Question

10
Particles of masses m, 2m, 3m ........... nm grams are placed on the same line at distances, I, 21, 31, ...... nl cm from a fixed point. The distance of centre of
mass of the particles from the fixed point in centimeters in
n(n' + )
и 41
(2n+1)/
ооо
n(n' +)/​

Answer 1

$\displaystyle\Large\boxed {\sf {\quad l\cdot\dfrac {2n+1}{3}\quad}}$

We have the formula to find the center of mass of the system,

$\displaystyle\longrightarrow\sf{\bar x=\dfrac{\displaystyle\sf{\sum_{i=1}^nm_ix_i}}{\displaystyle\sf{\sum_{i=1}^nm_i}}}$

Here,

• $\displaystyle\sf {m_i=im}$ where $\displaystyle\sf {i=1,\ 2,\ 3,\,\dots\,\ n}$

• $\displaystyle\sf {x_i=il}$ where $\displaystyle\sf {i=1,\ 2,\ 3,\,\dots\,\ n}$

Hence the position of the center of mass from the fixed point is,

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {\displaystyle\sf{\sum_{i=1}^nim\cdot il}}{\displaystyle\sf{\sum_{i=1}^nim}}}$

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {\displaystyle\sf{\sum_{i=1}^ni^2ml}}{\displaystyle\sf{\sum_{i=1}^nim}}}$

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {ml\displaystyle\sf{\sum_{i=1}^ni^2}}{m\displaystyle\sf{\sum_{i=1}^ni}}}$

$\displaystyle\longrightarrow\sf{\bar x=l\cdot\dfrac {\displaystyle\sf{\sum_{i=1}^ni^2}}{\displaystyle\sf{\sum_{i=1}^ni}}}$

$\displaystyle\longrightarrow\sf{\bar x=l\cdot\dfrac {\left (\dfrac {n(n+1)(2n+1)}{6}\right)}{\left (\dfrac {n(n+1)}{2}\right)}}$

$\displaystyle\longrightarrow\sf{\underline {\underline{\bar x=l\cdot\dfrac {2n+1}{3}}}}$

Or by direct application,

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {ml+(2m)(2l)+(3m)(3l)+\,\dots\,+(nm)(nl)}{m+2m+3m+\,\dots\,+nm}}$

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {ml+4ml+9ml+\,\dots\,+n^2ml}{m+2m+3m+\,\dots\,+nm}}$

$\displaystyle\longrightarrow\sf{\bar x=\dfrac {ml\left(1+4+9+\,\dots\,+n^2\right)}{m(1+2+3+\,\dots\,+n)}}$

$\displaystyle\longrightarrow\sf{\bar x=l\cdot\dfrac {1^2+2^2+3^2+\,\dots\,+n^2}{1+2+3+\,\dots\,+n}}$

$\displaystyle\longrightarrow\sf{\bar x=l\cdot\dfrac {\left (\dfrac {n(n+1)(2n+1)}{6}\right)}{\left (\dfrac {n(n+1)}{2}\right)}}$

$\displaystyle\longrightarrow\sf{\underline {\underline{\bar x=l\cdot\dfrac {2n+1}{3}}}}$