Four identical thin rods each of mass $M$ and length $l$, from a square frame. Moment of inertia of this…
Four identical thin rods each of mass $M$ and length $l$, from a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is :
$\frac{1}{3} \mathrm{M} l^2$
$\frac{4}{3} \mathrm{M} l^2$
$\frac{2}{3} \mathrm{M} l^2$
$\frac{13}{3} \mathrm{M} l^2$
Solution
Key Idea Apply theorem of parallel axis and the total moment of inertia will be the sum of moment of inertia of each rod.
Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod \(+\) (mass of rod) \(\times\) (perpendicular distance between two axes)
\(=\frac{\mathrm{Ml}^2}{12}+\mathrm{M}\left(\frac{1}{2}\right)^2=\frac{\mathrm{Ml}^2}{3}\)
Moment of inertia of the system \(=\frac{\mathrm{Ml}^2}{3} \times 4\)
\(=\frac{4}{3} \mathrm{Ml}^2\)
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