Ik heb nu exact dezelfde oplossing als ukster met Lagrange, dus dan zullen we het allebei wel juist hebben. Ik was gewoon vergeten te delen door 2 in de formule voor de kinetische energie.
Dit is dan de indrukwekkende Lagrangiaan met y(t) de hoogte van
\(m_2\)
$$L=- \frac{1.0 M m_{1}^{2} \cos^{2}{\left(\alpha \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}}{\left(M + m_{1} + m_{2}\right)^{2}} - g \left(- m_{1} \sin{\left(\alpha \right)} + m_{2}\right) - 0.5 m_{1} \left(\left(\frac{2 m_{1} \cos{\left(\alpha \right)}}{M + m_{1} + m_{2}} - 2 \cos{\left(\alpha \right)}\right) \left(\frac{m_{1} \cos{\left(\alpha \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}}{M + m_{1} + m_{2}} - \cos{\left(\alpha \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}\right) + 2 \sin^{2}{\left(\alpha \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}\right) - 0.5 m_{2} \cdot \left(\frac{2 m_{1}^{2} \cos^{2}{\left(\alpha \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}}{\left(M + m_{1} + m_{2}\right)^{2}} + 2 \frac{d^{2}}{d t^{2}} y{\left(t \right)}\right)$$
Code: Selecteer alles
from sympy import *
alpha, t, g, m1, m2, M, h = symbols('alpha t g m1 m2 M h')
x, y, ym2, xm1,xm2 = symbols('x y, ym2, xm1, ym1', cls=Function)
x = x(t)
y = y(t)
xm1 = xm1(t)
xm2 = xm2(t)
dydt = diff(y,t)
dxdt=(m1*cos(alpha))*dydt/(m1+m2+M)
T=0.5*(m1*((-dydt*cos(alpha) + dxdt)**2+(dydt*sin(alpha))**2))+0.5*m2*(dxdt**2+dydt**2)+0.5*M*dxdt**2
V=(m2*y+m1*(h-y*sin(alpha)))*g
L=T-V
LE1 = diff(L, y) - diff(diff(L,dydt), t)
sol=dsolve(LE1, y, ics={y.subs(t,0): 0, y.diff(t).subs(t, 0):0})
ym2=simplify(sol.rhs)
xm1 = -ym2*cos(alpha)
ym1 = h-ym2*sin(alpha)
afstand2 = xm1**2+ym1**2+ym2**2
dafstand2dt=afstand2.diff(t)
sol=solve(dafstand2dt,t)
oplossing=simplify(sol[2])
oplossing
