For given Atwood machine. Find displacement (in m) of centre of mass after 2 sec of release.
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In an Atwood machine, the heavier mass causes the system to accelerate in its direction. The center of mass moves accordingly, and the displacement can be calculated using kinematic equations.
Step 1: Understanding the Atwood Machine.
An Atwood machine consists of two masses connected by a light inextensible string passing over a pulley. Here, the two masses are \(2\,\text{kg}\) and \(1\,\text{kg}\). We need the displacement of the centre of mass after \(2\) seconds, not the displacement of an individual block.
Step 2: Find the acceleration of the blocks.
The acceleration of the system is given by:
\[
a = \frac{(m_1 - m_2)g}{m_1 + m_2}
\]
Taking \(m_1 = 2\,\text{kg}\) and \(m_2 = 1\,\text{kg}\), we get:
\[
a = \frac{(2 - 1)g}{2 + 1} = \frac{g}{3}
\]
If \(g = 10\,\text{m/s}^2\), then:
\[
a = \frac{10}{3}\,\text{m/s}^2
\]
So, in \(2\) seconds, each block moves:
\[
s = \frac{1}{2}at^2
\]
\[
s = \frac{1}{2}\times \frac{10}{3}\times (2)^2
\]
\[
s = \frac{20}{3}\,\text{m}
\]
Thus, the \(2\,\text{kg}\) block moves downward by \(\frac{20}{3}\,\text{m}\), and the \(1\,\text{kg}\) block moves upward by \(\frac{20}{3}\,\text{m}\).
Step 3: Find the displacement of the centre of mass.
The displacement of the centre of mass is:
\[
\Delta y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}
\]
Take downward displacement as positive. Then:
\[
y_1 = \frac{20}{3}, \qquad y_2 = -\frac{20}{3}
\]
So,
\[
\Delta y_{cm} = \frac{2\left(\frac{20}{3}\right) + 1\left(-\frac{20}{3}\right)}{2+1}
\]
\[
\Delta y_{cm} = \frac{\frac{40}{3} - \frac{20}{3}}{3}
\]
\[
\Delta y_{cm} = \frac{20}{9}\,\text{m}
\]
Since the result is positive in our sign convention, the centre of mass moves downward.
Step 4: Conclusion.
Therefore, the displacement of the centre of mass after \(2\) seconds is \( \frac{20}{9}\,\text{m} \) downward. Final Answer: \( \frac{20}{9}\,\text{m} \) (Downward).
