# Determining force of constraint

• Argelium
It doesn't, but forgive me but I do not understand at all what does that mean for my problem?Are you saying that you have learned about when the Lagrangian does not depend on one of the coordinates explicitly, but you have not seen the Beltrami identity?Yes, I think I understand what you are saying. In that case, the Lagrange equation for ##\lambda## would be$$\lambda = -mg+2bm(x\ddot x+\dot x^2)$$

## Homework Statement

Consider a particle moving over the curve ##z=a-bx^2## under the force of gravity. If the particle starts from rest at point ##(0,0)## (I'm guessing it means point ##(0,a)##), tell if the particle ever separates from the curve; if yes, find the point at which it does.

## Homework Equations

$$\frak{L} = T-U$$

$$\frac{\partial\frak{L}}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial\frak{L}}{\partial \dot q_i}\right)+\sum_{k=1}^n \lambda_k\frac{\partial f_k}{\partial q_i} = 0$$

## The Attempt at a Solution

Well, clearly it's a problem suited for Lagrangian mechanics. We have the coordinates to be ##x, z## and the Lagrangian to be

$$\frak{L} = \frac{m}{2}(\dot x^2+\dot z^2)-mgz$$

Then the Lagrange equations are:

$$m\ddot x+2b\lambda x=0$$

$$-mg-m\ddot z-\lambda = 0$$

Then applying the constraint we obtaint the equations:

$$m\ddot x+2b\lambda x=0$$

$$mg+2bm(x\ddot x+\dot x^2) = \lambda$$[/B]

How do I proceed to obtain ##\lambda##? I'm seriously stuck on here, so I'd appreciate if you could tell whether I am on the right track or not, and if yes, how to proceed.

Thanks!

You also need to use the constraint equation, which is the EL equation for ##\lambda##.

Yea, I wasn’t very explicit on the original post. But the constraint equation is

$$f(x,z)=a-z-bx^2=0$$

Using that and plugging them on the EL equations (of the first kind) we arrive at the last two equations I typed

You have two different differential equations, both involving ##\lambda##. You would typically eliminate ##\lambda## from those and solve the differential equations. Then you can find ##\lambda## by inserting the solution into one of the equations.

However, I suggest that you replace one of your EL equations by a suitable constant of motion.

Orodruin said:
You have two different differential equations, both involving ##\lambda##. You would typically eliminate ##\lambda## from those and solve the differential equations. Then you can find ##\lambda## by inserting the solution into one of the equations.

However, I suggest that you replace one of your EL equations by a suitable constant of motion.

However nothing is a constant of motion in this problem, right?

Argelium said:
However nothing is a constant of motion in this problem, right?
Why would you think so?

If I recall,##\dot q_i## is a constant of motion if ##\frac{\partial L}{\partial q_i}=0## and that is not the case here.

Argelium said:
If I recall,##\dot q_i## is a constant of motion if ##\frac{\partial L}{\partial q_i}=0## and that is not the case here.
This is only a sufficient condition for a constant of motion to exist, not a necessary one.

Hint: Does your Lagrangian depend explicitly on time?

Orodruin said:
This is only a sufficient condition for a constant of motion to exist, not a necessary one.

Hint: Does your Lagrangian depend explicitly on time?

It doesn't, but forgive me but I do not understand at all what does that mean for my problem?

Are you saying that you have learned about when the Lagrangian does not depend on one of the coordinates explicitly, but you have not seen the Beltrami identity?

More generally, both are manifestations of Noether's theorem.