Question

Masses \[ m,\ \left(\frac{1}{2}\right)\frac{1}{2}m,\ \left(\frac{1}{2}\right)^2\frac{1}{3}m,\ \ldots,\ \left(\frac{1}{2}\right)^{N-1}\frac{1}{N}m,\ \ldots \infty \] are placed at \[ x=1,2,3,\ldots,N,\ldots \infty \] respectively. If the total mass is \(M\), then the centre of mass of the system is

• A. \(\left(\frac{2m}{M},0,0\right)\)
• B. \(\left(\frac{m}{2M},0,0\right)\)
• C. \(\left(\frac{4m}{M},0,0\right)\)
• D. \(\left(\frac{m}{4M},0,0\right)\)

Hint:

For centre of mass, always compute \(\sum mx\) carefully. Here the factor \(\frac{1}{N}\) in mass cancels with the position \(x=N\), leaving a simple geometric series.

Answer 1

The Correct Option is A

Step 1: Identify the mass placed at position $x = N$.
The given sequence of masses placed at $x = 1, 2, 3, \ldots$ is: $m,\, \frac{1}{2} \cdot \frac{m}{2},\, \left(\frac{1}{2}\right)^2 \frac{m}{3}, \ldots$ The general term for the mass at position $x = N$ is: \[ m_N = \left(\frac{1}{2}\right)^{N-1} \frac{m}{N} \] This pattern shows each mass decreases rapidly as $N$ increases, ensuring the total mass converges to a finite value $M$.
Step 2: Set up the formula for the x-coordinate of the centre of mass.
Since all masses lie on the x-axis ($y = 0, z = 0$), the centre of mass coordinates are $\bar{y} = 0$ and $\bar{z} = 0$. For the x-coordinate: \[ \bar{x} = \frac{\displaystyle\sum_{N=1}^{\infty} m_N \cdot x_N}{\displaystyle\sum_{N=1}^{\infty} m_N} = \frac{\displaystyle\sum_{N=1}^{\infty} m_N \cdot N}{M} \] since each mass $m_N$ is placed at position $x_N = N$.
Step 3: Simplify the product $m_N \cdot N$.
\[ m_N \cdot N = \left(\frac{1}{2}\right)^{N-1} \frac{m}{N} \cdot N = m \left(\frac{1}{2}\right)^{N-1} \] Notice how the factor $N$ in the denominator of $m_N$ cancels exactly with the position $x_N = N$. This algebraic cancellation is what makes the numerator of the CoM formula tractable.
Step 4: Evaluate the infinite sum in the numerator.
\[ \sum_{N=1}^{\infty} m_N \cdot N = m \sum_{N=1}^{\infty} \left(\frac{1}{2}\right)^{N-1} \] This is an infinite geometric series with first term $a = 1$ and common ratio $r = 1/2$. Its sum is: \[ \sum_{N=1}^{\infty} \left(\frac{1}{2}\right)^{N-1} = \frac{1}{1 - \frac{1}{2}} = 2 \] Therefore: \[ \sum_{N=1}^{\infty} m_N \cdot N = 2m \]
Step 5: Compute the x-coordinate of the centre of mass.
Substituting back: \[ \bar{x} = \frac{2m}{M} \]
Step 6: State the full position of the centre of mass.
Since all masses are placed on the x-axis, the complete centre of mass position is: \[ \bar{x} = \frac{2m}{M}, \quad \bar{y} = 0, \quad \bar{z} = 0 \] \[ \boxed{\left(\frac{2m}{M},\, 0,\, 0\right)} \]