Particles of masses m, 2m, 3m, dots, nm grams | vedclass

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Question

(A) The formula for the centre of mass of a system of particles is given by $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.Here,the masses are $m_i = i \cdo

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

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(A) The formula for the centre of mass of a system of particles is given by $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.Here,the masses are $m_i = i \cdot m$ and their positions are $x_i = i \cdot l$ for $i = 1, 2, \dots, n$.Substituting these values:$X_{CM} = \frac{m(l) + 2m(2l) + 3m(3l) + \dots + nm(nl)}{m + 2m + 3m + \dots + nm}$$X_{CM} = \frac{ml(1^2 + 2^2 + 3^2 + \dots + n^2)}{m(1 + 2 + 3 + \dots + n)}$Using the standard summation formulas $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n(n+1)}{2}$:$X_{CM} = \frac{l \cdot \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$$X_{CM} = l \cdot \frac{n(n+1)(2n+1)}{6} \cdot \frac{2}{n(n+1)}$$X_{CM} = \frac{(2n + 1)l}{3}$

Particles of masses $m, 2m, 3m, \dots, nm$ grams are placed on the same line at distances $l, 2l, 3l, \dots, nl$ cm from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres is:

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