When n number of particles of masses m,2m,3m,……nm are at distances x1=1,x2=2,x3=3,……xn=n units respectively from origin on the x axis , then find the distance of centre of mass of the system from origin

xcm=m1x1+m2x2+m3x3+….+mnxnm1+m2+…..+mn=(m)(1)+(2m)(2)+(3m)(3)+……+(nm)(n)m+2m+3m+….+nm=m̸1+4+9+…n2m̸(1+2+3+…n)=n(n+1)(2n+1)6n(n+1)2=2n+13

When n number of particles of masses m,2m,3m,……nm are at distances x1=1,x2=2,x3=3,……xn=n units respectively from origin on the x axis , then find the distance of centre of mass of the system from origin

xcm=m1x1+m2x2+m3x3+….+mnxnm1+m2+…..+mn=(m)(1)+(2m)(2)+(3m)(3)+……+(nm)(n)m+2m+3m+….+nm=m̸1+4+9+…n2m̸(1+2+3+…n)=n(n+1)(2n+1)6n(n+1)2=2n+13

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When n number of particles of masses m,2m,3m,……nm are at distances $x_{1} = 1, x_{2} = 2, x_{3} = 3, \dots \dots x_{n} = n$ units respectively from origin on the x axis , then find the distance of centre of mass of the system from origin

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$\frac{2 n + 1}{3}$

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

$(2 n + 1)$

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

answer is A.

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Detailed Solution

$\begin{array}{l} x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + \dots . + m_{n} x_{n}}{m_{1} + m_{2} + \dots . . + m_{n}} \\ = \frac{(m) (1) + (2 m) (2) + (3 m) (3) + \dots \dots + (nm) (n)}{m + 2 m + 3 m + \dots . + nm} \\ = \frac{m ̸ (1 + 4 + 9 + \dots n^{2})}{m ̸ (1 + 2 + 3 + \dots n)} = \frac{\frac{n (n + 1) (2 n + 1)}{6}}{\frac{n (n + 1)}{2}} = \frac{2 n + 1}{3} \end{array}$