2003 Spring Classical Mechanics

# 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,

(1)
\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:

(2)
\begin{align} r=L \Leftrightarrow r-L = 0 \end{align}

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

Greg's solution:

(3)
\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).

(4)
\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

(5)
\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,

(6)
\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

(7)
\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

(8)
\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}


page revision: 2, last edited: 24 Aug 2006 19:15