Classical mechanics, motion of a particle over a helix

[fluidistic]

Homework Statement

I'm doing past a past exam (2003) and I'm stuck on the first exercise. Here it is:
Consider a helix centered in the z-axis, of radius R and fixed step [itex]a[/itex], given in cylindrical coordinates by [itex]z=\frac{a\theta }{2 \pi }[/itex], [itex]r=R[/itex].
A particle of mass m slides without rolling over the helix under the action of gravity [itex]-g \hat z[/itex], being at rest in [itex]t=0[/itex] and at a height [itex]h[/itex] over the x-y plane.
1)How many degrees of freedom does the system have? Does it has constraints? If so, of which kind? Write down the Lagrangian modified in order to use the method of Lagrange's multipliers.
2)Write down the conserved quantities.
3)Calculate the position of the particle for all [itex]t \geq 0[/itex].
4)Calculate the force of constraint over the particle for all [itex]t \geq 0[/itex].

Homework Equations

Coming in part 3)

The Attempt at a Solution

I was trying to solve the exercise until I realized I didn't use the constraints and I realized I don't understand something important.
Here is what I've done:
1)3 degrees of freedom, the motion is in 3d and requires 3 coordinates to be described (now I'm unsure of this since z is in function of theta, it would mean only 2 degrees of freedom, although since r is constant, maybe only 1 after all. Sigh)
Yes it has constraints, the motion of the particle must follow the helix. I don't know if the constraint is holonomic. I'd say it isn't, because the force applied to the mass will vary with its velocity, so I'd say the constraint is non-holonomic.
[itex]L=T-V[/itex]. [itex]T=\frac{ m \dot {\vec r}^2}{2}[/itex]. Writing [itex]\vec r =x \hat i +y \hat j + z\hat k[/itex], converting x, y and z into cylindrical coordinates and doing [itex]\vec r \cdot \vec r[/itex], I reached that [itex]T= \frac{m(R^2 \dot \theta ^2 + \dot z ^2)}{2}[/itex]. While [itex]V=mgz[/itex]. So that [itex] L=m \left ( \frac{R^2 \dot \theta ^2 +\dot z ^2 }{2} -gz \right )[/itex].
I calculated the generalized forces, [itex]F _r = \frac{d}{dt} \left ( \frac{\partial T }{\partial \dot r } \right )- \frac{\partial T }{\partial r}=-R \dot \theta ^2 m[/itex] which indeed has units of Newton in the SI.
I reached [itex]F_ \theta=R^2 \ddot \theta m[/itex] with units of Newton times meter (torque units) and [itex]F_z =m(\ddot z +g )[/itex] with units of Newton.
I had read that the values of Lagrange multipliers are worth the forces, so I'm guessing I can use the values I just calculated as values for [itex]\lambda[/itex]. Is what they ask for of the form [itex]L _{\text {modified} }=L- \lambda \Phi[/itex] where [itex]\Phi[/itex] is a constraint equation, like [itex]r=R[/itex], [itex]z=\frac{a \theta }{2 \pi}[/itex]?

After I had done this, I realized I never used the fact that [itex]z= \frac{a\theta }{2 \pi }[/itex]. Should I use it now? [itex]F_z[/itex] would disappear...
Also, I don't understand how can [itex]z=\frac{a\theta }{2 \pi }[/itex] describe a helix. First of all, in cylindrical coordinates theta has a range of [itex]2 \pi[/itex]. This seriously limits z. Does that mean that the helix is "cut", starting from [itex]z=0[/itex] to [itex]z=a[/itex]? So that [itex]0<h<a[/itex]. If not... then I do not understand how this equation (along with [itex]r=R[/itex]) describe a helix since theta is limited in cylindrical coordinates.
Any help is appreciated!

 

[fluidistic]

Ok I definitely believe the motion is over a truncated helix. The mass starts at a height h over the ground. This means [itex]z_0=\frac{a\theta _0}{2 \pi}[/itex]. There are 2 constraint equations, namely [itex]r-R=0[/itex] and [itex]z-\frac{a\theta }{2 \pi}=0[/itex]. These are holonomic constraints so I was wrong in my last post.
Taking the Lagrangian of the previous post and finding the Euler-Lagrange equation (there's only 1 because all coordinates can be put as dependent of theta), I reach the equation of motion [itex]\theta (t)=\underbrace{ \frac{2\pi h}{a}} _{\theta _0 }-\frac{2\pi ^2 gat^2}{4 \pi ^2 R^2+a^2}[/itex].
When they ask for the position of the particle for all t's greater or equal to 0, it's simply [itex]\vec r = R \cos \theta (t) \hat i + R \sin \theta (t) \hat j + \frac{a \theta (t) }{2 \pi } \hat k[/itex]. This expression is probably much nicer in terms of cylindrical coordinates unit vectors, but I'm not very comfortable with switching between the unit vectors (although any help on doing it is welcome!). So this method solves in part the problem but not entirely.

They ask for the Lagrange multipliers method. I've read some theory but I'm unsure about something. The modified Lagrangian should look like [itex]\tilde L =T-V-\lambda (\text{constraint equation})[/itex].
So since I have 2 constraints, should I have 2 Lagrange multipliers? In other words, is the following right:
[itex]\tilde L=T-V-\lambda _z \left ( z-\frac{a\theta }{2\pi } \right )-\lambda _r (r-R)[/itex]?
In which case [itex]T-V=m \left ( \frac{R^2 \dot \theta ^2 +\dot z ^2 }{2} -gz \right )[/itex] but I would NOT replace [itex]z[/itex] by [itex]\frac{a\theta }{2\pi}[/itex] and [itex]r[/itex] by [itex]R[/itex]; unlike what I've done in the first paragraph of this post.
Also, I don't really know how to get the force. Maybe its magnitude is [itex]\sum _{i =0} ^2 \lambda i Q _i[/itex] where the [itex]Q_i[/itex] are the constraint equations, but I don't know how to find its direction. Any help is really appreciated!

 

[mathfeel]

If you are using Lagrange multiplier, you need to keep r as a free variable and not to replace it with R. So your kinetic energy should be:
[tex]T = \frac{m}{2} \left(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2 \right)[/tex]

As for the force of constrain, answer this first: what are the dimension of the Lagrange multipliers?

 

[fluidistic]

If you are using Lagrange multiplier, you need to keep r as a free variable and not to replace it with R. So your kinetic energy should be:
[tex]T = \frac{m}{2} \left(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2 \right)[/tex]

As for the force of constrain, answer this first: what are the dimension of the Lagrange multipliers?

Ah ok thanks a lot, indeed I made an error. Also considering r as a variable rather than a constant, I had to derive the kinetic energy you provided (just did it).
Let me think about the dimensions of Lagrange multipliers. I know that a L. multiplier times the constraint must have units of energy/work, so joule in the SI. Since the constraint has units of distance, so meter in the SI, this means that lambda has units of kg/s² (it's a force units as the previous statement imply). Since work has units of distance times force, it means the L. multipliers have untis of force.
P.S.:I'm going to continue tomorrow (almost 2 am here).

 

[fluidistic]

I've reached the Euler-Langrage equations:
[itex]\frac{\partial L }{\partial \theta }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot \theta } \right ) =\frac{a \lambda _z }{2 \pi }-m (\ddot \theta r^2 +2 \dot \theta r)[/itex].
[itex]\frac{\partial L }{\partial r }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot r } \right ) = m (\dot \theta ^2 r +r)-\lambda _r[/itex].
And [itex]\frac{\partial L }{\partial z}-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot z} \right ) =-mg-\lambda _z -m \ddot z[/itex].
I'm not sure what they are equal to. Generalized forces? 0?

 

[mathfeel]

I've reached the Euler-Langrage equations:
[itex]\frac{\partial L }{\partial \theta }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot \theta } \right ) =\frac{a \lambda _z }{2 \pi }-m (\ddot \theta r^2 +2 \dot \theta r)[/itex].
[itex]\frac{\partial L }{\partial r }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot r } \right ) = m (\dot \theta ^2 r +r)-\lambda _r[/itex].
And [itex]\frac{\partial L }{\partial z}-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot z} \right ) =-mg-\lambda _z -m \ddot z[/itex].
I'm not sure what they are equal to. Generalized forces? 0?

You forgot the right-hand-side of the Euler-Lagrange equation:
[tex]\frac{\partial L}{\partial x_j} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_j} = 0[/tex]

You first two equations are also wrong (check the dimension of each term, they are not the same). For example:
[tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=m \left( r^2\ddot{\theta}+2r\dot{r} \dot{\theta} \right)[/tex]

Now that you have 5 equations and 5 unknown. Try to solve them.

 

[fluidistic]

I'm having a problem.
My modified Lagrangian is [itex]\tilde L= \frac{m}{2} (\dot \theta ^2 r^2 + \dot r ^2 + \dot z ^2)-mgz-\lambda _z \left ( z-\frac{a\theta }{2\pi } \right )- \lambda _r (r-R)[/itex].
The corresponding Euler-Lagrange equations are thus:
[tex]\frac{\partial L }{\partial \theta }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot \theta } \right ) =\frac{a\lambda _z}{2\pi}-m(\ddot \theta r^2 +2 \dot \theta r \dot r )[/tex]

[tex]\frac{\partial L }{\partial r}-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot r } \right ) =m(\dot \theta ^2 - \ddot r )-\lambda _r[/tex]

[tex]\frac{\partial L }{\partial z }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot z } \right ) =-m(g+\ddot z )-\lambda _z[/tex]

[tex]\frac{\partial L }{\partial \lambda _r }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot \lambda _r } \right ) =R-r[/tex]

[tex]\frac{\partial L }{\partial \lambda _z }-\frac{d}{dt} \left ( \frac{\partial L }{\partial \dot \lambda _z } \right ) =\frac{a \theta}{2 \pi }-z[/tex]
They are all worth [itex]0N[/itex] (the right hand side I was unsure of or "forgot" in my last post). All have units of Newton, except the first term in the first equation, which has units of [itex]N \cdot m[/itex] (torque). I'm guessing this is a huge problem, but I don't see what I did wrong.

 

[mathfeel]

You seems to think that the "derivative" of the Lagrangian:
[tex]\frac{\partial L}{\partial x_j} - \frac{d}{dt} \frac{\partial L}{\partial x}[/tex]
which, if [itex]x_j[/itex] has dimension length has dimension of force, is the force of constrain.

NO! You are trying to minimize the Lagrangian again some constrain. Just when you try to minimize something, you set the derivative to 0. So you should equate the first three equations to zero. Setting the last two equations to zero just amount to the two equations of constrain.

 

[fluidistic]

Ok thank you!
The first 3 equations equated to 0 yield (after using the constraint equations so that [itex]\dot r=0[/itex]):
[tex]\frac{\lambda _z a }{2 \pi}-m \ddot \theta r^2=0[/tex].
[tex]m \dot \theta ^2 r - \lambda _r =0[/tex]
[tex]m(\ddot z +g )+\lambda _z=0[/tex].
So I get [itex]\lambda _r = m \dot \theta ^2 r[/itex], [itex]\lambda _z =-m (\ddot z + g )[/itex] and [itex]\ddot \theta \theta = \frac{z (\ddot z +g )}{R^2}[/itex].

 

[fluidistic]

Little "bump". Are these 3 equations along with the 2 constraint equations correct? I'm just left with solving them, right?