If a system has $N$ coordinates and $M$ number of holonomic constraints then number of degree of freedom $=N-M$ and generalized coordinates $=N-M$ too. But if there are $k$ non-holonomic constraints then what will be no. of degree of freedom and generalized coordinates?

  • $\begingroup$ if there are K non-holonomic constraints and N degree of freedom, the numbers of the generalized coordinates are N-K $\endgroup$ – Eli Sep 24 '19 at 8:03
  • $\begingroup$ And degree of freedom? $\endgroup$ – Harsh Nigam Sep 24 '19 at 16:08
  • $\begingroup$ The degree of freedom are constraint, how many of them is dependent on your physical system $\endgroup$ – Eli Sep 24 '19 at 19:22
  1. Consider a classical point-mechanical system with $3N$ coordinates but only $n$ generalized coordinates $(q^1, \ldots,q^n)$, because of $3N-n$ holonomic constraints.

  2. Let us for simplicity assume that:

    • All constraints are 2-sided, i.e. we do not allow 1-sided constraints (= inequalities).

    • All non-holonomic constraints are semi-holonomic.

  3. If furthermore the system has $m$ semi-holonomic constraints, then we introduce $m$ Lagrange multipliers $(\lambda^1, \ldots,\lambda^m)$. The corresponding $n$ Lagrange equations are outlined in this Phys.SE post.

  4. The number of degrees of freedom (DOF) are conventionally defined as half the number of initial conditions needed for $(q^1, \ldots,q^n,\dot{q}^1, \ldots,\dot{q}^n)$, which is then $n\!-\!m/2$.


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