The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$

Question

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$ and length $L(\mathrm{R}<\mathrm{L})$ about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as $M$ ) :

Professor PC Thomas Classes · Answer

The mass of each cylinder is $m = M/4$. Two cylinders are parallel to the axis at a distance $L/2$. By the parallel axis theorem, their combined moment of inertia is $2 	imes [\frac{1}{2}mR^2 + m(\frac{L}{2})^2]$. The other two cylinders are perpendicular to the axis, and the axis passes directly through their centers. Their combined moment of inertia is $2 	imes [\frac{1}{12}mL^2 + \frac{1}{4}mR^2]$. Summing these gives $I_{total} = 2[\frac{MR^2}{8} + \frac{ML^2}{16}] + 2[\frac{ML^2}{48} + \frac{MR^2}{16}] = \frac{MR^2}{4} + \frac{ML^2}{8} + \frac{ML^2}{24} + \frac{MR^2}{8}$. Simplifying, $I_{total} = \frac{3}{8}MR^2 + \frac{1}{6}ML^2$.