problem involving inertia

Question

Update my lagrangian is

$$L=\frac{ml^2}2\left(\dot \theta_1^2+\frac 13\dot \theta_2^2+\dot \theta_1\dot \theta_2\cos\delta\right)+mgl(cos\theta_1+\frac 12\cos\theta_2)$$

and for my linearised equations of motion question (5) i got the following

$$(\ddot \theta_1) +\frac{2}{3}\ddot\theta_2=\frac{-2g}{l}\theta_2$$

and $$2\ddot \theta_1 +\ddot \theta_2=\frac{-2g}{l}\theta_1$$

These seem to give numerically unpleasant normal mode frequencies though, so I'm not entirely sure that they're right, maybe i went wrong on calculating the equations of motion. a futher solution would be ideal. Thanks

my differentials were

$$\frac{dL}{d\theta_1}=-\frac{ml^2}{2}\dot\theta_1\dot\theta_2\cos(\theta_1)sin(\theta_2)-\frac{ml^2}{2}\dot\theta_1\dot\theta_2\sin(\theta_1)\cos(\theta_2)-mgl\sin(\theta_1)$$

$$\frac{dL}{d\theta_2}=-\frac{ml^2}{2}\dot\theta_1\dot\theta_2\cos(\theta_1)\sin(\theta_2)+\frac{ml^2}{2}\dot\theta_1\dot\theta_2\sin(\theta_1)\cos(\theta_2)-\frac{mgl}{2}\sin(\theta_2)$$

$$\frac{dL}{d\dot\theta_1}=ml^2\dot\theta_1+\frac{ml^2}{2}\dot\theta_2\cos(\theta_2-\theta_1)$$

and finally $$\frac{dL}{d\dot\theta_2}=\frac{ml^2}{3}\dot\theta_2+\frac{ml^2}{2}\dot\theta_1\cos(\theta_2-\theta_1)$$

so the two equations of motion are

$$ml^2\ddot\theta_1+\frac{ml^2}{2}\dot\theta_1\dot\theta_2\cos(\theta_1)sin(\theta_2)+\frac{ml^2}{2}\dot\theta_1\dot\theta_2\sin(\theta_1)\cos(\theta_2)+mgl\sin(\theta_1)=0$$

and $$\frac{ml^2}{3}\ddot\theta_2+\frac{ml^2}{2}\dot\theta_1\dot\theta_2\cos(\theta_1)\sin(\theta_2)-\frac{ml^2}{2}\dot\theta_1\dot\theta_2\sin(\theta_1)\cos(\theta_2)+\frac{mgl}{2}\sin(\theta_2)=0$$

Answer 1

1. For an arbitrary point on the rod, let $p$ be its distance from the point of connection to the wire. Since the mass is uniformly distributed, then $$dm=\frac ml dp$$ and the distance of that point from the reference of rotation can be obtained from the law of cosines: $$r^2=l^2+p^2+2lp\cos\delta$$ Here, $\delta=\theta_2 - \theta_1\,$. The moment of intertia is defined by: $$dI=r^2 dm\implies I=\frac ml\int_0^l r^2 dp$$ Consequently $$I=\frac ml\int_0^l(l^2+p^2+2lp\cos\delta)dp=m l^2\left(\frac 43+\cos(\theta_2-\theta_1)\right)$$
2. The kinetic energy is calculated from a similar procedure. $$dK=\frac 12v^2dm\implies K=\frac m{2l}\int_l v^2 dp$$ where $$\begin{align}v^2&=v_x^2+v_y^2=[\partial_t(l\sin\theta_1+p\sin\theta_2)]^2+[\partial_t(l\cos\theta_1+p\cos\theta_2)]^2\\&= (l\dot{\theta_1}\cos\theta_1+p\dot{\theta_2}\cos\theta_2)^2+(l\dot{\theta_1}\sin\theta_1+p\dot{\theta_2}\sin\theta_2)^2\\&= (l\dot{\theta_1})^2+(p\dot{\theta_2})^2+2lp\dot{\theta_1}\dot{\theta_2}\cos\delta\end{align}$$ Thus $$K=\frac{ml^2}2\left(\dot{\theta_1}^2+\frac 13\dot{\theta_2}^2+\dot{\theta_1}\dot{\theta_2}\cos\delta\right)$$
3. And similarly, the potential energy is the result of an integral as well: $$dU=-gh\,dm\implies U=-\frac {mg}l\int_0^l hdp$$ where $$h=l\cos\theta_1+p\cos\theta_2$$ which results in $$U=-mgl(\cos\theta_1+\frac 12\cos\theta_2)$$
4. Lagrangian is given by $K-U$. Let $\omega_1=\dot\theta_1$ and $\omega_2=\dot\theta_2$ then $$L=\frac{ml^2}2\left(\omega_1^2+\frac 13\omega_2^2+\omega_1\omega_2\cos(\theta_2-\theta_1)+\frac gl(2\cos\theta_1+\cos\theta_2)\right)$$ and from Euler-Lagrange equations $$\frac d{dt}\left(\frac{\partial L}{\partial\omega_i}\right)=\frac{\partial L}{\partial\theta_i},\quad i=1,2$$ we have $$2\dot\omega_1+\dot\omega_2\cos(\theta_2-\theta_1)-\omega_2^2\sin(\theta_2-\theta_1)+2\frac gl\sin\theta_1=0\\ \frac 23\dot\omega_2+\dot\omega_1\cos(\theta_2-\theta_1)+\omega_1^2\sin(\theta_2-\theta_1)+\frac gl\sin\theta_2=0$$ Thus the state-space equations, knowing that the state variables are $\theta_1,\theta_2,\omega_1,\omega_2$ can be written as: $$\begin{align}\pmatrix{\dot\theta_1\\\dot\theta_2}&=\pmatrix{\omega_1\\ \omega_2}\\ \pmatrix{\dot\omega_1\\\dot\omega_2}&=\pmatrix{2&\cos(\theta_2-\theta_1)\\ \cos(\theta_2-\theta_1)&\frac 23}^{-1} \pmatrix{\omega_2^2\sin(\theta_2-\theta_1)-2\frac gl\sin\theta_1\\ -\omega_1^2\sin(\theta_2-\theta_1)-\frac gl\sin\theta_2} \end{align}$$ and you can linearize the system by calculating the Jacobian, $\frac{\partial f}{\partial x}|_{x=0}$ which is a straightforward but tedious process. $$x=(\theta_1,\theta_2,\omega_1,\omega_2)^T\longrightarrow\left[\frac{\partial f}{\partial x}\right]_{x=0}=\pmatrix{0&I_{2\times 2}\\J_{2\times 2}&0}$$ where $$J=\frac gl\pmatrix{-4&3\\6&-6}$$