Question

six identical particles each of mass m are arranged at the corners of a regular hexagon of side length a if theass of one of the particle is doubled the shift in the center of mass is​

Answer 1

Answer:

L/7cm will be the shift of the center of mass.

Explanation:

The total masses will be 6m + m = 7m. Since the six identical particles are present at six different positions of the hexagon. So, the moment will be  2m*L for that shifted particle. So, the total will be 2mL + mL/2 + m(-L/2) + mL +  m(-L/2) + m(L/2) / 7m which on solving we will get the value of the center of mass at L/7 from the position of the center.

Since, the masses are same for all positions except the shifted mass of 2m hence the moment can be added.