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Spinning Top |
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The rapidly spinning Symmetric Top exhibiting Precession under the force of gravity |
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is |
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a |
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classic |
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Physics |
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problem. |
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We
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assume
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that
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we
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have
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a
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symmetric
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top
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that
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can
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easily
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rotate
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about
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an
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axis
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containing
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its
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at
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a
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high
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velocity ω . If the top tilts at all from the vertical, so that its center of mass does not lie directly above the top's point of contact with the surface it's resting on, there will be a torque (single-axis)
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exerted
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on
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the
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top,
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which
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will
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act
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to
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change
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its
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.
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What
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will
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happen?
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Solution
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The Spinning Top is a |
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If the top is perfectly upright, with its center of mass directly over the point of contact with the surface it's spinning on, then the torque (single-axis) about the point of contact is zero. The force due to gravity (near-earth) pulls directly downward, and the vector r between the point of contact and the center of mass points directly upward, so r X F = 0 .
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If the top tilts from the perfectly vertical orientation by an angle θ , then there will be a non-zero torque (single-axis).
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The top is spinning about its axis with angular velocity ω The moment of inertia about its axis of rotation is I . For the case where the top has its axis perfectly vertical, the angular momentum is given by:
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* {cloak:id=sys}The Spinning Top is a [rigid body] with a large amount of [angular momentum about a single axis].{cloak} {toggle-cloak:id=int} *Interactions:* {cloak:id=int}[torque (single-axis)] due to [gravity (near-earth)] and normal force from the surface the top is spinning on.{cloak} {toggle-cloak:id=mod} *Model:* {cloak:id=mod} [Angular Momentum and External Torque].{cloak} {toggle-cloak:id=app} *Approach:* {cloak:id=app} {toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color} {cloak:id=diag} If the top is perfectly upright, with its [center of mass] directly over the point of contact with the surface it's spinning on, then the [torque (single-axis)] about the point of contact is zero. The force due to [gravity (near-earth)] pulls directly downward, and the vector *r* between the point of contact and the center of mass points directly upward, so *r X F = 0* . |!Spinning Top Upright.PNG!| If the top tilts from the perfectly vertical orientation by an angle θ , then there will be a non-zero torque. |!Spinning Top Tipped.PNG!| {cloak:diag} {toggle-cloak:id=math} {color:red} *Mathematical Representation* {color} {cloak:id=math} The top is spinning about its axis with [angular velocity] ω The [moment of inertia] about its axis of rotation is *I* . For the case where the top has its axis perfectly vertical, the angular momentum is given by: {latex}\begin{large}\[ \vec{L} = I \vec{\omega} \]\end{large}{latex} |
The
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direction
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associated
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with
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both
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the
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and
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the
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angular
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momentum
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is
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directly
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upwards.
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If
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the
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axis
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of
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the
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top
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is
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tipped
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from
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the
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vertical
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by
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an
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angle
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θ ,
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then
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the
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situation
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is
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different.
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The
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angular
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velocity
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and
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the
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angular
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momentum
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both
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point
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along
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this
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angle
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θ .
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We
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define
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the
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vector
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r
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that
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runs
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from
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the
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point
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of
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contact
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(between
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the
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top
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and
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the
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surface
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it's
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spinning
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on)
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and
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the
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of
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the
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top.
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The
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force
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of
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,
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F
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g
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,
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pulls
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on
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the
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center
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of
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mass,
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and
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this
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exerts
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a
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on
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the
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top.
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The
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torque,
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τ ,
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is
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given
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by
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the
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between
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r
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and
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F
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g :
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{~}* : {latex}\begin{large}\[ \vec{\tau} = \vec{r} \times \vec{F_{\rm g}} \]\end{large}{latex} where {latex} |
where
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\begin{large}\[ F_{\rm g} = mg \]\end{large}{latex}
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and,
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as
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the
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gravitational
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force,
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points
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straight
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downwards.
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The magnitude of the cross product is given by
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|!Spinning Top Precession.PNG!| The magnitude of the cross product is given by {latex}\begin{large}\[ \mid \vec{r} \times \vec{F_{\rm g}} \mid = rmg\; sin(\theta) \]\end{large}{latex} |
The
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direction
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associated
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with
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this
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torque
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is
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horizontal
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and
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perpendicular
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to
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the
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angular
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momentum
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vector
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L
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.
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As
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a
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result,
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the
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torque,
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which
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causes
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a
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change
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in
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the
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angular
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momentum
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vector,
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does
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not
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cause
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a
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change
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in
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the
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magnitude
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of
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the
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angular
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momentum,
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but
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only
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in
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its
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direction.
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The
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angular
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momentum
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vector
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remains
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tipped
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at
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the
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angle
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θ with
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respect
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to
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the
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vertical,
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but
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it
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begins
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to
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rotate
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around
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the
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vertical
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axis
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in
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a
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counter-clockwise
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direction
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with
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an angular velocity Ω . This motion is called precession.
This new rotational motion ought to contribute to the angular momentum of the system. in most cases of interest and practical importance, however, the small addition to the angular momentum this causes is negligible compared to the angular momentum of the top rotating at angular velocity ω , and so it is usually ignored in the gyroscopic approximation, which holds that
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angular velovitu Ω . This motion is called *precession*. This new rotational motion ought to contribute to the angular momentum of the system. in most cases of interest and practical importance, however, the small addition to the angular momentum this causes is negligible compared to the angular momentum of the top rotating at angular velocity ω , and so it is usually ignored in the *gyroscopic approximation*, which holds that {latex}\begin{large}\[ L = I \omega \]\end{large}{latex} |
as
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long
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as
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{latex}\begin{large}\[ \Omega \ll \omega \]\end{large}{latex} |
In
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that
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case,the
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change
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in
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angular
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momentum
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only
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affects
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the
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horizontal
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portion:
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and
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the
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change
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is
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{latex}\begin{large} \[ \left| \frac{\Delta \vec{L}}{\Delta t} \right| = L \Omega sin(\theta) = rmg \; sin(\theta) \]\end{large} |
and the precession angular velocity is
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{latex} and the precession angular velocity is {latex}\begin{large} \[ \Omega = \frac{rmg}{L} = \frac{rmg}{I\omega} \]\end{large}{latex} {cloak:math} {cloak:app} {td} {tr} {table} {live-template:RELATE license} |
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