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Friction & Circular Motion Static and Moving Friction Centripetal and Centrifugal Forces o Friction was introduced in last lecture, expand  Nature of.

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Presentation on theme: "Friction & Circular Motion Static and Moving Friction Centripetal and Centrifugal Forces o Friction was introduced in last lecture, expand  Nature of."— Presentation transcript:

1 Friction & Circular Motion Static and Moving Friction Centripetal and Centrifugal Forces o Friction was introduced in last lecture, expand  Nature of Friction, its origins  Static Friction  Friction in Motion o Circular motion using centripetal force  ‘Equivalence’ with gravity  Linear analogues

2 o Aim of the lecture  Concepts in Friction Depends on surfaces Depends on contact Forces Depends on motion Air resistance  Circular Motion Centrifugal and Centripetal Forces o Main learning outcomes  familiarity with  forces needed to sustain motion  static and moving friction, coefficients  Forces needed for circular motion  Centrifugal and Centripetal force Lecture 4

3 Newton’s First Law o Reminder: It appears that:  A force is needed to keep an object moving at constant speed.  An object is in its “natural state” when at rest. o These are wrong  friction creates this illusion. o” Where does friction come from? o The illustration shows two surfaces in contact – at the microscopic level Surfaces are NOT ‘flat’.  ‘locked’ together to make them slide will require force to ‘break the locking’ even when already moving to keep ‘breaking the locking’

4 Friction o How hard it is to break the ‘locking’ will depend on:  nature of the surfaces  smoother, less locking  rougher, more locking  (also other effects ‘sticky’ surfaces like rubber)  How hard the surfaces are pressed together  because the ‘true’ contact area depends on force:

5 o For objects sliding over each other o Can be described by simple equation: F =  R Where:  F is the force need to overcome friction  R is the force between the surfaces   is a constant which depends on the surfaces in contact Note: This does NOT depend on the area in contact! If the force BETWEEN the surfaces, R, is the same, then the area does not matter.

6 F =  R Why is the area not important?  Friction force depends on o Force per unit area o Total area  Double area, with same TOTAL force, means o force per unit area is halved  So (total area) × (force per unit area) is not changed Total force needed = const × (F/A) × A

7  is called the coefficient of friction o depends on BOTH surfaces o low for ice on metal o High for rubber on concrete o NOTE: o rolling friction is a different thing  high  for rubber tyre on concrete  Stops tyre sliding  Helps with braking  Does NOT resist the wheel rotating  because there is no sliding of wheel past concrete for rotation F =  R

8 F =  R It is easier to keep something sliding than to start it moving oThere are two coefficients of friction  One for static friction,  s  force needed to start a static object moving  One for dynamic friction  r  force needed to keep moving at constant speed  s >  r Easier to keep sliding than to get moving

9 Approximate coefficients of friction Materials Dry & clean AluminumSteel 0.61 CopperSteel 0.53 BrassSteel 0.51 Cast ironCopper 1.05 Cast ironZinc 0.85 Concrete (wet)Rubber careful driving in wet! 0.3 Concrete (dry)Rubber 1.0 CopperGlass 0.68 Glass 0.94 SteelTeflon Non-stick! Static friction, [ 1.0 is the biggest normal value. If >1.0, then ‘sticky’.

10 F =  R Often it is gravity that is creating the force o The force due to gravity is the weight of the top object o This is a ‘Reaction force’, hence use of ‘R’ in formula R = Force = mg

11 BRASS STEEL Brass-Steel  s = 0.51 W = 10kg x 9.81 ms -1 = 98N So F =  s x R = 98N x 0.51 = 50N This is the force it takes to get a 10kg steel block sliding over a brass surface. 10kg

12 The reaction force between the surfaces is given by R = mg cos(  ) The force down the slope is F = mg sin(  ) If F >  s R block will slide mgsin(  ) >  s mgcos(  ) ie if tan(  ) >  s 

13 o Note that there are other forms of friction  Rolling  Pulley  Air resistance o These do not all behave the ‘same way’ as simple sliding  eg for many situations the force due to air resistance is F = k v 2 o Where v is the velocity and k a constant which depends on  Air density  Shape of object moving through air. Supersonic plane k plane < k bus (at same altitude of flight)

14 Circular Motion It takes a force to change the direction of a moving object motion force New motion

15 oIf a constant force is always applied perpendicular to the motion then  the object will go in a Circle  Its speed will not change  Just the direction alters

16 The force of gravity does this, keeping planets in orbits  The force of the sun is always towards the sun  Which is perpendicular to the direction the plants travel  (not quite a circle – but don’t consider that here)

17 The force needed to make a circle is: Where m is the mass of the circling body v is its speed r is the radius of the circle the force is called the centripetal force v F r

18 If the body is made to circle by a string, then the string will feel a force pulling it into tension. This is the centrifugal force. Of course these two different names are really describing the same force. But from different perspectives o Centripetal  Force applied to make circular motion o Centrifugal  Force ‘felt’ by object being made to circle


20 Riders feel the centrifugal force holding them in the car. The rail applies a centripetal force to make the car circle. Riders ALSO feel gravity, but it is not enough to make them fall out of the car

21 It is common to use  instead of v, where  = v/2  r This is the angular speed - number of radians per second Then the centripetal force is: F =  2 r

22 With F =  R (concrete-rubber  = 1.0) and F = mv 2 /r can answer question: How fast can you drive your Ferarri round a corner? Answer: Depends on the speed limit Answer: I cant afford a FerarriAnswer: I cant afford a Ferarri, but in theory:

23 Consider a real car (Lotus Elise) v r o Force needed to make the car turn in circle  F circle = mv 2 /r o Force provided by friction between wheel and road  F friction =  R where R = mg o If F circle is greater than F friction then friction will not be enough  and Elise will slip (but designed with oversteer so will slip rear end first – chance to recover) Condition is mv 2 /r =  mg => v = √(  gr) Centripetal force

24 Answer: v = √(  gr) (BUT remember  is much smaller in the wet!)

25 Finally, when cornering: remember that F =  R o R can be bigger if the car has proper aerodynamics  The reason cars have ‘wings’ on them is to force the car down increasing R to a bigger value than mg, [it flies into the ground] o In formulae 1, the cars are designed so R is very large,  the cars stick like limpets round the corners,  but only if they go fast, so R is larger than mg!

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