Post on 18-Jan-2016
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Physics Chapter 6
Motion in Two Dimensions
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Physics
Turn in Chapter 5 Homework & Worksheet & Lab
Lecture Q&A
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1-D Review
Position:
t
dv
t
va
Displacement: Velocity:
– Instantaneous velocity = slope of Position-Time graph
Acceleration:
– Instantaneous acceleration = slope of Velocity-Time graph
dd = df – di
(unit: m)(unit: m)
(unit: m/s)
(unit: m/s2)
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1D Review (2)
The 3 Great Equations of constant acceleration motion:
fv
fd
2fv
i fv at
21
2i i f fd v t at
2 2iv a d
v
d
2v
iv at
21
2i id v t at
2 2iv a d
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Projectile Motion: 2-D
Choose coordinate so that the motion is confined in one plane 2-D motion
x
y
x y2 D
1 D
d yxv vx
vy
vi vix or vxi viy or vyi
a axay
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Package Dropped From Airplane
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Snapshots
Horizontal Direction: velocity is _______
Vertical Direction: velocity is __________
constant
increasing
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Ball projected straight up from truck
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Snapshots
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Projectile Motion: Horizontal and Vertical Displacements
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Velocity Components of Projectile
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Independence of Motion
From observation:– The horizontal motion and the vertical motion are
independent of each other; that is, neither motion affects the other.
Connection:– Both horizontal and vertical motions are functions of
time. Time connects the two independent motions. – Though these two motions are independent, they
are connected to each other by time.
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Initial Velocity
x
vi
y
i
ix
iy
v
v
Set up coordinate:
vix
viy
Initial velocity: vi at angle i with the horizontal
cosi iv
sini iv
Decompose vi into horizontal and vertical direction:
Horizontal: x direction Vertical: y direction
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Projectile Motion Breakdown
Projectile: Object launched into the air Horizontal:
ix
x ix
x v t
v v
0netF 0netFa
m Constant velocity
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Projectile Motion Breakdown (2)
Similarly, Vertical: Constant acceleration (ay = g,
downward) ay = g if downward is defined as +y direction
ay = -g if upward is defined as +y direction
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2i iy y
y iy y
y y v t a t
v v a t
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Symmetry of Trajectory
Trajectory: Path of projectile Upward motion and downward motion are
symmetric at same height.– Upward total time is equal to downward total time if
landing point is at same height as initial point.– At the same height, speed is the same.
vx stays unchanged.
vy remains at the same magnitude but changes in direction. vy is upward when the projectile is going up and
vy is downward when coming down
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Symmetry of Trajectory (2)
vx
vx
vy
vy
v
v
Velocity at any moment is tangent to the actual path. Velocity is horizontal at the top of the trajectory.
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Highest Point of Trajectory
y iyv v gt
2 2sin
2i iv
Hg
vy = 0
vx = vix = vi cos i
Maximum Height is
Minimum speed at top, but 0
0 t
y
siniy i iv v
g g
21
2iyv t gt 2
sin sin1sin
2i i i i
i i
v vv g
g g
2 2sin
2i iv
g
min cosi iv v
x
y
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Horizontal Range
Two angles (complementary) with the same initial speed give the same range.
Horizontal range is maximum when the launch angle is 45o. Valid only when the landing point and initial point are at the same
height.
x 2
sin 2ii
v
g
y
cosi iv t
21sin
2i iv t gt 02 sin
0 or i ivt
g
2
sin 2ii
vR
g
2 sincos i ii i
vv
g
x
y
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Equation of Path
No time involved. Can be used to find x or y when the other is given. Parabolic equation Trajectory is parabolic. Valid only when upward is defined as the +y direction and origin is
set at the initial point.
ix x cosi iv t iy y 21
sin2i iv t gt
cosi i
xtv
2
2tan2 cos
i
i i
gxy x
v
x
y
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Example: 150-1A stone is thrown horizontally at a speed of + 5.0 m/s from the top of a cliff 78.4 m high. a) How long does it take the stone to reach the bottom of the cliff? b) How far from the base of the cliff does the stone hit the ground? c) What are the horizontal and vertical components of the velocity of the stone just before it hits the ground?
vi = 5m/sx
y
0
y =
Set up coordinates as to the right. Then
ay =
y
x 0
a) t = ?
b) In x-direction:
g, yi = 0, xi = 0, vix = 5m/s, viy = 0
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2 ya t
t
5.0 4.0 20.ix
mv t s m
s
78.4m
21
2i iy yy v t a t
In y-direction:
i ixx v t
2
2 2 78.44.0
9.8 /y
y ms
a m s
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Continues …
c) Horizontal:
xv
yv
v
Vertical:
Speed:
Direction:
vi = 5m/sx
y
0
vvy
vx
5.0ix
mv
s
iyv ya t 29.8 4.0 39.m m
gt ss s
2 2 2 2(5.0 ) (39. ) 39.x y
m m mv v
s s s
1 139
tan tan 83 , below the horizontal5.0
y o
x
mv s
mvs
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Practice: A softball is tossed into the air at an angel of 50.0o with the vertical at an initial velocity of 11.0 m/s. What is its maximum height?
i
x
y
sin 11.0 sin 40.0 7.07oiy i i
m mv v
s s
yv t
maxy
Or Max. Height Eqn.
iy yv a t
iy 222
1 17.07 0.721 9.8 0.721 2.55
2 2iy y
m mv t a t s s m
s s
2
0 7.07 /0.721
9.8 /y iy
y
v v m ss
a m s
H
22
2 2
2
11.0 sin 40.0sin
2.552 2 9.8
o
i i
mv s
mmgs
11.0 / , 0, 0,i i iv m s x y
,max
max
0, , 0
?
x y ya a g v
y
90.0 50.0 40.0 ,o o o
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Practice: 152-10A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0o below the horizontal. How far does the ball move horizontally before it hits the ground?
x
y
0, 0, 28 , 0, ,
15.0 / , 20.0 , ?
i i x y
oi i
x y y m a a g
v m s x
cos 15.0 / cos 20.0 14.1 /oix i iv v m s m s
sin 15.0 / sin 20.0 5.13 /oiy i iv v m s m s
y
229.8 /
28 5.132
m m sm t t
s
x
1.92 or 2.97t s s
0
iy21
2i y yv t a t
ix 14.1 1.92 27.1ix
mv t s m
s
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Uniform Circular Motion
Uniform Circular Motion:
v
v
Magnitude of velocity:
Circular path or circular arc Uniform = Constant speed (Constant velocity? why?)
Direction of velocity:
Velocity:
constant
changing
changing
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Centripetal Acceleration
Direction of acceleration is always toward the center of circle (or circular arc)
Centripetal
2va
r
v: speed of particle
r: radius of circle or circular arc, where
a
a
a
v
v
v
for uniform circular motion at any time.a v
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Direction of Acceleration
a in the same direction as v:
a opposite to v:
a v:
Speed: increases
Speed: decreases
Speed: does not change
Direction of velocity: changing
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Uniform Circular Motion
Period: Time for one complete cycle
2 rT
v
t
1f
T
Unit: f
Frequency: Number of cycles per unit time
Hertz Hz
for uniform circular motion2
vf
r
D
v
2 r
v
1
T
1
s
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Centripetal Force
Centripetal force is in general not a single physical force; rather, it is in general the net force.
Do not draw centripetal force on force diagram (Free Body Diagram)
2
c c
vF ma m
r
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Example: 156-12A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. a) Find the centripetal acceleration of the runner. b) What agent exerts force on the runner (to round the bend)?
) 8.8 , 25 , ?c
ma v r m a
s
ca
b)
The friction the ground giving to the shoes provides the centripetal force.
22
2
8.8 /3.1 , towards the center
25
m sv m
r m s
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Practice: 156-13A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car’s centripetal acceleration. What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping?
) 22 , 56 , ?c
ma v r m a
s
ca
b) The friction provides the centripetal force, so
22
2
22 /8.6 , towards the center
56
m sv m
r m s
, and ?cf ma
N
f
W mg
N
f
N cma
mg
2
2
8.60.88
9.8
c
ma s
mgs
W
N
f
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What if?
If centripetal force is not provided or not large enough, the object will not be able to move in the circle it intends to move.
Centripetal force disappears. Insufficient Centripetal force.
• •
2
c
vF m
r
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Relative Velocity
Velocity of A relative to B Velocity of A measured by B Velocity of A at reference frame B
VA,B
, ,A C A Bv v
Bv
,C
1D or 2D 1D: Define positive direction 2D: vector addition (head-tail or parallelogram)
10m
s
+
2m
s
12m
s 2 10
m m
s s
,p gv ,t gv,p tv
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Example:159-22You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street?
Let direction bus moving = ”+” direction, also let
you = y, bus = b, street = s,
,y sv
then vb,s = 2.0m/s, Vy,b = 4.0 m/s, Vy,s =?
, ,y b b sv v 4.0 2.0 6.0m m m
s s s
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Example: 167-71A weather station releases a balloon to measure cloud conditions that rises at a constant 15 m/s relative to the air, but there is also a wind blowing at 6.5 m/s toward the west. What are the magnitude and direction of the velocity of the balloon?
Let west = +x, up = +y.
Balloon = b, air/wind = a, ground = g.
vb,a = 15 m/s, va,g = 6.5 m/s, vb,g = ?, = ?
x
y
vb,a
va,g
vb,g
,b gv
2 2, ,b a a gv v
2 2
15 6.5 16m m m
s s s
,1
,
tan b a
a g
v
v 1
15tan 67
6.5
o
msms
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Practice:159-24A boat is rowed directly upriver at a speed of 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat?
Let upriver = ”+” direction, also let
boat = b, river/water = w, shore = s,
,b sv then vb,w = 2.5 m/s, Vb,s = 0.5 m/s, Vw,s =?
, ,b w w sv v
,w sv , , 0.5 2.5 2.0b s b w
m m mv v
s s s against
,Or w sv , ,w b b sv v
, ,b w b sv v 2.5 0.5 2.0m m m
s s s