Summary Lecture 5Summary Lecture 5

5.4 Inertial/non-inertial reference frames

6.1-2 Friction

6.5 Taking a curve in the road

6.4 Drag force

Terminal velocity

Summary Lecture 5Summary Lecture 5

5.4 Inertial/non-inertial reference frames

6.1-2 Friction

6.5 Taking a curve in the road

6.4 Drag force

Terminal velocity

Problems Chap 6: 5, 14, 29 , 32, 33, Problems Chap 6: 5, 14, 29 , 32, 33,

Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

According to stationary observer

R

mgF = 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

R is reaction force

= reading on scales

Measured weight in an accelerating Reference Frame

accel a

Spring scales

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?

R

mg

R is reaction force

= reading on scales

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

mgmg

Why doesn’t Mick Doohan fall over?Why doesn’t Mick Doohan fall over?

Friction provides the central force

Friction provides the central force

In the rest reference frameIn the rest reference frame

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

The Source of Friction between two surfaces

fF

mg

Static Friction

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

F is now greater than f

and slipping begins

If no force F

No friction force fSurface with friction

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

fF

f depends on surface properties.

Combine these properties into a coefficient of friction

f N

is usually < 1

Static f < or = s N

Surface with friction

Kinetic f = k N

f

F

f < fmax (= kN )

Static friction

Kinetic friction

Coefficient of Kinetic friction < Coefficient of Static friction

Slipping begins (fmax = sN )fmax

mg

Ff

At crit

F = f

mg sin crit = f = S N

Independent of m, or g.

Property of surfaces only

S = tan crit

mg sin

mg cos

= S mg coscrit

crit

crit

cosmgsinmg

thus S =

F1

F2

Making the most of Friction

A F1 > F2

B F1 = F2

C F1 < F2

mg

N N

mg

fcrit1

fcrit2

fcrit = S N

Friction force does not depend on area!

fcrit = S mg

So why do Petrol Heads use fat tyres?

To reduce wear?

Tyres get hot and sticky which effectively increases .

The wider the tyre the greater the effect?

To reduce wear?

Tyres get hot and sticky which effectively increases .

The wider the tyre the greater the effect?

The tru

th!

Frictio

n is not a

s sim

ple as P

hysics 1

41 says!

The tru

th!

Frictio

n is not a

s sim

ple as P

hysics 1

41 says!tribophysics

Force of Tyre on road

Force of road on Tyre

acceleration

What force drives the car?

Driving Torque

Braking force

Friction road/tyres

v

d

f

v2 =vo2 + 2a(x-xo)

0 = vo2 + 2ad a

vd

2

20

F = ma

= smgMax value of a is when f is max.

Stopping Distance depends on friction

amax = - sg

-fmax = mamax -smg = mamax

vo

N

mg

fmax = sN

Thus since a

vd

2

20

maxmin a

vd

2

20

gv

ds

min

2

20

dmin depends on v2!! Take care!!

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

r

v Fcent

mg

N

rmv

F2

cent

Fcent is provided by friction.

If no slipping the limit is when

Fcent = fs(limit)= sN = smg

grμv

rmv

mgμ

s

2

s

So that

Does not depend on m

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!

Taking a curve on Flat surface

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.

Albert Park GP circuitCentral force provided Central force provided by friction.by friction.Central force provided Central force provided by friction.by friction.

mg

N

= v= v22/Rg/Rg

= 4.3= 4.3

= v= v22/Rg/Rg

= 4.3= 4.3

mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg

R = 70 m mv2/R

V=

55 m

s-1

for racing tyres is ~ 1 (not 4!).

How can the car stay on the road?

for racing tyres is ~ 1 (not 4!).

How can the car stay on the road?

Soft rubber

Grooved tread

Are these just for show, or advertising?

200 km/h

Another version of Newton #2

amF p= mv =momentum

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

t

pF

dt

vdm

dt

)vm(d

dt

pd

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

Distance travelled in 1 sec @ velocity v Distance travelled in 1 sec @ velocity v

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

Area A m2

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

Momentum of air hitting each spoiler in 1 secMomentum of air hitting each spoiler in 1 sec

If deflected by 900, mom change in 1 secIf deflected by 900, mom change in 1 sec

Newton says this is the resulting forceNewton says this is the resulting force

@ 200 kph v = 55 m s-1

A ~ 0.5 m2

~ 1 kg m-3

F ~ 3 x 104 N

~ 3 Tonne!

= v m

= v x A m3

= x v x A kg

= x v2 x A kg m s-1

mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg

mvmv22/R = /R = N = N = (m + 3000) g(m + 3000) gmvmv22/R = /R = N = N = (m + 3000) g(m + 3000) g

VISCOUS DRAG FORCEVISCOUS DRAG FORCEDRAG

VISCOUS DRAG FORCE

Assumptions

low viscosity (like air)

turbulent flow

What is it?

like fluid friction

a force opposing motion as fluid flows past object

What does the drag force depend on?D D velocity (v velocity (v22))

D D effective area (A) effective area (A)

D D fluid density ( fluid density (

D D A vA v22

D= ½ C A v2

D D velocity (v velocity (v22))

D D effective area (A) effective area (A)

D D fluid density ( fluid density (

D D A vA v22

D= ½ C A v2

C is the Drag coefficient.

It incorporates specifics like

shape, surface texture etc.

v

Fluid of density

V m

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

t

pF

Area A

In 1 sec a length of V metres hits the object

V

mg

mg

D

mg

D

V

V=0

F = mg - D

F = mg -1/2CAv2

D increases as v2

until F=0

i.e. mg= 1/2CAv2

AC

mg2v

AC

mg2v

term

term2

0mgAv1/2Cdt

dvm 2

F = mg –DD

mg

ma = mg -D

D- mgdt

dvm

2/1Ac

m2

)]e1(Ac

gm2[v

t

2/1]Ac

gm2[v

2/1Ac

m2

)]e1(Ac

gm2[v

t

When entertainment defies reality

D= ½ CAv2

Assume C = 1

v = 700 km h-1

Calculate:

Drag force on presidents wife

Compare with weight force

Could they slide down the wire?

D= ½ CAv2

Assume C = 1

v = 700 km h-1

Calculate:

The angle of the cable relative to horizontal.

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

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

2AvC2

1F