System of Particles and Rotational Motion - Result Question 40
43. Consider a system of two particles having masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the centre of mass of particles through a distance d, by what distance would the particle of mass $m_2$ move so as to keep the centre of mass of particles at the original position?[2004]
(a) $\frac{m_2}{m_1} d$
(b) $\frac{m_1}{m_1+m_2} d$
(c) $\frac{m_1}{m_2} d$
(d) $d$
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Answer:
Correct Answer: 43. (c)
Solution:
- (c) To maintain the position of centre of mass at the same position, the velocity of centre of mass $v _{cm}$ must be zero.
That means, $v _{cm}=\frac{m_1 v_1+m_2 v_2}{(m_1+m_2)}=0$
$\Rightarrow \frac{m_1 \frac{dr_1}{dt}+m_2 \frac{dr_2}{dt}}{(m_1+m_2)}=0 \quad \begin{cases} \because v_1=\frac{dr_1}{dt} \\ m_2=\frac{dr_2}{dt} \end{cases} $
where, $v_1$ and $v_2$ are the velocities of the particles and $dr_1$ and $dr_2$ are the change in dosplacement
of particles. Then, $dr_2=-(\frac{m_1}{m_2}) dr r_1$
As the centre of mass of a body is a point where the whole mass of the body is concentrated, therefore, sum of moments of masses of all the particles of the body about the centre of mass is
zero, i.e., $\sum _{i=1}^{i=n} m_i \overrightarrow{{}r_i}=0$
For two particle system
$ \begin{aligned} & m_1 \vec{r} _1+m_2 \vec{r} _2=0 \\ & \Rightarrow \overrightarrow{{}r_1}=\frac{m_1 \vec{r} _1}{m_2} \end{aligned} $
