The centre of mass of three particles of mass | vedclass

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(A) Let the masses be $m_1 = 1 \ kg, m_2 = 2 \ kg, m_3 = 3 \ kg$ with centre of mass $R_{CM} = (2, 2, 2)$.Let the sum of the moments of the first thre

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$(-3, -3, -3)$

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(A) Let the masses be $m_1 = 1 \ kg, m_2 = 2 \ kg, m_3 = 3 \ kg$ with centre of mass $R_{CM} = (2, 2, 2)$.Let the sum of the moments of the first three masses be $M_{123} = m_1 r_1 + m_2 r_2 + m_3 r_3$.The total mass of the first three particles is $M = 1 + 2 + 3 = 6 \ kg$.Using the formula $R_{CM} = \frac{M_{123}}{M}$,we have $M_{123} = M 	imes R_{CM} = 6 	imes (2, 2, 2) = (12, 12, 12)$.Now,we add a fourth mass $m_4 = 4 \ kg$ at position $r_4 = (x_4, y_4, z_4)$ such that the new centre of mass $R'_{CM} = (0, 0, 0)$.The new total mass is $M' = 6 + 4 = 10 \ kg$.The new centre of mass formula is $R'_{CM} = \frac{M_{123} + m_4 r_4}{M'}$.Substituting the values: $(0, 0, 0) = \frac{(12, 12, 12) + 4(x_4, y_4, z_4)}{10}$.This implies $(12, 12, 12) + 4(x_4, y_4, z_4) = (0, 0, 0)$.$4x_4 = -12 \implies x_4 = -3$.$4y_4 = -12 \implies y_4 = -3$.$4z_4 = -12 \implies z_4 = -3$.Therefore,the position of the fourth mass is $(-3, -3, -3)$.

The centre of mass of three particles of masses $1 \ kg, 2 \ kg$ and $3 \ kg$ is at $(2, 2, 2)$. The position of the fourth mass of $4 \ kg$ to be placed in the system so that the new centre of mass is at $(0, 0, 0)$ is:

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