Question

Moment of inertia of a thin rod of mass \(M\) and length \(L\) about an axis passing through its centre is \(\dfrac{ML^2}{12}\). Its moment of inertia about a parallel axis at a distance of \(\dfrac{L}{4}\) from this axis is given by:

• A. \(\dfrac{ML^2}{48}\)
• B. \(\dfrac{ML^3}{48}\)
• C. \(\dfrac{ML^2}{12}\)
• D. \(\dfrac{7ML^2}{48}\)

Hint:

Always use the \textbf{Parallel Axis Theorem} when the axis is shifted from the centre: \[ I = I_{\text{cm}} + Md^2 \] Never forget to square the distance \(d\).

Answer 1

The Correct Option is D

Step 1: Recall the Parallel Axis Theorem. The parallel axis theorem states: \[ I = I_{\text{cm}} + Md^2 \] where \(I_{\text{cm}}\) = moment of inertia about the centre of mass axis, \(d\) = distance between the two parallel axes.
Step 2: Substitute the given values. Given: \[ I_{\text{cm}} = \frac{ML^2}{12}, \quad d = \frac{L}{4} \] \[ I = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2 \]
Step 3: Simplify the expression. \[ I = \frac{ML^2}{12} + \frac{ML^2}{16} \] Taking LCM \(= 48\): \[ I = \frac{4ML^2 + 3ML^2}{48} \] \[ I = \frac{7ML^2}{48} \] Final Answer: \[ \boxed{I = \dfrac{7ML^2}{48}} \]