A bead on a spinning rod with magnetic field

[Loxias]

Homework Statement

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 $$B = B_0\hat{z}$$

Homework Equations

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

The Attempt at a Solution

I chose
$$x = rcos(wt) ,<br /> y = rsin(wt)$$

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 :)

 

[gabbagabbahey]

Hi Loxias, welcome to PF!

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

 

[Loxias]

Thanks

It is fixed.

 

[gabbagabbahey]

I chose
$$x = rcos(wt) ,<br /> y = rsin(wt)$$

Good...

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

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 $\textbf{v}$

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

good...

and $$L = \frac{1}{2}m V^2 + qV\vec{A}$$

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

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

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)

 

[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 don't 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

 

[gabbagabbahey]

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

It depends by what you mean by $(\dot{r}, r\omega)$...do you mean [itex]\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]?[/itex][/itex]

 

[Loxias]

$$\textbf{v}=\dot{r}\hat{\mathbf{e}}_r+r\omega\hat{\mathbf{e}}_{\theta}$$

 

[gabbagabbahey]

Okay, so hat do you get for $\textbf{v}\cdot\textbf{A}$ then?

 

[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 :

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

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

$$A_z\,=\,A_z$$

 

[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?

 

[gabbagabbahey]

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?

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

 

[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

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

 

[gabbagabbahey]

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

Good...

which results in

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

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

 

[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

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

you get my final answer

 

[gabbagabbahey]

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

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

Good...

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

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

 

[Loxias]

making it

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

seems good?

 

[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)

 

[Loxias]

Thanks for all the help

 

[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: