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Solving a complex differential equation for non-uniform circular motion

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BHAGYASHREE WAGHULE
BHAGYASHREE WAGHULE le 18 Juil 2018
Commenté : David Goodmanson le 24 Juil 2018
I have previously used ODE45 to solve a single-variable differential equation. I am not able to figure out how to solve this equation using ODE or other solvers. The context of the problem is to keep 'a' constant while changing 'r' with respect to time.
a = (r" - r.w^2) + (r.w' + 2.r'.w)
Conditions:
a = 9.81, r = 12 to 1000, w(initial) = 0.6
I am trying to find r(t) and \omega(t) to keep a constant. I am not sure if I am missing any other information. Thanks.
  2 commentaires
Abraham Boayue
Abraham Boayue le 19 Juil 2018
Modifié(e) : Abraham Boayue le 19 Juil 2018
You can't use a single equation to find two unknowns (r and w). You need a second equation in addition to the one given.
BHAGYASHREE WAGHULE
BHAGYASHREE WAGHULE le 23 Juil 2018
I understand. But I just have a set of data for values of w varying with time t. I am not sure if there is a way to use that information to solve this equation.

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Réponses (1)

David Goodmanson
David Goodmanson le 19 Juil 2018
Hello Bhagyashree,
although polar coordinates of course are not the most natural set of coordinates for this problem, they certainly do work. If u_r and u_th are unit vectors in the direction of r and theta, then
a = (r'' - r (theta')^2) u_r + (r theta'' + 2 r' theta') u_th
which is more or less what you have, only with unit vectors inserted. It would be pretty unusual if g = 9.81 did not mean a uniform gravitational field in the downward direction, so assuming that is the case, with theta measured ccw from the x axis as usual, and g pointing in the -y direction, then
x = r cos(theta) y = r sin(theta)
a = -g ( sin(theta) u_r + cos(theta) u_th )
and equating both sides gives
(r'' - r (theta')^2) = -g sin(theta)
(r theta'' + 2 r' theta') = -g cos(theta)
These two second order eqns can be turned into a first order eqn in variables r,r',theta,theta'=w and solved with ode45.
  2 commentaires
BHAGYASHREE WAGHULE
BHAGYASHREE WAGHULE le 23 Juil 2018
I appreciate your response. But, would this hold true for an object spinning in space (zero-gravity?). It seems the assumption you have made is that the object is spinning under a uniform gravitational field.
David Goodmanson
David Goodmanson le 24 Juil 2018
These equations describe the motion of a point particle through space, so there is no 'spinning' in the usual meaning of rotation of a solid object. In free space the equations are the same except g = 0 so the right hand side of each equation is 0. With those equations you can that the motion is constant velocity in a straight line, but it's a chore.

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