Question

Particles of \(1 \mathrm{gm}, 1 \mathrm{gm}, 2 \mathrm{gm}, 2 \mathrm{gm}\) are placed at the corners \(A, B, C, D\), respectively of a square of side \(6 \mathrm{~cm}\) as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. (A) \(1 \mathrm{~cm}\) (B) \(2 \mathrm{~cm}\) (C) \(3 \mathrm{~cm}\) (D) \(4 \mathrm{~cm}\)

Short answer

The distance of the center of mass of the system from the geometrical center of the square is (A) \(1 \mathrm{~cm}\).

Step-by-step solution

Find the coordinates of the corners.

We need to determine the coordinates of the corners of the square. Let's label them as follows: - A (0, 6) - B (0, 0) - C (6, 0) - D (6, 6)

Calculate the center of mass for both X and Y components.

The formula for the center of mass in the X component is given by: \(\frac{\sum m_ix_i}{\sum m_i}\). Let's calculate the X-component: \[\frac{(1)(0) + (1)(0) + (2)(6) + (2)(6)}{1+1+2+2} = \frac{24}{6} = 4\] Similarly, for the Y-component, we use the formula \(\frac{\sum m_iy_i}{\sum m_i}\): \[\frac{(1)(6) + (1)(0) + (2)(0) + (2)(6)}{1+1+2+2} = \frac{18}{6} = 3\] So the center of mass coordinates is (4, 3).

Calculate the distance between the geometrical center and the center of mass.

Now we need to determine the distance between the coordinates of the center of mass (4, 3) and the geometrical center of the square, which has coordinates (3, 3), using the Pythagorean theorem: Distance = \(\sqrt{(4-3)^2 + (3-3)^2} = \sqrt{1^2 + 0^2} = \sqrt{1} = 1\) Thus, the distance of the center of mass of the system from the geometrical center of the square is \(1 \mathrm{~cm}\). So, the correct answer is (A) \(1 \mathrm{~cm}\).

Key concepts

Coordinates of Mass Positions

To solve problems involving the center of mass, knowing the coordinates of mass positions is crucial. In this exercise, the masses are located at the corners of a square. Assign each corner a specified coordinate.
For example, each corner of a square may be labeled as follows:

• Point A at (0, 6)
• Point B at (0, 0)
• Point C at (6, 0)
• Point D at (6, 6)

This labeling helps in easily identifying and using these coordinates in calculations later.
Once you have these coordinates, it's easier to calculate the center of mass by considering the mass placed at each corner.

Square Geometry

Understanding the geometry of a square is essential to solving this problem. A square is a regular quadrilateral with equal-length sides and right angles between each. In this problem, the side length is 6 cm.
The geometrical center of a square is the point where both diagonals intersect, and it is equidistant from all corners. This center is important because it's used as a reference point for determining the distance to other points, such as the center of mass.

• Every side of the square is equal.
• Diagonals bisect each other at right angles.
• The formula to find the intersection point (geometric center) is \(\left( \frac{a}{2}, \frac{a}{2} \right)\) where \(a\) is the side length.

When the side of a square is 6 cm, the coordinates of its geometrical center will be (3, 3).

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle often used in geometry to find the distance between two points in a plane. It is given by the equation: \( c = \sqrt{a^2 + b^2} \), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides of a right triangle.
This theorem is particularly useful in this exercise to find the distance between the center of mass and the geometrical center of the square.
For instance, if you have two points with coordinates (4, 3) and (3, 3), the distance \(d\) between them can be calculated as follows:

• Subtract the x-coordinates: \(4-3 = 1\)
• Subtract the y-coordinates: \(3-3 = 0\)
• Apply the formula: \( d = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \)

Therefore, the distance is 1 cm, which helps determine how far the center of mass is from the geometric center.

Mass Distribution

In the context of finding a center of mass, understanding mass distribution involves both how much mass is at each point and where each mass is located. This concept helps balance the system of masses.
In our exercise, the distribution is given as follows:

• A: \(1\) gm
• B: \(1\) gm
• C: \(2\) gm
• D: \(2\) gm at different corners of the square

These different masses will affect where the center of mass is located because heavier masses will 'pull' the center of mass closer to them. To compute the center of mass, use the formula:
\( x_{cm} = \frac{\sum m_ix_i}{\sum m_i} \) and \(y_{cm} = \frac{\sum m_iy_i}{\sum m_i} \)
where \( m_i \), \( x_i \), and \( y_i \) are mass and coordinates of each point, respectively.
By calculating these, you get a clear picture of the system's balance.