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T = 1
S = NDSolve[{r''[t] == r[t] (θ'[t])^2, 
   m (2 r'[t] θ'[t] + r[t] θ''[t]) == T/r[t], 
   r[0] == 0.1, θ'[0] == 0,
   r'[0] == 0, θ[0] == 0}, {r, θ}, {t, 0, 100}]

Now relation of r[t] and θ is parametric equation but I want to plot my graph in polar by don't use ParametricPlot

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1
  • $\begingroup$ mis undefined! $\endgroup$ Nov 23, 2023 at 13:37

1 Answer 1

1
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Assuming m=1

T = 1;
m=1;
S = NDSolve[{r''[t] == 
r[t] (\[Theta]'[t])^2,  (2 r'[t] \[Theta]'[t] + 
r[t] \[Theta]''[t]) == T/r[t], r[0] == 0.1, \[Theta]'[0] == 0, 
r'[0] == 0, \[Theta][0] == 0}, {r, \[Theta]}, {t, 0, 100}][[1]]

we have to solve teta= \[Theta][t] for t

solt = Solve[(\[Theta][t] /. S) == teta, t][[1]]

and substitude into result r[t]

PolarPlot[Evaluate[r[t] /. S /. solt], {teta, 0 , 3.39},AspectRatio -> 1, PlotRange -> {{-120, 0}, {-10, 5}}]

enter image description here

The advantage compared to ParametricPlot is not clear to me.

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