Q: If the centre of mass of three particles of masses, of 1 kg, 2 kg, 3kg is at (2,2,2), then where should a fourth particle of mass 4kg be placed so that the combined centre of mass may be at (0, 0, 0).

Sol: Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the positions of masses 1 kg, 2 kg, 3 kg and let the co-ordinates of centre of mass of the three particle system is (xcm,ycm, zcm) respectively.

$\large x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} $

$\large 2 = \frac{1 \times x_1 + 2 \times x_2 + 3 \times x_3}{1 + 2 + 3} $

$\large x_1 + 2 x_2 + 3 x_3 = 12 $ …(i)

Suppose the fourth particle of mass 4kg is placed at (x4, y4, z4) so that centre of mass of new system shifts to (0, 0, 0). For x coordinate of new centre of mass we have

$\large 0 = \frac{1 \times x_1 + 2 \times x_2 + 3 \times x_3 + 4 \times x_4 }{1 + 2 + 3 + 4} $

$\large x_1 + 2 x_2 + 3 x_3 + 4 x_4 = 0 $ …(ii)

From equations (i) and (ii)

12 + 4x_{4} = 0 ⇒ x_{4} = -3

Similarly, y_{4} = -3 and z_{4} = -3

Therefore 4kg should be placed at (-3, -3, -3).

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