First slide

Location of centre on mass for multiple point masses

Question

Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1, 2, 3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1, 3, -2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system?

Moderate

A

(0, 0, 0)
B

$(1, 3, 2)$
C

$(- 1, 2, 3)$
D

$(3, 1, 8)$

Solution

According to the definition of centre of mass, we can imagine one particle of mass (1 + 2 + 3) kg at (1,2, 3); another particle of mass (2 + 3) kg at (-1,3,-2).
Let the third particle of mass 5 kg put at ( $x_{3}, y_{3}, z_{3}$ ) i.e.
$m_{1} = 6 kg, (x_{1}, y_{1}, z_{1}) = (1, 2, 3)$
$m_{2} = 5 kg, (x_{2}, y_{2}, z_{2}) = (- 1, 3, - 2)$
$m_{3} = 5 kg,$ let its position is $(x_{3,} y_{3 .} z_{3})$
Given, $(X_{CM}, Y_{CM}, Z_{CM}) = (1, 2, 3)$
1 = $\frac{6 \times 1 + 5 \times (- 1) + 5 x_{3}}{6 + 5 + 5}$
$5 x_{3} = 16 - 1 = 15 or x_{3} = 3$
Similarly, $y_{3} = 1 a n d z_{3} = 8$

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