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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 (onesingle-dimensionalaxis)] 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 (onesingle-dimensionalaxis)] 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 (onesingle-dimensionalaxis)] 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}

{cloak:id=math}

We ignore the distribution of tensions in the upper cables, and simply view the pendulum as a simple pendulum along either the plane of the drawing or perpendicular to it. In the plane perpendicular to the drawing (where the mass oscillates toward and away from the reader) the pendulum length is *L{~}2{~}* and the angular frequency of oscillation is given by the formula for the Simple Pendulum (see [Simple Harmonic Motion].


{latex}\begin{large}\[ \omega_{2} = \sqrt{\frac{g}{L_{\rm 2}}} \]\end{large}{latex}

Along the plane lying in the page, where the mass moves left and right, the pendulum length is the shorter *L{~}1{~}* and the angular frequency is

{latex}\begin{large}\[ \omega_{1} = \sqrt{\frac{g}{L_{\rm 1}}} \]\end{large}{latex}

the ratio of frequencies is thus:

{latex}\begin{large}\[ \frac{\omega_{2}}{\omega_{1}}= \frac{\sqrt{\frac{g}{L_{2}}}}{\sqrt{\frac{g}{L_{1}}}} = \sqrt{\frac{L_{1}}{L_{2}}} \]\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.

{cloak:math}
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