Vertical Circular Motion? Can someone please explain this to me.?

Question

Hey,

I missed a class on Vertical Circular Motion.

If someone could please give me a brief lesson on it. How to figure out minimum speeds in regards to staying in the circle? How forces work at the top of circles(hill crest)/bottom of circles(hill valleys)/loops(within the circle)?? When Fc = Fg...? Regards to T=Fc+Fg?

Even a link to a site explaining this would be great. Thanks,

Answer 1

If you understand Fc the rest is easy.
Fc = mv^2/r = mω^2r. Fc/m is Ac or centripetal acceleration, a term that can be more useful since m often "cancels out" of Fc vs. Fg problems and related horizontal-plane problems like Fc vs. Ffriction or vs. g on a banked curve. Distance r is not just the radius of the curve at the point where m is located, it's the line drawn from the center of rotation to that point and is perpendicular to the velocity. By convention we refer to Fc as centripetal, not centrifugal, force, defined as the force that is applied to a moving body to constrain it to a curved path when Newton says it would go in a straight path with no force acting on it. FC is always directed inward along r. Therefore when talking about Fc interacting with Fg, we consider Fg the same way, as the upward constraint force needed to resist the acceleration of gravity.
On a wheel rotating in a vertical plane (i.e., about a horizontal axis), or for anything rotating in a vertical plane, Fc acts against Fg when it's directed downward, which occurs when m is on the upper half of a rotation. Conversely Fc acts with Fg when m is on the lower half of a rotation. Since m rotates about the center, Fc rotates with it. This means Fc is aligned exactly along (against and with) Fg only when r is vertical, at the top and bottom of the curve.
The velocity needed to null Fg at the top of a rotation is found by setting Fc = Fg. Then mv^2/r = mg, so v = sqrt(gr). An object spinning on a string at that critical speed experiences a string tension varying from 0 at the top to 2mg at the bottom. The string won't go slack (or the water won't fall out of the pail) anywhere in the cycle.
I suggest the ref. below for further discussion.

Answer 2

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