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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 about a single axis. What will happen?

Solution

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 .

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

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:

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

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 (near-earth), 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 :

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

where

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

and, as the gravitational force, points straight downwards.

The magnitude of the cross product is given by

\begin{large}\[ \mid \vec{r} \times \vec{F_{\rm g}} \mid = rmg\; sin(\theta) \]\end{large}

The direction associated with this torque is horizontal and perpendicular to the angular momentum vector L. As a result, the torque, which causes a change in the angular momentum vector, does not cause a change in the magnitude of the angular momentum, but only in its direction. The angular momentum vector remains tipped at the angle θ with respect to the vertical, but it begins to rotate around the vertical axis in a counter-clockwise direction with 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

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

as long as

\begin{large}\[ \Omega \ll \omega \]\end{large}

In that case,the change in angular momentum only affects the horizontal portion:

and the change is

\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

\begin{large} \[ \Omega = \frac{rmg}{L} = \frac{rmg}{I\omega}  \]\end{large}

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