Question

Four identical particles each of mass 1kg are arranged at the corners of a square of side length 2√2m.If one of the particles is removed,the shift in the centre of mass is​

Answer 1

Suppose 4 partcles are at (0,2√2) (2√2,0) , (0,0) (2√2,2√2) as side of square is 1m

Centre of mass will be at (√2,√2)due to symmetry i.e. at centre of square

Suppose particle at (2√2,2√2) is removed, then coordinates of Centre of Mass are:

x-coordinate = (m1x1 +m2x2 +m3x3)/3

= 2√2/3

y-coordinate = 2√2/3

So shift = √ { (√2 - 2√2/3)^2 + (√2 - 2√2/3)^2}

= 2/3

Answer 2

Answer:

2/3

Explanation:

Before the removal of any one particle, all four are considered,

Xcom=M1X1+M2X2+M3X3+M4X4 / M1+M2+M3+M4

         =1*0+1*0+1*2root2+1*2root2 / 1+1+1+1

         =2root2+2root2 / 4 = 4root2 / 4 = root2

Ycom = M1Y1+M2Y2+M3Y3+M4Y4 / M1+M2+M3+M4

          = 1*0+1*0+1*2root2+1*2root2 / 1+1+1+1

          = 2root2+2root2 / 4 = 4root2 / 4 = root2

After removal of one point,

Xcom=M1X1+M2X2+M3X3 / M1+M2+M3

         =0*1+1*0+1*2root2 / 1+1+1

         =2root2/3

Ycom = M1Y1+M2Y2+M3Y3 / M1+M2+M3

          = 1*2root2+0*1+1*0 / 1+1+1

          = 2root2/3

The shift in the centre of the mass is,

$\sqrt{2-2\sqrt{2} /3 + \sqrt{2-2\sqrt{2} /3}$

=$\sqrt{2/9+2/9\\}$

=$\sqrt{4/9}$

=2/3