# A bead on a spinning rod with magnetic field

1. Nov 16, 2009

### Loxias

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

A bead of mass m and charge q is placed on a frictionless, rigid rod that is spun
about at one end at a constant rate w on the xy plane. There is a constant magnetic
field in space $$B = B_0\hat{z}$$

2. Relevant equations

Write the Lagrangian for the system, use the generalized coordinate r (the
distance of the bead from the origin).

3. The attempt at a solution

I chose
$$x = rcos(wt) , y = rsin(wt)$$

and from
$$v = rw$$
we get
$$v = (wrcos(wt), wrsin(wt))$$

assuming vector potential
$$\vec{A} = B_0(0,x,0)$$

and $$L = \frac{1}{2}m V^2 + qV\vec{A}$$
I get
$$L = qB_0wr^2cos(wt)sin(wt) + \frac{1}{2}m (\dot{r}^2 +r^2w^2)$$

deriving equations of motion:

$$m\ddot{r} = mrw^2 + 2rB_0qwcos(wt)sin(wt)$$

which is good unit-wise.

My question is, did I derive everything right or did I forget something or misused the potential of magnetic field?

Thanks :)

Last edited: Nov 16, 2009
2. Nov 16, 2009

### gabbagabbahey

Hi Loxias, welcome to PF!

Is the pivot end of the rod held fixed at the origin, or is it allowed to move?

3. Nov 17, 2009

### Loxias

Thanks

It is fixed.

4. Nov 17, 2009

### gabbagabbahey

Good...

But $v\neq r\omega$....$\textbf{v}=\frac{d\textbf{r}}{dt}$ and if the bead is allowed to slide along the rod , $r$ will dpend on time and you will need to use the product rule to calculate $\textbf{v}$

good...

I assume you mean $L=\frac{1}{2}mv^2+q\textbf{v}\cdot\textbf{A}$ ?

Where is the $\sin(\omega t)$ coming from, and shouldn't there be another term involving $\dot{r}$? (from the $q\textbf{v}\cdot\textbf{A}$ part)

5. Nov 17, 2009

### Loxias

The bead is allowed to slide.

It seems i have forgotten about the radial velocity, but $$v = wr$$ is the one tangent to it.

so should the speed be $$(\dot{r}, rw)$$ or am I still wrong?

you assume right about the lagrangian.

The $$sin(wt)$$ came from the velocity i chose earlier, and i dont have $$\dot{r}$$ because i forgot about the radial velocity.

so to sum up, the right this is $$L = \frac{1}{2}mv^2 + q(\dot{r},wr,0)(0,rcos(wt),0)$$

because if it is, i will still get the same result as earlier and it the radial velocity doesn't affect $$L$$ (though i think it should since its in a right angle to the field)

Thanks again

6. Nov 17, 2009

It depends by what you mean by $(\dot{r}, r\omega)$...do you mean $\textbf{v}=\dot{r}\hat{\mathbf{e}}_r+r\omeg\hat{\mathbf{e}}_{\theta}[/tex] or [itex]\textbf{v}=\dot{r}\hat{\mathbf{e}}_x+r\omega\hat{\mathbf{e}}_y[/tex]? 7. Nov 17, 2009 ### Loxias $$\textbf{v}=\dot{r}\hat{\mathbf{e}}_r+r\omega\hat{\mathbf{e}}_{\theta}$$ 8. Nov 17, 2009 ### gabbagabbahey Okay, so hat do you get for [itex]\textbf{v}\cdot\textbf{A}$ then?

9. Nov 17, 2009

### Loxias

Ok ok, I first thought i should work in spherical coordinates and then I should obviously transform $$A$$ to those coordinates as well, but I've had a lot of trouble with that.

Then, why not use cylindrical coordinates?

In that case, I found transformation equations and I get

$$A = (rcos(wt)sin(wt), rcos^2(wt), 0)$$
$$v = (\dot{r}, wr, 0)$$

and the product is straightforward.

the transformation equations :

$$A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A} \,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi$$

$$A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A} \,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi$$

$$A_z\,=\,A_z$$

10. Nov 17, 2009

### Loxias

following my previous post, I've just noticed something else and it might be where you were going for from the beginning :

$$v = (\dot{x}, \dot{y},0) = (\dot{r}cos(wt) - rwsin(wt), \dot{r}sin(wt) + rwcos(wt),0)$$

and then
$$Av = (0,rcos(wt),0)(\dot{r}cos(wt) - rwsin(wt), \dot{r}sin(wt) + rwcos(wt),0)$$

which will give the same result as the previous post.

Is that what you meant earlier?

11. Nov 17, 2009

### gabbagabbahey

That looks better....what does that give you for your Lagrangian?

12. Nov 17, 2009

### Loxias

$$L = \frac{1}{2}m(\dot{r}^2+r^2w^2) + qB_0(r\dot{r}cos(wt)sin(wt) + r^2wcos^2(wt)$$

which results in

$$m\ddot{r} = mrw^2 + qB_0rw + 2qB_0rwcos(2wt)$$

13. Nov 17, 2009

### gabbagabbahey

Good...

That looks a little off to me....why not show me what you get for your derivatives?

14. Nov 17, 2009

### Loxias

$$\frac{\partial L}{\partial r} = mrw^2 + qB_0\dot{r}cos(wt)sin(wt) + 2qB_0rwcos^2(wt)$$

$$\frac{\partial L }{\partial \dot{r}} = m\dot{r} + qB_0rwcos(wt)sin(wt)$$

$$\frac{d}{dt} \frac{\partial L }{\partial \dot{r}} = m\ddot{r} + qB_0\dot{r}cos(wt)sin(wt) - qB_0rwcos^2(wt) +qB_0rwsin^2(wt) = m\ddot{r} + qB_0\dot{r}cos(wt)sin(wt) - qB_0rwcos(2wt)$$

which results in

$$m\ddot{r} = mrw^2 + qB_0rwcos(2wt) + 2qB_0rwcos^2(wt)$$

and from

$$cos(2wt) = 1-2sin^2(wt)$$

15. Nov 17, 2009

### gabbagabbahey

Good...

Looks like you got your signs backwards; [itex]\frac{d}{dt}\sin(\omega t)=+\omega\cos(\omega t)[/tex]
and [itex]\frac{d}{dt}\cos(\omega t)=-\omega\sin(\omega t)[/tex]

16. Nov 17, 2009

### Loxias

making it

$$m\ddot{r} = mrw^2 + qB_0rw$$

seems good?

17. Nov 17, 2009

### gabbagabbahey

Looks good to me....you can check that you get the same thing from the Lorentz force law (just look at the radial component of the force, as there will of course be an azimuthal pseudo-force that keep the particle attached to the spinning rod)

18. Nov 17, 2009

### Loxias

Thanks for all the help

19. Nov 17, 2009

### Loxias

I am now trying to solve the equation of motion

$$\ddot{r} - Cr = 0$$ where $$C = \frac{mw^2+qB_0w}{m}$$

solving this will give

$$r = C_1 e^{\sqrt{C}t} + C_2e^{-\sqrt{C}t}$$

then i should make $$C_2 = 0$$, right? :shy: