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 $O{O}^{\prime }$ is
$I_{AC}=I_{BC}=\frac{m{\ell }^2}{3}si{n}^{2}60^{\circ }=\frac{m{\ell }^2}{4}$

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