WEBVTT

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Let us begin by admitting

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that the intuition almost everyone
brings to a spinning top is wrong.

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Spin does not manufacture an upward
force, and it does not cancel gravity.

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Gravity remains present, downward,

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with the same strength
whether the top spins or not.

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Here is the object we have to explain.

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One point touches the table at O,
the centre of mass C lies above

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and to one side, and the solid
spins rapidly about its tilted axle.

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This is a three-dimensional arrangement,

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so let the shape turn
once while we identify it.

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Without spin,
the expected motion is simple.

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The centre of mass drops and the top
rotates downward about the contact point.

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It is tempting to say that spin
must provide an opposite, upward effect.

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But draw the actual force.

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Gravity acts at the centre
of mass and points straight down.

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The vector r runs from the fixed
contact point to that centre.

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There is still no upward
force anywhere in the picture.

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The distinction between force
and torque matters here.

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Gravity pulls downward,

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but because its line of action misses O,
it also has a turning effect about O.

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That turning effect
is the torque r cross m g.

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In this pose the torque points sideways,

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perpendicular to the plane
containing r and gravity.

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That direction is not upward,

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and it is not the direction
in which the centre of mass would fall.

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Now add the vector created by rapid spin.

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The angular momentum L points
approximately along the axle.

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The crucial observation is that the
gravitational torque is sideways to L,

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not opposite to it.

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An opposite torque would reduce the spin
angular momentum and bring the top toward

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rest. A sideways torque
does something different:

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it turns the angular-momentum vector.

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Since the axle follows that vector
in a rapidly spinning symmetric top,

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the axle turns sideways too.

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That sideways turning is precession.

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The top is not held up by a hidden force.

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It is continually responding to gravity,

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but its first response is mainly
a change of direction around the vertical

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rather than an immediate
collapse toward the table.

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So our task is now precise.

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We need to see how a perpendicular torque
changes a vector without substantially

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changing its length, and then

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calculate how quickly that changing
direction travels around the vertical.

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Forget the physical top

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for a moment and watch angular
momentum in its own vector space.

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The green arrow has a fixed
vertical component and a fixed tilt.

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Its endpoint is the yellow point.

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Gravity supplies the magenta torque.

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At this instant it is tangent to the
horizontal circle through the tip of L.

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In particular, it is perpendicular to L,

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so it does not point inward
or outward along the green arrow.

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The torque continually changes
its direction as the top precesses,

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but it remains tangent to the same circle.

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Let the vector move.

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Its tip traces the blue circle
while the magenta arrow rides beside it,

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always pointing along the next small step.

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The equation governing
this motion is compact:

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the time derivative
of angular momentum equals torque.

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Torque is not merely
associated with a change in L.

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It is the rate
and direction of that change.

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Over a short time d t,

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the new angular momentum is the old
vector plus a small change d L.

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The change is torque times d t,

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so it points in the same
tangent direction as the torque.

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Picture that addition head to tail.

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The new vector points slightly
to one side of the old one.

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Repeating many tiny additions produces
a smooth turn rather than a sequence

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of visible corners.

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Because the small change is perpendicular
to L, their dot product is zero.

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Geometrically, adding a little
sideways vector turns the long vector.

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It changes the length only
by a second-order amount that disappears

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in the instantaneous limit.

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We can state the length result
exactly for steady precession.

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Differentiate L squared.

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The answer is twice L dot torque,

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and that dot product vanishes
because the two vectors are perpendicular.

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So the magnitude L stays constant
while its direction changes.

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This is the central correction
to the usual intuition.

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Gravity is not failing to act,
and the angular momentum is not frozen.

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Gravity is steering it.

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There is also no conflict with the idea
that angular momentum resists change.

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A larger L does not
make change impossible.

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It means that a given sideways change
represents a smaller angular turn.

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Watch another part of the circle.

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The torque points along the motion of the
tip, the tip moves around the vertical,

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and the length of the green
vector remains unchanged throughout.

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The remaining question is quantitative.

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A given torque moves
the vector tip at a given speed.

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The larger the circle

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and the larger L are, the slower
the corresponding angular sweep must be.

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Now return to the top and make
the approximation explicit.

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We consider a nearly
symmetric top spinning fast enough

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that its angular momentum lies
approximately along the axle.

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We also look at steady precession, where
the tilt angle remains almost constant.

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There are two angular rates,
and confusing them ruins the argument.

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Lowercase omega is the rapid
spin about the top's own axle.

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Capital omega
is the much slower precession

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of that axle around the vertical.

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The centre of mass
is a distance ell from the support.

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Gravity acts downward there, so r
cross m g supplies the sideways torque.

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The angle between the axle
and the upward vertical is theta.

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First calculate the torque magnitude.

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It is the magnitude of r cross m g.

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The angle between r
and downward gravity is pi minus theta,

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whose sine is the same as sine theta.

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Therefore the torque
magnitude is m g ell sine theta.

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That sine factor
has a direct geometric meaning.

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If the axle were vertical, gravity's line
of action would pass through the support

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and the torque would vanish.

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Tilting the axle gives gravity
a perpendicular lever arm ell sine theta.

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Now calculate how fast the tip
of L moves in angular-momentum space.

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Its horizontal circle
has radius L sine theta.

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A point moving around that circle
at angular rate capital omega has speed

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capital omega times L sine theta.

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This is ordinary circular-motion geometry,
but the circle lives in vector space.

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The radius is the horizontal component

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of L, and the speed around it
is the magnitude of d L by d t.

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The dynamical law
supplies that vector speed.

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Since d L by d t equals torque,

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the speed of the tip must equal
the gravitational torque magnitude.

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Equate them.

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Capital omega times L sine
theta equals m g ell sine theta.

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This line is the whole balance: the
required turning rate of L on the left,

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and the turning rate supplied
by gravity on the right.

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For a nonzero tilt, sine theta
appears on both sides and cancels.

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The steady-precession rate
is therefore m g ell divided by L.

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The units already make sense.

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Torque has units of angular momentum per
time, so torque divided by L has units

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of one over time, exactly
the units of an angular rate.

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For rapid spin about a symmetry axis,

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angular momentum
is approximately moment of inertia

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I times spin rate lowercase omega.

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A heavier rim gives a larger I,
and faster rotation gives a larger omega.

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Substitute that spin angular momentum.

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Capital omega is approximately m g
ell divided by I times lowercase omega.

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This is the standard rate for steady
precession of a fast symmetric top.

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The formula has a clean physical reading.

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More gravitational torque,

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from more weight or a longer
lever arm, turns L faster.

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More angular momentum, from a larger
moment of inertia or faster spin,

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makes the same torque
turn the vector more slowly.

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Notice what did not
survive the cancellation:

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the tilt angle.

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In this ideal steady, fast-top model,

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the leading precession
rate is independent of theta.

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That does not say every
real release is steady,

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or that a top may be tilted
arbitrarily without complication.

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Friction slowly reduces the spin,
large wobbling violates the steady

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approximation, and a badly asymmetric
body needs a more complete treatment.

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Within the model we stated,

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however, the inverse dependence on spin
is a sharp prediction that we can test.

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The derived formula makes a prediction
that initially sounds backward.

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Spin the top faster,

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and its slow motion around
the vertical becomes slower, not faster.

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Let the two tops be identical,
with the same mass, lever arm,

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and moment of inertia.

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Take m g ell divided by I
to be twenty per second squared.

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Only the spin rate changes.

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The first top spins at one
hundred radians per second,

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giving a precession rate
of zero point two radians per second.

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The second spins twice as fast,

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at two hundred, and the formula
predicts only zero point one.

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Run both predictions
over the same interval.

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The original top's angular-momentum
tip travels one full circuit.

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The faster-spinning top covers
only half a circuit in that time.

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The pictures are not claiming that the
faster top has less angular momentum.

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It has more.

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That larger vector is exactly
why its direction changes

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by a smaller angle each second.

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Gravity has not become weaker,
and the torque has not disappeared.

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For identical geometry it supplies the
same sideways change d L in each second.

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The same change attached to a vector twice
as large produces half the angular turn.

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This is the part

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that sounds backward only if we use the
word rotation for two different things.

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The top spins rapidly about its own axle,

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while the axle precesses
slowly around the vertical.

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Increasing the first
rate decreases the second.

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The inverse relationship fits in one line.

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Capital omega is a fixed top-dependent
constant divided by lowercase omega.

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Doubling lowercase omega
doubles the spin angular momentum.

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Because the gravitational
torque is essentially unchanged,

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it then takes twice as much time
to turn L through a given angle.

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A practical comparison must be made before
friction changes the spin very much,

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and with nearly equal tilt and geometry.

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Under those conditions, doubling
the spin rate should approximately

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halve the steady-precession rate.

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As a real top loses spin to friction,

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the same formula predicts
that its precession tends to speed up.

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Eventually the fast-top
approximation fails,

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the motion becomes more
complicated, and the top falls.

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The apparent stability
was never freedom from gravity.

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It was gravity steering
a large angular momentum slowly.

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A bicycle wheel suspended from one end
makes that steering unusually easy to see.

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The same argument becomes
striking with a bicycle wheel.

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Tie a string to one end of the axle,
hold the wheel with its axle horizontal,

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spin it, and release.

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The string supplies a fixed support
point O while the wheel's centre

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of mass hangs well to one side.

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If the wheel is not
spinning, there is no mystery.

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Gravity turns the axle downward about
the supported end, and the wheel drops.

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Now remove that imagined fall
and spin the wheel rapidly.

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Its angular momentum
points along the axle.

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The direction follows the right-hand rule,
so reversing the wheel's spin reverses L.

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Gravity still acts at C.

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The lever arm r runs
from the string's support to the centre,

00:12:56.295 --> 00:12:59.315
so gravity produces r cross m g.

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With r horizontal and gravity
downward, the torque points sideways.

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That sideways torque is perpendicular
to the wheel's spin angular momentum.

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It therefore changes the direction

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of L rather than rapidly
reducing its magnitude.

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The axle follows the turning vector.

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The unsupported centre does begin
to accelerate, but not simply downward.

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As L turns sideways,
the axle turns sideways with it.

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The continual response to gravity becomes
motion around the vertical support line.

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The wheel therefore swings
around the vertical line through O.

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This is the same precession
as the top, now exposed in a form

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where the single support point
and the long lever arm are easy to see.

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Write the chain once more.

00:13:49.466 --> 00:13:55.620
The gravitational torque is r
cross m g. Torque is d L by d t.

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For a fast wheel
in approximately steady precession,

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the angular rate is m g ell divided by L.

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The demonstration also checks direction.

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Reverse the spin and L reverses,

00:14:08.386 --> 00:14:11.940
while the gravitational torque
for the same pose does not.

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The sideways change then carries the axle
around the vertical in the opposite sense.

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Spin the wheel faster and L grows.

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The torque is almost unchanged,
so the wheel precesses more slowly.

00:14:25.066 --> 00:14:28.247
Let friction reduce the spin
and the precession generally

00:14:28.247 --> 00:14:32.415
quickens until the simple steady model
ceases to describe the motion well.

00:14:33.328 --> 00:14:38.251
Real tops usually add a small bobbing
of the tilt angle called nutation, caused

00:14:38.251 --> 00:14:42.710
by how they are released and by departures
from perfectly steady precession.

00:14:43.507 --> 00:14:45.283
We can now answer the opening question

00:14:45.283 --> 00:14:48.012
without invoking
a hidden supporting force.

00:14:48.209 --> 00:14:52.331
Gravity tries to turn the body
about its contact or support point.

00:14:52.783 --> 00:14:56.310
Rapid spin supplies a large
angular-momentum vector,

00:14:56.393 --> 00:15:00.167
and the gravitational torque
changes that vector mainly sideways.

00:15:00.964 --> 00:15:05.130
The vector tip therefore travels
around the vertical, the axle follows it,

00:15:05.213 --> 00:15:07.014
and we see precession.

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A top does not refuse gravity.

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Its spin changes the direction in which
gravity first makes the motion develop.

00:15:15.310 --> 00:15:19.908
That is why a rapidly spinning top stays
up for a while, why it circles slowly

00:15:19.908 --> 00:15:24.959
instead of simply falling, and why making
it spin faster slows that circle down.