Lot of works on simscape, IFF and DVF
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		@@ -3,13 +3,13 @@
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 (lambda ()
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   (TeX-add-to-alist 'LaTeX-provided-package-options
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                     '(("inputenc" "utf8") ("fontenc" "T1") ("ulem" "normalem") ("tcolorbox" "most") ("babel" "USenglish" "english")))
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    "latex2e"
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    "import"
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    "babel")
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   (LaTeX-add-labels
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    "sec:org8c48899"
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    "sec:org335669b"
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    "sec:introduction"
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    "sec:org60f23e3"
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    "sec:org8b756e7"
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    "sec:theory"
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    "sec:orgd677659"
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    "sec:orgbf4a596"
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    "fig:rotating_xy_platform"
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    "sec:orgaa8880a"
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    "eq:energy_inertial_frame"
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    "eq:lagrangian_inertial_frame"
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    "sec:org754b644"
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    "sec:org9cbf82a"
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    "sec:org8d24de3"
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    "sec:conclusion"
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    "sec:orgf333899")
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    "sec:orgb252937")
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    "ref"))
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 :latex)
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							@@ -58,11 +58,145 @@
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* Theory
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<<sec:theory>>
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** Rotating Positioning Stage
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# Description of the system
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- $k$: Actuator's Stiffness [N/m]
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- $m$: Payload's mass [kg]
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- $\omega_0 = \sqrt{\frac{k}{m}}$: Resonance of the (non-rotating) mass-spring system [rad/s]
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- $\omega_r = \dot{\theta}$: rotation speed [rad/s]
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#+name: fig:rotating_xy_platform
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#+caption: Figure caption
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#+attr_latex: :scale 1
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[[file:figs/rotating_xy_platform.pdf]]
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** Equation of Motion
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Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$ (neglecting the rotational energy):
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#+name: eq:energy_inertial_frame
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\begin{subequations}
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  \begin{align}
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    T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\
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    R & = \frac{1}{2} c \left( \dot{x}^2 + \dot{y}^2 \right) \\
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    V & = \frac{1}{2} k \left( x^2 + y^2 \right)
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  \end{align}
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\end{subequations}
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The Lagrangian is the kinetic energy minus the potential energy:
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#+name: eq:lagrangian_inertial_frame
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\begin{equation}
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L = T-V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - \frac{1}{2} k \left( x^2 + y^2 \right)
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\end{equation}
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The external forces applied to the mass are:
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\begin{subequations}
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  \begin{align}
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    F_{\text{ext}, x} &= F_u \cos{\theta} - F_v \sin{\theta}\\
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    F_{\text{ext}, y} &= F_u \sin{\theta} + F_v \cos{\theta}
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  \end{align}
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\end{subequations}
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From the Lagrange's equations of the second kind eqref:eq:lagrange_second_kind, the equation of motion eqref:eq:eom_mixed is obtained.
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#+name: eq:lagrange_second_kind
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\begin{equation}
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  \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) = \frac{\partial L}{\partial q_j}
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\end{equation}
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#+name: eq:eom_mixed
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\begin{subequations}
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  \begin{align}
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    m\ddot{x} + kx = F_u \cos{\theta} - F_v \sin{\theta}\\
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    m\ddot{y} + ky = F_u \sin{\theta} + F_v \cos{\theta}
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  \end{align}
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\end{subequations}
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Performing the change coordinates from $(x, y)$ to $(d_x, d_y, \theta)$:
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\begin{subequations}
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  \begin{align}
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    x & = d_u \cos{\theta} - d_v \sin{\theta}\\
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    y & = d_u \sin{\theta} + d_v \cos{\theta}
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  \end{align}
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\end{subequations}
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Gives
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#+name: eq:oem_coupled
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\begin{subequations}
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  \begin{align}
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    m \ddot{d_u} + (k - m\dot{\theta}^2) d_u &= F_u + 2 m\dot{d_v}\dot{\theta} + m d_v\ddot{\theta} \label{eq:du_coupled} \\
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    m \ddot{d_v} + (k \underbrace{-\ m\dot{\theta}^2}_{\text{Centrif.}}) d_v &= F_v \underbrace{-\ 2 m\dot{d_u}\dot{\theta}}_{\text{Coriolis}} \underbrace{-\ m d_u\ddot{\theta}}_{\text{Euler}} \label{eq:dv_coupled}
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  \end{align}
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\end{subequations}
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We obtain two differential equations that are coupled through:
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- *Euler forces*: $m d_v \ddot{\theta}$
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- *Coriolis forces*: $2 m \dot{d_v} \dot{\theta}$
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Without the coupling terms, each equation is the equation of a one degree of freedom mass-spring system with mass $m$ and stiffness $k- m\dot{\theta}^2$.
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Thus, the term $- m\dot{\theta}^2$ acts like a negative stiffness (due to *centrifugal forces*).
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** Constant Rotating Speed
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To simplify, let's consider a constant rotating speed $\dot{\theta} = \omega_r$ and thus $\ddot{\theta} = 0$.
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#+NAME: eq:coupledplant
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\begin{equation}
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\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
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\frac{1}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2}
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\begin{bmatrix}
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  ms^2 + (k-m{\omega_0}^2) & 2 m \omega_0 s \\
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  -2 m \omega_0 s          & ms^2 + (k-m{\omega_0}^2) \\
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\end{bmatrix}
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\begin{bmatrix} F_u \\ F_v \end{bmatrix}
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\end{equation}
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#+NAME: eq:coupled_plant
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\begin{equation}
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\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
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\frac{\frac{1}{k}}{\left( \frac{s^2}{{\omega_0}^2} + (1 - \frac{{\omega_r}^2}{{\omega_0}^2}) \right)^2 + \left( 2 \frac{{\omega_r} s}{{\omega_0}^2} \right)^2}
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\begin{bmatrix}
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  \frac{s^2}{{\omega_0}^2} + 1 - \frac{{\omega_r}^2}{{\omega_0}^2} & 2 \frac{\omega_r s}{{\omega_0}^2} \\
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  -2 \frac{\omega_r s}{{\omega_0}^2}          & \frac{s^2}{{\omega_0}^2} + 1 - \frac{{\omega_r}^2}{{\omega_0}^2} \\
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\end{bmatrix}
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\begin{bmatrix} F_u \\ F_v \end{bmatrix}
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\end{equation}
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When the rotation speed is null, the coupling terms are equal to zero and the diagonal terms corresponds to one degree of freedom mass spring system.
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#+NAME: eq:coupled_plant_no_rot
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\begin{equation}
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\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
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\frac{\frac{1}{k}}{\frac{s^2}{{\omega_0}^2} + 1}
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\begin{bmatrix}
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  1 & 0 \\
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  0 & 1
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\end{bmatrix}
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\begin{bmatrix} F_u \\ F_v \end{bmatrix}
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\end{equation}
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# Campbell Diagram
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When the rotation speed in not null, the resonance frequency is duplicated into two pairs of complex conjugate poles.
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As the rotation speed increases, one of the two resonant frequency goes to lower frequencies as the other one goes to higher frequencies (Figure [[fig:campbell_diagram]]).
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#+name: fig:campbell_diagram
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#+caption: Campbell Diagram
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[[file:figs/campbell_diagram.pdf]]
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# Bode Plots for different ratio wr/w0
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The magnitude of the coupling terms are increasing with the rotation speed.
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** Integral Force Feedback
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** Direct Velocity Feedback
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* Conclusion
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<<sec:conclusion>>
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@@ -1,4 +1,4 @@
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% Created 2020-06-08 lun. 11:15
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% Created 2020-06-08 lun. 11:40
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% Intended LaTeX compiler: pdflatex
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\documentclass{ISMA_USD2020}
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\usepackage[utf8]{inputenc}
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@@ -49,27 +49,56 @@
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}
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\section{Introduction}
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\label{sec:orgd787473}
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\label{sec:org335669b}
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\label{sec:introduction}
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\section{Theory}
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\label{sec:org808f338}
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\label{sec:org8b756e7}
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\label{sec:theory}
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\subsection{Rotating Positioning Stage}
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\label{sec:orgbf4a596}
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\begin{figure}[htbp]
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\centering
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\includegraphics[scale=1]{figs/rotating_xy_platform.pdf}
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\caption{\label{fig:figure_name}Figure caption}
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\caption{\label{fig:rotating_xy_platform}Figure caption}
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\end{figure}
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\subsection{Equation of Motion}
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\label{sec:orgaa8880a}
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Let's express the kinetic energy \(T\) and the potential energy \(V\) of the mass \(m\):
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\begin{align}
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\label{eq:energy_inertial_frame}
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T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\
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V & = \frac{1}{2} k \left( x^2 + y^2 \right)
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\end{align}
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The Lagrangian is the kinetic energy minus the potential energy.
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\begin{equation}
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\label{eq:lagrangian_inertial_frame}
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L = T-V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - \frac{1}{2} k \left( x^2 + y^2 \right)
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\end{equation}
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\subsection{Integral Force Feedback}
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\label{sec:org754b644}
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\subsection{Direct Velocity Feedback}
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\label{sec:org9cbf82a}
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\section{Conclusion}
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\label{sec:orgab2ddd2}
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\label{sec:org8d24de3}
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\label{sec:conclusion}
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\section{Acknowledgment}
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\label{sec:orga63f041}
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\label{sec:orgb252937}
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\bibliography{ref}
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