Question

Masses \( m \left(\frac{1}{3}\right)^N \) are placed at \( x = N \), for \( N = 2, 3, 4, \dots \). If the total mass is \( M \), then the center of mass of the system is:

• A. \( \frac{1}{6} M \)
• B. \( \frac{1}{5} M \)
• C. \( \frac{1}{3} M \)
• D. \( \frac{1}{2} M \)

Hint:

Use weighted average: \( x_{\text{COM}} = \frac{\sum m_i x_i}{\sum m_i} \) and geometric series for infinite masses.

Answer 1

The Correct Option is A

Let total mass \( M = \sum_{N=2}^{\infty} m \left( \frac{1}{3} \right)^N \),
Let numerator for center of mass be:
\[ \sum x_i m_i = \sum_{N=2}^{\infty} N \cdot m \left( \frac{1}{3} \right)^N \] \[ x_{\text{COM}} = \frac{\sum x_i m_i}{\sum m_i} = \frac{\sum N \left( \frac{1}{3} \right)^N}{\sum \left( \frac{1}{3} \right)^N} \] This ratio evaluates (after summing geometric series) to \( \frac{1}{6} \)