The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameters is `1/4 MR^(2)`. What is the moment of inertia of the disc about an axis passing through its centre and normal to the disc?

Options

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

Answer

To find the moment of inertia of a uniform circular disc about an axis passing through its center and normal to the disc, we can use the perpendicular axis theorem. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
We are given that the moment of inertia of a uniform circular disc about any of its diameters is \( I_d = \frac{1}{4} MR^2 \), where \( M \) is the mass of the disc and \( R \) is its radius.

### Step 2: Identify the Axes
- Let’s denote the moment of inertia about the x-axis (one of the diameters) as \( I_x \) and about the y-axis (the other diameter) as \( I_y \).
- According to the given information, both \( I_x \) and \( I_y \) are equal:
\[ I_x = I_y = \frac{1}{4} MR^2 \]

### Step 3: Apply the Perpendicular Axis Theorem
The perpendicular axis theorem states that for a planar body, the moment of inertia about an axis perpendicular to the plane (let's denote it as \( I_z \)) is the sum of the moments of inertia about two perpendicular axes in the plane of the body:
\[ I_z = I_x + I_y \]

### Step 4: Substitute the Values
Now, substituting the values of \( I_x \) and \( I_y \):
\[ I_z = \frac{1}{4} MR^2 + \frac{1}{4} MR^2 \]
\[ I_z = \frac{1}{4} MR^2 + \frac{1}{4} MR^2 = \frac{2}{4} MR^2 = \frac{1}{2} MR^2 \]

### Step 5: Conclusion
Thus, the moment of inertia of the disc about an axis passing through its center and normal to the disc is:
\[ I_z = \frac{1}{2} MR^2 \]