Question

Using Lagrangian multipliers, find the equations of motion for a block sliding down an inclined plane

Answer 1

let us presume that for now, we do not know if the particle can or cannot sink into the surface of the wedge. Hence the presence of the z coordinate. These two coordinates are the ones that really describe the particle motion. Then, if the z coordinate can vary, then the particle height above the ground will be rsinθ−zcosθ. Taking this into account, its potential energy would be:

V=mg(rsinθ−zcosθ)

But, in order to keep the particle on the surface, its zz coordinate must be zero. This leads to the holonomic constraint:

f(z)=z=0

This constraint being holonomic, it enables a constraint potential λf(z)λf(z)which can be included into our initial potential energy:

V(z)=mg(rsinθ−zcosθ)+λz

The Kinetic Energy of the particle is:

T(r˙,z˙)=12m(r˙2+z˙2),

so the motion Lagrangian would in our case be:

L(r,r˙,z,z˙)=12m(r˙2+z˙2)−mg(rsinθ−zcosθ)−λz

This Lagrangian leads to two equations of motion,

(d/dt)*(∂L∂r˙)−(∂L/∂r)=0⇔mr¨=mgsinθ

(d/dt )*(∂L/∂r)˙−(∂L/∂r)=0⇔mz¨+mgcosθ=λ

But, our initial f(z) constraint implies that z¨=0 ,So we end up with the following two equations of motion :

r¨=gsinθ

λ=mgcosθ

where in fact, the second equation of motion corresponds to the force of constraint, where λ is the normal force exerted by the wedge surface on the particle; the first equation corresponds to the sliding motion of the particle.

In the end, the lagrange multiplier λis the constraining force of the sliding motion: the one needed to keep the particle on the surface.