# A wheel rolls down a flat inclined surface

1. Oct 30, 2013

### alefisico

1. A wheel rolls down a flat inclined surface that makes an angle $\alpha$ with the horizontal. The wheel is constrained so that its plane is always perpendicular to the inclined plane, but it may rotate about the axis normal to the surface. Obtain the solution for two-dimensional motion of the wheel, using Lagrange's equations and the method of undetermined multipliers.

2. Relevant equations

3. The attempt at a solution

The kinetic energy has two components in a two dimensional motion:

T = T_{trans} + T_{rot} = \frac{m \dot x^2}{2} + \frac{m r^2 \dot \theta^2}{2} \label{eq:301}

and the potential energy:

V = mg y = mg x \sin \alpha \label{eq:302}

Therefore, the Lagrangian is given by:

L = \frac{m \dot x^2}{2} + \frac{m r^2 \dot \theta^2}{2} - -mg x \sin \alpha \label{eq:303}

The Lagrange's equations and the method of undefined multipliers use the relation:

\frac{d}{dt}\left( \frac{\partial L}{\partial \dot q} \right) - \frac{\partial L}{\partial q} = \lambda_q \frac{\partial f}{\partial q} \label{eq:304
}

Applying this in the Lagrangian :
$$m \ddot x - mg \sin \alpha = \lambda_x \frac{\partial f}{\partial x}$$
$$m r^2 \ddot \theta = \lambda_\theta \frac{\partial f}{\partial \theta}$$

where the constrain equation is $r \theta = x$ and therefore $f = r \theta - x = 0$. Then:
$$m \ddot x - mg \sin \alpha = - \lambda$$
$$m r^2 \ddot \theta = r \lambda$$
$$m r \ddot \theta = \lambda$$
But, from the constrain we can found: $r \ddot \theta = \ddot x$. Then,
$$\lambda = \frac{mg \sin \alpha}{2}$$
$$\ddot x = \frac{mg \sin \alpha}{2}$$
$$\ddot \theta = \frac{mg \sin \alpha}{2r}$$
The general solution for the equation of motion are:
$$x(t) = \frac{1}{4} g t^2 \sin \alpha + At + B$$
$$\theta(t) = \frac{1}{4r} g t^2 \sin \alpha + Ct + D$$
where $A,B,C,D$ are constants.

Apparently it is not the right answer. What am I doing wrong?

2. Nov 4, 2013

### GregoryS

Hi, alefisico!

I guess there was some kind of mishmash. So, the Lagrange's equations are $$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot q} \right) - \frac{\partial L}{\partial q} = 0.$$ These equations can be solved very simply without any "undetermined multipliers" which used when we try to find conditional extremums in non-linear programming theory.

3. Nov 9, 2013

### Judas503

In the answer, is $\ddot{\theta}=\frac{2g\sin{\alpha}}{3r}$?

4. Nov 9, 2013

### tiny-tim

hi alefisico! welcome to pf!

just wanted to say that on this forum you don't need to type all those ""s etc …

just type [noparse] before and after
(or ## before and after for inline-tex)[/noparse]