Angular acceleration

Radians per second squared
Unit system	SI derived unit
Unit of	Angular acceleration
Symbol	rad/s2

In physics, angular acceleration refers to the time rate of change of angular velocity. As there are two types of angular velocity, namely spin angular velocity and orbital angular velocity, there are naturally also two types of angular acceleration, called spin angular acceleration and orbital angular acceleration respectively. Spin angular acceleration refers to the angular acceleration of a rigid body about its centre of rotation, and orbital angular acceleration refers to the angular acceleration of a point particle about a fixed origin.

Angular acceleration is measured in units of angle per unit time squared (which in SI units is radians per second squared), and is usually represented by the symbol alpha (α). By convention, positive angular acceleration indicates an increase in counter-clockwise angular speed or decrease in clockwise angular speed, while negative angular acceleration indicates an increase in clockwise angular speed or decrease in counterclockwise angular speed. In three dimensions, angular acceleration is a pseudovector..

For rigid bodies, angular acceleration must be caused by a net external torque. However, this is not so for non-rigid bodies: For example, a figure skater can speed up her rotation (thereby obtaining an angular acceleration) simply by contracting her arms and legs inwards, which involves no external torque.

Orbital Angular Acceleration of a Point Particle

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

$\omega ={\frac {v_{\perp }}{r}},$

where $r$ is the distance from the origin and $v_{\perp }$ 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

$\alpha ={\frac {d}{dt}}({\frac {v_{\perp }}{r}}).$

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

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

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

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

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 ${\boldsymbol {\omega }}$ is defined by

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

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

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

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

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

Since ${\frac {dr}{dt}}$ is just the radial component $v_{\|}$ of the velocity vector, and $\mathbf {r} \times \mathbf {v}$ is just $r^{2}{\boldsymbol {\omega }}$, the second term may be rewritten more compactly as $-{\frac {2v_{\|}{\boldsymbol {\omega }}}{r}}$. In the case where the distance $r$ 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

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

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

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

Unlike in two dimensions, the angular acceleration in three dimensions need not be associated with a change in angular speed: If the particle's trajectory "twists" in space such that the instantaneous plane of angular velocity (i.e. the instantaneous plane in which the position vector "sweeps out" angle) continuously changes with time, then even if the angular speed (i.e. the speed at which the position vector "sweeps out" angle) is constant, there will still be a nonzero angular acceleration because the direction of the angular velocity vector changes with time. This cannot not happen in two dimensions because the position vector is restricted to a fixed plane so that any change in angular velocity must be through a change in its magnitude.

In three dimensions, angular acceleration is what is called a pseudovector. It has three components which transform under rotations the same way as a vector does, but which under reflections transforms differently than a vector.

Relation to Torque

The net torque ${\boldsymbol {\tau }}$ on a point particle is given by

${\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} ,$

where F is the net force on the particle. By Newton's Second Law, this may be rewritten as

${\boldsymbol {\tau }}=m(\mathbf {r} \times \mathbf {a} ),$

where $m$ is the mass of the particle. Dividing both sides by the moment of inertia $I=mr^{2}$ of the particle about the origin, this becomes

${\frac {\boldsymbol {\tau }}{I}}={\frac {\mathbf {r} \times \mathbf {a} }{r^{2}}}$

But from the previous section, it was derived that

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

where ${\boldsymbol {\alpha }}$ is the orbital angular acceleration of the particle and ${\boldsymbol {\omega }}$ is the orbital angular velocity of the particle.

Therefore, it follows that

${\frac {\boldsymbol {\tau }}{I}}={\boldsymbol {\alpha }}+{\frac {2}{r}}{\frac {dr}{dt}}{\boldsymbol {\omega }}.$

The above relationship is the rotational analog of Newton's Second Law; it connects the net applied torque on a point particle to the induced angular acceleration of the particle. Unlike the relationship between force and acceleration, the angular acceleration vector is not necessarily parallel or directly proportional to the torque vector, and does not depend solely on the net torque and moment of inertia. However, in the special case where the distance to the origin does not change with time, the second term in the above equation vanishes and the above equation simplifies to

${\boldsymbol {\tau }}=I{\boldsymbol {\alpha }}.$

See also

• Torque
• Angular momentum
• Angular speed
• Angular velocity