Differential equation after using Euler-Lagrange equations

Zaknife
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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


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

How to deal with such differential equation ?
 
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So \phi 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 \ddot\phi = c which gives a simple linear solution (which makes sense, because the only freedom you have is how fast you move along the curve).
 
So all i need to do is double integration ?
 
Zaknife said:
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.
 
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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