Torque criterion for constant areal velocity ($\ddot{A}={\tau }_n/(2m)$)

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

A path along a circle can have variable areal velocity. As an illustrative controlled trajectory, keep a point mass on a circle of radius $R$ with angleIts acceleration in polar coordinates and the required net force arewhere ${\mathbf{e}}_r$ and ${\mathbf{e}}_{\theta }$ are the radial and transverse unit vectors. A circular guide can supply the radial force, while a tangential drive supplies $mR\alpha$. HenceFor $R=m={\omega }_0=\alpha =1$ in illustrative units, the three equal time intervals $[0,0.5]$, $[0.5,1]$ and $[1,1.5]$ sweep areas $0.3125$, $0.4375$ and $0.5625$. Their difference comes from the transverse force's positive torque, even though the radius and path shape stay fixed. The area under each areal velocity graph segment equals the corresponding swept area.