DEpartment of physics

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\section{Example: Motion of a Single Particle in Two-Dimensional Polar Coordinates. } Here we are going to do the same problem we did earlier. Earlier, we considered the motion of a particle in 2D using Cartesian Coordinates We basically recovered $F=m\ddot{x}$ etc. In the previous classes, we proved that the velocity component of a particle in polar coordinates can be written as: $v_r=\dot{r}$ and $v_\phi=r\dot{\phi}$, so the kinetic energy can be written as : \begin{equation*} T=\frac{1}{2}mv^2=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2) \end{equation} And let's write down $U$ as \begin{equation*} U=U(r,\phi) \end{equation} So the Lagrangean for the system can be written as: \begin{equation*} L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2)=U(r,\phi) \end{equation} Let's write the Euler -Lagrangean equation w.r.t. $r$: \begin{equation*} \frac{\partial L}{\partial r}-\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}=0 \end{equation} \begin{equation*} mr\dot{\phi}^2-\frac{\partial U}{\partial r}-\frac{d}{dt}m\dot{r}=0 \end{equation} \begin{equation*} mr\dot{\phi}^2-\frac{\partial U}{\partial r}=m\ddot{r} \end{equation} Since we know $F_r=-\frac{\partial U}{\partial r}$, We will get \begin{equation*} F_r=m\ddot{r}-mr\dot{\phi}^2 \end{equation} (\textbf{NOTE} We got this equation in our chapter 1 work. Look at Page 29, Eq. 1.47) Let's look at the Euler Lagrangean Equation w.r.t. $\phi$. \begin{equation*} \frac{\partial L}{\partial \phi}-\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}=0 \end{equation} \begin{equation*} -\frac{\partial U}{\partial \phi}-\frac{d}{dt}mr^2\dot{\phi}=0 \end{equation} If we write $F=-\nabla U$ in polar coordinates: \begin{equation*} \nabla U=\frac{\partial U}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial U}{\partial \phi}\hat{\phi} \end{equation} So we got: \begin{equation*} -\frac{\partial U}{\partial \phi}=\frac{d}{dt}mr^2\dot{\phi} \end{equation} \begin{equation*} r F_\phi=\frac{d}{dt}mr^2\dot{\phi} \end{equation} That means, we can identify the quantity $mr^2\dot{\phi$ as the angular momentum of the system Do earlier we said $\frac{\partial L}{\partial q_j}=\text{Generalized Force}$ In the case of the $\hat{phi$ direction in the above problem, the generalized force turn out to be the torque. In the similar way, we defined $\frac{\partial L}{\partial \dot{q_j}}$ as the generalized momentum. It is clear here that why we call it \textit{generalized} momentum. It does not have to be momentum. In this problem $\frac{\partial L}{\partial \dot{\phi}}$ equals to the Torque in this problem. \section{More about Constrained Systems} We have already done some problems where the particle's motion is constrained. For example, Simple Pendulum, a particle attached to a Spring etc. Let's look at the generalized coordinates in detail. We are basically looking at a system of $N$ particles, with $\alpha=1,\cdots, N$ with positions $r_\alpha$. If we have a set of generalized coordinates as: $q_1,q_2,\cdots , q_n$ we can write the position coordinate of each particle as \begin{equation*} r_\alpha=r_\alpha(q_1,q_2, \cdots, q_n,t) \,\,\,\,\,\, \text{for}\,\,\,\,\, \alpha=1,2, \cdots N \end{equation} And also the generalized coordinates can also be expressed in terms of $r\alpha$ \begin{equation*} q_i=q_i(r_1,r_2,\cdots,r_N,t) \,\,\,\,\,\, \text{for} \,\,\,\,\,\,\, i=1,2,\cdots,n \end{equation} Let's look at the Simple Pendulum: We took the generalized coordinate as $\phi$. Now we can write the transformation equation (the cartesian coordinates in terms of generalized coordinates) as \begin{equation*} \vec{r}=(x,y)= (l \text{Sin} \phi, l \text{Cos} \phi) \end{equation} Let's look at the Double Pendulum Problem: \begin{figure}[!h] \vspace{-10pt} \begin{center} \includegraphics[scale =1.15]{./taylor_07f03.jpg} % No texting or web browsing in the class. % \caption{Two forces in projectile motion, the gravity \textbf{w} = m \textbf{g} and the drag force of air resistance \textbf{f} = - $f(v) \hat{ \textbf{v}}$} \end{center} % \label{fig:03} \vspace{-15pt} \end{figure} We can select the generalized coordinates as: $\phi_1$ and $\phi_2$. Then we can write the transformation equations as \begin{equation*} r_1=(x,y)=(l_1 \text{Sin} \phi_1, l_1 \text{Sin} \phi_2) \end{equation} Similarly, we get \begin{equation*} r_2=(x,y)=(l_1 \text{Sin} \phi_1 + l_2 \text{Sin} \phi_2, l_1 \text{Sin} \phi_2+ l_2 \text{Cos} \phi_2) \end{equation} In the previous two problems, Simple Pendulum and the Double Pendulum, the transformation equations are independent of time $t$. But there are cases where the transformation equations depend on the time $t$. Consider a simple pendulum suspended from the ceiling of a railroad car. \begin{equation*} r=(x,y)=(l \text{Sin} \phi + v_0 t +\frac{1}{2} at^2, - l \text{Cos} \phi)=r(\phi,t) \end{equation} \subsection{Example: Find the frequency of Small Oscillations of a Simple Pendulum placed in a rail road car, that has a constant acceleration $a$ in the $x-$ direction. } \begin{figure}[!h] \vspace{-10pt} \begin{center} \includegraphics[scale =0.75]{./RailRoadCar.png} % No texting or web browsing in the class. % \caption{Two forces in projectile motion, the gravity \textbf{w} = m \textbf{g} and the drag force of air resistance \textbf{f} = - $f(v) \hat{ \textbf{v}}$} \end{center} % \label{fig:03} \vspace{-15pt} \end{figure} We can write the transformation equations: \begin{equation*} x=v_0 t + \frac{1}{2} a t^2 + l \text{Sin} \theta \end{equation} \begin{equation*} y=-l \text{Cos} \theta \end{equation} We can write the velocity in $x-y$ coordinate system as: \begin{equation*} \dot{x}=v_0+at +l \dot{\theta} \text{Cos} \theta \end{equation} \begin{equation*} \dot{y}=l \dot{\theta} \text{Sin} \theta \end{equation} We can write the Kinetic and Potential energies as: \begin{equation*} T=\frac{1}{2} m (\dot{x}^2+\dot{y}^2) \,\,\,\,\, \text{and} \,\,\,\,\,\ V=-mg l \text{Cos}\theta \end{equation} So we can write the Lagrangean as: \begin{equation*} L=\frac{1}{2}m \left( \left(v_0+at +l \dot{\theta} \text{Cos} \theta \right)^2 + \left( l \dot{\theta} Sin \theta \right)^2\right)+mg l \text{Cos} \theta \end{equation} \begin{equation*} L=\frac{1}{2} \m (v_0+at)^2+m (v_0 + at) l \dot{\theta} \text{Cos} \theta + \frac{1}{2}ml^2\dot{\theta}^2+mgl \text{Cos} \theta \end{equation} Now we can write the Euler-Lagrangean Equation w.r.t. $\theta$, \begin{equation*} \frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=0 \end{equation} After few steps, we get: \begin{equation*} \ddot{\theta}+\frac{g}{l} \text{Sin} \theta + \frac{a}{l}\text{Cos} \theta=0 \end{equation} That's the result from Euler-Lagrangean Equation. Now we are going to solve this further for understanding small oscillations around the equilibrium point: Let's first look as the equilibrium point: At equilibrium point: \begin{equation*} \ddot{\theta}=0 \end{equation} \begin{equation*} \frac{g}{l} \text{Sin} \theta + \frac{a}{l}\text{Cos} \theta=0 \end{equation} \begin{equation*} \text{tan} \theta = -\frac{a}{g} \end{equation} We can look at the small oscillations around this equilibrium point. However, we are not going to do the small oscillations at this time. (Refer to Marion and Thornton Example 7.6) Notes can be found here

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