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I gave a mental ability test yesterday in which this question was asked.

A rolling coin on a flat surface has how many degrees of freedom?

I read about degrees of freedom but I haven't had much interaction with mechanics after high school(4 years ago), can someone please explain on how to solve this question?

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There is one condition, namely that the height coordinate $z$ of the lowest point of the coin is $z_{\rm lowest}=0$. This reduces the $3+3=6$ degrees of freedom of the coin by one (because there was one condition) to five. Note that those $3+3$ I started with are 3 degrees of freedom from a position of the object's center-of-mass; and 3 degrees of freedom parameterizing a rotation in $SO(3)$ determining the orientation (i.e. two angular coordinates to determine the axis; plus one angular coordinate to determine the right rotation around the axis).

Equivalently, one may parameterize these degrees of freedom in many ways. For example, there are 2 horizontal coordinates of the center-of-mass of the coin. And one may also determine the longitude and latitude (2) of the direction of the axis of the coin (directed from tails to heads) plus the right rotation angle $\phi$ around this axis. Again, $2+3=5$. Everything else (namely the right amount of "vertical translation") is determined by saying that the coin sits on the table at a given moment. (I've changed the counting from the first version of the answer, the newer one seems more logical to me.)

Note that these are five x-like degrees of freedom. Each of them has a velocity or a momentum so the "phase space" has $2 \times 5 = 10$ dimensions.

As the Ghost of Perdition mentioned in the comments, one reduces the degrees of freedom if the coin is preventing from slipping. For a general change of the 5 x-like coordinates above, the coin was slipping by $\Delta x$ and $\Delta y$ in the two horizontal directions. Setting them to zero removes 2 x-like coordinates and also their velocities. So one ends up with 3 degrees of freedom - and the phase space is locally 6-dimensional.

However, there's no natural way to extend this 6-dimensional space to all times because the precise location and orientation that is allowed at a given moment depends on the whole history. If one wants to include all possible positions and orientations of the coin into which the coin can get by a history, he still needs the 10-dimensional phase space. It's only true that for the dynamics around some point in time, the no-slipping constraint will effectively reduce the 10-dimensional phase space to a 6-dimensional one.

In other words, we need a 5-dimensional configuration space with 2 (no-slipping) constraints or Lagrange multipliers that could be used to reduce 5 to 3 if these constraints were simple and "integrable". However, they are not, so the reduction or the "simplification" of the problem (elimination of the constraints along with some coordinates) cannot be done.

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    $\begingroup$ Please correct me if I'm wrong, if the coin doesn't slip, doesn't it reduce the degrees of freedom by 1, because the angle of rotation of the coin and the horizontal distance covered aren't independent, also since the coin can rotate about one of the horizontal axes DOF should reduce by 1 more. Where am I wrong with my understanding? $\endgroup$ – Oswald May 29 '16 at 6:21
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    $\begingroup$ Yes, I agree with that. If one imposes the condition that the change in time $\Delta t$ involved no slipping, he removes 2 coordinates of the phase space - basically the two horizontal coordinates $x,y$ - because it's determined how far in $x,y$ the coin has gotten by analyzing the changes of the 3 angular coordinates. - However, this is just a local analysis at a moment of time. The actual set of positions+orientations that can be reached by some history of rolling is still the 10-dimensional phase space one obtains by neglecting the no-slipping condition. $\endgroup$ – Luboš Motl May 29 '16 at 6:23
  • $\begingroup$ So the configuration space is 5-dimensional with 2 constraints but the constraints can't be "integrated" so there's no natural or easy way to reduce the 5 coordinates down to 3 that could easily describe the coin at all times. $\endgroup$ – Luboš Motl May 29 '16 at 6:30

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