Rotational Motion Part 3

Concept:The moment of inertia of a system is the sum of the MIs of its components. For each cylinder, use the appropriate axis (transverse or longitudinal) and apply the parallel axis theorem if needed.Explanation:Mass of each cylinder: $m = M/4$.The square loop consists of four identical cylinders. Two cylinders (B and D) have their length perpendicular to the axis, and the other two (A and C) have their length parallel to the axis.For cylinders B and D (axis passes through centre, perpendicular to length):MI = $\frac{mL^2}{12} + \frac{mR^2}{4}$.Total for B and D: $2\left(\frac{mL^2}{12} + \frac{mR^2}{4}\right) = \frac{mL^2}{6} + \frac{mR^2}{2}$.For cylinders A and C (axis parallel to length, at distance $L/2$ from centre):MI about own longitudinal axis = $\frac{1}{2}mR^2$.Using parallel axis theorem: $I = \frac{1}{2}mR^2 + m\left(\frac{L}{2}\right)^2 = \frac{mR^2}{2} + \frac{mL^2}{4}$.Total for A and C: $2\left(\frac{mR^2}{2} + \frac{mL^2}{4}\right) = mR^2 + \frac{mL^2}{2}$.Total MI of the loop: $I_{\text{total}} = \left(\frac{mL^2}{6} + \frac{mR^2}{2}\right) + \left(mR^2 + \frac{mL^2}{2}\right) = \frac{4}{6}mL^2 + \frac{3}{2}mR^2 = \frac{2}{3}mL^2 + \frac{3}{2}mR^2$.Substitute $m = M/4$: $I_{\text{total}} = \frac{2}{3}\cdot\frac{M}{4}L^2 + \frac{3}{2}\cdot\frac{M}{4}R^2 = \frac{2}{12}ML^2 + \frac{3}{8}MR^2 = \frac{1}{6}ML^2 + \frac{3}{8}MR^2$.Answer:Option D: $\frac{3}{8}MR^2 + \frac{1}{6}ML^2$