Thursday 12 – 2 pm PPP “ Extension” lecture. Room 211 podium level Turn up any time. Summary Lecture 5 5.4Inertial/noninertial reference frames 6.12Friction 6.5Taking a curve in the road 6.4Drag 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
5.4Inertial/noninertial reference frames
6.12Friction
6.5Taking a curve in the road
6.4Drag force
Terminal velocity
Problems Chap6:5, 14,29 , 32, 33,
29
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
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!
there's a fraction too much....
Friction
Why doesn’t Mick Doohan fall over?
Friction provides the central force
mg
In the rest reference frame
What is Friction
The Source of Friction between two 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
fmN
m is usually < 1
Staticf < or = ms N
Kineticf = 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
AF1 > F2
fcrit = S N
BF1 = F2
fcrit = S mg
CF1 < 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
Force of road on Tyre
Force of Tyre on road
What force drives the car?
v
vo
N
d
f
mg
Stopping Distance depends on friction
Braking force
Friction road/tyres
v2 =vo2 + 2a(xxo)
0 = vo2 + 2ad
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 s1) 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 Formula1 driver on a GP circuit can be as high as 4.5 g
This is equivalent to that experienced by a jetfighter pilot in fastturn manoeuvres.
Albert Park GP circuit
N
mv2/R
mg
R = 70 m
V=55 m s1
Central force provided by friction.
mv2/R = N = mg
= v2/Rg
= 4.3
How can the car stay on the road?
Soft rubber
Grooved tread
Are these just for show, or advertising?
200 km/h
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 s1
= 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 m3
= x v2x A kg m s1
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
DRAG
That's all folks
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
When entertainment defies reality
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 h1
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 h1
In working out this problem you will prove the expression for the viscous drag force