Transcript No Slide Title

Summary for Lecture 3
4.5
Projectile motion
4.6
Range of projectile
Dynamics
4.8
Relative motion in 1D
4.9
Relative motion in 3-D
4.7
Circular motion (study at home)
5.2
Newton 1
5.3-5.5
Force, mass, and Newton 2
Problems Chap. 4
24, 45, 53,
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Notices
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Parabola
Running out of
“impetus”
2 angles for any range
Max range 45o
Gallileo
What do we know from experience?
The trajectory depends on:
Initial velocity
Projection angle
Anything
else? Ignore for now
Air
resistance
Projectile Motion
y
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).
r(t) = i x(t) + j y(t)
r(t)
y(t)
x(t)
x
What do we know?
-g
y
Usually we know the
initial vector velocity vo.
q
v0cosq
v0 = ivox + j voy
=
vo
v0sinq
i v0cosq + jv0sinq
We know acceleration is constant = -g.
accel is a vector
a = iax + jay
=i0 -jg
x
Finding an expression for Projectile motion
The vertical
component of
the projectile
motion is the
same as for a
falling object
The horizontal
component is
motion at
constant velocity
To define the trajectory, we need r(t)
That is:- we need x(t) and y(t)
Consider Horizontal motion
•velocity
vx(t) = v0x + a0xtsince ax=0
= v0 cosq  const
-g
y
vo
r
q
v0cosq
•displacement
Use x x(t)
- xo = v00xt t++½½a
t2
0 atx2since
ax=0
x(t) = v0 cosq t 
x
t
v o cos q
x
vertical component of displacement
Use y - yo = ut + ½ at2
ay= -g
y
vo
y(t) = v0yt + ½ ayt2
q
v0sinq
r
x
= v0 sinq t – ½ gt2
2
v 0 sinθ x 1
x
y(x)
 g 2
v 0 cosθ 2 v 0 cos2 θ
Recall
x
t
v o cos q
Re-arranging gives
y( x)  
g
2v 02 cos2 θ
x2 + tanθ x
Height (y) as a function of
distance (x) for a projectile.
g
2
y
x
+ tanq x
2
2
2v 0 cos q
y=
-k1
x2 + k2 x
y
y = kx2
x
y
g
2
x
+ tanq x
2
2
2v 0 cos q
y
vo
range
x
y = -k1x2 + k2x
Range
y
y
x
For what x values does y = 0?
where
i.e. where

g
2
x
+ tanq x  0
2
2
2v 0 cos q
 x(
g
2
2 v 0 cos q
x=0
x=0
sinθ
now tanθ 
cosθ
sine
0
g
x 2 + tanq x
2
2v 0 cos q
2
angle
or
2
x  tan q )  0
g
2 v 0 cos q
2
2
x  tanq  0
2 v 0 2 cos2 q sinθ
x
tan q
g
cos θ
v02
Rx  2 sinq cos q
g
v0
 R  sin2q
g
2
Maximum when
2q = 900.
i.e when q = 450
http://www.colorado.edu/physics/phet/web-pages/simulations-base.html
Range up a slope
y
g
2v 0 cos q
2
2
x 2 + tanq x
y
y  mx
Gradient of slope
q
R
x
y  mx mxy  tanq x 
g
2V0 cos q
2
2
x
2
g
2
0  tanq . x  m x 
x
2
2
2V cos q
g
0  x(tanq  m 
x)
2
2
2V cos q
True for x = 0 or tanq  m 
g
2Vo cos2 q
2
2Vo cos2 q
x  (tanq  m )
g
2
x
2Vo cos2 q
x'  (tanq  m )
g
2
y
R' 
x' 2 + y' 2
y’ = mx’
R'  x' 1 + m 2
x’
y’
q
R
If m = 0
(on level ground)
R’= x’ = R
x
For collision, x and y for dart and
monkey must be the same at instant t
d
t

Time to travel distance d is
v o cosq
d 1 122
d
y

v
sin
q
t

gt
 g(
)2
y for dart y -vyo 0osin
=qv0t + ½ at
v o cosq2 2 v o cosq
yo = 0
v sinθ d 1
d
y o
 g(
)2
v o cosθ
2 v o cosθ
1
d
vo
y  tanq d  g(
)2
2 v o cosq
q
h
1 d d
y  d  g(
)2
d
2 v o cosq
1
d
y  h  g(
)2
2 v o cosq
-g
h
y
y for dart at impact
(ie at time t 
d
)
v o cosq
1
d
y  h  g(
)2
2 v o cosq
How far has monkey fallen?
g
1
d 22

g
(
dist = v0monkt + ½ at )
2 v o cosq
h
vo
y
q
d
Therefore the height
of the monkey is
1
d
)2
y = h - g(
2 v o cosθ
Dynamics
Kinematics
HOW things move
Dynamics
WHY things move
P
Q
.
40%
R
In the absence of a FORCE
a body is at rest
A body only moves if it is driven.
350 BC
In the absence of a FORCE
A body at rest
WILL REMAIN AT REST
If it is moving with a constant velocity,
IT WILL CONTINUE TO DO SO
1660 AD
Dynamics
Aristotle
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!
???????
Newton clarified the mechanics of
motion in the “real world”.
He believed that there existed an absolute (not accelerating)
reference frame, and an absolute time.
Inertial
reference
His laws
applied only when measurements were made in this
frame
reference frame…..
…or in any other reference frame that was at rest or moving at a
constant velocity relative to this absolute frame.
The laws of physics are always the same
in any inertial reference frame.
Newton believed that there existed an absolute (not
accelerating) reference frame, and an absolute time.
Einstein recognised that all measurements of
position and velocity (and time) are relative.
There is no absolute reference frame.
The laws of physics are always the same
in any inertial reference frame.
Frames of Reference
In mechanics we need to specify position,
velocity etc. of an object or event. y’
The reference frames may have a
constant relative velocity
We need to be able to relate one
set of measurements to the other
p
z
z’
This requires a frame of reference
The reference frames for the
same object may be different
y
o
o
x
x’
The connection
between inertial
reference frames
is the “Gallilean
transformation”.
Ref. Frame G Ref. Frame P (my
seat in Plane)
(ground)
Gallilean
transformations
VPG(const)
xPG
xTP
T (Lunch Trolley)
xTG
xTG = xTP + xPG
Vel. = d/dt(xTG)  vTG = vTP + vPG
Accel = d/dt (vTG)  aTG = aTP + a0PG
In any inertial frame
the laws of physics are
the same
GROUND
AIR
N’
N
Looking
from above
P
vAG
E’
rPG = rPA + rAG
E
vPG =
d
dt (rPG) = vPA
aPG =
d
(vPG) = aPA +
dt
In any inertial
frame the laws
of physics are
a0AG the same
+ vAG
Ground Speed of Plane
AG
PA
VPG = VPA + VAG
v PG  v pa + v ag
2
VPA = 215 km/h to East
2
v PG  46225+ 4225
v PG  225 km
PA/ h
VAG = 65 km/h to North
Tan AG
= 65/215
 = 16.8o
Ground Speed of Plane
AG
PA
VPA = 215 km/h to East
VAG = 65 km/h to North
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?
PA
AG
Do this at home and then do
sample problem 4-11 (p. 74)
A motorboat with its engine running at a constant rate travels
across a river from Dock A, constantly pointing East.
Flow
X
N
A
Y
I will quiz
you on this
next lecture
Z
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.
Here
endeth
the
lesson
lecture
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
Force
The alteration of motion is ever proportional
to the motive force impressed; and is made in
If things dothenot
needof pushing
move
direction
the right lineto
in which
thatat
force is impressed.
what is the role of FORCE???
constant velocity,
An applied force changes the velocity of the body
aF
1
a = mF
Inertial a
mass
= F/m
F = ma
The more massive a body is, the less
it is accelerated by a given force.
Newton’s 2nd Law
a = SF/m
vector SUM of ALL
EXTERNAL forces
acting on the body
a = _______
m
SF = ma
vector SUM of ALL
EXTERNAL forces
acting on the body
= ma
Newton’s 2nd Law
SF=ma
F = iFx + jFy + kFz
a = iax + jay + kaz
SFx=max
SFy=may
SFz=maz
Applies to each component of the vectors
FA+ FB+ FC= 0 since a= SF/m and a = 0
SF = 0
SFx = 0
SFy = 0
220N
FCcos - FAcos47= 0
170N
170 cos - 220 cos47= 0
0
cos= 220
cos47

=28
170
SFy = 0
FA sin 47 + FCsin – FB = 0
(280)
FB = FA sin 47 +FCsin
= 220 sin 47 +170 sin28
= 160 +80  240 N
? N
T
M1
m1g
N
M2
m2g
T
For mass m1
Vertical (only)
Apply SF = ma to each body
i.e SFx = max and SFy = may
For m2
Vertically
SFy = -N + m2g = m2ay = 0
N = m2g
m1g - T = m1a
m1g - m2a = m1a
m1g = m2a + m1a = a(m2 + m1)
m1g
a
m1 + m 2
Horizontally
SFx = T = m2a
Analyse the equation!
2.5 x 103 N
2.5 x 103 N
41%
26%
33%
3.9 x 104 N
2.5 x 103 N
SF = ma
T
m is mass of road train
3.9 x 104 N
5.0 x 103 N
a
SF = ma
2.5 x 103 N
a = SF/m
a = (39 – 5) x 103/4 x 104
a = 0.85 m s-2
3.9 x 104 N
Don’t care!
a=0
SF = ma
2.5 x 103 N
T
2.5 x 103 N
m is mass of trailer!
SF is net force on TRAILER
T
2.5 x 103 N
a=0
So SF= 0
T - 2.5 x 103 = 0
T = 2.5 x 103 N
3.9 x 104 N
2.5 x 103 N
2.5 x 103 N
SF = ma
m is mass of roadtrain
No driving force so SF = -5.0 x 103 N
Use v2 = vo2 + 2a(x – x0)
v = 0 u = 20 m s-1,
 x – x0 = 400 m
a = - 5.0 x 103/4.0 x 104
= - 0.125 m s -2