## RE: [uk_jugglers] friction

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Message 1 of 2 , May 31, 2004
Wandering around the garden I had more thoughts on the problem, and was

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@...>
To: <uk_jugglers@yahoogroups.com>
Subject: [uk_jugglers] friction
Date: Mon, 31 May 2004 09:56:18 +0100

Hi,
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,
Owen
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
was
> > independent of contact area?
> No.
> My O level physics did involve some laws about boils ( well, it sounded
like
> that...) and getting your molecules excited, some random things involving
> 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
freely
> > against each other. The coefficient of friction (u) is a number that is
the
> > ratio of the resistive force of friction (Fr) divided by the normal or
> > perpendicular force (Fn) pushing the objects together. It is
represented
by
> > 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
and
> running....
>
> Or perhaps I should just gently point out the rather high likelihood of
Fr
> being greater when the contacting surfaces are larger...
> >
> >
> > Inflatedness (an interesting word)
> Hence my hyphenation...
>
> N.

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