Angular velocity

See also: Angular frequency

In physics, the angular velocity of a rotating body is defined as the rate of change of angular displacement, and is a vector quantity (more precisely, a pseudovector) that specifies the angular speed (rotational speed) of an object and the axis about which the object is rotating. This speed can be measured in the SI unit of angular velocity, radians per second, or in terms of degrees per second, degrees per hour, etc. Angular velocity is usually represented by the symbol omega (ω, rarely Ω).

The direction of the angular velocity vector is perpendicular to the plane of rotation, in a direction that is usually specified by the right-hand rule.

Angular velocity of a particle

Particle in two dimensions

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.

The angular velocity describes the speed of rotation and the orientation of the instantaneous axis about which the rotation occurs. The direction of the angular velocity pseudovector will be along the axis of rotation; in this case (counter-clockwise rotation) the vector points up.

The angular velocity of a particle is measured around or relative to a point, called the origin. As shown in the diagram (with angles ɸ and θ in radians), if a line is drawn from the origin (O) to the particle (P), then the velocity (v) of the particle has a component along the radius (radial component, v_‖) and a component perpendicular to the radius (cross-radial component, v_⊥). If there is no radial component, then the particle moves in a circle. On the other hand, if there is no cross-radial component, then the particle moves along a straight line from the origin.

A radial motion produces no change in the direction of the particle relative to the origin, so, for the purpose of finding the angular velocity, the radial component can be ignored. Therefore, the rotation is completely produced by the perpendicular motion around the origin, and the angular velocity is completely determined by this component.

In two dimensions, the angular velocity ω is given by

$\omega = \frac{d\phi}{dt}$

This is related to the cross-radial (tangential) velocity by:

$\mathrm{v}_\perp=r\,\frac{d\phi}{dt}$

An explicit formula for v_⊥ in terms of v and θ is:

$\mathrm{v}_\perp=|\mathrm{\mathbf{v}}|\,\sin(\theta)$

Combining the above equations gives a formula for ω:

$\omega=\frac{|\mathrm{\mathbf{v}}|\sin(\theta)}{|\mathrm{\mathbf{r}}|}$

In two dimensions, the angular velocity is a single number with an orientation but no direction. The angular velocity in two dimensions is a pseudoscalar, a quantity that changes its sign under a parity inversion (for example if one of the axes is inverted or if axes are swapped). The positive direction of rotation is taken, by convention, to be in the direction towards the y axis from the x axis. If the parity is inverted, but the orientation of a rotation is not, then the sign of the angular velocity changes.

There are three types of angular velocity involved in the movement on an ellipse corresponding to the three anomalies (true, eccentric and mean).

Particle in three dimensions

In three dimensions, the angular velocity becomes a bit more complicated. The angular velocity in this case is generally thought of as a vector, or more precisely, a pseudovector. It now has not only a magnitude, but a direction as well. The magnitude is the angular speed, and the direction describes the axis of rotation that Euler's rotation theorem guarantees must exist. The right-hand rule indicates the positive direction of the angular velocity pseudovector.

Let $\bold u$ be a unitary vector along the instantaneous rotation axis, so that from the top of the vector the rotation is counter-clock-wise. The angular velocity vector $\vec \omega$ is then defined as:

$\boldsymbol\omega = \frac{d\phi}{dt}\bold u$

Just as in the two dimensional case, a particle will have a component of its velocity along the radius from the origin to the particle, and another component perpendicular to that radius. The combination of the origin point and the perpendicular component of the velocity defines a plane of rotation in which the behavior of the particle (for that instant) appears just as it does in the two dimensional case. The axis of rotation is then a line normal to this plane, and this axis defines the direction of the angular velocity pseudovector, while the magnitude is the same as the pseudoscalar value found in the 2-dimensional case. Using the unit vector $\bold u$ defined before, the angular velocity vector may be written in a manner similar to that for two dimensions:

$\boldsymbol\omega=\frac{|\bold v|\sin(\theta)}{|\bold r|}\bold u$

which, by the definition of the cross product, can be written:

$\boldsymbol\omega=\frac{\bold r\times\bold v}{|\bold r|^2}$

and using this vector, it can be seen that the formula for the tangential speed of a point $\boldsymbol r(t)$ is:

$\frac {d \boldsymbol r(t)} {dt} = \boldsymbol{\omega} \times\boldsymbol{r}$

Addition of angular velocity vectors

Angular velocity can be defined as angular displacement for unit of time. If a point rotates with $\omega_2$ in a frame $F_2$ that itself rotates with an angular speed $\omega_1$ with respect to an external frame $F_1$, we can define the addition of $\omega_1 + \omega_2$ as the angular velocity vector of the point with respect to $F_1$.

With this operation defined like this, angular velocity, which is a pseudovector, also becomes a real vector because it has two operations:

• An internal operation (addition), which is associative, commutative, distributive and with zero and unity elements
• An external operation (external product), with the normal properties for an external product.

This is the definition of a vector space. The only property that presents difficulties to prove is the commutativity of the addition. This can be proven from the fact that the velocity tensor W (see below) is skew-symmetric. Therefore, $R=e^{Wt}$ is a rotation matrix and in a time dt is an infinitesimal rotation matrix. Therefore, it can be expanded as $R = I + W\cdot dt + {1 \over 2} (W \cdot dt)^2 + ...$

The composition of rotations is not commutative; but, when the rotations are infinitesimal, the first order approximation of the previous series can be taken and $(I+W_1\cdot dt)(I+W_2 \cdot dt)=(I+W_2 \cdot dt)(I+W_1\cdot dt)$ and therefore $\omega_1 + \omega_2 = \omega_2 + \omega_1$

Angular velocity vector for a frame

Schematic construction for addition of angular velocity vectors for rotating frames

Given a rotating frame composed by three unitary vectors, all the three must have the same angular speed in any instant. In such a frame, each vector is a particular case of the previous case (moving particle), in which the module of the vector is constant.

Though it just a particular case of a moving particle, this is a very important one for its relationship with the rigid body study, and special tools have been developed for this case. There are two possible ways to describe the angular velocity of a rotating frame: the angular velocity vector and the angular velocity tensor. Both entities are related and they can be calculated from each other.

In a consistent way with the general definition, the angular velocity of a frame is defined as the angular velocity of each of the three vectors of the frame (it will be the same for any of them). The addition of angular velocity vectors for frames is also defined by movement composition, and can be useful to decompose the movement as in a gimbal. Components of the vector can be calculated as derivatives of the parameters defining the moving frames (Euler angles or rotation matrices). As in the general case, addition is commutative: $\omega_1 + \omega_2 = \omega_2 + \omega_1$

It is known by Euler's rotation theorem that, for any rotating frame, there always exists an instantaneous axis of rotation in any instant. In the case of a frame, the angular velocity vector is over the instantaneous axis of rotation. Any transversal section of a plane perpendicular to this axis has to behave as a two dimensional rotation. Thus, the magnitude of the angular velocity vector at a given time t is consistent with the two dimensions case.

Components from the vectors of the frame

Substituting in the expression

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

any unitary vector e of the frame we obtain $\boldsymbol\omega=\frac{\boldsymbol e\times\dot{\boldsymbol e}}{|{\mathbf e}|^2}$, and therefore $\boldsymbol\omega = \mathbf{e}_1\times\dot{\mathbf{e}}_1=\mathbf{e}_2\times\dot{\mathbf{e}}_2=\mathbf{e}_3\times\dot{\mathbf{e}}_3.$

As the columns of the matrix of the frame are the components of its vectors, this allows also the calculation of $\omega$ from the matrix of the frame and its derivative.

Components from Euler angles

Diagram showing Euler frame in green

The components of the angular velocity pseudovector were first calculated by Leonhard Euler using his Euler angles and an intermediate frame made out of the intermediate frames of the construction:

• One axis of the reference frame (the precession axis)
• The line of nodes of the moving frame respect 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 over these three axes was the derivative of its associated angle (which is equivalent to decomposing the instant rotation in three instantaneous Euler rotations). Therefore:

$\boldsymbol\omega = \dot\alpha\bold u_1+\dot\beta\bold u_2+\dot\gamma \bold u_3$

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:

$\boldsymbol\omega = (\dot\alpha\sin\beta\sin\gamma+\dot\beta\cos\gamma)\bold I+(\dot\alpha\sin\beta\cos\gamma-\dot\beta\sin\gamma)\bold J +(\dot\alpha\cos\beta+\dot\gamma)\bold K$

where $\bold I,\bold J,\bold K$ 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.

Angular velocity tensor

See also: Skew-symmetric matrix

A similar way to describe the angular speed for a rotating frame is the angular velocity tensor. It can be introduced from rotation matrices. Any vector $\vec r$ that rotates around an axis with an angular speed vector $\vec \omega$ (as defined before) satisfies:

$\frac {d \vec r(t)} {dt} = \vec{\omega} \times\vec{r}$

We can introduce here the angular velocity tensor associated to the angular speed $\omega$:

$W(t) = \begin{pmatrix} 0 & -\omega_z(t) & \omega_y(t) \\ \omega_z(t) & 0 & -\omega_x(t) \\ -\omega_y(t) & \omega_x(t) & 0 \\ \end{pmatrix}$

Notice that this is an infinitesimal angular displacement divided by an infinitesimal time. This tensor W(t) will act as if it were a $(\vec \omega \times)$ operator :

$\vec \omega(t) \times \vec{r}(t) = W(t) \vec{r}(t)$

Calculation from the orientation matrix

Given the orientation matrix A(t) of a frame, defined for all t and derivable, we can obtain its instant angular velocity tensor W as follows. We know that:

$\frac {d \vec r(t)} {dt} = W \cdot \vec{r}$

As angular speed must be the same for the three vectors of a rotating frame, if we have a matrix A(t) whose columns are the vectors of the frame, we can write for the three vectors as a whole:

$\frac {dA(t)} {dt} = W \cdot A (t)$

And therefore the angular velocity tensor we are looking for is:

$W = \frac {dA(t)} {dt} \cdot A^{-1}(t)$

Properties of angular velocity tensors

See also: Infinitesimal rotation

In general, the angular velocity in an n-dimensional space is the time derivative of the angular displacement tensor, which is a second rank skew-symmetric tensor.

This tensor W will have n(n-1)/2 independent components and this number is the dimension of the Lie algebra of the Lie group of rotations of an n-dimensional inner product space.

Duality with respect to the velocity vector

In three dimensions, angular velocity can be represented by a pseudovector because second rank tensors are dual to pseudovectors in three dimensions. Notice that the tensor is a matrix with this structure:

$W(t) = \begin{pmatrix} 0 & -\omega_z(t) & \omega_y(t) \\ \omega_z(t) & 0 & -\omega_x(t) \\ -\omega_y(t) & \omega_x(t) & 0\\ \end{pmatrix}$

As it is a skew symmetric matrix, it has a Hodge dual vector, which is precisely the previous angular velocity vector $\vec \omega$:

$\boldsymbol\omega=[\omega_x,\omega_y,\omega_z]$

Exponential of W

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. As $\frac {dA(t)} {dt} = W\cdot A(t)$, this can be read as a differential equation that defines A(t) knowing W(t).

$\frac {dA(t)} {A} = W \cdot {dt}$

And if the angular speed is constant, then W is also constant and the equation can be integrated. The result is:

$A(t) = e^{W \cdot t}A(0)$

which shows a connection with the Lie group of rotations.

W is skew-symmetric

It is possible to prove that angular velocity tensor are skew symmetric matrices, which means that a $W = \frac {dR(t)}{dt}\cdot {R^T}$ satisfies $W^T= -W$.

To prove it we start taking the time derivative of $\mathcal{R}\mathcal{R}^T$ being R(t) a rotation matrix:

$\mathcal{I}=\mathcal{R}\mathcal{R}^T$ because R(t) is a rotation matrix

$0=\frac{d\mathcal{R}}{dt}\mathcal{R}^T+\mathcal{R}\frac{d\mathcal{R}^T}{dt}$

Applying the formula (AB)^T = B^TA^T:

$0 = \frac{d\mathcal{R}}{dt}\mathcal{R}^T+\left(\frac{d\mathcal{R}}{dt}\mathcal{R}^T\right)^T = W + W^T$

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

Coordinate-free description

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

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

where we omitted the $t$ parameter, and regard $\mathbf{v}$ and $\mathbf{r}$ as elements of the same 3-dimensional Euclidean vector space $V$.

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

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

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

bilinear form is skew-symmetric. (Here $\cdot$ stands for the scalar product). So we can apply the fact of exterior algebra that there is a unique linear form $L$ on $\Lambda^2 V$ that

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

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

Taking the dual vector L* of L we get

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

Introducing $\omega := *L^*$, as the Hodge dual of L*, and apply further Hodge dual identities we arrive at

$(W\mathbf{r}) \cdot \mathbf{s} = * ( *L^* \wedge \mathbf{r} \wedge \mathbf{s}) = * (\omega \wedge \mathbf{r} \wedge \mathbf{s}) = *(\omega \wedge \mathbf{r}) \cdot \mathbf{s} = (\omega \times \mathbf{r}) \cdot \mathbf{s}$

where

$\omega \times \mathbf{r} := *(\omega \wedge \mathbf{r})$

by definition.

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

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

Angular velocity as a vector field

For angular velocity tensor maps velocities to positions, it is a vector field. In particular, this vector field is a Killing vector field belonging to an element of the Lie algebra so(3) of the 3-dimensional rotation group SO(3). This element of so(3) can also be regarded as the angular velocity vector.

Rigid body considerations

See also: axes conventions

Position of point P located in the rigid body (shown in blue). R_i is the position with respect to the lab frame, centered at O  and r_i is the position with respect to the rigid body frame, centered at O'  . The origin of the rigid body frame is at vector position R from the lab frame.

The same equations for the angular speed can be obtained reasoning over a rotating rigid body. Here is not assumed that the rigid body rotates around the origin. Instead, it can be supposed rotating around an arbitrary point that is moving with a linear velocity V(t) in each instant.

To obtain the equations, it is convenient to imagine a rigid body attached to the frames and consider a coordinate system that is fixed with respect to the rigid body. Then we will study the coordinate transformations between this coordinate and the fixed "laboratory" system.

As shown in the figure on the right, the lab system's origin is at point O, the rigid body system origin is at O' and the vector from O to O' is R. A particle (i) in the rigid body is located at point P and the vector position of this particle is R_i in the lab frame, and at position r_i in the body frame. It is seen that the position of the particle can be written:

$\mathbf{R}_i=\mathbf{R}+\mathbf{r}_i$

The defining characteristic of a rigid body is that the distance between any two points in a rigid body is unchanging in time. This means that the length of the vector $\mathbf{r}_i$ is unchanging. By Euler's rotation theorem, we may replace the vector $\mathbf{r}_i$ with $\mathcal{R}\mathbf{r}_{io}$ where $\mathcal{R}$ is a 3x3 rotation matrix and $\mathbf{r}_{io}$ is the position of the particle at some fixed point in time, say t=0. This replacement is useful, because now it is only the rotation matrix $\mathcal{R}$ that is changing in time and not the reference vector $\mathbf{r}_{io}$, as the rigid body rotates about point O'. Also, since the three columns of the rotation matrix represent the three versors of a reference frame rotating together with the rigid body, any rotation about any axis becomes now visible, while the vector $\mathbf{r}_i$ would not rotate if the rotation axis were parallel to it, and hence it would only describe a rotation about an axis perpendicular to it (i.e., it would not see the component of the angular velocity pseudovector parallel to it, and would only allow the computation of the component perpendicular to it). The position of the particle is now written as:

$\mathbf{R}_i=\mathbf{R}+\mathcal{R}\mathbf{r}_{io}$

Taking the time derivative yields the velocity of the particle:

$\mathbf{V}_i=\mathbf{V}+\frac{d\mathcal{R}}{dt}\mathbf{r}_{io}$

where V_i is the velocity of the particle (in the lab frame) and V is the velocity of O' (the origin of the rigid body frame). Since $\mathcal{R}$ is a rotation matrix its inverse is its transpose. So we substitute $\mathcal{I}=\mathcal{R}^T\mathcal{R}$:

$\mathbf{V}_i = \mathbf{V}+\frac{d\mathcal{R}}{dt}\mathcal{I}\mathbf{r}_{io}$

$\mathbf{V}_i = \mathbf{V}+\frac{d\mathcal{R}}{dt}\mathcal{R}^T\mathcal{R}\mathbf{r}_{io}$

$\mathbf{V}_i = \mathbf{V}+\frac{d\mathcal{R}}{dt}\mathcal{R}^T\mathbf{r}_{i}$

or

$\mathbf{V}_i = \mathbf{V}+W\mathbf{r}_{i}$

where $W = \frac{d\mathcal{R}}{dt}\mathcal{R}^T$ is the previous angular velocity tensor.

It can be proved that this is a skew symmetric matrix, so we can take its dual to get a 3 dimensional pseudovector that is precisely the previous angular velocity vector $\vec \omega$:

$\boldsymbol\omega=[\omega_x,\omega_y,\omega_z]$

Substituting ω for W into the above velocity expression, and replacing matrix multiplication by an equivalent cross product:

$\mathbf{V}_i=\mathbf{V}+\boldsymbol\omega\times\mathbf{r}_i$

It can be seen that the velocity of a point in a rigid body can be divided into two terms – the velocity of a reference point fixed in the rigid body plus the cross product term involving the angular velocity of the particle with respect to the reference point. This angular velocity is the "spin" angular velocity of the rigid body as opposed to the angular velocity of the reference point O' about the origin O.

Consistency

We have supposed that the rigid body rotates around an arbitrary point. We should prove that the angular velocity previously defined is independent from the choice of origin, which means that the angular velocity is an intrinsic property of the spinning rigid body.

Proving the independence of 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 $\mathbf{v}_1$ and $\mathbf{v}_2$ respectively. Suppose the angular velocity with respect to O₁ and O₂ is $\boldsymbol{\omega}_1$ and $\boldsymbol{\omega}_2$ respectively. Since point P and O₂ have only one velocity,

$\mathbf{v}_1 + \boldsymbol{\omega}_1\times\mathbf{r}_1 = \mathbf{v}_2 + \boldsymbol{\omega}_2\times\mathbf{r}_2$

$\mathbf{v}_2 = \mathbf{v}_1 + \boldsymbol{\omega}_1\times\mathbf{r} = \mathbf{v}_1 + \boldsymbol{\omega}_1\times (\mathbf{r}_1 - \mathbf{r}_2)$

The above two yields that

$(\boldsymbol{\omega}_1-\boldsymbol{\omega}_2) \times \mathbf{r}_2=0$

Since the point P (and thus $\mathbf{r}_2$) is arbitrary, it follows that

$\boldsymbol{\omega}_1 = \boldsymbol{\omega}_2$

If the reference point is the instantaneous axis of rotation the expression of velocity of a point in the rigid body will have just the angular velocity term. This is because the velocity of instantaneous axis of rotation is zero. An example of 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.

See also

• Angular acceleration
• Angular frequency
• Angular momentum
• Areal velocity
• Isometry
• Orthogonal group
• Rigid body dynamics
• Vorticity

External links

• A college text-book of physics By Arthur Lalanne Kimball (Angular Velocity of a particle)