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

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

Thursday 12 – 2 pm PPP “ Extension” lecture. Room 211 podium level Turn up any time. Summary 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.

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Thursday 12 – 2 pm

PPP “Extension” lecture.

Room 211 podium level

Turn up any time

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

29

accela

R is reaction force

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

R is reaction force

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!

Why doesn’t Mick Doohan fall over?

Friction provides the central force

mg

In 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
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

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

Static f < or = ms N

Kinetic f = mk N

fmax

Slipping begins (fmax = sN )

f

f < fmax (= kN )

Coefficient of Kinetic friction < Coefficient of Static friction

F

Static friction

Kinetic 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

N

N

F1

F2

fcrit2

fcrit1

mg

mg

Making the most of Friction

A F1 > F2

fcrit = S N

B F1 = F2

fcrit = S mg

C F1 < F2

Friction force does not depend on area!

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!

acceleration

Driving Torque

What force drives the car?

v

vo

N

d

f

mg

Stopping Distance depends on friction

Braking force

v2 =vo2 + 2a(x-xo)

Max value of ais when f is max.

fmax = sN

= smg

F = ma

 -fmax= mamax

-smg = mamax

amax = - sg

Thus since

dmin depends on v2!! Take care!!

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

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

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 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?

Soft rubber

Are these just for show, or advertising?

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

@ 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

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

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

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

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

D

dv

=

m

mg

-

D

dt

mg

dv

+

r

-

=

2

m

1/2C

Av

mg

0

dt

F = mg –D

ma = mg -D

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

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