System Of Particles And Rotational Motion Ques 42
42. 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) $\frac{m_1}{m_2} d$
$d$
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Answer:
Correct Answer: 42.(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} \\ v_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 differential displacements of particles.
Then, $dr_2=-(\frac{m_2}{m_1}) d r_1$
As the centre of mass of a body is a point where the whole mass of the body is concentrated, therefore, the sum of moments of masses of all the particles of the body about the centre of mass is zero
zero, i.e., $\sum _{i=1}^{i=n} m_i \overrightarrow{{}r_i}=0$
For two-particle system
$ \begin{alignedat} & 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} $
BETA
