Question

Three identical thin rods, each of mass $m$ and length $\ell$ are joined to form an equilateral triangular frame. The moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex is

• A. $\frac{5}{2}m\ell^{2}$
• B. $\frac{5}{4} m \ell^{2}$
• C. $\frac{3}{2} m\ell^{2}$
• D. $\frac{5}{3} m\ell^{2}$

Answer 1

Answer: (B)

Moment of Inertia of each of rod $AC$ and $BC$ about the given axis $OO'$ is
$I_{AC}=I_{BC} =\frac{m\ell^{2}}{3} sin^{2}\,60^{\circ} =\frac{m \ell^{2}}{4}$

and M.I. o f rod AB about the given axis $OO'$ is
$I_{AB}=m \left(\frac{\ell\sqrt{3}}{2}\right)^{2} =\frac{3}{4}m\ell^{2} $
Hence,
$I=I_{AC}+I_{BC}+I_{AB}=\frac{m\ell^{2}}{4}+\frac{m\ell^{2}}{4}+\frac{3}{4}m\ell^{2} =\frac{5}{4}m\ell^{2}$