Conservation of angular momentum - tetherball vs charged particle

[gabdolce]

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Hey all,

I was wondering if one of you could help me out with a debacle I'm having.

I'm having trouble reconciling what exactly happens between an orbiting tetherball and the separate scenario of a orbiting particulate charge.

With the particulate charge: you can, given the velocity and magnetic field magnitude, determine the radius of the orbit of the charged particle.

F=ma=mv^2/r=qvB
r=mv/qB

In the case of varying B-field magnitudes, the only thing that changes is the radius of the orbit . The magnetic force is ALWAYS perpendicular to the velocity vector and thus can NEVER do work and NEVER (de)accelerate the spinning particle. Velocity of this particle must remain constant when the only perturbation can be altering the strength of the magnetic field.

Here is my confusion: As far as I can see, there is NO net torque on the system, therefore, there must be conservation of angular momentum.

Let's consider an INCREASING B field magnitude: According to L=mvr, since r is consequently getting smaller, shouldn't the velocity of this particle be increasing as well?!?!

This would make sense, especially since when looking at a combination of F=mv^2/r and mvr, you get F=L/r^3 which further corroborates that the centripetal force here is getting larger (as we expected from the increasing of B)...

Is anyone else confused by this as well?

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Let's now look at a mechanical scenario: a tetherball circling the pole and getting shorter and shorter.

Can I use the conservation of angular here too and say that since the radius of orbit is getting shorter and shorter, then the tetherball must also be accelerating as well?

Thanks in advance.

 

[Pisiform]

Here is what I think:

In case of increasing magnetic field, the charge will spiral and along its motion, as you mentioned, the radius will decrease. But the equation you gave L = mvr is more like L = m (omega) r ... there is a difference between translation velocity and angular velocity. So, yes, angular velocity will increase but still KE will remain constant because KE = 0.5 I (omega)^2 .... even though omega is increasing, I (moment of inertia) will decrease - since radius is decreasing - and so there wouldn't be any change in KE.

and also if you post question in the correct forum, you might get more responses 🙂

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum.

 

[MedPR]

Hey all,

I was wondering if one of you could help me out with a debacle I'm having.

I'm having trouble reconciling what exactly happens between an orbiting tetherball and the separate scenario of a orbiting particulate charge.

With the particulate charge: you can, given the velocity and magnetic field magnitude, determine the radius of the orbit of the charged particle.

F=ma=mv^2/r=qvB
r=mv/qB

In the case of varying B-field magnitudes, the only thing that changes is the radius of the orbit . The magnetic force is ALWAYS perpendicular to the velocity vector and thus can NEVER do work and NEVER (de)accelerate the spinning particle. Velocity of this particle must remain constant when the only perturbation can be altering the strength of the magnetic field.

Here is my confusion: As far as I can see, there is NO net torque on the system, therefore, there must be conservation of angular momentum.

Let's consider an INCREASING B field magnitude: According to L=mvr, since r is consequently getting smaller, shouldn't the velocity of this particle be increasing as well?!?!

This would make sense, especially since when looking at a combination of F=mv^2/r and mvr, you get F=L/r^3 which further corroborates that the centripetal force here is getting larger (as we expected from the increasing of B)...

Is anyone else confused by this as well?

--------

Let's now look at a mechanical scenario: a tetherball circling the pole and getting shorter and shorter.

Can I use the conservation of angular here too and say that since the radius of orbit is getting shorter and shorter, then the tetherball must also be accelerating as well?

Thanks in advance.

You lost me on the first example, but I think the acceleration on the teatherball would be decreasing.

a=w^2r. w=2pi/T. As you decrease the radius, you also decrease the angular velocity necessary to complete 1 revolution (circumference) in an equivalent amount of time. Thus, decreasing radius, and decreasing angular velocity = decreasing acceleration.

 

[MT Headed]

In the first example, you are correct that no work is performed. Work is F d cos theta, and theta is always 90 degrees. But then you are confusing linear acceleration with circular motion. The particle has a force on it. It must accelerate. Its velocity must be changing. Its speed might ot might not remain the same, but its velocity will change direction.

If you increase the B field then you will increase the force, and you will increase the acceleration, and the object will spin faster in a tighter loop.

 

[gabdolce]

...

If you increase the B field then you will increase the force, and you will increase the acceleration, and the object will spin faster in a tighter loop.

How can this be true? You agreed yourself that there cannot be any work done on the revolving particle in the direction of its motion (the acting force is always perpendicular).

If there is no work, there is no change in kinetic energy. So even if the particle is revolving in a smaller circle when B is increased (because r=mv/qB), the magnitude of this translational velocity cannot change, isn't that right?

The v in question must be translational velocity as well because the v in question comes from qvBsin(theta) or rather, q*(v X B) as I have shown in the derivation in the original post.

 

[gabdolce]

Here is what I think:

In case of increasing magnetic field, the charge will spiral and along its motion, as you mentioned, the radius will decrease. But the equation you gave L = mvr is more like L = m (omega) r ... there is a difference between translation velocity and angular velocity. So, yes, angular velocity will increase but still KE will remain constant because KE = 0.5 I (omega)^2 .... even though omega is increasing, I (moment of inertia) will decrease - since radius is decreasing - and so there wouldn't be any change in KE.

and also if you post question in the correct forum, you might get more responses 🙂

Is this even true?

Linear momentum is p=mv

Angular momentum is L=pr=mvr

The v in this equation is surely still the translational velocity

 

[gabdolce]

You lost me on the first example, but I think the acceleration on the teatherball would be decreasing.

a=w^2r. w=2pi/T. As you decrease the radius, you also decrease the angular velocity necessary to complete 1 revolution (circumference) in an equivalent amount of time. Thus, decreasing radius, and decreasing angular velocity = decreasing acceleration.

There are two competing variables here: if the radius of revolution becomes smaller, at a given starting velocity v, the period to complete a revolution will become necessary smaller, making w bigger.

Although it's impossible to say for sure since you have one growing variable (w) and one decreasing variable (r), I would venture a guess that w^2 beats out the contribution from the reduced r.

This would mean that the tetherball accelerates which is congruent with my postulation using the conservation of angular momentum L=mvr ...

Or am I way off base here?

Disregard all of the beginnings of this post.

The a=w^2/r equation is for the acceleration component in the radial direction. Any alterations or fluctuations in this value wouldn't alter the linear velocity of the positron in any way, making this equation irrelevant in this discussion.

 

[Ssina]

How can this be true? You agreed yourself that there cannot be any work done on the revolving particle in the direction of its motion (the acting force is always perpendicular).

If there is no work, there is no change in kinetic energy. So even if the particle is revolving in a smaller circle when B is increased (because r=mv/qB), the magnitude of this translational velocity cannot change, isn't that right?

The v in question must be translational velocity as well because the v in question comes from qvBsin(theta) or rather, q*(v X B) as I have shown in the derivation in the original post.

so as you mentioned the magnitude won't change, but since velocity is a vector and the direction is constantly changing, the velocity is also changing..... and since velocity is changing, there is acceleration (going back to the OP that you mentioned it can't de/accelerate)

KE is also scalar and only takes into account the magnitude and not the direction, so it'll be the same.

I'm now a little lost for the second part of your post 😕

 

[MT Headed]

OP, your question is a deep one, and I think I found the source of the confusion. It boils down to the question of 'if B doubles, does L change?' Argument one says that there is no external torque, so L doesn't change. Argument 2 says that L=mvr, and r changed while v and m remain, so therefore L does change.

Fundamentally circular motion must all be analyzed from one consistent point of reference, and I think you are subtly changing yours, hence the confusion. OP, as you analyze the circular motion, what is your frame of reference?

Imagine a charged particle going in a circle. You can measure L=mvr where r is the distance to the center of the circle. Then flip a switch and BLAM double B. What happens? The v does not change. The r of the circle changes, it goes down by half. So you are thinking AHA the L went down by half as well, in the absence of an external torque. You are like wtf? But did you notice that the center of the circle, i.e. the point of reference, also changed??? You can't do that.

If you choose a consistent point of reference, there will be no contradiction. In fact, depending on your reference, the B field can produce a torque (while simultaneously from a different point of reference it doesn't).

Interesting problem!

 

[MT Headed]

And for the record, OP, I was ambiguous when I said the particle would spin faster in a larger B field. V would be constant, and omega would increase (because r decreased). You are correct, and I should have been more clear.