Particles of masses m, 2m, 3m, ... , nm are placed on the same line at distances L, 2L, 3L, ... , nL from O. The distance of centre of mass from O is

$\frac{(2 n + 1) L}{4}$

$\frac{L}{(2 n + 1)}$

$\frac{n (n^{2} + 1) L}{2}$

$\frac{(2 n + 1) L}{3}$

Solution

$\frac{(2 n + 1) L}{3}$

Centre of mass formula

centre of mass = $\frac{sum of all (position \times mass)}{sum of all masses}$

$∵$ X_CM = $\frac{mL + 2 m \times 2 L + . . . . + nm \times nL}{m + 2 m + . . . . + nm}$

= $\frac{L (1^{2} + 2^{2} + . . . . . + n^{2})}{(1 + 2 + . . . . . + n)}$

= L $\frac{Σ n^{2}}{Σ n}$

= L $\frac{\frac{n (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}}$

X_CM = $\frac{(2 n + 1) L}{3}$

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