Редактирование: Dzenushko Dainis. Course project for theoretical mechanics "Double pendulum"

[[Дзенушко Дайнис. Курсовой проект по теоретической механике|Русская версия ]][[Файл:RUS.jpg]]
== Subject ==
Specification of double pendulum oscillations 

== Task ==
Rod is connected to the roof by cylindrical joint. The second rod is connected to the end of the first rod via cylindrical joint so that when both rods are vertical the angle between joint's axis equals to <math>\alpha</math>. Dissipative forces are not taken to account.<br>
'''System parameters:''' <br>
#Inertia tensor of first and second rods are equal. <math>\underline{\underline{\Theta}}_1</math> and <math>\underline{\underline{\Theta}}_2</math> respectively. 
#length of rods are a and b, their masses <math>m_1</math> and <math>m_2</math> for first and second rods.
#Angle between rotation axis of joints equals to <math>\alpha</math><br>
*<math>\varphi</math> - angle between first rod and vertical direction
*<math>\psi</math> - angle between axis of first and second rods that means the angle in second joint relative to vertical orientation of system.

<gallery widths=231px heights=319px perrow = 1>
Файл:2_oscillator.jpg‎
</gallery>
'''Main task:'''<br>
*find the equation of motion for this double pendulum

== Решение ==
'''Определимся с подходом к решению:''' Задачу будем решать при помощи уравнения Лагранжа имеющего следующий вид:<br>
<math>\frac{d}{dt}\frac{\partial T}{\partial \dot{q}_i}-\frac{\partial T}{\partial q_i} = -\frac{\partial \Pi}{\partial q_i}+Q_i</math><br>
*<math>T</math> - Кинетическая энергия системы <br>
*<math>\Pi</math> - Потенциальная энергия системы<br>
*<math>q_i</math> - Обобщенные координаты<br>
*<math>\dot{q}_i</math> - Обобщенные скорости<br>
*<math>Q_i</math> - Обобщенные непотенциальные силы<br>
<br>

'''Выберем обобщенные координаты:''' в качестве обобщенных координат возьмем углы <math>\varphi</math> и <math>\psi</math> <br>
*В нашем случае отсутствуют обощенные силы, соответствующие непотенциальным взаимодействиям.<br>

'''Найдем потенциальную и кинетическую энергии системы:''' <math>\Pi_1 , T_1 ; \Pi_2 , T_2 </math> соответственно первого и второго стержней.<br> <math>\Pi = \Pi_1 + \Pi_2</math> - Потенциальная энергия системы<br>
<math>T = T_1 + T_2</math> - Кинетическая энергия системы<br>
<math>T_1 = \frac{\underline{\omega}_1 \cdot \underline{\underline{\Theta}}_1 \cdot \underline{\omega}_1}{2} = \frac{\Theta_1 \omega_1^2}{2} = \frac{\Theta_1 \dot{\varphi}^2}{2}</math> - Кинетическая энергия первого стержня; Где 
<math>\qquad \Theta_1 = \frac{m_1 a^2}{3}</math> - момент инерции первого стержня<br>
<math>\Pi_1 = m_1 g \left( \frac{a}{2} - \frac{a}{2} \cos \varphi \right)</math> - Потенциальная энергия первого стержня<br>
<math>T_2 = \frac{\underline{\omega}_2 \cdot \underline{\underline{\Theta}}_2 \cdot \underline{\omega}_2}{2} + \frac{m_2 \vartheta_c^2}{2}</math> - Кинетическая энергия второго стержня<br>
<math>\underline{\omega}_2 = ?</math><br><br>

'''Finding second rod's angular velocity vector:''' <br>
For calculating <math>\underline{\omega}_2</math> we will find rotation tensors of both rods<br>
<math>\underline{\underline{P}}_1(\varphi,\underline{k}) = \underline{k}\underline{k} + (\underline{\underline{E}} - \underline{k}\underline{k})cos(\varphi) + \underline{k} \times \underline{\underline{E}}sin(\varphi)</math><br>
<math>\underline{\underline{P}}_2(\psi,\underline{e}) = \underline{e}\underline{e} + (\underline{\underline{E}} - \underline{e}\underline{e})cos(\psi) + \underline{e} \times \underline{\underline{E}}sin(\psi)</math><br>
Where:<br>
<math>\underline{e} = \underline{\underline {P}}_1 \cdot \underline{e}_0</math> - second rod's ongoing rotation axis<br>
<math>\underline{e}_0 = \cos(\alpha) \underline{k} + \sin(\alpha) \underline{i}</math> - second rod's initial rotation axis<br><br>

<math>\underline{\underline{P}} = \underline{\underline{P}}_2 \cdot \underline{\underline{P}}_1</math> - full rotation tensor for second rod<br><br>

But:<br>
<math> \underline{\underline{P}} = \underline{\underline{P}}_2 \cdot \underline{\underline{P}}_1 = \underline{\underline{P}}(\psi,\underline{e}) \cdot \underline{\underline{P}}_1 = \underline{\underline{P}}_1 \cdot \underline{\underline{P}}(\psi,\underline{e_0})\cdot \underline{\underline{P}}^T_1 \cdot \underline{\underline{P}}_1 = \underline{\underline{P}}_1 \cdot \underline{\underline{P}}(\psi,\underline{e_0})</math><br>

Now using the angular velocity addition formula we get:<br>
<math>\underline{\omega}_2 = \underline{\omega}_1 + \underline{\underline{P}}_1 \cdot \underline{\tilde{\omega}}_2; \qquad \underline{\tilde{\omega}}_2 = \dot{\psi}\underline{e_0}</math><br>

So we got the equation:<br>
<math>\underline{\omega}_2 =  \dot{\varphi} \underline{k} + \dot{\psi}\underline{e}</math><br><br>

'''Finding second rod's center of mass velocity'''<br>
<math>\underline{\vartheta}_c = \frac{1}{2}\underline{\omega}_2 \times \underline{b} + \dot{\varphi}\underline{k}\times \underline{a} ; \qquad \underline{a} = \underline{\underline{P}}_1 \cdot a\underline{j} ; \qquad \underline{b} = \underline{\underline{P}}_1 \cdot \underline{\underline{P}}(\psi,\underline{e_0}) \cdot b\underline{j}</math>
<br><br>

'''Finding second rod's kinetic energy'''<br>
Second rod's inertia tensor:<br>
<math>\underline{\underline{\Theta}}_2 = \frac{ml^2}{12}\left(\underline{\underline{E}} - \underline{\tilde{e}\tilde{e}} \right) ;\qquad \underline{\tilde{e}} = \underline{\underline{P}}_1 \cdot \underline{\underline{P}}(\psi,\underline{e_0}) \cdot \underline{j}</math><br><br>
Now we found all expressions required for second rod's kinetic energy formula:<br>
<math>T_2 = \frac{\underline{\omega}_2 \cdot \underline{\underline{\Theta}}_2 \cdot \underline{\omega}_2}{2} + \frac{m_2 \vartheta_c^2}{2}</math>
<br><br>

'''Finding second rod's potential energy'''<br>
<math>\Pi_2 = mg(a+b - \underline{r}_c \cdot \underline{j}); \qquad \underline{r}_c = \underline{a} + \frac{1}{2}\underline{b}</math> - radius-vector of second rod's center of mass<br><br>
'''Finding the motion equation'''<br>
We need to differentiate the equations for potential and kinetic energy, the way Lagrange equation requires and use the result in it. As the result we going to get the system of two differential equations which itself is equation of motion.<br>
We notice that using this method we can get the motion equation also for big angles, it can be linearized assuming that angles <math>\varphi,\psi</math> are small and dropping terms of second order.<br>

== Using this method for special case ==
We'll check mentionad method for special case when <math>\alpha = 0</math><br> 
In this case problem becomes 2D.<br>
'''Finding rotation tensors'''<br>
<math>\underline{\underline{P}}_1(\varphi,\underline{k}) = \underline{k}\underline{k} + (\underline{\underline{E}} - \underline{k}\underline{k})cos(\varphi) + \underline{k} \times \underline{\underline{E}}sin(\varphi)</math><br><br>
<math>\underline{e}_0 = \cos(\alpha) \underline{k} + \sin(\alpha) \underline{i} = \underline{k}</math><br>
<math>\underline{e} = \underline{\underline {P}}_1 \cdot \underline{e}_0 = \underline{k}</math><br>
<math>\underline{\underline{P}}_2(\psi,\underline{e})= \underline{\underline{P}}_2(\psi,\underline{k}) = \underline{k}\underline{k} + (\underline{\underline{E}} - \underline{k}\underline{k})cos(\psi) + \underline{k} \times \underline{\underline{E}}sin(\psi)</math><br><br>

'''Finding second rod's angular vielocity'''<br>
<math>\underline{\omega}_2 = (\dot{\varphi}+\dot{\psi})\underline{k}</math><br><br>

'''Finding second rod's center of mass velocity'''<br>
<math>\upsilon^2_c = \frac{1}{4} b^2 (\dot{\varphi}+\dot{\psi})^2 + ab\cos\psi(\dot{\varphi}+\dot{\psi})\dot{\varphi} + a^2\dot{\varphi}^2</math><br><br>

'''Finding second rod's kinetic energy'''<br>
<math>T_2 = \frac{1}{2} \left( \frac{m_2 b^2}{3}(\dot{\varphi}+\dot{\psi})^2 + m_2 ab\cos\psi(\dot{\varphi}+\dot{\psi})\dot{\varphi} + m_2 a^2\dot{\varphi}^2 \right)</math><br><br>

'''Finding second rod's potential energy'''<br>
<math>\Pi_2 = m_2 g \left[ a \left(1-\cos\varphi \right) + \frac{b}{2}\left(2 + \sin\varphi\sin\psi - \cos\varphi\cos\psi \right) \right]</math><br><br>

'''Finding first rod's kinetic and potential energy'''<br>
<math>T_1 = \frac{1}{2}\frac{m_1 a^2}{3}\dot{\varphi}^2</math><br>
<math>\Pi_1 = m_1 g \left( \frac{a}{2} - \frac{a}{2} \cos \varphi \right)</math><br><br>

'''Finding equation of motion for this special case'''<br>
Let's write equations for system's kinetic and potential energy:<br>
<math>T = \frac{1}{2}\frac{m_1 a^2}{3}\dot{\varphi}^2 + \frac{1}{2} \left( \frac{m_2 b^2}{3}(\dot{\varphi}+\dot{\psi})^2 + m_2 ab\cos\psi(\dot{\varphi}+\dot{\psi})\dot{\varphi} + m_2 a^2\dot{\varphi}^2 \right)</math><br>
<math>\Pi = m_1 g \left( \frac{a}{2} - \frac{a}{2} \cos \varphi \right) + m_2 g \left[ a \left(1-\cos\varphi \right) + \frac{b}{2}\left(2 + \sin\varphi\sin\psi - \cos\varphi\cos\psi \right) \right]</math><br>
Now we differentiate both equations and linearise results assuming that <math>\varphi , \psi</math> are small and taking to account only infinitesimals of the first order. As the result we find equation of motion:<br>
<math>
\begin{cases}
    \ddot{\varphi} \left( \frac{m_1 a^2}{3} + \frac{m_2 b^2}{3} + m_2 a (a+b) \right) + \ddot{\psi} \left( \frac{m_2 b^2}{3} + \frac{m_2 ab}{2} \right) + \varphi \frac{g}{2} \left((m_1+2m_2)a+m_2 b \right)+\psi \frac{g}{2}m_2 b = 0\\
    \ddot{\varphi} \left( \frac{m_2 b^2}{3} + \frac{m_2 ab}{2} \right) + \ddot{\psi} \frac{m_2 b^2}{3} + \varphi \frac{g}{2} m_2 b + \psi \frac{g}{2} m_2 b = 0\\
\end{cases}
</math>

== Discussing the results and conclusion ==
The algorithm of solving the problem about double pendulum where both joints are cylindrical was described in this work. After that this method was used for special case of two-dimential problem.

== Related links ==
*[http://en.wikipedia.org/wiki/Double_pendulum  Double pendulum 2D Wiki]<br>
*[http://vuz.exponenta.ru/PDF/koleb2s2.html  Двумерный случай двойного маятника]

== See also ==

* [[Справка:Содержание|Справочная информация]]
* [[Курсовые проекты ТМ 20510 (2012)|Список курсовых проектов]]
* [[Кафедра "Теоретическая механика"|Department for Theoretical Mechanics]]
* [[Dzenushko Dainis]]


[[Category: Студенческие проекты]]