A circular disc of radius `20cm` is cut from one edge of a larger circular disc of radius `50cm`. The shift of centre of mass is

Options

A. `5.7 cm`
B. `-5.7 cm`
C. `3.2 cm`
D. `-3.2 cm`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Radius of the larger disc, \( R_1 = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Radius of the smaller disc, \( R_2 = 20 \, \text{cm} = 0.2 \, \text{m} \)
- The mass of the larger disc can be expressed in terms of its density and area.

2. **Calculate the Area and Mass of the Discs:**
- Area of the larger disc, \( A_1 = \pi R_1^2 = \pi (0.5)^2 = \frac{\pi}{4} \, \text{m}^2 \)
- Area of the smaller disc, \( A_2 = \pi R_2^2 = \pi (0.2)^2 = \frac{\pi}{25} \, \text{m}^2 \)

3. **Assume the Density:**
- Let the density of the material be \( \rho \).
- Mass of the larger disc, \( m_1 = \rho A_1 = \rho \cdot \frac{\pi}{4} \)
- Mass of the smaller disc, \( m_2 = \rho A_2 = \rho \cdot \frac{\pi}{25} \)

4. **Calculate the Masses:**
- The mass of the larger disc is \( m_1 = \frac{\pi \rho}{4} \)
- The mass of the smaller disc is \( m_2 = \frac{\pi \rho}{25} \)

5. **Use the Concept of Negative Mass:**
- When the smaller disc is cut out, we can consider it as placing a negative mass \( -m_2 \) at its center.
- The center of the larger disc is at the origin (0,0), and the center of the smaller disc is at (30 cm, 0) or (0.3 m, 0).

6. **Calculate the Center of Mass:**
- The x-coordinate of the center of mass \( x_{cm} \) is given by:
\[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]
- Here, \( x_1 = 0 \) (for the larger disc) and \( x_2 = 0.3 \) m (for the smaller disc).
- Substitute the values:
\[ x_{cm} = \frac{m_1 \cdot 0 + (-m_2) \cdot 0.3}{m_1 - m_2} \]
- Substitute \( m_1 \) and \( m_2 \):
\[ x_{cm} = \frac{0 - \left(\frac{\pi \rho}{25}\right) \cdot 0.3}{\frac{\pi \rho}{4} - \frac{\pi \rho}{25}} \]

7. **Simplify the Expression:**
- The denominator becomes:
\[ \frac{\pi \rho}{4} - \frac{\pi \rho}{25} = \pi \rho \left(\frac{25 - 4}{100}\right) = \frac{21 \pi \rho}{100} \]
- The numerator becomes:
\[ -\frac{\pi \rho}{25} \cdot 0.3 = -\frac{0.3 \pi \rho}{25} \]
- Thus,
\[ x_{cm} = \frac{-\frac{0.3 \pi \rho}{25}}{\frac{21 \pi \rho}{100}} = -\frac{0.3 \cdot 100}{25 \cdot 21} = -\frac{30}{21} \text{ m} = -1.42857 \text{ m} \]

8. **Convert to Centimeters:**
- Converting to centimeters gives:
\[ x_{cm} = -142.857 \text{ cm} \approx -5.7 \text{ cm} \]

9. **Conclusion:**
- The center of mass shifts approximately \( 5.7 \, \text{cm} \) towards the left (negative x-direction).

### Final Answer:
The shift of the center of mass is approximately \( -5.7 \, \text{cm} \).