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Maybe this is a bit of a silly question, but let us pretend we have a pendulum in a ideal universe with no friction, drag, or anomalous forces there to affect it. Additionally, our pendulum is subjected to an ideal gravitation force, the force vector of which is unaffected by the position of the ball (hence the responsible gravitating body is so massive that its own spherical geometry does not affect the forces on the pendulum-ball. Now we have our pendulum spin around so that it experiences uniform circular motion and some angle theta is formed between the resultant orientation of the spinning pendulum string and its previous stationary form.

My question is, is this system in equilibrium? I know that no work is being done even though there is a force.

And that from a relativistic (non-Newtonian) point of view, there is no centrifugal force to cancel the centripetal.

So please, could you explain the "answer" from a Newtonian (high-school physics) POV and a relativistic POV.

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    $\begingroup$ "Uniform circular motion" by definition implies constant speed, something a pendulum does not have. $\endgroup$ – DJohnM Dec 6 '13 at 15:18
  • $\begingroup$ I only used the term pendulum to denote the idea of a ball held by a string - I am not referring to the motion of a pendulum - merely to the object $\endgroup$ – Just_a_fool Dec 6 '13 at 15:20
  • $\begingroup$ Acrually it will not have uniform orbit, but it will rise and drop tracing a wavy line around. Only if the initial conditions are just right it might do circular motion. $\endgroup$ – ja72 Dec 6 '13 at 21:42
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Carl's first paragraph answers your question (though I disagree with his second paragraph) so this is just an addendum to Carl's answer.

It sounds to me as if you are describing an ideal conical pendulum. You're correct that no work is done because the two forces, the string and gravity, act at right angles to the direction of motion so $\vec{F}.\vec{dr}$ is always zero. The centripetal/centrifugal force is just the tension in the string time $\sin\theta$ where $\theta$ is the angle the string makes to the vertical.

Assuming you are talking about a conical pendulum the bob can indeed move uniformly in a circle pace Carl.

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  • $\begingroup$ John, you are correct, so long as the initial condition's velocity (magnitude) meets the requirement v=2*pi*r/t (Wikipedia). I was thinking of a random launch velocity. $\endgroup$ – Carl Witthoft Dec 6 '13 at 19:24
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There is force exerted (along the string) thru the point at which the mass's string is attached to whatever is holding it up. If there weren't, yes the ball would fly off.

Gravity is a separate force, and will cause the mass to oscillate about the vertical axis. So in fact you won't be able to achive uniform circular motion.

Take a look at the toys you can buy which use a pendulum (free to oscillate in the plane) to draw figures on paper or a sandbox. That's pretty much what you're describing here.

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