Conversion of non-holonomic constraints to holonomic

Question

In the case of a disc rolling without slipping, we have a constraint $\dot{x}=a\dot{\theta}$ where $a$ is the radius of the disc. Note that I have considered $x$ and $\theta$ as the generalized coordinates. By definition, this is a non-holonomic constraint. However, on integrating the constraint, we arrive at $x=a\theta+\phi$ ($\phi$ is a numerical constant of integration), which turns out to be holonomic.

In the method of finding the equations of motion using Lagrangians with a Lagrangian multiplier, we have $\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q_i}}\right)-\frac{\partial L}{\partial q_i} + \lambda \frac{\partial f}{\partial q_i} = 0 $ where $f$ represents the constraint and $q_i$ is the $i$th generalised coordinate. In the case above, $q_i=\{x,\theta\}$. Now, had we used $f=\dot{x}-a\dot{\theta}$ as the constraint, the last term of the modified Euler Lagrange would have been zero. However, if we use the integrated version of the same constraint, we get non zero terms ($\lambda$ and $a\lambda$ for $x$ and $\theta$ respectively). Surprisingly, the latter is correct according to Goldstein. What am I missing here? (I'm specifically referring to the example of a hoop rolling down the incline in chapter two of Goldstein)

This brings me to the more general question: In the method of Lagrangian multipliers, how should I write the constraint relation? To illustrate what I mean, take the following constraint presented in words: the particle moves in a circle of radius $a$. If I denote the position of the particle by $r$ (generalised coordinate), then the constraint dictates $r-a=0$. Alternatively, I can write the same as $r^3-a^3=0$, whose partial derivate with respect to $r$ is not the same as in the $r-a=0$ case. What's going on here?

Answer 1

1. The type of non-holonomic constraint, that Ref. 1 is discussing at this point, is a so-called semi-holonomic constraint, which is a non-holonomic constraint given by a one-form $$ \omega~\equiv~\sum_{j=1}^na_j(q,t)~\mathrm{d}q^j+a_0(q,t)\mathrm{d}t~=~0. \tag{S}$$

2. If there exist (i) a holonomic constraint $$f(q,t)~=~0,\tag{H}$$ (ii) an integrating factor $\lambda(q,t)\neq 0$ and (iii) a one-form $\eta$ such that $$ \lambda\omega+ f\eta~\equiv~\mathrm{d}f , \tag{I}$$ then the constraint (S) is equivalent to the holonomic constraint (H). This e.g. the case with 1D rolling in Fig. 2.5, which OP mentions; but not with 2D rolling in Fig. 1.5. To clear up any confusion we should probably emphasize that a non-integrable semi-holonomic constraint cannot be converted into a holonomic constraint.

3. For some of OP's other questions, see also this, this, this this & this related Phys.SE posts.

References:

1. Herbert Goldstein, Classical Mechanics, Chapter 1 and 2.

Answer 2

I want to answer the question whats happened if you write the constraint equation for a circle path like this one

$$f_{1}=r-a=0\tag 1$$ or like this one $$f_{2}=r^3-a^3=0\tag 2$$

The E.L equations with vector notation are:

$$\frac{d}{dt}\left( \frac{\partial L}{\partial \vec{\dot{w}}}\right)^T-\left(\frac{\partial L}{\partial \vec{w}}\right)^T + \left(\frac{\partial \vec{f}}{\partial \vec{w}}\right)^T\,\vec{\lambda} = \vec{0} \tag 3$$

with polar coordinate is $\vec{w}=[r\,,\varphi]^T$, the vector of the degrees of freedom

you need additional equation to solve equation (3) for $\ddot{r}_i\,,\ddot{\varphi}_i$ and $\lambda_i$

$$\frac{d^2}{dt^2}\vec{f}=\left(\frac{\partial \vec{f}}{\partial \vec{w}}\right)\,\vec{\ddot{w}}+\frac{d}{dt}\left(\frac{\partial \vec{f}}{\partial \vec{w}}\,\vec{\dot{w}}\right)=\left(\frac{\partial \vec{f}}{\partial \vec{w}}\right)\,\vec{\ddot{w}}+\frac{d}{d\vec{w}}\left(\frac{\partial \vec{f}}{\partial \vec{w}}\,\vec{\dot{w}}\right)\,\vec{\dot{w}} =\vec{0}\tag 4$$

with equation (3), (4) and (1) you get:

$$\left[ \begin {array}{c} {\frac {d^{2}}{d{\tau}^{2}}}r \left( \tau \right) \\ {\frac {d^{2}}{d{\tau}^{2}}}\varphi \left( \tau \right) +2\,{\frac { \left( {\frac {d}{d\tau}}r \left( \tau \right) \right) {\frac {d}{d\tau}}\varphi \left( \tau \right) } {r \left( \tau \right) }}\end {array} \right] =\vec{0}\tag 5 $$

and $$\lambda=\left[ \begin {array}{c} -mr \left( \tau \right) \left( {\frac {d}{d \tau}}\varphi \left( \tau \right) \right) ^{2}\end {array} \right] $$

and with equation (3), (4) and (2) you get:

$$\left[ \begin {array}{c} {\frac {d^{2}}{d{\tau}^{2}}}r \left( \tau \right) -2\,{\frac { \left( {\frac {d}{d\tau}}r \left( \tau \right) \right) ^{2}}{r \left( \tau \right) }}\\{\frac {d^ {2}}{d{\tau}^{2}}}\varphi \left( \tau \right) +2\,{\frac { \left( { \frac {d}{d\tau}}r \left( \tau \right) \right) {\frac {d}{d\tau}} \varphi \left( \tau \right) }{r \left( \tau \right) }}\end {array} \right] =\vec{0}\tag 6 $$

and

$$\lambda=\left[ \begin {array}{c} -\frac{1}{3}\,{\frac {m \left( \left( r \left( \tau \right) \right) ^{2} \left( {\frac {d}{d\tau}}\varphi \left( \tau \right) \right) ^{2}-2\, \left( {\frac {d}{d\tau}}r \left( \tau \right) \right) ^{2} \right) }{ \left( r \left( \tau \right) \right) ^{3}}}\end {array} \right] $$

thus the equations of motions and the constriant forces are not equal!

for both constraint equations $f_1$ and $f_2$ is $r=a$, substitute r equal a in equation
(5) and (6) thus the EOM's are now equal:

$$\left[ \begin {array}{c} 0\\{\frac {d^{2}}{d{\tau} ^{2}}}\varphi \left( \tau \right) \end {array} \right] =\vec{0}$$

and

$$\vec{F}_{\lambda i}=\left(\frac{\partial \vec{f_i}}{\partial \vec{w}}\right)^T\,\vec{\lambda_i}=\left[ \begin {array}{c} -am \left( {\frac {d}{d\tau}}\varphi \left( \tau \right) \right) ^{2}\\ 0\end {array} \right] \quad, i=1,2 $$

are now equal