# Understanding Angular Velocity Qualitatively.

1. Feb 7, 2014

### The_Engineer

There seems to be different types of angular velocity to me...

1) A rigid body could be moving in a nonlinear path (ex. a pendulum)

2) A rigid body could be rotating about an axis on itself (ex. the earth rotating about an axis)

3) Both #1 and #2 combined (ex. the earth moving around the sun as an object AND rotating about an axis)

The well known angular velocity equation for a rigid body is

va = vb + w × rab

where w is the angular velocity of the rotation of the body about an axis that is on the body (like #2 above), correct?

Is there angular velocity associated with a body moving in a nonlinear path, such as a sphere on a pendulum that is NOT rotating about an axis on itself (like the earth's rotation axis in on the earth)?

I just want to qualitatively understand angular velocity/momentum better.

2. Feb 8, 2014

### Sunfire

Let's say a point mass moves in a curvilinear trajectory not containing (0,0)

then one can define angular velocity

$\omega=\frac{d\varphi}{dt}$ where $\varphi$ is the polar angle.

Is this what the question is about?

3. Feb 9, 2014

### mal4mac

Imagine a satellite rotating about the Earth in a circular orbit at constant velocity. What we really mean is "constant angular velocity". That is, using (r, φ) coordinates, over equal periods of time, the satellite moves an angle φ, and as r remains constant there is *only* angular velocity. The satellite then blasts off into space. There is *still* a change in φ, so there is an angular velocity , but it's now unlikely to be constant. There is also now a radial velocity (due to change in r direction). So any change in φ and there is an angular velocity, unless it is moving only in the r direction, when φ is zero.

Notice it is all dependent on coordinate systems, which can be defined to have different origins. For instance, the origin of an (r, φ) co-ordinate system could be the centre of the sun, or it could be centre of the earth. If it is the earth's centre, then the earth's centre has no angular velocity because it doesn't move relative to the coordinate system.

4. Feb 9, 2014

### voko

There is no difference between (1) and (2). A pendulum still rotates about itself - it is just that the axis of rotation is at its end, not in its middle.

Angular velocity can be defined for any kind of motion that changes direction.

5. Feb 9, 2014

### mal4mac

Can't it always be defined for any object undergoing any kind of motion? Just sometimes it's zero, in certain inertial frames.

6. Feb 9, 2014

### voko

Angular velocity is independent of the choice of inertial frames.

The reason why I singled out motions with a change of direction is that there is a few definitions of angular velocity, some of which do not work (easily) in pure translation.

7. Feb 10, 2014

### Sunfire

Yes, I was thinking of point mass moving in a straight line parallel to the x-axis. The angular velocity is nonzero because the polar angle changes;

But if the motion is on the x-axis itself, then the angular velocity is...? because the polar angle stays at zero.

Is this what you meant?

8. Feb 10, 2014

### Sunfire

voko, this is a very interesting statement. Could you elaborate?

9. Feb 10, 2014

### voko

Yeah, there is a similar phenomenon with some of the generic definitions of "angular velocity".

In most basic terms, a rigid body rotating with some angular velocity in one inertial frame, has the same angular velocity in another inertial frame.

For example, a wheel of a car moving with velocity $v$ relative to the ground has angular velocity $\omega$ in the car's frame; $v = \omega R$. In the ground frame, the wheel also has angular velocity $\omega$. The difference between two frames is that the instantaneous axes of rotation are different: it is stationary (the wheel's hub) in the car's frame, and it is moving (the point of contact with the ground) in the ground's frame.

10. Feb 10, 2014

### Sunfire

For argument's sake:

Let a point mass m move in a circular trajectory with angular velocity ω and radius r. Then let
$\vec{v}= \vec{\omega} \times \vec{r}$
be the linear speed of m in the rest frame.

$\vec{v}= \vec{v'} + \vec{u}$
where v' is the linear velocity of m in another inertial frame, and u is the velocity by which the 2 frames are moving with respect to each other.

Why is it not true that
$\vec{v'}=\vec{\omega'} \times \vec{r'}$
where the "primed" quantities are seen in the "primed" (moving) frame?

11. Feb 10, 2014

### voko

The problem with your argument is that "angular velocity" is not a very meaningful concept for a point mass because you cannot define naturally a frame co-rotating with it.

So you have to consider motion of a rigid body for the discussion to be meaningful.

12. Feb 10, 2014

### Khashishi

That's wrong. Even linear velocity depends on the reference frame, but angular velocity in addition depends on a choice of pivot point.

For the original poster, you can distinguish between rotation about the center of mass, and angular velocity of the center of mass around a pivot. This distinction is useful when dealing with angular momentum, because you can use the Parallel Axis Theorem to figure out the change in angular momentum when you move the pivot point. If you are not dealing with angular momentum, then there's no need for any distinction, since everything is measured relative to the pivot.

13. Feb 10, 2014

### Staff: Mentor

This is wrong too - at least as a general statement, if not for the specific example. An object in orbit, for example, need not necessarily also be rotating and if it is, the rotation need not necessarily be tied to the revolution.

A good example would be the hubble space telescope, orbiting the earth while remaining pointed in the same direction.

Indeed, it is also not always true of a pendulum either:

14. Feb 11, 2014

### voko

The proof that my statement is correct and yours is not can be found, for example, in Landau & Lifshitz, Course of Theoretical Physics, Vol. I, section 31.

What does this statement mean? You are comparing "rotation" with "angular velocity".

I did say that "Angular velocity can be defined for any kind of motion that changes direction." A more accurate statement is given by Chasles' theorem, where any motion is represented by a rotation about the (instantaneous) axis of rotation and a translation along this axis (also known as the screw axis). Oh, and I am (and was) talking about rigid body motion.

I am not familiar with the details of the Hubble telescope kinematics, but if what you say is true, then its motion is pure translation, or at least a good approximation.

If you refer to the "balcony", it is possible that in some regimes it approximates pure translation.

15. Feb 11, 2014

### voko

They give a rather terse proof, which also refers to another section of the book. Here is my version.

Let there be a laboratory reference frame $L$. Let there be a rigid body, with reference frame $B$ rigidly attached to it. The origin of $B$ is $O$. Consider point $o$ in the rigid body. Its velocity in $L$ is given by $\vec v_o^L = \vec v_O^L + \vec \omega_O \times \vec {Oo}$, where $\vec v_O^L$ is the velocity of $O$ and $\vec \omega_O$ is the angular velocity of the rigid body about $O$. Take some other point $O'$, then its velocity in $L$ is $\vec v_{O'}^L = \vec v_O^L + \vec \omega_O \times \vec {OO'}$. On the other hand, $\vec v_o^L = \vec v_{O'}^L + \vec \omega_{O'} \times \vec {O'o}$, so $\vec v_O^L + \vec \omega_O \times \vec {Oo} = \vec v_O^L + \vec \omega_O \times \vec {OO'} + \vec \omega_{O'} \times \vec {O'o}$, giving $\vec \omega_O \times \vec {Oo} = \vec \omega_O \times \vec {OO'} + \vec \omega_{O'} \times \vec {O'o}$. Let $\vec \omega' = \vec \omega_{O'} - \vec \omega_{O}$, then $\vec \omega_O \times \vec {Oo} = \vec \omega_O \times \vec {Oo} + \vec \omega' \times \vec {O'o}$, thus $\vec \omega' \times \vec {O'o} = 0$. Because $o$ is arbitrary, this must be true for any $\vec {O'o}$, so $\vec \omega' = 0$ and $\vec \omega_O = \vec \omega_{O'}$.

So angular velocity is independent of the "pivot" point.

The question remains, however, whether it depends on the choice of frame $L$. Let there be another frame $L'$, so that the velocity of any point $o$ is $\vec v_o^{L'} = \vec v_o^L + \vec V$, where $\vec V = \mathrm {constant}$. Then $\vec v_o^{L'} = \vec v_o^L + \vec V = \vec v_O^{L} + \vec V + \vec \omega_O \times \vec {OO'} = \vec v_O^{L'} + \vec \omega_O \times \vec {OO'}$, which is the same equation as we had in $L$. So angular velocity is independent of the choice of inertial frames.

In fact, as the constancy of $\vec V$ was not used, we could say that angular velocity is invariant under any purely translational motion.

16. Feb 11, 2014

### Sunfire

Hello voko,

Could you elaborate? I am only trying to understand what is it I am missing. A point mass moving on a (curvilinear) trajectory not containing (0,0,0) is ω=dθ/dt, θ is the polar angle, right?

17. Feb 11, 2014

### voko

A rigid body possesses three degrees of freedom that a material point does not. These three degrees of freedom can be loosely called "orientation" (or "attitude"). For a rigid body, angular velocity has to do with change of this "orientation". And it happens to have a very useful property of invariance under translational transformations as shown above.

Not (necessarily) so with a material point.

18. Feb 11, 2014

### Sunfire

If I understand you correctly, then the above definition of angular velocity for point mass is frame-dependent

19. Feb 11, 2014

### voko

That definition, yes. Note that it does not have the property $\vec v = \vec \omega \times \vec r$ when $\vec v$ and $\vec r$ are parallel.

For some reference frames and some motions angular velocity of a material point defined that way would work fine, however.

20. Feb 11, 2014

### The_Engineer

This is what I was trying to ask in my original post. The "balcony" is actually translating and therefore has no angular velocity, correct? This is because every point on the rigid body of the "balcony" has equal velocities.

In order to have angular velocity, the "balcony" would have to have unequal velocities on its body.

Someone also mentioned something about the Hubble space telescope. If this is actually pointing in the same direction while orbiting earth, then it is in translation (zero angular velocity) correct?

Correct me if I'm thinking about this wrong, but a rigid body has angular velocity if and only if all of the points on the body don't have equal velocities.

21. Feb 12, 2014

### voko

That is correct. A more accurate statement would include "in an inertial frame".