## Step 1: **Define the System and Write the Energies** Let \( x \) be the distance fallen by mass \( m_1 \) (so mass \( m_2 \) rises by the same amount, since the rope is inextensible). - **Kinetic Energy (T):** - \( m_1 \) moves down with speed \( v \) - \( m_2 \) moves up with speed \( v \) - Pulley rotates with angular speed \( \omega = v/r \), moment of inertia \( I \) So, \[ T = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{2} I \omega^2 \] Substitute \( v = \dot{x} \) and \( \omega = \dot{x}/r \): \[ T = \frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{r}\right)^2 = \frac{1}{2} (m_1 + m_2) \dot{x}^2 + \frac{1}{2} \frac{I}{r^2} \dot{x}^2 = \frac{1}{2} \left(m_1 + m_2 + \frac{I}{r^2}\right) \dot{x}^2 \] **Explanation:** All kinetic energies are written in terms of \( \dot{x} \) (the velocity of the masses), using the relation between linear and angular speed at the rim of the pulley. --- ## Step 2: **Write the Potential Energy and Lagrangian, then Apply Lagrange's Equation** - **Potential Energy (U):** Let the zero level be at the ground. At time \( t \), \( m_1 \) is at height \( L-x \), \( m_2 \) is at height \( x \). \[ U = m_1 g (L - x) + m_2 g x \] \[ U = m_1 g L - m_1 g x + m_2 g x = m_1 g L + (m_2 - m_1) g x \] (The constant \( m_1 g L \) can be ignored in the equations of motion.) - **Lagrangian \( \mathcal{L} = T - U \):** \[ \mathcal{L} = \frac{1}{2} \left(m_1 + m_2 + \frac{I}{r^2}\right) \dot{x}^2 - (m_2 - m_1) g x \] - **Apply Euler-Lagrange Equation:** \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} = \] Compute derivatives: \[ \frac{\partial \mathcal{L}}{\partial \dot{x}} = \left(m_1 + m_2 + \frac{I}{r^2}\right) \dot{x} \] \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \left(m_1 + m_2 + \frac{I}{r^2}\right) \ddot{x} \] \[ \frac{\partial \mathcal{L}}{\partial x} = - (m_2 - m_1) g \] Plug into Euler-Lagrange: \[ \left(m_1 + m_2 + \frac{I}{r^2}\right) \ddot{x} + (m_2 - m_1) g = \] \[ \boxed{ \ddot{x} = \frac{(m_1 - m_2) g}{m_1 + m_2 + \frac{I}{r^2}} } \] **Explanation:** The Lagrangian is constructed using the energies, and Lagrange’s equation gives the acceleration of the masses. The additional term \(\frac{I}{r^2}\) in the denominator accounts for the rotational inertia of the pulley. --- ## **Final Answer** **Equation of motion:** \[ \boxed{ \ddot{x} = \frac{(m_1 - m_2) g}{m_1 + m_2 + \frac{I}{r^2}} } \] where \(x\) is the downward displacement of mass \(m_1\).