# Kinetic energy of double pendulum

using cartesian coordinates as an intermediate step, the kinetic energy is calculated as such

why is it incorrect to just say that the kinetic energy for the first bob is $$T_1 = (1/2) m_1 (l_1\dot{\theta_1})^2$$

and for the second bob is $$T_2 = (1/2) m_2 (l_2\dot{\theta_2}+l_1 \dot{\theta_1}))^2$$

where is the $$cos(\theta_1 - \theta_2)$$ term coming from?

• The $\cos(\theta_1-\theta_2)$ is due to the articulation at point $1$, which leads to a non alignment of the two velocities. Mathematically, your two expressions are the same iff $\theta_1 = \theta_2$ ie the double pendulum is made rigid.
– LPZ
Commented Aug 21, 2022 at 11:38
• so what if $v_1$ (velocity of the top of second pendulum) and $v_2$(velocity of the bottom of second pendulum) do not align ? why not just add them, why do we need to consider the angle between them? Commented Aug 21, 2022 at 18:27
• You would be adding up the magnitude of two vectors that in generall are not pointing in the same direction in space - but the relative direction matters for the length of the resultant vector Commented Aug 22, 2022 at 16:23

The expression $$T_2 = (1/2) m_2 (l_2\dot{\theta_2}+l_1 \dot{\theta_1}))^2$$ is incorrect because you are just adding up two vector quantities algebraically.

The vector form the kinetic energy of a particle with mass $$m$$ is

$$T = \tfrac{1}{2} m (\vec{v} \cdot \vec{v}) = \tfrac{1}{2} m \| \vec{v} \|^2$$

$$T \overset{?}{=} \tfrac{1}{2} m_1 \| \vec{v}_1 \|^2 + \tfrac{1}{2} m_2 \| \vec{v}_2 - \vec{v}_1 \|^2$$
$$T = \tfrac{1}{2} m_1 \| \vec{v}_1 \|^2 + \tfrac{1}{2} m_2 \| \vec{v}_2 \|^2$$
Your first analysis is correct because it splits the velocity vector into components and considers the correct combination of $$\dot{\theta}_1$$ and $$\dot{\theta}_2$$ that yields kinetic energy.