The figure shows an isosceles triangular plat | vedclass

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(C) Let the height of the triangular plate be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ is half the base $L$,so

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the height of the triangular plate be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ is half the base $L$,so $h = L/2$.The area density of the plate is $\sigma = \frac{M}{	ext{Area}} = \frac{M}{\frac{1}{2} 	imes L 	imes h} = \frac{M}{\frac{1}{2} 	imes L 	imes (L/2)} = \frac{4M}{L^2}$.Consider a thin strip of thickness $dy$ at a distance $y$ from the apex. The length of the strip $l(y)$ at height $y$ is given by $l(y) = 2y$ (since the apex angle is $90^o$,the sides make $45^o$ with the vertical).The mass of this 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 base (which is at $y = h = L/2$) is $dI = dm 	imes (h - y)^2$.Integrating from $y = 0$ to $y = L/2$:$I = \int_0^{L/2} \frac{8M}{L^2} y (L/2 - y)^2 dy = \frac{8M}{L^2} \int_0^{L/2} (y \frac{L^2}{4} - Ly^2 + y^3) dy$.$I = \frac{8M}{L^2} [\frac{L^2}{4} \frac{y^2}{2} - L \frac{y^3}{3} + \frac{y^4}{4}]_0^{L/2} = \frac{8M}{L^2} [\frac{L^2}{4} \frac{L^2}{8} - L \frac{L^3}{24} + \frac{L^4}{64}] = \frac{8M}{L^2} [\frac{L^4}{32} - \frac{L^4}{24} + \frac{L^4}{64}] = \frac{8M}{L^2} [\frac{6L^4 - 8L^4 + 3L^4}{192}] = \frac{8M}{L^2} \frac{L^4}{192} = \frac{ML^2}{24}$.

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 its base parallel to the $x$-axis is

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