Question

All are cylindrical rods having radius of cross-section $R$ and mass of each rod $\dfrac{M}{4}$. Find the moment of inertia about $yy'$ axis:

• A. $I=\dfrac{1}{16}MR^2+\dfrac{1}{6}M\ell^2$
• B. $I=\dfrac{5}{16}MR^2+M\ell^2$
• C. $I=\dfrac{16}{5}MR^2+\dfrac{1}{6}M\ell^2$
• D. $I=\dfrac{3}{8}MR^2+\dfrac{M\ell^2}{6}$

Hint:

For composite bodies, always break the system into simple parts, find MOI of each about the required axis, and then add them carefully using the parallel axis theorem.

Answer 1

The Correct Option is D

Concept: The system consists of four identical cylindrical rods forming a square frame of side $\ell$. To find the moment of inertia about axis $yy'$:
Use standard MOI formulas of a cylindrical rod
Apply the parallel axis theorem wherever required
Add contributions from all four rods
Step 1: Vertical rods (parallel to $yy'$ axis) Each vertical rod has mass $\dfrac{M}{4}$ and is at a distance $\dfrac{\ell}{2}$ from the axis. MOI of a cylindrical rod about its own axis: \[ I_{\text{self}}=\frac{1}{2}mR^2 \] Using parallel axis theorem: \[ I = \frac{1}{2}\left(\frac{M}{4}\right)R^2 + \left(\frac{M}{4}\right)\left(\frac{\ell}{2}\right)^2 \] For two vertical rods: \[ I_v = 2\left[\frac{M R^2}{8} + \frac{M\ell^2}{16}\right] = \frac{M R^2}{4} + \frac{M\ell^2}{8} \]
Step 2: Horizontal rods (perpendicular to $yy'$ axis) Each horizontal rod has its center on the axis. MOI of a rod about an axis through its center and perpendicular to length: \[ I = \frac{1}{12}m\ell^2 \] Including rotational MOI due to finite radius: \[ I = \frac{1}{12}\left(\frac{M}{4}\right)\ell^2 + \frac{1}{2}\left(\frac{M}{4}\right)R^2 \] For two horizontal rods: \[ I_h = 2\left[\frac{M\ell^2}{48} + \frac{M R^2}{8}\right] = \frac{M\ell^2}{24} + \frac{M R^2}{4} \]
Step 3: Total moment of inertia \[ I = I_v + I_h \] \[ I = \left(\frac{M R^2}{4} + \frac{M\ell^2}{8}\right) + \left(\frac{M R^2}{4} + \frac{M\ell^2}{24}\right) \] \[ I = \frac{3}{8}MR^2 + \frac{M\ell^2}{6} \]