AIIMS 2007 Physics Moment of Inertia MCQ Question

To solve the problem of finding the moment of inertia of the system of three rods about rod B as the axis of rotation, we will first need to understand the concept of moment of inertia and how it applies to different shapes and configurations.

Step 1: Understanding Moment of Inertia

The moment of inertia $I$ of a body about a given axis is defined as:

$$I = \int r^2 \, dm$$

where $r$ is the distance from the axis of rotation to the infinitesimal mass element $dm$. For discrete masses, it can be calculated using:

$$I = \sum m_i r_i^2$$

Step 2: Analyzing the System

In this scenario, we have three rods of equal length $L$ and mass $M$:

1. Rod A: Horizontal rod positioned at the top (length $L$).
2. Rod B: Vertical rod positioned at the center (length $L$).
3. Rod C: Horizontal rod positioned at the bottom (length $L$).

Since rod B is the axis of rotation, we need to calculate the moment of inertia contributions from rods A and C with respect to this axis.

Step 3: Moment of Inertia Calculation

1. For Rod B: Since rod B is rotating about its own axis, we can use the standard formula for the moment of inertia of a rod about its own center:

   $$I_B = \frac{1}{12} ML^2$$

2. For Rod A: The distance from the axis (rod B) to the center of rod A is $L/2$. Using the parallel axis theorem, we can find the moment of inertia of rod A as follows:

   $$I_A = I_{A, \text{center}} + Md^2$$

   Where:

   • $I_{A, \text{center}} = \frac{1}{12} ML^2$ (moment of inertia about its center)
   • $d = \frac{L}{2}$ (distance from rod B to rod A)

   Therefore:

   $$I_A = \frac{1}{12} ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12} ML^2 + M\frac{L^2}{4} = \frac{1}{12} ML^2 + \frac{3}{12} ML^2 = \frac{4}{12} ML^2 = \frac{1}{3} ML^2$$

3. For Rod C: Similar to rod A, the distance from the axis (rod B) to the center of rod C is also $L/2$. Thus, the moment of inertia for rod C is calculated the same way:

   $$I_C = \frac{1}{12} ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12} ML^2 + \frac{1}{4} ML^2 = \frac{1}{12} ML^2 + \frac{3}{12} ML^2 = \frac{4}{12} ML^2 = \frac{1}{3} ML^2$$

Step 4: Total Moment of Inertia

Now, we can calculate the total moment of inertia $I_{\text{total}}$:

$$I_{\text{total}} = I_A + I_B + I_C$$

Substituting the values we found:

$$I_{\text{total}} = \frac{1}{3} ML^2 + \frac{1}{12} ML^2 + \frac{1}{3} ML^2$$

To add these together, we convert them to a common denominator (12):

$$I_{\text{total}} = \frac{4}{12} ML^2 + \frac{1}{12} ML^2 + \frac{4}{12} ML^2 = \frac{9}{12} ML^2 = \frac{3}{4} ML^2$$

However, we need to be careful; we missed the contributions from rods A and C which are indeed $\frac{1}{3} ML^2$ each, which adds up to:

\frac{1}{3} ML^2 + \frac{1}{3} ML^2 + \frac{1}{12} ML^2 = \frac{