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Chapter 12: Problem 1

A particle of mass \(m\) slides on the inside of a smooth cone of semivertical angle \(\alpha\), whose axis points vertically upwards. Obtain the Hamiltonian function, using the distance \(r\) from the vertex, and the azimuth angle \(\varphi\) as generalized co-ordinates. Show that stable circular motion is possible for any value of \(r\), and determine the corresponding angular velocity, \(\omega\). Find the angle \(\alpha\) if the frequency of small oscillations about this circular motion is also \(\omega\).

Short Answer

Expert verified
Based on the given step by step solution, create a short answer for the problem: To find the Hamiltonian function for a particle of mass m sliding on the inside surface of a smooth cone with semivertical angle α, we first write the expressions for kinetic energy and potential energy, then write the Lagrangian and find the conjugate momenta. Then, we calculate the Hamiltonian and write it in terms of r and φ. To check if stable circular motion is possible, we study the extremum of the Hamiltonian and check if it is a minimum. If it exists, we determine the angle α for the given condition if the frequency of small oscillations about the circular motion is also ω.

Step by step solution

01

1. Write the kinetic energy and potential energy expressions

Given a particle of mass \(m\) sliding on the inside surface of a smooth cone, let's first find its position vector \(\textbf{r}\). In a spherical coordinate system, the position vector can be expressed as: \(\textbf{r} = r\sin\alpha \cos\varphi\,\, \textbf{i} + r\sin\alpha \sin\varphi\,\, \textbf{j} + r\cos\alpha\,\, \textbf{k}\) To find the particle's velocity, we differentiate \(\text(r)\) with respect to \(t\): \(\textbf{v} = \frac{d\textbf{r}}{dt}\) To find the kinetic energy, we use \(T=\frac{1}{2}m\textbf{v}\cdot\textbf{v}\). The potential energy is given by \(V=mgh\), where \(h=r\cos\alpha\) is height.
02

2. Write the Lagrangian and find the conjugate momenta

The Lagrangian is given by the difference between the kinetic energy and potential energy: \(L = T - V\) We now find the conjugate momenta, \(p_r\) and \(p_{\varphi}\), from the Lagrangian: \(p_r = \frac{\partial L}{\partial \dot{r}}\) \(p_{\varphi} = \frac{\partial L}{\partial \dot{\varphi}}\)
03

3. Calculate the Hamiltonian and write it in terms of \(r\) and \(\varphi\)

The Hamiltonian is given by the Legendre transformation: \(H = p_r\dot{r} + p_{\varphi}\dot{\varphi} - L\) Now we should express \(\dot{r}\) and \(\dot{\varphi}\) in terms of \(p_r\) and \(p_{\varphi}\), and substitute these expressions back into the Hamiltonian. The resulting Hamiltonian function should be in terms of \(r\) and \(\varphi\).
04

4. Analyze the stable circular motion condition

To check if stable circular motion is possible for any value of \(r\), we need to study the extremum of the Hamiltonian and check if it is a minimum. To find the extremum, we need to differentiate \(H\) with respect to \(r\) and \(\varphi\), and set the result to 0: \(\frac{\partial H}{\partial r} = 0\) \(\frac{\partial H}{\partial \varphi} = 0\) Solving these equations will give us the condition for circular motion (if it exists). To check whether the solution corresponds to a minimum (stable condition), we take the second derivative of the Hamiltonian, and check if it is positive: \(\frac{\partial^2 H}{\partial r^2} > 0\) \(\frac{\partial^2 H}{\partial \varphi^2} > 0\)
05

5. Determine the angle \(\alpha\) for the given condition

Finally, we need to determine the angle \(\alpha\) if the frequency of small oscillations about the circular motion is also \(\omega\). For that, we use the fact that frequency is given by: \(\omega^2 = \frac{\partial^2 H}{\partial r^2}\) Solving for \(\alpha\), we'll get the required angle in terms of \(\omega\).

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Most popular questions from this chapter

To prove that the effective potential energy function \(U(\theta)\) of the symmetric top (see \(\S\) 12.4) has only a single minimum, show that the equation \(U(\theta)=E\) can be written as a cubic equation in the variable \(z=\cos \theta\), with three roots in general. Show, however, that \(f(z)\) has the same sign at both \(z=\pm 1\), and hence that there are either two roots or none between these points: for every \(E\) there are at most two values of \(\theta\) for which \(U(\theta)=E\)

A particle of mass \(m\) moves in three dimensions under the action of a central, conservative force with potential energy \(V(r)\). Find the Hamiltonian function in terms of spherical polar co-ordinates, and show that \(\varphi\), but \(\operatorname{not} \theta\), is ignorable. Express the quantity \(J^{2}=\) \(m^{2} r^{4}\left(\dot{\theta}^{2}+\sin ^{2} \theta \dot{\varphi}^{2}\right)\) in terms of the generalized momenta, and show that it is a second constant of the motion.

Show that the condition that Hamilton's equations remain unchanged under the transformation generated by \(G\) is \(\mathrm{d} G / \mathrm{d} t=0\) even in the case when \(G\) has an explicit time-dependence, in addition to its dependence via \(q(t)\) and \(p(t)\). Proceed as follows. The first set of Hamilton's equations, \((12.6)\), will be unchanged provided that $$\frac{\mathrm{d}}{\mathrm{d} t}\left(\delta q_{\alpha}\right)=\delta\left(\frac{\partial H}{\partial p_{\alpha}}\right)$$ Write both sides of this equation in terms of \(G\) and use \((12.33)\) applied both to \(\partial G / \partial p_{\alpha}\) and to \(G\) itself to show that it is equivalent to the condition $$\frac{\partial}{\partial p_{\alpha}}\left(\frac{\mathrm{d} G}{\mathrm{~d} t}\right)=0$$ Thus \(\mathrm{d} G / \mathrm{d} t\) is independent of each \(p_{\alpha} .\) Similarly, by using the other set of Hamilton's equations, \((12.7)\), show that it is independent of each \(q_{\alpha}\). Thus \(\mathrm{d} G / \mathrm{d} t\) must be a function of \(t\) alone. But since we can always add to \(G\) any function of \(t\) alone without affecting the transformation it generates, this means we can choose it so that \(\mathrm{d} G / \mathrm{d} t=0\).

To investigate the stability of the motion described in the preceding question, evaluate the second derivatives of \(U\) at \(\rho=a, z=0\), and show that they may be written \(\begin{gathered} \frac{\partial^{2} U}{\partial \rho^{2}}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} \\ \frac{\partial^{2} U}{\partial \rho \partial z}=0, \quad \frac{\partial^{2} U}{\partial z^{2}}=-\frac{q^{2}}{m}\left[B_{z} \rho \frac{\partial B_{z}}{\partial \rho}\right]_{\rho=a, z=0} \end{gathered}\) (Hint: You will need to use the \(\varphi\) component of the equation \(\boldsymbol{\nabla} \wedge \boldsymbol{B}=\mathbf{0}\), and the fact that, since \(B_{\rho}=0\) for all \(\rho, \partial B_{\rho} / \partial \rho=0\) also.) Given that the dependence of \(B_{z}\) on \(\rho\) near the equilibrium orbit is described by \(B_{z} \propto(a / \rho)^{n}\), show that the orbit is stable if \(0

A particle of mass \(m\) and charge \(q\) is moving in the equatorial plane \(z=0\) of a magnetic dipole of moment \(\mu\), described (see Appendix A, Problem 12) by a vector potential with the single non-zero component \(A_{\varphi}=\mu_{0} \mu \sin \theta / 4 \pi r^{2}\). Show that it will continue to move in this plane. Initially, it is approaching from a great distance with velocity \(v\) and impact parameter \(b\), whose sign is defined to be that of \(p_{\varphi}\). Show that \(v\) and \(p_{\varphi}\) are constants of the motion, and that the distance of closest approach to the dipole is \(\frac{1}{2}\left(\sqrt{b^{2} \mp a^{2}} \pm b\right)\), according as \(b>a\) or \(b
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