Lagrange equation of motion question

**Ed Quanta** · Mar22-04, 07:32 PM

A smooth wire is bent into the form of a helix the equations of which, in cylindrical coordinates, are z=a*beta and r=b , in which a and b are constants. The origin is a center of attractive force, , which varies directly as the distance, r. By means of Lagrange’s equations find the motion of a bead which is free to slide on the wire.

Ok, so my variables are r, beta, and z, right? But what is the attractive force? Is it gravity? I need to know how to account for the attractive force in this problem in order to know what the potential energy V is, which is necessary to solve for L=K-V, and thus derive the equations of motions. Any ideas?

**Ed Quanta** · Mar22-04, 08:46 PM

I had another idea for what the force could be? Can we assume F(r)= mr'' - (L^2)/mr^3 since this relates to central force? And if so, how would I write F(r) as a potential? It occurred to me to take the integral of F(r) but I wasn't sure how to treat the term mr''. I know in the Lagrangian method we work with r and r'' as if they are independent variables, but I was not sure if this would apply when simply trying to define what the potential is that we are working with.

~~pmb_phy~~ · Mar23-04, 06:27 AM

Originally posted by Ed Quanta
A smooth wire is bent into the form of a helix the equations of which, in cylindrical coordinates, are z=a*beta and r=b , in which a and b are constants. The origin is a center of attractive force, , which varies directly as the distance, r. By means of Lagrange’s equations find the motion of a bead which is free to slide on the wire.

Ok, so my variables are r, beta, and z, right? But what is the attractive force? Is it gravity? I need to know how to account for the attractive force in this problem in order to know what the potential energy V is, which is necessary to solve for L=K-V, and thus derive the equations of motions. Any ideas?

All you have to do is know the potential and kinetic energy in order to determine the Lagrangian. Since you were given the force then simply set F(r) = k/r for some k. You don't need to know more than that. Now find V(r) from this F(r) by F(r) = - grad V = -dV/dr. Solve for V.

So all you really need to know is the relationship between F and V and that is given by the definition of V as in F(r) = - grad V = -dV/dr.

**Ed Quanta** · Mar23-04, 10:46 AM

Thanks, so does this look right?

L=K-V= 1/2m(r')^2 + 1/2m(r)^2(beta') + 1/2m(z')^2 - kln[r]
(k is a constant here)

Now since r=b where b is a constant is given to us, wouldn't there be a constraint force necessary to keep this condition? Is there another constraint force keeping the condition z=a(beta) where a is constant? I'm supposing there is, so how do I solve for these constraints? I am used to just setting them to zero.

~~pmb_phy~~ · Mar23-04, 01:19 PM

Originally posted by Ed Quanta
Thanks, so does this look right?

L=K-V= 1/2m(r')^2 + 1/2m(r)^2(beta') + 1/2m(z')^2 - kln[r]
(k is a constant here)

Now since r=b where b is a constant is given to us, wouldn't there be a constraint force necessary to keep this condition? Is there another constraint force keeping the condition z=a(beta) where a is constant? I'm supposing there is, so how do I solve for these constraints? I am used to just setting them to zero.

Yup! That's correct.

That's the beauty of analytic mechanics. You don't have to worry about forces of constraints. You only have to worry about potential energy. The forces of constraint are what you find with Lagrangian dynanimcs. It's not something you need to know at the beginning. All you need to know is potential energy. There is no potential energy associated with forces of contraint.

To find the force of constraint simply find the differential equations buy plugging the Lagrangian into the Lagrange equations. Then solve the equation using the inertial conditions. Once you have the equation of motion use F = dr/dt in vector form and that will give you the total force on the particle.

**turin** · Mar23-04, 06:24 PM

Originally posted by Ed Quanta
The origin is a center of attractive force, , which varies directly as the distance, r.

This looks more like a Hooke's Law force, like there's a spring tying the particle to the origin. Also, it looks suspiciously like you've got two different r's here: the disance from the axis of the helix (traditionally denoted ρ) and the distance from the origin (denoted r). That's just two cents worth of crap though, something to consider.

~~pmb_phy~~ · Mar24-04, 06:53 AM

Originally posted by turin
This looks more like a Hooke's Law force, like there's a spring tying the particle to the origin. Also, it looks suspiciously like you've got two different r's here: the disance from the axis of the helix (traditionally denoted ρ) and the distance from the origin (denoted r). That's just two cents worth of crap though, something to consider.

Ah! Excellent point. I incorrectly thought that he wrote thbat the force varies inversely proiportional to r since both the EM and G force vary as such for an infinite line.

Thanks for the correction. In this case the potential is F(r) = kr^2

**turin** · Mar24-04, 07:21 PM

Originally posted by pmb_phy
All you have to do is know the potential and kinetic energy in order to determine the Lagrangian.

... and the equations of constraint when they are not meaningfully represented in terms of potential energy, as in this problem. Just knowing the kinetic and potential energy is fine for smoothly varying potentials, but, since this is ultimately based on the principle of least action, you have to pay attention to what freedoms your system has. To be more specific, it doesn't make any sense to have a Lagrangian with r's in it since one of the constraints is r = const. To be more specific, the Lagragian should be single variable, since there is only 1 degree of freedom.

Originally posted by pmb_phy
Thanks for the correction. In this case the potential is F(r) = kr^2

Well, it's hard to really tell what is meant by "directly." But, generally this qualifier is used to indicate the exclusion of inverse proportionality. Incidently, if you're Hooke's Law-ing as: force = -k r, then you should potential-ize as: F(r) = (1/2) k r². Just another little picky detail.

If the force is a central force, and if the bead is confined to the helix, then you can express both the kinetic and potential energies in terms of one of the Cartesian coordinates. Then, you get a single coordinate Lagrangian. If you don't figure it out soon, just let me know, and I'll tell you which coordinate.