The moment of inertia of a disc of mass `M` and radius `R` about an axis. Which is tangential to sircumference of disc and parallel to its diameter is.

Options

A. `(5)/(4)M R^2`
B. `(2)/(3) MR^2`
C. `(3)/(2) MR^2`
D. `MR^2`

Explanation

To find the moment of inertia of a disc of mass \( M \) and radius \( R \) about an axis that is tangential to the circumference of the disc and parallel to its diameter, we can use the parallel axis theorem. Here’s how to solve the problem step by step:

### Step 1: Moment of Inertia about the Center
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane (the central axis) is given by the formula:
\[ I_{\text{COM}} = \frac{1}{2} M R^2 \]

### Step 2: Identify the Distance from the Center to the New Axis
The new axis is tangential to the circumference of the disc. The distance \( d \) from the center of the disc to this new axis is equal to the radius \( R \) of the disc.

### Step 3: Apply the Parallel Axis Theorem
The parallel axis theorem states that if you know the moment of inertia about an axis through the center of mass, you can find the moment of inertia about any parallel axis by adding \( Md^2 \) to the moment of inertia about the center of mass:
\[ I = I_{\text{COM}} + Md^2 \]
Substituting the values we have:
\[ I = \frac{1}{2} M R^2 + M R^2 \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:
\[ I = \frac{1}{2} M R^2 + 1 M R^2 = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 5: Final Result
The moment of inertia of the disc about the given axis is:
\[ I = \frac{3}{2} M R^2 \]

### Summary
Thus, the moment of inertia of the disc about the axis that is tangential to its circumference and parallel to its diameter is \( \frac{3}{2} M R^2 \).