Originally Posted by Pioneer1
Ok. Net force is zero at two points but how do we know from this equation that the arm is stationary at points where net force is zero? We know that at the first point (r = 0.13) the net force was zero but the arm was in motion because it moved to the next zero point. How do know that it is stationary at the next point (r = .59)?
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For clarity, let's call the points where the net force equals zero the "zero-force" points. Who said anything about the small masses being
stationary at those points? The first zero-force point is a stable equilibrium point, since there is a restoring force about that position. (To understand the motion, you must examine the
net force as a function of position.)
Ok. We are finding where the torsion equals gravity, not the (stationary) equilibrium point because this equation, as far as I understand, does not give any information about the motion of the arm.
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Again, to understand the motion you must consider the net force on the small masses as a function of position as well as the initial energy of the system. If you examine the net force at points surrounding the equilibrium point at r = 0.13, you'll find a net restoring force directed towards the equilibrium point.
Sorry, I don't understand this language. Torsion never acts on the small masses. Torsion is kr and it does not have a mass term. And at r = 0 we assume that the torsion is zero.
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Come on. There are two forces acting on the small masses: Torsion (kr) and gravity. The torsion force doesn't require a "mass term".
Ok. Let’s say the arm comes to equilibrium at r=0. Once gravity starts to act, it exerts variable acceleration to the arm.
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You'll lessen your confusion by thinking of gravity as exerting a variable
force on the arm. (Acceleration depends on the
net force.)
(First differences in this spreadsheet.) The simple harmonic motion is caused by constant acceleration.
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Simple harmonic motion is caused by a restoring force proportional to the displacement from equilibrium.
In this case there is increasing acceleration. This suggests to me that the arm will always have acceleration towards the attracting weight and will collide with it. What is wrong with this reasoning?
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We been through this several times already. You still think--despite the detailed refutations in this thread--that the force of gravity is always greater that the torsion force. Nope. (Review the analysis given in post #22.)
What happens (in our toy model) is this: Assume that the small masses are initially at rest when the large masses are brought into position. The arm is accelerated from r = 0 to r = 0.13; it passes through the equilibrium position (r = 0.13) with some speed, then begins to slow due to the restoring force of the torsion; it stops and begins its return journey long before reaching the second zero-force point.
But in the case of Cavendish experiment attracting weight imposes variable acceleration which increases and it always pulls the arm with greater acceleration.
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Certainly the gravitational force always increases with r, but what determines the motion of the arm is the
net force on it, which
does not simply increase with r.
I think you would profit from the study of the simpler problem.