Acceleration of Centre of Mass of Pulley System – Rankers Physics

Motion of Center of Mass: Practice Problem & Solution

Two bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:
\( \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \)
\( \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \)
\( \frac{m_1 g}{(m_1 + m_2)} \)
\( \frac{m_2 g}{(m_1 + m_2)} \)

Solution Explained:

To solve this problem, we apply the core principles of Motion of Center of Mass. Understanding the underlying formula is key to arriving at the correct answer below:

The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).

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