This webpage derives the equations of motion for the double pendulum. To see their solution, obtained via the Runge-Kutta-Fehlberg fourth-order method with fifth-order error checking (RKF45) , go to this page .
The problem of the double pendulum has been widely analysed in the mechanics literature. Its most simplified version involves two pendulums, with the second attached to the first at the bob, the rods being massless and there being no friction and air resistance. Even this simplified version is subject to chaos and has equations of motion far more complicated than that of the simple pendulum. In this webpage, we will be deriving the equations of motion with dissipative forces and rods with mass.
Figure 1: Diagram of the double pendulum. Figure 1 depicts the double pendulum system being analysed. Most physicists treat θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2
To obtain our equations of motion, we must apply the Euler-Lagrange equations with generalized dissipation force (Q i Q_i Q i [1]
d d t ∂ L ∂ q i ˙ − ∂ L ∂ q i = Q i . \begin{aligned} \dfrac{d}{dt}\dfrac{\partial \mathcal{L}}{\partial \dot{q_i}} - \dfrac{\partial \mathcal{L}}{\partial q_i} &= Q_i. \end{aligned} d t d ∂ q i ˙ ∂ L − ∂ q i ∂ L = Q i . Where
L \mathcal{L} L
( q i ) = ( θ 1 , θ 2 ) (q_i)=(\theta_1,\theta_2) ( q i ) = ( θ 1 , θ 2 )
Q i = ∑ j F ⃗ D , j ⋅ e ^ j , i Q_i=\displaystyle \sum_j \vec{F}_{D,j} \cdot \hat{e}_{j,i} Q i = j ∑ F D , j ⋅ e ^ j , i
j = 1 b , 1 r , 2 b , 2 r j=1b,1r,2b,2r j = 1 b , 1 r , 2 b , 2 r
F ⃗ D , j \vec{F}_{D,j} F D , j j j j
r ⃗ j \vec{r}_j r j j j j
e ^ j , i = ∂ r ⃗ j ∂ q i \hat{e}_{j,i}=\dfrac{\partial \vec{r}_j}{\partial q_i} e ^ j , i = ∂ q i ∂ r j
The first pendulum bob would have the coordinates, velocity and generalized basis vector
x 1 b = l 1 cos θ 1 ⟹ x ˙ 1 b = − l 1 θ ˙ 1 sin θ 1 y 1 b = l 1 sin θ 1 ⟹ y ˙ 1 b = l 1 θ ˙ 1 cos θ 1 r ⃗ 1 b = [ x 1 b y 1 b ] ∴ v ⃗ 1 b = d r ⃗ 1 b d t = l 1 [ − sin θ 1 cos θ 1 ] = l 1 θ ˙ 1 [ − sin θ 1 cos θ 1 ] e ⃗ 1 b , θ 1 = ∂ r ⃗ 1 b ∂ θ 1 v 1 b 2 = x ˙ 1 b 2 + y ˙ 1 b 2 = l 1 [ − sin θ 1 cos θ 1 ] = ( − l 1 θ ˙ 1 sin θ 1 ) 2 + ( l 1 θ ˙ 1 cos θ 1 ) 2 e ⃗ 1 b , θ 2 = ∂ r ⃗ 1 b ∂ θ 2 = l 1 2 θ ˙ 1 2 = [ 0 0 ] . \begin{aligned} x_{1b} &= l_1 \cos{\theta_1} &\implies \dot{x}_{1b} &= -l_1\dot{\theta}_1 \sin{\theta_1}\\ y_{1b} &= l_1 \sin{\theta_1} &\implies \dot{y}_{1b} &= l_1 \dot{\theta}_1 \cos{\theta_1} \\ \vec{r}_{1b} &= \begin{bmatrix} x_{1b}\\ y_{1b} \end{bmatrix} &\therefore \vec{v}_{1b} &= \dfrac{d \vec{r}_{1b}}{dt}\\ &= l_1\begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix} & &=l_1\dot{\theta}_1 \begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix} \\ \vec{e}_{1b,\theta_1} &= \dfrac{\partial \vec{r}_{1b}}{\partial \theta_1}& v_{1b}^2 &= \dot{x}_{1b}^2 + \dot{y}_{1b}^2 \\ &= l_1 \begin{bmatrix} -\sin{\theta_1}\\ \cos{\theta_1} \end{bmatrix} & &= \left(-l_1\dot{\theta}_1 \sin{\theta_1}\right)^2 + \left(l_1 \dot{\theta}_1 \cos{\theta_1}\right)^2 \\ \vec{e}_{1b,\theta_2} &= \dfrac{\partial \vec{r}_{1b}}{\partial \theta_2} &&= l_1^2 \dot{\theta}_1^2\\ &= \begin{bmatrix} 0\\ 0 \end{bmatrix}. \end{aligned} x 1 b y 1 b r 1 b e 1 b , θ 1 e 1 b , θ 2 = l 1 cos θ 1 = l 1 sin θ 1 = [ x 1 b y 1 b ] = l 1 [ − sin θ 1 cos θ 1 ] = ∂ θ 1 ∂ r 1 b = l 1 [ − sin θ 1 cos θ 1 ] = ∂ θ 2 ∂ r 1 b = [ 0 0 ] . ⟹ x ˙ 1 b ⟹ y ˙ 1 b ∴ v 1 b v 1 b 2 = − l 1 θ ˙ 1 sin θ 1 = l 1 θ ˙ 1 cos θ 1 = d t d r 1 b = l 1 θ ˙ 1 [ − sin θ 1 cos θ 1 ] = x ˙ 1 b 2 + y ˙ 1 b 2 = ( − l 1 θ ˙ 1 sin θ 1 ) 2 + ( l 1 θ ˙ 1 cos θ 1 ) 2 = l 1 2 θ ˙ 1 2 The first pendulum rod would have the coordinates, velocity and generalized basis vector below. It is important to note that we are analysing its effect on the motion of the pendulum based on the centre of mass approach.
x 1 r = l 1 cos θ 1 2 ⟹ x ˙ 1 r = − l 1 θ ˙ 1 sin θ 1 2 y 1 r = l 1 sin θ 1 2 ⟹ y ˙ 1 r = l 1 θ ˙ 1 cos θ 1 2 r ⃗ 1 r = [ x 1 r y 1 r ] ∴ v ⃗ 1 r = d r ⃗ 1 r d t = l 1 2 [ cos θ 1 sin θ 1 ] = l 1 θ ˙ 1 2 [ − sin θ 1 cos θ 1 ] e ⃗ 1 r , θ 1 = ∂ r ⃗ 1 r ∂ θ 1 v 1 r 2 = x ˙ 1 r 2 + y ˙ 1 r 2 = l 1 2 [ − sin θ 1 cos θ 1 ] = ( − l 1 θ ˙ 1 sin θ 1 2 ) 2 + ( l 1 θ ˙ 1 cos θ 1 2 ) 2 e ⃗ 1 r , θ 2 = ∂ r ⃗ 1 r ∂ θ 2 = l 1 2 θ ˙ 1 2 4 = [ 0 0 ] . \begin{aligned} x_{1r} &= \dfrac{l_1\cos{\theta_1}}{2} &\implies \dot{x}_{1r} &= -\dfrac{l_1\dot{\theta}_1\sin{\theta_1}}{2}\\ y_{1r} &= \dfrac{l_1\sin{\theta_1}}{2} &\implies \dot{y}_{1r} &= \dfrac{l_1\dot{\theta}_1\cos{\theta_1}}{2}\\ \vec{r}_{1r} &= \begin{bmatrix} x_{1r} \\ y_{1r} \end{bmatrix} &\therefore \vec{v}_{1r} &= \dfrac{d\vec{r}_{1r}}{dt} \\ &= \dfrac{l_1}{2}\begin{bmatrix} \cos{\theta_1}\\ \sin{\theta_1} \end{bmatrix} & &= \dfrac{l_1\dot{\theta}_1}{2} \begin{bmatrix} -\sin{\theta_1}\\ \cos{\theta_1} \end{bmatrix} \\ \vec{e}_{1r,\theta_1} &= \dfrac{\partial \vec{r}_{1r}}{\partial \theta_1} & v_{1r}^2 &= \dot{x}_{1r}^2 + \dot{y}_{1r}^2 \\ &= \dfrac{l_1}{2} \begin{bmatrix} -\sin{\theta_1}\\ \cos{\theta_1} \end{bmatrix} & &= \left(-\dfrac{l_1\dot{\theta}_1 \sin{\theta_1}}{2}\right)^2 + \left(\dfrac{l_1 \dot{\theta}_1 \cos{\theta_1}}{2}\right)^2 \\ \vec{e}_{1r,\theta_2} &= \dfrac{\partial \vec{r}_{1r}}{\partial \theta_2} &&= \dfrac{l_1^2 \dot{\theta}_1^2}{4} \\ &= \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{aligned} x 1 r y 1 r r 1 r e 1 r , θ 1 e 1 r , θ 2 = 2 l 1 cos θ 1 = 2 l 1 sin θ 1 = [ x 1 r y 1 r ] = 2 l 1 [ cos θ 1 sin θ 1 ] = ∂ θ 1 ∂ r 1 r = 2 l 1 [ − sin θ 1 cos θ 1 ] = ∂ θ 2 ∂ r 1 r = [ 0 0 ] . ⟹ x ˙ 1 r ⟹ y ˙ 1 r ∴ v 1 r v 1 r 2 = − 2 l 1 θ ˙ 1 sin θ 1 = 2 l 1 θ ˙ 1 cos θ 1 = d t d r 1 r = 2 l 1 θ ˙ 1 [ − sin θ 1 cos θ 1 ] = x ˙ 1 r 2 + y ˙ 1 r 2 = ( − 2 l 1 θ ˙ 1 sin θ 1 ) 2 + ( 2 l 1 θ ˙ 1 cos θ 1 ) 2 = 4 l 1 2 θ ˙ 1 2 x 2 b = x 1 b + l 2 cos θ 2 x ˙ 2 b = − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 = l 1 cos θ 1 + l 2 cos θ 2 y ˙ 2 b = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 y 2 b = y 1 b + l 2 sin θ 2 ∴ v ⃗ 2 b = [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] = l 1 sin θ 1 + l 2 sin θ 2 ∴ v 2 b 2 = x ˙ 2 b 2 + y ˙ 2 b 2 r ⃗ 2 b = [ x 2 b y 2 b ] = ( − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 ) 2 + ( l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ) 2 = [ l 1 cos θ 1 + l 2 cos θ 2 l 1 sin θ 1 + l 2 sin θ 2 ] = l 1 2 θ ˙ 1 2 sin 2 θ 1 + l 2 2 θ ˙ 2 2 sin 2 θ 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ ˙ 1 2 cos 2 θ 1 + l 2 2 θ ˙ 2 2 cos 2 θ 2 + e ⃗ 2 b , θ 1 = ∂ r ⃗ 2 b ∂ θ 1 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos θ 1 cos θ 2 = l 1 [ − sin θ 1 cos θ 1 ] = l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) e ⃗ 2 b , θ 2 = ∂ r ⃗ 2 b ∂ θ 2 = l 2 [ − sin θ 2 cos θ 2 ] . \begin{aligned} x_{2b} &= x_{1b} + l_2 \cos{\theta_2} &\dot{x}_{2b} &= -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2}\\ &= l_1 \cos{\theta_1} + l_2 \cos{\theta_2} & \dot{y}_{2b} &= l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \\ y_{2b} &= y_{1b} + l_2 \sin{\theta_2} & \therefore \vec{v}_{2b} &= \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2}\\ l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \end{bmatrix}\\ &= l_1 \sin{\theta_1} + l_2 \sin{\theta_2} & \therefore v_{2b}^2 &= \dot{x}_{2b}^2 + \dot{y}_{2b}^2 \\ \vec{r}_{2b}&= \begin{bmatrix} x_{2b} \\ y_{2b} \end{bmatrix}& &= \left(-l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2}\right)^2 + \left(l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2}\right)^2 \\ &= \begin{bmatrix} l_1 \cos{\theta_1} + l_2 \cos{\theta_2}\\ l_1 \sin{\theta_1} + l_2 \sin{\theta_2} \end{bmatrix} & &= l_1^2 \dot{\theta}_1^2 \sin^2{\theta_1} + l_2^2 \dot{\theta}_2^2 \sin^2{\theta_2} + 2l_1 l_2 \dot{\theta}_1\dot{\theta}_2 \sin{\theta_1}\sin{\theta_2} + l_1^2 \dot{\theta}_1^2 \cos^2{\theta_1} + l_2^2 \dot{\theta}_2^2 \cos^2{\theta_2} + \\ \vec{e}_{2b,\theta_1} &= \dfrac{\partial \vec{r}_{2b}}{\partial \theta_1} && 2l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\theta_1}\cos{\theta_2} \\ &= l_1\begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix}& &= l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 + 2l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\left(\theta_1-\theta_2\right)}\\ \vec{e}_{2b,\theta_2} &= \dfrac{\partial \vec{r}_{2b}}{\partial \theta_2} \\ &= l_2 \begin{bmatrix} -\sin{\theta_2} \\ \cos{\theta_2} \end{bmatrix}. \end{aligned} x 2 b y 2 b r 2 b e 2 b , θ 1 e 2 b , θ 2 = x 1 b + l 2 cos θ 2 = l 1 cos θ 1 + l 2 cos θ 2 = y 1 b + l 2 sin θ 2 = l 1 sin θ 1 + l 2 sin θ 2 = [ x 2 b y 2 b ] = [ l 1 cos θ 1 + l 2 cos θ 2 l 1 sin θ 1 + l 2 sin θ 2 ] = ∂ θ 1 ∂ r 2 b = l 1 [ − sin θ 1 cos θ 1 ] = ∂ θ 2 ∂ r 2 b = l 2 [ − sin θ 2 cos θ 2 ] . x ˙ 2 b y ˙ 2 b ∴ v 2 b ∴ v 2 b 2 = − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 = [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] = x ˙ 2 b 2 + y ˙ 2 b 2 = ( − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 ) 2 + ( l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ) 2 = l 1 2 θ ˙ 1 2 sin 2 θ 1 + l 2 2 θ ˙ 2 2 sin 2 θ 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ ˙ 1 2 cos 2 θ 1 + l 2 2 θ ˙ 2 2 cos 2 θ 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos θ 1 cos θ 2 = l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) x 2 r = x 1 b + l 2 cos θ 2 2 x ˙ 2 r = − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 = l 1 cos θ 1 + l 2 cos θ 2 2 y ˙ 2 r = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 y 2 r = y 1 b + l 2 sin θ 2 2 ∴ v ⃗ 2 r = [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 2 ] = l 1 sin θ 1 + l 2 sin θ 2 2 ∴ v 2 r 2 = x ˙ 2 r 2 + y ˙ 2 r 2 r ⃗ 2 r = [ x 2 r y 2 r ] = ( − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 ) 2 + ( l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 2 ) 2 = [ l 1 cos θ 1 + l 2 cos θ 2 2 l 1 sin θ 1 + l 2 sin θ 2 2 ] = l 1 2 θ ˙ 1 2 sin 2 θ 1 + l 2 2 θ ˙ 2 2 sin 2 θ 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ ˙ 1 2 cos 2 θ 1 + l 2 2 θ ˙ 2 2 cos 2 θ 2 4 + e ⃗ 2 b , θ 1 = ∂ r ⃗ 2 b ∂ θ 1 l 1 l 2 θ ˙ 1 θ ˙ 2 cos θ 1 cos θ 2 = l 1 [ − sin θ 1 cos θ 1 ] = l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) e ⃗ 2 b , θ 2 = ∂ r ⃗ 2 b ∂ θ 2 = l 2 2 [ − sin θ 2 cos θ 2 ] . \begin{aligned} x_{2r} &= x_{1b} + \dfrac{l_2 \cos{\theta_2}}{2} &\dot{x}_{2r} &= -l_1\dot{\theta}_1 \sin{\theta_1}-\dfrac{l_2\dot{\theta}_2 \sin{\theta_2}}{2}\\ &= l_1 \cos{\theta_1} + \dfrac{l_2 \cos{\theta_2}}{2} & \dot{y}_{2r} &= l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \\ y_{2r} &= y_{1b} + \dfrac{l_2 \sin{\theta_2}}{2} & \therefore \vec{v}_{2r} &= \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-\dfrac{l_2\dot{\theta}_2 \sin{\theta_2}}{2}\\ l_1\dot{\theta}_1 \cos{\theta_1}+\dfrac{l_2\dot{\theta}_2 \cos{\theta_2}}{2} \end{bmatrix}\\ &= l_1 \sin{\theta_1} + \dfrac{l_2 \sin{\theta_2}}{2} & \therefore v_{2r}^2 &= \dot{x}_{2r}^2 + \dot{y}_{2r}^2 \\ \vec{r}_{2r}&= \begin{bmatrix} x_{2r} \\ y_{2r} \end{bmatrix}& &= \left(-l_1\dot{\theta}_1 \sin{\theta_1}-\dfrac{l_2\dot{\theta}_2 \sin{\theta_2}}{2}\right)^2 + \left(l_1\dot{\theta}_1 \cos{\theta_1}+\dfrac{l_2\dot{\theta}_2 \cos{\theta_2}}{2}\right)^2 \\ &= \begin{bmatrix} l_1 \cos{\theta_1} + \dfrac{l_2 \cos{\theta_2}}{2}\\ l_1 \sin{\theta_1} + \dfrac{l_2 \sin{\theta_2}}{2} \end{bmatrix} & &= l_1^2 \dot{\theta}_1^2 \sin^2{\theta_1} + \dfrac{l_2^2 \dot{\theta}_2^2 \sin^2{\theta_2}}{4} + l_1 l_2 \dot{\theta}_1\dot{\theta}_2 \sin{\theta_1}\sin{\theta_2} + l_1^2 \dot{\theta}_1^2 \cos^2{\theta_1} + \dfrac{l_2^2 \dot{\theta}_2^2 \cos^2{\theta_2}}{4} + \\ \vec{e}_{2b,\theta_1} &= \dfrac{\partial \vec{r}_{2b}}{\partial \theta_1} && l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\theta_1}\cos{\theta_2} \\ &= l_1\begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix}& &= l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\left(\theta_1-\theta_2\right)}\\ \vec{e}_{2b,\theta_2} &= \dfrac{\partial \vec{r}_{2b}}{\partial \theta_2} \\ &= \dfrac{l_2}{2} \begin{bmatrix} -\sin{\theta_2} \\ \cos{\theta_2} \end{bmatrix}. \end{aligned} x 2 r y 2 r r 2 r e 2 b , θ 1 e 2 b , θ 2 = x 1 b + 2 l 2 cos θ 2 = l 1 cos θ 1 + 2 l 2 cos θ 2 = y 1 b + 2 l 2 sin θ 2 = l 1 sin θ 1 + 2 l 2 sin θ 2 = [ x 2 r y 2 r ] = ⎣ ⎢ ⎢ ⎡ l 1 cos θ 1 + 2 l 2 cos θ 2 l 1 sin θ 1 + 2 l 2 sin θ 2 ⎦ ⎥ ⎥ ⎤ = ∂ θ 1 ∂ r 2 b = l 1 [ − sin θ 1 cos θ 1 ] = ∂ θ 2 ∂ r 2 b = 2 l 2 [ − sin θ 2 cos θ 2 ] . x ˙ 2 r y ˙ 2 r ∴ v 2 r ∴ v 2 r 2 = − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 = ⎣ ⎢ ⎢ ⎡ − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + 2 l 2 θ ˙ 2 cos θ 2 ⎦ ⎥ ⎥ ⎤ = x ˙ 2 r 2 + y ˙ 2 r 2 = ( − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 ) 2 + ( l 1 θ ˙ 1 cos θ 1 + 2 l 2 θ ˙ 2 cos θ 2 ) 2 = l 1 2 θ ˙ 1 2 sin 2 θ 1 + 4 l 2 2 θ ˙ 2 2 sin 2 θ 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ ˙ 1 2 cos 2 θ 1 + 4 l 2 2 θ ˙ 2 2 cos 2 θ 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos θ 1 cos θ 2 = l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) First, we will calculate the kinetic energy of the first pendulum bob.
T 1 b = m 1 b v 1 b 2 2 . \begin{aligned} T_{1b} &= \dfrac{m_{1b}v_{1b}^2}{2}. \end{aligned} T 1 b = 2 m 1 b v 1 b 2 . Substituting in what we previously found v 1 b 2 v_{1b}^2 v 1 b 2
T 1 b = m 1 b l 1 2 θ ˙ 1 2 2 . \begin{aligned} T_{1b} &= \dfrac{m_{1b} l_1^2 \dot{\theta}_1^2}{2}. \end{aligned} T 1 b = 2 m 1 b l 1 2 θ ˙ 1 2 . First we must determine the kinetic energy of the first rod, it is given by
T 1 r = 1 2 I c m ω c m 2 . \begin{aligned} T_{1r} &= \dfrac{1}{2} I_{\mathrm{cm}} \omega_{\mathrm{cm}}^2. \end{aligned} T 1 r = 2 1 I c m ω c m 2 . Where I c m I_{\mathrm{cm}} I c m ω c m \omega_{\mathrm{cm}} ω c m I c m = m 1 r l 1 2 12 I_{\mathrm{cm}} = \dfrac{m_{1r}l_1^2}{12} I c m = 1 2 m 1 r l 1 2 [2] ω c m = θ ˙ 1 \omega_{\mathrm{cm}} = \dot{\theta}_1 ω c m = θ ˙ 1
T 1 r = 1 2 m 1 r l 1 2 12 θ ˙ 1 2 = m 1 r l 1 2 θ ˙ 1 2 24 . \begin{aligned} T_{1r} &= \dfrac{1}{2}\dfrac{m_{1r} l_1^2}{12}\dot{\theta}_1^2 \\ &= \dfrac{m_{1r} l_1^2\dot{\theta}_1^2}{24}. \end{aligned} T 1 r = 2 1 1 2 m 1 r l 1 2 θ ˙ 1 2 = 2 4 m 1 r l 1 2 θ ˙ 1 2 . The kinetic energy of the second bob is
T 2 b = m 2 b v 2 b 2 2 = m 2 b 2 ( l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . \begin{aligned} T_{2b} &= \dfrac{m_{2b} v_{2b}^2}{2} \\ &= \dfrac{m_{2b}}{2} \left(l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 + 2l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\left(\theta_1-\theta_2\right)}\right). \end{aligned} T 2 b = 2 m 2 b v 2 b 2 = 2 m 2 b ( l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . The second rod should be the same except as rod 1 but with its own parameters
T 2 r = m 2 r l 2 2 θ ˙ 2 2 24 . \begin{aligned} T_{2r} &= \dfrac{m_{2r} l_2^2\dot{\theta}_2^2}{24}. \end{aligned} T 2 r = 2 4 m 2 r l 2 2 θ ˙ 2 2 . Hence the total kinetic energy is
T = ∑ j T j = T 1 r + T 1 b + T 2 r + T 2 b = m 1 r l 1 2 θ ˙ 1 2 24 + m 1 b l 1 2 θ ˙ 1 2 2 + m 2 r l 2 2 θ ˙ 2 2 24 + m 2 b 2 ( l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) = m 1 r l 1 2 θ ˙ 1 2 + m 2 r l 2 2 θ ˙ 2 2 24 + m 1 b + m 2 b 2 l 1 2 θ ˙ 1 2 + m 2 b 2 ( l 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) \begin{aligned} T &= \sum_j T_j \\ &= T_{1r} + T_{1b} + T_{2r} + T_{2b} \\ &= \dfrac{m_{1r}l_1^2\dot{\theta}_1^2}{24} + \dfrac{m_{1b} l_1^2 \dot{\theta}_1^2}{2} + \dfrac{m_{2r} l_2^2\dot{\theta}_2^2}{24} + \dfrac{m_{2b}}{2} \left(l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 + 2l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\left(\theta_1-\theta_2\right)}\right) \\ &= \dfrac{m_{1r} l_1^2 \dot{\theta}_1^2 + m_{2r}l_2^2 \dot{\theta}_2^2}{24} + \dfrac{m_{1b}+m_{2b}}{2}l_1^2 \dot{\theta}_1^2 + \dfrac{m_{2b}}{2} \left(l_2\dot{\theta}_2^2 + 2l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}\right) \end{aligned} T = j ∑ T j = T 1 r + T 1 b + T 2 r + T 2 b = 2 4 m 1 r l 1 2 θ ˙ 1 2 + 2 m 1 b l 1 2 θ ˙ 1 2 + 2 4 m 2 r l 2 2 θ ˙ 2 2 + 2 m 2 b ( l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) = 2 4 m 1 r l 1 2 θ ˙ 1 2 + m 2 r l 2 2 θ ˙ 2 2 + 2 m 1 b + m 2 b l 1 2 θ ˙ 1 2 + 2 m 2 b ( l 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) T = l 1 2 θ ˙ 1 2 2 ( m 1 r 12 + m 1 b + m 2 b ) + l 2 2 θ ˙ 2 2 2 ( m 2 r 12 + m 2 b ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) . \begin{aligned} T &= \dfrac{l_1^2 \dot{\theta}_1^2}{2} \left(\dfrac{m_{1r}}{12} + m_{1b} + m_{2b}\right) + \dfrac{l_2^2 \dot{\theta}_2^2}{2} \left(\dfrac{m_{2r}}{12} + m_{2b}\right) + m_{2b}l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}. \end{aligned} T = 2 l 1 2 θ ˙ 1 2 ( 1 2 m 1 r + m 1 b + m 2 b ) + 2 l 2 2 θ ˙ 2 2 ( 1 2 m 2 r + m 2 b ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) . Next, we must calculate the potential energy. In this problem, there is only one source of potential energy–-gravity. This means that the potential energy of each component of the pendulum will be V j = m j g y j V_j = m_jgy_j V j = m j g y j y j y_j y j y y y m j m_j m j
First, we have the gravitational potential energy of the first pendulum bob.
V 1 b = m 1 b g y 1 b = m 1 b g l 1 sin θ 1 . \begin{aligned} V_{1b} &= m_{1b} gy_{1b} \\ &= m_{1b}gl_1 \sin{\theta_1}. \end{aligned} V 1 b = m 1 b g y 1 b = m 1 b g l 1 sin θ 1 . Next, we will calculate the gravitational potential energy of the first rod. For it, we will use the centre of mass of the pendulum for the y y y y 1 r = 1 2 y 1 b y_{1r} = \dfrac{1}{2}y_{1b} y 1 r = 2 1 y 1 b
V 1 r = m 1 r g y 1 r = m 1 r g l 1 sin θ 1 2 . \begin{aligned} V_{1r} &= m_{1r} gy_{1r} \\ &= \dfrac{m_{1r}gl_1 \sin{\theta_1}}{2}. \end{aligned} V 1 r = m 1 r g y 1 r = 2 m 1 r g l 1 sin θ 1 . The potential energy of the second pendulum bob is
V 2 b = m 2 b g y 2 b = m 2 b g ( l 1 sin θ 1 + l 2 sin θ 2 ) . \begin{aligned} V_{2b} &= m_{2b} gy_{2b} \\ &= m_{2b} g \left(l_1 \sin{\theta_1} + l_2 \sin{\theta_2}\right). \end{aligned} V 2 b = m 2 b g y 2 b = m 2 b g ( l 1 sin θ 1 + l 2 sin θ 2 ) . The potential energy of the second pendulum rod is
V 2 r = m 2 r g y 2 r = m 2 r g ( l 1 sin θ 1 + l 2 sin θ 2 2 ) . \begin{aligned} V_{2r} &= m_{2r} gy_{2r} \\ &= m_{2r}g \left(l_1 \sin{\theta_1} + \dfrac{l_2\sin{\theta_2}}{2}\right). \end{aligned} V 2 r = m 2 r g y 2 r = m 2 r g ( l 1 sin θ 1 + 2 l 2 sin θ 2 ) . Therefore, the total potential energy is
V = ∑ j V j = V 1 r + V 2 r + V 1 b + V 2 b = m 1 r g l 1 sin θ 1 2 + m 2 r g ( l 1 sin θ 1 + l 2 sin θ 2 2 ) + m 1 b g l 1 sin θ 1 + m 2 b g ( l 1 sin θ 1 + l 2 sin θ 2 ) = g l 1 sin θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) + g l 2 sin θ 2 ( m 2 r 2 + m 2 b ) . \begin{aligned} V &= \sum_j V_j \\ &= V_{1r} + V_{2r} + V_{1b} + V_{2b} \\ &= \dfrac{m_{1r}gl_1 \sin{\theta_1}}{2} + m_{2r}g \left(l_1 \sin{\theta_1} + \dfrac{l_2\sin{\theta_2}}{2}\right) + m_{1b}gl_1 \sin{\theta_1} + m_{2b} g \left(l_1 \sin{\theta_1} + l_2 \sin{\theta_2}\right) \\ &= gl_1 \sin{\theta_1}\left(\dfrac{m_{1r}}{2} + m_{2r} + m_{1b} + m_{2b}\right) + gl_2 \sin{\theta_2}\left(\dfrac{m_{2r}}{2} + m_{2b}\right). \end{aligned} V = j ∑ V j = V 1 r + V 2 r + V 1 b + V 2 b = 2 m 1 r g l 1 sin θ 1 + m 2 r g ( l 1 sin θ 1 + 2 l 2 sin θ 2 ) + m 1 b g l 1 sin θ 1 + m 2 b g ( l 1 sin θ 1 + l 2 sin θ 2 ) = g l 1 sin θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) + g l 2 sin θ 2 ( 2 m 2 r + m 2 b ) . Hence the Lagrangian is
L = T − V = l 1 2 θ ˙ 1 2 2 ( m 1 r 12 + m 1 b + m 2 b ) + l 2 2 θ ˙ 2 2 2 ( m 2 r 12 + m 2 b ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) − g l 1 sin θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − g l 2 sin θ 2 ( m 2 r 2 + m 2 b ) . \begin{aligned} \mathcal{L} &= T - V \\ &= \dfrac{l_1^2 \dot{\theta}_1^2}{2} \left(\dfrac{m_{1r}}{12} + m_{1b} + m_{2b}\right) + \dfrac{l_2^2 \dot{\theta}_2^2}{2} \left(\dfrac{m_{2r}}{12} + m_{2b}\right) + m_{2b}l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)} - gl_1 \sin{\theta_1}\left(\dfrac{m_{1r}}{2} \right.\\ &\left.+ m_{2r} + m_{1b} + m_{2b}\right) - gl_2 \sin{\theta_2}\left(\dfrac{m_{2r}}{2} + m_{2b}\right). \end{aligned} L = T − V = 2 l 1 2 θ ˙ 1 2 ( 1 2 m 1 r + m 1 b + m 2 b ) + 2 l 2 2 θ ˙ 2 2 ( 1 2 m 2 r + m 2 b ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) − g l 1 sin θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − g l 2 sin θ 2 ( 2 m 2 r + m 2 b ) . Calculating the left-hand side terms of Equation (1 ) with regards to θ 1 \theta_1 θ 1 θ 1 \theta_1 θ 1
p θ 1 = ∂ L ∂ θ ˙ 1 = m 1 r l 1 2 θ ˙ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ˙ 1 + m 2 b l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) . \begin{aligned} p_{\theta_1} &= \dfrac{\partial \mathcal{L}}{\partial \dot{\theta}_1} \\ &= \dfrac{m_{1r}l_1^2 \dot{\theta}_1}{12} + (m_{1b}+m_{2b}) l_1^2 \dot{\theta}_1 + m_{2b}l_1 l_2 \dot{\theta}_2 \cos{\left(\theta_1-\theta_2\right)}. \end{aligned} p θ 1 = ∂ θ ˙ 1 ∂ L = 1 2 m 1 r l 1 2 θ ˙ 1 + ( m 1 b + m 2 b ) l 1 2 θ ˙ 1 + m 2 b l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) . Hence its time derivative is
p ˙ θ 1 = m 1 r l 1 2 θ ¨ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) . \begin{aligned} \dot{p}_{\theta_1} &= \dfrac{m_{1r} l_1^2 \ddot{\theta}_1}{12} + (m_{1b}+m_{2b})l_1^2 \ddot{\theta}_1 + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} - m_{2b}l_1 l_2 \dot{\theta}_2\left(\dot{\theta}_1 - \dot{\theta}_2\right)\sin{(\theta_1-\theta_2)}. \end{aligned} p ˙ θ 1 = 1 2 m 1 r l 1 2 θ ¨ 1 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) . As for the other term in Equation (1 )
F θ 1 = ∂ L ∂ θ 1 = − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 1 r g l 1 cos θ 1 2 − m 2 r g l 1 cos θ 1 − m 1 b g l 1 cos θ 1 − m 2 b g l 1 cos θ 1 = − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) . \begin{aligned} F_{\theta_1} &= \dfrac{\partial \mathcal{L}}{\partial \theta_1} \\ &= -m_{2b}l_1l_2\dot{\theta}_1\dot{\theta}_2 \sin{(\theta_1-\theta_2)} - \dfrac{m_{1r}gl_1 \cos{\theta_1}}{2} -m_{2r}gl_1 \cos{\theta_1} -m_{1b}gl_1 \cos{\theta_1} -m_{2b}gl_1 \cos{\theta_1} \\ &= -m_{2b}l_1l_2\dot{\theta}_1\dot{\theta}_2 \sin{(\theta_1-\theta_2)} -gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right). \end{aligned} F θ 1 = ∂ θ 1 ∂ L = − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − 2 m 1 r g l 1 cos θ 1 − m 2 r g l 1 cos θ 1 − m 1 b g l 1 cos θ 1 − m 2 b g l 1 cos θ 1 = − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) . Hence the LHS of Equation (1 ) for θ 1 \theta_1 θ 1
p ˙ θ 1 − F θ 1 = m 1 r l 1 2 θ ¨ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) = m 1 r l 1 2 θ ¨ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) = l 1 2 θ ¨ 1 ( m 1 r 12 + m 1 b + m 2 b ) + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) = l 1 2 θ ¨ 1 ( m 1 r 12 + m 1 b + m 2 b ) + m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) + g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) . \begin{aligned} \dot{p}_{\theta_1} - F_{\theta_1} &= \dfrac{m_{1r} l_1^2 \ddot{\theta}_1}{12} + (m_{1b}+m_{2b})l_1^2 \ddot{\theta}_1 + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} - m_{2b}l_1 l_2 \dot{\theta}_2\left(\dot{\theta}_1 - \dot{\theta}_2\right)\sin{(\theta_1-\theta_2)} + m_{2b}l_1l_2\dot{\theta}_1\dot{\theta}_2\sin{(\theta_1-\theta_2)} +gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right) \\ &= \dfrac{m_{1r} l_1^2 \ddot{\theta}_1}{12} + (m_{1b}+m_{2b})l_1^2 \ddot{\theta}_1 + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} + m_{2b}l_1 l_2 \dot{\theta}_2^2\sin{(\theta_1-\theta_2)} +gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right) \\ &= l_1^2 \ddot{\theta}_1 \left(\dfrac{m_{1r}}{12} + m_{1b} + m_{2b}\right) + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} + m_{2b}l_1 l_2 \dot{\theta}_2^2\sin{(\theta_1-\theta_2)} +gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right)\\ &= l_1^2 \ddot{\theta}_1 \left(\dfrac{m_{1r}}{12} + m_{1b} + m_{2b}\right) + m_{2b}l_1 l_2\left( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} + \dot{\theta}_2^2\sin{(\theta_1-\theta_2)}\right) +gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right). \end{aligned} p ˙ θ 1 − F θ 1 = 1 2 m 1 r l 1 2 θ ¨ 1 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) = 1 2 m 1 r l 1 2 θ ¨ 1 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) = l 1 2 θ ¨ 1 ( 1 2 m 1 r + m 1 b + m 2 b ) + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) = l 1 2 θ ¨ 1 ( 1 2 m 1 r + m 1 b + m 2 b ) + m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) + g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) . Hence θ ¨ 1 \ddot{\theta}_1 θ ¨ 1
θ ¨ 1 = 1 l 1 2 ( m 1 r 12 + m 1 b + m 2 b ) [ Q θ 1 − m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) ] . \begin{aligned} \ddot{\theta}_1 &= \dfrac{1}{l_1^2\left(\dfrac{m_{1r}}{12} + m_{1b} + m_{2b}\right)}\left[Q_{\theta_1} - m_{2b}l_1 l_2\left( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} + \dot{\theta}_2^2\sin{(\theta_1-\theta_2)}\right) -gl_1\cos{\theta_1}\left(\dfrac{m_{1r}}{2}+m_{2r}+m_{1b} + m_{2b}\right)\right]. \end{aligned} θ ¨ 1 = l 1 2 ( 1 2 m 1 r + m 1 b + m 2 b ) 1 [ Q θ 1 − m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) ] . As for the generalized dissipation force
Q θ 1 = F ⃗ D , 1 r ⋅ ∂ r ⃗ 1 r ∂ θ 1 + F ⃗ D , 1 b ⋅ ∂ r ⃗ 1 b ∂ θ 1 + F ⃗ D , 2 r ⋅ ∂ r ⃗ r o d , 2 ∂ θ 1 + F ⃗ D , 2 b ⋅ ∂ r ⃗ 2 b ∂ θ 1 \begin{aligned} Q_{\theta_1} &= \vec{F}_{D, 1r} \cdot \dfrac{\partial \vec{r}_{1r}}{\partial \theta_1} + \vec{F}_{D, 1b} \cdot \dfrac{\partial \vec{r}_{1b}}{\partial \theta_1}+ \vec{F}_{D, 2r} \cdot \dfrac{\partial \vec{r}_{rod, 2}}{\partial \theta_1} + \vec{F}_{D, 2b} \cdot \dfrac{\partial \vec{r}_{2b}}{\partial \theta_1} \end{aligned} Q θ 1 = F D , 1 r ⋅ ∂ θ 1 ∂ r 1 r + F D , 1 b ⋅ ∂ θ 1 ∂ r 1 b + F D , 2 r ⋅ ∂ θ 1 ∂ r r o d , 2 + F D , 2 b ⋅ ∂ θ 1 ∂ r 2 b The dissipation force applied to the first bob is
F ⃗ D , 1 b = − b 1 b v ⃗ 1 b − c 1 b ∣ v ⃗ 1 b ∣ v ⃗ 1 b . \begin{aligned} \vec{F}_{D,1b} &= -b_{1b}\vec{v}_{1b} - c_{1b}|\vec{v}_{1b}|\vec{v}_{1b}. \end{aligned} F D , 1 b = − b 1 b v 1 b − c 1 b ∣ v 1 b ∣ v 1 b . Hence the generalized dissipation force, canonical to θ 1 \theta_1 θ 1
F ⃗ D , 1 b ⋅ ∂ r ⃗ 1 b ∂ θ 1 = ( − b 1 b − c 1 b l 1 θ ˙ 1 ) [ − l 1 θ ˙ 1 sin θ 1 l 1 θ ˙ 1 cos θ 1 ] ⋅ [ − l 1 sin θ 1 l 1 cos θ 1 ] = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) ( l 1 2 θ ˙ 1 sin 2 θ 1 + l 1 2 θ ˙ 1 cos 2 θ 1 ) = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 . \begin{aligned} \vec{F}_{D, 1b} \cdot \dfrac{\partial \vec{r}_{1b}}{\partial \theta_1}&= (-b_{1b} - c_{1b}l_1\dot{\theta}_1)\begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1} \\ l_1\dot{\theta}_1 \cos{\theta_1} \end{bmatrix} \cdot \begin{bmatrix} -l_1 \sin{\theta_1} \\ l_1 \cos{\theta_1} \end{bmatrix} \\ &= -(b_{1b}+c_{1b}l_1\dot{\theta}_1) (l_1^2 \dot{\theta}_1 \sin^2{\theta_1}+l_1^2 \dot{\theta}_1 \cos^2{\theta_1}) \\ &= -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1^2 \dot{\theta}_1. \end{aligned} F D , 1 b ⋅ ∂ θ 1 ∂ r 1 b = ( − b 1 b − c 1 b l 1 θ ˙ 1 ) [ − l 1 θ ˙ 1 sin θ 1 l 1 θ ˙ 1 cos θ 1 ] ⋅ [ − l 1 sin θ 1 l 1 cos θ 1 ] = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) ( l 1 2 θ ˙ 1 sin 2 θ 1 + l 1 2 θ ˙ 1 cos 2 θ 1 ) = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 . Next we will calculate the dissipation force on the first rod. We will use a centre of mass approximation (as otherwise we would likely have to integrate over the rod, which would drastically complicate the calculation)
F ⃗ D , 1 r = − ( b 1 r + c 1 r ∣ v ⃗ 1 r ∣ ) v ⃗ 1 r F ⃗ D , 1 r = − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 2 [ − sin θ 1 cos θ 1 ] . \begin{aligned} \vec{F}_{D,1r} &= -\left(b_{1r} + c_{1r}|\vec{v}_{1r}|\right)\vec{v}_{1r} \\ \vec{F}_{D,1r} &= -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right)\dfrac{l_1 \dot{\theta}_1}{2} \begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix}. \end{aligned} F D , 1 r F D , 1 r = − ( b 1 r + c 1 r ∣ v 1 r ∣ ) v 1 r = − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 2 l 1 θ ˙ 1 [ − sin θ 1 cos θ 1 ] . Hence the generalized dissipation force, canonical to θ 1 \theta_1 θ 1
F ⃗ D , 1 r ⋅ ∂ r ⃗ 1 r ∂ θ 1 = − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 2 [ − sin θ 1 cos θ 1 ] ⋅ l 1 2 [ − sin θ 1 cos θ 1 ] = − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 2 θ ˙ 1 4 . \begin{aligned} \vec{F}_{D,1r} \cdot \dfrac{\partial \vec{r}_{1r}}{\partial \theta_1} &= -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right)\dfrac{l_1 \dot{\theta}_1}{2} \begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix} \cdot \dfrac{l_1}{2}\begin{bmatrix} -\sin{\theta_1} \\ \cos{\theta_1} \end{bmatrix} \\ &= -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1^2 \dot{\theta}_1}{4}. \end{aligned} F D , 1 r ⋅ ∂ θ 1 ∂ r 1 r = − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 2 l 1 θ ˙ 1 [ − sin θ 1 cos θ 1 ] ⋅ 2 l 1 [ − sin θ 1 cos θ 1 ] = − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 2 θ ˙ 1 . The dissipation force on the second bob would be
F ⃗ D , 2 b = − b 2 b v ⃗ 2 b − c 2 b ∣ v ⃗ 2 b ∣ v ⃗ 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] . \begin{aligned} \vec{F}_{D,2b} &= -b_{2b}\vec{v}_{2b} - c_{2b}|\vec{v}_{2b}|\vec{v}_{2b} \\ &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right) \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2} \\ l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \end{bmatrix}. \end{aligned} F D , 2 b = − b 2 b v 2 b − c 2 b ∣ v 2 b ∣ v 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] . Hence the generalized dissipation force, canonical to θ 1 \theta_1 θ 1
F ⃗ D , 2 b ⋅ ∂ r ⃗ 2 b ∂ θ 1 = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] ⋅ [ − l 1 sin θ 1 l 1 cos θ 1 ] = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 sin 2 θ 1 + l 1 l 2 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ 1 ˙ cos 2 θ 1 + l 1 l 2 θ ˙ 2 cos θ 1 cos θ 2 ) = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . \begin{aligned} \vec{F}_{D, 2b} \cdot \dfrac{\partial \vec{r}_{2b}}{\partial \theta_1} &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right) \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2} \\ l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \end{bmatrix} \cdot \begin{bmatrix} -l_1 \sin{\theta_1} \\ l_1 \cos{\theta_1} \end{bmatrix}\\ &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 \sin^2{\theta_1}+l_1l_2 \dot{\theta}_2 \sin{\theta_1}\sin{\theta_2} +l_1^2 \dot{\theta_1}\cos^2{\theta_1}+l_1l_2\dot{\theta}_2 \cos{\theta_1}\cos{\theta_2}\right) \\ &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + l_1l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right). \end{aligned} F D , 2 b ⋅ ∂ θ 1 ∂ r 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] ⋅ [ − l 1 sin θ 1 l 1 cos θ 1 ] = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 sin 2 θ 1 + l 1 l 2 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ 1 ˙ cos 2 θ 1 + l 1 l 2 θ ˙ 2 cos θ 1 cos θ 2 ) = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . The friction force applied to the second rod would be
F ⃗ D , 2 r = − ( b 2 r + c 2 r ∣ v ⃗ 2 r ∣ ) v ⃗ 2 r F ⃗ D , 2 r = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 2 ] . \begin{aligned} \vec{F}_{D,2r} &= -\left(b_{2r} + c_{2r}|\vec{v}_{2r}|\right)\vec{v}_{2r} \\ \vec{F}_{D,2r} &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\begin{bmatrix} -l_1 \dot{\theta}_1\sin{\theta_1} - \dfrac{l_2\dot{\theta}_2\sin{\theta_2}}{2}\\ l_1 \dot{\theta}_1 \cos{\theta_1} + \dfrac{l_2\dot{\theta}_2\cos{\theta_2}}{2} \end{bmatrix}. \end{aligned} F D , 2 r F D , 2 r = − ( b 2 r + c 2 r ∣ v 2 r ∣ ) v 2 r = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ⎣ ⎢ ⎢ ⎡ − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + 2 l 2 θ ˙ 2 cos θ 2 ⎦ ⎥ ⎥ ⎤ . Therefore the generalized dissipation force, canonical to θ 1 \theta_1 θ 1
F ⃗ D , 2 r ⋅ ∂ r ⃗ 2 r ∂ θ 1 = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 sin 2 θ 1 + l 1 l 2 θ ˙ 2 sin θ 1 sin θ 2 2 + l 1 2 θ ˙ 1 cos 2 θ 1 + l 1 l 2 θ ˙ 2 cos θ 1 cos θ 2 2 ) = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 2 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) . \begin{aligned} \vec{F}_{D,2r}\cdot \dfrac{\partial \vec{r}_{2r}}{\partial \theta_1} &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right) \left(l_1^2 \dot{\theta}_1 \sin^2{\theta_1} + \dfrac{l_1 l_2 \dot{\theta}_2 \sin{\theta_1}\sin{\theta_2}}{2} +l_1^2 \dot{\theta}_1\cos^2{\theta_1} + \dfrac{l_1l_2 \dot{\theta}_2 \cos{\theta_1}\cos{\theta_2}}{2}\right) \\ &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1^2 + \dfrac{l_1 l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right). \end{aligned} F D , 2 r ⋅ ∂ θ 1 ∂ r 2 r = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 sin 2 θ 1 + 2 l 1 l 2 θ ˙ 2 sin θ 1 sin θ 2 + l 1 2 θ ˙ 1 cos 2 θ 1 + 2 l 1 l 2 θ ˙ 2 cos θ 1 cos θ 2 ) = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 2 + 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . Hence Q θ 1 Q_{\theta_1} Q θ 1
Q θ 1 = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 2 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 2 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) . \begin{aligned} Q_{\theta_1} &= -\left(b_{1b} + c_{1b} l_1 \dot{\theta}_1\right)l_1^2 \dot{\theta}_1 -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + l_1l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right) -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1^2 \dot{\theta}_1}{4} \\ &-\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1^2 + \dfrac{l_1 l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right). \end{aligned} Q θ 1 = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 2 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 2 + 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) . The equation of motion for θ 1 \theta_1 θ 1
p ˙ θ 1 − F θ 1 = Q θ 1 \begin{aligned} \dot{p}_{\theta_1} - F_{\theta_1} &= Q_{\theta_1} \end{aligned} p ˙ θ 1 − F θ 1 = Q θ 1 m 1 r l 1 2 θ ¨ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) − ( − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 1 r g l 1 cos θ 1 2 − m 2 r g l 1 cos θ 1 − m 1 b g l 1 cos θ 1 − m 2 b g l 1 cos θ 1 ) = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 2 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) m 1 r l 1 2 θ ¨ 1 12 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + m 1 r g l 1 cos θ 1 2 + m 2 r g l 1 cos θ 1 + m 1 b g l 1 cos θ 1 + m 2 b g l 1 cos θ 1 = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 2 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) \begin{aligned} &\dfrac{m_{1r} l_1^2 \ddot{\theta}_1}{12} + (m_{1b}+m_{2b})l_1^2 \ddot{\theta}_1 + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} - m_{2b}l_1 l_2 \dot{\theta}_2\left(\dot{\theta}_1 - \dot{\theta}_2\right)\sin{(\theta_1-\theta_2)} - \left(-m_{2b}l_1l_2\dot{\theta}_1\dot{\theta}_2 \sin{(\theta_1-\theta_2)} - \dfrac{m_{1r}gl_1 \cos{\theta_1}}{2} -m_{2r}gl_1 \cos{\theta_1} -m_{1b}gl_1 \cos{\theta_1} \right.\\ &\left.-m_{2b}gl_1 \cos{\theta_1}\right) = -\left(b_{1b} + c_{1b} l_1 \dot{\theta}_1\right)l_1^2 \dot{\theta}_1 -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + l_1l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right) -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1^2 \dot{\theta}_1}{4} \\ & -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + \dfrac{l_1 l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right) \\ &\dfrac{m_{1r} l_1^2 \ddot{\theta}_1}{12} + (m_{1b}+m_{2b})l_1^2 \ddot{\theta}_1 + m_{2b}l_1 l_2 \ddot{\theta}_2\cos{(\theta_1-\theta_2)} + m_{2b}l_1 l_2 \dot{\theta}_2^2\sin{(\theta_1-\theta_2)} + \dfrac{m_{1r}gl_1 \cos{\theta_1}}{2} +m_{2r}gl_1 \cos{\theta_1} +m_{1b}gl_1 \cos{\theta_1} +m_{2b}gl_1 \cos{\theta_1} = -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1^2 \dot{\theta}_1 \\ &-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1^2 \dot{\theta}_1 + l_1l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}) -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1^2 \dot{\theta}_1}{4} \\ &-\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + \dfrac{l_1 l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right) \\ \end{aligned} 1 2 m 1 r l 1 2 θ ¨ 1 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) − m 2 b l 1 l 2 θ ˙ 2 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) − ( − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − 2 m 1 r g l 1 cos θ 1 − m 2 r g l 1 cos θ 1 − m 1 b g l 1 cos θ 1 − m 2 b g l 1 cos θ 1 ) = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 2 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 + 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) 1 2 m 1 r l 1 2 θ ¨ 1 + ( m 1 b + m 2 b ) l 1 2 θ ¨ 1 + m 2 b l 1 l 2 θ ¨ 2 cos ( θ 1 − θ 2 ) + m 2 b l 1 l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + 2 m 1 r g l 1 cos θ 1 + m 2 r g l 1 cos θ 1 + m 1 b g l 1 cos θ 1 + m 2 b g l 1 cos θ 1 = − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 2 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 + 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( m 1 r 12 + m 1 b + m 2 b ) l 1 2 θ ¨ 1 = − m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g l 1 cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 2 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) \begin{aligned} & \left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)l_1^2 \ddot{\theta}_1 = - m_{2b}l_1 l_2 \left( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} +\dot{\theta}_2^2\sin{(\theta_1-\theta_2)}\right) - gl_1 \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1^2 \dot{\theta}_1 \\ &-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + l_1l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right) -\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1^2 \dot{\theta}_1}{4} \\ & -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1^2 \dot{\theta}_1 + \dfrac{l_1 l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right) \\ \end{aligned} ( 1 2 m 1 r + m 1 b + m 2 b ) l 1 2 θ ¨ 1 = − m 2 b l 1 l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g l 1 cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 2 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 2 θ ˙ 1 + l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 2 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 2 θ ˙ 1 + 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) θ ¨ 1 = − 1 ( m 1 r 12 + m 1 b + m 2 b ) l 1 [ m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) + g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 + ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) + ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 + ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] \begin{aligned} \ddot{\theta}_1 = &\dfrac{-1}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)l_1} \left[m_{2b}l_2 ( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} +\dot{\theta}_2^2\sin{(\theta_1-\theta_2)}) + g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1 \right.\\ &\left.+\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}) +\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} \right.\\ &\left.+\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1 \dot{\theta}_1 + \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right)\right] \end{aligned} θ ¨ 1 = ( 1 2 m 1 r + m 1 b + m 2 b ) l 1 − 1 [ m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) + g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 + ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) + ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 + ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ⎦ ⎥ ⎤ Next, we will derive the equations for θ 2 \theta_2 θ 2
p θ 2 = ∂ L ∂ θ ˙ 2 = m 2 r l 2 2 θ ˙ 2 12 + m 2 b l 2 ( l 2 θ ˙ 2 + l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) ) p ˙ θ 2 = m 2 r l 2 2 θ ¨ 2 12 + m 2 b l 2 ( l 2 θ ¨ 2 + l 1 θ ¨ 1 cos ( θ 1 − θ 2 ) − l 1 θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) = l 2 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) F θ 2 = ∂ L ∂ θ 2 = m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 2 r g l 2 cos θ 2 2 − m 2 b g l 2 cos θ 2 = m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − g l 2 cos θ 2 ( m 2 r 2 + m 2 b ) . \begin{aligned} p_{\theta_2} &= \dfrac{\partial \mathcal{L}}{\partial \dot{\theta}_2} \\ &= \dfrac{m_{2r}l_2^2 \dot{\theta}_2}{12} + m_{2b}l_2 \left(l_2 \dot{\theta}_2 + l_1 \dot{\theta}_1\cos{(\theta_1-\theta_2)}\right)\\ \dot{p}_{\theta_2} &= \dfrac{m_{2r}l_2^2 \ddot{\theta}_2}{12} + m_{2b}l_2 \left(l_2 \ddot{\theta}_2 + l_1 \ddot{\theta}_1\cos{(\theta_1-\theta_2)}-l_1\dot{\theta}_1(\dot{\theta}_1-\dot{\theta}_2)\sin{(\theta_1-\theta_2)}\right) \\ &= l_2^2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}l_1l_2\left(\ddot{\theta}_1\cos{(\theta_1-\theta_2)}-\dot{\theta}_1(\dot{\theta}_1-\dot{\theta}_2)\sin{(\theta_1-\theta_2)}\right)\\ F_{\theta_2} &= \dfrac{\partial \mathcal{L}}{\partial \theta_2} \\ &= m_{2b} l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \sin{(\theta_1-\theta_2)}-\dfrac{m_{2r}gl_2 \cos{\theta_2}}{2} - m_{2b}gl_2\cos{\theta_2} \\ &= m_{2b} l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \sin{(\theta_1-\theta_2)}-gl_2\cos{\theta_2}\left(\dfrac{m_{2r}}{2} + m_{2b}\right). \end{aligned} p θ 2 p ˙ θ 2 F θ 2 = ∂ θ ˙ 2 ∂ L = 1 2 m 2 r l 2 2 θ ˙ 2 + m 2 b l 2 ( l 2 θ ˙ 2 + l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) ) = 1 2 m 2 r l 2 2 θ ¨ 2 + m 2 b l 2 ( l 2 θ ¨ 2 + l 1 θ ¨ 1 cos ( θ 1 − θ 2 ) − l 1 θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) = l 2 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) = ∂ θ 2 ∂ L = m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − 2 m 2 r g l 2 cos θ 2 − m 2 b g l 2 cos θ 2 = m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − g l 2 cos θ 2 ( 2 m 2 r + m 2 b ) . The LHS of our equation will be
p ˙ θ 2 − F θ 2 = l 2 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + g l 2 cos θ 2 ( m 2 r 2 + m 2 b ) = l 2 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 2 sin ( θ 1 − θ 2 ) ) + g l 2 cos θ 2 ( m 2 r 2 + m 2 b ) . \begin{aligned} \dot{p}_{\theta_2} - F_{\theta_2} &= l_2^2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}l_1l_2\left(\ddot{\theta}_1\cos{(\theta_1-\theta_2)}-\dot{\theta}_1(\dot{\theta}_1-\dot{\theta}_2)\sin{(\theta_1-\theta_2)}\right) - m_{2b} l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \sin{(\theta_1-\theta_2)}+gl_2\cos{\theta_2}\left(\dfrac{m_{2r}}{2} + m_{2b}\right) \\ &= l_2^2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}l_1l_2\left(\ddot{\theta}_1\cos{(\theta_1-\theta_2)}-\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}\right)+gl_2\cos{\theta_2}\left(\dfrac{m_{2r}}{2} + m_{2b}\right). \end{aligned} p ˙ θ 2 − F θ 2 = l 2 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + g l 2 cos θ 2 ( 2 m 2 r + m 2 b ) = l 2 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 2 sin ( θ 1 − θ 2 ) ) + g l 2 cos θ 2 ( 2 m 2 r + m 2 b ) . Q θ 2 = ∑ j F ⃗ D , j ⋅ ∂ r ⃗ j ∂ θ 2 = F ⃗ D , 1 r ⋅ ∂ r ⃗ 1 r ∂ θ 2 + F ⃗ D , 1 b ⋅ ∂ r ⃗ 1 b ∂ θ 2 + F ⃗ D , 2 r ⋅ ∂ r ⃗ 2 r ∂ θ 2 + F ⃗ D , 2 b ⋅ ∂ r ⃗ 2 b ∂ θ 2 . \begin{aligned} Q_{\theta_2} &= \sum_{j} \vec{F}_{D,j} \cdot \dfrac{\partial \vec{r}_j}{\partial \theta_2} \\ &= \vec{F}_{D,1r} \cdot \dfrac{\partial \vec{r}_{1r}}{\partial \theta_2} + \vec{F}_{D,1b} \cdot \dfrac{\partial \vec{r}_{1b}}{\partial \theta_2} + \vec{F}_{D,2r} \cdot \dfrac{\partial \vec{r}_{2r}}{\partial \theta_2} + \vec{F}_{D,2b} \cdot \dfrac{\partial \vec{r}_{2b}}{\partial \theta_2}. \end{aligned} Q θ 2 = j ∑ F D , j ⋅ ∂ θ 2 ∂ r j = F D , 1 r ⋅ ∂ θ 2 ∂ r 1 r + F D , 1 b ⋅ ∂ θ 2 ∂ r 1 b + F D , 2 r ⋅ ∂ θ 2 ∂ r 2 r + F D , 2 b ⋅ ∂ θ 2 ∂ r 2 b . The terms corresponding to the first rod and first bob will be zero as their position and velocities are independent of θ 2 \theta_2 θ 2
The dissipation force applied to the second rod is
F ⃗ D , 2 r = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 2 ] . \begin{aligned} \vec{F}_{D,2r} &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\begin{bmatrix} -l_1 \dot{\theta}_1\sin{\theta_1} - \dfrac{l_2\dot{\theta}_2\sin{\theta_2}}{2}\\ l_1 \dot{\theta}_1 \cos{\theta_1} + \dfrac{l_2\dot{\theta}_2\cos{\theta_2}}{2} \end{bmatrix}. \end{aligned} F D , 2 r = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ⎣ ⎢ ⎢ ⎡ − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + 2 l 2 θ ˙ 2 cos θ 2 ⎦ ⎥ ⎥ ⎤ . Hence the generalized dissipation force, canonical to θ 2 \theta_2 θ 2
F ⃗ D , 2 r ⋅ ∂ r ⃗ 2 r ∂ θ 2 = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 2 ] ⋅ l 2 2 [ − sin θ 2 cos θ 2 ] = − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 sin θ 1 sin θ 2 2 + l 2 2 θ ˙ 2 sin 2 θ 2 4 + l 1 l 2 θ ˙ 1 cos θ 1 cos θ 2 2 + l 2 2 θ ˙ 2 cos 2 θ 2 4 ) = − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . \begin{aligned} \vec{F}_{D,2r} \cdot \dfrac{\partial \vec{r}_{2r}}{\partial \theta_2} &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\begin{bmatrix} -l_1 \dot{\theta}_1\sin{\theta_1} - \dfrac{l_2\dot{\theta}_2\sin{\theta_2}}{2}\\ l_1 \dot{\theta}_1 \cos{\theta_1} + \dfrac{l_2\dot{\theta}_2\cos{\theta_2}}{2} \end{bmatrix} \cdot \dfrac{l_2}{2} \begin{bmatrix} -\sin{\theta_2} \\ \cos{\theta_2} \end{bmatrix} \\ &= -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(\dfrac{l_1l_2 \dot{\theta}_1 \sin{\theta_1}\sin{\theta_2}}{2} + \dfrac{l_2^2 \dot{\theta}_2\sin^2{\theta_2}}{4}+\dfrac{l_1l_2\dot{\theta}_1\cos{\theta_1}\cos{\theta_2}}{2}+\dfrac{l_2^2\dot{\theta}_2\cos^2{\theta_2}}{4}\right) \\ &= -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(2l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2). \end{aligned} F D , 2 r ⋅ ∂ θ 2 ∂ r 2 r = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ⎣ ⎢ ⎢ ⎡ − l 1 θ ˙ 1 sin θ 1 − 2 l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + 2 l 2 θ ˙ 2 cos θ 2 ⎦ ⎥ ⎥ ⎤ ⋅ 2 l 2 [ − sin θ 2 cos θ 2 ] = − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 l 2 θ ˙ 1 sin θ 1 sin θ 2 + 4 l 2 2 θ ˙ 2 sin 2 θ 2 + 2 l 1 l 2 θ ˙ 1 cos θ 1 cos θ 2 + 4 l 2 2 θ ˙ 2 cos 2 θ 2 ) = − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . The dissipation force applied to the second bob is
F ⃗ D , 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] . \begin{aligned} \vec{F}_{D,2b} &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right) \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2} \\ l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \end{bmatrix}. \end{aligned} F D , 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] . Therefore the generalized dissipation force, canonical to θ 2 \theta_2 θ 2
F ⃗ D , 2 b ⋅ ∂ r ⃗ 2 b ∂ θ 2 = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] ⋅ l 2 [ − sin θ 2 cos θ 2 ] = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 sin θ 1 sin θ 2 + l 2 2 θ ˙ 2 sin 2 θ 2 + l 1 l 2 θ ˙ 1 cos θ 1 cos θ 2 + l 2 2 θ ˙ 2 cos 2 θ 2 ) = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . \begin{aligned} \vec{F}_{D,2b} \cdot \dfrac{\partial \vec{r}_{2b}}{\partial \theta_2} &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right) \begin{bmatrix} -l_1\dot{\theta}_1 \sin{\theta_1}-l_2\dot{\theta}_2 \sin{\theta_2} \\ l_1\dot{\theta}_1 \cos{\theta_1}+l_2\dot{\theta}_2 \cos{\theta_2} \end{bmatrix} \cdot l_2\begin{bmatrix} -\sin{\theta_2} \\ \cos{\theta_2} \end{bmatrix} \\ &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right) \left(l_1 l_2 \dot{\theta}_1 \sin{\theta_1}\sin{\theta_2} + l_2^2 \dot{\theta}_2 \sin^2{\theta_2} + l_1l_2 \dot{\theta}_1 \cos{\theta_1}\cos{\theta_2} + l_2^2\dot{\theta}_2 \cos^2{\theta_2}\right) \\ &= -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2\right). \end{aligned} F D , 2 b ⋅ ∂ θ 2 ∂ r 2 b = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) [ − l 1 θ ˙ 1 sin θ 1 − l 2 θ ˙ 2 sin θ 2 l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ] ⋅ l 2 [ − sin θ 2 cos θ 2 ] = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 sin θ 1 sin θ 2 + l 2 2 θ ˙ 2 sin 2 θ 2 + l 1 l 2 θ ˙ 1 cos θ 1 cos θ 2 + l 2 2 θ ˙ 2 cos 2 θ 2 ) = − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . Q θ 2 = − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . \begin{aligned} Q_{\theta_2} &= -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(2l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2\right)\\ & -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2\right). \end{aligned} Q θ 2 = − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . l 2 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + m 2 r g l 2 cos θ 2 2 + m 2 b g l 2 cos θ 2 = − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) \begin{aligned} &l_2^2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}l_1l_2\left(\ddot{\theta}_1\cos{(\theta_1-\theta_2)}-\dot{\theta}_1(\dot{\theta}_1-\dot{\theta}_2)\sin{(\theta_1-\theta_2)}\right) - m_{2b} l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \sin{(\theta_1-\theta_2)} +\dfrac{m_{2r}gl_2 \cos{\theta_2}}{2}+ m_{2b}gl_2\cos{\theta_2}\\ &= -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(2l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2) -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}\right. \\ &\left.+ l_2^2 \dot{\theta}_2\right) \end{aligned} l 2 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b l 1 l 2 ( θ ¨ 1 cos ( θ 1 − θ 2 ) − θ ˙ 1 ( θ ˙ 1 − θ ˙ 2 ) sin ( θ 1 − θ 2 ) ) − m 2 b l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) + 2 m 2 r g l 2 cos θ 2 + m 2 b g l 2 cos θ 2 = − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) Moving everything on the left-hand side except the θ ¨ 1 \ddot{\theta}_1 θ ¨ 1
m 2 b l 1 l 2 cos ( θ 1 − θ 2 ) θ ¨ 1 = − l 2 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b l 2 ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . \begin{aligned} &m_{2b}l_1l_2\cos{(\theta_1-\theta_2)}\ddot{\theta}_1 = -l_2^2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}l_2(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(2l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} \\ &+ l_2^2 \dot{\theta}_2) -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1l_2 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2^2 \dot{\theta}_2\right). \end{aligned} m 2 b l 1 l 2 cos ( θ 1 − θ 2 ) θ ¨ 1 = − l 2 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b l 2 ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 2 θ ˙ 2 ) . Dividing both sides by m 2 b l 1 l 2 cos ( θ 1 − θ 2 ) m_{2b}l_1l_2 \cos{(\theta_1-\theta_2)} m 2 b l 1 l 2 cos ( θ 1 − θ 2 )
θ ¨ 1 = sec ( θ 1 − θ 2 ) m 2 b l 1 [ − l 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] . \begin{aligned} &\ddot{\theta}_1 = \dfrac{\sec{(\theta_1-\theta_2)}}{m_{2b}l_1} \left[-l_2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2})-\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2\right)\right.\\ &\left.-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2\right)\right]. \end{aligned} θ ¨ 1 = m 2 b l 1 sec ( θ 1 − θ 2 ) ⎣ ⎢ ⎡ − l 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] . Dividing both sides by m 2 l 1 l 2 cos ( θ 1 − θ 2 ) m_2l_1l_2\cos{(\theta_1-\theta_2)} m 2 l 1 l 2 cos ( θ 1 − θ 2 ) θ ¨ 1 \ddot{\theta}_1 θ ¨ 1 (2 ) yields
1 ( m 1 r 12 + m 1 b + m 2 b ) l 1 [ − m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] = sec ( θ 1 − θ 2 ) m 2 b l 1 [ − l 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] \begin{aligned} &\dfrac{1}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)l_1} [-m_{2b}l_2 ( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} +\dot{\theta}_2^2\sin{(\theta_1-\theta_2)}) - g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1 \\ &-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)})-\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(l_1 \dot{\theta}_1 \\ &+ \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2})] =\dfrac{\sec{(\theta_1-\theta_2)}}{m_{2b}l_1} [-l_2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\\ &(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}+ l_2 \dot{\theta}_2) -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2)]\\ \end{aligned} ( 1 2 m 1 r + m 1 b + m 2 b ) l 1 1 [ − m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ] = m 2 b l 1 sec ( θ 1 − θ 2 ) [ − l 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] Multiply both sides by l 1 m 2 b cos ( θ 1 − θ 2 ) l_1m_{2b}\cos{(\theta_1-\theta_2)} l 1 m 2 b cos ( θ 1 − θ 2 )
m 2 b cos ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) [ − m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] = − l 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) \begin{aligned} &\dfrac{m_{2b}\cos{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)} \left[-m_{2b}l_2 \left( \ddot{\theta}_2\cos{(\theta_1-\theta_2)} +\dot{\theta}_2^2\sin{(\theta_1-\theta_2)}\right) - g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1\right. \\ &\left.-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)})-\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1 \dot{\theta}_1\right.\right. \\ &\left.\left.+ \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right)\right] =-l_2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 + m_{2b}(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\\ &\left(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}+ l_2 \dot{\theta}_2\right) -\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2\right)\\ \end{aligned} ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b cos ( θ 1 − θ 2 ) [ − m 2 b l 2 ( θ ¨ 2 cos ( θ 1 − θ 2 ) + θ ˙ 2 2 sin ( θ 1 − θ 2 ) ) − g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ] = − l 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2 + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) Adding l 2 ( m 2 r 12 + m 2 b ) θ ¨ 2 l_2 \left(\dfrac{m_{2r}}{12} + m_{2b}\right)\ddot{\theta}_2 l 2 ( 1 2 m 2 r + m 2 b ) θ ¨ 2
θ ¨ 2 ( ( m 2 r 12 + m 2 b ) l 2 − m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) ) + m 2 b cos ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) [ − m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) − g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] = m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) θ ¨ 2 ( ( m 2 r 12 + m 2 b ) l 2 − m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) ) = − m 2 b cos ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) [ − m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) − g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 − ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) \begin{aligned} &\ddot{\theta}_2\left(\left(\dfrac{m_{2r}}{12} + m_{2b}\right)l_2 - \dfrac{m_{2b}^2l_2\cos^2{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}\right) + \dfrac{m_{2b}\cos{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}\left[-m_{2b}l_2\dot{\theta}_2^2\sin{(\theta_1-\theta_2)} - g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1\right. \\ &\left.-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right)-\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1 \dot{\theta}_1\right.\right. \\ &\left.\left.+ \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right)\right] = m_{2b}\left(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}\right) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}+ l_2 \dot{\theta}_2)\\ &-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2\right)\\\\\\ &\ddot{\theta}_2\left(\left(\dfrac{m_{2r}}{12} + m_{2b}\right)l_2 - \dfrac{m_{2b}^2l_2\cos^2{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}\right) = -\dfrac{m_{2b}\cos{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}\left[-m_{2b}l_2\dot{\theta}_2^2\sin{(\theta_1-\theta_2)} - g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1 \right.\\ &\left.-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}\right)-\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} -\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \right.\right.\\ &\left.\left.+ \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right)\right] + m_{2b}\left(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}\right) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}+ l_2 \dot{\theta}_2\right)\\ &-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2\right)\\\\\\ \end{aligned} θ ¨ 2 ⎝ ⎜ ⎛ ( 1 2 m 2 r + m 2 b ) l 2 − ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ + ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b cos ( θ 1 − θ 2 ) [ − m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) − g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ] = m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) θ ¨ 2 ⎝ ⎜ ⎛ ( 1 2 m 2 r + m 2 b ) l 2 − ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ = − ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b cos ( θ 1 − θ 2 ) [ − m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) − g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) − ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 − ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ] + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) θ ¨ 2 = 1 ( m 2 r 12 + m 2 b ) l 2 − m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) [ m 2 b cos ( θ 1 − θ 2 ) ( m 1 r 12 + m 1 b + m 2 b ) [ m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g cos θ 1 ( m 1 r 2 + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 + ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) + ( b 1 r + c 1 r l 1 θ ˙ 1 2 ) l 1 θ ˙ 1 4 + ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) 2 ) ] + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 1 4 ( b 2 r + c 2 r l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 4 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] . \begin{aligned} &\ddot{\theta}_2 = \dfrac{1}{\left(\dfrac{m_{2r}}{12} + m_{2b}\right)l_2 - \dfrac{m_{2b}^2l_2\cos^2{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}}\left[\dfrac{m_{2b}\cos{(\theta_1-\theta_2)}}{\left(\dfrac{m_{1r}}{12} + m_{1b}+m_{2b}\right)}\left[m_{2b}l_2\dot{\theta}_2^2\sin{(\theta_1-\theta_2)} + g \cos{\theta_1}\left(\dfrac{m_{1r}}{2} +m_{2r} +m_{1b} + m_{2b}\right) -(b_{1b} + c_{1b} l_1 \dot{\theta}_1)l_1 \dot{\theta}_1 \right.\right.\\ &\quad\left.\left.+\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1 l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 + l_2 \dot{\theta}_2 \cos{(\theta_1-\theta_2)})+\left(b_{1r} + \dfrac{c_{1r}l_1 \dot{\theta}_1}{2}\right) \dfrac{l_1 \dot{\theta}_1}{4} +\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)\left(l_1 \dot{\theta}_1 \right.\right.\right. \\ &\quad\left.\left.\left.+ \dfrac{l_2\dot{\theta}_2 \cos{\left(\theta_1 - \theta_2\right)}}{2}\right)\right] + m_{2b}(l_1\dot{\theta}_1^2\sin{(\theta_1-\theta_2)}-g\cos{\theta_2}) -\dfrac{1}{4}\left(b_{2r} + c_{2r}\sqrt{l_1^2 \dot{\theta}_1^2 + \dfrac{l_2^2 \dot{\theta}_2^2}{4} + l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1 -\theta_2)}}\right)(2l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)}+ l_2 \dot{\theta}_2)\right.\\ &\quad\left.-\left(b_{2b}+c_{2b}\sqrt{l_1^2 \dot{\theta}_1^2 + l_2^2 \dot{\theta}_2^2 +2l_1l_2\dot{\theta}_1 \dot{\theta}_2 \cos{(\theta_1-\theta_2)}}\right)(l_1 \dot{\theta}_1 \cos{(\theta_1-\theta_2)} + l_2 \dot{\theta}_2)\right]. \end{aligned} θ ¨ 2 = ( 1 2 m 2 r + m 2 b ) l 2 − ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b 2 l 2 cos 2 ( θ 1 − θ 2 ) 1 ⎣ ⎢ ⎡ ( 1 2 m 1 r + m 1 b + m 2 b ) m 2 b cos ( θ 1 − θ 2 ) [ m 2 b l 2 θ ˙ 2 2 sin ( θ 1 − θ 2 ) + g cos θ 1 ( 2 m 1 r + m 2 r + m 1 b + m 2 b ) − ( b 1 b + c 1 b l 1 θ ˙ 1 ) l 1 θ ˙ 1 + ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 + l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) + ( b 1 r + 2 c 1 r l 1 θ ˙ 1 ) 4 l 1 θ ˙ 1 + ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( l 1 θ ˙ 1 + 2 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ] + m 2 b ( l 1 θ ˙ 1 2 sin ( θ 1 − θ 2 ) − g cos θ 2 ) − 4 1 ⎝ ⎜ ⎛ b 2 r + c 2 r l 1 2 θ ˙ 1 2 + 4 l 2 2 θ ˙ 2 2 + l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ⎠ ⎟ ⎞ ( 2 l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) − ( b 2 b + c 2 b l 1 2 θ ˙ 1 2 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ) ( l 1 θ ˙ 1 cos ( θ 1 − θ 2 ) + l 2 θ ˙ 2 ) ] . At this point, we could rewrite one of our θ ¨ 1 \ddot{\theta}_1 θ ¨ 1 θ ¨ 2 \ddot{\theta}_2 θ ¨ 2 (3 ) . We will not do this as this will add even more potential for errors to creep in, and Equations (2 ) and (3 ) are already suitable for numerical integration.
