# Rotating rod

The rod of length 2a swings and it rotates $\dot{\phi}[/tex]. Find kinetic energy. ## The Attempt at a Solution I know how to find kinetic energy if we have rotation just in xy plane, but I'm having a problem to understand how to do it with rotation around y axis. If we have rotation in xy plane: $$x_{cm}=asin(\theta)$$ $$y_{cm}=acos(\theta)$$ $$\dot{x}_{cm}=acos(\theta)\dot{\theta}$$ $$\dot{y}_{cm}=-asin(\theta)\dot{\theta}$$ $$E_{kin}=\frac{1}{2}m(\dot{x}_{cm}^2+\dot{y}_{cm}^2)+\frac{1}{2}m\cdot\frac{1}{12}(2a)^2\dot{\theta}^2$$ $$E_{kin}=\frac{2}{3}m\dot{\theta}^2$$ For rotation around y, I assume that we have translation component in z direction, and rotation component. $$z_{cm}=asin(\theta)sin(\phi)$$ Is this correct? (look at the picture) $$\dot{z}_{cm}=a(cos(\theta)sin(\phi)\dot{\theta}+cos(\phi)sin(\theta)\dot{\phi})$$ $$E_{z}=\frac{1}{2}m\dot{z}_{cm}^2+\frac{1}{2}I_{cm}\dot{\phi}^2$$. Rotation part of [itex]E_{z}$ bothers me, maybe I have to multiply it by $(lsin(\theta))^2$.

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Is this ok:

$$E=\frac{1}{2}(\dot{x}_{cm}^2+\dot{y}_{cm}^2+\dot{z}_{cm}^2)+\frac{1}{2}I_{cm}(\dot{\theta}^2+\dot{\phi}^2)$$

?

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I have successfully solved the problem by my self, so no need for an answer.