Angular velocity

The angular velocity of the particle at P with respect to the origin O is determined by the perpendicular component of the velocity vector v.

In the simplest case of circular motion at radius ${\displaystyle r}$, with position given by the angular displacement ${\displaystyle \phi (t)}$ from the x-axis, the orbital angular velocity is the rate of change of angle with respect to time: ${\textstyle \omega ={\frac {d\phi }{dt}}}$. If ${\displaystyle \phi }$ is measured in radians, the arc-length from the positive x-axis around the circle to the particle is ${\displaystyle \ell =r\phi }$, and the linear velocity is ${\textstyle v(t)={\frac {d\ell }{dt}}=r\omega (t)}$, so that ${\textstyle \omega ={\frac {v}{r}}}$.

In the general case of a particle moving in the plane, the orbital angular velocity is the rate at which the position vector relative to a chosen origin "sweeps out" angle. The diagram shows the position vector ${\displaystyle \mathbf {r} }$ from the origin ${\displaystyle O}$ to a particle ${\displaystyle P}$, with its polar coordinates ${\displaystyle (r,\phi )}$. (All variables are functions of time ${\displaystyle t}$.) The particle has linear velocity splitting as ${\displaystyle \mathbf {v} =\mathbf {v} _{\|}+\mathbf {v} _{\perp }}$, with the radial component ${\displaystyle \mathbf {v} _{\|}}$ parallel to the radius, and the cross-radial (or tangential) component ${\displaystyle \mathbf {v} _{\perp }}$ perpendicular to the radius. When there is no radial component, the particle moves around the origin in a circle; but when there is no cross-radial component, it moves in a straight line from the origin. Since radial motion leaves the angle unchanged, only the cross-radial component of linear velocity contributes to angular velocity.

The angular velocity ω is the rate of change of angular position with respect to time, which can be computed from the cross-radial velocity as:

Here the cross-radial speed ${\displaystyle v_{\perp }}$ is the signed magnitude of ${\displaystyle \mathbf {v} _{\perp }}$, positive for counter-clockwise motion, negative for clockwise. Taking polar coordinates for the linear velocity ${\displaystyle \mathbf {v} }$ gives magnitude ${\displaystyle v}$ (linear speed) and angle ${\displaystyle \theta }$ relative to the radius vector; in these terms, ${\displaystyle v_{\perp }=v\sin(\theta )}$, so that

These formulas may be derived doing ${\displaystyle \mathbf {r} =(r\cos(\varphi ),r\sin(\varphi ))}$, being ${\displaystyle r}$ a function of the distance to the origin with respect to time, and ${\displaystyle \varphi }$ a function of the angle between the vector and the x axis. Then ${\textstyle {\frac {d\mathbf {r} }{dt}}=({\dot {r}}\cos(\varphi )-r{\dot {\varphi }}\sin(\varphi ),{\dot {r}}\sin(\varphi )+r{\dot {\varphi }}\cos(\varphi ))}$. Which is equal to ${\displaystyle {\dot {r}}(\cos(\varphi ),\sin(\varphi ))+r{\dot {\varphi }}(-\sin(\varphi ),\cos(\varphi ))={\dot {r}}{\hat {r}}+r{\dot {\varphi }}{\hat {\varphi }}}$. (See Unit vector in cylindrical coordinates). Knowing ${\textstyle {\frac {d\mathbf {r} }{dt}}=\mathbf {v} }$, we conclude that the radial component of the velocity is given by ${\displaystyle {\dot {r}}}$, because ${\displaystyle {\hat {r}}}$ is a radial unit vector; and the perpendicular component is given by ${\displaystyle r{\dot {\varphi }}}$ because ${\displaystyle {\hat {\varphi }}}$ is a perpendicular unit vector.

In two dimensions, angular velocity 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 radius vector turns counter-clockwise, and negative if clockwise. Angular velocity 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.

The orbital angular velocity vector encodes the time rate of change of angular position, as well as the instantaneous plane of angular displacement. In this case (counter-clockwise circular motion) the vector points up.

In three-dimensional space, we again have the position vector r of a moving particle. Here, orbital angular velocity is a pseudovector whose magnitude is the rate at which r sweeps out angle, and whose direction is perpendicular to the instantaneous plane in which r sweeps out angle (i.e. the plane spanned by r and v). However, as there are two directions perpendicular to any plane, an additional condition is necessary to uniquely specify the direction of the angular velocity; conventionally, the right-hand rule is used.

Let the pseudovector ${\displaystyle \mathbf {u} }$ be the unit vector perpendicular to the plane spanned by r and v, so that the right-hand rule is satisfied (i.e. the instantaneous direction of angular displacement is counter-clockwise looking from the top of ${\displaystyle \mathbf {u} }$). Taking polar coordinates ${\displaystyle (r,\phi )}$ in this plane, as in the two-dimensional case above, one may define the orbital angular velocity vector as:

${\displaystyle {\boldsymbol {\omega }}=\omega \mathbf {u} ={\frac {d\phi }{dt}}\mathbf {u} ={\frac {v\sin(\theta )}{r}}\mathbf {u} ,}$

where θ is the angle between r and v. In terms of the cross product, this is:

${\displaystyle {\boldsymbol {\omega }}={\frac {\mathbf {r} \times \mathbf {v} }{r^{2}}}.}$[4]

From the above equation, one can recover the tangential velocity as:

${\displaystyle \mathbf {v} _{\perp }={\boldsymbol {\omega }}\times \mathbf {r} }$

Consider a rigid body rotating about a fixed point O. Construct a reference frame in the body consisting of an orthonormal set of vectors ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}$ fixed to the body and with their common origin at O. The angular velocity vector of both frame and body about O is then

${\displaystyle {\boldsymbol {\omega }}=\left({\dot {\mathbf {e} }}_{1}\cdot \mathbf {e} _{2}\right)\mathbf {e} _{3}+\left({\dot {\mathbf {e} }}_{2}\cdot \mathbf {e} _{3}\right)\mathbf {e} _{1}+\left({\dot {\mathbf {e} }}_{3}\cdot \mathbf {e} _{1}\right)\mathbf {e} _{2},}$

where ${\displaystyle {\dot {\mathbf {e} }}_{i}={\frac {d\mathbf {e} _{i}}{dt}}}$ is the time rate of change of the frame vector ${\displaystyle \mathbf {e} _{i},i=1,2,3,}$ due to the rotation.

Note that this formula is incompatible with the expression

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

as that formula defines only the angular velocity of a single point about O, while the formula in this section applies to a frame or rigid body. In the case of a rigid body a single ${\displaystyle {\boldsymbol {\omega }}}$ has to account for the motion of all particles in the body.

Diagram showing Euler frame in green

The components of the spin angular velocity pseudovector were first calculated by Leonhard Euler using his Euler angles and the use of an intermediate frame:

• One axis of the reference frame (the precession axis)
• The line of nodes of the moving frame with respect to the reference frame (nutation axis)
• One axis of the moving frame (the intrinsic rotation axis)

Euler proved that the projections of the angular velocity pseudovector on each of these three axes is the derivative of its associated angle (which is equivalent to decomposing the instantaneous rotation into three instantaneous Euler rotations). Therefore:[5]

${\displaystyle {\boldsymbol {\omega }}={\dot {\alpha }}\mathbf {u} _{1}+{\dot {\beta }}\mathbf {u} _{2}+{\dot {\gamma }}\mathbf {u} _{3}}$

This basis is not orthonormal and it is difficult to use, but now the velocity vector can be changed to the fixed frame or to the moving frame with just a change of bases. For example, changing to the mobile frame:

${\displaystyle {\boldsymbol {\omega }}=({\dot {\alpha }}\sin \beta \sin \gamma +{\dot {\beta }}\cos \gamma ){\hat {\mathbf {i} }}+({\dot {\alpha }}\sin \beta \cos \gamma -{\dot {\beta }}\sin \gamma ){\hat {\mathbf {j} }}+({\dot {\alpha }}\cos \beta +{\dot {\gamma }}){\hat {\mathbf {k} }}}$

where ${\displaystyle {\hat {\mathbf {i} }},{\hat {\mathbf {j} }},{\hat {\mathbf {k} }}}$ are unit vectors for the frame fixed in the moving body. This example has been made using the Z-X-Z convention for Euler angles.

A vector ${\displaystyle \mathbf {r} }$ undergoing uniform circular motion around a fixed axis satisfies:

${\displaystyle {\frac {d\mathbf {r} }{dt}}={\boldsymbol {\omega }}\times \mathbf {r} =W\cdot \mathbf {r} }$

Given the orientation matrix A(t) of a frame, whose columns are the moving orthonormal coordinate vectors ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}$, we can obtain its angular velocity tensor W(t) as follows. Angular velocity must be the same for the three vectors ${\displaystyle \mathbf {r} =\mathbf {e} _{i}}$, so arranging the three vector equations into columns of a matrix, we have:

${\displaystyle {\frac {dA}{dt}}=W\cdot A.}$

(This holds even if A(t) does not rotate uniformly.) Therefore the angular velocity tensor is:

${\displaystyle W={\frac {dA}{dt}}\cdot A^{-1}={\frac {dA}{dt}}\cdot A^{\mathrm {T} },}$

since the inverse of the orthogonal matrix ${\displaystyle A}$ is its transpose ${\displaystyle A^{\mathrm {T} }}$.

In three dimensions, angular velocity can be represented by a pseudovector because second rank tensors are dual to pseudovectors in three dimensions. Since the angular velocity tensor W = W(t) is a skew-symmetric matrix:

${\displaystyle W={\begin{pmatrix}0&-\omega _{z}&\omega _{y}\\\omega _{z}&0&-\omega _{x}\\-\omega _{y}&\omega _{x}&0\\\end{pmatrix}},}$

its Hodge dual is a vector, which is precisely the previous angular velocity vector ${\displaystyle {\boldsymbol {\omega }}=[\omega _{x},\omega _{y},\omega _{z}]}$.

If we know an initial frame A(0) and we are given a constant angular velocity tensor W, we can obtain A(t) for any given t. Recall the matrix differential equation:

${\displaystyle {\frac {dA}{dt}}=W\cdot A.}$

This equation can be integrated to give:

${\displaystyle A(t)=e^{Wt}A(0),}$

which shows a connection with the Lie group of rotations.

We prove that angular velocity tensor is skew symmetric, i.e. ${\displaystyle W={\frac {dA(t)}{dt}}\cdot A^{\text{T}}}$ satisfies ${\displaystyle W^{\text{T}}=-W}$.

A rotation matrix A is orthogonal, inverse to its transpose, so we have ${\displaystyle I=A\cdot A^{\text{T}}}$. For ${\displaystyle A=A(t)}$ a frame matrix, taking the time derivative of the equation gives:

${\displaystyle 0={\frac {dA}{dt}}A^{\text{T}}+A{\frac {dA^{\text{T}}}{dt}}}$

Applying the formula ${\displaystyle (AB)^{\text{T}}=B^{\text{T}}A^{\text{T}}}$,

${\displaystyle 0={\frac {dA}{dt}}A^{\text{T}}+\left({\frac {dA}{dt}}A^{\text{T}}\right)^{\text{T}}=W+W^{\text{T}}}$

Thus, W is the negative of its transpose, which implies it is skew symmetric.

At any instant ${\displaystyle t}$, the angular velocity tensor represents a linear map between the position vector ${\displaystyle \mathbf {r} (t)}$ and the velocity vectors ${\displaystyle \mathbf {v} (t)}$ of a point on a rigid body rotating around the origin:

${\displaystyle \mathbf {v} =W\mathbf {r} .}$

The relation between this linear map and the angular velocity pseudovector ${\displaystyle {\boldsymbol {\omega }}}$ is the following.

Because W is the derivative of an orthogonal transformation, the bilinear form

${\displaystyle B(\mathbf {r} ,\mathbf {s} )=(W\mathbf {r} )\cdot \mathbf {s} }$

is skew-symmetric. Thus we can apply the fact of exterior algebra that there is a unique linear form ${\displaystyle L}$ on ${\displaystyle \Lambda ^{2}V}$ that

${\displaystyle L(\mathbf {r} \wedge \mathbf {s} )=B(\mathbf {r} ,\mathbf {s} )}$

where ${\displaystyle \mathbf {r} \wedge \mathbf {s} \in \Lambda ^{2}V}$ is the exterior product of ${\displaystyle \mathbf {r} }$ and ${\displaystyle \mathbf {s} }$.

Taking the sharp L^♯ of L we get

${\displaystyle (W\mathbf {r} )\cdot \mathbf {s} =L^{\sharp }\cdot (\mathbf {r} \wedge \mathbf {s} )}$

Introducing ${\displaystyle {\boldsymbol {\omega }}:={\star }(L^{\sharp })}$, as the Hodge dual of L^♯, and applying the definition of the Hodge dual twice supposing that the preferred unit 3-vector is ${\displaystyle \star 1}$

${\displaystyle (W\mathbf {r} )\cdot \mathbf {s} ={\star }({\star }(L^{\sharp })\wedge \mathbf {r} \wedge \mathbf {s} )={\star }({\boldsymbol {\omega }}\wedge \mathbf {r} \wedge \mathbf {s} )={\star }({\boldsymbol {\omega }}\wedge \mathbf {r} )\cdot \mathbf {s} =({\boldsymbol {\omega }}\times \mathbf {r} )\cdot \mathbf {s} ,}$

where

${\displaystyle {\boldsymbol {\omega }}\times \mathbf {r}$ :={\star }({\boldsymbol {\omega }}\wedge \mathbf {r} )}

by definition.

Because ${\displaystyle \mathbf {s} }$ is an arbitrary vector, from nondegeneracy of scalar product follows

${\displaystyle W\mathbf {r} ={\boldsymbol {\omega }}\times \mathbf {r} }$

Since the spin angular velocity tensor of a rigid body (in its rest frame) is a linear transformation that maps positions to velocities (within the rigid body), it can be regarded as a constant vector field. In particular, the spin angular velocity is a Killing vector field belonging to an element of the Lie algebra SO(3) of the 3-dimensional rotation group SO(3).

Also, it can be shown that the spin angular velocity vector field is exactly half of the curl of the linear velocity vector field v(r) of the rigid body. In symbols,

${\displaystyle {\boldsymbol {\omega }}={\frac {1}{2}}\nabla \times \mathbf {v} }$

We have supposed that the rigid body rotates around an arbitrary point. We should prove that the spin angular velocity previously defined is independent of the choice of origin, which means that the spin angular velocity is an intrinsic property of the spinning rigid body. (Note the marked contrast of this with the orbital angular velocity of a point particle, which certainly does depend on the choice of origin.)

Proving the independence of spin angular velocity from choice of origin

See the graph to the right: The origin of lab frame is O, while O₁ and O₂ are two fixed points on the rigid body, whose velocity is ${\displaystyle \mathbf {v} _{1}}$ and ${\displaystyle \mathbf {v} _{2}}$ respectively. Suppose the angular velocity with respect to O₁ and O₂ is ${\displaystyle {\boldsymbol {\omega }}_{1}}$ and ${\displaystyle {\boldsymbol {\omega }}_{2}}$ respectively. Since point P and O₂ have only one velocity,

${\displaystyle \mathbf {v} _{1}+{\boldsymbol {\omega }}_{1}\times \mathbf {r} _{1}=\mathbf {v} _{2}+{\boldsymbol {\omega }}_{2}\times \mathbf {r} _{2}}$

${\displaystyle \mathbf {v} _{2}=\mathbf {v} _{1}+{\boldsymbol {\omega }}_{1}\times \mathbf {r} =\mathbf {v} _{1}+{\boldsymbol {\omega }}_{1}\times (\mathbf {r} _{1}-\mathbf {r} _{2})}$

The above two yields that

${\displaystyle ({\boldsymbol {\omega }}_{2}-{\boldsymbol {\omega }}_{1})\times \mathbf {r} _{2}=0}$

Since the point P (and thus ${\displaystyle \mathbf {r} _{2}}$) is arbitrary, it follows that

${\displaystyle {\boldsymbol {\omega }}_{1}={\boldsymbol {\omega }}_{2}}$

If the reference point is the instantaneous axis of rotation the expression of the velocity of a point in the rigid body will have just the angular velocity term. This is because the velocity of the instantaneous axis of rotation is zero. An example of the instantaneous axis of rotation is the hinge of a door. Another example is the point of contact of a purely rolling spherical (or, more generally, convex) rigid body.