Differential equation after using Euler-Lagrange equations

  • Thread starter Zaknife
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  • #1
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Homework Statement


Particle is moving along the curve parametrized as below (x,y,z) in uniform gravitational field. Using Euler- Lagrange equations find the motion of the particle.

The Attempt at a Solution


[tex]\begin{array}{ll} x=a \cos \phi & \dot{x}= -\dot{\phi} a \sin \phi \\
y=a \sin \phi & \dot{y}=\dot{\phi} a \cos \phi \\
z=b \phi & \dot{z}= b \dot{\phi} \\
\end{array}
[/tex]
Lagrangean will be :
[tex] L=T-V=\frac{m}{2}\dot{\phi}^{2}(a^{2}+b^{2})-mgb\phi[/tex]
Using Euler-Lagrange equations we obtain:
[tex]\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q_{l}}}\right)-\frac{\partial \mathcal{L}}{\partial q_{l}}=0[/tex]
[tex]m\ddot{\phi}(a^{2}+b^{2})+mgb=0[/tex]

How to deal with such differential equation ?
 

Answers and Replies

  • #2
CompuChip
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So [itex]\phi[/itex] is the only generalised co-ordinate, and all the other letters are constants, right?

In that case, this is a differential equation of the form [tex]\ddot\phi = c[/tex] which gives a simple linear solution (which makes sense, because the only freedom you have is how fast you move along the curve).
 
  • #3
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So all i need to do is double integration ?
 
  • #4
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So all i need to do is double integration ?
That's correct. Write the equation in the form suggested by CompuChip and then integrate twice with respect to time. Since the right side is a constant is pretty simple to do.
 
  • #5
dextercioby
Science Advisor
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The particle is moving along a helix (on a circular cylinder) with constant angular velocity. Therefore the angle in the plane xOy depends quadratically on time.
 

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