Moment of inertia of a solid cylinder of length L and diameter D about an axis passing through its centre of gravity and perpendicular to its geometric axis is

Options

A. `M((D^(2))/(4)+(L^(2))/(12))`
B. `M((L^(2))/(16)+(D^(2))/(8))`
C. `M((D^(2))/(4)+(L^(2))/(6))`
D. `M((L^(2))/(12)+(D^(2))/(16))`

Explanation

To find the moment of inertia of a solid cylinder of length \( L \) and diameter \( D \) about an axis passing through its center of gravity and perpendicular to its geometric axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Length of the cylinder \( L \)
- Diameter of the cylinder \( D \)
- Radius \( R \) of the cylinder is given by \( R = \frac{D}{2} \).

2. **Understand the Moment of Inertia Formula**:
- The moment of inertia \( I \) for a solid cylinder about an axis through its center of gravity and perpendicular to its geometric axis can be derived from the general formula for a cylinder.

3. **Use the Moment of Inertia Formula**:
- The moment of inertia \( I \) of a solid cylinder about the axis perpendicular to its length (geometric axis) is given by:
\[ I = \frac{1}{12} m (3R^2 + L^2) \]
- Here, \( m \) is the mass of the cylinder, \( R \) is the radius, and \( L \) is the length.

4. **Express Mass in Terms of Volume and Density**:
- The mass \( m \) of the cylinder can be expressed as:
\[ m = \rho V = \rho \cdot \pi R^2 L \]
- Where \( \rho \) is the density of the material.

5. **Substitute the Radius**:
- Substitute \( R = \frac{D}{2} \) into the moment of inertia formula:
\[ I = \frac{1}{12} m \left(3 \left(\frac{D}{2}\right)^2 + L^2\right) \]

6. **Final Expression**:
- After substituting \( m \) and simplifying, we can express the moment of inertia in terms of \( D \) and \( L \):
\[ I = \frac{1}{12} \left(\rho \cdot \pi \left(\frac{D}{2}\right)^2 L\right) \left(3 \left(\frac{D}{2}\right)^2 + L^2\right) \]
- This gives us the moment of inertia of a solid cylinder about the specified axis.

### Final Result:
The moment of inertia of a solid cylinder of length \( L \) and diameter \( D \) about an axis passing through its center of gravity and perpendicular to its geometric axis is:
\[ I = \frac{1}{12} m (3R^2 + L^2) \]