The moment of inertia of a uniform circular disc of radius `R` and mass `M` about an axis passing from the edge of the disc and normal to the disc is.

Options

A. `1/2 MR^(2)`
B. `MR^(2)`
C. `7/2 MR^(2)`
D. `3/2 MR^(2)`

Answer

To find the moment of inertia of a uniform circular disc of radius \( R \) and mass \( M \) about an axis passing from the edge of the disc and normal to the disc, 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 circular disc about an axis passing through its center and perpendicular to the plane of the disc is given by the formula:
\[ I_{cm} = \frac{1}{2} M R^2 \]

### Step 2: Use the Parallel Axis Theorem
To find the moment of inertia about an axis that is parallel to the one through the center of mass but located at the edge of the disc, we can use the Parallel Axis Theorem. The theorem states:
\[ I = I_{cm} + M d^2 \]
where \( d \) is the distance between the two axes. In this case, the distance \( d \) is equal to the radius \( R \) of the disc.

### Step 3: Substitute the Values
Substituting the values into the equation:
\[ I = I_{cm} + M R^2 \]
\[ I = \frac{1}{2} M R^2 + M R^2 \]

### Step 4: Combine the Terms
Now, we can combine the terms:
\[ I = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 5: Final Result
Thus, the moment of inertia of the uniform circular disc about the axis passing from the edge of the disc and normal to the disc is:
\[ I = \frac{3}{2} M R^2 \]

### Summary
The final answer is:
\[ \boxed{\frac{3}{2} M R^2} \]