first arrangement. Consider the entire 6m at the center, then add the two masses. since the one triangle now has two masses on each base corner, and 6m at the vertix...the triangle is all sides the same (think about that), and each angle is 60 deg.
so the cm will shift exactly 2m*cos30/6m
Six identical particles each of mass ‘m’ are arranged at the corners of a regular
hexagon of side length ‘L’. If the masses of any two adjacent particles are doubled.
The shift in the centre of mass is???
1 answer