A small disc of radius 2cm is cut from a disc of radius 6cm. If the distance their centers is 3.2cm, what is the shift in the center of mass of the disc?

Options

A. ` -0.4 cm`
B. ` -0.5 cm`
C. ` -0.24 cm`
D. ` -0.6 cm`

Explanation

To solve the problem of finding the shift in the center of mass when a small disc is cut from a larger disc, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Radii**:
- Let the mass of the larger disc (radius = 6 cm) be \( M \).
- Let the mass of the smaller disc (radius = 2 cm) be \( m \).
- The area of the larger disc \( A_1 = \pi (6^2) = 36\pi \) cm².
- The area of the smaller disc \( A_2 = \pi (2^2) = 4\pi \) cm².
- Assuming uniform density, the mass is proportional to the area, so:
\[ M \propto 36\pi \quad \text{and} \quad m \propto 4\pi \]
- Thus, we can write:
\[ M = k \cdot 36\pi \quad \text{and} \quad m = k \cdot 4\pi \]
where \( k \) is the density.

2. **Determine the Center of Mass Positions**:
- Let the center of the larger disc be at the origin (0, 0).
- The center of the smaller disc is at a distance of 3.2 cm from the center of the larger disc, so its position is at (3.2, 0).

3. **Calculate the Shift in Center of Mass**:
- The center of mass \( x_{cm} \) of the system (larger disc with the smaller disc removed) can be calculated using the formula:
\[ x_{cm} = \frac{M \cdot x_M - m \cdot x_m}{M - m} \]
where:
- \( x_M = 0 \) (position of the larger disc's center),
- \( x_m = 3.2 \) cm (position of the smaller disc's center).

Substituting the values:
\[ x_{cm} = \frac{M \cdot 0 - m \cdot 3.2}{M - m} = \frac{-m \cdot 3.2}{M - m} \]

4. **Substituting the Masses**:
- Substitute \( M = k \cdot 36\pi \) and \( m = k \cdot 4\pi \):
\[ x_{cm} = \frac{-k \cdot 4\pi \cdot 3.2}{k \cdot 36\pi - k \cdot 4\pi} = \frac{-4 \cdot 3.2}{36 - 4} = \frac{-12.8}{32} = -0.4 \text{ cm} \]

5. **Interpret the Result**:
- The negative sign indicates that the center of mass has shifted 0.4 cm towards the center of the larger disc from its original position.

### Final Answer:
The shift in the center of mass of the disc is **0.4 cm towards the center of the larger disc**.