problems chap 4 45, 53, chap 5: 13, 19, 27, 45, 50, 53 summary lecture 4 dynamics 4.8relative motion...
Post on 19-Dec-2015
216 views
TRANSCRIPT
Problems Chap 4 45, 53, Chap 5: 13, 19, 27, 45, 50, 53Problems Chap 4 45, 53, Chap 5: 13, 19, 27, 45, 50, 53
Summary Lecture 4Summary Lecture 4
Dynamics4.84.8 Relative motion in 1DRelative motion in 1D4.94.9 Relative motion in 3-DRelative motion in 3-D4.74.7 Circular motion (study at home) Circular motion (study at home) 5.25.2 Newton 1Newton 15.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 25.75.7 Newton 3 Newton 3 5.85.8 Some examples Some examples 6.1-26.1-2 FrictionFriction6.46.4 Drag forceDrag force
Summary Lecture 4Summary Lecture 4
Dynamics4.84.8 Relative motion in 1DRelative motion in 1D4.94.9 Relative motion in 3-DRelative motion in 3-D4.74.7 Circular motion (study at home) Circular motion (study at home) 5.25.2 Newton 1Newton 15.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 25.75.7 Newton 3 Newton 3 5.85.8 Some examples Some examples 6.1-26.1-2 FrictionFriction6.46.4 Drag forceDrag force
Labs and Tutes start this week.
Check times and places on notice board: level 2
physics podium.
Thursday 12 – 2 pm
“Extension” lecture.
Room 211 podium level
Turn up any time
If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO
In the absence of a FORCE
a body is at rest
1660 AD
A body only moves if it is driven.
In the absence of a FORCE
A body at rest WILL REMAIN AT REST
350 BC
DynamicsDynamicsAristotle
For an object to MOVE
we need a force.
Newton
For an object to CHANGE its motion
we need a force
Newtons mechanics applies for motion in
an inertial frame of reference! ???????
He believed that there existed an absolute (not accelerating) reference frame, and an absolute time.
The laws of physics are always the same in any inertial reference frame.
His laws applied only when measurements were made in this reference frame…..
Newton clarified the mechanics of motion in the “real world”.
…or in any other reference frame that was at rest or moving at a constant velocity relative to this absolute frame.
Inertial
reference
frame
Inertial
reference
frame
Inertial reference
frame
Newton believed that there existed an absolute (not accelerating) reference frame, and an absolute time.
The laws of physics are always the same in any inertial reference frame.
Einstein recognised that all measurements of position and velocity (and time) are relative.
There is no absolute reference frame.
Frames of Reference
The reference frames may have a constant relative velocity
We need to be able to relate one set of measurements to the other
The connection between inertial reference frames is the “Gallilean transformation”.
In mechanics we need to specify position, velocity etc. of an object or event.
This requires a frame of reference
x
y
z
o
p
The reference frames for the same object may be different
x’
y’
z’
o
Ref. Frame P (my seat in plane)
T (Lunch Trolley)xTP
Ref. Frame G (ground)
xPG
xTG
xTG = xTP + xPG
Vel. = d/dt(xTG) vTG = vTP + vPG
Accel = d/dt (vTG) aTG = aTP + aPG
In any inertial frame the laws of physics are the same
VPG(const
)
Gallilean transformations
0
N
E
GROUND
N’
E’
AIR
r PG
r AG
r PA
aPG = (vPG) = aPA + aAGdtd
vAG
rPG = rPA + rAG
dtd
vPG = (rPG) = vPA + vAG
PLooking
from above
0
In any inertial frame the laws of physics are the same
AG
PA
PA AG
VPG = VPA + VAG VPA = 215 km/h to East
VAG = 65 km/h to North
hkmv
v
vvv
PG
PG
agpaPG
/225
422546225
22
Tan = 65/215
= 16.8o
Ground Speed of Plane
AG
PA
PA AG
Ground Speed of Plane
In order to travel due East, in what direction relative to the air must the plane travel? (i.e. in what direction is the plane pointing?), and what is the plane’s speed rel. to the ground?
Do this at home and then do sample problem 4-11 (p. 74)
VPA = 215 km/h to East
VAG = 65 km/h to North
Isaac Newton
1642-1727
Newton’s 1st Law
If a body is at rest, and no force acts on it, IT WILL REMAIN AT REST.
If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO
If a body is moving at constant velocity, we can always find a reference frame where it is AT REST.
At rest moving at constant velocity
An applied force changes the velocity of the body
a F
ForceForceIf things do not need pushing to move at constant velocity, what is the role of FORCE???
Inertial mass
The more massive a body is, the less it is accelerated by a given force.
a = F 1m
F = am
The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line
in which that force is impressed.
a = F/m
?
How to measure the Inertial Mass
m
kfreq
mgweightextensionF
Balances the gravitational weight
Which would work in a space ship away from gravity?
mmeasures inertial mass
Newton’s 2nd Law
a = F/m
aa = = vector SUM of ALL EXTERNAL forces acting on the body
vector SUM of ALL EXTERNAL forces acting on the body
mm______________
F = ma
vector SUM of ALL EXTERNAL forces acting on the body
vector SUM of ALL EXTERNAL forces acting on the body
= ma= ma
Newton’s 2nd Law
F=maF = iFx + jFy + kFz a = iax + jay + kaz
Fx=max
Fy=may
Fz=maz
Applies to each component of the vectors
FA+ FB+ FC= 0 since a= F/m and a = 0
F = 0
Fx = 0 Fy = 0
FCcos - FAcos47= 0
= 220 sin 47 +170 sin 28
= 160 +80 240 N
FA sin 47 + FCsin – FB = 0
FB = FA sin 47 +FCsin
220N
? N
170N
Fy = 0
cos= cos47 =280220170
170 cos - 220 cos47= 0
(280)
A (M2 – M1)g
B (M1 – M2)g
C M1g
D Unknown
m2g
N
TM2
m1g
TM1
Apply F = ma to each body
i.e Fx = max and Fy = may
For m2
Vertically
Fy = -N + m2g = m2ay = 0
N = m2g
Horizontally
Fx = T = m2a
21
1
mmgm
a
For mass m1
Vertical (only)
m1g - T = m1a
Analyse the equation!
m1g - m2a = m1a
m1g = m2a + m1a = a(m2 + m1)
41%
26%
33%
2.5 x 103 N 2.5 x 103 N 3.9 x 104 N
2.5 x 103 N 2.5 x 103 NT
3.9 x 104 N
F = ma m is mass of road train
5.0 x 103 N
3.9 x 104 N
F = ma
a = (39 – 5) x 103/(4 x 104)
a = F/m
a
a = 0.85 m s-2
2.5 x 103 N 2.5 x 103 N Don’t care!T
F = ma m is mass of trailer!
F is net force on TRAILER
2.5 x 103 N
Ta = 0 So F= 0
T - 2.5 x 103 = 0
T = 2.5 x 103 N
a = 0
2.5 x 103 N 2.5 x 103 N
3.9 x 104 N
F = ma m is mass of roadtrain
No driving force so F = -5.0 x 103 N a = - 5.0 x 103/4.0 x 104
= - 0.125 m s -2
Use v2 = vo2 + 2a(x – x0)
x – x0 = 400 mv = 0 u = 20 m s-1,
Newton’s 3rd Law
Forces come in pairs
To every action there is an equal and opposite reaction.
The action-reaction pairs
ALWAYS act on DIFFERENTDIFFERENT bodies
NAG
mg
What is reaction pair to the weight force mg?
Is it N (normal reaction)?
N
The reaction pair is the force of the apple on the Earth
N is the force of the ground on the apple NGA
The reaction pair to NGA (or N) is the force of the apple on the ground, NAG
NGA
NO! . N and mg act on the SAME body
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
Man in lift
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 as lift accelerates?
R
mg
R is reaction force
= reading on scales
Only if it is an inertial frame of reference! The accelerating lift is NOT!
Didn’t we say the laws of physics do not depend on the frame of reference?
Here endeth
the lessonlecture
No. IV
Summary Lecture 5Summary Lecture 5
6.1-2 Friction
6.4 Drag force
Terminal velocity
6.5 Taking a curve in the road
Summary Lecture 5Summary Lecture 5
6.1-2 Friction
6.4 Drag force
Terminal velocity
6.5 Taking a curve in the road
Problems Chap 6: 5, 14, 29 , 32, 33, Problems Chap 6: 5, 14, 29 , 32, 33,
Staff-Student Liaison Committee
This committee meets 3 times each semester, to consider reports from lectures, labs. and tutorials. The meetings are after 5 pm, and last ~1 hour.
We need a rep. from this class to report comments from this lecture group.
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
NN
mg
f1 f2
f = S N
Friction force does not depend on area!
f = 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
AC
mg2vterm
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