Question

41. Four particles, each of mass ikg, are placed at the corners of a square OABC of side Im.
'O' is at the origin of the coordinate system. OA and OC are aligned along positive
x-axis and positive y-axis respectively. The position vector of the centre of mass is
(in 'm') (2006 M)
Dit
6 (1+)
3) (-1)
6-1)​

Answer 1

Answer:

The answer will be l/2(i+j)

Explanation:

According to the problem the mass of each particle is 1 kg

Let the center of mass o this system is at coordinate(x,y)

Therefore, the center of mass for x = m1x1+m2x2+m3x3+m4x4/m1+m2+m3+m4

As all masses are equal therefore m1,m2,m3,m4 = 1

and x1 = 0, x2 = l, x3 = l, x4 = 0 with respect to coordinates.

Therefore,center of mass for x = l/2

Now, center of mass for y = m1y1+m2y2+m3y3+m4y4/m1+m2+m3+m4

      As all masses are equal therefore m1,m2,m3,m4 = 1

and y1 = l, y2 = 0, y3 = 0, y4 = l with respect to coordinates.

Therefore,center of mass for y = l/2

Therefore ,(x,y) = (l/2 , l/2)

Now let the position vector is r

r = xi + yj

  = l/2i +l/2j

  = l/2(i+j)