Particle in two dimensions

In two dimensions, the orbital angular acceleration is the rate at which the two-dimensional orbital angular velocity of the particle about the origin changes. The instantaneous angular velocity ω at any point in time is given by

where ${\displaystyle r}$ is the distance from the origin and ${\displaystyle v_{\perp }}$ is the cross-radial component of the instantaneous velocity (i.e. the component perpendicular to the position vector), which by convention is positive for counter-clockwise motion and negative for clockwise motion.

Therefore, the instantaneous angular acceleration α of the particle is given by^[2]

Expanding the right-hand-side using the product rule from differential calculus, this becomes

${\displaystyle \alpha ={\frac {1}{r}}{\frac {dv_{\perp }}{dt}}-{\frac {v_{\perp }}{r^{2}}}{\frac {dr}{dt}}.}$

In the special case where the particle undergoes circular motion about the origin, ${\displaystyle {\frac {dv_{\perp }}{dt}}}$ becomes just the tangential acceleration ${\displaystyle a_{\perp }}$, and ${\displaystyle {\frac {dr}{dt}}}$ vanishes (since the distance from the origin stays constant), so the above equation simplifies to

${\displaystyle \alpha ={\frac {a_{\perp }}{r}}.}$

In two dimensions, angular acceleration is a number with plus or minus sign indicating orientation, but not pointing in a direction. The sign is conventionally taken to be positive if the angular speed increases in the counter-clockwise direction or decreases in the clockwise direction, and the sign is taken negative if the angular speed increases in the clockwise direction or decreases in the counter-clockwise direction. Angular acceleration then may be termed a pseudoscalar, a numerical quantity which changes sign under a parity inversion, such as inverting one axis or switching the two axes.

Particle in three dimensions

In three dimensions, the orbital angular acceleration is the rate at which three-dimensional orbital angular velocity vector changes with time. The instantaneous angular velocity vector ${\displaystyle {\boldsymbol {\omega }}}$ at any point in time is given by

${\displaystyle {\boldsymbol {\omega }}={\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}},}$

where ${\displaystyle \mathbf {r} }$ is the particle's position vector, ${\displaystyle r}$ its distance from the origin, and ${\displaystyle \mathbf {v} }$ its velocity vector.^[2]

Therefore, the orbital angular acceleration is the vector ${\displaystyle {\boldsymbol {\alpha }}}$ defined by

${\displaystyle {\boldsymbol {\alpha }}={\frac {d}{dt}}\left({\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}}\right).}$

Expanding this derivative using the product rule for cross-products and the ordinary quotient rule, one gets:

${\displaystyle {\begin{aligned}{\boldsymbol {\alpha }}&={\frac {1}{r^{2}}}\left(\mathbf {r} \times {\frac {d\mathbf {v} }{dt}}+{\frac {d\mathbf {r} }{dt}}\times \mathbf {v} \right)-{\frac {2}{r^{3}}}{\frac {dr}{dt}}\left(\mathbf {r} \times \mathbf {v} \right)\\\\&={\frac {1}{r^{2}}}\left(\mathbf {r} \times \mathbf {a} +\mathbf {v} \times \mathbf {v} \right)-{\frac {2}{r^{3}}}{\frac {dr}{dt}}\left(\mathbf {r} \times \mathbf {v} \right)\\\\&={\frac {\mathbf {r} \times \mathbf {a} }{r^{2}}}-{\frac {2}{r^{3}}}{\frac {dr}{dt}}\left(\mathbf {r} \times \mathbf {v} \right).\end{aligned}}}$

Since ${\displaystyle \mathbf {r} \times \mathbf {v} }$ is just ${\displaystyle r^{2}{\boldsymbol {\omega }}}$, the second term may be rewritten as ${\displaystyle -{\frac {2}{r}}{\frac {dr}{dt}}{\boldsymbol {\omega }}}$. In the case where the distance ${\displaystyle r}$ of the particle from the origin does not change with time (which includes circular motion as a subcase), the second term vanishes and the above formula simplifies to

${\displaystyle {\boldsymbol {\alpha }}={\frac {\mathbf {r} \times \mathbf {a} }{r^{2}}}.}$

From the above equation, one can recover the cross-radial acceleration in this special case as:

${\displaystyle \mathbf {a} _{\perp }={\boldsymbol {\alpha }}\times \mathbf {r} .}$

Unlike in two dimensions, the angular acceleration in three dimensions need not be associated with a change in the angular speed ${\displaystyle \omega =|{\boldsymbol {\omega }}|}$: If the particle's position vector "twists" in space, changing its instantaneous plane of angular displacement, the change in the direction of the angular velocity ${\displaystyle {\boldsymbol {\omega }}}$ will still produce a nonzero angular acceleration. This cannot not happen if the position vector is restricted to a fixed plane, in which case ${\displaystyle {\boldsymbol {\omega }}}$ has a fixed direction perpendicular to the plane.

The angular acceleration vector is more properly called a pseudovector: It has three components which transform under rotations in the same way as the Cartesian coordinates of a point do, but which do not transform like Cartesian coordinates under reflections.

Relation to torque

The net torque on a point particle is defined to be the pseudovector

where ${\displaystyle \mathbf {F} }$ is the net force on the particle.^[3]

Torque is the rotational analogue of force: it induces change in the rotational state of a system, just as force induces change in the translational state of a system. As force on a particle is connected to acceleration by the equation , one may write a similar equation connecting torque on a particle to angular acceleration, though this relation is necessarily more complicated.^[4]

First, substituting ${\displaystyle \mathbf {F} =m\mathbf {a} }$ into the above equation for torque, one gets

${\displaystyle {\boldsymbol {\tau }}=m\left(\mathbf {r} \times \mathbf {a} \right)=mr^{2}\left({\frac {\mathbf {r} \times \mathbf {a} }{r^{2}}}\right).}$

From the previous section:

${\displaystyle {\boldsymbol {\alpha }}={\frac {\mathbf {r} \times \mathbf {a} }{r^{2}}}-{\frac {2}{r}}{\frac {dr}{dt}}{\boldsymbol {\omega }},}$

where ${\displaystyle {\boldsymbol {\alpha }}}$ is orbital angular acceleration and ${\displaystyle {\boldsymbol {\omega }}}$ is orbital angular velocity. Therefore:

${\displaystyle {\boldsymbol {\tau }}=mr^{2}\left({\boldsymbol {\alpha }}+{\frac {2}{r}}{\frac {dr}{dt}}{\boldsymbol {\omega }}\right)=mr^{2}{\boldsymbol {\alpha }}+2mr{\frac {dr}{dt}}{\boldsymbol {\omega }}.}$

In the special case of constant distance ${\displaystyle r}$ of the particle from the origin (${\displaystyle {\tfrac {dr}{dt}}=0}$), the second term in the above equation vanishes and the above equation simplifies to

${\displaystyle {\boldsymbol {\tau }}=mr^{2}{\boldsymbol {\alpha }},}$

which can be interpreted as a "rotational analogue" to ${\displaystyle \mathbf {F} =m\mathbf {a} }$, where the quantity ${\displaystyle mr^{2}}$ (known as the moment of inertia of the particle) plays the role of the mass ${\displaystyle m}$. However, unlike ${\displaystyle \mathbf {F} =m\mathbf {a} }$, this equation does not apply to an arbitrary trajectory, only to a trajectory contained within a spherical shell about the origin.