Rotational Motion Question 177
Question: Particles of masses m, 2m, 3m,………….nm grams are placed on the same line at distances $l,2l,3l,…….nl$ cm from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeters is
Options:
A) $\frac{(2n+1)l}{3}$
B) $\frac{1}{n+1}$
C) $\frac{n(n^{2}+1)l}{2}$
D) $\frac{2l}{n(n^{2}+1)}$
Show Answer
Answer:
Correct Answer: A
Solution:
[a]
$x _{cm}=\frac{m\times \ell +2m\times 2\ell +3m\times 3\ell +…+nm\times n\ell }{m+2m+3m+..+nm}$
$=\frac{(1+4+9+..+n^{2})\ell }{1+2+..+n}$
$=\frac{\ell n(n+1)(2n+1)/6}{n(n+1)/2}=\frac{(2n+1)\ell }{3}$
