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Moment of inertia of three uniform rods of mass M and length l joined to form an equilateral triangle, about an axis passing through one of its sides.


\[\begin {array} {1 1} (a)\;\frac{Ml^2}{12} & \quad (b)\;\frac{2}{3}Ml^2 \\ (c)\;\frac{Ml^2}{6} & \quad  (d)\;3Ml^2 \end {array}\]

1 Answer

M.I of AB=0
M.I of $BC= \frac{1}{3}Ml^2 \sin^2 30$
$\qquad=\large \frac{Ml^2}{4} \times \frac{1}{3}$
$\qquad= \large\frac{Ml^2}{12}$
M.I of AC $=\large\frac{Ml^2}{12}$
M.I.$= \large\frac{Ml^2}{6}$
answered Nov 26, 2013 by meena.p
edited Jun 18, 2014 by lmohan717

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