Chapter 13, Problem 13.21P

(a)

To determine

The Hamiltonian using generalized coordinates $X,Y,r,\varphi$.

Answer to Problem 13.21P

The Hamiltonian is $H=\frac{p_x{}^2+p_y{}^2}{2M}+\frac{1}{2\mu }\left(p_r{}^2+\frac{p_{\varphi }{}^2}{r^2}\right)+\frac{1}{2}k{\left(r-l_0\right)}^2$

Explanation of Solution

Write the kinetic energy of the system

  $T=\frac{1}{2}M{\dot{R}}^2+\frac{1}{2}\mu {\dot{r}}^2$

Here, $R$ is the position of the center of the mass, and $r$ the relative position of theses mass

Write the potential energy of the system

  $U=\frac{1}{2}k{\left(r-l_0\right)}^2$        (I)

Here, $k$ is the spring constant.

The lagrangian of the spring – mass system is,

  $\begin{array}{c}L=T-U\\ =\frac{1}{2}M{\dot{R}}^2+\frac{1}{2}\mu {\dot{r}}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)\end{array}$        (II)

Write the center of mass position for rectangular coordinates,

  ${\dot{R}}^2=X^2+Y^2$        (III)

Write the center of mass position for rectangular coordinates,

  ${\dot{r}}^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$        (IV)

Substitute expression (III) and (IV) in equation (II) and solve for $L$ 

  $L=\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2\right)}^2+\frac{1}{2}\mu {\left({\dot{r}}^2+r^2{\dot{\varphi }}^2\right)}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)$        (V)

Write the Canonical momenta along the $X$ coordinates,

  $\begin{array}{c}{p}_x=\frac{\partial L}{\partial \dot{X}}\\ =-\frac{\partial }{\partial \dot{X}}\left(\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2\right)}^2+\frac{1}{2}\mu {\left({\dot{r}}^2+r^2{\dot{\varphi }}^2\right)}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)\right)\\ =M\dot{X}\\ \dot{X}=\frac{p_x}{M}\end{array}$        (VI)

Write the Canonical momenta along the $Y$ coordinates,

  $\begin{array}{c}{p}_y=\frac{\partial L}{\partial \dot{Y}}\\ =-\frac{\partial }{\partial \dot{Y}}\left(\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2\right)}^2+\frac{1}{2}\mu {\left({\dot{r}}^2+r^2{\dot{\varphi }}^2\right)}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)\right)\\ =M\dot{Y}\\ \dot{Y}=\frac{p_y}{M}\end{array}$        (VII)

Write the canonical velocity is,

  $\begin{array}{c}{p}_r=\frac{\partial L}{\partial \dot{r}}\\ =-\frac{\partial }{\partial \dot{r}}\left(\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2\right)}^2+\frac{1}{2}\mu {\left({\dot{r}}^2+r^2{\dot{\varphi }}^2\right)}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)\right)\\ =\mu \dot{r}\\ \dot{r}=\frac{p_r}{\mu }\end{array}$        (VII)

Write the another canonical velocity is,

  $\begin{array}{c}{p}_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}\\ =-\frac{\partial }{\partial \dot{\varphi }}\left(\frac{1}{2}M{\left({\dot{X}}^2+{\dot{Y}}^2\right)}^2+\frac{1}{2}\mu {\left({\dot{r}}^2+r^2{\dot{\varphi }}^2\right)}^2-\left(\frac{1}{2}k{\left(r-l_0\right)}^2\right)\right)\\ =\mu r^2\dot{\varphi }\\ \dot{\varphi }=\frac{p_{\varphi }}{\mu r^2}\end{array}$        (VIII)

Conclusion:

Write the Hamiltonian of the system,

  $\begin{array}{c}H={\displaystyle \sum _i{p}_i{\dot{q}}_i-L}\\ =p_x\dot{X}+p_y\dot{Y}+p_r\dot{r}+p_{\varphi }\dot{\varphi }-L\\ =\left\{\begin{array}{l}\left(p_x\frac{p_x}{M}+p_y\frac{p_y}{M}+p_r\frac{p_r}{\mu }+p_{\varphi }\frac{p_{\varphi }}{\mu r^2}\right)\\ -\left(\frac{1}{2}M\left({\dot{X}}^2+{\dot{Y}}^2\right)+\frac{1}{2}\mu \left({\dot{r}}^2+{\dot{r}}^2{\dot{\varphi }}^2\right)-\frac{1}{2}k{\left(r-l_0\right)}^2\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\frac{p_x}{M}+\frac{p_y}{M}+\frac{p_r}{\mu }+\frac{p_{\varphi }}{\mu r^2}-\frac{1}{2}M\left(\frac{p_x{}^2}{M^2}+\frac{p_y{}^2}{M^2}\right)\\ +\frac{1}{2}\mu \left(\frac{p_r{}^2}{M^2}+r^2\frac{p_{\varphi }{}^2}{{\mu }^2{r}^4}\right)+\frac{1}{2}k{\left(r-l_0\right)}^2\end{array}\right\}\\ =\frac{p_x{}^2+p_y{}^2}{2M}+\frac{1}{2}\mu \left(p_r{}^2+r^2\frac{p_{\varphi }{}^2}{{\mu }^2{r}^4}\right)+\frac{1}{2}k{\left(r-l_0\right)}^2\end{array}$        (IX)

The Hamiltonian is $H=\frac{p_x{}^2+p_y{}^2}{2M}+\frac{1}{2\mu }\left(p_r{}^2+\frac{p_{\varphi }{}^2}{r^2}\right)+\frac{1}{2}k{\left(r-l_0\right)}^2$

(b)

To determine

The $\epsilon$ will oscillate about zero.

Answer to Problem 13.21P

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 13.21P

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 13.21P

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.