WEBVTT

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Kinematics of a rigid body
begins with one promise:

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distances inside the body never change.

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That promise will give us
its complete velocity law.

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Here is the body itself, a collection
of material points locked together.

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Pick any two of them
and call them A and B.

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The vector r runs from A to B, and
rigidity says its length never changes.

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Write that fixed length as r dotted with
itself, equal to the constant L squared.

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Differentiate. The constant disappears,

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leaving r dotted with its own
rate of change equal to zero.

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So that rate points square to r.

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Here is the whole family.

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Change the length and even reverse
the arrow; it stays perpendicular.

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Every member can be written
as omega crossed with r.

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For a rigid body,

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one angular velocity omega works
for every pair of points at once.

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It belongs to the body, not to A or B.

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Now compose the positions.

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Start at the position of A,
add r, and land at the position of B.

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Differentiate that addition.

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The first velocity is v A.

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The second term is the rate of r, which
becomes omega cross r. Their sum is v B.

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A translates with v A while the body
turns about A with angular velocity omega.

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B carries both effects, so its green
arrow follows the sum we just built.

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Now let the body move through the
translation and rotation we just composed.

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Translation plus rotation is the entire
instantaneous freedom of a rigid body.

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Every construction that follows is this
one relation read in a different way.

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In the plane, the rigid-body relation
has a consequence you can watch.

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Here is a wheel rolling along
the ground without slipping.

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The material point touching
the ground is not sliding.

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At this instant C has velocity
zero, so it is momentarily at rest.

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That is the instantaneous centre.

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Now take A in the relation to be C.

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Its velocity term is zero.

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What remains says that every
point P moves as omega crossed

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with the vector from C to P.

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The hub is one radius
from C, so it moves at omega R.

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The top material point is two radii away,
so its speed is twice the hub speed.

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At Q, the same rule gives
a velocity perpendicular to C Q.

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Its length grows in direct
proportion to Q's distance from C.

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At one instant, then, the whole velocity
field looks exactly like a wheel pinned

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at C and rotating about it.

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But C is not one fixed material point.

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Watch C zero leave
the ground as the wheel rolls.

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A moment later, a different
material point is touching down,

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and that new point is the one at rest.

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The same idea gives a ruler construction.

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Here is a moving bar with the velocity
direction already attached at A and B.

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Each point circles C, so draw
a perpendicular to each velocity.

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The centre must lie on both lines.

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The two lines meet here, at C.

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From that one point,

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every speed is omega times
the distance out to the material point.

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There is one limiting case. If the two
velocities are equal and parallel,

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their perpendiculars are parallel too.

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They never meet, so C is at infinity
and the body is purely translating.

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Lift the body into three dimensions.

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This solid extends through all
three coordinate directions,

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with its material points
locked into one shape.

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Which of those points,
if any, are momentarily at rest?

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Ask the rigid-body relation.

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Choose A on the body,

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draw its velocity, and keep the body's
one angular velocity omega in view.

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We are hunting for a point
P whose velocity is zero.

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The cross-product term
must cancel v A exactly.

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It can do that only
with the part square to omega,

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because every omega cross r
is perpendicular to omega.

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Here the magenta arrow
is that exact cancellation.

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For the moment v
A is entirely square to omega.

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Solving the cancellation gives
this particular displacement r zero.

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But r zero is only one answer.

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Add any multiple lambda omega
and the cross product does not change.

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Watch lambda sweep through its values:

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every point it reaches is another
solution, so the solutions fill a line.

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That tilted line
is the instantaneous axis of rotation.

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Every green location on it
has zero velocity at this instant.

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A point off the axis swings around it
in a circle lying square to omega.

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Its speed is omega times
its perpendicular distance from the axis.

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The same construction
at B uses the same omega.

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Turn the view
and the geometry separates cleanly:

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the red axis is one tilted line

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in space, and both green
velocities stand square to it.

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Omega belongs to the whole body.

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A and B do not get
different angular velocities.

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What changes from point to point is r, and
therefore the cross-product contribution.

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Now remove the assumption we just used.

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If v A is not square to omega,

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split it into a perpendicular
piece and a parallel piece.

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Let those two pieces land.

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Complete their parallelogram,

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and its green diagonal
is the original velocity v A.

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The cross product can cancel the
perpendicular piece, exactly as before.

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It can never touch the parallel piece,
whose projection formula is this.

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So the axis still exists, but its points
now slide along omega instead of resting.

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This yellow arrow is the velocity
shared by every point on it.

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Take a point away from the axis.

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Its velocity is the sum of a turn around
the axis and that same slide along it.

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Follow one material point.

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The combined motion
traces this yellow helix,

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winding around the tilted
axis while advancing along it.

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This is Chasles' theorem.

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At every instant,
rigid-body motion is a screw:

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a rotation about an axis together
with a translation along that axis.

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Watch the numbered
point perform both parts.

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The slide per unit turn is the pitch.

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First limiting case:
let the slide shrink to nothing.

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The advancing helix closes into a circle,

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the axis velocity vanishes,
and the motion becomes pure rotation.

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Second limiting case: restore the slide
and let omega shrink to nothing.

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The winding opens into a straight path,

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the red omega arrow disappears, and equal
green velocities show pure translation.

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Put turn and slide back together.

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The tilted axis, the helix and the
travelling point return as the one picture

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that contains the general case.

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Three ideas carry the lecture.

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Keep the moving screw beside
them while we read the list.

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First, any two points share one omega,

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and their velocities differ
by omega cross the vector between them.

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Second, plane motion
has an instantaneous centre,

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but the material point occupying
it changes as the body moves.

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Third, spatial motion
has an instantaneous axis.

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Rotation about it

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plus translation along it is the
general screw motion, every instant.