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) \(1 \mathrm{~cm}\).