Question

A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

• A. $\frac{ML^2}{6}$
• B. $\frac{\sqrt{2}ML^2}{24}$
• C. $\frac{ML^2}{24}$
• D. $\frac{ML^2}{12}$

Answer 1

The Correct Option is D

The problem at hand is to determine the moment of inertia of a bent rod about an axis passing through its bending point and perpendicular to the plane of the bend. The rod, originally of length L and mass M, is bent at its midpoint such that each half forms an angle of 45^\circ with the axis of bending.

Firstly, since the rod is bent at its midpoint, each segment of the rod is of length \frac{L}{2}. The mass of each segment is \frac{M}{2}.

To find the moment of inertia of the complete bent rod about the specified axis, we need to calculate the contribution from both segments.

Each half can be treated like a thin rod rotating about an end. The moment of inertia I of a single thin rod of length l and mass m about an axis through one end perpendicular to its length is given by:

I = \frac{1}{3} m l^2

Substituting m = \frac{M}{2} and l = \frac{L}{2}, the moment of inertia for one half about its end is:

I_{\text{half}} = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \cdot \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{24}

Both halves of the rod have the same moment of inertia about the bending point because they are symmetric and identical. Hence, the total moment of inertia I_{\text{total}} is twice that of one half:

I_{\text{total}} = 2 \times \frac{ML^2}{24} = \frac{ML^2}{12}

Thus, the correct answer is \frac{ML^2}{12}, which matches the given correct answer.