Question

The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4 \sqrt2} x$ , where the value of x is ___________

Answer 1

Given three identical spheres of mass $2M$ placed at the corners of a right-angled triangle. The sides of the triangle are 4 m each. Let the point of intersection of the two sides be the origin $(0, 0)$. The position vectors of the masses are:

• $m_1 = 2M$, $r_1 = (0, 0)$
• $m_2 = 2M$, $r_2 = (4, 0)$
• $m_3 = 2M$, $r_3 = (0, 4)$

The position vector of the center of mass is given by:

$r_{\text{com}} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3}{m_1 + m_2 + m_3}$

Substituting the values:

$r_{\text{com}} = \frac{2M \times (0, 0) + 2M \times (4, 0) + 2M \times (0, 4)}{6M} = \left(\frac{4}{3}, \frac{4}{3}\right)$

Magnitude of $r_{\text{com}}$:

$|r_{\text{com}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3}$

Thus, $x = 3$.