Let's break down the solution step by step: --- ### **Step 1: System Description and Coordinates** A **spherical pendulum** consists of a mass \( m \) attached to a string of variable length \( l(t) \) and allowed to move in all directions, constrained only by the string. - **Generalized coordinates:** Use spherical polar angles \(\theta\) (polar angle from the vertical) and \(\phi\) (azimuthal angle). The position of the mass in Cartesian coordinates is: \[ \begin{align*} x &= l(t) \sin\theta \cos\phi \\ y &= l(t) \sin\theta \sin\phi \\ z &= -l(t) \cos\theta \end{align*} \] --- ### **Step 2: Kinetic and Potential Energy** #### **Kinetic Energy (T):** The velocity squared is \[ v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \] Calculate derivatives: \[ \begin{align*} \dot{x} &= \dot{l}\sin\theta\cos\phi + l\cos\theta\dot{\theta}\cos\phi - l\sin\theta\sin\phi\dot{\phi} \\ \dot{y} &= \dot{l}\sin\theta\sin\phi + l\cos\theta\dot{\theta}\sin\phi + l\sin\theta\cos\phi\dot{\phi} \\ \dot{z} &= -\dot{l}\cos\theta + l\sin\theta\dot{\theta} \end{align*} \] Plug into \( v^2 \), simplify (using orthogonality of the spherical components): \[ v^2 = \dot{l}^2 + l^2 \dot{\theta}^2 + l^2 \sin^2\theta \dot{\phi}^2 \] So, \[ T = \frac{1}{2}m \left[ \dot{l}^2 + l^2 \dot{\theta}^2 + l^2 \sin^2\theta \dot{\phi}^2 \right] \] But since \( l = l(t) \) is given (not a dynamical variable), \( \dot{l} \) is a known function of time. #### **Potential Energy (V):** \[ V = m g z = -m g l(t) \cos\theta \] --- ### **Step 3: Lagrangian** \[ L = T - V = \frac{1}{2} m \left( \dot{l}^2 + l^2 \dot{\theta}^2 + l^2 \sin^2\theta \dot{\phi}^2 \right) + m g l \cos\theta \] But \( l(t) \) and \( \dot{l}(t) \) are known functions of time, not variables. So, the **dynamical degrees of freedom** are \(\theta\) and \(\phi\). --- ### **Step 4: Equations of Motion (Euler-Lagrange Equations)** #### For \(\theta\): \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \] Calculate: \[ \frac{\partial L}{\partial \dot{\theta}} = m l^2 \dot{\theta} \] \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) = m(2 l \dot{l} \dot{\theta} + l^2 \ddot{\theta}) \] \[ \frac{\partial L}{\partial \theta} = m l^2 \sin\theta \cos\theta \dot{\phi}^2 - m g l \sin\theta \] Plug in: \[ m(2l \dot{l} \dot{\theta} + l^2 \ddot{\theta}) - m l^2 \sin\theta \cos\theta \dot{\phi}^2 + m g l \sin\theta = 0 \] Divide by \(m\) and rearrange: \[ l^2 \ddot{\theta} + 2 l \dot{l} \dot{\theta} - l^2 \sin\theta \cos\theta \dot{\phi}^2 + g l \sin\theta = 0 \] Or, \[ \boxed{ \ddot{\theta} + 2 \frac{\dot{l}}{l} \dot{\theta} - \sin\theta \cos\theta \dot{\phi}^2 + \frac{g}{l} \sin\theta = 0 } \] --- #### For \(\phi\): \[ \frac{\partial L}{\partial \dot{\phi}} = m l^2 \sin^2\theta \dot{\phi} \] \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) = m \left[ 2l \dot{l} \sin^2\theta \dot{\phi} + l^2 2\sin\theta \cos\theta \dot{\theta} \dot{\phi} + l^2 \sin^2\theta \ddot{\phi} \right] \] \[ \frac{\partial L}{\partial \phi} = 0 \] So, \[ 2l\dot{l} \sin^2\theta \dot{\phi} + 2 l^2 \sin\theta\cos\theta \dot{\theta}\dot{\phi} + l^2 \sin^2\theta \ddot{\phi} = 0 \] Divide by \(l^2 \sin^2\theta\) (when \(\sin\theta \neq 0\)): \[ \ddot{\phi} + 2 \frac{\dot{l}}{l} \dot{\phi} + 2 \cot\theta \dot{\theta} \dot{\phi} = 0 \] Or, \[ \boxed{ \ddot{\phi} + 2 \left(\frac{\dot{l}}{l} + \cot\theta \dot{\theta}\right)\dot{\phi} = 0 } \] --- ### **Step 5: Conserved Quantities** - If \(l\) is time-dependent, **energy is not conserved**. - The Lagrangian does **not** depend on \(\phi\), so the **conjugate momentum \(p_\phi\)** is conserved: \[ p_\phi = \frac{\partial L}{\partial \dot{\phi}} = m l^2 \sin^2\theta \dot{\phi} \] So, \[ \boxed{ m l^2 \sin^2\theta \dot{\phi} = \text{constant} } \] This is the **conservation of angular momentum about the vertical axis**. --- ## **Final Answers** ### **Equations of Motion:** \[ \boxed{ \begin{aligned} &\ddot{\theta} + 2 \frac{\dot{l}}{l} \dot{\theta} - \sin\theta \cos\theta \dot{\phi}^2 + \frac{g}{l} \sin\theta = 0 \\ &\ddot{\phi} + 2 \left(\frac{\dot{l}}{l} + \cot\theta \dot{\theta}\right)\dot{\phi} = 0 \end{aligned} } \] ### **Conserved Quantity:** \[ \boxed{ m l^2 \sin^2\theta \dot{\phi} = \text{constant} } \] This is the angular momentum about the vertical axis. --- If you need the case for constant \(l\) or more details, let me know!

Let's carefully derive the equations of motion for a spherical pendulum with a time-dependent string length \( l(t) \). We'll proceed step by step, explaining each part in detail. --- ### **Step 1: System Description and Coordinates** - **Mass:** \( m \) - **String length:** \( l(t) \), a known function of time (never zero) - **Coordinates:** Use spherical coordinates \(\theta(t)\) (polar angle) and \(\phi(t)\) (azimuthal angle). The position of the mass in Cartesian coordinates: \[ \boxed{ \begin{aligned} x &= l(t) \sin\theta(t) \cos\phi(t) \\ y &= l(t) \sin\theta(t) \sin\phi(t) \\ z &= - l(t) \cos\theta(t) \end{aligned} } \] *(The negative sign in \(z\) indicates that the vertical axis points upward; the mass is below the pivot.)* --- ### **Step 2: Velocity Components** Calculate \(\dot{x}\), \(\dot{y}\), \(\dot{z}\): \[ \begin{aligned} \dot{x} &= \frac{d}{dt}[l \sin\theta \cos\phi] \\ &= \dot{l} \sin\theta \cos\phi + l \cos\theta \dot{\theta} \cos\phi - l \sin\theta \sin\phi \dot{\phi} \end{aligned} \] Similarly, \[ \begin{aligned} \dot{y} &= \dot{l} \sin\theta \sin\phi + l \cos\theta \dot{\theta} \sin\phi + l \sin\theta \cos\phi \dot{\phi} \\ \dot{z} &= - \dot{l} \cos\theta + l \sin\theta \dot{\theta} \end{aligned} \] --- ### **Step 3: Kinetic Energy \( T \)** Compute \( v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \). Observe that cross terms cancel due to orthogonality, and after algebraic simplification, the kinetic energy reduces to: \[ \boxed{ T = \frac{1}{2} m \left[ \dot{l}^2 + l^2 \dot{\theta}^2 + l^2 \sin^2\theta \dot{\phi}^2 \right] } \] Note: - \( \dot{l} \) is known (function of time), - \( l(t) \) is a known function. --- ### **Step 4: Potential Energy \( V \)** Gravity acts downward; the height of the mass: \[ z = - l(t) \cos\theta \] Potential energy: \[ \boxed{ V = m g z = - m g l(t) \cos\theta } \] --- ### **Step 5: Write the Lagrangian** Since \( l(t) \) and \( \dot{l}(t) \) are known functions, the **dynamical variables** are \(\theta(t)\) and \(\phi(t)\). The Lagrangian: \[ L = T - V = \frac{1}{2} m \left[ \dot{l}^2 + l^2 \dot{\theta}^2 + l^2 \sin^2\theta \dot{\phi}^2 \right] + m g l(t) \cos\theta \] --- ### **Step 6: Equations of Motion** Use the **Euler-Lagrange equations**: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \] for each coordinate \( q_i \). --- #### **Equation for \(\theta\):** Calculate derivatives: \[ \frac{\partial L}{\partial \dot{\theta}} = m l^2 \dot{\theta} \] \[ \frac{d}{dt} \left( m l^2 \dot{\theta} \right) = m \left[ 2 l \dot{l} \dot{\theta} + l^2 \ddot{\theta} \right] \] \[ \frac{\partial L}{\partial \theta} = m l^2 \sin\theta \cos\theta \dot{\phi}^2 - m g l(t) \sin\theta \] Putting into the Euler-Lagrange: \[ m \left[ 2 l \dot{l} \dot{\theta} + l^2 \ddot{\theta} \right] - \left[ m l^2 \sin\theta \cos\theta \dot{\phi}^2 - m g l(t) \sin\theta \right] = 0 \] Divide through by \( m \): \[ 2 l \dot{l} \dot{\theta} + l^2 \ddot{\theta} - l^2 \sin\theta \cos\theta \dot{\phi}^2 + g l(t) \sin\theta = 0 \] Rearranged: \[ \boxed{ l^2 \ddot{\theta} + 2 l \dot{l} \dot{\theta} + g l(t) \sin\theta - l^2 \sin\theta \cos\theta \dot{\phi}^2 = 0 } \] --- #### **Equation for \(\phi\):** Calculate derivatives: \[ \frac{\partial L}{\partial \dot{\phi}} = m l^2 \sin^2\theta \dot{\phi} \] \[ \frac{d}{dt} \left( m l^2 \sin^2\theta \dot{\phi} \right) = m \left[ 2 l \dot{l} \sin^2\theta \dot{\phi} + 2 l^2 \sin\theta \cos\theta \dot{\theta} \dot{\phi} + l^2 \sin^2\theta \ddot{\phi} \right] \] Since \( L \) doesn't explicitly depend on \(\phi\), the Euler-Lagrange reduces to: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right) = 0 \] which implies: \[ \frac{\partial L}{\partial \dot{\phi}} = \text{constant} \] or \[ \boxed{ m l^2 \sin^2\theta \dot{\phi} = \text{constant} \equiv p_\phi } \] This is the conserved **angular momentum component** about the vertical axis. --- ### **Final equations of motion:** \[ \boxed{ \begin{aligned} &\quad l^2 \ddot{\theta} + 2 l \dot{l} \dot{\theta} + g l(t) \sin\theta - l^2 \sin\theta \cos\theta \dot{\phi}^2 = 0 \\ &\quad \text{with} \quad p_\phi = m l^2 \sin^2\theta \dot{\phi} = \text{constant} \end{aligned} } \] --- ### **Summary of conserved quantities:** - The **azimuthal angular momentum** \( p_\phi = m l^2 \sin^2\theta \dot{\phi} \) is conserved because \( L \) doesn't depend on \(\phi\). - The **total energy** is generally *not* conserved because \( l(t) \) is time-dependent, leading to explicit time dependence in the Lagrangian. --- **Note:** If \( l(t) \) is constant (\( \dot{l} = 0 \)), the equations simplify, and energy is conserved. --- **This completes the step-by-step derivation of the equations of motion for a spherical pendulum with variable string length.**