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  4. The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg be placed, so that the centre of mass of the system of all particles shifts to a point (1, 1, 1)?

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Q. The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg be placed, so that the centre of mass of the system of all particles shifts to a point (1, 1, 1)?

Jharkhand CECEJharkhand CECE 2010System of Particles and Rotational Motion

Solution:

Centre of mass of a solid body is given by
$ {{x}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{x}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}},{{y}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{y}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}} $
$ {{z}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{z}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta \Mu }_{j}} $
$ 1\times {{x}_{1}}+2\times {{x}_{2}}+3\times {{x}_{3}}=(1+2+3)3 $ ..(i)
and $ {{x}_{1}}={{x}_{2}}={{x}_{3}}=3 $
$ {{x}_{CM}}={{y}_{CM}}={{z}_{CM}}=1\,(given) $
$ 1(1+2+3+4)=1{{x}_{1}}+1{{x}_{2}}+3{{x}_{3}}+4{{x}_{4}} $ ..(ii)
Solving Eqs, (i) and (ii),
we get $ 4{{x}_{4}}=10-18\Rightarrow {{x}_{4}}=-2 $
Similarly, $ {{y}_{4}}=-2,{{z}_{4}}=-2 $
The fourth particle must be placed at the point
$ (-2,-2,-2). $