Question

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

• A. (3/4)MR² + (1/6)ML²
• B. (3/8)MR² + (7/12)ML²
• C. (3/8)MR² + (1/6)ML²
• D. (3/4)MR² + (7/12)ML²

Hint:

Always ensure you apply the parallel axis theorem to the transverse moment of inertia for the side rods, as they are a distance $L/2$ away from the central axis.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The square loop is formed by four identical cylinders. If the total mass is \(M\), each cylinder has mass \(m = M/4\). The axis of rotation passes through the midpoints of two opposite sides. This means the axis coincides with the longitudinal axis of two cylinders and is transverse (parallel) to the other two.

Step 2: Key Formula or Approach:
1. M.I. of cylinder about its own axis: \(I_{own} = \frac{1}{2}mR^2\).
2. M.I. of cylinder about transverse axis through center: \(I_{trans} = \frac{1}{4}mR^2 + \frac{1}{12}mL^2\).
3. Parallel axis theorem: \(I = I_{cm} + md^2\).
Step 3: Detailed Explanation:
Let the axis be the x-axis.
For the two cylinders parallel to the x-axis: \[ I_1 = 2 \times \left( \frac{1}{2}mR^2 \right) = mR^2 \] For the two cylinders perpendicular to the x-axis, the axis is at a distance \(d = L/2\) from their centers: \[ I_2 = 2 \times \left[ \left( \frac{1}{4}mR^2 + \frac{1}{12}mL^2 \right) + m\left(\frac{L}{2}\right)^2 \right] \] \[ I_2 = 2 \times \left[ \frac{1}{4}mR^2 + \frac{1}{12}mL^2 + \frac{1}{4}mL^2 \right] = 2 \times \left[ \frac{1}{4}mR^2 + \frac{4}{12}mL^2 \right] = \frac{1}{2}mR^2 + \frac{2}{3}mL^2 \] Total \(I = I_1 + I_2 = \frac{3}{2}mR^2 + \frac{2}{3}mL^2\).
Substitute \(m = M/4\): \[ I = \frac{3}{2}\left(\frac{M}{4}\right)R^2 + \frac{2}{3}\left(\frac{M}{4}\right)L^2 = \frac{3}{8}MR^2 + \frac{1}{6}ML^2 \]
Step 4: Final Answer:
The moment of inertia is (3/8)MR² + (1/6)ML².