The centre of mass of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ is at…
The centre of mass of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ is at $(2,2,2)$. The position of the fourth mass of $4 \mathrm{~kg}$ to be placed in the system as that the new centre of mass is at $(0,0,0)$ is.
$(-3,-3,-3)$
$(-3,3,-3)$
$(2,3,-3)$
$(2,-2,3)$
Solution
$m_1=1 \mathrm{~kg}, m_2=2 \mathrm{~kg}, m_3=3 \mathrm{~kg}$
Position of centre of mass $(2,2,2)$
$m_4=4 \mathrm{~kg}$
New position of centre of mass $(0,0,0)$.
For initial position,
$\begin{aligned}
& \qquad X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
& \qquad 2=\frac{m_1 \times x_1+m_2 x_2+m_3 x_3}{1+2+3} \\
& m_1 x_1+m_2 x_2+m_3 x_3=12 \\
\end{aligned}$
Similarly,
$\begin{aligned}
& m_1 y_1+m_2 y_2+m_3 y_3=12 \\
& m_1 z_1+m_2 z_2+m_3 z_3=12
\end{aligned}$
and
$m_1 z_1+m_2 z_2+m_3 z_3=12$
For new position,
$\begin{gathered}
X_{\mathrm{CM}}^{\prime}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4} \\
0=\frac{12+4+x_4}{1+2+3+4} \\
4 x_4=-12 \\
x_4=-3
\end{gathered}$
Similarly, $y_4=-3$
$z_4=-3$
$\therefore$ Position of fourth mass $(-3,-3,-3)$.