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?


2 Answers 2

  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.


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

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


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

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:

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


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

are now equal


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.