A circular disc is rotating about an axis through its circumference in a vertical plane. If its mass is 0.1 kg and radius 0.5 m, its M.I. In kg. `m^(2)` units about this axis will be

Options

A. (a)0.0375
B. (b)0.025
C. (c)0.5
D. (d)0.75

Answer

To find the moment of inertia (M.I.) of a circular disc rotating about an axis through its circumference, we can follow these steps:

### Step 1: Understand the Moment of Inertia Formula
The moment of inertia of a solid disc about its center is given by the formula:
\[ I_{CM} = \frac{1}{2} M R^2 \]
where:
- \( I_{CM} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the disc,
- \( R \) is the radius of the disc.

### Step 2: Use the Parallel Axis Theorem
Since the disc is rotating about an axis through its circumference, we need to use the parallel axis theorem. The parallel axis theorem states:
\[ I = I_{CM} + M d^2 \]
where:
- \( d \) is the distance from the center of mass to the new axis of rotation. For a disc, this distance is equal to the radius \( R \).

### Step 3: Calculate \( I_{CM} \)
Substituting the values of mass \( M = 0.1 \, \text{kg} \) and radius \( R = 0.5 \, \text{m} \) into the formula for \( I_{CM} \):
\[ I_{CM} = \frac{1}{2} (0.1) (0.5)^2 \]
\[ I_{CM} = \frac{1}{2} (0.1) (0.25) \]
\[ I_{CM} = \frac{1}{2} (0.025) \]
\[ I_{CM} = 0.0125 \, \text{kg m}^2 \]

### Step 4: Calculate \( I \) using the Parallel Axis Theorem
Now, we apply the parallel axis theorem:
\[ I = I_{CM} + M R^2 \]
Substituting the values:
\[ I = 0.0125 + (0.1) (0.5)^2 \]
\[ I = 0.0125 + (0.1) (0.25) \]
\[ I = 0.0125 + 0.025 \]
\[ I = 0.0375 \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia of the disc about the axis through its circumference is:
\[ I = 0.0375 \, \text{kg m}^2 \]