## Step 1: Write the Lagrangian Let's choose coordinates: - θ: angle of the rod (with the vertical) - x: distance between masses \(M\) and \(m\) along thehorizontal) long rod ### Kinetic Energy - **Mass \(M\)**: Moves in a circle of radius \(l\), so its velocity is \(l\dot{\theta}\). \[ T_M = \frac{1}{2} M (l\dot{\theta})^2 \] - **Mass \(m\)**: Its velocity has two components: - Along the rod: \(\dot{x}\) - Due to rotation: \((l + x)\dot{\theta}\) (perpendicular to the rod) \[ T_m = \frac{1}{2} m \left[ \dot{x}^2 + ((l+x)\dot{\theta})^2 \right] \] Thus, \[ T = \frac{1}{2} M l^2 \dot{\theta}^2 + \frac{1}{2} m \left[ \dot{x}^2 + (l + x)^2 \dot{\theta}^2 \right] \] ### Potential Energy - **Mass \(M\)**: Height is \(-l\cos\theta\) \[ V_M = -M g l \cos\theta \] - **Mass \(m\)**: Height is \(-(l+x)\cos\theta\) \[ V_m = -m g (l + x) \cos\theta \] So, \[ V = -M g l \cos\theta - m g (l + x) \cos\theta \] ### **Lagrangian** \[ \mathcal{L} = T - V = \frac{1}{2} (M l^2 + m (l + x)^2) \dot{\theta}^2 + \frac{1}{2} m \dot{x}^2 + (M l + m(l + x))g\cos\theta \] --- **Explanation:** We have written the full Lagrangian in terms of the coordinates \(\theta\) and \(x\), accounting for both kinetic and potential energies of the masses. --- ## Step 2: Equations of Motion Use the Euler-Lagrange equations for both \(x\) and \(\theta\): ### For \(x\): \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = \] Compute: - \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = m \dot{x}\) - \(\frac{d}{dt} (m \dot{x}) = m \ddot{x}\) - \(\frac{\partial \mathcal{L}}{\partial x} = m (l + x) \dot{\theta}^2 - m g \cos\theta\) So, \[ m \ddot{x} - m (l + x) \dot{\theta}^2 + m g \cos\theta = \] \[ \boxed{\ddot{x} = (l + x)\dot{\theta}^2 - g\cos\theta} \] --- **Explanation:** We applied the Euler-Lagrange equation to \(x\) to get its equation of motion, which includes coupling to the angular variable \(\theta\). --- ### For \(\theta\): \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta} = \] Compute: - \(\frac{\partial \mathcal{L}}{\partial \dot{\theta}} = (M l^2 + m (l + x)^2)\dot{\theta}\) - \(\frac{d}{dt} = (M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta}\) - \(\frac{\partial \mathcal{L}}{\partial \theta} = -(M l + m(l + x))g \sin\theta\) So, \[ (M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta} + (M l + m(l + x))g\sin\theta = \] --- **Explanation:** We applied the Euler-Lagrange equation to \(\theta\) to get its equation of motion, which is coupled to both \(x\) and their derivatives. --- ## **Final Answers** ### a) **Equations of Motion:** \[ \boxed{ \begin{aligned} &\ddot{x} = (l + x)\dot{\theta}^2 - g\cos\theta \\ &(M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta} + (M l + m(l + x))g\sin\theta = \end{aligned} } \] --- ### b) **Normal Modes (when \(x\) and \(\theta\) are small)** When \(x \ll l\), \(\theta \ll 1\): - \(\sin\theta \approx \theta\), \(\cos\theta \approx 1\) - \(l + x \approx l\) Linearize the equations: 1. \(\ddot{x} = l\dot{\theta}^2 - g\) - But for small \(\dot{\theta}\), \(\dot{\theta}^2 \approx \), so: \[ \ddot{x} \approx -g\theta \] 2. \((M l^2 + m l^2)\ddot{\theta} + (M l + m l) g \theta = \) - Or: \[ (M + m)l^2\ddot{\theta} + (M + m)l g \theta = \] \[ \ddot{\theta} + \frac{g}{l} \theta = \] --- **Final Linearized Equations (Normal Modes):** \[ \boxed{ \begin{aligned} &\ddot{x} = -g\theta \\ &\ddot{\theta} + \frac{g}{l}\theta = \end{aligned} } \] --- **Summary:** - The normal mode for \(\theta\) is a simple pendulum with frequency \(\omega^2 = \frac{g}{l}\). - The \(x\) coordinate is driven by the tilt \(\theta\). The two are weakly coupled in the small angle/small \(x\) approximation.

## Step 1: Write the Lagrangian Let's define the coordinates for the system: - **\(\theta\)**: angle of the rod with the vertical - **\(x\)**: distance along the horizontal rod between masses \(M\) and \(m\) ### Kinetic Energy 1. **For Mass \(M\)**: - Moves in a circle of radius \(l\); its velocity: \(l\dot{\theta}\). \[ T_M = \frac{1}{2} M (l\dot{\theta})^2 \] 2. **For Mass \(m\)**: - Velocity components: - Along the rod: \(\dot{x}\) - Due to rotation: \((l + x)\dot{\theta}\) (perpendicular to the rod) \[ T_m = \frac{1}{2} m \left[ \dot{x}^2 + ((l+x)\dot{\theta})^2 \right] \] Combining these: \[ T = \frac{1}{2} M l^2 \dot{\theta}^2 + \frac{1}{2} m \left[ \dot{x}^2 + (l + x)^2 \dot{\theta}^2 \right] \] ### Potential Energy 1. **For Mass \(M\)**: - Height: \(-l\cos\theta\) \[ V_M = -M g l \cos\theta \] 2. **For Mass \(m\)**: - Height: \(-(l+x)\cos\theta\) \[ V_m = -m g (l + x) \cos\theta \] Total potential energy: \[ V = -M g l \cos\theta - m g (l + x) \cos\theta \] ### Lagrangian Combining kinetic and potential energies: \[ \mathcal{L} = T - V = \frac{1}{2} (M l^2 + m (l + x)^2) \dot{\theta}^2 + \frac{1}{2} m \dot{x}^2 + (M l + m(l + x))g\cos\theta \] --- **Explanation:** The Lagrangian \(\mathcal{L}\) is expressed in terms of the coordinates \(\theta\) and \(x\), capturing the dynamics of both masses through their kinetic and potential energies. --- ## Step 2: Equations of Motion We will apply the Euler-Lagrange equations for both coordinates \(x\) and \(\theta\). ### For \(x\): The Euler-Lagrange equation is: \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = 0 \] Calculating the components: - \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = m \dot{x}\) - Taking the time derivative: \(\frac{d}{dt} (m \dot{x}) = m \ddot{x}\) - \(\frac{\partial \mathcal{L}}{\partial x} = m (l + x) \dot{\theta}^2 - m g \cos\theta\) Putting it all together: \[ m \ddot{x} - m (l + x) \dot{\theta}^2 + m g \cos\theta = 0 \] This simplifies to: \[ \ddot{x} = (l + x)\dot{\theta}^2 - g\cos\theta \] \[ \boxed{\ddot{x} = (l + x)\dot{\theta}^2 - g\cos\theta} \] --- **Explanation:** The equation for \(x\) shows how the motion is influenced by both the angular motion of the rod and gravitational effects. --- ### For \(\theta\): Using the Euler-Lagrange equation: \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta} = 0 \] Calculating: - \(\frac{\partial \mathcal{L}}{\partial \dot{\theta}} = (M l^2 + m (l + x)^2)\dot{\theta}\) - Taking the derivative: \[ \frac{d}{dt} \left((M l^2 + m (l + x)^2)\dot{\theta}\right) = (M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta} \] - \(\frac{\partial \mathcal{L}}{\partial \theta} = -(M l + m(l + x))g \sin\theta\) Thus, we have: \[ (M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta} + (M l + m(l + x))g\sin\theta = 0 \] --- **Explanation:** This equation for \(\theta\) indicates that the angular motion is coupled to the linear motion of \(x\) and is influenced by gravity. --- ## **Final Answers** ### a) **Equations of Motion:** \[ \boxed{ \begin{aligned} &\ddot{x} = (l + x)\dot{\theta}^2 - g\cos\theta \\ &(M l^2 + m (l + x)^2)\ddot{\theta} + 2m(l + x)\dot{x}\dot{\theta} + (M l + m(l + x))g\sin\theta = 0 \end{aligned} } \] --- ### b) **Normal Modes (when \(x\) and \(\theta\) are small)** For small angles \(x \ll l\) and \(\theta \ll 1\): 1. Approximating \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1\): - The equations simplify to: - For \(x\): \[ \ddot{x} = -g\theta \] 2. For \(\theta\): \[ (M + m)l^2\ddot{\theta} + (M + m)g \theta = 0 \] --- **Final Linearized Equations (Normal Modes):** \[ \boxed{ \begin{aligned} &\ddot{x} = -g\theta \\ &\ddot{\theta} + \frac{g}{l}\theta = 0 \end{aligned} } \] --- **Summary:** - The normal mode for \(\theta\) behaves like a simple pendulum with a frequency \(\omega^2 = \frac{g}{l}\). - The motion of \(x\) is driven by the angle \(\theta\), revealing a coupled system in small-angle approximation.