0 votes
134 views
in Discussion by (98.9k points)
Four particles of mass 1kg are placed on four coners of a square of side 1m. What is the center of mass of this system?

1 Answer

0 votes
by (98.9k points)

Given:

  • Mass of each particle, m = 1 kg
  • Side length of the square, a = 1 m

The coordinates of the four corners of the square are: A(0, 0), B(1, 0), C(0, 1), D(1, 1)

The x-coordinate of the center of mass (Xcm) is given by: Xcm = (m1x1 + m2x2 + m3x3 + m4x4) / (m1 + m2 + m3 + m4)

Similarly, the y-coordinate of the center of mass (Ycm) is given by: Ycm = (m1y1 + m2y2 + m3y3 + m4y4) / (m1 + m2 + m3 + m4)

Here, since all particles have the same mass, m1 = m2 = m3 = m4 = 1 kg.

Let's plug in the values: Xcm = (0 + 1 + 0 + 1) / 4 = 0.5 Ycm = (0 + 0 + 1 + 1) / 4 = 0.5

So, the center of mass of the system is located at the coordinates (0.5, 0.5), which corresponds to the intersection point of the diagonals of the square.

Related questions

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

559 users

...