# Step-by-Step Solution Using Lagrange's Equations We are to derive the equations of motion for the given system using **Lagrange's equations** with generalized coordinates \( x \) (horizontal displacement of the mass \( m \)) and \( \theta \) (angle of the slender bar). --- ## **1. Identify Generalized Coordinates** - \( x \): Displacement of mass \( m \) - \( \theta \): Angular displacement of the slender bar --- ## **2. Kinetic Energy (\( T \))** ### **A. Mass \( m \) (Slider)** - Moves horizontally: velocity \( \dot{x} \) - Kinetic energy: \[ T_1 = \frac{1}{2} m \dot{x}^2 \] ### **B. Slender Bar (mass \( 2m \), length \( L \))** - The bar's center of mass has both translational and rotational motion. #### **i. Position of Center of Mass (COM)** - Relative to the slot: \( (x, ) \) - COM position: \[ x_{com} = x + \frac{L}{2} \sin\theta \] \[ y_{com} = -\frac{L}{2} \cos\theta \] #### **ii. Velocity of COM** \[ \dot{x}_{com} = \dot{x} + \frac{L}{2} \dot{\theta} \cos\theta \] \[ \dot{y}_{com} = \frac{L}{2} \dot{\theta} \sin\theta \] \[ v_{com}^2 = (\dot{x}_{com})^2 + (\dot{y}_{com})^2 = \left(\dot{x} + \frac{L}{2} \dot{\theta} \cos\theta\right)^2 + \left(\frac{L}{2} \dot{\theta} \sin\theta\right)^2 \] \[ = \dot{x}^2 + L \dot{x} \dot{\theta} \cos\theta + \frac{L^2}{4} \dot{\theta}^2 (\cos^2\theta + \sin^2\theta) \] But \( \cos^2\theta + \sin^2\theta = 1 \): \[ v_{com}^2 = \dot{x}^2 + L \dot{x} \dot{\theta} \cos\theta + \frac{L^2}{4} \dot{\theta}^2 \] #### **iii. Rotational Kinetic Energy about COM** \[ I_{com} = \frac{1}{12} (2m) L^2 = \frac{1}{6} m L^2 \] \[ T_{rot} = \frac{1}{2} I_{com} \dot{\theta}^2 = \frac{1}{12} m L^2 \dot{\theta}^2 \] #### **iv. Total Kinetic Energy of Bar** \[ T_2 = \frac{1}{2} (2m) v_{com}^2 + \frac{1}{2} I_{com} \dot{\theta}^2 \] \[ = m v_{com}^2 + \frac{1}{12} m L^2 \dot{\theta}^2 \] \[ = m \left(\dot{x}^2 + L \dot{x} \dot{\theta} \cos\theta + \frac{L^2}{4} \dot{\theta}^2 \right) + \frac{1}{12} m L^2 \dot{\theta}^2 \] \[ = m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \frac{L^2}{4} \dot{\theta}^2 + \frac{1}{12} m L^2 \dot{\theta}^2 \] \[ = m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \left(\frac{1}{4} + \frac{1}{12}\right) L^2 \dot{\theta}^2 \] \[ = m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \frac{1}{3} L^2 \dot{\theta}^2 \] ### **C. Total Kinetic Energy** \[ T = T_1 + T_2 = \frac{1}{2} m \dot{x}^2 + m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \frac{1}{3} L^2 \dot{\theta}^2 \] \[ = \frac{3}{2} m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + \frac{2}{3} m L^2 \dot{\theta}^2 \] --- ## **3. Potential Energy (\( V \))** ### **A. Springs** - Left and right springs (each \( k \)), attached to mass \( m \): \[ V_1 = \frac{1}{2}k x^2 + \frac{1}{2}k x^2 = kx^2 \] - Spring attached to the bar (2\( k \)), extension is \( y_{end} = x + L \sin\theta \): \[ V_2 = \frac{1}{2} (2k) (x + L \sin\theta)^2 = k(x + L \sin\theta)^2 \] \[ V = kx^2 + k(x + L \sin\theta)^2 \] --- ## **4. Lagrangian (\( L \))** \[ L = T - V \] \[ = \left[\frac{3}{2} m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + \frac{2}{3} m L^2 \dot{\theta}^2 \right] - [kx^2 + k(x + L \sin\theta)^2] \] --- ## **5. Lagrange’s Equations** For each coordinate \( q_i \): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = \] Let’s compute for \( x \) and \( \theta \): --- ### **A. For \( x \):** #### **i. Compute \( \frac{\partial L}{\partial \dot{x}} \):** \[ \frac{\partial L}{\partial \dot{x}} = 3m \dot{x} + m L \dot{\theta} \cos\theta \] #### **ii. Compute \( \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) \):** \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) = 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta \] (using product rule on \( m L \dot{\theta} \cos\theta \)) #### **iii. Compute \( \frac{\partial L}{\partial x} \):** \[ \frac{\partial L}{\partial x} = -2k x - 2k(x + L \sin\theta) = -2k x - 2k x - 2k L \sin\theta = -2k x - 2k L \sin\theta \] #### **iv. Lagrange’s equation for \( x \):** \[ 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 2k x + 2k (x + L \sin\theta) = \] \[ 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 2k x + 2k x + 2k L \sin\theta = \] \[ 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 4k x + 2k L \sin\theta = \] --- ### **B. For \( \theta \):** #### **i. Compute \( \frac{\partial L}{\partial \dot{\theta}} \):** \[ \frac{\partial L}{\partial \dot{\theta}} = m L \dot{x} \cos\theta + \frac{4}{3} m L^2 \dot{\theta} \] #### **ii. Compute \( \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) \):** \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) = m L \ddot{x} \cos\theta - m L \dot{x} \dot{\theta} \sin\theta + \frac{4}{3} m L^2 \ddot{\theta} \] #### **iii. Compute \( \frac{\partial L}{\partial \theta} \):** \[ \frac{\partial L}{\partial \theta} = - m L \dot{x} \dot{\theta} \sin\theta - 2k L (x + L \sin\theta) \cos\theta \] #### **iv. Lagrange’s equation for \( \theta \):** \[ m L \ddot{x} \cos\theta - m L \dot{x} \dot{\theta} \sin\theta + \frac{4}{3} m L^2 \ddot{\theta} + m L \dot{x} \dot{\theta} \sin\theta + 2k L (x + L \sin\theta) \cos\theta = \] \[ m L \ddot{x} \cos\theta + \frac{4}{3} m L^2 \ddot{\theta} + 2k L (x + L \sin\theta) \cos\theta = \] --- ## **6. Final Equations of Motion** \[ \boxed{ \begin{aligned} &3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 4k x + 2k L \sin\theta = \\ & m L \ddot{x} \cos\theta + \frac{4}{3} m L^2 \ddot{\theta} + 2k L (x + L \sin\theta) \cos\theta = \end{aligned} } \] --- ### **Summary** - These are the coupled nonlinear differential equations (in \( x \) and \( \theta \)) describing the dynamics of the system using Lagrange's equations, as required.

# Step-by-Step Solution Using Lagrange's Equations We will derive the equations of motion for the given system using **Lagrange's equations** with generalized coordinates \( x \) (horizontal displacement of the mass \( m \)) and \( \theta \) (angle of the slender bar). --- ## 1. Identify Generalized Coordinates - **\( x \)**: Displacement of mass \( m \) - **\( \theta \)**: Angular displacement of the slender bar --- ## 2. Kinetic Energy (\( T \)) ### A. Mass \( m \) (Slider) - Moves horizontally: velocity \( \dot{x} \) - Kinetic energy: \[ T_1 = \frac{1}{2} m \dot{x}^2 \] ### B. Slender Bar (mass \( 2m \), length \( L \)) - The bar's center of mass has both translational and rotational motion. #### i. Position of Center of Mass (COM) - Relative to the slot: \( (x, 0) \) - COM position: \[ x_{com} = x + \frac{L}{2} \sin\theta \] \[ y_{com} = -\frac{L}{2} \cos\theta \] #### ii. Velocity of COM \[ \dot{x}_{com} = \dot{x} + \frac{L}{2} \dot{\theta} \cos\theta \] \[ \dot{y}_{com} = \frac{L}{2} \dot{\theta} \sin\theta \] \[ v_{com}^2 = \left(\dot{x} + \frac{L}{2} \dot{\theta} \cos\theta\right)^2 + \left(\frac{L}{2} \dot{\theta} \sin\theta\right)^2 \] \[ = \dot{x}^2 + L \dot{x} \dot{\theta} \cos\theta + \frac{L^2}{4} \dot{\theta}^2 \] #### iii. Rotational Kinetic Energy about COM \[ I_{com} = \frac{1}{12} (2m) L^2 = \frac{1}{6} m L^2 \] \[ T_{rot} = \frac{1}{2} I_{com} \dot{\theta}^2 = \frac{1}{12} m L^2 \dot{\theta}^2 \] #### iv. Total Kinetic Energy of Bar \[ T_2 = m v_{com}^2 + \frac{1}{12} m L^2 \dot{\theta}^2 \] \[ = m \left(\dot{x}^2 + L \dot{x} \dot{\theta} \cos\theta + \frac{L^2}{4} \dot{\theta}^2 \right) + \frac{1}{12} m L^2 \dot{\theta}^2 \] Combining terms: \[ T_2 = m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \left(\frac{1}{4} + \frac{1}{12}\right) L^2 \dot{\theta}^2 \] \[ = m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \frac{1}{3} L^2 \dot{\theta}^2 \] ### C. Total Kinetic Energy \[ T = T_1 + T_2 = \frac{1}{2} m \dot{x}^2 + m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + m \frac{1}{3} L^2 \dot{\theta}^2 \] \[ = \frac{3}{2} m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + \frac{2}{3} m L^2 \dot{\theta}^2 \] --- ## 3. Potential Energy (\( V \)) ### A. Springs - Left and right springs (each \( k \)), attached to mass \( m \): \[ V_1 = kx^2 \] - Spring attached to the bar (2\( k \)), extension is \( y_{end} = x + L \sin\theta \): \[ V_2 = (k)(x + L \sin\theta)^2 \] Total potential energy: \[ V = kx^2 + k(x + L \sin\theta)^2 \] --- ## 4. Lagrangian (\( L \)) \[ L = T - V \] \[ = \left[\frac{3}{2} m \dot{x}^2 + m L \dot{x} \dot{\theta} \cos\theta + \frac{2}{3} m L^2 \dot{\theta}^2 \right] - [kx^2 + k(x + L \sin\theta)^2] \] --- ## 5. Lagrange’s Equations For each coordinate \( q_i \): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0 \] Let’s compute for \( x \) and \( \theta \): ### A. For \( x \): #### i. Compute \( \frac{\partial L}{\partial \dot{x}} \): \[ \frac{\partial L}{\partial \dot{x}} = 3m \dot{x} + m L \dot{\theta} \cos\theta \] #### ii. Compute \( \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) \): \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) = 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta \] #### iii. Compute \( \frac{\partial L}{\partial x} \): \[ \frac{\partial L}{\partial x} = -2kx - 2k(x + L \sin\theta) = -4kx - 2kL \sin\theta \] #### iv. Lagrange’s equation for \( x \): \[ 3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 4k x + 2kL \sin\theta = 0 \] --- ### B. For \( \theta \): #### i. Compute \( \frac{\partial L}{\partial \dot{\theta}} \): \[ \frac{\partial L}{\partial \dot{\theta}} = m L \dot{x} \cos\theta + \frac{4}{3} m L^2 \dot{\theta} \] #### ii. Compute \( \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) \): \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}}\right) = m L \ddot{x} \cos\theta - m L \dot{x} \dot{\theta} \sin\theta + \frac{4}{3} m L^2 \ddot{\theta} \] #### iii. Compute \( \frac{\partial L}{\partial \theta} \): \[ \frac{\partial L}{\partial \theta} = -m L \dot{x} \dot{\theta} \sin\theta - 2kL (x + L \sin\theta) \cos\theta \] #### iv. Lagrange’s equation for \( \theta \): \[ m L \ddot{x} \cos\theta - m L \dot{x} \dot{\theta} \sin\theta + \frac{4}{3} m L^2 \ddot{\theta} + 2kL (x + L \sin\theta) \cos\theta = 0 \] --- ## 6. Final Equations of Motion The final equations of motion are: \[ \boxed{ \begin{aligned} &3m \ddot{x} + m L \ddot{\theta} \cos\theta - m L \dot{\theta}^2 \sin\theta + 4k x + 2k L \sin\theta = 0 \\ &\quad \\ &m L \ddot{x} \cos\theta + \frac{4}{3} m L^2 \ddot{\theta} + 2k L (x + L \sin\theta) \cos\theta = 0 \end{aligned} } \] --- ### Summary These are the coupled nonlinear differential equations (in \( x \) and \( \theta \)) describing the dynamics of the system using Lagrange's equations, as required.