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Can someone please show the steps for deriving the total energy of a particle from the Polar form of Newton's Second Law?

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The total energy is $T+V$, with $T$ kinetic and $V$ potential, and by energy conservation $0=\dot{T}+\dot{V}$. Taking $T$ to depend only on $x$ and $V$ only on $\dot{x}$, $0=\ddot{x}\cdot\nabla_\dot{x}T+\dot{x}\cdot\nabla_xV$. We want this to be equivalent to $0=m\ddot{x}+\nabla_xV$, so $\nabla_\dot{x}T=m\dot{x}$. The convention that $T(0)=0$ gives $T=\frac{m\dot{x}^2}{2}$.

This argument doesn't commit us to Cartesian coordinates; scalar products, including squared lengths, are unchanged when we switch to polar coordinates. In three dimensions, $$T=\frac{m}{2}\left(\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2\right).$$The last term is deleted in two dimensions, to which we typically switch if radial forces' angular momentum conservation implies the motion is confined to a plane.

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