Moment of inertia of a thin rod of mass m and length l about an axis passing through a point l/4 from one end and perpendicular to the rod is

Options

A. `(ml^(2))/(12)`
B. `(ml^(2))/(3)`
C. `(7 ml^(2))/(48)`
D. `(ml^(2))/(9)`

Explanation

To find the moment of inertia of a thin rod of mass \( m \) and length \( l \) about an axis passing through a point \( \frac{l}{4} \) from one end and perpendicular to the rod, 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 thin rod about an axis through its center of mass is given by the formula:
\[ I_{CM} = \frac{1}{12} m l^2 \]

### Step 2: Determine the distance from the center of mass to the new axis
The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from one end. The new axis is located at a distance of \( \frac{l}{4} \) from the same end. Therefore, the distance \( d \) from the center of mass to the new axis is:
\[ d = \frac{l}{2} - \frac{l}{4} = \frac{l}{4} \]

### Step 3: Apply the parallel axis theorem
The parallel axis theorem states that:
\[ I = I_{CM} + m d^2 \]
where \( I \) is the moment of inertia about the new axis, \( I_{CM} \) is the moment of inertia about the center of mass, \( m \) is the mass of the rod, and \( d \) is the distance between the two axes.

### Step 4: Substitute the values into the equation
Substituting the values we have:
\[ I = \frac{1}{12} m l^2 + m \left(\frac{l}{4}\right)^2 \]

Calculating \( \left(\frac{l}{4}\right)^2 \):
\[ \left(\frac{l}{4}\right)^2 = \frac{l^2}{16} \]

Thus, substituting this back into the equation gives:
\[ I = \frac{1}{12} m l^2 + m \cdot \frac{l^2}{16} \]

### Step 5: Find a common denominator and simplify
To combine the terms, we need a common denominator, which is 48:
\[ I = \frac{4}{48} m l^2 + \frac{3}{48} m l^2 = \frac{7}{48} m l^2 \]

### Final Result
The moment of inertia of the thin rod about the specified axis is:
\[ I = \frac{7}{48} m l^2 \]