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Summary Lecture 5 5.4Inertial/non-inertial reference frames 6.1-2Friction PowerPoint PPT Presentation


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Thursday 12 – 2 pm PPP “ Extension” lecture. Room 211 podium level Turn up any time. Summary Lecture 5 5.4Inertial/non-inertial reference frames 6.1-2Friction 6.5Taking a curve in the road 6.4Drag force Terminal velocity.

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Summary Lecture 5 5.4Inertial/non-inertial reference frames 6.1-2Friction

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Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

Summary Lecture 5

5.4Inertial/non-inertial reference frames

6.1-2Friction

6.5Taking a curve in the road

6.4Drag force

Terminal velocity

Problems Chap6:5, 14,29 , 32, 33,

29


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

accela

R is reaction force

= reading on scales

R

mg

Spring scales

Measured weight in an accelerating Reference Frame

According to stationary observer

F = ma

Taking “up” as +ve

R - mg = ma

R = m(g + a)

If a = 0  R = mg normal weight

If a is +ve R = m(g + a) weight increase

If a is -ve R = m(g - a) weight decrease


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

R is reaction force

= reading on scales

R

mg

According to traveller

F = ma

R - mg = ma

BUT in his ref. frame a = 0!

so R = mg!!

How come he still sees R changing when lift accelerates?

Didn’t we say the laws of physics do not depend on the frame of reference?

Only if it is an inertial frame of reference! The accelerating lift is NOT!


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

there's a fraction too much....

Friction


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Why doesn’t Mick Doohan fall over?

Friction provides the central force

mg

In the rest reference frame


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

What is Friction

  • Surfaces between two materials are not even

  • Microscopically the force is atomic

  •  Smooth surfaces have high friction

  • Causes wear between surfaces

  •  Bits break off

  • Lubrication separates the surfaces


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

The Source of Friction between two surfaces


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Static Friction

If no force F

No friction forcef

N

Surface with friction

F

f

mg

As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero does not move

As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly.

Fis now greater thanf

and slipping begins


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

N

Surface with friction

F

f

f depends on surface properties.

Combine these properties into a coefficient of frictionm

fmN

m is usually < 1

Staticf < or = ms N

Kineticf = mk N


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

fmax

Slipping begins (fmax = sN )

f

f < fmax (= kN )

Coefficient of Kinetic friction < Coefficient of Static friction

F

Static friction

Kinetic friction


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

N

f

F

q

mg cosq

q

mg sinq

mg

thus mS =

At qcrit

F = f

mg sin qcrit = f = mS N

= mS mg cosqcrit

mS = tan qcrit

Independent of m, or g.

Property of surfaces only


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

N

N

F1

F2

fcrit2

fcrit1

mg

mg

Making the most of Friction

AF1 > F2

fcrit = S N

BF1 = F2

fcrit = S mg

CF1 < F2

Friction force does not depend on area!


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

So why do Petrol Heads use fat tyres?

tribophysics

To reduce wear?

Tyres get hot and sticky which effectively increases .

The wider the tyre the greater the effect?

The truth!

Friction is not as simple as Physics 141 says!


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

acceleration

Driving Torque

Force of road on Tyre

Force of Tyre on road

What force drives the car?


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

v

vo

N

d

f

mg

Stopping Distance depends on friction

Braking force

Friction road/tyres

v2 =vo2 + 2a(x-xo)

0 = vo2 + 2ad

Max value of ais when f is max.

fmax = sN

= smg

F = ma

 -fmax= mamax

-smg = mamax

amax = - sg


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Thus since

dmin depends on v2!!Take care!!

If v0 = 90 kph (24 m s-1) and m = 0.6 ==> d = 50 m!!


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Taking a curve on Flat surface

N

v

Fcent

r

mg

Fcent is provided by friction.

If no slippingthe limit is when

Fcent = fs(limit)

= sN

= smg

So that

So for a given s (tyre quality) and given r there is a maximum vel. for safety.

If s halves, safe v drops to 70%….take care!

Does not depend on m


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Lateral Acceleration of 4.5 g

The lateral acceleration experienced by a Formula-1 driver on a GP circuit can be as high as 4.5 g

This is equivalent to that experienced by a jet-fighter pilot in fast-turn manoeuvres.


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Albert Park GP circuit

N

mv2/R

mg

R = 70 m

V=55 m s-1

Central force provided by friction.

mv2/R = N = mg

 = v2/Rg

  = 4.3

  • for racing tyres is ~ 1 (not 4!).

    How can the car stay on the road?


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Soft rubber

Grooved tread

Are these just for show, or advertising?


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

200 km/h


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

p= mv =momentum

Another version of Newton #2

Momentum p transferred over a time t gives a force:-

F is a measure of how much momentum is transferred in time Dt


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

@ 200 kph v = 55 m s-1

= v m

Distance travelled in 1 sec @ velocity v

= v x A m3

Volume of air hitting each spoiler (area A) in 1 sec

Area A m2

A ~ 0.5 m2

= x v x A kg

mass of air (density )hitting each spoiler in 1 sec

 ~ 1 kg m-3

= x v2x A kg m s-1

Momentum of air hitting each spoiler in 1 sec

If deflected by 900, mom change in 1 sec

F ~ 3 x 104 N

~ 3 Tonne!

Newton says this is the resulting force

mv2/R = N = mg

mv2/R = N =  (m + 3000) g


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

VISCOUS DRAG FORCE

DRAG


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

That's all folks


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

VISCOUS DRAG FORCE

What is it?

like fluid friction

a force opposing motion as fluid flows past object

Assumptions

low viscosity (like air)

turbulent flow


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

v

C is the Drag coefficient.

It incorporates specifics like shape, surface texture etc.

What does the drag force depend on?

D  velocity (v2)

D  effective area (A)

D  fluid density (r)

D  rA v2

D= ½ C rA v2


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Vm

Area A

Fluid of density

In 1 sec a length of V metres hits the object

Volume hitting object in 1 sec. =AV

Mass hitting object in 1 sec. =  AV

momentum (p) transferred to object in 1 sec. = ( AV)V

Force on object = const AV2


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

V=0

mg

D

V

mg

D

V

mg

SF = mg - D

SF = mg -1/2CrAv2

D increases as v2

until SF=0

i.e. mg= 1/2CrAv2


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

D

dv

=

m

mg

-

D

dt

mg

dv

+

r

-

=

2

m

1/2C

Av

mg

0

dt

F = mg –D

ma = mg -D


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

When entertainment defies reality


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Calculate:

Drag force on presidents wife

Compare with weight force

Could they slide down the wire?

D= ½ CrAv2

Assume C = 1

v = 700 km h-1


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

Calculate:

The angle of the cable relative to horizontal.

Compare this with the angle in the film (~30o)

D= ½ CrAv2

Assume C = 1

v = 700 km h-1


Summary lecture 5 5 4 inertial non inertial reference frames 6 1 2 friction

In working out this problem you will prove the expression for the viscous drag force


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