Three point masses,each of mass m,are placed | vedclass

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(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the axis of rotation be along the side $BC$.The masses at $B$ and $C$ lie on

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

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(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the axis of rotation be along the side $BC$.The masses at $B$ and $C$ lie on the axis of rotation,so their perpendicular distance from the axis is $0$. Thus,their moment of inertia is $I_{B} = m(0)^{2} = 0$ and $I_{C} = m(0)^{2} = 0$.The mass at $A$ is at a perpendicular distance $h$ from the side $BC$. In an equilateral triangle of side $\ell$,the height $h$ is given by $h = \ell \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \ell$.The moment of inertia of the mass at $A$ about the axis $BC$ is $I_{A} = m h^{2} = m \left( \frac{\sqrt{3}}{2} \ell \right)^{2} = m \left( \frac{3}{4} \ell^{2} \right) = \frac{3}{4} m \ell^{2}$.The total moment of inertia of the system is $I = I_{A} + I_{B} + I_{C} = \frac{3}{4} m \ell^{2} + 0 + 0 = \frac{3}{4} m \ell^{2}$.

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $\ell$. The moment of inertia of the system about an axis along any one side of the triangle is:

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