Finding the value of the holonomic constraint forces

Question

So let's say I have a Lagrangian augmented with some holonomic constraints.

$$L' = L + \sum_i \lambda_i(t) f_i(q,t).\tag{i}$$

The solutions is the system of differential equations:

$$\frac{\partial L}{\partial q_k} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k} = \sum_i \lambda_i(t) \frac{\partial f_i(q,t)}{\partial q_k},\tag{ii}$$

$$f_i(q,t) = 0.\tag{iii}$$

I am not sure about the value of the constraint forces. I think they are:

$$F = \frac{\partial f_i(q,t)}{\partial q_k}.\tag{iv}$$

But I'm far from sure.

Answer 1

Finding the value of the holonomic constraint forces is done as follows.

1. The equations are:
   \begin{align*} &\frac{\partial L}{\partial\vec{q}}-\frac{d}{dt}\frac{\partial L}{\partial\vec{\dot{q}}}=\sum_{i}^{n_c}\lambda_i\frac{\partial f_i(\vec{q},t)}{\partial\vec{q}}\quad,\text{($n_q$ equations )}\tag{1}\\ & f_i(\vec{q},t)=0\quad i=1\,,\ldots\,,n_c\quad\text{($n_c$ constraints equations) }\tag{2}\\\\ &\text{If we evaluate equation (1) we get:}\\ &\vec{f}_q(\ddot{\vec{q}}\,,\dot{\vec{q}}\,,\vec{q}\,,\vec{\lambda)}=0 \tag{3}\\ &f_i(\vec{q},t)=0 \tag{4} \end{align*}

Equations (3) and (4) are $n_q+n_c$ equations for $\ddot{\vec{q}}$ and $\vec{\lambda}$ unknowns.

How to solve:

2. Example: Pendulum

\begin{align*} &T=\frac{1}{2} m\,\dot{x}^2+\frac{1}{2} m\,\dot{y}^2 \qquad V=\,m\,g\,y\ \qquad L=T-V\\ &\text{with:}\quad q_1=x\quad q_2=y\quad \Rightarrow\\ &\frac{\partial L}{\partial\vec{q}}=[0\,,-m\,g]\, ,\qquad \frac{d}{dt}\frac{\partial L}{\partial\vec{\dot{q}}}= [m\,\ddot{q}_1\,,m\,\ddot{q}_2]\\\\ &\text{constraint equation (where $l$ is the pendulum length) :}\\\ &f_1=x^2+y^2=l^2=q_1^2+q_2^2-l^2=0\tag{5}\\ & \sum_{i}^{n_c}\lambda_i\frac{\partial f_i(\vec{q},t)}{\partial\vec{q}}=\lambda[2\,q_1\,,2\,q_2]\\\\ &\Rightarrow\quad\text{Insert into equation (3)}\\ &[0\,,m\,g]- [m\,\ddot{q}_1\,,m\,\ddot{q}_2]-\lambda[2\,q_1\,,2\,q_2]=0\quad &\tag{6}\\\\ &\text{if we differentiate twice the constraint equation (5) we get: }\\ &\frac{d^2}{dt^2}f_1=\frac{d^2}{dt^2}(q_1^2+q_2^2-l^2)= 2(\dot{q}_1^2\,\ddot{q}_1+\dot{q}_2^2\ddot{q}_2)=0\tag{7} \end{align*} We can now solve equation (6) and (7) to get $\ddot{q}_1\,,\ddot{q}_2$ and the generalized constraint force $\lambda$: \begin{align*} &\ddot{q}_1=-g\,\frac{q_1\,\dot{q}_2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\tag{8}\\ &\ddot{q}_2=-g\,\frac{q_1\,\dot{q}_1^2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\tag{9}\\ &\lambda=\frac{1}{2}\,m\,g\frac{\dot{q}_2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\\\\ &\text{The constraint force $F_x$ [N] toward the $x$-axis is:} |F_x|=|\lambda\,2\,q_1|\\ &\text{and toward the $y$-axis: }\qquad \qquad \qquad \qquad \quad\, |F_y|=|\lambda\,2\,q_2| \end{align*}

Answer 2

1. The $i$'th holonomic constraint $$f_i(q,t)~=~0\tag{A}$$ can be turned into an (integrable) semi-holonomic constraint $$ \sum_{k=1}^na_{ik}(q,t)\dot{q}^k + a_{i0}(q,t)~=~0\tag{B}$$ by differentiation wrt. time $t$, and by using the identifications $$ a_{ik}~=~\frac{\partial f_i}{\partial q^k}, \qquad a_{i0}~=~\frac{\partial f_i}{\partial t}. \tag{C}$$

2. The generalized forces serve as sources for Lagrange equations (in the same way that forces serve as sources for Newton's second law), cf. e.g. my Phys.SE answer here. The $k$'th generalized constraint force, say $Q^{(c)}_k$, is $$ Q^{(c)}_k~=~\sum_i \lambda^i a_{ik}~\stackrel{(C)}{=}~\sum_i \lambda^i \frac{\partial f_i}{\partial q^k}, \tag{D}$$ where $\lambda^i$ is the $i$'th Lagrange multiplier. This answers OP's question.

References:

1. H. Goldstein, Classical Mechanics; Section 2.4 (Warning).