Summary for Lecture 3
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Transcript of Summary for Lecture 3
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4.54.5 Projectile motionProjectile motion
4.64.6 Range of projectileRange of projectile
Dynamics
4.84.8 Relative motion in 1DRelative motion in 1D
4.94.9 Relative motion in 3-DRelative motion in 3-D
4.74.7 Circular motion (study at home) Circular motion (study at home)
5.25.2 Newton 1Newton 1
5.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 2
4.54.5 Projectile motionProjectile motion
4.64.6 Range of projectileRange of projectile
Dynamics
4.84.8 Relative motion in 1DRelative motion in 1D
4.94.9 Relative motion in 3-DRelative motion in 3-D
4.74.7 Circular motion (study at home) Circular motion (study at home)
5.25.2 Newton 1Newton 1
5.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 2
Summary for Lecture 3Summary for Lecture 3
Problems Chap. 4 24, 45, 53, Problems Chap. 4 24, 45, 53,
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First-year Learning Centre
Tutors are available to help you
from 12.30 pm to 2.30 pm
on Tuesday – Friday
In the First-year Learning Centre
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Notices
NoticesRepresentative for
SSLC(Staff-Student Liaison Cttee)
Representative forSSLC
(Staff-Student Liaison Cttee)
Physics Phrequent PhlyersClarify the physics
Extend your knowledge Ask the lecturer
Physics Phrequent PhlyersClarify the physics
Extend your knowledge Ask the lecturer
Room pp211Physics Podium
Thursdays 12 – 2 pm
Room pp211Physics Podium
Thursdays 12 – 2 pm
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Gallileo
Parabola
Max range 45o
2 angles for any range
Parabola
Max range 45o
2 angles for any range
Running out of “impetus”
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What do we know from experience?
The trajectory depends on:
Initial velocity
Projection angle
Anything else?Air resistance Ignore for now
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Projectile Motion
To specify the trajectory we need to specify every point (x,y) on the curve.
That is, we need to specify the displacement vector r at any time (t).
x
y
r(t)
r(t) = i x(t) + j y(t)
x(t)
y(t)
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v0 = ivox + j voy
-g
v0sin
v0cos
Usually we know the
initial vector velocity vo.
We know acceleration is constant = -g.
accel is a vector a = iax + jay
= i 0 - j g
= i v0cosjv0sin
What do we know?
vo
x
y
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The vertical component of the projectile motion is the same as for a falling object
The horizontal component is motion at constant velocity
Finding an expression for Projectile motion
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To define the trajectory, we need r(t)
That is:- we need x(t) and y(t)
vx(t) = v0x + axt
cosvx
to
x(t) = v0 cos t x(t) = v0xt + ½axt2
0 since ax=0
0 since ax=0
v0cos x
y
r
vo
Consider Horizontal motion
•velocity
= v0 cos•displacement
-g
const
Use x - xo = v0t + ½ at2
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vertical component of displacement
ay= -gUse y - yo = ut + ½ at2
y(t) = v0yt + ½ ayt2
θcosv
xg
2
1
θcosv
xθsinvy(x)
220
2
0
0 Recall
cosvx
to
v0sin
x
y
r
vo
= v0 sin t – ½ gt2
xθtanxθcosv2
g)x(y 2
220
Re-arranging gives
Height (y) as a function of distance (x) for a projectile.
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y = -k1 x2 + k2 x
x
y y = kx2
xtanxcosv
gy
2
2202
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y = -k1x2 + k2x
xtanxcosv
gy
2
2202
x
y
range
vo
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Range
For what x values does y = 0?
i.e. where 0)tanxcosv2
g(x
220
g
vcossin2x
20
0tancos2
2
22
0
xxv
g where
Maximum when 2 = 900.
i.e when = 450cosθ
sinθtanθnow
g
v2sinR
20
x = 0 or 0tanxcosv2
g22
0
tangcosv2
x22
0x=0cosθ
sinθ
R
x
yxx
v
gy
tan
cos22
22
0
x
y
x
yxx
v
gy
tan
cos22
22
0
sine0 angle
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http://www.colorado.edu/physics/phet/web-pages/simulations-base.html
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R
R’
xxv
gy
tan
cos 2
2202
mxy
y
x
Gradient of slope
Range up a slope
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2
2202
xV
gxy
costan mxy mx
)cos
(tan xV
gmx
2220
xV
gm
o
222 costan
2222
0 xV
gxmx
cos.tan
True for x = 0 or
g
Vmx o
222 cos)(tan
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y’
If m = 0 (on level ground)
g
Vmx o
22cos2
)(tan' 22 'y'x'R
y’ = mx’
2m1x'R'
R’= x’ = R
R
R’
y
x
x’
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y
vo
For collision, x and y for dart and monkey must be the same at instant t
Time to travel distance d is cosov
dt
d
y for dart y - y0 = v0t + ½ at2 2
2
1gttvy o sin 2
2
1)
cos(
cossin
ooo v
dg
v
dvy
2
2
1)
cos(tan
ov
dgdy
2
2
1)
cos(
ov
dgd
d
hy
2
2
1)
cos(
ov
dghy
h2
oo
o )cosθv
dg(
2
1
cosθv
dsinθvy
-g
yo = 0
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y
h
vo
d
How far has monkey fallen?
2
2
1)
cos(
ov
dghy y for dart at impact
(ie at time )cosov
dt
Therefore the height of the monkey is
2
o)
cosθv
dg(
2
1 y = h -
dist = v0monkt + ½ at2 2
2
1)
cos(
ov
dg
g
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Dynamics
KinematicsKinematicsHOW things moveHOW things move
DynamicsDynamicsWHY things moveWHY things move
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40%
QP
. R
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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
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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! ???????
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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
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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.
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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
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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
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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
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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
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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
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A motorboat with its engine running at a constant rate travels across a river from Dock A, constantly pointing East.
FlowN
A
Compare the times taken to reach points X, Y, or Z when the river is flowing and when it is not. Your answer should include an explanation of what might affect the time taken to cross the river, and a convincing justification of your conclusion.
I will quiz you on this next lecture
X
Y
Z
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Here endeth
the lessonlecture
No. III
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Isaac Newton
1642-1727
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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
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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
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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
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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
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FA+ FB+ FC= 0 since a= F/m and a = 0
F = 0
Fx = 0 Fy = 0
FCcos - FAcos47= 0
= 220 sin 47 +170 sin28
= 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)
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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)
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41%
26%
33%
2.5 x 103 N 2.5 x 103 N 3.9 x 104 N
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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
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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
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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,