degree of freedom of a rigid body 5 or 6? I'm confused here. I have a three particle (rigid) system. What would be the degree of freedom? I found out five. 3 coordinates for center of mass and 2 for describing orientation.
But we have only three constraints, i.e. three equations that reduce 9 coordinates by 3, 9 - 3 = 6, which gives 6 degrees of freedom?? Did I miss something above?
 A: Every rigid body has 3 translational dof. In addition, there are 0, 2, or 3 rotational dof, depending on the geometry, giving a total of 3, 5, or 6 dof.
A spherically symmetric rigid body has no rotational dof.
A rigid body with rotational symmetry around an axis has 2 rotational dof, namely two angles for orienting the symmetry axis along a direction.
All other rigid bodies have 3 rotational dof, namely two angles with respect to an arbitrary axis attached to the body, and an angle for rotationg around this axis. This gives the Euler angle parameterization of the manifold of orientations (algebraically an $SO(3)$.) An important alternative parameterization is the quaternion parameterization, especially useful in computational geometry. It has a parameter vector $u$ with 4 components whose length is 1, leaving 3 dof. ($u$ and $-u$ describe the same rotation.)
A: You missed that to specify orientation, you not only need an axis, but how far the body rotated around the axis. The two axis angles and the angle of rotation is the Euler angle parametrization, and I find it unweildy because the relation between this and position involves transcendental functions.
The nicest way to give the rotation part is to specify a rotation matrix R which has the property that $R^TR=I$. This has 3 parameters, since you have three orthogonal unit vectors inside, which is 2 components for the first (it's unit length), one component for the second (it's perpendicular to the first and unit length) and none for the third. This is most convenient for pencil and paper and computer calculations both, this is why it is hardly ever presented in textbooks.
