{L^2}{4}$.The mass per unit area is $\sigma = \frac{M}{A} = \frac{4M}{L^2}$.The mass of the strip is $dm = \sigma 	imes l(y) 	imes dy = \frac{4M}{L^2} 	imes 2y 	imes dy = \frac{8M}{L^2} y dy$.The moment of inertia of this strip about the $y$-axis is $dI_y = \frac{dm 	imes l(y)^2}{12} = \frac{1}{12} 	imes \left( \frac{8M}{L^2} y dy ight) 	imes (2y)^2 = \frac{1}{12} 	imes \frac{8M}{L^2} y dy 	imes 4y^2 = \frac{8M}{3L^2} y^3 dy$.Integrating from $y=0$ to $y=h=L/2$:$I_y = \int_0^{L/2} \frac{8M}{3L^2} y^3 dy = \frac{8M}{3L^2} \left[ \frac{y^4}{4} ight]_0^{L/2} = \frac{8M}{3L^2} 	imes \frac{1}{4} 	imes \left( \frac{L}{2} ight)^4 = \frac{2M}{3L^2} 	imes \frac{L^4}{16} = \frac{ML^2}{24}$.

The figure shows an isosceles triangular plate of mass $M$ and base $L$. The angle at the apex is $90^o$. The apex lies at the origin and the base is parallel to the $X$-axis. The moment of inertia of the plate about the $y$-axis is:

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