Let us consider a spinning “top”, spinning about an axis of rotational symmetry with angular velocity *ω*; the mass of the top is *m*, and the moment of inertia for the top about this axis is *I*. The top is supported by a surface below, which it contacts at a single “pivot” point on the rotation axis; the distance between this pivot point and the center of gravity is *l*.

For all following analysis, we ignore dissipative forces.

The only external forces on the top are gravity and the normal force at the pivot.

If the rotational axis is perfectly vertical, these forces cancel and produce no torques, so that the top continues its simple rotation unchanged.

But what if the top is tilted from the vertical by an angle *θ*?

We choose as our origin the pivot point. We now have a nonzero torque which has magnitude , and is directed perpendicular to the vertical plane containing the rotation axis.

Now, we note that the top has angular momentum *L*=*Iω* directed along the rotation axis. Now, we recall that ; thus the angular momentum will change as a result of the torque. We see that **τ** is horizontal, and so the vertical component of *L*, *L*_{z}, does not change, and as **τ** and *L* are perpendicular, neither does the magnitude of *L*. Thus, we see that the angular momentum vector **L** will be rotating uniformly about the vertical axis. Due to the object’s symmetry, the permanent axis must remain along **L**, and so will rotate about the axis. This is classical torque-driven precession.

If the angular velocity of the precession is *Ω*, then

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Tags: Angular Momentum, Friday Physics, physics, Precession, Torque

This entry was posted on February 27, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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March 13, 2009 at 9:04 am |

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