Three identical balls each of radius 10 cm and mass 1 kg each are placed touching one another on a horizontal surface. Where is their centre of mass located?

Options

A. On the horizontal surface
B. At the point of contact of any two spheres
C. At the centre of one ball
D. none of these

Answer

To find the center of mass of three identical balls placed touching one another on a horizontal surface, we can follow these steps:

### Step 1: Understand the Setup
We have three identical balls, each with a radius of 10 cm and a mass of 1 kg. They are arranged in a way that they touch each other.

### Step 2: Position the Balls
Let's position the balls in a coordinate system for easier calculation. We can place the centers of the balls as follows:
- Ball 1: Center at (0, 0)
- Ball 2: Center at (20 cm, 0) — since the radius is 10 cm, the distance between the centers of the two balls is 20 cm.
- Ball 3: Center at (10 cm, 20 cm) — this ball will be placed above the midpoint between the first two balls.

### Step 3: Calculate the Center of Mass
The center of mass (CM) can be calculated using the formula:
\[ x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \]
\[ y_{cm} = \frac{\sum m_i y_i}{\sum m_i} \]
Where \( m_i \) is the mass of each ball and \( (x_i, y_i) \) are the coordinates of their centers.

For our three balls:
- \( m_1 = m_2 = m_3 = 1 \, \text{kg} \)
- Coordinates:
- Ball 1: \( (0, 0) \)
- Ball 2: \( (20, 0) \)
- Ball 3: \( (10, 20) \)

Calculating \( x_{cm} \):
\[ x_{cm} = \frac{(1 \cdot 0) + (1 \cdot 20) + (1 \cdot 10)}{1 + 1 + 1} = \frac{0 + 20 + 10}{3} = \frac{30}{3} = 10 \, \text{cm} \]

Calculating \( y_{cm} \):
\[ y_{cm} = \frac{(1 \cdot 0) + (1 \cdot 0) + (1 \cdot 20)}{1 + 1 + 1} = \frac{0 + 0 + 20}{3} = \frac{20}{3} \approx 6.67 \, \text{cm} \]

### Step 4: Conclusion
The center of mass of the three balls is located at the coordinates \( (10 \, \text{cm}, \frac{20}{3} \, \text{cm}) \), which is approximately \( (10 \, \text{cm}, 6.67 \, \text{cm}) \) above the horizontal surface.