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

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Question

(A) The position of the centre of mass $X_{CM}$ is given by the formula:$X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$Substituting the given values:$X_{CM}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The position of the centre of mass $X_{CM}$ is given by the formula:$X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$Substituting the given values:$X_{CM} = \frac{m(l) + 2m(2l) + 3m(3l) + \dots + nm(nl)}{m + 2m + 3m + \dots + nm}$Factoring out $m$ and $l$:$X_{CM} = \frac{ml(1^2 + 2^2 + 3^2 + \dots + n^2)}{m(1 + 2 + 3 + \dots + n)}$Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:$X_{CM} = \frac{l \cdot \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$Simplifying the expression:$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 cm) is

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