Suppose a point particle is constrained to move on the curve $y=x^2$. This would then be a non-holonomic geometric constraint since the particle has one degree of freedom and requires two coordinates ($x$ and $y$) to describe its position fully (clarification about aforementioned statement is needed). My question is (if) this constraint is non-holonomic, how come $y$ can be expressed in terms of $x$ as a finite relation (i.e. $y=x^2$)?
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Using a Cartesian coordinate system, the particle has two degrees of freedom because you need two variables to define its position. You use the holonomic contraint to eliminate a coordinate and so reduce the coordinates and degree of freedom by one, but transforming the coordinates in such a way to make this obvious. I.e along the path it's constrained to move along! Clever eh? $$\begin{align*}dt &= \sqrt{\left(\frac{dy}{dx}\right)^2 + 1}= \sqrt{\left(2x\right)^2 + 1}\\ t &= \frac 1 4\left(2\sqrt{4x^2 + 1}x+ \text{sinh}^{-1}(2x)\right) \end{align*}$$ Holonomic was introduced by Hertz in his book on mechanics, and comes from Greek to mean whole law. A holonomic constraint is therefore expressed in terms of the coordinates and not differentials that usually can't be integrated. This is equivalent to $f(q_1,q_2...,q_n,t) = 0$ making your constraint $y - x^2 = 0$ also holonomic. |
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