# Problem on Lagrangian

1. Nov 16, 2013

### halukalutalu

1. The problem statement, all variables and given/known data

Consider a bead of mass m moving on a spoke of a rotating bicycle wheel. If there are no forces other than the constraint forces, then find the Lagrangian and the equation of motion in generalised coordinates. What is the possible solution of this motion?

2. Relevant equations

Lagrange's equation of motion

3. The attempt at a solution

Use cylindrical coordinates for the problem.

Let the origin be at the centre of the wheel and let the wheel lie on the rθ-plane such that θ=0 at the top of the wheel and θ>0 in the clockwise direction.

So, $T = 0.5m(\dot{r}^{^2} + r^{2}\dot{\theta}^{2})$ and $U = mgrcosθ$.

So, $\frac{∂L}{∂r} - \frac{d}{dt}(\frac{∂L}{∂\dot{r}}) = 0$

$\Rightarrow (mr\dot{\theta}^{2} - mgcosθ) - m\ddot{r} = 0$

$\Rightarrow \ddot{r} = r\dot{\theta}^{2} - gcosθ$

and $\frac{∂L}{∂θ} - \frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = 0$

$\Rightarrow (-mgrsinθ) - mr^{2}\ddot{θ} = 0$

$\Rightarrow \ddot{θ} = - \frac{g}{r}sinθ$

Do you think I've got the equations of motion right?

I have no idea how to work out the possible solution of the motion. Thoughts?

2. Nov 17, 2013

### ehild

The last one is not correct. Take care whit the time derivative.

ehild

3. Nov 17, 2013

### halukalutalu

To be honest, I don't see how I've got things wrong with the time derivative. Of course, $\dot{θ}$ differentiated wrt time becomes $\ddot{θ}$, right?

Also, isn't r constrained to be smaller than the radius of the wheel. Doesn't this have an effect on the solution?

4. Nov 17, 2013

### ehild

$\frac{d}{dt}(\frac{∂L}{∂\dot{θ}}) = \frac{d}{dt}(mr^2\dot{θ})=2mr\dot r\dot {θ}+mr^2 \ddot {θ}$

You ignored the time dependence of r.

The text of the problem is not clear.

Does that wheel rotate in the horizontal or in the vertical plane? mg is an outside force, not force of constraint.

Is there something making the wheel rotate with a given angular speed, or does the wheel rote freely about a fixed axis?

I think you can assume the radius of the wheel arbitrary long.

ehild

5. Nov 18, 2013

### halukalutalu

Thanks for the pointer. :)

I think the wheel rotates in the horizontal plane because there are no forces apart from the given force of constraint and because the force of gravity is not a force of constraint. What do you think? Isn't this also why we can safely say that the wheel rotates freely?

So, if we assume that the radius of the wheel is arbitrarily large, would you say that under those assumptions, my solution is correct?

6. Nov 18, 2013

### ehild

If the wheel is free to rotate, it is part of the system. You have to include it into the Lagrangian.

Are you sure you copied the problem correctly?

How do you include the constraint?

ehild

Last edited: Nov 18, 2013