Solving a complex differential equation for non-uniform circular motion
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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 Commenti
Abraham Boayue
il 19 Lug 2018
Modificato: Abraham Boayue
il 19 Lug 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.
Risposte (1)
David Goodmanson
il 19 Lug 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 Commenti
David Goodmanson
il 24 Lug 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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