Caculate the reaction force of a radially moving object

[losbellos]

Hello,

An object moves outward in shaft radially due to centrifugal force.
lest say it weights 1kg . in a timestep=0.0001 it travels 0.0017 meters
this is actually the difference of two radius,
let say first radius r1 = 0.25 meters
second r2 = 0.25 + 0.0017 meters
rpm of the rotating system is: 500.0

velocity difference by the distance it traveled from r1 to r2 vd = ((500 / 60.0 * r1*2.0*pi) - (500 / 60.0 * r2*2.0*pi))
now this velocity have to be delivered to the system by decelerating it.
What would be the torque?
v=a*t
a=v/t = vd/timestep
f = m*a
T = f * r2 (its ok to use the higher one not much difference)
the problem is that it makes pretty big anti torque, it seems that it is impossible even for the timesteps..
in a system 0.5kg moving outwards with the same velocity would create like 100+ NM anti torque which seems too much work for such a small object.

My question would be then that is it valid to calc this way or any other valid ways?

Thanks.

 

[A.T.]

...100+ NM anti torque which seems too much work...

Torque is not work, even though they have the same units.

My question would be then that is it valid to calc this way or any other valid ways?

Use the rotating frame where there is only radial motion, compute radial velocity based on centrifugal force, then Coriolis force to get the tangential contact force required to counter it.

 

[A.T.]

Coriolis force is a perpendicular upwards force

Coriolis force is perpendicular to the frame rotation axis and the object's velocity in that frame. In the frame with radial motion only, Coriolis is tangential and gives the required force by the shaft wall.

 

[losbellos]

maybe you misunderstood.
Velocity dir = centrifugal force dir, object moving outward or inwards would cause the system decelerate or accelerate but the rotating frame remains the same...

In physics, the Coriolis force is an inertial force[1] that acts on objects that are in motion relative to a rotating reference frame. In a reference frame with clockwise rotation, the force acts to the left of the motion of the object. In one with anticlockwise rotation, the force acts to the right.

So it is not valid right, maybe in one scenario. This should be the same vice-versa.
I calculate and see anyways thanks.

Well in some motors this force as the weight moves outwards and inward periodically creating a summa zero anti torque-torque . not feeling like Coriolis force

also if I use cross the force will be out of plane upwards or downwards.

well I calculated after and seen the using the d = 0.5 *a*t^2 , f=m*a gives the same as the Coriolis. So this two actually gives different result which seems more valid then the velocity based.
So its okay thanks.

((rpm/60.0)*(r1*2*pi))-((rpm/60.0)*(r2*2*pi))
d = 0.00785446*0.0001 as double
7.85446d-007
timestep = 0.0001 as double
0.0001d0
d/timestep^2/0.5
157.089d0 m/s2
power = (1570.89 * 0.5) * 52.3599 // torque * angvel
41125.8 watts much less than a megawatts for this simple things. (plus some more because there are more weights in the system.)

 

[jbriggs444]

In physics, the Coriolis force is an inertial force[1] that acts on objects that are in motion relative to a rotating reference frame. In a reference frame with clockwise rotation, the force acts to the left of the motion of the object. In one with anticlockwise rotation, the force acts to the right.

An object which is moving radially in the rotating reference frame is moving in the rotating reference frame and is subject to the Coriolis force.

 

[A.T.]

also if I use cross the force will be out of plane upwards or downwards.

What plane are you talking about? The Coriolis force has no component perpendicular to the plane of rotation.

 

[losbellos]

What plane are you talking about? The Coriolis force has no component perpendicular to the plane of rotation.

Okay, so vector force = [0,1,0] vector angvel = [-50,0,0]
In math.
cross [0,1,0] [-50,0,0] = [0,0,50] that is out of the plane.
Fc = [0,v,0] × [angvel,0,0] * 2.0 * m
(min or plus depends on rot dir
okay so how is it for you that it is in plane?
So this is an official calculus
Freactionforce would be (lets not call it coriolis) = (angvel*v) * 2.0 * m

 

[jbriggs444]

Okay, so vector force = [0,1,0] vector angvel = [-50,0,0]
In math.
cross [0,1,0] [-50,0,0] = [0,0,50] that is out of the plane.

That angular velocity. You specified 500 rpm. That's not -50 radians per second and it should not be in the plane. Angular velocity is perpendicular to the plane of rotation.

 

[losbellos]

That angular velocity. You specified 500 rpm. That's not -50 radians per second and it should not be in the plane. Angular velocity is perpendicular to the plane of rotation.

angular velocity is not perpendicular to the rotation it is the rotation itself.

https://www.symbolab.com/solver/vec...trix}\times\begin{pmatrix}-5&0&0\end{pmatrix}

x/y plane the velocity and the angular velocity the cross is out of plane.
Wrong description.

 

[losbellos]

That angular velocity. You specified 500 rpm. That's not -50 radians per second and it should not be in the plane. Angular velocity is perpendicular to the plane of rotation.

It doesn't matter that is an other example and its ok, the point is that by cross calculation you can't get it right in terms of direction but only values.
Its really getting annoying and stupid that every second thing have wrong units then we say it doesn't have units or direction that it is not the direction but the values ok. So it is this.

 

[jbriggs444]

It doesn't matter that is an other example and its ok, the point is that by cross calculation you can't get it right in terms of direction but only values.

The angular rotation vector is perpendicular to the plane of rotation. The cross product of the angular rotation vector with any vector in the plane of rotation will produce a result that is in the plane of rotation.

Its really getting annoying and stupid

Not much I can do about that.

 

[A.T.]

angular velocity is not perpendicular to the rotation

https://en.wikipedia.org/wiki/Angular_velocity

When the angular velocity is represented as a vector, its direction is perpendicular to the plane of rotation

 

[losbellos]

I found that out that this is actually incorrect.
Just imagine a large radius disc where the two radial distance is same (on the radius)
the first pair is close to the center, the second is further.
The circum velocity of the pairs have different ratios. There the force that will accel the object to match the circumference velocity will different.
the 2 *ω*v*m ≠ the forces requires to accel at different circumferences but same distance...
This is a very good news by the way. Stick to Newton system that will work always. :D

 

[jbriggs444]

I had a great deal of difficulty making sense of #13.

I found that out that this is actually incorrect.

What are you saying is incorrect?

Just imagine a large radius disc where the two radial distance is same (on the radius)
the first pair is close to the center, the second is further.

So we have two pairs of points, with the two members of one pair separated from each other by the same radial distance as the two members of the other pair?

The circum velocity of the pairs have different ratios.

Yes, that would be correct. For instance, there could be one pair at r=1 and r=2 and a second pair at r=10 and r=11. The tangential velocities for the first pair would have a ratio of 2 to 1 and the second pair, a ratio of 1.1 to 1.

There the force that will accel the object to match the circumference velocity will different.

It is not clear what force is being mentioned here.

the 2 *ω*v*m ≠ the forces requires to accel at different circumferences but same distance...

The $2m \omega \times v$ gives Coriolis -- the force in addition to centrifugal which seems to deflect a free falling mass m moving with velocity v from the point of view of a frame of reference rotating with angular velocity omega. The Coriolis pseudo-force is independent of distance from the center of rotation.

 

[A.T.]

2 *ω*v*m ≠ the forces requires to accel at different circumferences but same distance...

If you mean that Coriolis force is not the total inertial force in the rotating frame, then that's correct. It's only one component, along with Centrifugal force and (if the frame has variable angular velocity ) Euler force:

https://en.wikipedia.org/wiki/Rotating_reference_frame#Newton's_second_law_in_the_two_frames

 

[losbellos]

Hej it doesn't matter what force is driving that mass outwards. CF or other doesn't matter! In this case the velocity that moves the object out is constant!

Anyways I found it out that in my example the results were close enough initially and later I found it out that with bigger systems it is wrong.
Coriolis force is a force that points upwards in a frame and originally is not related to this. But the calculus seemed to be the same and this is why the guy posted it. So summa-summarium this method cannot be used to calculate the force requires to accel an object to the next circum velocity.Only Newton's s = .5 * a *t^2 and v = a*t can be used in this model.

 

[A.T.]

Coriolis force is a force that points upwards in a frame

"Upwards" means nothing unless you define it.

 

[jbriggs444]

Hej it doesn't matter what force is driving that mass outwards. CF or other doesn't matter! In this case the velocity that moves the object out is constant!

The angular velocity that results in the outward motion of the object is constant, yes. The resulting radial velocity is not constant, no. It seems to me that the easiest way to approach this problem is to adopt the rotating frame of reference.

In this frame of reference there are three forces that act on the object: centrifugal force, Coriolis force, and the tangential force of the shaft on the object. Both the Coriolis force and the tangential force of the shaft on the object are perpendicular to the object's direction of travel in this frame. So they do no work. Only the centrifugal force does work.

The centrifugal force is a central force with a magnitude that depends only on the distance from the center. That means that it is a conservative force. That is, it gives rise to a potential. The force has a magnitude of $m \omega^2 r$. Its potential is [the additive inverse of] the integral of this over r: $PE=-\frac{m \omega^2 r^2}{2}$.

We have two forces that do no work and a third force that has an associated potential. That means that mechanical energy is conserved and can be written as PE+KE=constant. We derived a formula for PE above. We know that [in this frame] tangential velocity is zero. So KE is given by $\frac{1}{2}mv_r^2$ where $v_r$ is the radial velocity.

Given the object's starting radial velocity (and hence, starting KE) and starting radius (and hence, starting PE), one can easily calculate the object's radial velocity at any other radius based on conservation of mechanical energy. That yields a formula for current radial velocity as a function of current radius.

With that formula for radial velocity in hand, one can then easily calculate the Coriolis force as a function of current radius.

The Coriolis force times the current radius is [the additive inverse of] the torque required to keep the object on its radial path down the shaft. That torque is what we wanted to find.

Edit:

However, reading back I see that:

in a system 0.5kg moving outwards with the same velocit

So it seems that there are four forces involved. Something is restraining the object, preventing it from free falling outward. So its radial velocity is constant after all and Coriolis is much more easily calculated based on the fixed velocity.