Chapter 13, Problem 13.17P

(a)

To determine

The corresponding value of the height $z_0$

Answer to Problem 13.17P

The corresponding value of the height is $z_0={\left(\frac{p_{\varphi }{}^2}{m^2{c}^{2}g}\right)}^{\frac{1}{3}}$

Explanation of Solution

Write the Hamiltonian equation for mass on a cone

  $\begin{array}{c}H=T+U\\ =\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+\frac{p_{\varphi }{}^2}{c^2{z}^2}\right)+mgz\end{array}$

Write the Hamiltonian equation with respect to $p_z$ is,

  $\begin{array}{c}\dot{z}=\frac{\partial H}{\partial p_z}\\ =\frac{\partial }{\partial p_z}\left(\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+\frac{p_{\varphi }{}^2}{c^2{z}^2}\right)+mgz\right)\\ =\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+0\right)+0\\ =\frac{p_z{}^2}{m\left(c^2+1\right)}\end{array}$        (I)

The Hamiltonian equation with respect to $p_{\varphi }$ is,

  $\begin{array}{c}\dot{\varphi }=\frac{\partial H}{\partial p_{\varphi }}\\ =\frac{\partial }{\partial p_{\varphi }}\left(\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+\frac{p_{\varphi }{}^2}{c^2{z}^2}\right)+mgz\right)\\ =\frac{1}{2m}\left(0+\frac{2p_{\varphi }{}^2}{c^2{z}^2}\right)+0\\ =\frac{p_{\varphi }{}^2}{m\left(c^2{z}^2\right)}\end{array}$        (II)

The Hamiltonian equation for generalized coordinate $z$ is,

  $\begin{array}{c}{\dot{p}}_z=-\frac{\partial H}{\partial z}\\ =-\frac{\partial }{\partial z}\left(\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+\frac{p_{\varphi }{}^2}{c^2{z}^2}\right)+mgz\right)\\ =\frac{1}{2m}\left(0+\frac{-2p_{\varphi }{}^2}{c^2{z}^3}\right)+mg\\ =\frac{p_{\varphi }{}^2}{m\left(c^2{z}^3\right)}-mg\end{array}$        (III)

The Hamiltonian equation for generalized coordinate $\varphi$ is,

  $\begin{array}{c}{\dot{p}}_z=-\frac{\partial H}{\partial \varphi }\\ =-\frac{\partial }{\partial \varphi }\left(\frac{1}{2m}\left(\frac{p_z{}^2}{c^2+1}+\frac{p_{\varphi }{}^2}{c^2{z}^2}\right)+mgz\right)\\ =0\end{array}$        (IV)

Conclusion:

Write the fixed height of mass $m$ is given by

  $z=z_0$        (V)

Hence, the value of $\dot{z}$ is,

    $\dot{z}=z_0$

  $\begin{array}{c}{p}_z=m\dot{z}\\ =m\left(0\right)\\ =0\end{array}$        (VI)

The value of ${\dot{p}}_z$ is also zero. Thus, the equation (III) can be written as,

  $0=\frac{p_{\varphi }{}^2}{m\left(c^2{z}_0{}^3\right)}-mg$        (VII)

Write the expression (VII) in terms of $z_0$

  $\begin{array}{c}{z}_0{}^3=\frac{p_{\varphi }{}^2}{m\left(c^{2}mg\right)}\\ z_0={\left(\frac{p_{\varphi }{}^2}{m\left(c^{2}mg\right)}\right)}^{\frac{1}{3}}\\ z_0={\left(\frac{p_{\varphi }{}^2}{m^2{c}^{2}g}\right)}^{\frac{1}{3}}\end{array}$

(b)

To determine

The $\epsilon$ will oscillate about zero.

Answer to Problem 13.17P

The $\epsilon$ will oscillate about zero when the orbit has, $z=z_0+\epsilon$ with $\epsilon$ small.

Explanation of Solution

Write the Hamiltonian equation with respect to $p_z$ is,

  $\dot{z}=\frac{p_z{}^2}{m\left(c^2+1\right)}$        (I)

Differentiate the equation on both sides,

  $\ddot{z}=\frac{{\dot{p}}_z}{m\left(c^2+1\right)}$        (II)

Substitute $\frac{p_{\varphi }{}^2}{m\left(c^2{z}^3\right)}-mg$ for ${\dot{p}}_z$, and $z_0+\epsilon$ for $z$ in expression (II)

  $\begin{array}{c}\ddot{z}=\frac{{\dot{p}}_z}{m\left(c^2+1\right)}\left(\frac{p_{\varphi }{}^2}{m{c}^2\left(z^3\right)}-mg\right)\\ \ddot{z}=\frac{{\dot{p}}_z}{m\left(c^2+1\right)}\left(\frac{p_{\varphi }{}^2}{m{c}^2{\left(z_0+\epsilon \right)}^3}-mg\right)\\ \ddot{z}=\frac{{\dot{p}}_z}{m\left(c^2+1\right)}\left(\frac{p_{\varphi }{}^2}{m{c}^2{\left(1+\frac{\epsilon }{z_0}\right)}^3}-mg\right)\\ \ddot{z}=\frac{1}{m\left(c^2+1\right)}\left(\frac{p_{\varphi }{}^2}{m{c}^2{z}_0{}^3}{\left(1+\frac{\epsilon }{z_0}\right)}^{-3}-mg\right)\end{array}$        (II)

Conclusion:

Re-write the expression (II) by using binomial theorem to reduce the term ${\left(1+\frac{\epsilon }{z_0}\right)}^{-3}$ as $\left(1-\frac{3\epsilon }{z_0}\right)$

  $\begin{array}{c}\ddot{z}=\frac{1}{m\left(c^2+1\right)}\left(\frac{p_{\varphi }{}^2}{m{c}^2{z}_0{}^3}\left(1-\frac{3\epsilon }{z_0}\right)-mg\right)\\ \ddot{z}=\frac{p_{\varphi }{}^2}{m^2{c}^2\left(c^2+1\right)z_0{}^3}-\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}-\frac{g}{c^2+1}\\ =\left(\frac{p_{\varphi }{}^2}{m^2{c}^2\left(c^2+1\right)z_0{}^3}-\frac{g}{c^2+1}\right)-\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}\end{array}$        (III)

Write the fixed height of mass $m$ is given by

  $z=z_0+\epsilon$        (IV)

Differentiate the equation on both the sides,

    $\begin{array}{c}\dot{z}=0+\dot{\epsilon }\\ =\dot{\epsilon }\end{array}$

Similarly differentiate the equation on both the sides,

  $\ddot{z}=\ddot{\epsilon }$        (V)

Substitute expression (III) in above expression (V),

  $\ddot{\epsilon }=\left(\frac{p_{\varphi }{}^2}{m^2{c}^2\left(c^2+1\right)z_0{}^3}-\frac{g}{c^2+1}\right)-\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}$$\begin{array}{l}H=\frac{p^2}{2m}+\frac{1}{2}k{q}^2\\ H=\frac{1}{2}\left(q^2+p^2\right)\end{array}$

Here, neglect the term $\frac{p_{\varphi }{}^2}{m^2{c}^2\left(c^2+1\right)z_0{}^3}-\frac{g}{c^2+1}$ in the above equation,

  $\ddot{\epsilon }=-\left(\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}\right)\epsilon$        (VI)

Therefore, the above result indicates that the $\epsilon$ will oscillate about zero when the orbit has  $z=z_0+\epsilon$ with $\epsilon$ small.

(c)

To determine

The angular frequency of the oscillations.

Answer to Problem 13.17P

The angular frequency of the oscillations $\omega =\sqrt{3}\dot{\varphi }\sin \alpha$

Explanation of Solution

Write the expression for $\ddot{\epsilon }$ from subpart (b),

  $\begin{array}{l}\ddot{\epsilon }=-\left(\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}\right)\epsilon \\ \frac{\ddot{\epsilon }}{\epsilon }=-\left(\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}\right)\end{array}$        (I)

Write the relation between normal frequencies and $\ddot{\epsilon }$ is,

  ${\ddot{\epsilon }}_i\left(t\right)=-{\omega }_i{}^2{\epsilon }_i\left(t\right)$

Write the frequency of the system is,

  $\begin{array}{c}{\omega }^2=-\frac{\ddot{\epsilon }}{\epsilon }\\ =-\left(\frac{3p_{\varphi }{}^2\epsilon }{m^2{c}^2\left(c^2+1\right)z_0{}^4}\right)\\ =3\left(\frac{p_{\varphi }}{m{c}^2{z}_0{}^2}\right)\left(\frac{c^2}{c^2+1}\right)\end{array}$        (II)

Conclusion:

Substitute $\tan \alpha$ for $c$, in expression (II)

  $\begin{array}{c}{\omega }^2=3\left(\frac{p_{\varphi }}{m{c}^2{z}_0{}^2}\right)\left(\frac{\tan {\alpha }^2}{{\tan }^2\alpha +1}\right)\\ {\omega }^2=3\left(\frac{p_{\varphi }}{m{c}^2{z}_0{}^2}\right)\left(\frac{\frac{\sin {\alpha }^2}{\cos {\alpha }^2}}{\frac{1}{\cos {\alpha }^2}}\right)\\ {\omega }^2=3\left(\frac{p_{\varphi }}{m{c}^2{z}_0{}^2}\right)\sin {\alpha }^2\end{array}$        (III)

Substitute $\dot{\varphi }$ for $\frac{p_{\varphi }}{m{c}^2{z}^2}$, in expression (III)

  $\begin{array}{c}{\omega }^2=3\left(\frac{p_{\varphi }}{m{c}^2{z}_0{}^2}\right)\sin {\alpha }^2\\ {\omega }^2=3\left({\dot{\varphi }}^2\right){\sin }^2\alpha \\ =\sqrt{3}\dot{\varphi }\sin \alpha \end{array}$

The angular frequency of the oscillations $\omega =\sqrt{3}\dot{\varphi }\sin \alpha$

(d)

To determine

The value for $\alpha$.

Answer to Problem 13.17P

The value for $\alpha$ is $35.3°$

Explanation of Solution

Write the expression for the angular frequency of the oscillations

  $\omega =\sqrt{3}\dot{\varphi }\sin \alpha$        (I)

Here, it is given that the angular frequency of the oscillation is equal to the orbital angular velocity,

  $\omega =\dot{\varphi }$

Conclusion:

Therefore the angular frequency equation can be re-written as,

  $\begin{array}{c}\dot{\varphi }=\sqrt{3}\dot{\varphi }\sin \alpha \\ 1=\sqrt{3}\sin \alpha \\ \alpha ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ =35.26°\end{array}$

The value for $\alpha$ is $35.3°$. Here, the height $z$ returns to its initial value for one complete circle. Thus it means the orbital must be closed one.