First slide

Location of centre on mass for multiple point masses

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 centimetre is

Easy

A

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

$\large \frac{l}{{n + l}}$
C

$\large \frac{{n({n^2} + 1)l}}{2}$
D

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

Solution

$\large {x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}......... + {m_n}{x_n}}}{{{m_1} + {m_2} + {m_3} + ......... + {m_n}}}$
$\large = \frac{{m\ell + 2m\left( {2\ell } \right) + 3m\left( {3\ell } \right) + ......nm\left( {n\ell } \right)}}{{m + 2m + 3m + ..... + nm}}$
$\large = \frac{{m\ell \left[ {1 + {2^2} + {3^2} + .....{n^2}} \right]}}{{m\left[ {1 + 2 + 3 + - - - + n} \right]}}$
$\large = \ell .\frac{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{\frac{{n\left( {n + 1} \right)}}{2}}} = \frac{{\left( {2n + 1} \right)\ell }}{3}$

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