Question

Particles of masses m, 2m, 3m, …, nm are placed on the same line at distances L, 2L, 3L, …, nL from O. The distance of centre of mass from O is
${\text{A}}{\text{. }}\dfrac{{\left( {2n + 1} \right)L}}{4} \\\ {\text{B}}{\text{. }}\dfrac{L}{{\left( {2n + 1} \right)}} \\\ {\text{C}}{\text{. }}\dfrac{{n\left( {{n^2} + 1} \right)L}}{2} \\\ {\text{D}}{\text{. }}\dfrac{{\left( {2n + 1} \right)L}}{3} \\\$

Answer 1

Hint: The centre of mass of a body contains a distribution of masses which equal to the ratio of sum of products of individual masses and their distances from origin to the sum of the masses or the total mass of the body.

Formula used:

Centre of mass:
$C = \dfrac{{\sum {{m_i}r_i^2} }}{{\sum {{m_i}} }}$
Identities used:

$\begin{gathered} \sum\limits_{i = 1}^n i = \dfrac{{n\left( {n + 1} \right)}}{2} \\\ \sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\\ \end{gathered}$

Detailed step by step solution:

Centre of mass: For a given distribution of masses in a space, we can define a centre about which the masses have been distributed in the space which is called the centre of mass of a body or a system.

It is defined as the ratio of sum of products of individual masses and their distances from origin to the sum of the masses or the total mass of the body. Mathematically, it can be represented as follows:

$C = \dfrac{{\sum {{m_i}r_i^2} }}{{\sum {{m_i}} }}$

where ${m_i}$ is the mass of the ith particle of the body and ${r_i}$ is the distance of the ith particle from the origin.

We are given the following masses: m, 2m, 3m, …, nm
Their distance from origin: L, 2L, 3L, …, nL

Using these values, we can calculate the position of centre of mass as follows:

$

C = \dfrac{{\sum\limits_{i = 1}^n {\left( {im} \right)} \left( {iL} \right)}}{{\sum\limits_{i = 1}^n {im} }} \\

\Rightarrow C = \dfrac{{mL\sum\limits_{i = 1}^n {{i^2}} }}{{m\sum\limits_{i = 1}^n i }} \\
$

Using the identities, we get

$C = L\dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{\dfrac{{n\left( {n + 1} \right)}}{2}}} = \dfrac{{\left( {2n + 1} \right)}}{3}L$

Hence, the correct answer is option D.

Note: The origin can be chosen arbitrarily for calculation of centre of mass. We get the distance of the centre of mass from the origin which can vary with the choice of origin but the actual position of centre of mass remains the same irrespective of the choice of origin.