John2020

Well, then I got it all wrong. Let us close the discussion here.

Yes this is it. Please show me what doesn't match. The first expression (as also in the rotating frame where the tension is labelled as centrifugal (see @swansont comment on centripetal) shows it is equal or greater and never less than the centripetal. When the mass m reaches the stop then, tension again becomes equal to the centripetal (stop=infinite static friction coefficient). How it is external to the rod when nothing applies externally to the rod. The rod has an angular velocity and when this changes then it acquires an angular acceleration. Thus, the tangential acceleration cannot be attributed to an external force but just from the y-component tension of the rod (or call it whatever you like it). I have already answered this. As I mentioned above since the centripetal is the static friction force, we could express the tension proportional to the angular velocity. At Δω = 0, we have the tension being equal to the centripetal force (uniform circular motion). For Δω ≠ 0, results a non-uniform circular motion that implies Tension > Centripetal. That is all.

The idea behind this is that the centripetal equals to the static friction force initially that means the mass m is not fixed on the rod. If it was fixed then the tension would be always equal to the centripetal force (like having an infinite friction coefficient) at any angular velocity e.g. ω+Δω. From the moment mass m is not fixed, when the angular velocity increases the centripetal force remains the same during the transition (or better it becomes equal to the kinetic friction (which is much smaller than the static friction, something that is not shown in the equations) we may formulate the tension as being proportional to Δω. They describe the drawing with the rotating mass m. I don't know how should I call it. What is certain is, it is not an external force.

I admit I didn't backed my arguments properly. This is what I wanted to show you (I hope now you may agree with this): \[\text{Inertial frame} \\ T_{x_{(\omega + \Delta \omega)}} = m (\omega + \Delta \omega )^2(r + \Delta r) \\ F_{cp_{(\omega)}} = F_{fr} = \mu_s\cdot T = m \omega^2r \\ \\ x-Axis: \overbrace{T_{x_{(\omega + \Delta \omega)}}}^{\longrightarrow} \overbrace{- F_{cp_{(\omega)}}}^{\longleftarrow} = \Delta p_m / \Delta t \\ \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{x_{(\omega + \Delta \omega)}} = F_{cp_{(\omega)}} = m \omega^2r \Rightarrow \Delta p_m / \Delta t = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{x_{(\omega + \Delta \omega)}} > F_{cp_{(\omega)}} \Rightarrow \Delta p_m / \Delta t \neq 0 \\ \\ y-Axis: T_{y_{(\omega + \Delta \omega)}} = m (\Delta \omega / \Delta t)\cdot (\Delta r) \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} \neq 0 \\ \\ \\ \\ \\ \text{Rotating frame} \\ F_{cf_{(\omega + \Delta \omega)}} = m (\omega + \Delta \omega )^2(r + \Delta r) \\ F_{cp_{(\omega)}} = F_{fr} = \mu_s\cdot T = m \omega^2r \\ \\ x-Axis: \overbrace{F_{cf_{(\omega + \Delta \omega)}}}^{\longrightarrow} \overbrace{- F_{cp_{(\omega)}}}^{\longleftarrow} = \Delta p_m / \Delta t \\ \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow F_{cf_{(\omega + \Delta \omega)}} = F_{cp_{(\omega)}} = m \omega^2r \Rightarrow \Delta p_m / \Delta t = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow F_{cf_{(\omega + \Delta \omega)}} > F_{cp_{(\omega)}} \Rightarrow \Delta p_m / \Delta t \neq 0 \\ \\ y-Axis: T_{y_{(\omega + \Delta \omega)}} = m (\Delta \omega / \Delta t)\cdot (\Delta r) \\ \Delta \omega = 0 \Rightarrow \Delta r = 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} = 0 \\ \Delta \omega \neq 0 \Rightarrow \Delta r \neq 0 \Rightarrow T_{y_{(\omega + \Delta \omega)}} \neq 0\] T: Tension of the rigid rod Fcp: Centripetal force Fcf: Centrifugal force Ffr: Static friction force μs: Static friction coefficient Consequently, for Δω≠0 and from the conservation of linear momentum on x-Axis and y-Axis, the system (M+m) will start to accelerate (following a curve) during the transition of angular velocity from ω to ω+Δω

Not yet. I have still some tricks under my sleeve.

This is my fault, I will come again on this later.

OK. I will find a simpler example to explain what is going on in the inertial frame. A little bit later. I will present a little bit later the simple classical case of the centripetal acceleration. OK. I answered to his post, however wrongly because I assumed that he was speaking for the rotating frame.

This is all I can. It will be the same from wherever I start. I would suggest to present your version for both frames and to show me where I am wrong.

I am not assuming something else. This is the reason I take the difference that equals to Fnet that represents the effective force that applies to mass m. In other words, friction remains where it is but due to the increase in angular velocity, the centrifugal force will be greater than friction, thus it will start to accelerate mass m. So you indirectly claim there are no centrifugal forces in rotational frame. Who is violating the laws of physics? You or me. Could you please show your version for both frames? Just to have a reference for what we are talking about as also how it is correctly expressed mathematically.

You speak about the inertial frame. Well, this is the part I couldn't otherwise justify in the inertial frame in order to agree with what I wrote about the rotating frame (here the centripetal equals to friction at ω1. Afterwards, the centrifugal becomes greater than the friction while up to ω2).

If you agree with what I wrote about the rotational frame (centrifugal is present and cannot be removed) then I missed something in the inertial frame or vice versa (here you have to justify the centrifugal does not exist but in rotational frames there is always one therefore it has to be taken into consideration).

Scroll up this page and about 4 posts from here there you may find the expression for the inertial frame. Then show me your version for both frames.

This is valid in the inertial frame as I show above but not in the rotating frame.

It depends how one interprets it. Normally, I should have placed a negative sign in front of the centripetal force for the inertial frame in order to be consistent with the forces directions in all frames. In the inertial frame, it should appear with an opposite sign from that in the rotating frame. We shouldn't argue about this. The issue here is do we have an acceleration upon the mass m or not? OK.

You are right my mistake (regarding the sum). The difference is the transition from one to another angular velocity. \[\text{rotating frame} \\ F_{net(\omega))} = F_{cf} - F_{cp} = F_{cf} - F_{fr} = m \omega^2r - F_{fr} \\ \\ \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow F_{cf} = F_{fr} \Rightarrow \\ F_{net(\omega_1))} = 0 \Rightarrow a = F_{net(\omega_1))} / m = 0 \\ \\ \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow F_{cf} > F_{fr} \Rightarrow \\ F_{net} = F_{net(\omega_2))} - F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ a = F_{net} / m > 0 \\ \\ \\ \\ \text{inertial frame} \\ F_{net(\omega))} = F_{cp} = F_{fr} = m \omega^2r \\ \\ \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow F_{cp} = F_{fr} \Rightarrow \\ F_{net(\omega_1))} > 0 \Rightarrow a = F_{net(\omega_1))} / m > 0 \\ \\ \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow F_{cp} > F_{fr} \Rightarrow \\ F_{net} = F_{net(\omega_2))} - F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ a = F_{net} / m > 0 \\ \] Correct. I would say for the time frame of the analysis, from above ω1 up to ω2 motion is not circular anymore.

I will show a solution for each frame. Please see below: \[\text{rotating frame} \\ \sum F_{net(\omega))} = F_{cf} - F_{cp} = F_{cf} - F_{fr} = m \omega^2r - F_{fr} \\ \\ \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow F_{cf} = F_{fr} \Rightarrow \\ \sum F_{net(\omega_1))} = 0 \Rightarrow a = \sum F_{net(\omega_1))} / m = 0 \\ \\ \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow F_{cf} > F_{fr} \Rightarrow \\ \sum F_{net} = \sum F_{net(\omega_2))} - \sum F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ a = \sum F_{net} / m > 0 \\ \\ \\ \\ \text{inertial frame} \\ \sum F_{net(\omega))} = F_{cp} = F_{fr} = m \omega^2r \\ \\ \text{Initially: } \omega = \omega_1 \Rightarrow r = r_1 \Rightarrow F_{cp} = F_{fr} \Rightarrow \\ \sum F_{net(\omega_1))} > 0 \Rightarrow a = \sum F_{net(\omega_1))} / m > 0 \\ \\ \text{Later: } \omega = \omega_2 \Rightarrow r = r_2 \Rightarrow F_{cp} > F_{fr} \Rightarrow \\ \sum F_{net} = \sum F_{net(\omega_2))} - \sum F_{net(\omega_1))} = m (\omega_2^2r_2 - \omega_1^2r_1) > 0 \Rightarrow \\ a = \sum F_{net} / m > 0 \\ \]

OK, I will present them within the next 30 min or so.

Show me yours first.

OK. Let us see what may come out from this now.

OK.

Then we have just one reading that should be outwards.

Actually, I don't know how you assume an acceleration from the inertial frame perspective since the acceleration can be measured only when the accelerometer is in the rotating frame. I am again confused with the reference frames. In the rotating frame is for instance +2G (outwards) and in the inertial frame is -2G (inwards) again during the transition. For constant angular velocity there is no radial acceleration (accelerometer will show zero). This is how I understand it.

The same applies for my example and I am still speaking for the rotating frame. Nobody said the laws of physics change because of that. When you say the acceleration will be always towards the center in the rotating frame, it is like you deny the existence of the centrifugal force in the rotating frame (of course there is none in the inertial frame). Again, I speak for the transition time to a greater angular velocity.

I speak always for the case of increase of angular velocity and not while the angular velocity is constant (radial acceleration would be then zero). Inside a car you are influenced by the centrifugal (outwards acceleration), otherwise you wouldn't be pushed against the door of the car.

No. In the inertial frame the accelerometer will show an inward (e.g. minus) and in the rotating frame an outward (plus).