Question

Centre of mass of $3$ particles $5\text{kg},10\text{kg}$ and $30\text{kg}$ is at $(0,0,0)$. Where should a mass of $20\text{kg}$ be placed, so that the centre of mass of the combination is at $(2\text{m},2\text{m},2\text{m})$ ?

• A. ($6.5\text{m},6.5\text{m},6.5\text{m}$)
• B. $(3.5\text{m},2.5\text{m},6.5\text{m})$
• C. $(4\text{m},4\text{m},4\text{m})$
• D. $(5\text{m},6.6\text{m},2\text{m})$

Answer 1

The correct option is A ($6.5\text{m},6.5\text{m},6.5\text{m}$)

Replacing the $3$ masses $5\text{kg},10\text{kg}$ and $30\text{kg}$ by their $\text{CM}$ at $(0,0,0)$.
$\therefore$ Mass of $\text{CM}=5+10+30=45\text{kg}$

Now, let us take
$m_1=m_{\text{CM}}=45\text{kg}$
$m_2=20\text{kg}$

So, applying the formula for coordinates of new $\text{CM}$, for the above combination and equating it to $(2\text{m},2\text{m},2\text{m})$:

$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\Rightarrow 2=\frac{(45\times 0)+(20\times x_2)}{65}$
$\therefore x_2=6.5\text{m}$

$y_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
$\Rightarrow 2=\frac{(45\times 0)+20y_2}{65}$
$\therefore y_2=6.5$

$z_{CM}=\frac{m_1{z}_1+m_2{z}_2}{m_1+m_2}$
$\Rightarrow 2=\frac{(45\times 0)+20\times z_2}{65}$
$\therefore z_2=6.5$

Hence, the mass $20\text{kg}$ has to be kept at ($6.5\text{m},6.5\text{m},6.5\text{m}$)