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 $C$,each with mass $m$. The axis passes through vertex $A$ and is parallel to the side

Vedclass · Accepted Answer

$\frac{3mL^2}{2}$

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(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each with mass $m$. The axis passes through vertex $A$ and is parallel to the side $BC$.The moment of inertia $I$ of a system of particles is given by $I = \sum mr^2$,where $r$ is the perpendicular distance of each mass from the axis.$1$. The mass at vertex $A$ lies on the axis,so its perpendicular distance $r_A = 0$. Thus,its contribution to the moment of inertia is $m(0)^2 = 0$.$2$. The masses at vertices $B$ and $C$ are at a perpendicular distance $h$ from the axis passing through $A$. In an equilateral triangle of side $L$,the height $h$ is given by $h = L \sin 60^{\circ} = L 	imes \frac{\sqrt{3}}{2}$.$3$. The total moment of inertia $I$ is the sum of the moments of inertia of the three masses:$I = m(r_A)^2 + m(r_B)^2 + m(r_C)^2$$I = 0 + m(h)^2 + m(h)^2 = 2mh^2$Substituting the value of $h$:$I = 2m \left( L \frac{\sqrt{3}}{2} ight)^2$$I = 2m \left( \frac{3L^2}{4} ight)$$I = \frac{3mL^2}{2}$

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $L$. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining the other two vertices will be

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