# Meaning of Lagrange multiplier for a particle in a cylinder with a spring My procedure is the following in cylindrical coordinates ($$\rho,\theta,z$$):

The kinetic energy, $$\begin{equation} T=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\dot{p}^{2}+\rho^{2} \theta^{2}+\dot{z}^{2}\right) \end{equation}$$

The potential energy for OP being the distance from the particle to the spring/origin($$OP=\sqrt{x^2+y^2+z^2}=\sqrt{z^2+\rho^2}$$), $$\begin{equation} V=m g z+\frac{1}{2} k {O P^{2}}=m gz+\frac{1}{2} k\left(z^{2}+\rho^{2}\right) \end{equation}$$

The constraints, $$z=h\quad \rho=R$$ So the Lagrangian is: $$L=T-V=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\dot{p}^{2}+\rho^{2} \theta^{2}+\dot{z}^{2}\right)-m gz-\frac{1}{2} k\left(z^{2}+\rho^{2}\right)+\lambda_\rho(\rho-R)+\lambda_z(z-h)$$ The equations of motion: $$\begin{equation} \rho \rightarrow m \ddot{\rho} -m \rho \dot\theta^{2}+k p=\lambda_\rho \end{equation}$$ $$\begin{equation} \theta \rightarrow \frac{d}{d t}\left(m\rho^2 \dot{\theta}\right)=0\rightarrow \dot\theta=\frac{L_z}{\rho^2m}(\text {L_z angular momentum for z axis }) \end{equation}$$ $$\begin{equation} z \rightarrow m \ddot{z}+m g+k z=\lambda_z \end{equation}$$ Since $$\ddot \rho=0$$ and $$\ddot z=0$$, we have: $$\lambda_\rho=m\ddot\rho-m\rho(\frac{L_z}{\rho^2m})^2+k\rho=k\rho-\frac{L_z^2}{\rho^3m}$$ $$\lambda_z=m\ddot z+mg+kz=k+mg$$ So finally we have:

$$\lambda_\rho=k\rho-\frac{L_z^2}{\rho^3m} \quad \lambda_z=k+mg$$ The problem and my question is when I get the generalized forces, I don't understand what kind of force I'm working with. I mean, if I have a strong spring($$k_\rho>\frac{L_z^2}{\rho^3m}$$), so the constraint force is positive at $$\hat{\rho}$$, is this still the reaction force? Shouldn't be negative why it's positive, specially if I'm dealing with a particle that starts without velocity. And the same goes for $$\lambda_z$$.

I don't really understand what I'm getting and their physical meaning.

Any help will be very much appreciated.

• The physical meaning of Lagrange multipliers is always a "rate" of change. They describe a change in the thing you're optimizing (in this case the Lagrangian) with respect to the constraint. The units work out to force in this case, so it physically makes sense. The Lagrangian is in units of energy and establishing the rate of change with respect to $\rho$ or $z$ will give you something in units of force. The multipliers tell you how much a small change in those dimensions will change the optimized value of the Lagrangian. – Michael Burt Jun 2 at 14:14
• the generalized constraint forces haven’t the force unit [N] you obtain the constraint forces unit [N] with these equations \begin{aligned}\lambda _p\left( \rho-R\right) \\ \lambda_ z\left( z-h\right) \end{aligned} – Eli Jun 3 at 15:49

In this case, your particle needs its spring force in the radial direction canceled out, because we can't have it accelerating in the radial direction, but then it also needs an extra term which is a centripetal force, because it needs to be kept in a circular motion about the $$z$$ axis. So given that you derive this extra force-term, you can immediately understand that there are these two aspects of alternate sign, if the spring is very strong compared to the spin rate, then the cylinder needs to pull the mass out from its natural tendency to fall into the center: but as you spin it faster around the center, the constraint starts to need to push it in towards the center instead, otherwise it will want to fly off in a tangent plane.
You should not have a Lagrange multiplier for the $$z$$-direction because that is not a constrained dimension. Unless, maybe, you are only investigating the case where the particle is stuck at the top of a finite cylinder and so the kinematics become one-dimensional? If so, then the answer is very easy as you have seen, angular momentum is conserved and $$\theta(t) =\theta_0+\omega~t$$ for some $$\theta_0,\,omega.$$ But more usually you should also get that $$z(t)=z_0+A \cos(\Omega~t+\phi)$$ or so, a simple harmonic motion in the $$z$$-direction.
Just as one further tip, it is overkill to use Lagrange multipliers in a Lagrangian mechanics calculation if your generalized coördinates already implement the constraint. The nice thing about lagrangian analysis is that it allows for generalized coordinates. You can just say “due to my constraints $$\rho=R,\dot\rho=0$$” at the outset and save quite a bit of paper, heh.