Fall 2003 11 Mech

A point mass slides without friction inside a surface of revolution described by $z(r) = \alpha \sin r / R$. The mass is subject to a uniform gravitational field $-g \hat{z}$.

  1. Construct the Lagrangian in terms of the coordinates $r$ and $\phi$.
  2. Find the horizontal circular orbits.
  3. Which of these orbits is linearly stable under perturbations applied transverse to the direction of motion? What is the small oscillations frequency around the stable orbits?

*

Answer

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In cylindrical coordinates, $T = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\phi}^2 + \dot{z}^2)$, For $z(r)$ defined as in the problem, we get $\dot{z} = \frac{\alpha \cos r}{R}\dot{r}$, so

(1)
\begin{eqnarray} T &=& \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\phi}^2 + \frac{\alpha^2 \cos^2 r}{R^2}\dot{r}^2 ) \\ &=& \frac{1}{2}m\dot{r}^2 \left(1 + \frac{\alpha^2 \cos^2 r}{R^2} \right) + \frac{1}{2}mr^2\dot{\phi}^2 \end{eqnarray}

Since we just have a gravitational force, the potential energy is $V = mgz = \frac{mg\alpha}{R}\sin r$, Hence the Lagrangian is

(2)
\begin{align} L = \frac{1}{2}m\dot{r}^2 \left(1 + \frac{\alpha^2 \cos^2 r}{R^2} \right) + \frac{1}{2}mr^2\dot{\phi}^2 - \frac{mg\alpha}{R}\sin r \end{align}

The horizontal circular orbits are when $\dot{r} = 0$. First we must find the equations of motion. For $\phi$, we get

(3)
\begin{align} \frac{d}{dt}(mr^2\dot{\phi}) = 0 \end{align}

which is just conservation of angular momentum. For $r$, we get

(4)
\begin{align} \frac{d}{dt} \left(m \left(1+\frac{\alpha^2}{R^2}\cos^2 r \right) \dot{r} \right) = mr\dot{\phi}^2 - \frac{1}{2}m\dot{r}^2 \left( \frac{2\alpha^2}{R^2}\cos r \sin r \right) - \frac{mg\alpha}{R} \cos r \end{align}

Setting $\dot{r} = 0$, we get

(5)
\begin{align} mr\dot{\phi}^2 - \frac{mg\alpha}{R}\cos r = 0 \end{align}

or

(6)
\begin{align} \dot{\phi} = \sqrt{\frac{g\alpha}{rR} \cos r} \end{align}

We have horizontal circular orbits when $\dot{\phi}$ is real. This is when $0<r<\frac{\pi}{2}$.

For the last part, we originally satisy $\dot{\phi}_0 = sqrt{\frac{g\alpha}{r_0 R} \cos r_0}$, a give a bump transverse to motion (i.e. essentially in the $\hat{r}$ direction). We take $r \rightarrow r_0 + \delta r(t)$, $\phi \rightarrow \phi_0 + \delta \phi(t)$, and look at the resulting equation of motion to first order in $\delta r$. Expanding out the derivative in the EOM for $r$, we get

(7)
\begin{align} m \ddot{r} \left(1 + \frac{\alpha^2}{R^2} \cos^2 r \right) - m\dot{r}^2 \left(\frac{2\alpha^2}{R^2}\cos r \sin r \right) = mr\dot{\phi}^2 - \frac{1}{2}m\dot{r}^2 \left(2\alpha^2}{R^2}\cos r \sin r\right) - \frac{mg\alpha}{R} \cos r \end{align}

The other equation of motion (conservation of angular momentum) says

(8)
\begin{align} m r^2 \dot{\phi} = K \end{align}

where $K$ is a constant. Plugging in to the second equation our expansion of $r$ and $\phi$, we get

(9)
\begin{align} m(r_0 + \delta r)^2 (\dot{\phi}_0 + \delta \dot{\phi} ) \approx m r_0^2 \dot{\phi}_0 + m r_0^2 \delta\dot{\phi} + 2m r_0 \delta r_0 \dot{\phi}_0 = K \end{align}

to first order in $\delta r$ and $\delta \dot{\phi}$. Since initially $m r_0^2 \dot{\phi}_0 = K$ and we don't add any angular momentum, then

(10)
\begin{align} \delta \dot{\phi} = -2\frac{\dot{\phi}_0}{r_0}\delta r \end{align}

Expanding the EOM for $r$ to first order, we get

(11)
\begin{eqnarray} m\delta\ddot{r}\left(1+\frac{\alpha^2}{R^2}\cos^2 r_0 \right) &=& m r_0\dot{\phi}_0^2 + m\delta r \dot{\phi}_0^2 + 2mr_0 \dot{\phi}_0 \delta\dot{\phi} - \frac{mg\alpha}{R}(\cos r_0 - \sin r_0 \delta r ) \\ &=& m\frac{g\alpha}{r_0 R} \cos r_0 \delta r - 4m \frac{g\alpha}{r_0 R} \cos r_0 \delta r + \frac{mg\alpha}{R} \sin r_0 \delta r \\ &=& \left( -\frac{3mg\alpha}{r_0 R} \cos r_0 + \frac{mg\alpa}{R} \sin r_0 \right) \delta r \end{eqnarray}

The coefficient in front of $\delta r$ must be negative in order to have stable oscillations. This happens when

(12)
\begin{align} -\frac{3}{r_0} \cos r_0 + \sin r_0 < 0 \end{align}

or

(13)
\begin{align} \tan r_0 < \frac{3}{r_0} \end{align}
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