# Expansion of velocity in polar coordinates using kinetic equations

Why does total energy

$$E=\dfrac{1}{2}\mu\mathbf{v}^2+U(r)$$ expand to

$$E=\dfrac{1}{2}\mu(\dot{r}^{2}+r^2\dot{\phi}^2)+U(r)?$$

It's some form of the kinetic equations in polar coordinates. I know it's a stupid question but I'm totally stuck.

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## migrated from physics.stackexchange.comJul 12 '13 at 5:42

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Migrated because as such the physics content of the question is resolved. – Manishearth Jul 12 '13 at 5:44

Using the polar coordinate transformation you have

$$x = r \cos \phi$$ $$y = r \sin \phi$$

differentiating you get

$$\dot{x} = \dot{r} \cos\phi - r \dot{\phi} \sin\phi$$ $$\dot{y} = \dot{r} \sin\phi + r \dot{\phi} \cos\phi$$

Now the kinetic energy is

$$K = \frac{1}{2}\mu ( \dot{x}^2 + \dot{y}^2 )$$ $$= \frac{1}{2}\mu ( \dot{r}^2 + r^2 \dot{\phi}^2 )$$

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The position vector is $\mathbf{r}=r\mathbf{e}_r$ do the velocity is $\mathbf{v}=\frac{d\mathbf{r}}{dt}=\dot{\mathbf{v}}=\dot{r}\mathbf{e}_r+r\frac{d\mathbf{e}_r}{d\phi}\frac{d\phi}{dt}=\dot{r}\mathbf{e}_r+r\dot{\phi}\mathbf{e}_{\phi}$

$\mathbf{v}^2=v^2$ is calculated using the Pythagorean theorem:

$v^2=\dot{r}^2+\left(r\dot{\phi}\right)^2$

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