Rigid-body dynamics in 3d

Rigid-body dynamics is quite easy as long as everything lies in the plane:

all the mass is concentrated on the plane, all the forces are in the plane;

all the vectors separating the point of application of the force to the point of rotation are in the plane. As we look down the plane, the only possible rotation is clockwise (negative) or counterclockwise (positive) about an axis perpendicular to the plane. The torque isτ = F⊥r= r⊥F where r is the distance to the axis from the point of application of the force.

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That was then. But now we have to deal with the fact that in real life objects like tops and potatoes are not just planar and move in three dimen-sions. This is one case where not all the essential ideas can be conveyed in anything fewer than three dimensions.

So we will go to d= 3 but stay with a single point mass for a while.

More complicated bodies can be built out of point masses. What is the analog ofτ = Iα when the mass is running around in d = 3? How do you define torque and angular momentum in d= 3?

Suppose I have a point mass m in d= 3 at a location r with respect to some origin, as shown in the left half of Figure 11.4. Let F be the force on it. As in d=2, we want to combine the two vectors r and F and get a torque out of it. We want the magnitude of the torque to be rF sinγ as in d = 2 because the two vectors still lie in one plane, which could have been our old d= 2 plane. But now we need to specify the orientation of this common plane in d= 3 to specify the direction of the torque. A natural choice is the normal to the plane. But there are two possible orientations of the normal!

We break the tie by defining the torque to point in the direction in which a screw will advance if turned from r to F. This completely defines the torque vectorτ. What we have arrived at is called the cross product of r and F, written as

τ = r × F. (11.30)

In general C, the cross product of two vectors A and B,

C= A × B, (11.31)

points in the direction in which a screw would advance if you turned it from A to B, and it has a magnitude

C= ABsin γ , (11.32)

whereγ is the angle measured from A to B in their common plane, as shown in the right half of Figure 11.4.

This rule for fixing the direction of the cross product is called the right-hand rule for the following reason. If you grab the z-axis with your right hand and your fingers curl around it from A to B as shown by the big arrow, the torque points along the thumb, or the positive z-axis. This is

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Figure 11.4 Left: The unit vectors i, j, and k shown for later use. We consider the torqueτ due to force F with respect to a point (the origin) from which it is separated by r. The vectors F and (the continuation of) r (dotted line) define the shaded plane in which angleγ is measured from r to F. The magnitude of τ is assigned as in d= 2: τ = Fr sin γ . The vector τ = r × F, called the cross product of r and F, is oriented perpendicular to the plane in the sense dictated by the right-hand rule. Right: The cross product of A and B (whose common plane is chosen to be the x− y plane for convenience) points in the direction a typical screw would advance if turned from A to B, as indicated by the curved arrow wrapping around the z-axis. It has a magnitude AB sinγ , where γ is the angle measured from A to B. Instead of the screw rule, one can invoke the right-hand rule: if the fingers of the right hand curl from A to B, the thumb points along the cross product C.

why we homo sapiens can teach our children the cross product and lower primates can’t.

Let us briefly digress to gain familiarity with the cross product.

Whereas the dot product of two vectors that yields a scalar is defined in any number of dimensions, the cross product of two vectors that yields a vector makes sense only in d= 3. This trick of using two vectors to define a unique direction perpendicular to the plane they define works only in d= 3. In d = 4, there will be two independent directions perpendicular to the plane defined by any two vectors.

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Back to the cross product. If we reverse the order of the factors in the cross product it changes sign:

B× A = −A × B (11.33)

because the screw will advance the opposite way in B× A, or equivalently, γ will change sign. If we set A = B, then we find

A× A = −A × A = 0 (11.34)

in agreement with the result C= AB sin γ , which also yields 0 when γ = 0. Indeed, the cross product vanishes even if B is any scalar η times A:

the cross product of two parallel vectors is zero. This is good because the parallel vectors do not define a plane.

Let us compute the cross product for the basis vectors depicted in the left half of Figure 11.4. Consider i× j. It points along k by the right-hand rule and has unit length because|i| · |j| sin^π₂ = 1. Here is a table of the nine possible cross products of the unit vectors:

i× i = j × j = k × k = 0 (11.35)

i× j = k = −j × i (11.36)

j× k = i = −k × j (11.37)

k× i = j = −i × k. (11.38)

Recall that we wrote the dot product in two equivalent ways:

A· B = AB cos θ = AxBx+ AyBy+ AzBz. (11.39)

We now want to write the cross product in terms of components instead of lengths and angles. We can guess the formula for A× B in terms of components by placing them both in the x− y plane or choosing the x − y plane so the two of them lie on it. If the vectors make anglesγA andγB