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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?

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  • $\begingroup$ That's not what they are doing at all. You have to compute $v_1^2$ and $v_2^2$ separately. $\frac{1}{2}m_1l_1^2\dot\theta_1^2$ is the KE of the first bob, everything else is the KE of the second bob. $\endgroup$
    – Troy
    Nov 17, 2016 at 7:43
  • $\begingroup$ Bob 1 has velocity $l_1 \dot{\theta_1}$. Bob 2 has velocity magniture of $\lVert l_1 \dot{\theta_1}\hat{\theta_1} + l_2 \dot{\theta_2}\hat{\theta_2}\rVert$. To combine these two vectors, we can use the law of cosines. $\endgroup$
    – sequence
    Nov 17, 2016 at 7:57
  • $\begingroup$ I'm not sure how you computed the velocity magnitude of Bob 2, but leaving that aside for now, why are you computing the magnitude of the sum of the vectors? Their masses are different, so you can't combine them together anyway $\endgroup$
    – Troy
    Nov 17, 2016 at 8:05
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    $\begingroup$ Okay let us call the velocity of the first bob relative to the origin $v_1$, that of the second bob $v_2$, and the relative velocity of bob 2 to be $v_{\text{rel}}$. So we have: $\vec{v_2} = \vec{v_1}+\vec{v_{\text{rel}}}$. If you draw it out and label the angles, you'll see that the angle between $v_1$ and $v_2$ is $\cos(\pi-\theta_2+\theta_1)$. This rearranges to $\color{red}{-}\cos(\theta_1-\theta_2)$ $\endgroup$
    – Troy
    Nov 17, 2016 at 8:55
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    $\begingroup$ Sorry, you are right. I meant to say the angle between $\vec{v_1}$ and $\vec{v_\text{rel}}$. The result still stands though. $\endgroup$
    – Troy
    Nov 18, 2016 at 2:20

1 Answer 1

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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.

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