Let's solve the given problem step by step: --- ## 1) **Equations of Motion Using the Lagrangian Method** ### **System Description** - **\( m_1 \):** Mass of the cart - **\( m_2 \):** Mass of the pendulum - **\( x \):** Displacement of the cart - **\( \theta \):** Angle of pendulum from vertical - **\( k \):** Spring constant - **\( b \):** Damping coefficient - **\( L \):** Length of the pendulum - **\( f \):** External force applied to cart ### **Step 1: Kinetic Energy (T)** - **Cart:** \( T_1 = \frac{1}{2}m_1\dot{x}^2 \) - **Pendulum:** Both translational and rotational motion. The position of \( m_2 \): - \( x_{m_2} = x + L\sin\theta \) - \( y_{m_2} = -L\cos\theta \) Velocities: - \( \dot{x}_{m_2} = \dot{x} + L\dot{\theta}\cos\theta \) - \( \dot{y}_{m_2} = L\dot{\theta}\sin\theta \) So, \[ T_2 = \frac{1}{2}m_2 \left[ (\dot{x} + L\dot{\theta}\cos\theta)^2 + (L\dot{\theta}\sin\theta)^2 \right] \] \[ = \frac{1}{2}m_2 \left[ \dot{x}^2 + 2L\dot{x}\dot{\theta}\cos\theta + L^2\dot{\theta}^2\cos^2\theta + L^2\dot{\theta}^2\sin^2\theta \right] \] \[ = \frac{1}{2}m_2 \left[ \dot{x}^2 + 2L\dot{x}\dot{\theta}\cos\theta + L^2\dot{\theta}^2(\cos^2\theta + \sin^2\theta) \right] \] \[ = \frac{1}{2}m_2 \left[ \dot{x}^2 + 2L\dot{x}\dot{\theta}\cos\theta + L^2\dot{\theta}^2 \right] \] **Total Kinetic Energy:** \[ T = \frac{1}{2}m_1\dot{x}^2 + \frac{1}{2}m_2\left[ \dot{x}^2 + 2L\dot{x}\dot{\theta}\cos\theta + L^2\dot{\theta}^2 \right] \] \[ = \frac{1}{2}(m_1 + m_2)\dot{x}^2 + m_2 L\dot{x}\dot{\theta}\cos\theta + \frac{1}{2}m_2 L^2\dot{\theta}^2 \] ### **Step 2: Potential Energy (V)** - **Spring:** \( V_\text{spring} = \frac{1}{2}kx^2 \) - **Pendulum:** \( V_\text{pendulum} = -m_2gL\cos\theta \) (taking the lowest point as zero potential) \[ V = \frac{1}{2}kx^2 - m_2gL\cos\theta \] ### **Step 3: Lagrangian** \[ \mathcal{L} = T - V \] \[ = \frac{1}{2}(m_1 + m_2)\dot{x}^2 + m_2 L\dot{x}\dot{\theta}\cos\theta + \frac{1}{2}m_2 L^2\dot{\theta}^2 - \frac{1}{2}kx^2 + m_2gL\cos\theta \] --- ### **Step 4: Equations of Motion (Euler-Lagrange Equations)** For \( x \): \[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = Q_x \] where \( Q_x = f - b\dot{x} \) (external force minus damping) For \( \theta \): \[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta} = 0 \] --- #### **For \( x \):** \[ \frac{\partial \mathcal{L}}{\partial \dot{x}} = (m_1 + m_2)\dot{x} + m_2 L\dot{\theta}\cos\theta \] \[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) = (m_1 + m_2)\ddot{x} + m_2 L[\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta] \] \[ \frac{\partial \mathcal{L}}{\partial x} = -kx \] So: \[ (m_1 + m_2)\ddot{x} + m_2 L[\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta] + kx = f - b\dot{x} \] --- #### **For \( \theta \):** \[ \frac{\partial \mathcal{L}}{\partial \dot{\theta}} = m_2 L \dot{x}\cos\theta + m_2 L^2\dot{\theta} \] \[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) = m_2 L \ddot{x}\cos\theta - m_2 L\dot{x}\dot{\theta}\sin\theta + m_2 L^2\ddot{\theta} \] \[ \frac{\partial \mathcal{L}}{\partial \theta} = -m_2 L \dot{x}\dot{\theta}\sin\theta - m_2gL\sin\theta \] So: \[ m_2 L \ddot{x}\cos\theta - m_2 L\dot{x}\dot{\theta}\sin\theta + m_2 L^2\ddot{\theta} + m_2 L \dot{x}\dot{\theta}\sin\theta + m_2gL\sin\theta = 0 \] Simplifies to: \[ m_2 L \ddot{x}\cos\theta + m_2 L^2\ddot{\theta} + m_2gL\sin\theta = 0 \] --- ### **Summary of Equations of Motion:** \[ \boxed{ \begin{align*} &(m_1 + m_2)\ddot{x} + m_2 L[\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta] + kx + b\dot{x} = f \\ &m_2 L \ddot{x}\cos\theta + m_2 L^2\ddot{\theta} + m_2gL\sin\theta = 0 \end{align*} } \] --- ## 2) **Find \( \xi \), \( W_n \), and \( W_d \) for the System** Assume **small angles** (\( \theta \approx 0 \)), so \( \sin\theta \approx \theta \), \( \cos\theta \approx 1 \), and neglect nonlinear terms. For the **cart only** (**if pendulum is not swinging**, i.e., \( m_2 = 0 \)), the system is a **damped mass-spring system**: \[ m_1 \ddot{x} + b\dot{x} + kx = f \] For the **cart plus pendulum mass**, the effective mass is \( m_1 + m_2 \): \[ (m_1 + m_2) \ddot{x} + b\dot{x} + kx = f \] This is a standard **second-order system**: \[ M\ddot{x} + B\dot{x} + Kx = f \] Where \( M = m_1 + m_2 \), \( B = b \), and \( K = k \). - **Damping Ratio (\( \xi \)):** \[ \xi = \frac{B}{2\sqrt{MK}} \] - **Natural Frequency (\( W_n \)):** \[ W_n = \sqrt{\frac{K}{M}} \] - **Damped Natural Frequency (\( W_d \)):** \[ W_d = W_n \sqrt{1 - \xi^2} \] **Plug in the values:** - \( M = m_1 + m_2 \) - \( B = b \) - \( K = k \) --- ## **Final Answers** ### **1. Equations of Motion:** \[ \boxed{ \begin{align*} (m_1 + m_2)\ddot{x} + m_2 L[\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta] + kx + b\dot{x} = f \\ m_2 L \ddot{x}\cos\theta + m_2 L^2\ddot{\theta} + m_2gL\sin\theta = 0 \end{align*} } \] --- ### **2. System Parameters:** \[ \boxed{ \begin{align*} \xi &= \frac{b}{2\sqrt{(m_1 + m_2)k}} \\ W_n &= \sqrt{\frac{k}{m_1 + m_2}} \\ W_d &= W_n \sqrt{1 - \xi^2} \end{align*} } \] --- If you need the equations for the coupled cart-pendulum system (i.e., including \( \theta \)), you would need to consider both equations together. Otherwise, for just the cart, use the simplified forms above.