Question 353: Centre of Mass and Rotational Motion

Question

In system of two particles of masses $m_1$ and $m_2$, the first particle is moved by a distance $d$ towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

Accepted Answer

Let $x_1$ and $x_2$ be the position of masses $m_1$ and $m_2$, respectively.
The position of centre of mass is
$$x_{\mathrm{CM}}=\frac{x_1 m_1+x_2 m_2}{m_1+m_2}$$
If $\Delta x_1$ and $\Delta x_2$ be the changes in positions, then change in the position of centre of mass,
$$\Delta x_{\mathrm{CM}}=\frac{\Delta x_1 m_1+\Delta x_2 m_2}{m_1+m_2}$$
Given that, the centre of mass remains unchanged e., $\Delta x_{C M}=0$ and $\Delta x_1=d$
$\Rightarrow \quad 0=\frac{d m_1+m_2 \Delta x_2}{m_1+m_2}$
or $$\Delta x_2=-\frac{m_1}{m_2} d$$
Here, negative sign shows that the second particle should be moved towards the centre of mass.

Answer

$\frac{m_2}{m_1} d$, towards the centre of mass

Answer

$\frac{m_1}{m_2} d$, away from the centre of mass

Answer

$\frac{m_1}{m_2} d$, towards the centre of mass

Answer

$\frac{m_2}{m_1} d$, away from the centre of mass

Centre of Mass and Rotational Motion

In system of two particles of masses m_1 and m_2, the first particle is moved by a distance d towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

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