A disc of radius r is removed from the centre of a disc of mass M and radius R. The M.I. About the axis through the centre and normal to the plane will be:

Options

A. `M(R^(2)-r^(2))//2`
B. `M(R^(2)+r^(2))//2`
C. `M(R^(4)-r^(4))//2R^(2)`
D. `M(R^(2))//2`

Answer

To find the moment of inertia (M.I.) of a disc with a smaller disc removed from its center, we can follow these steps:

### Step 1: Understand the Moment of Inertia of a Disc
The moment of inertia \( I \) of a solid disc about an axis through its center and perpendicular to its plane is given by the formula:
\[ I = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the disc and \( R \) is its radius.

### Step 2: Determine the Mass of the Larger Disc
Let the larger disc have a mass \( M \) and radius \( R \). The mass per unit area \( \sigma \) of the larger disc can be expressed as:
\[ \sigma = \frac{M}{\pi R^2} \]

### Step 3: Calculate the Moment of Inertia of the Larger Disc
Using the mass per unit area, the moment of inertia \( I_1 \) of the larger disc can be calculated as:
\[ I_1 = \frac{1}{2} M R^2 \]

### Step 4: Determine the Mass of the Smaller Disc
Let the smaller disc that is removed have a radius \( r \). The mass of the smaller disc \( m \) can be expressed as:
\[ m = \sigma \cdot \pi r^2 = \left(\frac{M}{\pi R^2}\right) \cdot \pi r^2 = \frac{M r^2}{R^2} \]

### Step 5: Calculate the Moment of Inertia of the Smaller Disc
The moment of inertia \( I_2 \) of the smaller disc about the same axis is:
\[ I_2 = \frac{1}{2} m r^2 = \frac{1}{2} \left(\frac{M r^2}{R^2}\right) r^2 = \frac{1}{2} \frac{M r^4}{R^2} \]

### Step 6: Calculate the Net Moment of Inertia
The net moment of inertia \( I \) of the system after removing the smaller disc from the larger disc is:
\[ I = I_1 - I_2 = \frac{1}{2} M R^2 - \frac{1}{2} \frac{M r^4}{R^2} \]
Factoring out \( \frac{1}{2} M \):
\[ I = \frac{1}{2} M \left( R^2 - \frac{r^4}{R^2} \right) \]

### Step 7: Simplify the Expression
To simplify the expression, we can write it as:
\[ I = \frac{1}{2} M \left( \frac{R^4 - r^4}{R^2} \right) \]
Thus, the final expression for the moment of inertia of the system is:
\[ I = \frac{M}{2} \left( R^2 - \frac{r^4}{R^2} \right) \]

### Final Answer
The moment of inertia about the axis through the center and normal to the plane is:
\[ I = \frac{M}{2} \left( R^2 - \frac{r^4}{R^2} \right) \]

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