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Chapter 4: Problem 1

A particle of unit mass moves under gravity on a smooth surface given in cylindrical polar coordinates \(z, r, \theta\) by \(z=f(r)\). Show that the motion is governed by the Lagrangian $$ L=\frac{1}{2} \dot{r}^{2}\left(1+f^{\prime}(r)^{2}\right)+\frac{1}{2} r^{2} \dot{\theta}^{2}-g f(r) . $$ Show that \(\theta\) is an ignorable coordinate. Write down the conserved conjugate momentum and give its physical interpretation.

Short Answer

Expert verified
The conserved conjugate momentum in this problem is \(p_\theta = r^2\dot{\theta}\). It represents the angular momentum of the particle in the plane of motion (r, \(\theta\)).

Step by step solution

01

Kinetic energy in cylindrical coordinates

First, the velocity components in cylindrical coordinates are $$ v_r = \dot{r},\quad v_\theta = r\dot{\theta},\quad v_z = \frac{dz}{dt} = \frac{df(r)}{dr} \dot{r} = f^{\prime}(r) \dot{r}. $$ The total kinetic energy (T) of a particle of unit mass is given by $$ T = \frac{1}{2} (v_r^2 + v_\theta^2 + v_z^2). $$ Substitute the velocity components found above into the kinetic energy expression $$ T = \frac{1}{2} \left(\dot{r}^2 + r^2\dot{\theta}^2 + f^{\prime}(r)^2\dot{r}^2\right). $$
02

Potential energy in cylindrical coordinates

The potential energy (V) of a particle of unit mass in a gravitational field is given by $$ V = -mgz = -mg f(r), $$ where \(g\) is the acceleration due to gravity.
03

Write the Lagrangian

The Lagrangian (L) of a system is defined as the difference between the kinetic and potential energies $$ L = T - V $$ Substitute the kinetic and potential energies found in steps 1 and 2 into the Lagrangian definition $$ L = \frac{1}{2} \dot{r}^{2}\left(1+f^{\prime}(r)^{2}\right)+\frac{1}{2} r^{2} \dot{\theta}^{2}-g f(r) . $$ This matches the given expression for the Lagrangian, as required.
04

Show that \(\theta\) is an ignorable coordinate

The coordinate \(\theta\) is ignorable if the Lagrangian is independent of \(\theta\). Taking the partial derivative of the Lagrangian with respect to \(\theta\) gives $$ \frac{\partial L}{\partial\theta} = 0, $$ so \(\theta\) is an ignorable coordinate.
05

Calculate the conserved conjugate momentum and discuss the physical interpretation

Since \(\theta\) is an ignorable coordinate, the conjugate momentum \(p_\theta\) is conserved. The conjugate momentum is defined as $$ p_\theta = \frac{\partial L}{\partial \dot{\theta}}. $$ Differentiating the Lagrangian with respect to \(\dot{\theta}\) we get $$ p_\theta = r^2\dot{\theta}. $$ The conserved conjugate momentum \(p_\theta\) represents the angular momentum of the particle in the plane of motion. This result is expected, as the motion is happening on a smooth surface, and the effect of gravity is entirely along the \(z\)-axis, leading to conservation of angular momentum in the plane (r, \(\theta\)).

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Most popular questions from this chapter

An astronaut is floating in empty space at rest relative to an inertial frame with her arms by her side. Explain how it is that by waving her arms and then returning them to their original position, she can rotate her body, but cannot move her centre of mass.

A system of particles with masses \(m_{\alpha}\) moves in a gravitational field \(\boldsymbol{g}\). Show that if \(\boldsymbol{F}_{\alpha}=-m_{\alpha} \boldsymbol{g}\) where \(\boldsymbol{g}\) is constant, then \(\sum_{\alpha} \boldsymbol{F}_{\alpha}=m \boldsymbol{g} \quad\) and \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}=m \boldsymbol{c} \wedge \boldsymbol{g}\), where \(c\) is the position vector of the centre of mass and \(m\) is the total mass. Deduce that the effect of a uniform gravitational fieldon the total linear momentum and angular momentum is the same as that of a single force \(m g\) acting through the centre of mass. Show by counter-example that this is not true for a non-uniform field.

Let \(\rho_{s}: C \times \mathbb{R} \rightarrow C \times \mathbb{R}\) be a dynamical symmetry of a system with Lagrangian \(L\), with generators \(u_{a}\). Show that under the action of \(\rho_{s}\) for small \(s\), the change in \(\partial L / \partial v_{a}\) along a kinematic trajectory is $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=s u_{b} \frac{\partial^{2} L}{\partial v_{a} \partial q_{b}}+s w_{b} \frac{\partial^{2} L}{\partial v_{a} \partial v_{b}} $$ to the first order in \(s\), where $$ w_{a}=\frac{\partial u_{a}}{\partial q_{b}} v_{b}+\frac{\partial u_{a}}{\partial t} $$ Deduce that $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ and $$ \delta\left(\frac{\partial L}{\partial q_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial q_{b}}-s \frac{\partial w_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ Show that $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial u_{b}}{\partial q_{a}}\right)=\frac{\partial w_{b}}{\partial q_{a}} $$ Hence show that if \(q_{a}=q_{a}(t)\) is a solution of Lagrange's equations, then so is \(q_{a}=q_{a}(t)+s u_{a}(q(t), t)\), to the first order in \(s\). Deduce that if \(\rho_{s}\) is a dynamical symmetry, then it maps dynamical trajectories in \(C \times \mathbb{R}\) to dynamical trajectories.

The configuration space for a particle of mass \(m\) moving in space is Euclidean space, with Cartesian coordinates \(x, y, z\). Show that the generator of the one-parameter group of rotations about the \(x\)-axis is $$ u=(0,-z, y) \text {. } $$ Hence write down an expression for the \(x\)-component of angular momentum in terms of the spherical polar coordinates defined by $$ \begin{aligned} &x=r \sin \theta \cos \varphi \\ &y=r \sin \theta \sin \varphi \\ &z=r \cos \theta . \end{aligned} $$
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