Homework Help: Falling disk with string

1. Dec 1, 2016

Vrbic

1. The problem statement, all variables and given/known data
Let's have a disk of mass $m$ and radius $a$ and massless rope tangled in it. One end of rope is tied to the ceiling and the disk is falling freely down. System has one degree of freedom. As a coordinate we can choose angle $\phi$ which says an angle of rotation from the start position. Find from the Lagrange equations of the first kind equation of motion and binding force.

2. Relevant equations
$$m\ddot{x}=F_x+\lambda\frac{\partial G}{\partial x}$$
$$m\ddot{y}=F_y+\lambda\frac{\partial G}{\partial y}$$
where G is binding function.

3. The attempt at a solution
I see that kinetic translation and rotation energy is connected. Probably by $y=-a \phi$ ($\dot{y}=-a\dot{\phi}$ ) but generally I don't know what is my G function. I have seen a problem with pendulum or problem falling from ball where this functions are easy to guess. Please give me a kick.

Thank you.

2. Dec 1, 2016

TSny

For this problem, you are working with the two variables $\phi$ and $y$ rather than $x$ and $y$.

You can rearrange your constraint condition as $y + a \phi = 0$. Thus, you can choose $G( \phi, y) = y + a \phi$.

3. Dec 1, 2016

haruspex

Do you mean wrapped around it?

4. Dec 2, 2016

Yes.

5. Dec 2, 2016

Vrbic

Ok I have some attemp of solution but I mean the result is wrong:
$G=y+a\phi=0$ ,+ because of it goes against y direction
$m\ddot{y}=F_g+\lambda\frac{\partial G}{\partial y}=-mg +\lambda$
$m\ddot{\phi}=0+\lambda\frac{\partial G}{\partial \phi}=\lambda a$
Now I want find out $\lambda$ =>
$\ddot{y}+a\ddot{\phi}=0$
$\ddot{y}=-g+\frac{\lambda}{m}$
$a\ddot{\phi}=\frac{\lambda a^2}{m}$
________________________________
$\ddot{y}+a\ddot{\phi}=0=-g+\frac{\lambda (a^2+1)}{m}$ => $\lambda=\frac{gm}{a^2+1}$
$m\ddot{y}=-mg +\frac{gm}{a^2+1}$
$m\ddot{\phi}=\frac{gm}{a^2+1} a$

6. Dec 2, 2016

TSny

The last equation (for $\ddot {\phi})$ is not correct. $\phi$ is an angular coordinate rather than a Cartesian coordinate, like $x$ or $y$.
Do you know how to get the equation of motion for $\phi$ starting from the Lagrangian?

7. Dec 5, 2016

Vrbic

It seems very reasonable. I believe I can manipulate with lagrangian :) For me is more familiar deal with Lagrange equation of second kind. So in this point of view I take a Lagrangian $L=T+V=T_t+T_{rot} - V=\frac{1}{2}m\dot{y}^2 + \frac{1}{2}J\dot{\phi}^2 - mgy=\frac{1}{2}m\dot{y}^2 + \frac{1}{4}m a ^2\dot{\phi}^2 - mgy$ =>
$\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}} - \frac{\partial L}{\partial \phi} =\frac{1}{2}m a ^2\ddot{\phi}=0$. $\phi$ isn't there but I feel it should be there probably due the constrain...or I don't know.
On the other hand in problem is written they want Lagrange equation of 1. kind...

8. Dec 5, 2016

TSny

Lagrange's equation for $\phi$ with constraint $G(\phi, y) = 0$ is $$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi} - \frac{\partial L}{\partial \phi} = \lambda \frac{\partial G}{\partial \phi}$$

9. Dec 5, 2016

Vrbic

I see... so my equations of 1. kind are
$m\ddot{y}+mg=\lambda$
$\frac{ma^2\ddot{\phi}}{2}=\lambda a$
$y+a\phi=0$ do you agree?

10. Dec 5, 2016

Vrbic

$\ddot{y}=-g+\lambda/m$
$a\ddot{\phi}=2\lambda/m$
$\ddot{y}+a\ddot{\phi}=-g+3\lambda/m=0$=>$\lambda=gm/3$, $\ddot{\phi}=\frac{2g}{3a}$ and $\ddot{y}=-\frac{2g}{3}$.
A binding force is $\vec{F}_B=\lambda(\frac{\partial G}{\partial y},\frac{\partial G}{\partial \phi})$ ?