2003 Spring Classical Mechanics
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1. Lagrangian Motion with Constraints

As shown in the figure, two identical disks ofm ass M are connected by a massless rigid rod of length L. They are resting on a frictionless, horizontal table. As you will read below, the final goal of this problem is to use the method of Lagrange multipliers to determine the tension in the rod as a function of generalized coordinates and their derivatives.

A. [5 points] Choose a system of generalized coordinates, including any coordinate(s) that you will need for describing the constraint(s). You must carefully describe your coordinates both with a sketch and with a brief written description. You are strongly advised to read the entire problem and think carefully about your choice of coordinates, as some systems of coordinates may greatly simplify your work while others may make subsequent calculations intractable.

Greg's Solution:

Let $\vec{r_1},\vec{r_2}$ be the vector position of the two discs. Then we can define relative and center of mass coordinates,

\begin{aligned} \vec{r} &:= \vec{r_2}-\vec{r_1} \\ \vec{R} &:= \frac{\vec{r_2}+\vec{r_1}}{2} \\ \end{aligned}

Furthermore, we can express $\vec{r}$ in terms of polar coordinates, $r,\theta$.

Thus, we have chosen four coordinates: two in $\vec{R}$, which is the center of mass of the bar, one for the length of the rod, $r$, and one for the orientation of the rod, $\theta$.

B. [5 points] Write the equation(s) of constrain neede to enforce the condition that the rod have length L.

Greg's Solution:

Since the coordinate $r$ gives the length of the rod, the constraint is simple:

\begin{align} r=L \Leftrightarrow r-L = 0 \end{align}

C. [10 points] Write the Lagrangian for the system.

Greg's solution:

\begin{align} L = \frac{M}{2}(\dot \vec{r_1}^2 + \dot \vec{r_2}^2) \equiv \frac{M}{4}\dot \vec{r}^2 + M\dot\vec{R}^2 \equiv \frac{M}{4}(\dot r^2 + r^2\dot\theta^2) + M\dot \vec{R}^2 \end{align}

D. [10 points]] Write Lagrange's equations of motion for the system, including the terms due to the Lagrange multiplier(s).

\begin{aligned} \frac{M}{2} \ddot r &= \frac{M}{2}r\dot\theta^2 - \lambda\quad\left(\text{since }\frac{d}{dr}(r-L)=1\right) \\ \drv{}{t} \left[\frac{M}{2} r^2\dot\theta\right] &= 0\\ \drv{}{t} \left[2M \dot\vec{R}\right] &= 0\\ \end{aligned}

E. [10 points] Solve the equations of motion to the extent needed to determine the undetermined multipler(s). Determine the tension in the rod.

The constraint equation gives us that $\ddot r = 0$, so we have that

\begin{align} \lambda = \frac{M}{2}r\dot\theta^2. \end{align}

$\lambda$ is the force needed to keep a disc with the reduced mass, $\mu=M/2$, at a constant radius, so the tension needed to keep a disc with mass $M$ must be,

\begin{align} \text{tension} = 2\lambda = Mr\dot\theta^2 \end{align}

This is the answer that we expected, since in uniform circular motion we have that

\begin{align} F_{\text{centripedal}} = Mr\dot\theta^2 \end{align}

2. Poisson Brackets and Anisotropic Oscillators

Recall that the Poisson bracket is defined by

\begin{align} \{a,b\}_{PB} := \sum_k \left(\frac{\partial a}{\partial q_k}\frac{\partial b}{\partial p_k}-\frac{\partial a}{\partial p_k}\frac{\partial b}{\partial q_k}\right) \end{align}

A. [10 points] Let $f=f(\vec{q},\vec{p},t$)$. Prove that

\begin{align} \frac{df}{dt} = \{f,H}_{PB}+\frac{\partial f}{\partial t} \end{align}

Greg's Solution:

By the chain rule, we have that

\begin{align} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \frac{df}{dt} = \pdrv{f}{\vec{q}}\cdot \drv{\vec{q}}{t}+\pdrv{f}{\vec{p}}\cdot \drv{\vec{p}}{t} + \pdrv{f}{t} \end{align}

However, recall that

\begin{align} \newcommand{\drv}[2]{\frac{d#1}{d#2}} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \begin{aligned} \drv{\vec{q}}{t} &= \pdrv{H}{\vec{p}} \\ \drv{\vec{p}}{t} &= -\pdrv{H}{\vec{q}}, \\ \end{aligned} \end{align}

so that

\begin{align} \newcommand{\drv}[2]{\frac{d#1}{d#2}} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \begin{aligned} \frac{df}{dt} = \pdrv{f}{\vec{q}}\cdot \pdrv{H}{\vec{p}}-\pdrv{f}{\vec{p}}\cdot \pdrv{H}{\vec{q}} + \pdrv{f}{t} \equiv \{f,H\}_{PB} + \pdrv{f}{t} \end{aligned} \end{align}


A two-dimensional oscillator has kinetic and potential enegies

\begin{aligned} T(x,y) &= \frac{1}{2}m(\dot x^2+\dot y^2) \\ V(x,y) &= \frac{1}{2}K(x^2 + y^2) + Cxy \\ \end{aligned}

i. [10 points] Show by coordinate transform that this oscillator is equivalent to an anisotropic harmonic oscillator with Lagrangian $L=\frac{1}{2}m(\dot\eta^2+\dot\xi^2)-\frac{1}{2}A\eta^2-\frac{1}{2}B\dot\xi^2,$ where $\eta$ and $\xi$ are the transformed coordinates and A and B are functions of K and C. Express A and B in terms of K and C.

Greg's Solution:


\begin{aligned} \eta &:= \frac{x-y}{\sqrt{2}}\\ \xi &:= \frac{x+y}{\sqrt{2}}\\ \end{aligned}

Then observe that,

\begin{align} A\eta^2 + B\xi ^2 = \frac{1}{2} (A\dot x^2 -2A\dot x\dot y + A\dot y^2 + B\dot x^2 + 2B\dot x\dot y + B\dot y^2) = \frac{A+B}{2}(x^2+y^2) + \frac{B-A}{2}xy\right]. \end{align}

So in particular, we see immediately that $\frac{m}{2}(\dot\eta^2+\dot\xi^2) \equiv \frac{m}{2}(\dot x^2+\dot y^2)$. Furthermore, if we set $A+B=K$ and $\frac{B-A}{2}=C$, then we will have that $A\eta^2+B\xi^2 = \frac{K}{2}(x^2+y^2) + Cxy$ as desired. Solving these two simultaneous equations, we see that

\begin{aligned} A&=\frac{K}{2}-C \\ B&=\frac{K}{2}+C \\ \end{aligned}

With these choices, we have that

\begin{align} L = \frac{1}{2}m(\dot x^2+\dot y^2) - \frac{1}{2}K(x^2+y^2) - Cxy \equiv \frac{1}{2}m(\dot\eta^2+\dot\xi^2) -\frac{A}{2}\eta^2-\frac{B}{2}\xi^2 \end{align}

as desired.

ii. [5 points] Use a Legendre transform to derive the Hamiltonian for the transformed problem.

The canonical momenta are given by,

\begin{aligned} p_\eta &= \frac{\partial L}{\dot\eta} = m\dot\eta \\ p_\xi &= \frac{\partial L}{\dot\xi} = m\dot\xi \\ \end{aligned}

The Hamiltonian is given by,

\begin{aligned} H &= p_\eta \dot\eta + p_\xi \dot\xi - L \\ &= \frac{p_\eta^2}{2m} + \frac{p_\xi^2}{2m} +\frac{A}{2}\eta^2+\frac{B}{2}\xi^2 \\ \end{aligned}

iii. [10 points] Find two independent constants of motion for the problem, and verify this fact using the result of part A.

Greg's Solution:

There are two constants of the motion: $E_\eta:=\frac{p_\eta^2}{2m}+\frac{A}{2}\eta^2$ and $E_\xi:=\frac{p_\xi^2}{2m}+\frac{B}{2}\xi^2$.

Observe that the Hamiltonian is actually just the sum of these two energies,

\begin{align} H = \underbrace{\frac{p_\eta^2}{2m} +\frac{A}{2}\eta^2}_{E_\eta}+ \underbrace{\frac{p_\xi^2}{2m} +\frac{B}{2}\xi^2}_{E_\xi} = E_\eta + E_\xi \end{align}

In this form, we see that

\begin{align} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \begin{aligned} \pdrv{H}{\eta} &= \pdrv{E_\eta}{\eta}\\ \pdrv{H}{p_\eta} &= \pdrv{E_\eta}{p_\eta}\\ \pdrv{H}{\xi} &= \pdrv{E_\xi}{\xi}\\ \pdrv{H}{p_\xi} &= \pdrv{E_\xi}{p_\xi}\\ \end{aligned} \end{align}


\begin{align} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \begin{aligned} \{E_\xi,H\}_{PB} &= \pdrv{E_\xi}{\eta}\pdrv{H}{p_\eta}-\pdrv{E_\xi}{\eta}\pdrv{H}{p_\eta} + \pdrv{E_\xi}{\xi}\pdrv{H}{p_\xi}-\pdrv{E_\xi}{p_\xi}\pdrv{H}{\xi} \\ &= \pdrv{E_\xi}{\xi}\pdrv{E_\xi}{p_\xi}-\pdrv{E_\xi}{p_\xi}\pdrv{E_\xi}{\xi} \\ &= 0, \end{aligned} \end{align}

and similarly $\{E_\eta,H\}_{PB}=0$. Thus,

\begin{aligned} \frac{d E_\eta}{dt} &= \frac{\partial E_\eta}{\partial t} = 0, \\ \frac{d E_\xi}{dt} &= \frac{\partial E_\xi}{\partial t} = 0, \\ \end{aligned}

and so $E_\xi,E_\eta$ are indeed constants of the motion as claimed.

iv. [10 points] If $C=0$, find a third independent constatn of motion. Again verify this fact using the result of part A.

Greg's Solution:

If $C=0$, then $A=B=\frac{K}{2}$ and so

\begin{aligned} H &= \frac{1}{2m}\left(p_\eta^2+p_\xi^2}\right) +\frac{K}{4}\left(\eta^2+\xi^2\right) \\ \end{aligned}

A constant of motion in this system is the angular momentum, $L=(\vec{r}\times m\vec{v})_3=m(\eta p_\xi - \xi p_\eta),$ since

\begin{align} \newcommand{\pdrv}[2]{\frac{\partial #1}{\partial #2}} \begin{aligned} \{L,H\}_{PB} &= \pdrv{L}{\eta}\pdrv{H}{p_\eta}-\pdrv{L}{p_\eta}\pdrv{H}{\eta}+\pdrv{L}{\xi}\pdrv{H}{p_\xi}-\pdrv{L}{p_\xi}\pdrv{H}{\xi}\\ &= m^2p_\xi p_\eta + m^2 \xi\eta - m^2p_\eta p_\xi -m^2 \eta\xi \\ &= 0, \\ \end{aligned} \end{align}

and so $\frac{dL}{dt}=\frac{\partial L}{\partial t}=0$.

3. [15 points] Three cubes

A torsion pendulum consists of a vertical wire attached to a mass which may rotate about the vertical. Consider three torsion pendulums which consist of identical wries from which identical homogenous solid cubes are hung. Cube A is hung from a corner, cube B from midway along an edge, and cube C from the middle of a face. What are the ratios of periods of the three pendulums, $T_A:T_B:T_C$? Briefly explain or derive your answer.

Greg's Solution:

We may treat each cube as a point mass located at the center of mass of the cube, which is located at the center of each cube. Thus, we may use the formula for the period of a simple pendulum, $\sqrt{\frac{l}{g}}$, in which l is the distance of the center of the cube from the attached point of the wire.

Assume that the length of a side of the cube is some value L. For cube C, the pivot is $l_C=L/2$ away from the center, for cube B $l_B=\sqrt{\frac{L^2}{4}+\frac{L^2}{4}}=\frac{L}{\sqrt{2}}$, and for cube A $l_A=\sqrt{\frac{L^2}{4}+\frac{L^2}{4}+\frac{L^2}{4}}=L\frac{\sqrt{3}}{2}$. Thus, since $T\propto \sqrt{l}$, we conclude that the ratio of periods is $\sqrt{\frac{\sqrt{3}}{2}}:\frac{1}{\sqrt[4]{2}}:\frac{1}{\sqrt{2}}$

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