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Spinning Top |
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 (single-axis) exerted on the top, which will act to change its [angular momentum (single-axis)]. What will happen?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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)
Mathematical Representation
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
[ \vec
= I \vec
]\end
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], Fg , pulls on the center of mass, and this exerts a torque (single-axis) on the top. The torque is given by the cross product between r and Fg :
\begin
[ \omega_
= \sqrt{\frac
{L_
}} ]\end
the ratio of frequencies is thus:
\begin
[ \frac{\omega_{2}}{\omega_{1}}= \frac{\sqrt{\frac
{L_{2}}}}{\sqrt{\frac
{L_{1}}}} = \sqrt{\frac{L_{1}}{L_
}} ]\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.