Three point masses m are placed at the corner | vedclass

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(C) Let the equilateral triangle have vertices at $A, B,$ and $C$. Let the axis of rotation pass through the side $AB$.The moment of inertia $I$ of a

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$\frac{3}{4}m\ell^2$

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(C) Let the equilateral triangle have vertices at $A, B,$ and $C$. Let the axis of rotation pass through the side $AB$.The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.$1$. The two masses at vertices $A$ and $B$ lie on the axis of rotation,so their perpendicular distances are $r_1 = 0$ and $r_2 = 0$.$2$. The third mass at vertex $C$ is at a perpendicular distance $h$ from the side $AB$. In an equilateral triangle of side $\ell$,the height $h$ is given by $h = \ell \sin(60^\circ) = \frac{\sqrt{3}}{2}\ell$.$3$. The moment of inertia of the system is $I = m(0)^2 + m(0)^2 + m(h)^2 = m\left(\frac{\sqrt{3}}{2}\ell\right)^2 = m\left(\frac{3}{4}\ell^2\right) = \frac{3}{4}m\ell^2$.

Three point masses $m$ are placed at the corners of an equilateral triangle of side length $\ell$. What is the moment of inertia of the system about an axis passing through one side of the triangle?

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