Question

The center of mass of a system of three particles of masses $1g,2g$and $3g$is taken as the origin of a coordinate system. The position vector of a fourth particle of mass $4g$ such that the center of mass of the four particle system lies at the point $\left( {1,2,3} \right)$ is $\alpha \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$, where $\alpha$is a constant . The value of $\alpha$ is
A. $\dfrac{5}{2}$
B. $\dfrac{{10}}{3}$
C. $\dfrac{2}{5}$
D. $\dfrac{1}{2}$

Answer 1

In order to answer this question we have to find the center of mass of any one of the axes . Center of mass along x axis is calculated by
$x = \dfrac{{{m_1}{x_1} + m{2_{}}{x_2} + {m_3}{x_3} + - - - - - - - - - {m_n}{x_n}}}{{m{1_{}} + {m_2} + {m_3} + - - - - - - - - {m_n}}}$

Complete step-by-step solution:
Center of masses is the individual point where the weighted relative situation of the distributed mass is null.
The coordinate of x of masses $1g,2g,3g\& 4g$ respectively-
$\left( {{x_1} = 0,{x_2} = 0,{x_3} = 0,{x_4} = \alpha } \right)$
Therefore, ${x_{cm}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + - - - - - - - - - {m_n}{x_n}}}{{{m_1} + {m_2} + {m_3} + - - - - - - - - {m_n}}}$
${x_{cm}} = \dfrac{{4\alpha }}{{1 + 2 + 3 + 4}}$
Now $1 = \dfrac{{4\alpha }}{{10}}$
$\therefore \alpha = \dfrac{5}{2}$
Therefore, option A.) $\dfrac{5}{2}$ is the right answer.

Note: For a single rigid body, center of mass is fixed corresponding to the body and if a body has uniform thickness, it will be situated at centroid. COM of a body is discovered by the scientist Archimedes.