Question

It has been mentioned that two bodies of mass $1kg$ and $3kg$ will be having position vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-3\hat{i}-2\hat{j}+\hat{k}$ respectively. What will be the position vector of the centre of mass of this system?
$\begin{align}
  & A.-\hat{i}+\hat{j}+\hat{k} \\
 & B.-2\hat{i}+2\hat{k} \\
 & C.-2\hat{i}-\hat{j}+\hat{k} \\
 & D.2\hat{i}-\hat{j}-2\hat{k} \\
\end{align}$

Answer 1

Hint: The position of the centre of mass can be found by taking the ratio of the sum of the product of the mass of the first body and its position vector and the product of the mass of the second body and its position vector to the total mass of the system. Substitute the values in this equation. This will help you in answering this question.

Complete answer:
The position of the centre of mass can be found by taking the ratio of the sum of the product of the mass of the first body and its position vector and the product of the mass of the second body and its position vector to the total mass of the system. This can be written as an equation given as,
$\vec{r}=\dfrac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
The position vector of the first body has been given as,
${{\vec{r}}_{1}}=\hat{i}+2\hat{j}+\hat{k}$
The mass of the first body has been mentioned as,
${{m}_{1}}=1kg$
The position vector of the second body has been given as,
${{\vec{r}}_{2}}=-3\hat{i}-2\hat{j}+\hat{k}$
The mass of the second body can be mentioned as,
${{m}_{2}}=3kg$
Now let us substitute the values in the equation given as,
$\vec{r}=\dfrac{1kg\times \hat{i}+2\hat{j}+\hat{k}+3kg\times -3\hat{i}-2\hat{j}+\hat{k}}{1kg+3kg}$
Simplifying this equation can be shown as,
$\begin{align}
  & \vec{r}=\dfrac{1}{4}\left( -8\hat{i}-4\hat{j}+4\hat{k} \right) \\
 & \therefore \vec{r}=-2\hat{i}-\hat{j}+\hat{k} \\
\end{align}$
Therefore the position vector of the centre of mass of the system has been found.

The correct answer has been mentioned as option C.

Note:
The centre of mass of a system is defined as the point of distribution of the whole mass in space. This the specific imaginary point where the whole mass is found to be concentrated. This will be the point to which a force may be acted for producing a linear acceleration without any angular acceleration.