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|!559px-Spinning_top.jpg!|
|Spinning Top
from Wikimedia Commons: Image by User:Lacen|



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{excerpt}The rapidly spinning Symmetric Top exhibiting Precession under the force of [Gravity] is a classic Physics problem.{excerpt}

We assume that we have a symmetric top that can easily rotate about an axis containing its [center of mass] at a high [angular 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)] exerted on the top, which will act to change its [angular momentum (single-axis)]. What will happen?

 



h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}The Spinning Top is a [rigid body] with a large amount of [angular momentum (single-axis)].{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}[torque (single-axis)] due to [gravity] 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:*  

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{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

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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] 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* .

(insert Drawing of perfectly vertical top)

If the top tilts from the perfectly vertical orientation by an angle θ , then there will be a non-zero torque.

(Insert Drawing showing tipped top with torque)


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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

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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:

{latex}\begin{large}\[ \vec{L} = I \vec{\omega} \]\end{large}{latex}

The direction associated with both the angular velocity and the angular momentum is directly upwards.

If the axis of the top is tipped from the vertical by an angle θ , then the situation is different. The angular velocity and the angular momentum both point along this angle θ . We define the vector *r* that runs from the point of contact (between the top and the surface it's spinning on) and the [center of mass] of the top. The force of [gravity], *F{~}g{~}* , pulls on the center of mass, and this exerts a [torque (single-axis)] on the top. The torque, τ , is given by the [cross product] between *r* and *F{~}g{~}* :

{latex}\begin{large}\[ \vec{\tau} = \vec{r} \times \vec{F_{\rm g}} \]\end{large}{latex}

where

{latex}\begin{large}\[ F_{\rm g} = mg \]\end{large}{latex}

and, as the gravitational force, points straight downwards.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}


in order to have a ratio of 1:2, one thus needs pendulum lengths of ratio 1:4. In order to get a ratio of 3:4 (as in the figure at the top of the page), the lengths must be in the ration 9:16.

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