Force on a block placed on a rotating disc, calculated in different reference frames

[Aurelius120]

Homework Statement

Equation of force acting on a particle placed on a disc in the rotating frame of reference

Relevant Equations

$$|\vec F_{centri}|=m\omega^2r$$

How is the equation for the force derived?
$\vec F_{rot}$ should be sum of $\vec F_{in} \pm \text{ pseudo force }$ which is the centrifugal force.
SO $$\vec F_{centri}=2m(\vec v_{rot}\times \vec \omega)+m(\vec \omega\times \vec r)\times \vec\omega$$ which does not seem incorrect
Why is it epressed in such a way?
What do the $2m(\vec v_{rot}\times \vec \omega)$ and $m(\vec \omega\times \vec r)\times \vec\omega$ terms represent??
Isn't the only pseudo force acting on the particle the centrifugal force $=m\vec \omega \times (\vec r \times \vec \omega) \text{ ?}$

 

[Aurelius120]

This is the derivation of centripetal acceleration but I can't see those terms here

 

[BvU]

Isn't the only pseudo force acting on the particle the centrifugal force

No.

The other one is the Coriolis force

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[Orodruin]

Isn't the only pseudo force acting on the particle the centrifugal force $=m\vec \omega \times (\vec r \times \vec \omega) \text{ ?}$

To expand just a little bit on the reply above: The centrifugal force is the only pseudo force acting on a particle at rest in the rotating frame, since the Coriolis force is linearly dependent on velocity in the rotating frame.

 

[kuruman]

This section of the Wikipedia article should clarify things for you.

 

[Aurelius120]

No.

The other one is the Coriolis force

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To expand just a little bit on the reply above: The centrifugal force is the only pseudo force acting on a particle at rest in the rotating frame, since the Coriolis force is linearly dependent on velocity in the rotating frame.

The Centripetal force is responsible for changing direction in circular motion.
What does the Coriolis force do in the inertial frame?

 

[Orodruin]

The Centripetal force is responsible for changing direction in circular motion.
What does the Coriolis force do in the inertial frame?

I think you are confusing things. The centripetal force is not a pseudo force. It exists in any frame. By definition, neither centrifugal or Coriolis forces exist in an inertial frame.

 

[Aurelius120]

I think you are confusing things. The centripetal force is not a pseudo force. It exists in any frame. By definition, neither centrifugal or Coriolis forces exist in an inertial frame.

Yes but there must be an
$$\text{Centripetal}:\text{Centrifugal}::\text{X}:\text{Coriolis}$$ What does X do in inertial frame?

 

[jbriggs444]

Yes but there must be an
$$\text{Centripetal}:\text{Centrifugal}::\text{X}:\text{Coriolis}$$ What does X do in inertial frame?

I am not convinced that the relationship you propose is at all logical.

However, one answer is that $X$ is the force from the rails pushing on the sliding block, thereby constraining it to follow a straight path in the rotating frame or constraining it to follow a spiral path in the inertial frame.

That sideways force from slot on block is a real interaction force. Like the centripetal force, it exists in all frames.

What does the $X$ force do in this situation from the point of view of the inertial frame? It adds energy to the sliding block, thus explaining its increase in kinetic energy.


In a different scenario with tangential motion as seen from the rotating frame, the Coriolis pseudo force can act either to add to (if the velocity is spinward) or to subtract from (if the velocity is anti-spinward) the centrifugal pseudo-force. As such, the corresponding real force would be the centripetal force.

In yet another different scenario, suppose that we eliminate the slot and allow the block to slide frictionlessly on the surface of the disc. Now there are no interaction forces of any kind. Now the combination of centrifugal and Coriolis force explains the spiral (or circular) path of the block as viewed from the rotating frame, even though the path in the inertial frame is uniform motion along a line (or a state of continuing rest).

[And now the negative potential energy in the centrifugal force field explains the gain in kinetic energy of the outward spiralling block. Or the loss in kinetic energy of an inbound block before it eventually spirals back out]

 

[kuruman]

To @Aurelius120:
A few years ago I wrote this article that investigated how a spaceship with front, rear, left and right thrusters can enter and exit a uniformly rotating tunnel without bumping into the walls. In other words, what equations must be programmed into the thrusters to compensate for the inertial forces so that the spaceship moves in a straight line without wall constraints in the non-inertial frame. You may find it informative and to the point.

 

[Orodruin]

Yes but there must be an
$$\text{Centripetal}:\text{Centrifugal}::\text{X}:\text{Coriolis}$$ What does X do in inertial frame?

No, this is incorrect. The centripetal force is whatever radial force keeps an object in circular motion in an inertial frame or, equivalently, the force keeping that very same object stationary in a rotating frame. It is not the Newton 3 partner of the centrifugal force, which does not exist in an inertial frame, but is a pseudo force without a 3rd law partner in a rotating frame. The only relationship between the two is that they are of equal magnitude and opposite direction on an object at rest in the rotating frame. Note that 3rd law partners never act on the same object.

Similarly, there is no third law partner of the Coriolis force, which also is a pseudo force in a rotating reference frame.

The nomenclature is somewhat complicated bybthe fact that the 3rd law partner of the centripetal force is sometimes referred to as the reactive centrifugal force. However, this is not a force on the circulating object, nor is it a pseudo force. It is a force exerted by the circulating object on whatever is providing the centripetal force. It is a real force that exists in all reference frames.

 

[Orodruin]

A few years ago I wrote this article that investigated how a spaceship with front, rear, left and right thrusters can enter and exit a uniformly rotating tunnel without bumping into the walls. In other words, what equations must be programmed into the thrusters to compensate for the inertial forces so that the spaceship moves in a straight line without wall constraints in the non-inertial frame. You may find it informative and to the point.

As a previous long-time player of Elite: Dangerous and an flight-assist off affectionado (essentially turning the computer compensating for rotational and translational thruster effects), I have many times implemented this practically. Luckily, with some training, this is actually quite easy to do manually.

 

[kuruman]

This link may work better. [I'll delete this post if I see that the link in #10 is working. #10 seems to link to a preview rather than to the actual Insight].

Thanks, but the original link work fine with my browser. Maybe it doesn't work with all browsers.

On edit
Problem fixed. See post #15 by @Orodruin.

 

[Orodruin]

Thanks, but the original link work fine with my browser. Maybe it doesn't work with all browsers.

You linked to the preview page instead of the published page. You have permission to view it because you are the author. The rest of us do not have that permission.

 

[kuruman]

Good catch, thanks. I corrected a typo I saw in an equation and lost track of what window was what. I should have read @jbriggs444's post more carefully. Problem fixed.

 

[Aurelius120]

Since Coriolis Force has a bending/deflecting effect, in this problem, it has no effective effect?

 

[jbriggs444]

Since Coriolis Force has a bending/deflecting effect, in this problem, it has no effective effect?

It is hard to say. What is an "effective effect"?

The Coriolis force is introduced by choosing to use rotating coordinates. That is to say that it is conjured into existence with pencil and paper. In some sense, the only effect it can have is changing the squiggles on a piece of paper. Or in a textbook.

Any measurable physical effect that is attributed to the Coriolis force must be attributed instead to something else if we choose to work purely in an inertial frame.

How would you go about running a physical experiment with the Coriolis force turned off?

 

[Orodruin]

This reminds me of my favorite Limerick:

On a merry-go-round in the night,
Coriolis was shaken with fright.
Despite how he walked,
'Twas like he was stalked,
By some fiend always pushing him right.

https://www.physics.harvard.edu/undergrad/limericks

 

[Aurelius120]

It is hard to say. What is an "effective effect"?

Bad choice of words on my part.
What I meant was since the block is constrained to move in a slot, the Coriolis force can't really have it's bending\deflecting effect?

 

[BvU]

Your thread started out with a question about forces acting .

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[jbriggs444]

Bad choice of words on my part.
What I meant was since the block is constrained to move in a slot, the Coriolis force can't really have it's bending\deflecting effect?

Ok. I can agree with that. In the same sense that pushing on a wall has no effect.

 

[Aurelius120]

Your thread started out with a question about forces acting .

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So would it be ok if I posted a continuation of the problem in the picture (There's one or two on this setup about position and all) here or should I start a new thread for that?

 

[Aurelius120]

@jbriggs444 There are a few questions in continuation of the problem in the picture regarding position of block, reaction force and all. Should I post those in this thread or create a new one?

 

[jbriggs444]

@jbriggs444 There are a few questions in continuation of the problem in the picture regarding position of block, reaction force and all. Should I post those in this thread or create a new one?

If they are continuation questions about the same situation then posting them here would be OK. If it is a new set up then a new thread would be more appropriate. Just my opinion, of course.

Post away and let us see what you have.

 

[Aurelius120]

If they are continuation questions about the same situation then posting them here would be OK. If it is a new set up then a new thread would be more appropriate. Just my opinion, of course.

Post away and let us see what you have.

Since the block is constrained to move in the slot, (and like you said the Coriolis Force didn't affect its movement) I was expecting an equation of SHM. $$\dfrac{d^2r}{dt^2}=k\omega^2r$$ where k is constant of proportionality.
I was expecting one of the trigonometric options.

But the option C is marked correct.

 

[jbriggs444]

Since the block is constrained to move in the slot, (and like you said the Coriolis Force didn't affect its movement) I was expecting an equation of SHM.

The net force in the rotating frame is always radially outward. The outward force increases as the outward position increases.

That is not negative feedback. That is positive feedback. You should expect runaway behavior.

It is almost the same second order differential equation as for simple harmonic motion. But a sign is reversed.

$$\dfrac{d^2r}{dt^2}=k\omega^2r$$ where k is constant of proportionality.
I was expecting one of the trigonometric options.

But the option C is marked correct.

A second order linear homogeneous differential equation like this one will have a characteristic equation that is a quadratic. If it has two real solutions, you get two exponentials as solutions to the original equation. If it has two complex solutions, you get trig functions as solutions. Complex exponentials such as $e^{i\theta}$ and trig functions are intimately related.

If the middle coefficient in the characteristic quadratic is zero (as in this case), it is the difference between solving $x^2 = 1$ (and getting two real roots) and solving $x^2 = -1$ (and getting two imaginary roots).

You have been exposed to solving linear homogeneous differential equations by finding roots of their associated characteristic polynomials, I hope?