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Old 02-13-2012, 01:24 PM
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Take a baton, it has a heavy and light end. place it on a finger. When it is balanced the light end is farther from your finger and the heavy end is closer. That demonstrates the center of Rotational balance, not the physical center of the baton. The beads will travel to the greatest distance from the Rotational balance center which will then move the center of rotational balance closer to the center of rotation. The beads acting as a fluid will continue to adjust as long as a rotational force greater than their individual mass is present. That is a simple explanation of how it works.

Someone said their memory of physics say it won't work,

After some Googling I came across the physics formula that explains why it does work:

Think about it like this:

x = \alpha \cos \frac{s}{\alpha} \ ;   y=\alpha \sin\frac{s}{\alpha} \ .

Then:

x^2+y^2 = \alpha^2 \ ,

which can be recognized as a circular path around the origin with radius α. The position s = 0 corresponds to [α, 0], or 3 o'clock. To use the above formalism the derivatives are needed:

y'(s) = \cos \frac{s}{\alpha}\ ; \ x'(s) = -\sin \frac{s}{\alpha} \
y''(s) = -\frac{1}{\alpha}\sin\frac{s}{\alpha} \ ; \ x''(s) = -\frac{1}{\alpha}\cos \frac{s}{\alpha} \ .

With these results one can verify that:

x'(s)^2 + y'(s)^2 = 1 \  ;  \frac{1}{\rho} = y''(s)x'(s)-y''(s)x''(s) = \frac{1}{\alpha}\ .

The unit vectors also can be found:

\mathbf{u}_t(s) = \left[-\sin\frac{s}{\alpha},\ \cos\frac{s}{\alpha} \right]\  ;  \mathbf{u}_n(s) = \left[\cos\frac{s}{\alpha},\ \sin\frac{s}{\alpha} \right] \ ,

which serve to show that s = 0 is located at position [ρ, 0] and s = ρπ/2 at [0, ρ], which agrees with the original expressions for x and y. In other words, s is measured counterclockwise around the circle from 3 o'clock. Also, the derivatives of these vectors can be found:

\frac{d}{ds}\mathbf{u}_t(s) = -\frac{1}{\alpha} \left[\cos\frac{s}{\alpha},\ \sin\frac{s}{\alpha} \right]\   = -\frac{1}{\alpha}\mathbf{u}_n(s) \ ;

\ \frac{d}{ds}\mathbf{u}_n(s)   = \frac{1}{\alpha} \left[-\sin\frac{s}{\alpha},\ \cos\frac{s}{\alpha} \right]   = \frac{1}{\alpha}\mathbf{u}_t(s) \ .

To obtain velocity and acceleration, a time-dependence for s is necessary. For counterclockwise motion at variable speed v(t):

s(t) = \int_0^t \ dt' \ v(t') \ ,

where v(t) is the speed and t is time, and s(t=0) = 0. Then:

\mathbf{v} = v(t)\mathbf{u}_t(s) \ ,

\mathbf{a} = \frac{dv}{dt}\mathbf{u}_t(s)+v\frac{d}{dt}\mathbf{ u}_t(s)   = \frac{dv}{dt}\mathbf{u}_t(s)-v\frac{1}{\alpha}\mathbf{u}_n(s)\frac{ds}{dt}

=\frac{dv}{dt}\mathbf{u}_t(s)-\frac{v^2}{\alpha}\mathbf{u}_n(s),


where it already is established that α = ρ. This acceleration is the standard result for non-uniform circular motion.
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