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The example is a thin disk rotating of an inclined plane. The disk can roll not only down the plane, but also "sideways". Let $(x,y)$ be the position of the CM, where the $y$ axis is down the slope and the $x$ axis is horizontal on the plane, the angle the disk has rolled w.r.t its symmetry axis be $\phi$, and the angle of the disk plane made with the $y$ axis be $\theta$. Then the constraints are $$\dot{x}=R\dot{\phi}\sin\theta,\quad \dot{y}=R\dot{\phi}\cos\theta$$ or $$dx=R\sin\theta d\phi,\quad dy=R\cos\theta d\phi$$ Then the author argued that the constraints are non-integrable as follows:

Suppose there were a constraint of the form $$f(x,y,\phi,\theta,t)=0$$ then $$0=df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial \phi}d\phi+\frac{\partial f}{\partial \theta}d\theta+\frac{\partial f}{\partial t}dt$$ Then $$dx=R\sin\theta d\phi, \quad dy=R\cos\theta d\phi$$ implies $$\frac{\partial f}{\partial x}=1,\quad \frac{\partial f}{\partial \phi}=R\sin\theta, \quad\frac{\partial f}{\partial y}=0,\quad \frac{\partial f}{\partial \theta}=0.$$

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I can't see how to get the last conclusion above.

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  • $\begingroup$ $\uparrow$ Which author? $\endgroup$ – Qmechanic May 21 '14 at 14:48
  • $\begingroup$ I am reading the lecture notes by Sunil Golwala from Caltech. $\endgroup$ – velut luna May 21 '14 at 15:01
  • $\begingroup$ Related: physics.stackexchange.com/q/21170/2451 $\endgroup$ – Qmechanic May 21 '14 at 15:06
  • $\begingroup$ I understand the argument in the related post. But still don't understand that in my post. $\endgroup$ – velut luna May 21 '14 at 15:51

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