Does radial velocity affect centripetal force

[Ry122]

If for example someone is walking along a radial line of a spinning disc at constant velocity
would they experience a different centripetal force than if they were standing still at that particular location (I know the person would experience jerk while moving along, but would they experience a different acceleration at a particular point)?

 

[S_Happens]

So they would experience jerk if there was a change of acceleration...

Did you make the assumption there's a change of acceleration, but it's not centripetal? You basically answered your own question, and it surprises me that you understand jerk, but don't show the simple relationship between centripetal acceleration (or force if multiplied by mass) and magnitude of the velocity (constantly changing direction).

Edit- It's not meant to be condescending. You're already all over the answer.

 

[Ry122]

Are you saying that the person experiences a centripetal acceleration as well as an acceleration due to the change of direction of the radial velocity vector?

How do you calculate that type of acceleration?

 

[S_Happens]

I didn't say that at all. It seemed like you were making that assumption, which is why I asked why you would consider them to be different.

How is centripetal acceleration defined? How do you calculate centripetal acceleration (in terms of velocity since that's the relationship you want)?

If the person moves (in any direction), then either the path or the velocity will change. If these change, can the centripetal acceleration be constant (equivalent to the stationary case)?

 

[Physicist1231]

That is an awesome topic. He is right that you did kind of answer your own question but may not have known all the forces involved in your setup.

In your set up you have a person walking on a disk (assumed circular) and walking in such a way as to stay the same distance from the center.

You have the centripital force of the spinning disk pusing you out, the velocity that your mass is trying to go and also the friction of your feet keeping you on the disk. With all this factored in the individual may not feel any change in acceleration.

Change anyone of those three factors and you will experience that 'jerk' however minute.

Say you stop walking (your mass no longer moving on its own) now your overall velocity has changed to that of the centripital force counter acted by the friction between you and the surface (similar to the planets with centripital force countered by Gravity).

Yes you experience a change in your over all velocity because you change one of the factors to be calculated in the over all velocity. You can experience the same thing if you are walking on an escalator and then you stop walking. Almost the same exact thing.

 

[AlephZero]

This could be a question about forces, or it could be a question about what the words mean. The force called the "centripetal force" is always taken to be $mr\omega^2$, i.e. it depends only on the position of the object and not on its velocity.

However the person would experience a change in the total force acting on them at a particular point, depending on their velocity along a radial line. The change is in the tangential (sideways) component of force acting on them, not on the radial force. The tangential force component is called the Coriolis force.

If they were moving tangentially around a circle on the disk, the radial force would change, because the angular velocity of the person is not the same as the angular velocity of the disk. Again, the "extra" radial force (which could be acting either towards the center or away from it) is called the Coriolis force.

So, the standard way to describe this is to split the force on into two parts. The centripetal force is independent of the person's velocity and always acts radially. The Coriolis always acts at right angles to the person's relative velocity moving on the disk and its magnitude depends on the relative velocity.

 

[Disinterred]

Not that it''s important to the discussion at all, but to be technically correct, the object or person moving in a circle experiences a centrifugal force not a "centripetal force".

To be even more nit-picky, there is no such thing as a "centripetal force" (there is only four forces in nature). The correct way of stating "centripetal force" is: the force which provides the centripetal acceleration.

However, the centrifugal force does exist as a fictitious force in the objects frame of view. Thus we can correctly call it a new force, the centrifugal force.

 

[rcgldr]

Ignoring any jerk factors, and assuming the disk is held to a constant rate of rotation, if a person is walking along a radial line on spinning disk, that person will experience a tangental acceleration as well as a change in centripetal acceleration. If they walk towards the center, they will experience tangental deceleration and a decrease in centripetal acceleration over time, and vice versa.

If the disk is not held to constant rate of rotation, and there are no losses in the person plus spinning disk system, then angular momentum is preserved and the disk's rate of rotation will increase as the person walks inwards, and vice versa. If the ratio of mass of person versus disk is high enough the increase in the rate of disk rotation results in an increase in centripetal acceleration of the person as that person walks inwards versus outwards.

The extreme case of this would be a massless disk or as another anlaogy, a person sliding on a frictionless plane while holding a massless rope attached to a frictionless bearing on a post. As the person pulls or relaxes on the rope to move inwards or outwards, his speed changes so angular momentum is kept constant. If the radius is halved, the centripetal acceleration increases by a factor of 8.

 

[S_Happens]

The force called the "centripetal force" is always taken to be $mr\omega^2$, i.e. it depends only on the position of the object and not on its velocity.

It can be written as (mv^2)/r as well, although I can't bring myself to not consider it, when written in terms of angular velocity, as not being dependant on velocity. A change in position would be a change in r or omega (or both) of the person.

Not that it''s important to the discussion at all, but to be technically correct, the object or person moving in a circle experiences a centrifugal force not a "centripetal force".

To be even more nit-picky, there is no such thing as a "centripetal force" (there is only four forces in nature). The correct way of stating "centripetal force" is: the force which provides the centripetal acceleration.

However, the centrifugal force does exist as a fictitious force in the objects frame of view. Thus we can correctly call it a new force, the centrifugal force.

You're right, it's not important. Pick a reference frame, then pick the direction. To me the implied frame was not that of the person, although the answer is simply the magnitude so frame doesn't matter.

 

[Ry122]

Okay, thanks now I understand that the person's total acceleration is determined not only from his centripetal acceleration but also his tangential acceleration since his radial position is changing and
V_tangential = radialPosition*AngularVelocity
Meaning V_tangential changes as he walks meaning there's A_tangential.

After looking up how to determine A_tangential I found that it's determined from 2*w*v.
Where v is the radial velocity and w is the angular velocity.
I don't really understand why this is the case.

it would make sense to me if A_tangential = 3*w (assuming v_radial=3m/s)
because he is 3m from the center of the circle after 1 second and so this would indicate the change in his velocity in a 1 second interval. ie. his acceleration.

 

[A.T.]

The centripetal force is independent of the person's velocity and always acts radially.

It can be written as (mv^2)/r as well, although I can't bring myself to not consider it, when written in terms of angular velocity, as not being dependant on velocity. A change in position would be a change in r or omega (or both) of the person

The centripetal force is independent of the velocity relative to the disc. It is not independent of the position on the disc. There is no contradiction here. Objects with different velocities relative to the disc at the same position on the disc will experience the same centripetal force.

 

[S_Happens]

Maybe the issue is just conceptual. When I am considering "the person" to move relative to the disc with constant r, I am seeing a change in omega (or tangential velocity, however you want to think of it). This corresponds to a change in acceleration.

I don't see how this is any different if we were to consider the disc to be stationary, such as a car in a turn or similar. The acceleration required to maintain that path at a larger velocity is also larger. Any velocity you consider in this situation would be relative to the stationary "disc," would it not?

Where is there a miscommunication/misunderstanding?

 

[BruceW]

The centripetal force depends on the frequency of rotation of the disk and the radial position of the person.
The centripetal force does NOT depend on the radial velocity of the person, but obviously if the person has radial velocity, then his radial position is changing, which does change the centripetal force.

If the person spins with the same rotational frequency of the disk, then the centripetal force is greater when the person is further from the centre.

 

[S_Happens]

The centripetal force depends on the frequency of rotation of the disk and the radial position of the person.

Frequency of rotation is easily written as angular frequency and then in terms of tangential velocity.

My conceptualization of the person on the disk represents a change in frequency of rotation. When the person moves, I am no longer considering constant angular frequency. The rotating disc with constant angular frequency that the peson is on is no longer my main concern.

The centripetal force does NOT depend on the radial velocity of the person, but obviously if the person has radial velocity, then his radial position is changing, which does change the centripetal force.

How would this not change angular frequency OF THE PERSON? I am considering a reference frame that is outside of both the disc and the person. In this case, the person has a different angular frequency of the disc. How is it that only the constant angular frequency of the disc is what matters? How is this not comparable to a car in a turn of constant radius, where the acceleration IS obvious and has to be accounted for by friction and/or banked turns?


Edit- I'm pretty sure we don't need any more repetition of the basic formulas unless followed by something substantive. The question really boils down to, "is omega held constant in this case?" I don't see how it is.

 

[BruceW]

Hopefully, now I understand how latex works, I've written the equations for this problem.
The capital F's represent real forces acting on the person (i.e. due to the friction between his feet and the surface). And the omega is the person's rotational frequency times 2Pi

$$F_{\theta} \ = \ 2m\omega\frac{dr}{dt} + mr^2\frac{d\omega}{dt}$$
$$F_{r} \ = \ m\frac{d^2r}{dt^2} - mr\omega^2$$

 

[BruceW]

I believe Ry122's question was what happens to the person if he tries to keep omega constant. I'll try to explain what would happen:

From the first equation, he wants to keep $\omega$ constant, so the rate of change of omega with time is zero, so he would have to apply a tangential force (using friction of his feet) to be able to have radial motion while keeping $\omega$ constant.

In the second equation, if he is walking steadily outwards, then the twice derivative of radius is zero, and we have the usual equation for centripetal motion. But if the person wants to accelerate outwards, he must apply extra force.

 

[S_Happens]

Yes the original question was about radial motion, and it diverged because I must not have read AlephZero's post in full. I must have started to comment on the first couple statements and missed the rest.

I was just considering the magnitude of net acceleration and stuck on that first statement about it not being dependant on angular frequency (specific to centripetal). I was looking at the centripetal and coriolis forces together, while everyone else treated them independantly. When I ignored the disk itself and treated it simply as a particle on a curved path, the Coriolis force disappeared.

It's mostly semantics, and I didn't understand the need to designate the Coriolis force.