Analytical mechanics: Why do the surfaces of constraint need non-zero gradients?

[apec45]

Hi,
In my course in analytical mechanics, it is said that for a system of n particles subjected to r constraint equations, it is necessary to impose regularity conditions on the constraint surface defined by G = 0 where G is a function of the position of the position of the particles and time, the condition is that the gradient of G is non-zero on the surface G = 0.

I don't understand why we're asking this?

Thanks for your help (and sorry if i made mistakes)

 

[wrobel]

don't understand why we're asking this?

because this condition guarantees that the equations of motion will be correct

For example if you have a particle and the constraint $G=x^2-y^6=0$ on the plane, then the particle will not know where to move from $(0,0)$: it can move along the curve $x=y^3$ or along the curve $x=-y^3$. In this example $\nabla G\mid_{(0,0)}=0$

 

[apec45]

Thanks you for the answer! but i don't understand it very well :(

First, I don't see the link between "the gradiant is zero somewhere" and "there is several ways to move the particule somewhere" (I am ok with your example but i don't see the link).

Then, i don't know what's the problem to have several ways to move the particle, for example if there was no constraints (G = 0 = 0 if i can say that), there would be infinitely many ways to move the particule (anywhere on the plane here) but i don't see the problem in this case, in fact i don't see the link with "the equation of movement has exactly one solution"

Thank you for your help!

 

[wrobel]

A motion of the particle must be determined uniquely by its initial position and velocity. In my example for the initial position take (0,0) and for the initial velocity take $\boldsymbol {e}_y$. So where will the particle move then?

 

[apec45]

Oh okey, i got it! Thank you so much! We don't know where the particule will go (x = y³ or x = -y³).

And could you give me a little bit explanation about the link with the gradiant? :) I really don't see the link with that, or maybe it is that : if G(x, y) = 0 is the equation of the surface then the particule will have, on a point (x, y) of the surface, a velocity proportional to the gradiant of G on (x, y) ? (i mean if we consider a gradiant like a vector field then if we let a particule follow the vector field, we can see its trajectory)

ps: according to me if the gradiant on a point is zero then the particule will stay there and keep that position, it doesn't seem a problem, or am i wrong?

 

[wrobel]

And could you give me a little bit explanation about the link with the gradiant?

Solve the following problem. A particle of mass m slides without friction on a surface $\{f(x,y,z)=0\}$. The function $f$ is given. For simplicity assume that there is no other forces except the force of normal reaction $\boldsymbol{N}.$ Find $$\boldsymbol {N}=\boldsymbol {N}(\boldsymbol {r},\boldsymbol {\dot r}).$$
Here $\boldsymbol r=x\boldsymbol {e}_x+ y\boldsymbol {e}_y+z\boldsymbol {e}_z$ is the radius-vector of the particle in the standard inertial frame $Oxyz$

 

[apec45]

Uhm, i only see that $N = m . a$ (by Newton's second law) and $f(x, y, z) = 0$ (constraint)

 

[apec45]

I also see that if i do that: $N.t = ma.t$ with$t$ tangent to the surface, i get $a*t = 0$