[Physics] Degree of freedom in Lagrange’s formalism

classical-mechanicsdegrees of freedomlagrangian-formalismrigid-body-dynamics

Degrees of freedom $=3K-N$ where $K$ is number of particles and $N$ is number of constraints. How to find the number of degrees of freedom for a rigid body which has both translation and rotation, since the rigid body has infinite particles as Degree of freedom is defined for particles and there are infinite number of particles in rigid body?

Best Answer

The degree of freedom of a rigid body in 3-dimensional space is always SIX. I will try to explain this logically by a simple example.

First of all, we need to know what actually is a rigid body. The most basic definition is that "rigid body is an object in which all the constituent particles are at the same relative separation." This means that all the particles are (rigidly) fixed at their relative positions.

Now let us start with a Cartesian coordinate system and a single particle, and follow these steps:

  1. Fix the particle in the origin of the coordinate system. For this single particle, the no. of independent parameters is 3, corresponding to motion along the three directions.
  2. Now we bring a second particle and attach it with the first particle at a constant separation (say along the x- axis). Now the second particle cannot move along the x- axis as it is rigidly fixed along the first particle along that direction. For this second particle, the no. of independent parameters is (3-1)=2, corresponding to motion only along the y- and z- directions.
  3. Now we bring a third particle and attach it rigidly with the other two particles (say in the xy- plane). Now the motion of this particle is restricted along the lines joining the other two particles (i.e. in the xy- plane where the rigid 3-body structure is formed). For this third particle, the no. of independent parameters is (3-2)=1, corresponding to motion only along the z- direction.

Thus, the d.o.f. of this 3-body structure is (3+2+1)=6. These 6 d.o.f. actually consists of 3 translational d.o.f. and 3 rotational d.o.f. of the structure as a whole.

Notice that I hadn't considered the rotational degrees of freedom. This is due to the fact that I had considered the individual particles which are actually constituents of a rigid body and those individual constituents cannot have rotational motion. Rotational motion exists only for the rigid body as a whole. On the contrary, translational degrees of freedom are required to specify the position of the constituent particles.

Till now I had considered only 3-body system. So what about a rigid body that actually consists of infinite no. of particles?

The answer to this question lies in the fact that when I discussed the 3-body structure, I had fixed the first particle at the origin, and the positions of the other two particles are completely arbitrary. In this way, in any rigid structure we can choose a single particle as the reference particle and all other particles can be reached by 3-body structures and evaluation of the d.o.f. for all such 3-body structures shows that only one of them contributes to the d.o.f., all the others being redundant.

So, the d.o.f. of a rigid body is 6; translational d.o.f.=3 and rotational d.o.f.=3.

Related Question