Classical Mechanics (Lagrangian)

[firemarsh]

Homework Statement

A ball is sitting on a frictionless seesaw with no inclination at the beginning, and a constant angular velocity $$\phi$$. Find the position of the ball as a function of time

Homework Equations

L=T-V, T=(m$$\dot{}x$$²+m$$\dot{}y$$²)/2, V=mgy

The Attempt at a Solution

The first problem I run into is the dimension of this system, I thought it was a 1 dimension system at first but then suspect it is 2 dimension.

Anyway I take r (distance from pivot pt. to the ball) and $$\theta$$ as the generalized coordinates.
So I have x=rcos$$\theta$$, y=rsin$$\theta$$, which makes T=m($$\dot{}r$$²+$$\dot{}\theta$$²r², and V=mglsin$$\theta$$, and results in Lagrangian:
L=m($$\dot{}r$$²+$$\dot{}\theta$$²r²-mglsin$$\theta$$

obviously $$\dot{}\theta$$ is the constant $$\phi$$, so...I am confused right here, how do I continue?

 

[kuruman]

Homework Statement

A ball is sitting on a frictionless seesaw with no inclination at the beginning, and a constant angular velocity $$\phi$$. Find the position of the ball as a function of time

I suspect the constant angular velocity is $$\dot{\phi}$$ (phi-dot) not just $$\phi$$.

 

[firemarsh]

I suspect the constant angular velocity is $$\dot{\phi}$$ (phi-dot) not just $$\phi$$.

Does that matter? I thought the useful information here is that augular velocity is constant?

 

[kuruman]

Does that matter? I thought the useful information here is that augular velocity is constant?

It doesn't really matter, you can call the constant angular velocity "Fred" if you so prefer. It's just that denoting an angular velocity with a symbol that is traditionally an angle might be confusing. So let's call the constant angular velocity ω and move on. You need to derive the equation(s) of motion by using the Euler-Lagrange equation.

 

[firemarsh]

Ok, using $$r$$ and $$\theta$$ as the coordinates, I get

$$x=-rcos \theta , y=-rsin \theta,$$
and

$$T=m(\dot{r}$$²+r²$$\dot{\theta}$$²)/2

$$V=-mgrsin\theta$$

$$L=m(\dot{r}$$²$$+r$$²$$\dot{\theta}$$²)$$/2+mgrsin\theta$$

and $$\frac{\partial L}{\partial r}-\frac{d}{dt} \frac{\partial L}{\partial\dot{r}}=mr\dot{\theta}$$²+$$mgsin\theta-m\ddot{r}=0$$

and $$\frac{\partialL}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=mgrcos\theta-mr$$²$$\ddot{\theta}+2m\dot{r}\dot{\theta}=0$$

and substitute $$\dot{\theta}$$ with $$\omega$$ and $$\ddot{\theta}$$ with 0

Is that how I do it?

 

[firemarsh]

But I am still not sure if I really need 2 degrees of freedom in this system

 

[kuruman]

But I am still not sure if I really need 2 degrees of freedom in this system

You do need two degrees of freedom because you need two numbers to describe the position of the ball uniquely as a function of time, i.e. find r(t) and θ(t). Also, your theta equation is incorrect. You are given that $$\dot{\theta}=\omega=constant$$, so what is θ(t)?

 

[firemarsh]

$$\theta$$(t) would be $$\omega$$t

You are right, the theta equation is wrong,

For theta, $$mgr\dot{\theta}cos\theta-2mr^2\ddot{\theta}+2mr\dot{r}\dot{\theta}=0$$, which can be simplified to $$gr\omega cos(\omega t)+2r\dot{r}\alpha=0$$

For r, $$r\omega^2+gsin(\omega t)=0$$

Is it possible to further simplify the equations? Say eliminate the $$\dot{r}$$ term?

 

[kuruman]

$$\theta$$(t) would be $$\omega$$t

You are right, the theta equation is wrong,

For theta, $$mgr\dot{\theta}cos\theta-2mr^2\ddot{\theta}+2mr\dot{r}\dot{\theta}=0$$, which can be simplified to $$gr\omega cos(\omega t)+2r\dot{r}\alpha=0$$

For r, $$r\omega^2+gsin(\omega t)=0$$

Is it possible to further simplify the equations? Say eliminate the $$\dot{r}$$ term?

What do you mean "for theta"? You already know that θ=ωt. There is nothing more to be done with that. Just replace θ with ωt in the radial equation and solve to find r(t).