# Homework Help: Force of Constraint for Particle in a Paraboloid

1. Oct 1, 2014

### Izzhov

1. The problem statement, all variables and given/known data
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 $(1+4r^2/a^2)^{-3/2}$

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

3. 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?

2. Oct 3, 2014

### 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).

3. Oct 5, 2014

### 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!)

4. Oct 5, 2014

### cwasdqwe

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