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
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
Show 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.
Updated On: Jun 26, 2026
- \(\left(\frac{2m}{M},0,0\right)\)
- \(\left(\frac{m}{2M},0,0\right)\)
- \(\left(\frac{4m}{M},0,0\right)\)
- \(\left(\frac{m}{4M},0,0\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the mass placed at \(x=N\).
At position \[ x=N, \] the mass is \[ m_N=\left(\frac{1}{2}\right)^{N-1}\frac{m}{N} \]
Step 2: Use the formula for centre of mass.
Since all masses are placed on the \(x\)-axis, \[ \bar{x}=\frac{\sum m_N x_N}{\sum m_N} \] Given total mass is \[ M \] So, \[ \bar{x}=\frac{\sum_{N=1}^{\infty}m_N x_N}{M} \]
Step 3: Calculate the numerator.
Now, \[ m_Nx_N= \left(\frac{1}{2}\right)^{N-1}\frac{m}{N}\cdot N \] \[ =m\left(\frac{1}{2}\right)^{N-1} \] Therefore, \[ \sum_{N=1}^{\infty}m_Nx_N = m\sum_{N=1}^{\infty}\left(\frac{1}{2}\right)^{N-1} \] This is a geometric series with first term \[ 1 \] and common ratio \[ \frac{1}{2} \] So, \[ \sum_{N=1}^{\infty}\left(\frac{1}{2}\right)^{N-1} = \frac{1}{1-\frac{1}{2}} \] \[ =2 \] Thus, \[ \sum_{N=1}^{\infty}m_Nx_N=2m \]
Step 4: Find the centre of mass.
Hence, \[ \bar{x}=\frac{2m}{M} \] Since all particles lie on the \(x\)-axis, \[ \bar{y}=0,\quad \bar{z}=0 \] Therefore, the centre of mass is \[ \left(\frac{2m}{M},0,0\right) \]
Step 5: Final conclusion.
Hence, the required centre of mass is \[ \boxed{\left(\frac{2m}{M},0,0\right)} \]
At position \[ x=N, \] the mass is \[ m_N=\left(\frac{1}{2}\right)^{N-1}\frac{m}{N} \]
Step 2: Use the formula for centre of mass.
Since all masses are placed on the \(x\)-axis, \[ \bar{x}=\frac{\sum m_N x_N}{\sum m_N} \] Given total mass is \[ M \] So, \[ \bar{x}=\frac{\sum_{N=1}^{\infty}m_N x_N}{M} \]
Step 3: Calculate the numerator.
Now, \[ m_Nx_N= \left(\frac{1}{2}\right)^{N-1}\frac{m}{N}\cdot N \] \[ =m\left(\frac{1}{2}\right)^{N-1} \] Therefore, \[ \sum_{N=1}^{\infty}m_Nx_N = m\sum_{N=1}^{\infty}\left(\frac{1}{2}\right)^{N-1} \] This is a geometric series with first term \[ 1 \] and common ratio \[ \frac{1}{2} \] So, \[ \sum_{N=1}^{\infty}\left(\frac{1}{2}\right)^{N-1} = \frac{1}{1-\frac{1}{2}} \] \[ =2 \] Thus, \[ \sum_{N=1}^{\infty}m_Nx_N=2m \]
Step 4: Find the centre of mass.
Hence, \[ \bar{x}=\frac{2m}{M} \] Since all particles lie on the \(x\)-axis, \[ \bar{y}=0,\quad \bar{z}=0 \] Therefore, the centre of mass is \[ \left(\frac{2m}{M},0,0\right) \]
Step 5: Final conclusion.
Hence, the required centre of mass is \[ \boxed{\left(\frac{2m}{M},0,0\right)} \]
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