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Constraint Forces and Conservation of energy

  1. Jul 6, 2012 #1
    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 lets 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?
  2. jcsd
  3. Jul 10, 2012 #2
    I'm not so sure what you mean exactly because apparently N is the normal force but then you are adding it in x-direction(if by x you mean horizontal line) . I also think you must differentiate between what you mean by normal force N and ground reaction force R. Ground reaction force R is the sum of normal force N and friction f which is tangent to the moving surface. Therefore, N and f are just components of R in y and x directions (or n and t directions etc).

    The conservation energy is valid only if the sum of the works done by non-conservative forces like friction is zero. That means if R≠N conservation of energy is not working. For it to work we must have R=N. That is, work done by friction is zero.
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