2002 Autumn Classical Mechanics

The following questions were taken from [http://www.phys.washington.edu/quals/02AutCM.pdf]; click that link in order to see the associated diagrams which are not included below.

1. A point mass m moves in a circular orbit of radius r under the influence of a central force with potential $-km/r^n$.

A. [20 points] Find the conditions on n such that the circular orbit is stable under small perturbations (i.e., the mass will oscillate about the circular orbit).

Since the force is central, we know that angular momentum is conserved, and therefore that the motion will always be constrained to move in a plane. Thus, we write down the Lagrangian in the polar coordinate form,

(1)
\begin{align} L = \underbrace{\frac{m}{2}(\dot r^2 + r\dot \theta^2)}_{T} + \underbrace{\frac{km}{r^n}}_{-U} \end{align}

From here, we obtain the Euler-Lagrange equation for r,

(2)
\begin{align} m\ddot r = m\dot\theta^2r-nkmr^{-n-1} \end{align}

We can get rid of $\dot\theta$ in the above equation by expressing it in terms of the angular momentum,

(3)
\begin{align} L = mr^2\dot\theta, \end{align}

which is a conserved quantity. Thus obtain,

(4)
\begin{align} \ddot r = \frac{L^2}{m^2}r^{-3}-nkr^{-n-1} \end{align}

A circular orbit is a solution of the above such that $r$ is a constant — that is,

(5)
\begin{align} m\ddot r = 0 = \frac{L^2}{m^2}r_0^{-3}-nkr_0^{-n-1} \end{align}

Assume that we are at nearly such an orbit — that is, $r = r_0 +\delta r$, with $\delta r \ll r_0$. Plugging this into (and dividing m from both sides), we obtain

(6)
\begin{align} \ddot r_0 + \delta \ddot r = \delta \ddot r = \frac{L^2}{m^2}(r_0+\delta r)^{-3}-nk(r_0+\delta r)^{-n-1} \end{align}

Expanding about small $\delta r$, the above becomes

(7)
\begin{aligned} \delta \ddot r & \underbrace{\approx \frac{L^2}{m^2}r_0^{-3} - nk r_0^{-n-1}}_{=0} - \left[3\frac{L^2}{m^2}r_0^{-4} - kn(n+1) \right]\delta r \\ &= -\left[3\frac{L^2}{m^2}r_0^{-4} - kn(n+1)r^{-n-2} \right]\delta r \\ \end{aligned}

Finally, we note that since the original trajectory was uniform circular motion, we had that

(8)
\begin{align} m\dot\theta^2r_0 = \frac{kmn}{r_0^{n+1}} \Rightarrow \dot\theta^2 = \frac{kn}{r^{n+2}}, \end{align}

and so

(9)
\begin{aligned} \delta \ddot r &= -\left[3\frac{L^2}{m^2}r_0^{-4} - kn(n+1)r^{-n-2} \right]\delta r \\ &= -\left[3\frac{knm^2 r_0^4/r^{n+2}}{m^2}r_0^{-4} - kn(n+1)r^{-n-2} \right]\delta r \\ &= -\left[3n - n(n+1) \right]\frac{k}{r_0^{n+2}} \delta r \\ &= \frac{kn(n-2)}{r_0^{n+2}} \delta r \\ \end{aligned}

In order for perturbations about the circular orbit to be stable, we need to have the coefficient in front of $\delta r$ be negative; this will be the case as long as $0<n<2$.

B. [10 points] Find the frequency of small oscillations about stable circular orbits and express this frequency in terms of the angular velocity of the mass moving in these orbits.

Observe that Eq. (9) is already the harmonic oscillator form,

(10)
\begin{align} \delta \ddot r = -\omega^2 \delta r, \end{align}

where the angular frequency $\omega$ is given by

(11)
\begin{align} \omega = \sqrt{\frac{kn(2-n)}{r_0^{n+2}}}. \end{align}

Recalling from before that

(12)
\begin{align} \dot\theta^2 = \frac{kn}{r_0^{n+2}}, \end{align}

we see that

(13)
\begin{align} \omega = \sqrt{(2-n)}\dot\theta \end{align}

2. [40 points total] A rigid uniform bar of mass M and length L is supported in equilibrium in a horizontal position by two massless springs attached at each end. The identical springs have the force constant k. the motion of the center of mass is constrained to move parallel to the vertical, x, axis. Furthermore the motion of the bar is constrained to lie in the xz-plane.

A. [5 points] Show that the moment of inertia of a bar about the y axis through its center of mass is $ML^2/12$.

Assume that the bar is thin — that is, that its radius is much less than its length. In that case, we have that

(14)
\begin{aligned} I &= \int_{-L/2}^{L/2} r^2 \frac{dm}{L/M} = 2\frac{M}{L}\frac{1}{3} \left(\frac{L}{2}\right)^3 = \frac{ML^2}{12} \end{aligned}

B. [15 points] Construct the Lagrangian for this bar-spring arrangement assuming only small deviations from equilibrium.

Greg's Solution:

Since the bar is constrained to move in the xz-plane, and since the center of mass is constrained to move along the x-axis, there are really only two degrees of freedom in this problem: the height of the bar's center of mass, $x$, and the angle of the bar to the horizontal, $\theta$.

When $\theta=0$, the bar is level. When $\theta$ is increased, the right side of the bar is moved up by $\frac{L}{2}\sin\theta$ and to the left by $\frac{L}{2}(1-\cos\theta)$. Thus,

(15)
\begin{align} L = \frac{M}{2}\dot x^2 + \frac{ML^2}{12}\dot\theta^2 -\frac{k}{2}\left[\underbrace{\left(x-\frac{L}{2}\sin\theta\right)^2+\frac{L^2}{4}(1-\cos\theta)^2}_{\text{left side of bar}} + \underbrace{\left(x+\frac{L}{2}\sin\theta\right)^2+\frac{L^2}{4}(1-\cos\theta)^2}_{\text{right side of bar}}\right] \end{align}

Note that we have neglected gravity; this is because the only effect of gravity is to shift the effective equilibrium length of the springs. You can see this from the fact that we can always absorb the gravitational energy into the spring energy term by completing the square, that is $\frac{k}{2}x^2 - gx \equiv \frac{k}{2}(x-l_0)^2 + C$, where the constant $C$ does not effect the motion and the constant $l_0$ can be eliminated by shifting the coordinate system.

For convenince, for the rest of the problem we shall work in units in which $M=1$,$L=2$, and $k=2$ (which corresponding to choosing respectively a unit mass, length, and time) so that the Lagrangian simplifies to

(16)
\begin{align} L = \frac{1}{2}\dot x^2 + \frac{1}{3}\dot\theta^2 -(x-\sin\theta)^2 -(x+\sin\theta)^2-2(1-\cos\theta)^2 \end{align}

We expand the potential energy part of the Lagrangian to second order in $\theta$ and x: (note that to do this we need only expand to first order inside the squares)

(17)
\begin{aligned} U &= (x-L\theta)^2+ (x+\theta)^2+2(1-1)^2\right] \\ &= (x-L\theta)^2 + (x+\theta)^2 \\ &= x^2-2x\theta + \theta^2 + x^2+2x\theta + \theta^2 \\ &= 2x^2 + 2\theta^2 \\ \end{aligned}

Thus the full Lagrangian to second order is,

(18)
\begin{align} L = \frac{1}{2}\dot x^2 + \frac{1}{3}\dot\theta^2 - 2x^2 - 2\theta^2 \end{align}

C. [15 points] Calculate the vibration frequencies of the nromal modes for small amplitude oscillations.

The Euler-Lagrange equations for this system are,

(19)
\begin{aligned} \ddot x &= -4x \\ \ddot \theta &= -6\theta \\ \end{aligned}

If we were to write this in matrix form, the matrix would be diagonal; thus we conclude that we already have our frequencies, 2 and $\sqrt{6}$, for the two normal modes. Of course, these are expressed in units such that a single unit of time is equal to $\sqrt{k/2m}$, so in normal units the frequencies are given by $\sqrt{2k/m}$ and $\sqrt{3k/m}$.

Note that the first frequency is exactly what we'd have expected; two springs in parallel is equivalent to doubling the spring constant, which in turn is equivalent to raising the frequency by a factor of $\sqrt{2}$.

D. [5 points] Describe the normal modes of oscillation.

The $\sqrt{2k/m}$ angular frequency mode consists of the bar moving up and down with a constant angle with respect to the horizontal.

The $\sqrt{3k/m}$ angular frequency mode consists of the bar pivoting back and forth (i.e. changing angle with respect to the horizontal) about its (fixed) center of mass.

3. [30 points total] A symmetric top of mass M is in a uniform gravitational field along the z axis and has one fixed point at the tip. The principal axes of the top are labeled 1, 2, and 3. The diagonal components of the inertia tensor in the pricnicpal axes are I1, I2, and I3, with $\textbf{I}_1=\textbf{I}_3$ because of axial symmetry. The center of mass is located at the point labled G, a distance of l from the tip.

A. [3 points] Show that the moment of inertia of the top about the 1-axis in the above figure is $\textbf{I}=\textbf{I}_1+Ml^2$.

Greg's Solution:

Since the 1-axis is parallel to the center of mass and a distance $l$ away, we conclude that the moment of inertia about the 1-axis is given by $\textbf{I}=\textbf{I}_1+Ml^2$.

B. [17 points] Show that the Lagrangian can be written in terms of the Euler angles $\theta, \varphi, \psi$ as

(20)
\begin{align} L = \frac{1}{2} \textbf{I}\left(\dot\varphi^2\sin^2\theta +\dot\theta^2\right)+\frac{1}{2}\textbf{I}_3\left(\dot\varphi\cos\theta+\dot\psi\right)^2 - Mgl\cos\theta \end{align}

Greg's Solution:

The angular momentum vector takes the form,

(21)
\begin{align} \vec{\om} = \dot\varphi \hat e_\varphi + \dot\theta \hat e_\theta + \dot\psi \hat e_\psi \end{align}

This form is inconvenient because it is not expressed in terms of an orthogonal basis and because the basis vectors due not coincide with the principal axes. Thus, we shall take each of the basis vectors above and express them in terms of the 1-, 2-, and 3- axes, which are an orthonormal basis coinciding with axes that have known moments of inertia.

First, we observe that $\psi$ just rotates about the $\hat e_3$ axis, so that

(22)
\begin{align} \hat e_\psi = \hat e_3 \end{align}

Next, we observe that $\theta$ rotates about a vector in the 1-2 plane — specifically, it rotates about where the $\hat e_1$ vector was before it was rotated by an angle $\psi$ about $\hat e_3$; from this fact we see that

(23)
\begin{align} \hat e_\theta = \hat e_1\cos\psi + \hat e_2\sin\psi \end{align}

Finally, $\phi$ rotates about the $\hat e_z$ vector, which is where the $\hat e_3$ vector was located before it was rotated an angle $\theta$ about the vector $\hat e_\theta$; we thus see that $\hat e_z$ is in the plane spanned by $\hat e_3$ and $\hat e_3\times \hat e_\theta$, so that

(24)
\begin{align} \hat e_\varphi = \hat e_3\cos\theta + \hat e_3\times \hat e_\theta\sin\theta = \hat e_3\cos\theta + \hat e_2\sin\theta \cos\psi - \hat e_1 \sin\thetat \sin\psi \end{align}

Collecting components, we see that

(25)
\begin{align} \vec{\omega} = \left(\dot\theta\cos\psi -\dot\varphi\sin\theta\sin\psi)\hat e_1 + \left(\dot\theta\sin\psi + \dot\varphi\sin\theta\cos\psi\right)\hat e_2 + \left(\dot\psi+\dot\varphi\cos\theta\right)\hat e_3 \end{align}

Since the origin is fixed in this problem, all the kinetic energy is rotational kinetic energy,

[[math almost-at-kinetic]]
T = \frac{\tilde I_1}{2}\left(\dot\theta\cos\psi -\dot\varphi\sin\theta\sin\psi)^2 + \frac{\tilde I_2}{2}\left(\dot\theta\sin\psi + \dot\varphi\sin\theta\cos\psi\right)^2 + \frac{I_3}{2}\left(\dot\psi+\dot\varphi\cos\theta\right)^2,
[[/math]]

where $\tilde I_1$ and $\tilde I_2$ are the moments of inertia about the origin. Since, as shown in the previous part, $I:=\tilde I_1 = \tilde I_2 = I_1+Ml^2$, we may rewrite in the simpler form,

(26)
\begin{align} T = \frac{I}{2} \left(\dot\varphi^2\sin^2\theta + \dot\theta^2\right)^2 + \frac{I_3}{2}\left(\dot\varphi\cos\theta + \dot\psi\right)^2 \end{align}

Last but not least, we note that the gravitational potential energy is just proportional to the height of the center of mass — $l\cos\theta$ — so that our full Lagrangian is

(27)
\begin{align} L = \frac{I}{2} \left(\dot\varphi^2\sin^2\theta + \dot\theta^2\right)^2 + \frac{I_3}{2}\left(\dot\varphi\cos\theta + \dot\psi\right)^2 - Mgl\cos\theta. \end{align}

C. [10 points] Show that components of the angular momentum about two axes are conserved.

Note that there is no explicit dependence in the Lagrangian on $\varphi$ or $\psi$. Thus, the canonical momenta associated with these coordinates,

(28)
\begin{aligned} p_\psi &= \pdrv{L}{\dot\psi} = I_3\dot\psi \\ p_\varphi &= \pdrv{L}{\dot\varphi} = I\dot\varphi(\cos^2\theta+\sin^2\theta) = I\dot\varphi, \end{aligned}

and so we see that the momenta about the $\hat e_\varphi$ and $\hat e_\psi$ axes are conserved.

page revision: 5, last edited: 24 Aug 2006 00:19