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RE: [uk_jugglers] friction

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  • Naomi Sajeri
    Wandering around the garden I had more thoughts on the problem, and was about to add just the point made by Owen about rotation. I had indeed only considered
    Message 1 of 2 , May 31, 2004
      Wandering around the garden I had more thoughts on the problem, and was
      about to add just the point made by Owen about rotation.

      I had indeed only considered straight line motion in the effect that the
      equation has on the friction, and was headed for the keyboard to revise, nay
      reverse , my stance. Owen got there before me.

      Contact radius becomes important when spinning because the frictional
      forces are applied as torques. For straight line friction the total
      frictional force is a simple summation of all the forces at each point of
      contact. What I had missed, and I feel rather silly about it, is that for
      rotation we are summing, for each point the product of the force times its
      distance from the centre of rotation to produce a total torque to slow the
      spin. The problem for a surface that distorts to support its weight becomes
      non trivial. Weight supported ( or friction force contributed ) 'per square
      millimetre' becomes less, the further out on the contact area it is, but is
      effectively applied using a longer lever.

      So Nigel, my apologies, for your intuition/observation is correct although
      not for reasons quite as simple as I suspect you may have thought.

      I stupidly missed a vital factor.

      Nao....wandering off to sharpen a nail to a point.

      ----Original Message Follows----
      From: "Owen Jones" <oj206@...>
      Reply-To: uk_jugglers@yahoogroups.com
      To: <uk_jugglers@yahoogroups.com>
      Subject: [uk_jugglers] friction
      Date: Mon, 31 May 2004 09:56:18 +0100

      I think you're both right. Naomi is correct when she says the friction
      is independent of contact area. This doesn't mean you can build a perpetual
      motion machine, it means that if you're dragging something, it doesn't
      matter how much of it is in contact with the ground. However I think this is
      true for linear motion, ie dragging something, but not for rotational
      motion, ie something spinning on your finger. The general idea is that the
      amount of friction acting on a small area is proportional to the force of
      gravity on that area. So doubling the area means each small area has half
      the 'weight' and hence half the friction, but there is twice as much of it,
      and hence the same amount of friction. This is also assuming that every part
      of the contact area is going at the same speed.
      In the end you have to make sure what you are saying is consistent with
      experience and certainly if you put your whole hand on a ball it slows down
      a lot faster than if you balance it on your finger tip. I believe this is
      because the part in contact with your fingertip is not moving very fast
      (being near the axis of rotation), but further out from the axis of rotation
      it is going a lot faster.
      Hope that helps,
      PS Am not claiming that this is definitely right...

      > Date: Sun, 30 May 2004 02:22:11 +0100
      > From: Nigel <n.schaay@...>
      > Subject: Re: ballspinning
      > In your last message, you said:
      > > But Nigel, did your 'O' level physics (GCSE?) not say that friction
      > > independent of contact area?
      > No.
      > My O level physics did involve some laws about boils ( well, it sounded
      > that...) and getting your molecules excited, some random things involving
      > pointing your index finger, sticking your thumb up and your middle finger
      > out at 90degrees to the other two then turning your wrist to help you
      > remember something I've since forgotten, and possibly how to wire a plug.
      > I did take it (very nearly) 21 years ago though, so they probably hadn't
      > invented friction then...
      > >
      > > "Friction is a resistive force that prevents two objects from sliding
      > > against each other. The coefficient of friction (u) is a number that is
      > > ratio of the resistive force of friction (Fr) divided by the normal or
      > > perpendicular force (Fn) pushing the objects together. It is
      > > the equation:
      > >
      > > u = Fr / Fn."
      > >
      > > Hence dynamic friction is independent of contact area.
      > Righty ho then, let us all know when your perpetual motion machine is up
      > running....
      > Or perhaps I should just gently point out the rather high likelihood of
      > being greater when the contacting surfaces are larger...
      > >
      > >
      > > Inflatedness (an interesting word)
      > Hence my hyphenation...
      > N.

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