Rod with Current on Rails in Magnetic Field

[domephilis]

Homework Statement

5.16 The cylindrical rod of length w straddles two parallel rails with spacing w. It carries a current I and is bathed in a field B perpendicular to the plane of the rod and rails. Assume the magnetic force is concentrated on the axis of the rod. If it starts at rest and rolls without slipping for a distance L, show that the velocity will be $$v = \sqrt{\frac{4BLIw}{3M}}$$. (First find the torque on a tiny segment dw of the rod. Draw a diagram.) Notice that the radius a of the cylinder is not given. (Source: Fundamentals of Physics II by R. Shankar)

Relevant Equations

$$v = \sqrt{\frac{4BLIw}{3M}}$$

I can't see what is the problem with my derivation but the answer is incorrect. Please help.

We assume here that $\omega_0 = 0$ and $v_0 = 0$, hence it immediately rolls without slipping without any transitional phase. Hence $v = \omega R$. Thus, $v(L) = R\omega(L)$. Since our static friction at the interface between the rail and the rod counteracts other forces (so that that point is static), $$f = IBw$$ which exerts a torque of $$\tau = IBwR$$. Then our angular acceleration is $$\alpha = \frac{\tau}{I_{moment}} = \frac{2IBw}{MR}$$. We may then apply the formula $$\omega^2 = 2\alpha \Delta \theta = 2 \alpha \frac{L}{R}$$ to find that $$\omega^2(L) = \frac{4IBwL}{MR^2}$$ Thus $$v(L) = \sqrt{\frac{4IBwL}{M}}$$.

Thank you all very much in advance if any of you may look over this. I've always had problems with the subtleties in rigid-body-dynamics. Also, I don't see why we have to evaluate the infinitesimal segments.

 

[domephilis]

Also, I am aware that energy conservation should give the correct result. I just don’t know why this way is wrong. Torque is torque.

 

[kuruman]

Torque is torque.

Torque about what point? You seem to have written the torque about the axis of the rod. If that's the case, you need to include the torque generated by static friction. The problem tells you to "First find the torque on a tiny segment dw of the rod. Draw a diagram." You don't seem to have done either of these.

 

[domephilis]

Torque about what point? You seem to have written the torque about the axis of the rod. If that's the case, you need to include the torque generated by static friction. The problem tells you to "First find the torque on a tiny segment dw of the rod. Draw a diagram." You don't seem to have done either of these.

Torque generated by static friction is IBwR. I wrote it up there. I was wondering it asked that about dw. What’s wrong with just considering the body as a whole? It’s not like there’s a continuum of forces/torques that we need to integrate over. The only torque is from the friction at two very much discrete points.

 

[haruspex]

Torque generated by static friction is IBwR.

Since the rod accelerates, horizontal forces are not in balance. The force of static friction must be less than the electromagnetic force.
You can consider rotation about the point of contact, so avoid worrying about friction. The torque is then $IB\omega R$, but what is the moment of inertia?

I was wondering it asked that about dw. What’s wrong with just considering the body as a whole? It’s not like there’s a continuum of forces/torques that we need to integrate over.

In principle you would integrate over the area, but by symmetry and the linear dependence of the torque on the vertical distance from the rail it comes to be the same as taking the force as acting at the rod axis.
One thing bothers me… once the rod is rolling, are there other effects to consider? Wouldn't that movement generate an emf? (I'm ok on mechanics, but no expert on electricity and magnetism.)

 

[kuruman]

Wouldn't that movement generate an emf?

It would, but I don't think that the author expects us to go there. If there is an induced motional emf, there will be magnetic braking and a first order ODE to solve that will have a time-dependent current flowing in the rod. Judging from the answer that we are supposed to "show", current $I$ is constant.

I think here one should treat the Lorentz force $IwB$ as constant acting all along the axis of the cylinder, no different from a wheel being pulled by a constant force acting at its center.

 

[Steve4Physics]

@domephilis, I think your Post #1 method is ok but you have used the wrong moment of inertia (MoI).

The torque you have calculated is about the instantaneous point of contact. But you have used the MoI for rotation about the cylinder’s axis. So you need a different MoI.

 

[domephilis]

Thanks to haruspex, Steve4Physics, and kuruman. I realize my error. If I take the contact point to be static then the torque should be about that point and MoI should be $\frac{3}{2} MR^2$, hence I’m off by 1/sqrt(3) in the final answer. Regarding the issue of motional emf generated by the moving rod (raised by haruspex), I was under the impression that current/charges cannot experience their own magnetic/electric field. Otherwise, you’d have a charge generating a field which moves it causing it to generate a different field which then moves it causing further motion when no forces are moving it. Does this still apply to currents? Should retardation be considered in this?

 

[domephilis]

Since the rod accelerates, horizontal forces are not in balance. The force of static friction must be less than the electromagnetic force.
You can consider rotation about the point of contact, so avoid worrying about friction. The torque is then $IB\omega R$, but what is the moment of inertia?

In principle you would integrate over the area, but by symmetry and the linear dependence of the torque on the vertical distance from the rail it comes to be the same as taking the force as acting at the rod axis.
One thing bothers me… once the rod is rolling, are there other effects to consider? Wouldn't that movement generate an emf? (I'm ok on mechanics, but no expert on electricity and magnetism.)

I see my error. The moment of inertia should be taken about the contact point. We can use the parallel axis theorem for that. That does yield the right answer. On your question about the emf generated by the moving rod, my feeling is that unless retardation is significant in this case, the current cannot experience its own field. At least, that was true for electrostatics. I’m not sure if that still applies for magnetism.

 

[renormalize]

On your question about the emf generated by the moving rod, my feeling is that unless retardation is significant in this case, the current cannot experience its own field. At least, that was true for electrostatics. I’m not sure if that still applies for magnetism.

No, your reasoning is not correct. Even with no imposed external current, a conductive rod rolling down a incline is slowed when it enters a perpendicular magnetic field due to induced eddy currents: https://sites.google.com/nd.edu/ndp...netic-induction/lenzs-law-and-the-rolling-rod. In your problem with the external current flowing, the induced eddy currents will still oppose the rod motion, thereby reducing its speed from what you calculated.

 

[domephilis]

No, your reasoning is not correct. Even with no imposed external current, a conductive rod rolling down a incline is slowed when it enters a perpendicular magnetic field due to induced eddy currents: https://sites.google.com/nd.edu/ndp...netic-induction/lenzs-law-and-the-rolling-rod. In your problem with the external current flowing, the induced eddy currents will still oppose the rod motion, thereby reducing its speed from what you calculated.

I see. I didn’t quite get to Lenz’s Law yet. Although I have seen some demonstrations on YouTube to that effect. I had this confusion because in electric fields a charge cannot move on their own (i.e. experience their own field). I thought it might carry over to magnetism. Perhaps this will be a bit clearer to me when I read the next chapter.

 

[renormalize]

I had this confusion because in electric fields a charge cannot move on their own (i.e. experience their own field). I thought it might carry over to magnetism.

The free charges in the rolling conductor are not "experiencing their own field". Because they translate at the speed $v$ in a direction perpendicular to the magnetic field $B$, they experience a Lorentz force proportional to $v\,B$ along the direction of the rod. If the support rails are shorted together somewhere to form a complete a circuit, this electromotive force causes the charges to continuously flow as an "eddy current" $I_e$. In turn, that current flowing in the length-$l$ rod as it moves through the magnetic field generates a drag force $I_e\,l\,B$ on the rod, thus slowing it down. Note that this drag current and opposing force are present even in the case where the rails are "shorted together" through a battery which is supplying its own current to accelerate the rod forward; i.e., the eddy-current opposes the battery-current.

 

[domephilis]

The free charges in the rolling conductor are not "experiencing their own field". Because they translate at the speed $v$ in a direction perpendicular to the magnetic field $B$, they experience a Lorentz force proportional to $v\,B$ along the direction of the rod. If the support rails are shorted together somewhere to form a complete a circuit, this electromotive force causes the charges to continuously flow as an "eddy current" $I_e$. In turn, that current flowing in the length-$l$ rod as it moves through the magnetic field generates a drag force $I_e\,l\,B$ on the rod, thus slowing it down. Note that this drag current and opposing force are present even in the case where the rails are "shorted together" through a battery which is supplying its own current to accelerate the rod forward; i.e., the eddy-current opposes the battery-current.

Thank you very much for the explanation. It helps a lot.