Equations of motion - lagrange

Question

A mass point of mass m moves on the circle $x^2+y^2=R^2$ and $z=0$. No external forces are acting. Solve the equation of motions and determine the constraint force with the lagrange equations of first kind.

I'm studying for my exams and since our professor was on an excursion for a week we had to learn about the lagrange formalism by ourselves. I can't say that it's easy since I'm stuck on the lagrange equations of first and second kind. Just reading the definitions didn't make it any clearer for me, so I tried myself on some of the problems involving the use of those equations but I'm already stuck on the first one.

How do I apply the definitions of the first kind to solve this type of question?

I was thinking that since there are no external forces that a whole bunch of the equations of the first kind would be $0$, but I still don't how to go from there.

Could anyone help me out here?

Answer 1

Here is a (way too) detailed walkthrough of this problem where I try to explain all the steps that you need to apply in order to solve such a problem using the Lagrange method. I will in this answer use the formalism laid out here where the constraints are separated out from the Lagrangian. This is different than what I mentioned in the comments above, but it's exactly the same (it's just a different packaging).

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We always start by defining the Lagrangian and the constraints in the 'language' of the Lagrange method. The free Lagrangian for a point particle in 3D is given by

$$L = \frac{1}{2}m\left[\dot{x}(t)^2 + \dot{y}(t)^2 + \dot{z}(t)^2)\right]$$

where $\dot{x} \equiv \frac{dx}{dt}$. The constraint that the motion takes place in the plane, $x^2+y^2 = R^2$ with $z=0$, can be denoted by $H(x,y,z) = G(x,y,z) = 0$ where

$$G(x,y,z) = x^2 + y^2 - R^2$$ $$H(x,y,z) = z$$

The last constraint is trival in the sense that we can deal with it straight away since $z=0\to \dot{z} = 0$ so the $z$ dependence falls out and we can exclude $z(t)$ from the Lagrangian and effectively reducing the problem to a 2D problem with a single constraint $G(x,y) = x^2 +y^2 - R^2 = 0$.

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After we have made the setup of the problem we need to calculate the Lagrange equations. The Lagrange's equations of the first kind are given by

$$\frac{dL}{dx} - \frac{d}{dt}\left[\frac{dL}{d\dot{x}}\right] - \lambda \frac{dG}{dx} = 0$$ $$\frac{dL}{dy} - \frac{d}{dt}\left[\frac{dL}{d\dot{y}}\right] - \lambda \frac{dG}{dy} = 0$$

To evaluate it we need the following terms $$\frac{dL}{dx} = \frac{dL}{dy} = 0,~~~~\frac{dL}{d\dot{x}} = m\dot{x},~~~~\frac{dL}{d\dot{y}} = m\dot{y},~~~~\frac{dG}{dx} = 2x,~~~~\frac{dG}{dy} = 2y$$

Note that $\dot{x},\dot{y},x$ and $y$ are all threated as independent variables in the derivation so for example $\frac{d\dot{x}}{dx} = 0$ (which might seems strange at first).

Inserted into the Lagrange equations we get equations of motion for the particle

$$m\ddot{x} + 2\lambda x = 0$$ $$m\ddot{y} + 2\lambda y = 0$$

We can phrase this as a vector equation, by introducing $\vec{r} = (x,y)$, as

$$m\vec{a} = -2\lambda \vec{r}$$

where $\vec{a} = \frac{d^2\vec{r}}{dt^2} = (\ddot{x},\ddot{y})$ is the acceleration vector. From this form, and remembering Newton's second law, we see that the constraint $x^2 + y^2 = R^2$ that we imposed have given rise to a central-force $\vec{F} = -2\lambda\vec{r}$ on the particle. We still need to solve for $x,y$ and determine $\lambda$.

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To solve for the evolution $(x(t),y(t))$ of the system we need to solve the Lagrange equations we derived above. The differential equations for $x$ and $y$ should be well known as those of a harmonic oscillator. The solutions reads

$$x(t) = A_x\sin(\omega t) + B_x\cos(\omega t)$$ $$y(t) = A_y\sin(\omega t) + B_y\cos(\omega t)$$

where $A_x,A_y,B_x,B_y$ are constants and $\omega^2 = \frac{2\lambda}{m}$. Note that $\lambda\geq 0$ is enforced (as we shall see below). To complete the solution we must enforce the constraint $x^2 + y^2 = R^2$ and put in the initial conditions.

Since the initial conditions are not given we simply pick our coordinate-system such that $x(0) = R$ giving $y(0) = 0$ (this is always possible). If you apply the constraint equation you will find $\dot{x}(0) = 0$ and find that the solution is characterized by only one free parameter $\dot{y}(0) \equiv v$. The constraints gives us $v = R\omega \implies 2\lambda = \frac{mv^2}{R^2}$. The full solution can now be written

$$x(t) = R\cos(\omega t)$$ $$y(t) = R\sin(\omega t)$$

which describes a particles moving around the circle with constant angular velocity. Since we have determined $\lambda$ in terms of $v$ we can also write the constraint-force in terms of $v$ as

$$\vec{F} = -\frac{mv^2}{R}\frac{\vec{r}}{R}$$

This is the well known expression for the centripetal force that makes a particles follow a circular path.

Answer 2

picking up from where Winther left off and noting that $\vec{F} = (-\lambda x, -\lambda y)$ we can use the DE and the constraint to recover the result

$$\vec F = -\frac{mv^2}{R} \hat r = -\frac{mv^2}{R^2}(x,y) $$

solving the DE's gives $$x(t) = A \sin \left (\sqrt{\frac{\lambda}{m}}t+\delta_x \right) $$

$$y(t) = B \cos \left (\sqrt{\frac{\lambda}{m}}t+\delta_y \right) $$

the constraint $x^2 + y^2 = R^2$ gives us that $A=B=R$ and that $\delta_x=\delta_y \equiv \delta$ where $\delta = \sin^{-1}\left(\frac{x(0)}{R} \right)= \cos^{-1}\left(\frac{y(0)}{R} \right) $

now we can write

$$ \dot x(t) = R\sqrt{\frac{\lambda}{m}} \sin \left (\sqrt{\frac{\lambda}{m}}t+\delta \right) $$

$$ \dot y(t) = R\sqrt{\frac{\lambda}{m}} \cos \left (\sqrt{\frac{\lambda}{m}}t+\delta \right) $$

so that

$$ {\dot x}^2(t) + {\dot y}^2(t) = \frac{R^2 \lambda}{m} \equiv v^2 $$

which is a constant of motion $v^2 \equiv {\dot x}^2(0) + {\dot y}^2(0)$

So now we can solve to get $\lambda = \frac{mv^2}{R^2}$ which is what is required for an $\frac{mv^2}{R}$ central force.