Rotational Motion Question 56
Question: Consider a system of two particles having masses $ m _1 $ and $ m _2 $ . If the particle of mass $ m _1 $ is pushed towards the mass centre of particles through a distance d, by what distance would the particle of mass $ m _2 $ move so as to keep the mass centre of particles at the original position?
[AIPMT (S) 2004]
Options:
A) $ \frac{m _1}{m _1+m _2}d $
B) $ \frac{m _1}{m _2}d $
C) $ d $
D) $ \frac{m _2}{m _1}d $
Show Answer
Answer:
Correct Answer: B
Solution:
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The system of two given particles of masses $ m _1 $ and $ m _2 $.
Initially the centre of mass
$ r _{CM}=\frac{m _1r _1+m _2r _2}{m _1+m _2} $ …(1)
When mass $ m _1 $ moves towards centre of mass by a distance d, then let mass $ m _2 $ moves a distance d away from CM to keep the CM in its initial position.
So, $ r _{CM}=\frac{m _1(r _1-d)+m _2(r _2+d’)}{m _1+m _2} $
Equating Eqs.
(i) and (ii), we get
$ \frac{m _1r _1+m _2r _2}{m _1+m _2}=\frac{m _1(r _1-d)+m _2(r _2+d’)}{m _1+m _2} $
$ \Rightarrow $ $ -m _1d+m _2d’=0 $
$ \Rightarrow $ $ d’=\frac{m _1}{m _2}d $
NOTE: If both the masses are equal i.e., $ m _1=m _2, $ then second mass will move a distance equal to the distance at which first mass is being displaced.
