Rotating frames - Apparent gravity

[dyn]

Hi
On the Earth , apparent gravity comes from the vector addition of the gravitational force directed towards the centre and the outward centrifugal force. It means that for a pendulum at rest , the direction the bob hangs downwards is not directly towards the centre of the Earth but there is a small deflection.
But what happens from the perspective of an observer in an inertial frame away from the Earth ? In an inertial frame there are no fictitious forces , so there is no centrifugal force. I presume the bob must also be slightly deflected in the same way ; but what causes this deflection ?
Thanks

 

[Halc]

But what happens from the perspective of an observer in an inertial frame away from the Earth ? In an inertial frame there are no fictitious forces , so there is no centrifugal force. I presume the bob must also be slightly deflected in the same way ; but what causes this deflection ?

I assume the bob is attached to something rotating, in which case the bob points away from the axis of rotation. This is due to centripetal force of the string accelerating the bob always towards the axis.

 

[kuruman]

The inertial observer sees the pendulum support go around in a circle about the Earth's axis of rotation. The pendulum bob also goes around a similar circle. The bob needs a centripetal acceleration to go around that circle and that can come only from the string which has to be tipped relative to the radial position so that there will be an inward tension component. In short, what is centrifugal in the non-inertial frame is centripetal in the inertial frame.

 

[anuttarasammyak]

In the IFR equation of motion for the rotating bob is
$$\mathbf{T}+m\mathbf{g}=m\mathbf{a}$$
where RHS is centripetal force and T is torsion working on the bob from the code.
In the frame of rotating Earth surface this same equation is interpreted as balancing of the forces, i.e.,
$$\mathbf{T}+m\mathbf{g}+(-m\mathbf{a})=0$$
,the third LHS term of which is centrifugal force.

 

[A.T.]

In the frame of rotating Earth surface this same equation is interpreted as balancing of the forces, i.e.,
$$\mathbf{T}+m\mathbf{g}+(-m\mathbf{a})=0$$

Why should the forces on the bob be balanced in the rotating frame of the Earth? The bob is accelerating in that frame as well.

 

[anuttarasammyak]

On the Earth , apparent gravity comes from the vector addition of the gravitational force directed towards the centre and the outward centrifugal force.

Why should the forces on the bob be balanced in the rotating frame of the Earth? The bob is accelerating in that frame as well.

I mean OP's "On the Earth" frame above quoted.

 

[A.T.]

I mean OP's "On the Earth" frame above quoted.

Sorry, I missed that we a talking about a pendulum just hanging down, so basically a plummet.

 

[dyn]

In the rotating frame , the centrifugal force deflects apparent gravity away from the rotation axis. In an inertial frame the centripetal force would deflect apparent gravity towards the rotation axis. So the two deflections would be in opposite directions ?

 

[anuttarasammyak]

In the rotating frame , the centrifugal force deflects apparent gravity away from the rotation axis. In an inertial frame the centripetal force would deflect apparent gravity towards the rotation axis. So the two deflections would be in opposite directions ?

I said in post #4 that :
In the rotating Earth frame, torsion and gravity BALANCE with centrifugal force.
In the inertial frame, torsion and gravity MAKE centripetal force.
The deflections are same.

 

[dyn]

In the rotating Earth frame apparent gravity is given by the vector addition of the centripetal acceleration and g. This produces a deflection away from g away from the Earths rotation axis. I don't understand how this same deflection occurs in an inertial frame as any centripetal force acts inwards ?

 

[A.T.]

I don't understand how this same deflection occurs in an inertial frame as any centripetal force acts inwards ?

The coordinate accelerations are different in the two frames, even though the deflection is the same. That's why the centrifugal force is introduced in the rotating frame, to match the frame dependent coordinate acceleration.

 

[kuruman]

Consider a pendulum moving around in a horizontal circle.

In the inertial (lab) frame fixed on a non-rotating Earth you see the bob moving around in a circle with the string tipped away from the vertical. There are two forces acting on the bob, tension and gravity. Which one do you think provides the centripetal acceleration? Answer: Tension because the centripetal acceleration is horizontal while gravity is vertical. So tension must have a vertical component that prevents the mass from falling vertically and a horizontal component pointing towards the center that makes the mass go around the circle. It is a fallacy to think that just because a force is in a given direction the mass acted upon by that force must also move in that direction.

Now consider being an observer at rest with respect to the bob in the non-inertial frame. You see the string at an angle relative to the vertical but not moving. You deduce the existence of a "centrifugal" force that pulls the bob which when added to gravity tilts the string away from the vertical.

 

[anuttarasammyak]

In the rotating Earth frame apparent gravity is given by the vector addition of the centripetal acceleration and g. This produces a deflection away from g away from the Earths rotation axis. I don't understand how this same deflection occurs in an inertial frame as any centripetal force acts inwards ?

I draw figures for post #9. You fill find same deflection angle $\theta$ here.

 

[dyn]

I'm sorry , but I am just not seeing this. For example in the book Classical Mechanics 5th Ed. by Kibble on P113 there is a diagram of the deflection and g* = g - w x (w x r ). This equation doesn't even mention tension
The centripetal acceleration acts in the opposite direction to the centrifugal acceleration so to me it gives a deflection in the opposite direction

 

[dyn]

I think I've got it now. Let me check :
In the rotating frame , the apparent gravity is the resultant vector from the addition of g and the centrifugal acceleration , so g* = g + (-w x (w x r) )

In the inertial frame the resultant vector is the centripetal acceleration from the addition of g and the negative of apparent g , so w x (w x r) = g + (-g*)

resulting in the same deflection from both frames of reference. Is that right ?

One last question ; I know in inertial frames there should be no mention of fictitious forces. Can there be centripetal forces in rotating frames ?
Thanks

 

[anuttarasammyak]

In the rotating Earth frame apparent gravity is given by the vector addition of the centripetal acceleration and g. This produces a deflection away from g away from the Earths rotation axis.

In the rotating Earth frame, not centripetal but centrifugal acceleration you mean, I read.

I don't understand how this same deflection occurs in an inertial frame as any centripetal force acts inwards ?

In IFR NO centripetal force works on free moving bodies. They keep moving with a constant speed, and do free fall if Earth gravity exists. Say a pendulum code is torn, a bob does a horizontally motion of straight and tangential to the rotating FR and does a vertical free fall. You will see the bob goes OUTWARD in IFR as well as in the rotating Earth frame. Corioli force should also be considered for a correct analysis.

In order centripetal force to act on body we need arrangements to fix it to the rotating Earth FR, for an example pendulum and torsion force.

In the rotating frame , the apparent gravity is the resultant vector from the addition of g and the centrifugal acceleration , so g* = g + (-w x (w x r) )

So In IFR, force applied on body without any constraint, e.g. torsion from pendulum codem, is just $m\mathbf{g}$.

 

[A.T.]

In the inertial frame the resultant vector is the centripetal acceleration from the addition of g and the negative of apparent g ...

There are no apparent forces in the inertial frame.

...resulting in the same deflection from both frames of reference.

You seem to have the misconception that the net forces in both frames must be equal in order to get the same deflection. Or that the string is parallel to the effective gravity in both frames. Neither is the case:

The net forces must not be equal, because the accelerations are different. The string is only parallel to effective gravity (mass attraction + centrifugal) in the rotating frame, where the bob is in equilibrium . In the inertial frame the bob is not in equilibrium and the string is not parallel to effective gravity (mass attraction).

 

[dyn]

So , for a mass dropped from rest before it starts to move so any Coriolis force is zero.
In the rotating frame . the initial acceleration is g* = g + (-w x (w x r )) which is directed slightly away from the centre of the Earth.
In an IF the initial acceleration is just g directed towards the centre of the Earth ?

 

[anuttarasammyak]

Yes, I think so.

 

[jbriggs444]

So , for a mass dropped from rest before it starts to move so any Coriolis force is zero.
In the rotating frame . the initial acceleration is g* = g + (-w x (w x r )) which is directed slightly away from the centre of the Earth.
In an IF the initial acceleration is just g directed towards the centre of the Earth ?

There is a good drawing in the final post of the thread here.