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

 

[paranormal]

First, let's review the concept of force of constraint. In this scenario, the particle is constrained to move along the surface of the paraboloid, meaning it cannot move in any other direction. This constraint is represented by the function f(r,z) = r^2-az = 0, which describes the paraboloid surface. The force of constraint is then the force required to keep the particle on this surface.

To find the force of constraint, we can use the Lagrange multiplier method, which involves taking the partial derivatives of f(r,z) with respect to r and z, and setting them equal to some constant value, denoted by λ in this case. This allows us to incorporate the constraint into the equations of motion.

Following this method, we get F_r = λ(2r) and F_z = λ(-a). The total force of constraint is then F = √(F_r^2+F_z^2) = √(4r^2λ^2+a^2λ^2) = √(λ^2(4r^2+a^2)) = λ√(4r^2+a^2).

Now, we can factor out an a from under the square root, giving us F = λa√((4r^2+a^2)/a^2) = λa√(1+(4r^2/a^2)).

Finally, taking the inverse square root and multiplying by a, we get F = λa(1+4r^2/a^2)^{-3/2}, which matches the desired form of the force of constraint. Therefore, we have successfully shown that the force of constraint is proportional to (1+4r^2/a^2)^{-3/2}.