Question

six identical particles each of mass 'm' are arranged at the corners of a regular hexagon of side L if the mass of one of the particle is doubled then what will be the shift in centre of mass

Answer 1

Answer: 6L/7

Explanation:

In this condition the centre of mass will be at the centre of the hexagon as there are 6 particles at 6 corners this makes the entire mass as 6m. Now the 3 diagonals will create 6 triangles and angles between each of them will be 60° and the triangles will be equilateral. Thus all the sides of it will be L as the sides of hexagon are L. Now centre of mass being in centre when one mass will be doubled then precisely one more mass m is added at corner. So the total mass now will be 7m

For the shift in centre of mass which is X

X*m= 6m(L-X)

X= 6L/7