From a disc of radius R, a concentric circular portion of radius r is cut out so as to leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

Options

A. `(1)/(2)M(R^(2)+r^(2))`
B. `(1)/(2)M(R^(2)-r^(2))`
C. `(1)/(2)M(R^(4)+r^(4))`
D. `(1)/(2)M'(R^(4)-r^(4))`

Answer

To find the moment of inertia of an annular disc about an axis perpendicular to its plane and passing through its center of gravity, we can follow these steps:

### Step 1: Define the Problem
We have a disc of radius \( R \) with a concentric circular portion of radius \( r \) cut out, leaving an annular disc of mass \( M \). We need to calculate the moment of inertia \( I \) of this annular disc.

### Step 2: Calculate the Mass per Unit Area
The area of the annular disc is given by the difference in the areas of the two circles:
\[ \text{Area} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \]
The mass per unit area \( \sigma \) is then:
\[ \sigma = \frac{M}{\pi (R^2 - r^2)} \]

### Step 3: Consider an Elemental Ring
We will consider a thin concentric ring at a distance \( A \) from the center with thickness \( dA \). The circumference of this ring is \( 2\pi A \), and its area \( dA \) is given by:
\[ dA = 2\pi A \, dA \]

### Step 4: Find the Elemental Mass
The elemental mass \( dM \) of the ring can be calculated using the mass per unit area:
\[ dM = \sigma \cdot dA = \sigma \cdot (2\pi A \, dA) = \frac{M}{\pi (R^2 - r^2)} \cdot (2\pi A \, dA) = \frac{2MA}{R^2 - r^2} \, dA \]

### Step 5: Calculate the Moment of Inertia of the Elemental Ring
The moment of inertia \( dI \) of this elemental ring about the axis is given by:
\[ dI = dM \cdot A^2 = \left(\frac{2MA}{R^2 - r^2} \, dA\right) A^2 = \frac{2MA^3}{R^2 - r^2} \, dA \]

### Step 6: Integrate to Find Total Moment of Inertia
Now, we integrate \( dI \) from \( r \) to \( R \):
\[ I = \int_{r}^{R} dI = \int_{r}^{R} \frac{2MA^3}{R^2 - r^2} \, dA \]
This can be simplified as:
\[ I = \frac{2M}{R^2 - r^2} \int_{r}^{R} A^3 \, dA \]

### Step 7: Evaluate the Integral
The integral of \( A^3 \) is:
\[ \int A^3 \, dA = \frac{A^4}{4} \]
Evaluating from \( r \) to \( R \):
\[ \int_{r}^{R} A^3 \, dA = \left[\frac{A^4}{4}\right]_{r}^{R} = \frac{R^4}{4} - \frac{r^4}{4} = \frac{R^4 - r^4}{4} \]

### Step 8: Substitute Back to Find \( I \)
Substituting back into the equation for \( I \):
\[ I = \frac{2M}{R^2 - r^2} \cdot \frac{R^4 - r^4}{4} = \frac{M(R^4 - r^4)}{2(R^2 - r^2)} \]

### Step 9: Simplify the Result
We can simplify \( R^4 - r^4 \) using the difference of squares:
\[ R^4 - r^4 = (R^2 - r^2)(R^2 + r^2) \]
Thus, the moment of inertia becomes:
\[ I = \frac{M(R^2 + r^2)}{2} \]

### Final Result
The moment of inertia of the annular disc about the axis perpendicular to its plane and passing through its center of gravity is:
\[ \boxed{\frac{M(R^2 + r^2)}{2}} \]