$$f(q,q^\prime, t) = 0, ~\mathrm df = \frac{\partial f}{\partial q}~\mathrm dq + \frac{\partial f}{\partial q^\prime}~\mathrm dq^\prime+ \frac{\partial f}{\partial t}~\mathrm dt = 0$$

I really want to know whether this constraint is holonomic or non-holonomic.

(As far as I know, Non-holonomic constraint has a term of velocity and do non-integrable. But this formula does not dependent on a path, because it is a total differential form.)

  • prime is a time derivative.
  • $\begingroup$ Is prime a time derivative? $\endgroup$
    – Qmechanic
    Commented Jul 31, 2016 at 14:39
  • $\begingroup$ yes, it is a time derivative. q and q' are dependent variable of time $\endgroup$
    – J.ahn
    Commented Jul 31, 2016 at 14:41

2 Answers 2

  1. Firstly, recall that virtual displacements don't change time $\delta t=0$. Time is fixed, say $t=t_0$.

  2. Secondly, let $M$ be the position space, often call the configuration space, with generalized position coordinates $q^i$.

  3. OP is right that the constraint $\delta f(q,v,t) \approx 0$ doesn't depend on virtual displacements $(\delta q,\delta v)$ in the tangent bundle $TM$ (with appropriate boundary conditions imposed). However, a velocity-dependent constraint $f(q,v,t) \approx 0$ is not well-defined on the position manifold $M$ itself. And it is down in the base manifold $M$ where the applications, such as, e.g., d'Alembert's principle take place.

  4. TL;DR: A holonomic constraint has by definition no velocity dependence.

  • $\begingroup$ There's nothing in here that says the constraint has velocity dependence. For all we know, $df/dq'=0$. (Also, isn't f(q, q', t) just the form of a 1-D functional?) $\endgroup$ Commented Jul 31, 2016 at 16:10

If $f$ is defined such that $f=0$ at all points, then it is a holonomic constraint. One example is this:


which constrains motion to the surface of a sphere of radius $r$.


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