Transcript Lec 4 - Mr. Lee at Hamilton High School

AP Physics Chapter 4
Motion in Two and
Three Dimensions
1
AP Physics


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2
Turn in Chapter 3 Homework, Worksheet, &
Lab
Lecture
Q&A
1-D Review



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3
Position: x, (m)
Displacement: x, (m)
Velocity: v, (m/s)
x
dx
v
v
t
dt
Acceleration: a, (m/s2)
v
dv
a
a
t
dt

The 3 Great Equations
of constant
acceleration motion:
 v  vi  at

1 2

 x  xi  vi t  at
2

 v2  v 2  2ax
i
Position



 3D: r or r
1D: x

r = x i +y j + z k or r  xiˆ  yˆj  zkˆ
–
–
–
4
x: scalar component of r in i direction
y: scalar component of r in j direction
z: scalar component of r in k direction
Position is a 3D vector
k
 r  x2  y2  z 2


1 y
  tan x

2
2

x

y
  tan 1
z
z

y
x
i
5
r

x2  y2
j
Displacement

r = r2 – r1  rf – ri

r = (xf – xi)i + (yf – yi)j + (zf – zi)k
or
 r   x f  xi  iˆ   y f  yi  ˆj   z f  zi  kˆ
6
Velocity
r
Average: v 
t
Instantaneous:

dr d ˆ ˆ
v

xi  yj  zkˆ
dt dt
dx ˆ dy ˆ dz ˆ
 i
j k
dt
dt
dt
 vx iˆ  v y ˆj  vz kˆ
7


dx
 vx 
dt

dy

  vy 
dt

dz

v

 z dt

 Instantaneous velocity
in the x direction
depends on the
displacement only in the
x direction.
 The instantaneous
velocity v of a particle is
always tangent to the
path of the particle.
 Instantaneous acceleration in
the x direction depends on the
change of velocity only in the x
direction.
Acceleration
Average:
v
a
t
Instantaneous:
dv
d
a

vx iˆ  v y ˆj  vz kˆ
dt dt
dvx ˆ dv y ˆ dvz ˆ

i
j
k
dt
dt
dt
 a iˆ  a ˆj  a kˆ

x
8

y
z

dvx
 ax 
dt

dv y

  ay 
dt

dvz

 az  dt

Huh?

9
Most often, we find the derivative to find
acceleration when the position is given as an
equation, but then it is more like a pure math
problem.
Example: Pg77-7
An ion’s position vector is initially r = 5.0i – 6.0j + 2.0k,
and 10 s later it is r = -2.0i + 8.0j –2.0k, all in meters. In
unit-vector notation, what is its vavg during the 10 s?
Given: ri = 5.0i – 6.0j + 2.0k, rf = -2.0i + 8.0j –2.0k, t = 10s,
vavg = ?
r  rf  ri
 2.0iˆ  8.0 ˆj  2.0kˆ  5.0iˆ  6.0 ˆj  2.0kˆ

 
  7.0iˆ  14.0 ˆj  4.0kˆ  m
vavg
10
r


t
 7.0iˆ  14.0 ˆj  4.0kˆ  m 
10s



m
0.70iˆ  1.40 ˆj  0.40kˆ
s
Practice: Pg77-11
The position r of a particle moving in an xy plane is given by r =
(2.00t3 – 5.00t)i + (6.00 – 7.00t4)j, with r in meters and t in seconds.
In unit-vector notation, calculate (a) r, (b) v, and (c) a for t = 2.00 s.
(d) What is the angle between the positive direction of the x axis and
a line tangent to the particle’s path a t = 2.00 s?
a)
rt  2.00 s   2.00  2.003  5.00  2.00  iˆ   6.00  7.00  2.004  ˆj


 6.00iˆ  106 ˆj m
b)
v
dr
  6.00t 2  5.00  iˆ   28.00t 3  ˆj
dt
vt 2.00 s   6.00  2.002  5.00  iˆ   28.00  2.003  ˆj


 19.0iˆ  224. ˆj m / s
11
Practice: Pg77-11
The position r of a particle moving in an xy plane y
is given by r = (2.00t3 – 5.00t)i + (6.00 – 7.00t4)j,
with r in meters and t in seconds. In unit-vector
notation, calculate (a) r, (b) v, and (c) a for t = 2.00
s. (d) What is the angle between the positive
direction of the x axis and a line tangent to the
particle’s path a t = 2.00 s?
tan
c)
a
•
dv
 12.00t  iˆ  84.00t 2 ˆj
dt
at  2.00 s 
 24.0iˆ  336 ˆj  m / s
tan 1
y
x
Velocity is tangent to the path, so what is asked for is the angle
between v and +x axis.
  tan
12
224
 tan
 85o
vx
19
vy
1
not tan 1
vy
vx
x
2
d)
1
1
y
x
Projectile Motion: 2-D

For projectile motion, choose coordinates so that
the motion is confined in one plane  2D motion
r  xiˆ  yjˆ
y

v0
v0y
v  vxiˆ  v y ˆj

Initial velocity
vo  voxiˆ  voy ˆj
13

 v0 x  v0 cos 0

v0 y  v0 sin 0


0
v0x
x
Package Dropped From Airplane
(Snapshot every second)
14
Horizontal Direction: velocity is _______
constant
Vertical Direction: velocity is __________
increasing
Snapshots
15
Ball projected straight up from
truck (Snapshot every second)
16
Snapshots
17
Projectile Motion: Horizontal and
Vertical Displacements
18
Velocity Components of Projectile
19
Independence of Motion

From observations and experiments:
–

Connection:
–
–
20
The horizontal motion and the vertical motion are
independent of each other; that is, neither motion
affects the other.
Both horizontal and vertical motions are functions of
time. Time connects the two independent motions.
Though independent, these two motions are related
to each other by time.
Projectile Motion Breakdown

Horizontal: constant velocity
 vx  vox
 x  x v t
o
ox


Vertical: Constant acceleration (ay = g, downward)
 ay = g if downward is defined as +y direction
 ay = -g if upward is defined as +y direction
 vy  voy  ayt
 y  y  v t  1 a t2
o
oy
y

2
 vy 2  voy 2  2ay y

21

 v0 x  v0 cos 0

v0 y  v0 sin 0


Example: 77-20
A small ball rolls horizontally off the edge of a tabletop that is 1.2 m
high. It strikes the floor at a point 1.52 m horizontally from the table
edge. a) How long is the ball in the air? b) What is its speed at the
instant it leaves the table?
Define coordinates as to the right.
v0x = ?, y = 1.2m, x = 1.52m, x0 = 0, y0 = 0,
vx = const, ay = g = 9.8m/s2, v0y = 0
a) Constant acceleration in y direction.
t ?
1
1
y  y0  voy t  a y t 2  a y t 2
2
2
t
2y

ay
2  1.2m
 0.49s
m
9.8 2
s
b) Constant velocity in x direction.
x  x0 1.52m  0
m
x  x0  v0x t  v0 x 

 3.1
t
0.49s
s
22
0
v0
y = 1.2m
y
x
x = 1.52m
Practice: 78-30
You throw a ball with a speed of 25.0 m/s at an angle of 40.0o above the
horizontal directly toward a wall. The wall is 22.0 m from the release point of the
ball. A) How long does the ball take to reach the wall? B) How far above the
release point does the ball hit the wall? C) What are the horizontal and vertical
components of its velocity as it hits the wall? D) When it hits, has it passed the
highest point on its trajectory?
Set up the coordinates as to the left. Then
vx=const, ay=-g, vox=v0cos40.0o=19.2m/s,
voy = v0sin40.0o = 16.1m/s,x0=0, y0=0,
x=22.0m
y
a) t = ?
v0
Constant v in horizontal direction.
40o
0
x
x  x0  v0 x t  v0 xt
t 
23
x
22.0m

 1.15s
v0 x 19.2m / s
Practice: 78-30
You throw a ball with a speed of 25.0 m/s at an angle of 40.0o above the
horizontal directly toward a wall. The wall is 22.0 m from the release point of the
ball. A) How long does the ball take to reach the wall? B) How far above the
release point does the ball hit the wall? C) What are the horizontal and vertical
components of its velocity as it hits the wall? D) When it hits, has it passed the
highest point on its trajectory?
b) y = ?
Constant acceleration in y direction.
1 2
m
1
m 
y  y0  voy t  a y t  0  16.1  1.15 s   9.8 2  1.15 s
2
2
s
s 
c) vx = ? and vy = ?


2
 12.m
vx  vox  19.2m / s
vy  voy  ayt  16.1


m
m
m
 9.8 2  1.15 s  4.8
s
s
s
d) It has not yet passed the highest point on its trajectory since it is
still going up because vy = 4.8 m/s > 0.
24
Highest Point of Trajectory



Trajectory: path of projectile
Minimum speed at top, but  0
 vy = 0
 vx = v0x = v0 cos 0  vmin  vo cos  o
Maximum Height is
vy  voy  gt  0  t 
voy
g
v02 sin 2 0
H
2g

v0 sin 0
g
2
25
2
2
1 2
 v0 sin  0  1  v0 sin  0 
v
sin
0
0
y  voy t  gt   v0 sin  0  
 g



2
g
g
2g

 2 

Symmetry of Trajectory

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 but at
different directions.


vx stays unchanged.
vy remains unchanged at the same magnitude but changes
in direction.


26
vy is up when object is on its way up.
vy is downward when object is on its way down.
Symmetry of Trajectory (2)
vy
v
vx
vx
vy
v
 At the same height, the speed is the same.
 It takes the same amount of time to go up as to fall down.
27
Equation of Path
x  x0   v0 cos  0   t  t 
1
y  y0   v0 sin  0   t  gt 2
2




28
x
v0 cos 0





y   tan 0  x 
gx 2
2  v0 cos 0 
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 as origin is
set at the initial point.
2
Horizontal Range
2v0 sin 0
1 2
y   v0 sin  0  t  gt  0  t  0 or
g
2
2vo sin o v0 2

sin  20 
x   v0 cos0  t   vo cos o 
g
g
v0 2
 R
sin  2 0 
g





29
R is a distance and, therefore, a scalar.
Independent of frame of reference. (Same result would arrive if different frames of
references were chosen.)
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.
Example:82-77
A projectile is fired with an initial speed vo = 30.0 m/s from level
ground at a target on the ground, at distance R = 20.0 m, as shown
in Fig. 4-51. What are the (a) least and (b) greatest launch angles
that will allow the projectile to hit the target?
m
v0  30.0 , R  20.0m,  0  ?
s
gR
v02
R
sin  20   sin  2o   2
v0
g
1 1 gR 1 1
 o  sin
 sin
2
2
v0
2
m
 20.0m 1
2
1
s

sin
 0.2178
2
2
m

 30.0 
s

9.8
1
  12.6o or 167.4o   6.3o or 84.o
2
30
Practice: 78-26
During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball
leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away
and 0.90 m high. When the ball reaches the net, (a) does the ball clear it and (b) what
is the distance between the center of the ball and the top of the net? Suppose that,
instead, the ball is served as before but now it leaves the racquet at 5.00o below the
horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now
is the distance between the center of the ball and the top of the net?
Set up the coordinates as in the diagram.
a) 0 = 0, vox = vo = 23.6m/s, voy = 0,
ax = 0, ay = -g, x0 = 0, y0 = 2.37m,
x = 12m, y = ?
y
0
Constant velocity in x direction:
x  x0  vox t  voxt  t 
x
12m

 0.508s
vox 23.6m / s
0
Constant acceleration in y direction:
y  y0  voy t 
31
1 2
1
m
2
a y t  2.37m   9.8 2   0.508s   1.11m  0.90m  Yes
2
2
s 
b) y  0.90m  1.11m  0.90m  0.21m
x
Practice: 78-26
During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball
leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away
and 0.90 m high. When the ball reaches the net, (a) does the ball clear it and (b) what
is the distance between the center of the ball and the top of the net? Suppose that,
instead, the ball is served as before but now it leaves the racquet at 5.00o below the
horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now
is the distance between the center of the ball and the top of the net?
c) o = -5.00o, voy = vosin o = 23.6m/s·sin(-5.00o)= -2.06m/s,
vox = vocos o=23.5m/s, ax = 0, ay = -g, xo = 0, yo = 2.37m,x = 12m,
y=?
Constant velocity in x-direction:
x  xo  vox t  vox t  t 
x
12m

 0.511s
vox 23.5m / s
Constant acceleration in y-direction:
1 2
y  yo  voy t  a y t
2

 1
m
m 
 2.37m   2.06  0.511 s    9.8 2  0.511 s
s
s 

 2

32
The ball will not clear the net now.

2
 0.038m  0.90m
Decomposing motion
x
More than one way to decompose a vector.
y
v
Fall at same
time
x
x: constant v
y: constant a, 
33
y
v
x: constant v
y: constant a, 
Uniform Circular Motion
Uniform Circular Motion:


Circular path or circular arc
Uniform = Constant speed (Constant velocity? why?)
v
Magnitude of velocity: constant
Direction of velocity:
v
34
Velocity: changing
changing
Centripetal Acceleration
v2
a
r


, where

v: speed of particle
r: radius of circle or circular arc
v
Direction of acceleration is
always toward the center of
a
a
circle (or circular arc)
a
 Centripetal
v
a  v for uniform circular
motion at any time.
35
v
Direction of Acceleration

a in the same direction as v:
 Speed: increases

a opposite to v:
 Speed: decreases

a  v:
 Speed: does not change
 Direction of velocity: changing
36

 Tangential Acceleration
  Change: speed



 Radial Acceleration
of
  Change: direction
velocity
Direction of Acceleration (2)



37
Tangential acceleration
changes only the speed.
Radial acceleration changes
only direction of velocity.
Both tangential and radial
components of acceleration
are needed to change both
the direction and magnitude
of a velocity.
Uniform Circular Motion
Period: Time for one complete cycle
2 r
D 2 r
T
t  
v
v
v
Frequency: Number of cycles per unit time
1  f  v for uniform circular motion
f 
2 r
T
Unit:  f  
38
1
1
  Hertz  Hz
T  s
Example: 85-116
A) What is the centripetal acceleration of an object on the Earth’s
equator owing to the rotation of the Earth? B) What would the period of
rotation of the Earth have to be for objects on the equator to have a
centripetal acceleration equal to 9.8 m/s2?
a)r  6.37  106 m, T  24hr 
3600s
 86400s, a  ?
hr
6
2 r 2  6.37 10 m 
m
v

 463
T
86400s
s
2
m

463 
2

v
m
s

0.0337
ac 
 
6.37 106 m
s2
r
b)ac  9.8m / s2 , T  ?
m
m
v2
6
ac 
 v  ac r  9.8 2   6.37 10 m   7901
s
s
r
6
 hr 
2 r 2  6.37 10 m 
2 r
T 

 5066s  
v
  1.4hr
3600
s
m


v
T
7901
39
s
Practice: 79-48
A rotating fan completes 1200 revolutions every minute. Consider
the tip of a blade, at a radius of 0.15 m. (a) Through what distance
does the tip move in one revolution? What are (b) the tip’s speed
and (c) the magnitude of its acceleration? (d) What is the period of
the motion?
r  0.15m, n  1200
a) D  ?
D  2 r  2  0.15m  0.942m
b)v  ?
v
x 1200  0.942m
m

 18.8
t
60s
s
c)a  ?
2
m

18.8

v 2 
m
s
a 
 2360 2
r
0.15m
s
d )T  ?
T
40
2 r 2  0.15m

 0.050s
m
v
18.8
s
Relative Motion

vAB:
–
–
–
–
velocity of A with respect of B,
velocity of A relative to B,
velocity of A measured by B, or
velocity of A in frame of reference B.
vAC  vA B  v B C
–
–
–
41
1D
2D,
or 3D?
v AB  vBA
No real cancellation, just helps us to remember.
Inertial Frame of Reference
42

Inertial Frame of Reference: Frame that stays
at rest or moves at constant velocity

Law of Physics: aPA = aPB
•
aPA: Acceleration of particle P as measured by
observer A (or in Frame A).
•
Observers on different inertial frames of references
will measure the same acceleration for a moving
particle.
Example: 80-64
A cameraman on a pickup truck is traveling westward at 20 km/h while he videotapes a
cheetah that is moving westward 30 km/h faster than the truck. Suddenly, the cheetah
stops, turns, and then runs at 45 km/h eastward, as measured by a suddenly nervous crew
member who stands alongside the cheetah’s path. The change in the animal’s velocity took
2.0 s. What are the (a)magnitude and (b) direction of the animal’s acceleration according to
the cameraman and (c) magnitude and (d) direction according to the nervous crew member?
T: Truck (or cameraman), G: ground (or nervous crew member), C: cheetah
Let eastward be the positive direction. Then
vTG = -20km/h, vCT.i = -30km/h, vCG.f = 45km/h, t = 2.0s
aCT  ?,aCG  ?
vCT . f
 vCG. f  vGT  vCG. f  vTG  45
vCG.i  vCT .i  vTG  30
a) aCT 
c) aCG 
43
vCT . f  vCT .i
t
vCG . f  vCG .i
t
km
km
km
 20
 50
h
h
h
65

km 
km 
  30

km
h 
h 
 48
2.0s
h s
45

km 
km 
km
  20

65

h 
h 
h
km 
km 
  50

km
h 
h 
 48
2.0s
h s
b) East
d) East
Practice: 80-56
A boat is traveling upstream in the positive direction of an x axis at 14 km/h with
respect to the water of a river. The water is flowing at 9.0 km/h with respect to the
ground. What are the (a) magnitude and (b) direction of the boat’s velocity with
respect to the ground? A child on the boat walks from front to rear at 6.0 km/h with
respect to the boat. What are the (c) magnitude and (d) direction of the child’s
velocity with respect to the ground?
Upstream has been defined as + direction. vbw  14
km
km
, vwg  9.0
h
h
a)vbg  ?
vbg  vbw  vwg
c)vcb  6.0
vcg 
km
km
km
 14
 9.0
 5.0
h
h
h
b) upstream
km
, vcg  ?
h
km
km
km
vcb  vbg  6.0
 5.0
 1.0
h
h
h
C
B
44
W
+x
d) downstream
Practice: 80-59
Snow is falling vertically at a constant speed of 8.0 m/s. At
what angle from the vertical do the snowflakes appear to be
falling as viewed by the driver of a car traveling on a
straight, level road with a speed of 50 m/s?
Set up frame of reference, then
m
m
v sg  8 j , v dg  50 i, v sd
s
s
vsg
?
We have
v sd vsg  v gd  vsg  vdg 
vdg
8
m
m
m
m
j  50 i  50 i  8 j
s
s
s
s
Then we can find
8
  tan 1  9.1o
50
   90o    90o  9.1o  81o

v sd
x


y
8
 9.1o  9.1o  180o  171o
50
     90o  171o  90o  81o
or   tan 1
45