User:Egm6321.f12.team7.parikh.a/HW14.6

=HW - Deriving the Equations of Motion for Coupled Pendulums=

Given
The equations of motion for the coupled pendulums shown above are $$m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (1) $$m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (2)

Find
1. Derive the equation of motion (1)-(2). 2. Write equations (1)-(2) in matrix form $$\mathbf{\dot{x}}(t) = \mathbf{A}(t)\,\mathbf{x}(t) + \mathbf{B}(t) \,\mathbf{u}(t)$$ (3) with $$\mathbf{x} := \left \lfloor \theta_1 \ \dot \theta_1 \ \theta_2 \ \dot \theta_2 \right \rfloor ^T \in \mathbb R^{4 \times 1}$$ $$\mathbf{u} := \begin{Bmatrix} u_1l \\ u_2l \end{Bmatrix} \in \mathbb R^{2 \times 1}$$

Solution
This problem was solved without referring to past solutions.

Deriving the Equations of Motion
Using Newton's Second Law as it relates to rotation
 * $$\displaystyle \sum \tau := I \alpha$$ (4)

Where,
 * $$\displaystyle \tau $$ = torque about a singular principal axis.
 * $$\displaystyle I = r l^2$$ = Moment of Inertia.
 * $$\displaystyle \alpha = \ddot \theta$$ = angular acceleration

Using the definition of torque.
 * $$\displaystyle \tau := \mathbf r \ x \ \mathbf F = r F sin(\theta)$$(4)

Where r and F are the magnitudes of $$ \mathbf r$$ and $$ \mathbf F$$ respectively. We can use equation (4) to obtain the effect of each force, and then sum them up to equal the total torque. Hooke's Law,
 * $$\displaystyle \mathbf F = -k x$$

where x is the displacement of the spring's end from its equilibrium position and k is the spring constant.
 * $$\displaystyle \tau_{s1} = -k a^2 (sin(\theta_1) - sin(\theta_2)) = -k a^2 (\theta_1 - \theta_2)$$

Since the spring force acting on the second pendulum is of equal magnitude and opposite direction of the spring force acting on the first pendulum the torque created by the spring for the second pendulum is,
 * $$\displaystyle \tau_{s2} = -k a^2 (\theta_2 - \theta_1)$$

The torque due to gravity,
 * $$\displaystyle \tau_{g1} = - m_1 g l sin(\theta_1) = - m_1 g l \theta_1$$
 * $$\displaystyle \tau_{g2} = - m_2 g l sin(\theta_2) = - m_2 g l \theta_2$$

The torque due to the applied forces,
 * $$\displaystyle \tau_{u1} = u_1 l cos(\theta_1) = u_1 l$$
 * $$\displaystyle \tau_{u2} = u_2 l cos(\theta_2) = u_2 l$$

Summing all the forces,
 * $$\displaystyle \sum \tau_1 = -k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l$$
 * $$\displaystyle \sum \tau_2 = -k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l$$

Using equation (1),
 * $$\displaystyle m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (5)
 * $$\displaystyle m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (6)

Which are equivalent to equations (1) and (2).

Write the Equations of Motion in Matrix Form
We need to find an equation that takes the form: $$\begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} ... \ \ \ \\ \ ... \ \ \\ \ \ ... \ \\ \ \ \ ... \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} ... \ \\ \ \ \\ \ ... \\ \ \ \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7) Reordering equations (5) and (6) we obtain,
 * $$\displaystyle \ddot \theta_1 = \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \theta_1 + \frac{k a^2}{m_1 l^2} \theta_2 + \frac{1}{m_1 l^2} u_1 l $$ (8)


 * $$\displaystyle \ddot \theta_2 = \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \theta_2 + \frac{k a^2}{m_2 l^2} \theta_1 + \frac{1}{m_2 l^2}u_2 l $$ (9)

Entering equations (8) and (9) into equation (7) we obtain: $$\displaystyle \begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} 0 \ \ 1 \ \ 0 \ \ 0 \\ \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \ 0 \ \frac{k a^2}{m_1 l^2} \ 0 \\ \ 0 \ \ 0 \ \ 0 \ \ 1 \\ \frac{k a^2}{m_2 l^2} \ 0 \ \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \ 0 \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} 0 \ \ 0 \\ \frac{1}{m_1 l^2} \ 0 \\ 0 \ 0 \\ 0 \ \frac{1}{m_2 l^2} \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7)