Coriolis and Centrifugal effects in balance

[DaTario]

Hi all,

Consider a platform with angular velocity $\omega$. A particle on top of it has a velocity with only $\hat \theta$ component (no radial $\hat r$ velocity). In this case, the inertial forces read:
$$
F_{in} = 2m\omega V_\theta \, \hat r + m \omega^2 r \, \hat r
$$
If $V_\theta = -(\omega r)/2$ the intertial forces apparently cancel out. Is it possible? I am a bit confuse with this situation. The particle seems not to be even at rest wrt the lab reference frame.
Any comment is wellcome.

Best wishes,

DaTario

 

[A.T.]

Is it possible that, in a platform with angular velocity $\omega$, if the particle has no radial velocity, the Coriolis and the centrifugal force cancel out for a given and specific velocity, yielding no inertial force at all?

Yes. But if there are no other forces, the particle would tend to move on a straight line in the rotating frame, so it wouldn't maintain zero radial velocity, and the two inertial forces wouldn't stay anti-parallel.

You can additionally apply a centripetal force, to have the particle going in circles at the same rate but opposite directions in both frames. This requires the same net force in the inertial and non-inertial frame, therefore inertial forces must cancel each other.

The particle seems not to be even at rest wrt the lab reference frame.

If it was at rest in the inertial frame, it would go in circles in the rotating frame, and need a centripetal force for that (provided by Coriolis).

 

[DaTario]

Yes. But if there are no other forces, the particle would tend to move on a straight line in the rotating frame, so it wouldn't maintain zero radial velocity, and the two inertial forces wouldn't stay anti-parallel.

Does it mean that the balance between Coriolis and Centrifugal forces, as I described, must necessarily occur intantaneously?

 

[jbriggs444]

Does it mean that the balance between Coriolis and Centrifugal forces, as I described, must necessarily occur intantaneously?

There are caveats.

An object subject to no other forces and with Coriolis and centrifugal force in balance will maintain a straight-line trajectory at constant velocity in the uniformly rotating frame. If that constant velocity is non-zero, the Coriolis force remains constant while the Centrifugal force will change. So the balance can only apply for an instant. If the velocity is zero then Coriolis and centrifugal forces could not have been balanced in the first place.

However, as @A.T. pointed out, if the object is subject to a real centripetal force, it can maintain a circular path with just the right tangential velocity so that centrifugal and Coriolis remain in balance continuously.

In a similar vein, an arbitrary external force that just happens to direct the same object tangentially past the same radial coordinate in the same direction at the same speed will naturally result in a second instant when Coriolis force matches centrifugal force.

A final caveat is rather stupid: If the "rotating" frame has an angular velocity of zero then Coriolis and centrifugal forces are always zero and are always in balance regardless.

 

[DaTario]

So, speaking from the non intertial frame of reference, by adjusting the velocity to $(\omega r/2) \, \hat \theta$, we manage do balance Coriolis and centrifugal, but as a side effect we see the appearance of a centripetal force, is it? Sorry but I am still a bit confused, :).

 

[jbriggs444]

So, speaking from the non intertial frame of reference, by adjusting the velocity to $(\omega r/2) \, \hat \theta$, we manage do balance Corilois and centrifugal, but as a side effect we see the appearance of a centripetal force, is it? Sorry but I am still a bit confused, :).

One cannot manipulate real forces by changing frames of reference. [And apologies in advance because you did not quite suggest that one could].

You have to actually physically apply the centripetal force to maintain the object in the trajectory. That trajectory is accelerating as viewed both by the inertial frame and by the rotating frame. The proper acceleration of the object is invariant.

 

[A.T.]

Sorry but I am still a bit confused, :).

Analyze the forces for a stone circling on a string, from the inertial frame, and a frame with twice the angular velocity of the stone.

 

[DaTario]

I guess now I understand, and two cases seem to be the most natural ones.
1) the particle has $v_\theta = (\omega r /2) \, \hat \theta$ at some given instant $t_0$ and the only non zero forces acting on it are the Coriolis and centrifugal. In this case the particle behaves as a free particle, for the two forces cancel out, and its movement is uniform along a straight line. But as it walks any given length after $t_0$ the equilibrium condition doesn't match any more and the centrifugal and Coriolis forces now yield a non zero resultant force.
2) the particle has always $v = v_\theta = (\omega r /2) \, \hat \theta$ and it implies that, in the non intertial reference frame, although the Coriolis and the centrifugal forces keep canceling out indefinitely, a centripetal force (real force) is present and the particle has its dynamics described by this force alone effectively.

 

[A.T.]

I guess now I understand, and two cases seem to be the most natural ones.
1) the particle has $v_\theta = (\omega r /2) \, \hat \theta$ at some given instant $t_0$ and the only non zero forces acting on it are the Coriolis and centrifugal. In this case the particle behaves as a free particle, for the two forces cancel out, and its movement is uniform along a straight line. But as it walks any given length after $t_0$ the equilibrium condition doesn't match any more and the centrifugal and Coriolis forces now yield a non zero resultant force.
2) the particle has always $v = v_\theta = (\omega r /2) \, \hat \theta$ and it implies that, in the non intertial reference frame, although the Coriolis and the centrifugal forces keep canceling out indefinitely, a centripetal force (real force) is present and the particle has its dynamics described by this force alone effectively.

Yes, but note that the real centripetal force in the 2nd case is the same in all frames. In the inertial frame and your rotating frame the particle goes in circles at the same rate, but opposite directions.

 

[vanhees71]

A nice example is to create an equipotential surface of a rotating system in the homogeneous gravitational field of the earth.

https://dspace.mit.edu/bitstream/ha...Sciences/12-003Fall-2007/Labs/detail/lab4.htm

There the 2nd law holds for the special case of a resting particle: Put a particle at rest relative to the parabolic turntable surface, and it will stay at rest. Of course if it comes to motion along the table, the 2nd law doesn't hold anymore but you get some fascinating trajectories, which can be calculated (some cases even analytically).

 

[DaTario]

Thanks, A.T., vanhees71 and jbriggs444.