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Spinning Top
from Wikimedia Commons: Image by User:Lacen

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

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 (one-dimensional)] exerted on the top, which will act to change its [angular momentum (one-dimensional)]. What will happen?

Solution

System:

Interactions:

[torque

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due to [gravity] and normal force from the surface the top is spinning on.

Model:

Approach:

Diagrammatic Representation

First, consider the Y support for the pendulum:

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The pendulum effectively has length L2 when swinging in the horizontal plane in and out of the page, but length L1 along the horizontal direction in the plane of the page.

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Mathematical Representation

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 L2 and the angular frequency of oscillation is given by the formula for the Simple Pendulum (see Simple Harmonic Motion.

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[ \omega_

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= \sqrt{\frac

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{L_

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Along the plane lying in the page, where the mass moves left and right, the pendulum length is the shorter L1 and the angular frequency is

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the ratio of frequencies is thus:

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[ \frac{\omega_{2}}{\omega_{1}}= \frac{\sqrt{\frac

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{L_{2}}}}{\sqrt{\frac

{L_{1}}}} = \sqrt{\frac{L_{1}}{L_

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}} ]\end

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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