Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular

Question

Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be x m . The value of x is

TestHub · Accepted Answer

The diagram represents the locations of the masses as mentioned in the question.
For point $1$ , $\vec{r_{1}} = 0 \hat{i} + 0 \hat{j}$ for point $2$ , $\vec{r_{2}} = 3 \hat{i} + 0 \hat{j}$ and for point $3$ , $\vec{r_{3}} = 0 \hat{i} + 3 \hat{j}$ .
The formula for centre of mass is
${\vec{r}}_{c o m} = \frac{m_{1} \vec{r_{1}} + m_{2} \vec{r_{2}} + m_{3} \vec{r_{3}}}{m_{1} + m_{2} + m_{3}}$
$\Rightarrow {\vec{r}}_{com} = \frac{M (0 \hat{i} + 0 \hat{j}) + M (3 \hat{i}) + M (3 \hat{j})}{3 M}$
$\Rightarrow {\vec{r}}_{com} = \hat{i} + \hat{j}$
$|{\vec{r}}_{com}| = \sqrt{1^{2} + 1^{2}} = \sqrt{2} = \sqrt{x}$
$\Rightarrow x = 2$

Physics - Center of Mass Question with Solution | TestHub

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