Question

Two bodies of mass $1{\text{ kg}}$ and ${\text{3 kg}}$ have position vectors $\widehat i + 2\widehat j + \widehat k$ and $- 3\widehat i - 2\widehat j + \widehat k$ respectively. The centre of mass of this system has a position vector:
(A) $- \widehat i + \widehat j + \widehat k$
(B) $- 2\widehat i + 2\widehat k$
(C) $- 2\widehat i - \widehat j + \widehat k$
(D) $2\widehat i - \widehat j - 2\widehat k$

Answer 1

In order to answer this question, we need to understand that the centre of mass of the system of particles is defined as the position where all the mass of the system of particles is considered to be. It is the point where if linear force is applied on the system, then there will not be any angular acceleration.

Formula Used:
The formula of the centre of mass is given by,
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{{m_1}{{\overrightarrow r }_1} + {m_2}{{\overrightarrow r }_2} + ...... + {m_n}{{\overrightarrow r }_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
Where centre of mass is ${\overrightarrow r _{cm}}$ and mass of different particles are ${m_1},{m_2},etc.$ and the different vector positions are ${\overrightarrow r _1},{\overrightarrow r _2}{\text{ and etc}}{\text{.}}$

Complete answer:
It is given in the problem that Two bodies of mass $1{\text{ kg}}$ and ${\text{3 kg}}$ are placed at position vector $\widehat i + 2\widehat j + \widehat k$ and $- 3\widehat i - 2\widehat j + \widehat k$respectively and we need to find the position of the centre of mass of the particles.
The mass of two bodies is ${{\text{m}}_1} = 1{\text{ kg, }}{{\text{m}}_2} = 3{\text{ kg}}$ also the position vector of the two bodies is ${\overrightarrow r _1} = \widehat i + 2\widehat j + \widehat k$ and ${\overrightarrow r _2} = - 3\widehat i - 2\widehat j + \widehat k.$
The formula of the centre of mass is given by,
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{{m_1}{{\overrightarrow r }_1} + {m_2}{{\overrightarrow r }_2} + ...... + {m_n}{{\overrightarrow r }_n}}}{{{m_1} + {m_2} + ..... + {m_n}}}$
Where centre of mass is ${\overrightarrow r _{cm}}$ and mass of different particles are ${m_1},{m_2},etc.$ and the different vector positions are ${\overrightarrow r _1},{\overrightarrow r _2}{\text{ and etc}}{\text{.}}$
Replacing the mass and the position vector in the above formula we get,
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{{m_1}{{\overrightarrow r }_1} + {m_2}{{\overrightarrow r }_2}}}{{{m_1} + {m_2}}}$
Now substitute the values,
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{1 \times \left( {\widehat i + 2\widehat j + \widehat k} \right) + 3 \times \left( { - 3\widehat i - 2\widehat j + \widehat k} \right)}}{{1 + 3}}$
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{\widehat i + 2\widehat j + \widehat k - 9\widehat i - 6\widehat j + 3\widehat k}}{4}$
Rearrange the terms for further calculations,
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{\widehat i - 9\widehat i + 2\widehat j - 6\widehat j + \widehat k + 3\widehat k}}{4}$
$\Rightarrow {\overrightarrow r _{cm}} = \dfrac{{ - 8\widehat i - 4\widehat j + 4\widehat k}}{4}$
Now, divide it be 4 we get
$\Rightarrow {\overrightarrow r _{cm}} = - 2\widehat i - \widehat j + \widehat k$
So, the centre of mass of the system of particles is equal to
$\Rightarrow {\overrightarrow r _{cm}} = - 2\widehat i - \widehat j + \widehat k$
Hence the correct option is (C).

Note:
The students are advised to understand and remember the formula of the centre of mass as it is very useful in solving problems like this. The centre of mass can be present outside the physical bodies. The centre of mass is an inertial frame and it is at rest compared to the origin of the coordinate system.