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Force of Constraint for Particle in a Paraboloid

  1. Oct 1, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle is sliding inside a frictionless paraboloid defined by [itex]r^2 = az[/itex] with no gravity. We must show that the force of constraint is proportional to [itex](1+4r^2/a^2)^{-3/2}[/itex]

    2. Relevant equations
    [itex]f(r,z) = r^2-az = 0[/itex]
    [itex]F_r = \lambda \frac{\partial f}{\partial r}[/itex] (and similarly for [itex]F_z[/itex])

    3. The attempt at a solution
    [itex]F_r = \lambda \frac{\partial f}{\partial r} = 2r\lambda[/itex] and [itex]F_z = \lambda \frac{\partial f}{\partial z} = -a\lambda[/itex] and hence the total force of constraint is [itex]F = \sqrt{F_r^2+F_z^2} = (4r^2+a^2)^{1/2}\lambda[/itex].

    So you can factor out the a from the square root and get [itex](1+4r^2/a^2)^{1/2} a \lambda[/itex] but the exponent is supposed to be -3/2, not 1/2. What am I missing? Am I supposed to find a factor of [itex](1+4r^2/a^2)^{-2}[/itex] in the [itex]\lambda[/itex] somewhere? How would I go about doing that?
     
  2. jcsd
  3. Oct 3, 2014 #2
    Hello izzhov,
    You calculated the module of F. Perhaps the force of constrain is proportional to [itex](1+4r^2/a^2)^{-3/2}[/itex] since [itex]\hat{F}=\frac{\vec{F}}{|F|}[/itex]. ([itex]\lambda[/itex] should be a number, independent of r).
     
  4. Oct 5, 2014 #3

    BvU

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    The ##\lambda## is the Lagrange multiplier in the Euler-Lagrange equations, extended with the constraints (see a treatise on EL / Hamilton's principle, e.g Goldstein, Classical Mechanics -- or, as I suspect, your lecture notes) $${d\over dt}{\partial L\over \partial \dot q_k} - {\partial L\over \partial q_k}=\lambda a_k.$$That's three eqations in four variables (the coordinates and ##\lambda##). The fourth equation is ##{df\over dt} = 0##. ##f## is the constraint equation and df = ##\sum {\partial f\over \partial q_k}\dot q + {\partial f\over \partial t}## (the very last one being 0).

    You need to solve for ##\lambda## only, since you aren't interested in the coordinates. Perhaps knowing that ##L = T## and choosing a nice set of ##q_k## makes it easier. Your choice is perhaps fine, but you do need to work out ##T##.

    I do not understand the cwasd post (nice password candidate, this cwasdwqe combination!)
     
  5. Oct 5, 2014 #4
    Yep, I was definitely thinking outside the box, feeling sleepy. You're right with Euler-Lagrange.
    Cwasdqwe's too weak for a proper password... :w
     
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