Question

Consider a circular disc of radius 20 cm with center located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of the center of mass of the residual or remaining disc from the origin will be:

• A. 2.0 cm
• B. 0.5 cm
• C. 1.5 cm
• D. 1.0 cm

Hint:

When calculating the center of mass of a system with a removed portion, use the formula: \[ X_{\text{com}} = \frac{m_{\text{1}} x_1 + m_{\text{2}} x_2}{m_{\text{1}} + m_{\text{2}}} \] where \( x_1 \) and \( x_2 \) are the distances of the centers of mass of the original and removed portions, and \( m_1 \) and \( m_2 \) are their respective masses.

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the center of mass of the remaining part of the circular disc after a smaller circular hole has been cut out. 

1. Initially, we have a circular disc with radius \( R = 20 \) cm and its center located at the origin \((0, 0)\).
2. A hole with radius \( r = 5 \) cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc.
3. The distance from the center of the hole to the center of the original disc is \( R - r = 20 \, \text{cm} - 5 \, \text{cm} = 15 \, \text{cm} \).

We apply the concept of center of mass for composite bodies:

\(X_{\text{cm}} = \frac{A_1 X_1 + A_2 X_2}{A_1 + A_2}\)

• \(A_1\) = area of the original disc = \(\pi R^2 = \pi (20)^2 = 400\pi \, \text{cm}^2\)
• \(A_2\) = area of the hole = \(\pi r^2 = \pi (5)^2 = 25\pi \, \text{cm}^2\)
• The center of mass of the original disc (\(X_1\)) is at the origin (0, 0).
• The center of mass of the hole (\(X_2\)) is at \((15, 0)\) as it touches the edge of the main disc.

Now, calculating the X-coordinate of the center of mass of the remaining region:

\(X_{\text{cm}} = \frac{400\pi \cdot 0 - 25\pi \cdot 15}{400\pi - 25\pi}\)

\(X_{\text{cm}} = \frac{-375\pi}{375\pi} = -1 \, \text{cm}\)

The center of mass is 1 cm towards the negative x-direction (since we took the left side as negative direction from the center to the origin).

Thus, the distance of the center of mass from the origin is 1.0 cm, which matches option:

1.0 cm

 

Answer 2

To determine the center of mass of the remaining disc after removing a smaller disc, we follow these steps:

1. Problem Setup:
- Original disc radius (R) = 20 cm (centered at origin)
- Removed disc radius (r) = 5 cm (edge touches original disc's edge)
- Center of removed disc is at x = 15 cm (since 20 - 5 = 15 cm)

2. Center of Mass Concept:
The center of mass of the remaining portion can be calculated by:
- Treating the original disc as a positive mass
- Treating the removed disc as a negative mass
- Using the weighted average formula for center of mass

3. Mass Proportionality:
Since the disc is uniform, mass ∝ area:
- Original disc area (A₁) = πR² = 400π cm²
- Removed disc area (A₂) = πr² = 25π cm²

4. Center of Mass Calculation:
Using the center of mass formula:

\[ X_{cm} = \frac{A_1x_1 + (-A_2)x_2}{A_1 - A_2} = \frac{(400π)(0) + (-25π)(15)}{400π - 25π} \]

Simplifying:

\[ X_{cm} = \frac{-375π}{375π} = -1 \text{ cm} \]

5. Interpretation:
- The negative sign indicates the COM is 1 cm to the left of the origin
- This makes physical sense as we removed mass from the right side
- The y-coordinate remains 0 due to symmetry

Final Answer:
The center of mass of the remaining disc is \(\boxed{1 \text{ cm}}\) from the origin.