Curl of velocity vector in rotational motion

[brotherbobby]

TL;DR

In rotational motion, $\boxed{\vec v = \vec\omega\times\vec r}$, irrespective of whether the angular velocity $\vec\omega$ is constant or not (I hope I am correct in saying this). However, the curl of velocity vector, $\mathbf{\vec\nabla\times\vec v=2\vec\omega}$ comes out only to be true if the angular velocity of the rotating body is $\vec\omega$ constant. Is this true? I show the details below.

Attempt : I draw the diagram of the problem to the right. A circular plate of radius $a$ rotates with an angular velocity $\vec\omega = \omega_0\hat k$, assumed constant for the moment. A particle P at rest with the plate lies on the rim along the $x$ axis, as shown. Its velocity is $\vec v= v\hat j$ and its position vector $\vec r = a\hat i$. Using the right hand rule, these vectors make sense : $\vec v = \vec\omega_0\times\vec r\Rightarrow v=\omega_0 a$, remembering that $\hat j = \hat k\times\hat i$. Calculating the curl of this vector using index notation, $\small{\vec\nabla\times\vec v = \vec\nabla\times(\vec\omega_0\times\vec r)=\epsilon_{ijk}\partial_j(\vec\omega_0\times\vec r)_k\hat e_i = \epsilon_{kij}\epsilon_{klm}\partial_j\,\omega_{0_l}\,x_m \hat e_i=(\delta_{il}\delta_{jm} -\delta_{im}\delta_{jl})\partial_j\,\omega_{0_l}\,x_m \hat e_i}$, which reduces to $\small={\partial_j\,\omega_{0_i}\,x_j \hat e_i-\partial_j\,\omega_{0_j}\,x_i \hat e_i=3\vec\omega_0-\vec\omega_0=2\omega_0}\Rightarrow \boxed{\vec\nabla\times\vec v=2\vec\omega_0}$, where I have used the fact that $\vec\omega_0$ is uniform and $\partial_i x_j = \delta_{ij}$. This can also be shown using the method of determinants for the cross product.

Question : Clearly, from my derivation, the angular velocity $\vec\omega_0$ is constant. Does it have to be? In which case, I am mistaken in what I have shown. Can it not be that the curl of the velocity vector at a given time $t$ equals twice the angular velocity at that time : $\vec\nabla\times\vec v(t)=2\vec\omega(t)$?

 

[A.T.]

However, the curl of velocity vector, $\mathbf{\vec\nabla\times\vec v=2\vec\omega}$ comes out only to be true if the angular velocity of the rotating body is $\vec\omega$ constant. Is this true?

Constant with respect to what? Time or position? If $\omega$ varies with respect to r, then curl is different than when it doesn't. If $\omega$ varies with respect to t, then so will the curl.

See also:

 

[haruspex]

Clearly, from my derivation, the angular velocity ω→0 is constant

It is not clear to me. What step breaks if it is not constant wrt time?

 

[Orodruin]

By definition, $\vec \omega$ is constant is space - or you don’t have rotational motion. The curl does not contain derivatives wrt time so clearly letting $\vec \omega$ have a time dependence does not affect the result here.

 

[brotherbobby]

Constant with respect to what?

Space.

So if the angular velocity of the body varies with time, say $\vec\omega(t)$, I have the following two questions. The first one is more basic and I suppose more important :

1. Can we write $\vec v(t)=\vec\omega(t)\times\vec r(t)$? Of course the position vector and linear velocity of the point P, viz. $\vec r(t)\,,\;\vec v(t)$ on the body varies with time even if the angular velocity $\vec\omega$ was some constant ($\vec\omega_0$).
2. Is the curl of the linear velocity twice the angular velocity, i.e. $\vec\nabla\times\vec v(t)=2\vec\omega(t)$? Given my derivation above, I just don't see how that is.

 

[brotherbobby]

It is not clear to me. What step breaks if it is not constant wrt time?

Thank you, and a bit silly on my part. My derivation up there in post#1 remains the same even if the the angular velocity vector $\vec\omega(t)$ changed with time. Hence, I suppose it would be ok for us to write that $\vec\nabla\times\vec v(t)=2\vec\omega(t)$.

 

[brotherbobby]

By definition, $\vec \omega$ is constant is space - or you don’t have rotational motion. The curl does not contain derivatives wrt time so clearly letting $\vec \omega$ have a time dependence does not affect the result here.

Yes I agree with the second sentence of yours. It is the first that is more interesting.

Let me write it again, despite being repititive.

The angular velocity vector of a rotating body $\vec\omega$ is necessarily a constant in space. Thus it would be quite nonsense to write $\vec\omega(\vec r)$.

What if a body was precessing - i.e. the rotating axis changing with time. Can we write its angular velocity $\vec\omega$ to be a function of position $\vec r$ or angle $\theta$?

 

[Orodruin]

Yes I agree with the second sentence of yours. It is the first that is more interesting.

Let me write it again, despite being repititive.

The angular velocity vector of a rotating body $\vec\omega$ is necessarily a constant in space. Thus it would be quite nonsense to write $\vec\omega(\vec r)$.

What if a body was precessing - i.e. the rotating axis changing with time. Can we write its angular velocity $\vec\omega$ to be a function of position $\vec r$ or angle $\theta$?

No. It is a function of time as it changes with time, but it does not depend on spatial coordinates. If it did the object would necessarily change shape.

 

[wrobel]

The angular velocity of a rigid body is defined by the Euler formula:
$$\boldsymbol{v}_A=\boldsymbol v_B+\boldsymbol\omega\times\boldsymbol{BA}.\qquad(*)$$
The vector $\boldsymbol\omega$ can depend on time, but it is independent of any spatial variables. And, indeed, if $\boldsymbol v$ is a velocity field in a rigid body, then $\boldsymbol \omega=\frac{1}{2}\mathrm{curl}\,\boldsymbol v$. This fact follows directly from (*).
To check it consider $\boldsymbol{BA}=(x-x_B(t),y-y_B(t),z-z_B(t))$ and
$\boldsymbol v(x,y,z)=\boldsymbol v_A$

 

[Vincf]

In continuum mechanics, one defines the vector field
$$
\boldsymbol{\omega} = \frac{1}{2}\,\nabla \times \mathbf{v}.
$$ It measures the local rotation of a material particle.
In the case of a deformable medium, this vector field may vary in space.
However, a defining property of a rigid (undeformable) body is that the vector field $\boldsymbol{\omega}$ is uniform throughout the body. All particles in a continuous, undeformable medium rotate on their own axis in the same way.
You are free to consider a velocity field of the form
$$
\mathbf{v} = \mathbf{A} \times \mathbf{OM},
$$
with $\mathbf{A}$ a non-uniform vector field. But in that case, this velocity field cannot correspond to that of a rigid body.
For example, in cylindrical coordinates, one could take: $\mathbf{A} = r\,\mathbf{e}_z,$
which gives :
$$
\mathbf{v} = (r\,\mathbf{e}_z) \times \mathbf{OM} = r^2\,\mathbf{e}_\theta.
$$
As one moves away from the origin, the velocity increases faster than in the rigid-body case.
In this case, if you compute
$$
\boldsymbol{\omega} = \frac{1}{2}\,\nabla \times \mathbf{v},
$$
you will find that $\boldsymbol{\omega} \neq \mathbf{A}.$