Two bodies of mass $$1$$ kg and $$3$$ kg have position vectors $$i^+2j^+k^$$ and $$-3i^-2j^+k^$$, respectively. The centre of mass of this system has a position vector.

The position vector of centre of mass r→ $$=$$ r→ $$=$$ $$m1r1$$→$$+m2r2$$→$$m1+m2$$ $$=$$ $$1(i^+2j^+k^$$ $$)+3(-3i^-2j^+k^)1+3$$ $$=$$ $$14(-8i^-4j^+4k^)$$ $$=$$ $$-2i^-j^+k^$$

Location of centre on mass for multiple point masses

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Question

Two bodies of mass 1 kg and 3 kg have position vectors $\hat{i} + 2 \hat{j} + \hat{k} and - 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector.

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Solution

The position vector of centre of mass
$\vec{r} = \vec{r} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}}}{m_{1} + m_{2}}$
= $\frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$
= $\frac{1}{4} (- 8 \hat{i} - 4 \hat{j} + 4 \hat{k})$
= $- 2 \hat{i} - \hat{j} + \hat{k}$

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Location of centre on mass for multiple point masses

Remember concepts with our Masterclasses.

Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k^ and -3i^-2j^+k^, respectively. The centre of mass of this system has a position vector.

The position vector of centre of mass r→ = r→ = m1r1→+m2r2→m1+m2 = 1(i^+2j^+k^ )+3(-3i^-2j^+k^)1+3 = 14(-8i^-4j^+4k^) = -2i^-j^+k^

Ready to Test Your Skills?

free study material

Two bodies of mass 1 kg and 3 kg have position vectors $\hat{i} + 2 \hat{j} + \hat{k} and - 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector.