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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
Options:
$\frac{1}{N} m \ldots \infty$ 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
Solution:
1043 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{2 m}{M}, 0,0\right)$
According to given distribution of masses, centre of mass is given as
$$
\begin{aligned}
& X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots . .+m_n x_n \ldots \infty}{M} \\
& m \times 1+\left(\frac{1}{2}\right) \frac{m}{2} \times 2+\left(\frac{1}{2}\right)^2 \frac{m}{3} \times 3+\ldots . . \\
& =\frac{\left(\frac{1}{2}\right)^{N-1} \cdot \frac{m}{N} \times N+\ldots \infty}{M} \\
& =\frac{m+\left(\frac{1}{2}\right) m+\left(\frac{1}{2}\right)^2 m+\ldots \ldots+\left(\frac{1}{2}\right)^{N-1} m \ldots+\ldots \infty}{M} \\
& =\frac{m}{M}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\ldots .+\left(\frac{1}{2}\right)^{N-1}+\ldots . \infty\right] \\
& =\frac{m}{M} \cdot \frac{1}{1-1 / 2} \\
& =\frac{m}{M} \cdot \frac{1}{1 / 2}=\frac{2 m}{M} \\
&
\end{aligned}
$$
( $\because$ sum of infinite series in G. P, $\mathrm{S}=\frac{a}{1-r}$ )
$$
=\frac{m}{M} \cdot \frac{1}{1 / 2}=\frac{2 m}{M}
$$
Since, masses are distributed only along $X$-axis, hence
$$
Y_{\mathrm{CM}}=0 \text { and } Z_{\mathrm{CM}}=0
$$
Position of centre of mass $=\left(\frac{2 m}{M}, 0,0\right)$
$$
\begin{aligned}
& X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots . .+m_n x_n \ldots \infty}{M} \\
& m \times 1+\left(\frac{1}{2}\right) \frac{m}{2} \times 2+\left(\frac{1}{2}\right)^2 \frac{m}{3} \times 3+\ldots . . \\
& =\frac{\left(\frac{1}{2}\right)^{N-1} \cdot \frac{m}{N} \times N+\ldots \infty}{M} \\
& =\frac{m+\left(\frac{1}{2}\right) m+\left(\frac{1}{2}\right)^2 m+\ldots \ldots+\left(\frac{1}{2}\right)^{N-1} m \ldots+\ldots \infty}{M} \\
& =\frac{m}{M}\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\ldots .+\left(\frac{1}{2}\right)^{N-1}+\ldots . \infty\right] \\
& =\frac{m}{M} \cdot \frac{1}{1-1 / 2} \\
& =\frac{m}{M} \cdot \frac{1}{1 / 2}=\frac{2 m}{M} \\
&
\end{aligned}
$$
( $\because$ sum of infinite series in G. P, $\mathrm{S}=\frac{a}{1-r}$ )
$$
=\frac{m}{M} \cdot \frac{1}{1 / 2}=\frac{2 m}{M}
$$
Since, masses are distributed only along $X$-axis, hence
$$
Y_{\mathrm{CM}}=0 \text { and } Z_{\mathrm{CM}}=0
$$
Position of centre of mass $=\left(\frac{2 m}{M}, 0,0\right)$
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