How to explain my Physics project?

Question

I've built a Physics project, where a vertical circle moves around its axis (Y-axis), and there's a small ring on the circle. When I start rotating the circle (using an eletric motor), at a certain velocity, the ring starts to ascend, and as the speed increases the ring ascends higher and higher, until it reaches the middle (height of half a circle, of one radius). From this speed on the ring stays at its height until I slow the circle down and the rings descends to its starting position.

How can I explain this phenomena to my classmates?
(I'm in 11th grade)

Answer 1

The rotation creates a force on the ring that pushes it out from the center of the circle. Since the circle is vertical the rotational motion must overcome gravity.

Looking at a free body diagram of the ring while it is subtended by an arc from an angle, th, that is greater than 0 and less than pi/2, it has gravity downward in the vertical as m*g. Decomposing these forces into the part that radiates from the center of the circle and the tangential portion, we find that the radial portion is (using outward as positive)
m*g*cos(th)
and the tangential (using upward as positive)
-m*g*sin(th)

The other force on the ring is the rotational force that is horizontal and equal to
m*r*w^2
where r is the distance from the vertical axis of the rotation and w is the rotational speed in radians per second
note that the distance r can be expressed in terms of th and R, the radiaus of the vertical circle, using a bit of trig:
r/R=sin(th)
or r=R*sin(th)

The rotational motion may also be decomposed into the radial portion and tangential portion:

Radial:
m*r*w^2*sin(th)
Tangential
m*r*w^2*cos(th)

Summing the forces in the tangential, the must equal 0, so
m*r*w^2*cos(th)-m*g*sin(th)=0

or
m*r*w^2*cos(th)=m*g*sin(th)
divide out m
r*w^2*cos(th)=g*sin(th)
substitute r with R*sin(th)
R*w^2*cos(th)=g
or
cos(th)=g/(R*w^2)

Now, knowing the angular velocity, you can calculate th.
Note that the angle asymptotically approaches 90 degrees and never exceeds it no matter how fast you rotate the circle.

Note that the equation is invalid until R*w^2 is greater than or equal to g since cos(th) is bounded between 0 and 1. I am having difficulty with this result since it implies very high speed rotation before the ring gains an angle greater than 0. You did observe this in your experiment, too, so I'm trying to figure out the equation that proves it. Using this result, though you can compute the minimum rpm that satisfies the g=R*w^2, which is when the ring will start rising I think but cannot prove yet.

Another interesting outcome is that it appears that the angle gets bigger as R gets bigger. This is not really the case since w is related to R as w=v/R, where v is the linear speed.

You can convert the w to rpm if you want to relate the experiment to terms your high school class will be more comfortable with. One revolution is 2*pi radians
so 120*pi*w=1 rpm


j

Answer 2

tell them your a techno-warlock and youve' learned how to levitate objects