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 string passing over a frictionless pulley. When the system is released, the heavier mass accelerates downward and the lighter mass moves upward. Therefore, the center of mass of the system shifts in the direction of the heavier mass.
Step 2: Acceleration of the system.
The acceleration of the masses in an Atwood machine is given by:
\[
a = \frac{(m_2 - m_1)g}{m_1 + m_2}
\]
Where:
- \( m_1 = 1 \, \text{kg} \) (lighter mass),
- \( m_2 = 2 \, \text{kg} \) (heavier mass),
- \( g = 10 \, \text{m/s}^2 \).
Substituting the given values:
\[
a = \frac{(2 - 1)\times 10}{1 + 2}
\]
\[
a = \frac{10}{3} \, \text{m/s}^2
\]
Step 3: Acceleration of the center of mass.
The acceleration of the center of mass of the system is determined using the relation:
\[
a_{cm} = \frac{(m_2 - m_1)}{m_1 + m_2} \times g \times \frac{1}{3}
\]
Thus,
\[
a_{cm} = \frac{10}{9} \, \text{m/s}^2
\]
Step 4: Displacement of the center of mass after 2 seconds.
Since the system starts from rest, the displacement of the center of mass can be calculated using the kinematic equation:
\[
s = \frac{1}{2} a_{cm} t^2
\]
Where \( t = 2 \) seconds.
\[
s = \frac{1}{2} \times \frac{10}{9} \times (2)^2
\]
\[
s = \frac{10}{9} \, \text{m}
\]
The displacement is downward because the heavier mass moves downward, causing the center of mass of the system to move in the same direction.
Final Answer: \( \frac{10}{9} \, \text{m} \) (Downward).
