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

Correct Answer: 3

Three identical spheres, each with mass \(2M\), are positioned at the vertices of a right-angled triangle with sides measuring 4 m. The origin \((0, 0)\) is set at the intersection of the two sides forming the right angle. The position vectors for these masses are:

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

The formula for the position vector of the center of mass is:

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

Substituting the given values yields:

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

The magnitude of \(r_{\text{com}}\) is calculated as:

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

Therefore, \(x = 3\).