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The energy of a particle under the action of a conservative force is given by

$$E = \frac{1}{2}mv^2 + \frac{L^2}{2mr^2} + U(r),$$

where the last two terms provide the effective potential energy. This is derived using:

$$v^2 = \left(\frac{dr}{dt}\right)^2 + \left(r\frac{d\theta}{dt}\right)^2,$$

and then substituting this expression into $E= \frac{1}{2}mv^2+U(r).$

Given that $L^2/(2mr^2)$ comes from the kinetic energy, why is it considered part of the potential energy of the system? Doesn't it directly relate to the velocity of the particle from the derivation? And isn't $\left(r\frac{d\theta}{dt}\right)^2$ is in the same tangential direction as the velocity?

I'm a first year calculus student, so this may be a simple misunderstanding.

The energy of a particle under the action of a radial conservative force is given by

$$E = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2+ \frac{L^2}{2mr^2} + U(r),$$

where the last two terms provide the effective potential energy. This is derived using:

$$v^2 = \left(\frac{dr}{dt}\right)^2 + \left(r\frac{d\theta}{dt}\right)^2,$$

and then substituting this expression into $E= \frac{1}{2}mv^2+U(r).$

Given that $L^2/(2mr^2)$ comes from the kinetic energy, why is it considered part of the potential energy of the system? Doesn't it directly relate to the velocity of the particle from the derivation? And isn't $\left(r\frac{d\theta}{dt}\right)^2$ is in the same tangential direction as the velocity?

I'm a first year calculus student, so this may be a simple misunderstanding.