*Euler-Lagrange equation*). These differential equations are the heart of Lagrangian mechanics, and indeed really what one tries to get to when applying the methods (it's essentially a way of getting Newton's 2nd Law for complicated systems). And why it is awesome. This is, like the previous part, going to be a more technical than visual, but we'll get back to that in Part 4.

Recall from Part 2 that for the double pendulum,

\[

L(\theta,\dot \theta) = T - V = \frac 1 2 (m_1+m_2) \ell_1^2 \dot{\theta}_1^2 +\frac 1 2 m_2\left( 2 \dot{\theta}_1 \dot{\theta}_2\ell_1 \ell_2 \cos(\theta_2 - \theta_1) + \ell_2^2 \dot{\theta}_2^2\right)\] \[+ (m_1+m_2) g \ell_1 \cos \theta_1 + m_2 g \ell_2 \cos \theta_2,

\]

where, $\theta_1$, $\theta_2$ are the angles from vertical of the two pendulums, $\dot \theta_1$, $\dot \theta_2$ are their angular speeds, $\ell_1$, $\ell_2$ are the length of the rods, and $m_1$, $m_2$ are their masses. This looks very imposing, so let's remark a little on the structure of that equation. In simpler situations, one often sees different squared speed terms, but here we have the

*interaction*of the two parts of the system, in terms with products like $\dot \theta_1 \dot \theta_2$, as well as the $\cos(\theta_2 - \theta_1)$. This distinguishes the system from two completely isolated (uncoupled) pendulums; the Lagrangian already captures the aspect that energy gets transferred back and forth between the two pendulums.

The equations of motion are derived from considering an

*optimization problem*, namely, what

*paths $\gamma$ through state space*connecting two states optimizes the

*action integral*

\[

\int_a^b L(\gamma(t),\dot\gamma(t)) dt.

\]

In a process quite similar to what's done in first-year calculus, we "take a derivative and set it to zero", which derives the equations

\[

\frac{\partial L}{\partial \theta_i} - \frac d {dt} \frac{\partial L}{\partial \dot \theta_i} = 0

\]

where $\partial L/\partial \dot \theta_i$ means the partial derivative with respect to $\dot \theta_i$

*as if it were*an independent variable. In fact, one way is to start off with a formula involving two independent variables $\theta$ and $\omega$; only after solving for the extremal path do we actually find $\omega = \dot \theta$ (this is a generalization called the Hamilton-Pontryagin principle).

#### $\partial L/\partial \dot {\boldsymbol \theta}$, angular momentum, and the moment of inertia

Let's now do the computation $\frac{\partial L}{\partial \dot \theta_1}$ (this quantity actually is the*angular momentum*of the first pendulum, $p_1$)—treating everything except $\dot \theta_1$ as a constant:

\[

p_1 = \frac{\partial L}{\partial \dot \theta_1} = (m_1 + m_2) \ell_1^2 \dot \theta_1 + m_2 \ell_1\ell_2

\cos(\theta_2-\theta_1)\dot \theta_2 .

\]

Note that unlike the more basic examples of angular momentum, this isn't simply dependent only on the (angular) velocity $\dot \theta_1$, but also on $\dot \theta_2$ and nonlinearly on $\theta_1$ and $\theta_2$. Similarly,

\[

p_2 = \frac{\partial L}{\partial \dot \theta_2} = m_2 \ell_2^2 \dot \theta_2 + m_2\ell_1\ell_2 \cos(\theta_2 - \theta_1) \dot \theta_1.

\]

This is more compactly expressed as

\[

\frac{\partial L}{\partial \dot{\boldsymbol{\theta}}}=\begin{pmatrix}p_1\\p_2\end{pmatrix} = \begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\dot \theta_1 \\ \dot\theta_2 \end{pmatrix}.

\]

This helps keep the unruliness under one umbrella. The square matrix in the above will be important, so let's call it $I$ (it is something like a

*moment of inertia*for our system—but it's more complicated, principally because there is an interaction between its two component parts... It's not even a rigid body!). It should also be noted that the kinetic energy is $T = \frac 1 2 \dot{\boldsymbol{\theta}}^t I \dot{\boldsymbol \theta}$, so it still is quadratic in $\dot {\boldsymbol \theta}$, just like good ol' $\frac{1}{2} mv^2$.

#### The forces, $\partial L/\partial \boldsymbol{\theta}$

Due to the nonlinear dependence of the Lagrangian on the angular variables, the derivatives are trickier. But we forge on, because sticking to just the simplest examples won't help us gain as much insight into how things are used in practice. We have

\[

\frac{\partial L}{\partial \theta_1} = m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - (m_1 + m_2) g \ell_1 \sin \theta_1

\]

\[

\frac{\partial L}{\partial \theta_1} = m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - (m_1 + m_2) g \ell_1 \sin \theta_1

\]

and

\[

\frac{\partial L}{\partial \theta_2} = -m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - m_2 g \ell_2 \sin \theta_2.

\]

\[

\frac{\partial L}{\partial \theta_2} = -m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - m_2 g \ell_2 \sin \theta_2.

\]

#### Putting it together

Now we take the time derivatives of the momenta we derived above and subtract and set to zero to derive the Euler-Lagrange equations. That's straightfoward calculus; but different from simpler examples in that, of course, the chain rule must be applied to the $\cos(\theta_2 - \theta_1)$, so there will be an extra $\dot \theta_1$ or $\dot \theta_2$ term:

\[

\frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol{\theta}}} = \begin{pmatrix} 0 & -m_2\ell_1\ell_2\sin(\theta_2-\theta_1)(\dot \theta_2 - \dot \theta_1) \\ -m_2 \ell_1 \ell_2 \sin(\theta_2 -\theta_1)(\dot \theta_2 - \dot \theta_1) & 0\end{pmatrix} \begin{pmatrix}\dot \theta_1 \\ \dot\theta_2 \end{pmatrix}\]

\[+\begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\ddot \theta_1 \\ \ddot\theta_2 \end{pmatrix}.

\]

The nice thing about it is that when it is subtracted from $\partial L/\partial \boldsymbol \theta$, the cross terms with the $\dot \theta_1 \dot\theta_2$ cancel. Also, the matrix multiplying the $\ddot{\theta}_i$'s is none other than our moment-of-inertia matrix $I$. This finally gives us

\[

0=\frac{\partial L}{\partial \boldsymbol\theta} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol\theta}} =\begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix} -I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix}

\]

as the Euler-Lagrange equations. As the final equation of motion, we bring the $\ddot {\boldsymbol \theta}$ term to the other side:

\[

I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix} = \begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix}

\]

(keep in mind that $I$ varies with $\theta_1$ and $\theta_2$, however). We discretize and solve this next time. For now, simply take note of the interesting nonlinear interactions: in the positions as the sine of the difference, and quadratic nonlinearity in the derivatives (note the reversal: the $\ddot\theta_1$ gets a $\dot \theta_2^2$ and vice versa).

\frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol{\theta}}} = \begin{pmatrix} 0 & -m_2\ell_1\ell_2\sin(\theta_2-\theta_1)(\dot \theta_2 - \dot \theta_1) \\ -m_2 \ell_1 \ell_2 \sin(\theta_2 -\theta_1)(\dot \theta_2 - \dot \theta_1) & 0\end{pmatrix} \begin{pmatrix}\dot \theta_1 \\ \dot\theta_2 \end{pmatrix}\]

\[+\begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\ddot \theta_1 \\ \ddot\theta_2 \end{pmatrix}.

\]

The nice thing about it is that when it is subtracted from $\partial L/\partial \boldsymbol \theta$, the cross terms with the $\dot \theta_1 \dot\theta_2$ cancel. Also, the matrix multiplying the $\ddot{\theta}_i$'s is none other than our moment-of-inertia matrix $I$. This finally gives us

\[

0=\frac{\partial L}{\partial \boldsymbol\theta} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol\theta}} =\begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix} -I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix}

\]

as the Euler-Lagrange equations. As the final equation of motion, we bring the $\ddot {\boldsymbol \theta}$ term to the other side:

\[

I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix} = \begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix}

\]

(keep in mind that $I$ varies with $\theta_1$ and $\theta_2$, however). We discretize and solve this next time. For now, simply take note of the interesting nonlinear interactions: in the positions as the sine of the difference, and quadratic nonlinearity in the derivatives (note the reversal: the $\ddot\theta_1$ gets a $\dot \theta_2^2$ and vice versa).

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