Finding the movement equation (non intertial system)

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velvetmist
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Homework Statement


A particle of a mass ##m## is embedded in a circular rail, (radius: ##R##), without any friction. In a given moment, the particle finds itselfs without velocity at point C, and a force is applied on the rail, which starts moving with an ## \vec A ## constant acceleration. Use a non-inertial system fixed to the rail to solve the problem.

Arrange the Newton's equations, and find the movement equation of the particle.

(It was originally in spanish that's why I only screenshoted the graph)
Diagram of a particle m on a frictionless circular rail of radius R, resting at the bottom point C on a horizontal line; g points down, A points right, and angle phi is measured from the vertical at the center.

Homework Equations


## \vec F' = m\vec a - m \vec A ##

The Attempt at a Solution


## \vec F_v = bending force ##

So I manage to integrate ##\ddot \varphi## so i get ##\dot \varphi (\varphi)##, but I guess i have to find ##\varphi (t)## to get the movement equation, and I'm really stuck at this point.
Handwritten physics sketch: a circular rail with a bead m at angle phi from the bottom point C, with axes x and r-hat and phi-hat, a free-body diagram showing tension T, normal force and weight mg, and initial conditions phi0 = phi0, t0 = 0, phi-dot0 = 0.

Handwritten physics derivation on lined paper: a force balance in a rotating rail frame, then separation of variables integrating phi-dot against phi, giving phi-dot squared over 2 equals g(cos phi - cos phi0)/R minus A(sin phi - sin phi0).
 
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  • Handwritten physics derivation on lined paper: a force balance in a rotating rail frame, then separation of variables integrating phi-dot against phi, giving phi-dot squared over 2 equals g(cos phi - cos phi0)/R minus A(sin phi - sin phi0).
    84664.jpg
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  • Handwritten physics sketch: a circular rail with a bead m at angle phi from the bottom point C, with axes x and r-hat and phi-hat, a free-body diagram showing tension T, normal force and weight mg, and initial conditions phi0 = phi0, t0 = 0, phi-dot0 = 0.
    4646846.jpg
    11.2 KB · Views: 456
  • Diagram of a particle m on a frictionless circular rail of radius R, resting at the bottom point C on a horizontal line; g points down, A points right, and angle phi is measured from the vertical at the center.
    46416541651.JPG
    19.4 KB · Views: 483
What does your equation for ##\ddot{\varphi}## look like? Where is ##\varphi## measured from? I assume that the mass is instantaneously at rest relative to the circle's reference frame. It may help your thinking to consider that in a non-inertial frame accelerating with constant acceleration, there is an "effective" acceleration of gravity ##g_{eff.}## and find its magnitude and direction.
 
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