Out of a disc of mass M and radius R a concentric disc of mass m and radius r is removed. The M.I. of the remaining part about the symmetric axis will be :

Options

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

Answer

To find the moment of inertia of the remaining part of the disc after removing a concentric disc, we can follow these steps:

### Step 1: Understand the Problem
We have a larger disc of mass \( M \) and radius \( R \). From this disc, a smaller concentric disc of mass \( m \) and radius \( r \) is removed. We need to find the moment of inertia of the remaining part about the symmetric axis.

### Step 2: Moment of Inertia of the Full Disc
The moment of inertia \( I_1 \) of a full disc about its symmetric axis is given by the formula:
\[ I_1 = \frac{1}{2} M R^2 \]

### Step 3: Moment of Inertia of the Removed Disc
The moment of inertia \( I_2 \) of the smaller disc that is removed is given by:
\[ I_2 = \frac{1}{2} m r^2 \]

### Step 4: Moment of Inertia of the Remaining Part
The moment of inertia \( I \) of the remaining part after removing the smaller disc is calculated as:
\[ I = I_1 - I_2 \]
Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ I = \frac{1}{2} M R^2 - \frac{1}{2} m r^2 \]

### Step 5: Factor Out Common Terms
We can factor out \( \frac{1}{2} \):
\[ I = \frac{1}{2} (M R^2 - m r^2) \]

### Step 6: Conclusion
Thus, the moment of inertia of the remaining part about the symmetric axis is:
\[ I = \frac{1}{2} (M R^2 - m r^2) \]