From: relativity@paulba.no   
      
   Den 22.01.2026 23:46, skrev Ross Finlayson:   
   > On 01/22/2026 12:57 PM, Paul B. Andersen wrote:   
   >> Den 21.01.2026 16:11, skrev Maciej Woźniak:   
   >>> On 1/21/2026 3:31 PM, Paul B. Andersen wrote:   
   >>>>   
   >>>> You are swinging an object around you in a string.   
   >>>> Is the object accelerating towards you, or away   
   >>>> from you? Or is it not accelerating at all?   
   >>>   
   >>> Towards me.   
   >>   
   >> Right, well done.   
   >>   
   >>>>   
   >>>> I bet you [Maciej Woźniak] won't give a sensible answer.   
   >>>> You never do.   
   >>   
   >> I was wrong about that.   
   >> You gave a very sensible answer.   
   >>   
   >> So you have understood that the object is accelerating   
   >> in the direction the force acting on it is pulling it.   
   >> This direction is along the string towards your hand.   
   >> The acceleration is:   
   >>    a = F/m   
   >> where F is the tension in the string and m is the mass   
   >> of the object.   
   >>   
   >> Let us look at a concrete example.   
   >> The length of the string is L = 1.4142 m, and the mass m is 1 kg.   
   >> You swing the object such that it takes t = two seconds to make   
   >> a full turn.   
   >> You observe that the angle of the string to the ground is 45⁰.   
   >> That means that the radius of the circle the object is moving   
   >> along is:   
   >>    r = L/√(2) = 1 m   
   >> It is then easy to calculate that the speed of the object is   
   >>    v = 2⋅π⋅r/t = 3.14 m/s   
   >> The horizontal centripetal acceleration is:   
   >>    aₕ = v²/r = 9.8 m/s²  (towards centre of circle)   
   >> The horizontal component of the tension in the string is:   
   >>    Fₕ = m⋅aₕ = 9.8 N   
   >> Since the angle of the string to the ground is 45⁰, the vertical   
   >> component of of the tension in the string is:   
   >>    Fᵥ = Fₕ   
   >> So the vertical acceleration is:   
   >>    aᵥ = Fᵥ/m = 9.8 m/s²  (upwards)   
   >> The tension i the string is:   
   >>    F = √(Fₕ² + Fᵥ²) = 13.86 N   
   >> And the acceleration of the object is:   
   >>    a = F/m = 13.86 m/s² (towards your hand)   
   >>   
   >> I am sure you will understand this, and   
   >> will give a sensible response.   
   >>   
   >   
   > Ah, ye olde "swinging the bucket".   
      
   Did you mean þe old "swinging the bucket"?   
      
   >   
   > The upwards acceleration must be according to   
   > greater-than or g.t.e. than 1G to keep the string taut.   
      
   The upwards acceleration _is_ 1G (1 g).   
      
   >   
   >   
   > About, "the true centrifugal",   
   > and schoolroom accounts of the centripetal,   
   > since the tournos or lathe and the foot-wheel,   
   > it's ideal that 1) the axis of rotation is   
   > an immovable object, that in the true centrifugal   
   > the centripetal is fictitious, as in the ideal   
   > centripetal the centrifugal is fictitious.   
   >   
   > Either (and both) of which depending where one lets go   
   > is in effect.   
   >   
   > Everybody sat the standard linear curriculum,   
   > including the day when it was explained why   
   > it was going to be "centripetal" instead of   
   > "centrifugal" for the course.   
   >   
   > Then the true centrifugal is still a thing,   
   > "wheels within wheels" as it where.   
   >   
   > For example capillary action up the blood platelet   
   > centrifuge.   
   >   
   >   
   > "Leibnitz" and "Newton" differ their explanations   
   > the "swinging the bucket", about the "centripetal"   
   > how many foot-kicks it takes to get it going.   
   >   
   > Or, you know, wrist-flips.   
   >   
      
   Amazing! :-D   
      
      
   --   
   Paul   
      
   https://paulba.no/   
      
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