Double pendulum lagrangian

Question

could someone help me with the following problem, for part $b$ i know that $$\Bbb L=T-U$$ but i've forgotten how i can calculate the kinetic energy and potential energies

Answer 1

I think you only need a hint. Let $r_{1},r_{2}$ be the positions of the particles. Then, $$r_{1}=l(\sin\phi_{1},\cos\phi_{1})$$ $$r_{2}=l(\sin\phi_{1}+\sin\phi_{2},\cos\phi_{1}+\cos \phi_{2}) $$ It follows that $T=\frac{1}{2}m\|\dot{r}_{1}\|^{2}+\frac{1}{2}m\|\dot{r}_{2}\|^{2}$.

Another aproach is using complex numbers: $$r_{2}=l(e^{i\phi_{1}}+e^{i\phi_{2}})$$ Then, $\frac{1}{2}m\|\dot{r}_{2}\|^{2}=\frac{1}{2}m|il(\dot{\phi}_{1}e^{i\phi_{1}}+\dot{\phi_{2}}e^{i\phi_{2}}))|^{2}$, etc.