Tangential movement in a waterslide

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haruspex said:
Yes. Can you write your equations in terms of that angle plus a perturbation, then make the small angle approximation?
I am still stuck...I don't know what equations you are referring to.
 
Fibo112 said:
I am still stuck...I don't know what equations you are referring to.
Re-derive your acceleration equation in post #1, but this time do not make any small angle approximations. Use the energy equation to substitute for v in that.
 
This is as far as I get...I would appreciate it if you could tell me your solution so I can see what I am missing.
 
Attachments
  • Handwritten derivation on white paper with a black notebook behind it: text states the remaining force on a particle after accounting for centripetal acceleration, then an equation (mv^2/(R sin(a))·cos(a) − mg)·sin(a) = a·m, followed by a solved expression for a.
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Fibo112 said:
This is as far as I get...I would appreciate it if you could tell me your solution so I can see what I am missing.
Ok. Now use your energy equation to substitute for v2.
When you have done that, write a as the sum of the equilibrium angle and a small angle perturbation.

Offline for the next four hours.
 
The solution to this problem is given as a''-ga/r+v^2/rR=0. Can someone please tell me how they would arrive at this solution so I can see which part I don't understand.
 
Attachments
  • Handwritten derivation on paper, rotated sideways: PE = mr(1 - cos α)g, K = ½mv², then energy conservation giving v² = constant - 2r(1 - cos α)g, then α̈ = (C - 2r(1 - cos α)g) / (R sin α) · (cos α - g/r) sin α, and finally ä = C cos α - 2(1 - cos α)g cos α / r - g sin α / r.
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