Question

Moment of inertia of a rod of mass \( M \) and length \( L \) about an axis passing through its center and normal to its length is \( \alpha \). Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to the plane containing cross is:

• A. \( \alpha \)
• B. \( \frac{\alpha}{4} \)
• C. \( \frac{\alpha}{8} \)
• D. \( \frac{\alpha}{2} \)

Hint:

To calculate the moment of inertia of composite shapes, use the parallel axis theorem and add the individual moments of inertia.

Answer 1

The Correct Option is B

Let the moment of inertia of the rod about the axis passing through its center and normal to its length be \( \alpha = \frac{ML^2}{12} \), where \( M \) is the mass and \( L \) is the length.  
Now, the rod is cut into two equal parts, each having mass \( \frac{M}{2} \) and length \( \frac{L}{2} \). 
Each part has a moment of inertia \( \alpha' \). 
For the cross shape, the total moment of inertia will be the sum of the moments of inertia of the two parts, considering the distance from the center of the rod. 
After using the parallel axis theorem, we get: \[ \alpha' = 2 \times \frac{M L^2}{48} = \frac{\alpha}{4} \] Thus, the correct option is \( \frac{\alpha}{4} \). 

Answer 2

We are asked to find the moment of inertia of a cross shape, formed by two equal parts of a rod, about an axis passing through its center and normal to its plane. The moment of inertia of the original rod of mass \( M \) and length \( L \) about its center is given as \( \alpha \).

Concept Used:

The moment of inertia of a thin uniform rod of mass \( m \) and length \( l \) about an axis passing through its center and perpendicular to its length is given by the formula:

\[ I_{\text{center}} = \frac{ml^2}{12} \]

The principle of superposition for moment of inertia states that the total moment of inertia of a system of rigid bodies about a given axis is the algebraic sum of the moments of inertia of the individual bodies about the same axis.

\[ I_{\text{system}} = \sum_{i} I_i = I_1 + I_2 + \dots \]

Step-by-Step Solution:

Step 1: Define the moment of inertia of the original rod. The moment of inertia of the rod of mass \( M \) and length \( L \) about an axis through its center and normal to its length is given as \( \alpha \).

\[ \alpha = \frac{ML^2}{12} \]

Step 2: Determine the properties of the two smaller rods. The original rod is cut into two equal parts. For each new, smaller rod:

The mass, \( m \), will be half of the original mass:

\[ m = \frac{M}{2} \]

The length, \( l \), will be half of the original length:

\[ l = \frac{L}{2} \]

Step 3: Calculate the moment of inertia of one small rod about the required axis. The two smaller rods are arranged to form a cross. The axis of rotation passes through the center of the cross and is perpendicular to its plane. This axis passes through the center of each of the two smaller rods and is perpendicular to their lengths. The moment of inertia of one small rod (\( I_{\text{small}} \)) about this axis is:

\[ I_{\text{small}} = \frac{ml^2}{12} \]

Substituting the values of \( m \) and \( l \):

\[ I_{\text{small}} = \frac{(\frac{M}{2})(\frac{L}{2})^2}{12} = \frac{\frac{M}{2} \cdot \frac{L^2}{4}}{12} = \frac{ML^2}{8 \times 12} = \frac{ML^2}{96} \]

Step 4: Calculate the total moment of inertia of the cross shape. The cross is formed by two identical small rods. Using the principle of superposition, the total moment of inertia of the cross (\( I_{\text{cross}} \)) is the sum of the moments of inertia of the two small rods about the same axis.

\[ I_{\text{cross}} = I_{\text{small}} + I_{\text{small}} = 2 \times I_{\text{small}} \] \[ I_{\text{cross}} = 2 \times \left( \frac{ML^2}{96} \right) = \frac{ML^2}{48} \]

Step 5: Express the result in terms of \( \alpha \). We know from Step 1 that \( \alpha = \frac{ML^2}{12} \). We can rearrange our result for \( I_{\text{cross}} \) to substitute this expression.

\[ I_{\text{cross}} = \frac{ML^2}{48} = \frac{1}{4} \left( \frac{ML^2}{12} \right) \]

Substituting \( \alpha \) into the equation:

\[ I_{\text{cross}} = \frac{\alpha}{4} \]

Therefore, the moment of inertia of the cross about an axis passing through its center and normal to its plane is \( \frac{\alpha}{4} \).