Let's solve the problem step by step. ### (a) **Lagrangian and Equation of Motion** #### **System description** - A bead of mass \( m \) moves without friction on a hoop of radius \( R \). - The **hoop rotates with constant angular velocity \( \omega \)** about its **vertical diameter** (the \( z \)-axis). - Gravity acts vertically downward. Let’s use an angle \( \theta \): - \( \theta \) is the angle from the lowest point of the hoop to the bead’s position, measured in the plane of the hoop. #### **Coordinates** Let the center of the hoop be at the origin. Let the **vertical axis (z)** be the rotation axis. Let the hoop’s plane be perpendicular to the axis at \( t = 0 \). But since the hoop is rotating, let's describe the position of the bead in terms of \( \theta \) (its position on the hoop) and the time-dependent rotation. - The bead's position in the lab frame is given by: \[ \mathbf{r} = R \left( \sin\theta\,\cos(\omega t),\, \sin\theta\,\sin(\omega t),\, -\cos\theta \right) \] (We take \( \theta = 0 \) at the **bottom** of the hoop.) #### **Kinetic energy** The bead's velocity has two contributions: 1. **Motion along the hoop:** \( \dot{\theta} \) 2. **Motion due to rotation of the hoop about the vertical axis** (\( \omega \)) Let’s calculate the velocity \( \mathbf{v} \): \[ \mathbf{v} = \frac{d\mathbf{r}}{dt} \] There are two terms by the chain rule: \[ \frac{d}{dt}[\text{function of } \theta(t), t] = \frac{\partial \mathbf{r}}{\partial \theta} \dot{\theta} + \frac{\partial \mathbf{r}}{\partial t} \] Let's compute both: - \( \frac{\partial\mathbf{r}}{\partial\theta} = R(\cos\theta \cos(\omega t),\, \cos\theta \sin(\omega t),\, \sin\theta) \) - \( \frac{\partial\mathbf{r}}{\partial t} = R(\sin\theta \cdot -\sin(\omega t) \cdot \omega,\, \sin\theta \cdot \cos(\omega t) \cdot \omega,\, 0) \) \[ = R\omega \sin\theta ( -\sin(\omega t),\ \cos(\omega t),\ 0 ) \] So, \[ \mathbf{v} = R\dot{\theta}(\cos\theta \cos(\omega t),\ \cos\theta \sin(\omega t),\ \sin\theta) + R\omega \sin\theta ( -\sin(\omega t),\ \cos(\omega t),\ 0 ) \] Now, let's compute \( v^2 \): - The two terms are orthogonal except possibly in the \( x \) and \( y \) components, so let's expand: \[ v^2 = |\mathbf{v}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2\mathbf{A}\cdot\mathbf{B} \] where \(\mathbf{A} = R\dot{\theta}(\cos\theta \cos\omega t,\ \cos\theta \sin\omega t,\ \sin\theta)\) and \(\mathbf{B} = R\omega \sin\theta ( -\sin\omega t,\ \cos\omega t,\ 0 )\). Calculate each term: 1. **|\(\mathbf{A}\)|²:** \[ (R\dot{\theta})^2 [\cos^2\theta (\cos^2\omega t + \sin^2\omega t) + \sin^2\theta ] = (R\dot{\theta})^2 [\cos^2\theta (1) + \sin^2\theta ] = (R\dot{\theta})^2 (\cos^2\theta + \sin^2\theta) = (R\dot{\theta})^2 \] 2. **|\(\mathbf{B}\)|²:** \[ (R\omega \sin\theta)^2 [\sin^2\omega t + \cos^2\omega t] = (R\omega \sin\theta)^2 (1) = (R\omega \sin\theta)^2 \] 3. **Cross Term \( 2\mathbf{A}\cdot\mathbf{B} \):** \[ 2R\dot{\theta} \cdot R\omega \sin\theta [\cos\theta \cos\omega t \cdot (-\sin\omega t) + \cos\theta \sin\omega t \cdot \cos\omega t + 0 ] \] \[ = 2R^2\dot{\theta}\omega \sin\theta \cos\theta [ -\cos\omega t \sin\omega t + \sin\omega t \cos\omega t ] = 0 \] (the two terms cancel) So: \[ v^2 = (R\dot{\theta})^2 + (R\omega \sin\theta)^2 \] So, \[ T = \frac{1}{2} m [ (R\dot{\theta})^2 + (R\omega \sin\theta)^2 ] \] #### **Potential energy** The height of the bead is \( z = -R\cos\theta \), so \[ V = mgz = -mgR\cos\theta \] #### **Lagrangian** \[ L = T - V = \frac{1}{2} m [ (R\dot{\theta})^2 + (R\omega \sin\theta)^2 ] + mgR\cos\theta \] --- #### **Equation of motion** Lagrange's equation: \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0 \] Calculate: - \( \frac{\partial L}{\partial \dot{\theta}} = mR^2\dot{\theta} \) - \( \frac{d}{dt}(\ldots) = mR^2\ddot{\theta} \) - \( \frac{\partial L}{\partial \theta} = mR^2\omega^2 \sin\theta \cos\theta - mgR\sin\theta \) So, \[ mR^2\ddot{\theta} - [mR^2\omega^2 \sin\theta \cos\theta - mgR\sin\theta] = 0 \] \[ mR^2\ddot{\theta} - mR^2\omega^2 \sin\theta \cos\theta + mgR\sin\theta = 0 \] Divide by \( mR^2 \): \[ \ddot{\theta} - \omega^2 \sin\theta \cos\theta + \frac{g}{R} \sin\theta = 0 \] \[ \boxed{ \ddot{\theta} + \left( \frac{g}{R} - \omega^2 \cos\theta \right) \sin\theta = 0 } \] --- ### (b) **Critical Angular Velocity \( \Omega \) for Stability at Bottom** The bottom corresponds to \( \theta = 0 \). Stability requires the equilibrium at \( \theta = 0 \) to be stable. Let's linearize around \( \theta = 0 \): - \( \sin\theta \approx \theta \) - \( \cos\theta \approx 1 \) So: \[ \ddot{\theta} + \left( \frac{g}{R} - \omega^2 \right) \theta = 0 \] The equilibrium is **stable** if the coefficient of \( \theta \) is **positive**: \[ \frac{g}{R} - \omega^2 > 0 \implies \omega^2 < \frac{g}{R} \] So, \[ \boxed{ \Omega = \sqrt{\frac{g}{R}} } \] is the **critical angular velocity**. For \( \omega < \Omega \), the bottom is stable; for \( \omega > \Omega \), it is unstable. --- ### (c) **Stable Equilibrium for \( \omega > \Omega \) (i.e., \( \omega > \sqrt{\frac{g}{R}} \))** We look for stationary points of the effective potential, i.e., set \( \ddot{\theta} = 0 \): From the equation of motion: \[ \left( \frac{g}{R} - \omega^2 \cos\theta \right) \sin\theta = 0 \] So, either \( \sin\theta = 0 \) (i.e., \( \theta = 0 \) or \( \pi \)), or \[ \frac{g}{R} - \omega^2 \cos\theta = 0 \implies \cos\theta = \frac{g}{R\omega^2} \] For \( \omega > \Omega \), \( \frac{g}{R\omega^2} < 1 \), so there is a solution. - The equilibrium at \( \theta = 0 \) is unstable for \( \omega > \Omega \). - The new **stable equilibrium** is at \[ \boxed{ \theta_0 = \arccos\left( \frac{g}{R\omega^2} \right) } \] (and symmetrically at \( -\theta_0 \)). This is **above the bottom** of the hoop, forming a "necklace" around the vertical axis. --- ## **Summary** ### (a) **Lagrangian and Equation of Motion** \[ L = \frac{1}{2} m [ (R\dot{\theta})^2 + (R\omega \sin\theta)^2 ] + mgR\cos\theta \] \[ \boxed{ \ddot{\theta} + \left( \frac{g}{R} - \omega^2 \cos\theta \right) \sin\theta = 0 } \] --- ### (b) **Critical Angular Velocity** \[ \boxed{ \Omega = \sqrt{\frac{g}{R}} } \] --- ### (c) **Stable Equilibrium Position for \( \omega > \Omega \)** \[ \boxed{ \theta_0 = \arccos\left( \frac{g}{R\omega^2} \right) } \] ---

Let's carefully analyze and solve the problem step by step. --- ### **Part (a): Set up the Lagrangian and find equations of motion** #### **System Description** - Bead of mass \( m \) constrained to move on a hoop of radius \( R \). - The hoop rotates with **constant angular velocity** \( \omega \) **about a vertical diameter** (the \( z \)-axis). - Gravity acts downward along the \( z \)-axis. --- ### **Choosing coordinates** Let: - \( \theta \) be the angle of the bead measured **from the lowest point** of the hoop (equator), so: - \( \theta = 0 \) at the bottom, - \( \theta = \pi/2 \) at the top. - The hoop rotates around the \( z \)-axis with angular velocity \( \omega \). In the inertial lab frame, the position of the bead at time \( t \): \[ \boxed{ \mathbf{r}(\theta, t) = R \left( \sin\theta \cos(\omega t),\, \sin\theta \sin(\omega t),\, -\cos\theta \right) } \] --- ### **Kinetic energy \( T \)** Compute the velocity \( \mathbf{v} = \frac{d\mathbf{r}}{dt} \): \[ \frac{d}{dt} \left[ R \sin\theta \cos(\omega t) \right] = R \left( \cos\theta\, \dot{\theta} \cos(\omega t) - \sin\theta\, \omega \sin(\omega t) \right) \] Similarly for the \( y \)-component: \[ \frac{d}{dt} \left[ R \sin\theta \sin(\omega t) \right] = R \left( \cos\theta\, \dot{\theta} \sin(\omega t) + \sin\theta\, \omega \cos(\omega t) \right) \] And for \( z \) (which depends only on \( \theta \)): \[ \frac{d}{dt}(- R \cos\theta) = R \sin\theta\, \dot{\theta} \] Now, the velocity components: \[ \mathbf{v} = \left( \frac{d x}{dt},\, \frac{d y}{dt},\, \frac{d z}{dt} \right) \] Calculating \( v^2 = |\mathbf{v}|^2 \): \[ v^2 = \left[ R \left( \cos\theta\, \dot{\theta} \cos(\omega t) - \sin\theta\, \omega \sin(\omega t) \right) \right]^2 + \left[ R \left( \cos\theta\, \dot{\theta} \sin(\omega t) + \sin\theta\, \omega \cos(\omega t) \right) \right]^2 + \left( R \sin\theta\, \dot{\theta} \right)^2 \] Let’s expand: 1. The first two terms: \[ A_x = R \left( \cos\theta\, \dot{\theta} \cos(\omega t) - \sin\theta\, \omega \sin(\omega t) \right) \] \[ A_y = R \left( \cos\theta\, \dot{\theta} \sin(\omega t) + \sin\theta\, \omega \cos(\omega t) \right) \] Calculate \( A_x^2 + A_y^2 \): \[ A_x^2 + A_y^2 = R^2 \left[ \left( \cos\theta\, \dot{\theta} \cos(\omega t) \right)^2 + \left( \sin\theta\, \omega \sin(\omega t) \right)^2 - 2 \cos\theta\, \dot{\theta} \sin\theta\, \omega \cos(\omega t) \sin(\omega t) \right] \] \[ + R^2 \left[ \left( \cos\theta\, \dot{\theta} \sin(\omega t) \right)^2 + \left( \sin\theta\, \omega \cos(\omega t) \right)^2 + 2 \cos\theta\, \dot{\theta} \sin\theta\, \omega \sin(\omega t) \cos(\omega t) \right] \] Adding: - Cross terms cancel because they are equal and opposite. - The remaining terms: \[ A_x^2 + A_y^2 = R^2 \left[ \cos^2\theta\, \dot{\theta}^2 (\cos^2 (\omega t) + \sin^2 (\omega t)) + \sin^2\theta\, \omega^2 (\sin^2 (\omega t) + \cos^2 (\omega t)) \right] \] \[ = R^2 \left[ \cos^2\theta\, \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right] \] Add \( A_z^2 = R^2 \sin^2\theta\, \dot{\theta}^2 \): \[ v^2 = R^2 \left[\cos^2\theta\, \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right] + R^2 \sin^2\theta\, \dot{\theta}^2 \] \[ = R^2 \left[ (\cos^2\theta + \sin^2\theta)\, \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right] \] \[ = R^2 \left[ \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right] \] --- ### **Kinetic energy:** \[ \boxed{ T = \frac{1}{2} m v^2 = \frac{1}{2} m R^2 \left( \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right) } \] --- ### **Potential energy:** \[ V = m g z = - m g R \cos\theta \] --- ### **Lagrangian:** \[ \boxed{ L = T - V = \frac{1}{2} m R^2 \left( \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right) + m g R \cos\theta } \] --- ### **Equation of motion:** Use Euler-Lagrange: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \] Compute derivatives: \[ \frac{\partial L}{\partial \dot{\theta}} = m R^2 \dot{\theta} \] \[ \frac{d}{dt} \left( m R^2 \dot{\theta} \right) = m R^2 \ddot{\theta} \] and \[ \frac{\partial L}{\partial \theta} = m R^2 \sin\theta\, \cos\theta\, \omega^2 - m g R \sin\theta \] Putting into the Euler-Lagrange: \[ m R^2 \ddot{\theta} - \left( m R^2 \sin\theta\, \cos\theta\, \omega^2 - m g R \sin\theta \right) = 0 \] Divide through by \( m R^2 \): \[ \boxed{ \ddot{\theta} + \sin\theta\, \cos\theta\, \omega^2 - \frac{g}{R} \sin\theta = 0 } \] --- ### **Summary for (a):** \[ \boxed{ \textbf{Lagrangian:} \quad L = \frac{1}{2} m R^2 \left( \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right) + m g R \cos\theta } \] \[ \boxed{ \text{Equation of motion:} \quad \ddot{\theta} + \sin\theta\, \cos\theta\, \omega^2 - \frac{g}{R} \sin\theta = 0 } \] --- ### **Part (b): Find the critical angular velocity \( \Omega \) below which the bottom is stable** **At \( \theta = 0 \)** (bottom point), examine the stability. Linearize for small \( \theta \): \[ \sin\theta \approx \theta,\quad \cos\theta \approx 1 \] Equation becomes: \[ \ddot{\theta} + \left( \omega^2 - \frac{g}{R} \right) \theta = 0 \] This is a simple harmonic oscillator with effective "spring constant" \( \left( \frac{g}{R} - \omega^2 \right) \). - **Stable** if this coefficient is positive: \[ \frac{g}{R} - \omega^2 > 0 \implies \omega^2 < \frac{g}{R} \] - **Unstable** if \( \omega^2 > \frac{g}{R} \). Thus, the **critical angular velocity**: \[ \boxed{ \Omega = \sqrt{\frac{g}{R}} } \] - For \( \omega < \Omega \), the bottom is stable. - For \( \omega > \Omega \), the bottom becomes unstable. --- ### **Part (c): Stable equilibrium position for \( \omega > \Omega \)** When \( \omega > \Omega \), the bottom point (\( \theta=0 \)) is unstable. Find equilibrium points where \( \ddot{\theta} = 0 \): \[ \sin\theta \left( \cos\theta\, \omega^2 - \frac{g}{R} \right) = 0 \] Solutions: - \( \sin\theta = 0 \implies \theta = 0, \pi \), i.e., bottom and top points. - Or, \[ \cos\theta = \frac{g}{R \omega^2} \] Since \( \cos\theta \leq 1 \), a real solution exists only if: \[ \frac{g}{R \omega^2} \leq 1 \implies \omega^2 \geq \frac{g}{R} \] which is consistent with the assumption \( \omega > \Omega \). - The point \( \theta = 0 \) (bottom) is **unstable** in this regime. - The **new stable equilibrium** is at: \[ \boxed{ \theta_{s} = \arccos \left( \frac{g}{R \omega^2} \right) } \] which is **away from the bottom**, at a fixed angle determined by \( \omega \). --- ### **Final answers:** **(a)** \[ \boxed{ L = \frac{1}{2} m R^2 \left( \dot{\theta}^2 + \sin^2\theta\, \omega^2 \right) + m g R \cos\theta } \] with equations of motion: \[ \boxed{ \ddot{\theta} + \sin\theta\, \cos\theta\, \omega^2 - \frac{g}{R} \sin\theta = 0 } \] --- **(b)** \[ \boxed{ \text{Critical angular velocity:} \quad \Omega = \sqrt{\frac{g}{R}} } \] below which the bottom (\( \theta=0 \)) is stable. --- **(c)** For \( \omega > \Omega \), the stable equilibrium position is: \[ \boxed{ \theta_{s} = \arccos \left( \frac{g}{R \omega^2} \right) } \] This describes the bead sitting at some angle away from the bottom, stable when \( \omega > \Omega \).