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Angular Oscillator

by SgrA*
Tags: angular, oscillator
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SgrA*
#1
Dec11-13, 11:58 AM
P: 11
Hi,

I'm considering this set up: A vessel with a curved (parabolic?) inner surface is rotating at an angular speed ω. Two heavy balls are placed near the axis of rotation of the vessel. Due to centrifugal force, the balls move outwards towards the edge. This increases the moment of inertia, and the angular speed decreases. This reduces the centrifugal force, and the heavy balls on the curved surface move back down towards the center. The moment of inertia decreases again as the masses move closer to the axis of rotation.

Is such an oscillator feasible? If yes, how would I calculate the curvature and masses of the balls?

Thanks!
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K^2
#2
Dec11-13, 12:27 PM
Sci Advisor
P: 2,470
Here is the cool thing about this problem. The potential of each sphere depends only on deflection. In other words, distance from center. So what you have is a central potential problem. You can forget about the tube and simply have a curved surface on which spheres are free to move. Of course, that neglects rolling inertia, but lets treat the masses as point objects for starters.

In central potential, angular momentum is conserved. Same as with your constrained problem. So you don't need to restrict angular motion to get the same trajectory. If the inner curvature is parabolic, potential is r. That's harmonic potential. That means each mass will oscillate as if attached to center by a spring. If surface is elliptical, for small deflections it's still going to work like a spring. Same as pendulum. For large deflections you'll start noticing a difference.

Back to rolling. Again, for small oscillations, it won't matter. The masses will simply act as if they are heavier, but otherwise, have the same oscillating motion. If there is a lot of movement, then it will matter whether spheres roll in a rotating tube or on a curved surface. But again, problem becomes much more complicated to solve.
SgrA*
#3
Dec17-13, 09:18 AM
P: 11
Hi,

Thanks for the reply and sorry, took me long to follow up. I have been trying it, and in the equations, the mass of the ball would directly be the mass on the spring, correct? What would the spring constant be? I know it has something to do with the curvature of the surface, but I'm not sure what to plug in.

Thanks!

K^2
#4
Dec17-13, 09:30 AM
Sci Advisor
P: 2,470
Angular Oscillator

The potential energy of the spring is (1/2)kx. Potential energy of the mass in gravitational field is mgh. So a bowl who's shape is h = (1/2)cr would correspond to a spring constant k = cm.


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