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???

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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???

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Answer 1 · 2018-05-15T00:39:06

$\frac{\sqrt{3}}{8} L$ $sqrt3/8 L$

Explanation:

The center of mass of the original arrangement (six equal masses $m$ $m$ on the six corners of a regular hexagon) is at the center of the hexagon.

The new arrangement (after two of the adjacent masses are doubled) is equivalent to the older one, with the addition of two masses equal to $m$ $m$ masses added to two adjacent corners of the hexagon.

As far as the center of mass is concerned, this is equivalent to a mass of $6 m$ $6m$ at the center of the hexagon, and a mass of $2 m$ $2m$ placed on the midpoint of the edge joining the two corners. The distance between the two points being $\frac{\sqrt{3}}{2} L$ $sqrt3/2 L$ .

The shift in the center of mass is

$\frac{0 \times 6 m + \frac{\sqrt{3}}{2} L \times 2 m}{6 m + 2 m} = \frac{\sqrt{3}}{8} L$ $(0times 6m+sqrt3/2Ltimes 2m)/(6m+2m) = sqrt3/8 L$

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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

Explanation:

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