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## Energy of a particle

 Quote by Reshma I don't have a clue. The force can be shown as $\vec F = -\vec \nabla V$. The work done will be: $W = \int \vec F \cdot dr$. Will this equal the energy?
$$T=(1/2)mv^2$$
Now, what is this is polar coordinates? Well, $$v^2=\dot{r^2}+r^2 \dot{\theta^2}$$

And, of course, E = T + V(r)...

-Dan

BTW, you can get rid of the $$\dot{\theta}$$ by using angular momentum conservation.