- #1
tut_einstein
- 31
- 0
Suppose you are trying the solve the equation of motion of say a particle constrained to move on a surface f(x[itex]\vec{}[/itex],t)=0. The equation of motion is:
mx[itex]\ddot{}[/itex] = F[itex]\vec{}[/itex] + N[itex]\vec{}[/itex], where F is an known external force and N is the unknown constraint force.
Now, when you assume that N always perpendicular to the surface, all classical mechanics books motivate that assumption by saying that it's for calculational convenience because N can in principle have any component parallel to the surface without violating the constraint. So, we just get rid of that degree of freedom by saying N = [itex]\lambda[/itex](t) *grad(f), where lambda is an arbitrary lagrange multiplier. this also let's us solve for four unknowns using four equations.
However, we also know that the assumption that N is always perpenidcular to the surface has a physical interpretation that energy is always conserved if F is derivable from a time independent potential and the surface doesn't have any explicit time dependence.
My question is whether it is possible to have energy conservation without assuming N is always perpendicular to the surface. Or did the assumption just happened to correspond to what actually happens?
mx[itex]\ddot{}[/itex] = F[itex]\vec{}[/itex] + N[itex]\vec{}[/itex], where F is an known external force and N is the unknown constraint force.
Now, when you assume that N always perpendicular to the surface, all classical mechanics books motivate that assumption by saying that it's for calculational convenience because N can in principle have any component parallel to the surface without violating the constraint. So, we just get rid of that degree of freedom by saying N = [itex]\lambda[/itex](t) *grad(f), where lambda is an arbitrary lagrange multiplier. this also let's us solve for four unknowns using four equations.
However, we also know that the assumption that N is always perpenidcular to the surface has a physical interpretation that energy is always conserved if F is derivable from a time independent potential and the surface doesn't have any explicit time dependence.
My question is whether it is possible to have energy conservation without assuming N is always perpendicular to the surface. Or did the assumption just happened to correspond to what actually happens?