The position vector of three particles of mass m1 - 3 kg- m2 - 4 kg and m3 - 1 kg are r1 - 2-3k- m- r2 - -3-2k- m and r3 - 3-2-k- m respectively- Find the position vector of centre of mass of the system of particles-

The position vector of COM of the system of three particles will be given by: r_CM= m_1 r_1+m_2 r_2+m_3 r_3m_1+m_2+m_3...(i) r_1 = (2î+ĵ+3k̂) m r_2 = (î 3ĵ+2k ...

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Standard XII

Physics

Center of Mass as an Average Point

The position ...

Question

The position vector of three particles of mass $m_{1} = 3 kg, m_{2} = 4 kg$ and $m_{3} = 1 kg$ are $_{1} = (2^i +^j + 3^k) m,_{2} = (^i - 3^j + 2^k) m$ and $_{3} = (3^i - 2^j -^k) m$ respectively. Find the position vector of centre of mass of the system of particles.

\frac{1}{8} (13^i - 12^j + 15^k) m

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\frac{1}{8} (11^i - 13^j + 16^k) m

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\frac{1}{8} (16^i + 11^j - 13^k) m

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\frac{1}{8} (13^i - 11^j + 16^k) m

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Solution

The correct option is D $\frac{1}{8} (13^i - 11^j + 16^k) m$
The position vector of COM of the system of three particles will be given by:

${\to r}_{CM} = \frac{m_{1}_{1} + m_{2}_{2} + m_{3}_{3}}{m_{1} + m_{2} + m_{3}} . . . (i)$

$_{1} = (2^i +^j + 3^k) m$
$_{2} = (^i - 3^j + 2^k) m$ $_{3} = (3^i - 2^j -^k) m$

Substituting the values in equation $(i)$

$\to r_{CM} = \frac{(3) (2^i +^j + 3^k) + (4) (^i - 3^j + 2^k) + (1) (3^i - 2^j -^k)}{3 + 4 + 1}$

$\Rightarrow \to r_{CM} = \frac{(6^i + 3^j + 9^k) + (4^i - 12^j + 8^k) + (3^i - 2^j -^k)}{8}$

$∴ \to r_{CM} = \frac{1}{8} (13^i - 11^j + 16^k) m$

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