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

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(A) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis pass through vertex $A$ and be parallel to the side $BC$.The mass at

Vedclass · Accepted Answer

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

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(A) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis pass through vertex $A$ and be parallel to the side $BC$.The mass at vertex $A$ lies on the axis,so its perpendicular distance from the axis is $0$. Its contribution to the moment of inertia is $m(0)^2 = 0$.The masses at vertices $B$ and $C$ are at a perpendicular distance $h$ from the axis,where $h$ is the altitude of the equilateral triangle.The altitude $h$ is given by $h = \ell \sin 60^{\circ} = \ell 	imes \frac{\sqrt{3}}{2}$.The moment of inertia $I$ of the system is the sum of the moments of inertia of the individual masses:$I = m(0)^2 + m(h)^2 + m(h)^2 = 2mh^2$.Substituting the value of $h$:$I = 2m \left( \ell 	imes \frac{\sqrt{3}}{2} ight)^2 = 2m \left( \frac{3}{4} \ell^2 ight) = \frac{3}{2} m \ell^2$.

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 passing through one of the vertices and parallel to the side joining the other two vertices is:

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