Solveeit Logo
Login

Question

Physics Question on Centre of mass

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)?

A

(−1,−1,−1)(-1,-1,-1)

B

(−2,−2,−2)(-2,-2,-2)

C

(2, 2, 2)(2,\,2,\,2)

D

(1, 1, 1)(1,\,1,\,1)

Answer

(−2,−2,−2)(-2,-2,-2)

Explanation

Solution

Centre of mass of a solid body is given by
xCM=∑i=1nΔmixi∑j=1nΔMj,yCM=∑i=1nΔmiyi∑j=1nΔMj{{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}}}}
zCM=∑i=1nΔmizi∑j=1nΔMj{{z}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{z}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta \Mu }_{j}}
1×x1+2×x2+3×x3=(1+2+3)31\times {{x}_{1}}+2\times {{x}_{2}}+3\times {{x}_{3}}=(1+2+3)3 ..(i)
and x1=x2=x3=3{{x}_{1}}={{x}_{2}}={{x}_{3}}=3
xCM=yCM=zCM=1 (given){{x}_{CM}}={{y}_{CM}}={{z}_{CM}}=1\,(given)
1(1+2+3+4)=1x1+1x2+3x3+4x41(1+2+3+4)=1{{x}_{1}}+1{{x}_{2}}+3{{x}_{3}}+4{{x}_{4}} ..(ii)
Solving Eqs, (i) and (ii),
we get 4x4=10−18⇒x4=−24{{x}_{4}}=10-18\Rightarrow {{x}_{4}}=-2
Similarly, y4=−2,z4=−2{{y}_{4}}=-2,{{z}_{4}}=-2
The fourth particle must be placed at the point
(−2,−2,−2).(-2,-2,-2).