Mathematica Asked by bob the legend on December 23, 2020
I know that there are similar questions on stack exchange. However, I am relatively new to mathematica, so I couldn’t really understand the answers. Sorry.
Here’s my code:
k = 50000000;
c = 37;
g = 9.81;
[Mu] = 0.3;
i = 10^-5;
l = 0.1;
m = 0.35;
NDSolve[
{m y''[t] == -k z[t]^(3/2) - c z'[t] - m g,
[CurlyPhi]''[
t] == ((-k z[t]^(3/2) - c z'[t]) l Sin[[CurlyPhi][
t]] + [Mu] (-k z[t]^(3/2) - c z'[t]) Cos[[CurlyPhi][t]])/i,
z[t] == y[t] - l Cos[[CurlyPhi][t]],
y[0] == 0, y'[0] == -3, [CurlyPhi][0] == 0, [CurlyPhi]'[0] == 0,
z[0] == 0, z'[0] == 0}, {y[t], z[t], [CurlyPhi][t]}, {t, 0, 1}]
The error message is NDSolve::icfail: Unable to find initial conditions that satisfy the residual function within specified tolerances. Try giving initial conditions for both values and derivatives of the functions.
Is there any way to solve the equations numerically and plot graphs of them as a function of time. Thanks
z[t] == y[t] - l Cos[[CurlyPhi][t]]
is an algebraic equation, Delete it and define instead: z[t_] = y[t] - l Cos[[CurlyPhi][t]]
and delete the initial conditions for z:
k = 50000000;
c = 37;
g = 9.81;
[Mu] = 0.3;
i = 10^-5;
l = 0.1;
m = 0.35;
NDSolve[
{m y''[t] == -k z[t]^(3/2) - c z'[t] - m g,
[CurlyPhi]''[
t] == ((-k z[t]^(3/2) - c z'[t]) l Sin[[CurlyPhi][
t]] + [Mu] (-k z[t]^(3/2) - c z'[t]) Cos[[CurlyPhi][t]])/i,
z[t] == y[t] - l Cos[[CurlyPhi][t]],
y[0] == 0, y'[0] == -3, [CurlyPhi][0] == 0, [CurlyPhi]'[0] == 0,
z[0] == 0, z'[0] == 0}, {y[t], z[t], [CurlyPhi][t]}, {t, 0, 1}]
Correct answer by Daniel Huber on December 23, 2020
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