The position vectors of three particles of ma | vedclass

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(A) The position vector of the centre of mass for a system of particles is given by the formula:$\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_

Vedclass · Accepted Answer

$\frac{1}{2}(3\hat{i} + \hat{j} - \hat{k})\,m$

Vedclass · Answer

(A) The position vector of the centre of mass for a system of particles is given by the formula:$\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$Given:$m_1 = 1\,kg, \vec{r}_1 = (\hat{i} + 4\hat{j} + \hat{k})\,m$$m_2 = 2\,kg, \vec{r}_2 = (\hat{i} + \hat{j} + \hat{k})\,m$$m_3 = 3\,kg, \vec{r}_3 = (2\hat{i} - \hat{j} - 2\hat{k})\,m$Substituting the values:$\vec{R}_{cm} = \frac{1(\hat{i} + 4\hat{j} + \hat{k}) + 2(\hat{i} + \hat{j} + \hat{k}) + 3(2\hat{i} - \hat{j} - 2\hat{k})}{1 + 2 + 3}$$\vec{R}_{cm} = \frac{(\hat{i} + 4\hat{j} + \hat{k}) + (2\hat{i} + 2\hat{j} + 2\hat{k}) + (6\hat{i} - 3\hat{j} - 6\hat{k})}{6}$$\vec{R}_{cm} = \frac{(1 + 2 + 6)\hat{i} + (4 + 2 - 3)\hat{j} + (1 + 2 - 6)\hat{k}}{6}$$\vec{R}_{cm} = \frac{9\hat{i} + 3\hat{j} - 3\hat{k}}{6}$$\vec{R}_{cm} = \frac{3(3\hat{i} + \hat{j} - \hat{k})}{6} = \frac{1}{2}(3\hat{i} + \hat{j} - \hat{k})\,m$

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

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