Moment of inertia of a uniform rod of length `L` and mass `M`, about an axis passing through `L//4` from one end and perpendicular to its length is

Options

A. `ML^2/3`
B. `(ML^2)`/12
C. `(ML^2)`/4
D. `(7ML^2)`/48

Answer

To find the moment of inertia of a uniform rod of length \( L \) and mass \( M \) about an axis passing through \( \frac{L}{4} \) from one end and perpendicular to its length, we can follow these steps:

### Step 1: Identify the Moment of Inertia about the Center of Mass
The moment of inertia \( I_{CM} \) of a uniform rod about its center of mass is given by the formula:
\[ I_{CM} = \frac{1}{12} ML^2 \]

### Step 2: Determine the Distance from the Center of Mass to the Axis
The center of mass of the rod is located at its midpoint, which is at a distance of \( \frac{L}{2} \) from one end (let's call it end A). The axis we are interested in is at a distance of \( \frac{L}{4} \) from end A.

To find the distance \( x \) from the center of mass to the axis, we calculate:
\[ x = \text{Distance from A to CM} - \text{Distance from A to the axis} = \frac{L}{2} - \frac{L}{4} \]
\[ x = \frac{L}{4} \]

### Step 3: Apply the Parallel Axis Theorem
The parallel axis theorem states that the moment of inertia about any axis parallel to the axis through the center of mass is given by:
\[ I = I_{CM} + Md^2 \]
where \( d \) is the distance between the two axes. In our case, \( d = x = \frac{L}{4} \).

### Step 4: Substitute Values into the Parallel Axis Theorem
Substituting the values we have:
\[ I = I_{CM} + M \left( \frac{L}{4} \right)^2 \]
\[ I = \frac{1}{12} ML^2 + M \left( \frac{L^2}{16} \right) \]

### Step 5: Simplify the Expression
Now we need to simplify the expression:
\[ I = \frac{1}{12} ML^2 + \frac{1}{16} ML^2 \]
To add these fractions, we need a common denominator. The least common multiple of 12 and 16 is 48. Thus, we convert each term:
\[ I = \frac{4}{48} ML^2 + \frac{3}{48} ML^2 \]
\[ I = \frac{7}{48} ML^2 \]

### Final Answer
Thus, the moment of inertia of the rod about the specified axis is:
\[ I = \frac{7}{48} ML^2 \]