Three identical spheres, each of mass M , are placed at the corners of a right angle triangle with mutually perpendicular sides equal to 2 m (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of centre of mass.

Question

Aicharya · Accepted Answer

First, recall the formula of the center of mass of $n$ particle system,

$r_{c m}=\frac{m_1 r_1+m_2 r_2+m_3 r_3+\ldots+m_n r_n}{m_1+m_2+m_3+. .+m_n}$

$\Rightarrow x_{com} = \frac{M \times 0 + M \times 2 + M \times 0}{3 M} = \frac{2}{3}$

$\Rightarrow y_{com} = \frac{M \times 0 + M + 2 + M \times 0}{3 M} = \frac{2}{3}$

So the position vector of center of mass is, $(\frac{2}{3} \hat{i} + \frac{2}{3} \hat{j})$ .

Three identical spheres, each of mass M , are placed at the corners of a right angle triangle with mutually…

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