# Mechanics Goldstein, chpt 1 exercise 11, Lagrangian of rolling disk

1. Feb 18, 2012

### dracobook

1. The problem statement, all variables and given/known data
I apologize if this is not the right place to put this. If it is not please redirect me for future reference.

11. Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk.

a. Derive Lagrange's equations and find the generalized force.

2. Relevant equations
$$\mathbb{L} = T- U$$
$$\frac{d}{dt}( \frac{\partial T}{\partial \dot q_j}) - \frac{\partial T}{\partial q_j} = Q_j$$

$$\frac{d}{dt}(\frac{\partial \dot L}{\partial \dot q_j}) - \frac{\partial L}{\partial q_j} = 0$$
$$Q_j$$ is the generalized forces
3. The attempt at a solution

$$\dot x = \dot\theta R$$, where R is radius of disk.
$$T = \frac{1}{2} m v^2 + \frac{1}{2}I\dot\theta^2 \text{, }V=0\text{, } I=\frac{mR^2}{2} \Rightarrow \mathbb{L} = \frac{3}{4}m \dot x^2$$
I would assume then that I would just need to plug in $$\mathbb{L}$$ into the euler Lagrange equation...however it seems that
$$\frac{d}{dt}(\frac{\partial \dot L}{\partial \dot q_j}) - \frac{\partial L}{\partial q_j} = Q \neq 0$$
So my question is:
1) Why is this so? Is this because the potential is zero? Hence the generalized force cannot be derivable from a scalar potential? and hence we can only reduce it to
$$\frac{d}{dt}(\frac{\partial \dot T}{\partial \dot q_j}) - \frac{\partial T}{\partial q_j} = Q_j$$
And so does that mean that whenever potential is zero we can only reduce the problem with the equation above?
2) Where does $$\frac{1}{2}I\dot\theta^2$$ come from in the equation for kinetic energy??

Last edited: Feb 19, 2012
2. Feb 18, 2012

### fluidistic

I can't really help you for this problem, however I've spotted an error (maybe just a typo?) when you wrote the generalized forces: $\frac{d}{dt}( \frac{\partial \dot T}{\partial \dot q_j}) - \frac{\partial T}{\partial q_j} = Q_j$. The first kinetic energy term shouldn't have a "dot"; or if you dot it, it should have the d/dt term in front of it.
Edit: My Goldstein's book 1st edition doesn't have such an exercise.
Your Lagrangian $\mathbb{L} = \frac{3}{4}m \dot x^2$ is clearly wrong. It should be worth T.
It's a rotating rigid body. Hence its kinetic energy is $\frac{mv^2 _{\text{center of mass }}}{2}+\frac{I\omega ^2}{2}$. Are you sure you didn't make a typo? I think it should be $\frac{I \dot \theta ^2}{2}$.

Last edited: Feb 18, 2012
3. Feb 18, 2012

### dracobook

Thanks for pointing on the mistakes and your response! My Lagrangian is actually not one and is equal to T if you plug everything in and make proper substitutions

4. Feb 18, 2012

### fluidistic

You're welcome. I see for your Lagrangian, it's probably right then.
About your question 1), I don't really know. (I'm a student like you who needs to take the final exam of classical mechanics).
$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} \neq 0$ if you don't include the forces involved into the potential energy (as you know, $F=-\nabla V$).
However if you include the forces into the potential energy then I think you'll reach $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} = 0$. However I suggest you to wait for someone else to reply... more experienced than me. :)

5. Feb 19, 2012

### sgd37

you need to include a potential due to the applied force. Do you know anything about the nature of this force i.e. is it constant?

6. Feb 19, 2012

### dracobook

nope. No other information is given .

7. Feb 19, 2012

### vela

Staff Emeritus
The generalized force is given by
$$Q_j = \sum_i \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j}$$ If the forces Fi can be written in terms of a potential V, the generalized force can be written as
$$Q_j = -\sum_i \nabla_i V \cdot \frac{\partial \vec{r}_i}{\partial q_j} = -\frac{\partial V}{\partial q_j}.$$ Therefore, you have
$$\frac{d}{dt}\left( \frac{\partial T}{\partial \dot q_j}\right) - \frac{\partial T}{\partial q_j} = -\frac{\partial V}{\partial q_j}$$ When you move everything to the lefthand side, you get
$$\frac{d}{dt}\left( \frac{\partial T}{\partial \dot q_j}\right) - \frac{\partial (T-V)}{\partial q_j} = 0$$ Since V doesn't depend on the velocities, you can replace T in the first term by T-V, which then yields the familiar Euler-Lagrange equations with L = T-V.

8. Feb 19, 2012

### dracobook

Thanks for the derivation. However I think I am mostly confused about whether the Euler-Lagrange holds in this problem, that is whether the LHS of your last equation with L in place of T is equal to Q or 0, and if it is Q why? If it is Q (which I have been led to believe by other people) I wonder if it has to do with the potential being zero..and if that implies that the force cannot be written in terms of a potential. I think this is a nonholonomic problem...not exactly sure why though) Can the forces be written in terms of a potential in this problem?

9. Feb 19, 2012

### sgd37

E-L equations always equal to zero, and remember these are differential equations just the same as x' = 0 means x is a constant

10. Feb 19, 2012

### vela

Staff Emeritus
It has to be Q because the applied force hasn't been incorporated into the potential.

11. Feb 19, 2012

### sgd37

Vela's last equation has to equal to zero since it is the euler lagrange equation basically your lagrangian should read

$L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 - V(x)$

which give you simply

$\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}) = \frac{\partial L}{\partial x} = - \frac{d V(x)}{d x} = Q$

so now you can find the generalized force

12. Feb 19, 2012

### dracobook

Ah I see thank you. Would it be safe to assume that the force cannot be expressed as a potential in this case? and is therefore not conservative?

13. Feb 19, 2012

### vela

Staff Emeritus

14. Feb 19, 2012

### fluidistic

I don't want to hijack this thread so you can ignore me...
I understand this vela.
However I have a doubt now. Why do we call the Lagrangian in this case to T-V, regardless if V has the potential of the force included or not?
This is like... if the Lagrangian is L=T then the Euler-Lagrange equations yield $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} =Q_i$ while if the Lagrangian takes into account all forces into the potential, L=T-V, the Euler-Lagrange equations yield $\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} = 0$. So all in all the Lagrangian isn't uniquely defined, up not to a constant but a function (potential function)?

15. Feb 19, 2012

### vela

Staff Emeritus
It's a question about how you want to deal with the external forces. If a force is non-conservative, you have no choice. It has to show up on the righthand side. If it is conservative, you can keep it on the righthand side and omit it from the potential, but what's the point?

It's similar to when you solve problems using energy. If a force is conservative, you have a potential you can work with, but you can also not use the potential and calculate the work done by the force explicitly. Either way, you'll get the same answer. It's just that if you have a conservative force, using the potential typically makes things easier.

16. Feb 19, 2012

### fluidistic

I see, thank you very much vela for the clarification.

17. Feb 19, 2012

### dracobook

I would like to thank both of you for clarifying this up for me.

18. Sep 23, 2012

### Burkay

Hi there,*
I want to reply to your force in this question. I hope I am not too late to answer :) since i didnt check the day of the post. To express the force you apply you must convert it to your degree of freedom. E.g. if dof is θ then it should be like F*r where r is the radius of the disk. If your dof is x, then it is simply x. Simply you must carry the Equilavent forces to your d.o.f. of the system.

19. Sep 23, 2012

### Burkay

I checked your equation again. Since your equations give Force equilibrium, not moment. It is simply F, whatever your F is.