Force of Constraint for Particle in a Paraboloid

[Izzhov]

Homework Statement

A particle is sliding inside a frictionless paraboloid defined by r^2 = az with no gravity. We must show that the force of constraint is proportional to (1+4r^2/a^2)^{-3/2}

Homework Equations

f(r,z) = r^2-az = 0
F_r = \lambda \frac{\partial f}{\partial r} (and similarly for F_z)

The Attempt at a Solution

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

So you can factor out the a from the square root and get (1+4r^2/a^2)^{1/2} a \lambda but the exponent is supposed to be -3/2, not 1/2. What am I missing? Am I supposed to find a factor of (1+4r^2/a^2)^{-2} in the \lambda somewhere? How would I go about doing that?

 

[cwasdqwe]

Hello izzhov,
You calculated the module of F. Perhaps the force of constrain is proportional to (1+4r^2/a^2)^{-3/2} since \hat{F}=\frac{\vec{F}}{|F|}. (\lambda should be a number, independent of r).

 

[BvU]

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!)

 

[cwasdqwe]

I do not understand the cwasd post (nice password candidate, this cwasdwqe combination!)

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