Law of cosines in double pendulum

Question

I can't yet explain why equation (8) at Double Pendulum implies that the square of the magnitude of $\vec{v_1}+\vec{v_2}$ (where $\vec{v_1}=l_1\dot{\theta_1}\hat{\theta}_1$ and $\vec{v_2}=l_1\dot{\theta_2}\hat{\theta}_2$) is $v_1^2+v_2^2+2v_1v_2\cos(\theta_1-\theta_2)$ instead of $v_1^2+v_2^2-2v_1v_2\cos(\theta_1-\theta_2)$, which should be the case by the law of cosines. How can this be explained?

Answer 1

In the usual formulation of the Law of Cosines, in the formula $c^2=a^2+b^2-2ab\cos\theta,$ the angle $\theta$ is the interior angle opposite side $c$. But in this question, the angle $\theta_1-\theta_2$ is the exterior angle of the relevant triangle. Its cosine is equal in magnitude to the cosine of the interior angle, but has the opposite sign. If you let $\theta=\pi-(\theta_1-\theta_2)$ then $\theta$ is the interior angle opposite the resultant vector when you construct a triangle for the sum of the two velocity vectors, $\cos\theta=-\cos(\theta_1-\theta_2),$ and the formula with $\cos\theta$ will have the sign you expected.

If you still doubt the formula, try it with some simple numbers. Let $\theta_1=\theta_2=0$. Then the velocity of the second mass at that instant will be $l_1\dot\theta_1+l_2\dot\theta_2$, and $\cos(\theta_1-\theta_2)=1$. You should find the formula in the question gives the correct kinetic energy at that instant.