A circular disc of radius `(R)/(4)` is cut from a uniform circular disc of radius R.
The centre of cut portion is at a distance of `(R)/(2)` from the centre of the disc from which it is removed. Calculate the centre of mass of remaining portion of the disc.

Answer

To find the center of mass of the remaining portion of the disc after a smaller disc has been cut out, we can follow these steps:

### Step 1: Define the System
We have a larger disc of radius \( R \) and a smaller disc of radius \( \frac{R}{4} \) that is cut out from the larger disc. The center of the smaller disc is at a distance of \( \frac{R}{2} \) from the center of the larger disc.

### Step 2: Calculate the Masses
Assuming the mass per unit area of the larger disc is uniform, we can denote:
- The mass of the larger disc \( M \).
- The area of the larger disc \( A_1 = \pi R^2 \).
- The area of the smaller disc \( A_2 = \pi \left(\frac{R}{4}\right)^2 = \frac{\pi R^2}{16} \).

The mass of the smaller disc \( M_1 \) can be calculated as:
\[ M_1 = \text{(mass per unit area)} \times A_2 = \frac{M}{A_1} \times A_2 = \frac{M}{\pi R^2} \times \frac{\pi R^2}{16} = \frac{M}{16} \]

The mass of the remaining portion \( M_2 \) is:
\[ M_2 = M - M_1 = M - \frac{M}{16} = \frac{15M}{16} \]

### Step 3: Determine the Position Vectors
- The center of the larger disc (original disc) is at the origin, \( (0, 0) \).
- The center of the smaller disc, which is cut out, is at \( \left(0, \frac{R}{2}\right) \).

### Step 4: Calculate the Center of Mass of the Remaining Portion
The center of mass \( X_{cm} \) of the remaining portion can be calculated using the formula:
\[ X_{cm} = \frac{M_2 \cdot X_2 - M_1 \cdot X_1}{M_2 - M_1} \]
Where:
- \( X_1 \) is the position of the center of mass of the smaller disc, which is \( \frac{R}{2} \).
- \( X_2 \) is the position of the center of mass of the larger disc, which is \( 0 \).

Substituting the values:
\[ X_{cm} = \frac{\left(\frac{15M}{16}\right) \cdot 0 - \left(\frac{M}{16}\right) \cdot \left(\frac{R}{2}\right)}{\frac{15M}{16} - \frac{M}{16}} \]
\[ = \frac{-\frac{MR}{32}}{\frac{14M}{16}} \]
\[ = \frac{-\frac{R}{32}}{\frac{14}{16}} = -\frac{R}{32} \cdot \frac{16}{14} = -\frac{R}{28} \]

### Step 5: Conclusion
Thus, the center of mass of the remaining portion of the disc is located at:
\[ \text{Center of Mass} = -\frac{R}{28} \text{ along the y-axis} \]