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In physics, the angular velocity of a particle is the time rate of change of its angular displacement relative to the origin. The SI unit of angular velocity is radians per second. Angular velocity is usually represented by the symbol omega (ω, sometimes Ω).

In three dimensions, the angular velocity of a particle can be represented as a vector, with its direction pointing perpendicular to both the position and velocity vectors, in a fashion specified (conventionally) by the right-hand rule.[1]

## 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 rate of change of the angular position, and the orientation of the instantaneous plane of angular displacement. If the particle is revolving about the origin, then 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 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 tangent to a circle centered about the origin. On the other hand, if there is no cross-radial component, then the particle moves tangent to a straight line that passes through 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, only the cross-radial (tangential) component of the velocity contributes to the angular velocity.

In two dimensions, the angular velocity ω is given by

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

This is related to the cross-radial velocity by:[1]

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

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

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

Combining the above equations gives a formula for ω:

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

In two dimensions, the angular velocity is a single number with an orientation but no vectorial direction. In other words, in two dimensions, it 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 direction of angular velocity is taken, by convention, to be positive if the position vector turns counterclockwise, and negative if the position vector turns clockwise. If the parity is inverted, but the orientation of the angular displacement is not, then the sign of the angular velocity changes.

### Particle in three dimensions

In three-dimensional space angular velocity is a pseudovector quantity that specifies the rate at which the position vector "sweeps out" angle, as well as the orientation of the plane of angular displacement. Angular velocity in three dimensions has a magnitude, and a direction. The right-hand rule indicates the positive direction of the angular velocity pseudovector.

Let ${\displaystyle {\mathbf {u}}}$ be a unitary vector perpendicular to the plane of angular displacement, so that from the top of the vector the angular displacement is counter-clock-wise. The angular velocity vector ${\displaystyle {\vec {\omega }}}$ is then defined as:

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

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 angular displacement in which the behavior of the particle (for that instant) appears just as it does in the two dimensional case. The direction normal to this plane is defined to be 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 ${\displaystyle {\mathbf {u}}}$ defined before, the angular velocity vector may be written in a manner similar to that for two dimensions:

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

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

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

and using this vector, it can be seen that the formula for the tangential velocity of the particle is:

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

#### Addition of angular velocity vectors

Angular velocity can be defined as angular displacement per unit time. If a point rotates with ${\displaystyle \omega _{2}}$ in a frame ${\displaystyle F_{2}}$ that itself rotates with an angular velocity ${\displaystyle \omega _{1}}$ with respect to an external frame ${\displaystyle F_{1}}$, we can define the addition of ${\displaystyle \omega _{1}+\omega _{2}}$ as the angular velocity vector of the point with respect to ${\displaystyle F_{1}}$.

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, ${\displaystyle R=e^{Wt}}$ is a rotation matrix and in a time dt is an infinitesimal rotation matrix. Therefore, it can be expanded as ${\displaystyle R=I+W\cdot dt+{1 \over 2}(W\cdot dt)^{2}+\ldots }$

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 ${\displaystyle (I+W_{1}\cdot dt)(I+W_{2}\cdot dt)=(I+W_{2}\cdot dt)(I+W_{1}\cdot dt)}$ and therefore ${\displaystyle \omega _{1}+\omega _{2}=\omega _{2}+\omega _{1}}$.

## 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: ${\displaystyle \omega _{1}+\omega _{2}=\omega _{2}+\omega _{1}}$

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

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

any unitary vector e of the frame we obtain ${\displaystyle {\boldsymbol {\omega }}={\frac {{\boldsymbol {e}}\times {\dot {\boldsymbol {e}}}}{{\mathbf {e} }^{2}}}}$, and therefore ${\displaystyle {\boldsymbol {\omega }}=\mathbf {e} _{1}\times {\dot {\mathbf {e} }}_{1}=\mathbf {e} _{2}\times {\dot {\mathbf {e} }}_{2}=\mathbf {e} _{3}\times {\dot {\mathbf {e} }}_{3}}$.

As the columns of the matrix of the frame are the components of its vectors, this allows also the calculation of ${\displaystyle \omega }$ 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:[2]

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

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:

${\displaystyle {\boldsymbol {\omega }}=({\dot {\alpha }}\sin \beta \sin \gamma +{\dot {\beta }}\cos \gamma ){\mathbf {I} }+({\dot {\alpha }}\sin \beta \cos \gamma -{\dot {\beta }}\sin \gamma ){\mathbf {J} }+({\dot {\alpha }}\cos \beta +{\dot {\gamma }}){\mathbf {K} }}$

where ${\displaystyle {\mathbf {I} },{\mathbf {J} },{\mathbf {K} }}$ 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.[3]

## Angular velocity tensor

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 ${\displaystyle {\vec {r}}}$ that rotates around an axis with an angular speed vector ${\displaystyle {\vec {\omega }}}$ (as defined before) satisfies:

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

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

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

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

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

### 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:

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

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:

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

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

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

## Properties of angular velocity tensors

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.[4]

### 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:

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

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

${\displaystyle {\boldsymbol {\omega }}=[\omega _{x},\omega _{y},\omega _{z}]}$

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

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

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

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

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 ${\displaystyle W={\frac {dR(t)}{dt}}\cdot R^{\text{T}}}$ satisfies ${\displaystyle W^{\text{T}}=-W}$.

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

${\displaystyle {\mathcal {I}}={\mathcal {R}}{\mathcal {R}}^{\text{T}}}$ because R(t) is a rotation matrix
${\displaystyle 0={\frac {d{\mathcal {R}}}{dt}}{\mathcal {R}}^{\text{T}}+{\mathcal {R}}{\frac {d{\mathcal {R}}^{\text{T}}}{dt}}}$

Applying the formula (AB)T = BTAT:

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

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

### Coordinate-free description

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

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

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

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

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

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

bilinear form is skew-symmetric. (Here ${\displaystyle \cdot }$ stands for the scalar product). So 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 ${\displaystyle \omega :={\star }(L^{\sharp })}$, as the Hodge dual of L, and applying the definition of the Hodge dual twice supposing that the preferred unit 3-vector is ${\displaystyle \star 1}$

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

where

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

by definition.

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

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

### 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

Position of point P located in the rigid body (shown in blue). Ri is the position with respect to the lab frame, centered at O and ri 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 Ri in the lab frame, and at position ri in the body frame. It is seen that the position of the particle can be written:

${\displaystyle \mathbf {R} _{i}=\mathbf {R} +\mathbf {r} _{i}}$

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 ${\displaystyle \mathbf {r} _{i}}$ is unchanging. By Euler's rotation theorem, we may replace the vector ${\displaystyle \mathbf {r} _{i}}$ with ${\displaystyle {\mathcal {R}}\mathbf {r} _{io}}$ where ${\displaystyle {\mathcal {R}}}$ is a 3×3 rotation matrix and ${\displaystyle \mathbf {r} _{io}}$ 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 ${\displaystyle {\mathcal {R}}}$ that is changing in time and not the reference vector ${\displaystyle \mathbf {r} _{io}}$, 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 ${\displaystyle \mathbf {r} _{i}}$ 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:

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

Taking the time derivative yields the velocity of the particle:

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

where Vi 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 ${\displaystyle {\mathcal {R}}}$ is a rotation matrix its inverse is its transpose. So we substitute ${\displaystyle {\mathcal {I}}={\mathcal {R}}^{\text{T}}{\mathcal {R}}}$:

${\displaystyle \mathbf {V} _{i}=\mathbf {V} +{\frac {d{\mathcal {R}}}{dt}}{\mathcal {I}}\mathbf {r} _{io}}$
${\displaystyle \mathbf {V} _{i}=\mathbf {V} +{\frac {d{\mathcal {R}}}{dt}}{\mathcal {R}}^{\text{T}}{\mathcal {R}}\mathbf {r} _{io}}$
${\displaystyle \mathbf {V} _{i}=\mathbf {V} +{\frac {d{\mathcal {R}}}{dt}}{\mathcal {R}}^{\text{T}}\mathbf {r} _{i}}$

or

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

where ${\displaystyle W={\frac {d{\mathcal {R}}}{dt}}{\mathcal {R}}^{\text{T}}}$ 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 ${\displaystyle {\vec {\omega }}}$:

${\displaystyle {\boldsymbol {\omega }}=[\omega _{x},\omega _{y},\omega _{z}]}$

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

${\displaystyle \mathbf {V} _{i}=\mathbf {V} +{\boldsymbol {\omega }}\times \mathbf {r} _{i}}$

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 ${\displaystyle \mathbf {v} _{1}}$ 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

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

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 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.