# Rotational Motion Quantities

1. Dec 10, 2012

### Bashyboy

Hello,

Of the different rotational quantities of motion--angular velocity, angular acceleration, torque, angular momentum, etc.--, which of them has their direction perpendicular to the plane of rotation (not sure if I worded that correctly)?

I am pretty certain that torque has its direction perpendicular to the plane of rotation. I've tried to research why that is, but most people say that it is simply mathematical convention, and that there is no physical reason why, or something to that effect. But I feel like if angular velocity, a quantity that is well defined, has its direction perpendicular to the plane of rotation, then we can use this fundamental quantity to reason about the others, which may give us more insight as to why they have their direction perpendicular to the plane of rotation.

I'm sorry if this isn't exactly coherent; this is just a stream of thoughts, and is the only way I can seem to articulate it.

2. Dec 10, 2012

### Staff: Mentor

All of them do (the ones that you named). That is, all of them are described using vectors parallel to the axis of rotation, in the direction given by the "right-hand rule": curl the fingers of your right hand in the direction of rotation, and your thumb points in the direction of the vector.

Rotational kinetic energy is not a vector (similarly to translational kinetic energy).

3. Dec 10, 2012

### Bashyboy

What vectors, that are parallel to the axis of rotation, are used to describe these different rotation quantities?

4. Dec 10, 2012

### Staff: Mentor

The vectors have their direction along the axis of rotation, and their magnitudes calculated according to the usual formulas: $\omega = v/r$ for angular velocity, $\alpha = a/r$ for angular acceleration, $\tau = rF \sin \theta$ for torque, etc.

Or have I misunderstood your question?

5. Dec 10, 2012

### Bashyboy

Well, I know how their magnitudes are quantified, but you said that you use vectors to describe the direction of the different rotational quantities, what vectors are you using?

6. Dec 10, 2012

### DocZaius

3 dimensional vectors. They point in a direction perpendicular to the plane of rotation. If a disc was rotating counterclockwise on a table, its angular momentum would be pointing up.

7. Dec 10, 2012

### Khashishi

There is no physical reason why torque or any other rotational quantity follows the right hand rule. This is purely a convention. Note that we are allowed to have angular momentum in 2D space, where angular momentum is described by a scalar, not a vector.

The truth is that these angular quantities do not live in the same "space" as our regular vector quantities, but in a 2-form, where the basis vectors are formed with pairs of the original directions. For example: the basic directions in regular 3D space are x, y, and z. The basic directions for the space for angular momentum are x^y, x^z, and y^z. Rotation in the x-y plane is given by a vector in the x^y direction. (We still need to use a sign convention (the right hand rule), but we can regard x^y as -y^x.) (^ should be read as the wedge product, not an exponent).

Notice that if the "regular" space is 3D, the angular momentum space is also 3D. This makes it handy to "relabel" the basis vectors using regular vectors. Refer to x^y as the z direction, y^z as x, z^x (or -x^z) as y. We say that the regular 3D space is "dual" to the angular 3D space.

But if the "regular" space is 2D, the angular momentum space is 1D (the only direction is the x^y direction). In 4D (useful for relativity), there are 6 angular momentum directions.

8. Dec 10, 2012

### Staff: Mentor

I don't know if this is what you're looking for, but anyway... suppose we have a vinyl phonograph record (remember those?) rotating at 33 1/3 rpm = (approximately) 3.49 radians/sec. That's the magnitude of the angular velocity. To get the angular velocity vector, we multiply this by a unit vector, that is, a vector with magnitude 1, in the desired direction. I'll call a "generic" unit vector $\hat n$.

Suppose our record is rotating in a horizontal plane, counterclockwise as viewed from above. Then the direction of the angular velocity is upwards, in the +z direction. In this case, $\hat n = \hat z$, the unit vector in the +z direction, and

$$\vec \omega = (3.49 \text{ rad/s}) \hat n = (3.49 \text{ rad/s}) \hat z$$

Suppose we now tilt the turntable that the record is rotating on, downwards by 30° in the +x direction. The unit vector now points 30° away from the +z direction, towards the +x direction. Using a little trigonometry we find

$$\hat n = \left( \frac{1}{2} \hat x + \frac{\sqrt{3}}{2} \hat z \right)$$

and the new angular velocity is

$$\vec \omega = (3.49 \text{ rad/s}) \hat n = (3.49 \text{ rad/s}) \left( \frac{1}{2} \hat x + \frac{\sqrt{3}}{2} \hat z \right) = (1.75 \text{ rad/s}) \hat x + (3.02 \text{ rad/s}) \hat z$$

9. Dec 11, 2012

### D H

Staff Emeritus
Imagine a rigid body rotating in empty space. The object will continue rotating with constant angular momentum if no torque is applied. Is a torque of orthogonal to the plane of rotation? The zero vector is orthogonal to any vector in the sense that the inner product between the zero vector and any other vector is zero. However, another way to look at it is that the concept of orthogonality doesn't apply to zero vectors.

Suppose you apply a non-zero torque to the rotating body. There's nothing forbidding you from applying a torque about some axis other than the axis of rotation. You can apply a torque about any axis, not just the axis of rotation. The body isn't going to suddenly jump to this new axis of rotation just because you applied a torque about that new axis.

In general, the only thing amongst angular velocity, angular momentum, and torque that is perpendicular to the plane of rotation is angular velocity. As discussed above, torque can be pointing anywhere. Angular momentum is parallel to angular velocity only in the special case where the body is rotating about a principal axis. Otherwise, it's pointing somewhere else (and then you get some rather funky classical physics).

Rotation in 3 dimensional space is quite unique. 3D space is the only space where angular velocity looks like a vector in that space. (Better: It's a pseudovector.) In 2D space, angular velocity, angular momentum, and torque are just scalars. (Better: they're pseudoscalars.) In 4D space, angular velocity has six degrees of freedom.

As Khashishi mentioned, angular velocity is a 2-form (a skew symmetric matrix). In general, the number of degrees of freedom in an NxN skew symmetric matrix is N*(N-1)/2. To look like a vector in N-space, we must have N*(N-1)/2=N. The only non-trivial solution is N=3. This is what makes 3D rotation so unique.

Orientation and rotation are difficult concepts, much harder to grasp than position and translation. High school and freshman physics classes keep rotation simple by only presenting problems where rotation is restricted to a single axis (single plane). The sophomore/junior level classical physics class starts to investigate rotation in 3D space, but even then it doesn't quite get into the whys and hows. There's quite a bit of handwaving with regard to orientation and rotation in all but the honors versions of those sophomore/junior level classical physics classes.

Group theory is needed to get a better understanding of orientation and rotation. Physics students typically aren't (or weren't) introduced to group theory in their undergrad classes. That's done in introductory graduate level physics classes, and even group theory doesn't quite cut it. To truly understand orientation and rotation you need to know a bit about Lie groups and Lie algebras.