phy ch 3 - wbm

advertisement
Motion in a Plane
Chapter 3
Physics Chapter 3
1
Position Vector
• Goes from the origin to the object – point p
y
P

r
Physics Chapter 3
x
2
Average Velocity Vector
• Change in displacement over change in
time

v av 

r
t
Physics Chapter 3
3
Instantaneous velocity vector

v  lim
t  0

r
t
Physics Chapter 3
4
Instantaneous velocity
• In two dimensions

v v
2
vx  vy
tan  
2
vy
vx
Physics Chapter 3
5
Average acceleration vector

a av 

v
t
Physics Chapter 3
6
Instantaneous acceleration
vector

a  lim
t  0

v
t
Physics Chapter 3
7
Example
• A particle has x and y coordinates
(4.0 m, 2.0 m) at time t1 = 2.0 s and
coordinates (7.0 m, 6.0 m) at time t2 = 2.5
s. Find the components of the average
velocity and the magnitude and direction of
the average velocity during this time
interval.
Physics Chapter 3
8
Projectile Motion
• Projectile – any body given an initial
velocity which then follows a path
(trajectory) based on gravitational
acceleration and air resistance
• Thrown ball
• Bullet
• Dropped package
Physics Chapter 3
9
Projectile motion
• We will neglect air resistance
• We will also neglect curvature and rotation
of the earth
Physics Chapter 3
10
Projectile Motion
• Consists of both horizontal and vertical
motion
• We will break these problems into x and y
components to make them easier to solve
Physics Chapter 3
11
Projectile motion
• Two dimensional – gravity only acts
vertically
• We assign y as the vertical direction
• We assign x as the horizontal direction
ax  0
ay  g
Physics Chapter 3
12
Projectile Motion
0
v x  v0 x  a x t
v x  v0 x
v y  v 0 y  gt
x  x0  v0 xt 
1
2
0
a xt
2
x  x0  v xt
y  y0  v0 yt 
Physics Chapter 3
1
gt
2
2
13
Projectile Motion
• If we express initial velocity in terms of its
magnitude and angle with the x-axis
v 0 x  v 0 cos  0
v 0 y  v 0 sin  0
Physics Chapter 3
14
Projectile Motion
• We can calculate the projectile’s speed and
the direction of its velocity
v
2
vx  vy
tan  
2
vy
vx
Physics Chapter 3
15
Trajectory shape
• Projectiles always travel in parabolas
Physics Chapter 3
16
Example
• A policeman chases a thief across city
rooftops. They are both running at 5 m/s
when they come to a gap between
buildings that is 4 m wide and has a drop
of 3 m.
• The thief leaps at 5 m/s at an angle of 45°.
Does he clear the gap?
• The policeman leaps at 5 m/s horizontally.
Does he clear the gap?
Physics Chapter 3
17
Example
x0
y0
Physics Chapter 3
18
Example – the thief
x  x0  v xt
x0  0
x  v xt
v0  5
m
s
@ 45

 m

vx   5
cos
45

 s 
 m 

x   5
 cos 45  t
 Chapter 3 
  sPhysics
19
Example – the thief
y0  0
v0  5
m
@ 45

v0 y
s
0
y  y0  v0 yt 
y  v0 y t 
1
1
 m

 5
sin
45

 s 
gt
2
2
gt
2
2
Physics Chapter 3
20
Example – the thief
1
2
gt  v 0 y t  y  0
2
when he hits the roof,
y  3 m
m  2  m 


 4 .9 2  t    5
 sin 45  t  3 m  0
s 

 s 

Physics Chapter 3
21
Quadratic formula
ax  bx  c  0
2
x
b
b  4 ac
2
2a
Physics Chapter 3
22
Example – the thief
m  2  m 


 4 .9 2  t    5
 sin 45  t  3 m  0
s 

 s 

t

m

   5
sin
45


s 


2

m
m


 3 m 

5
sin
45

4
4
.
9


2 


s 
s 



m

2  4 .9 2 
s 

Physics Chapter 3
23
Example – the thief
t
m

 3.54

s 

2
2

m  
m 
 12 . 5 2    58 . 8 2 
s  
s 

m

 9 .8 2 
s 

m 
m

 3.54
   8.44

s  
s 

t
m

 9 .8 2 
s 

Physics Chapter 3
24
Example – the thief
11 . 98
t
m
 4 .9
m
s or t 
s
m
m


 9 .8 2 
 9 .8 2 
s 
s 


t  1 . 22 s
Physics Chapter 3
25
Example – the thief
 m 

x   5
 cos 45  t
 s 

 m 

x   5
 cos 45  1 . 22 s 
 s 

x  4 . 32 m
• Yes, he clears the gap
Physics Chapter 3
26
Example – the policeman
• Everything is the same, except that
v0  5
m
@0

s
 m
vx   5

 s 
v0 y  0
Physics Chapter 3
27
Example – the policeman
 m
x  5
t
 s 
y
1
gt
2
2
y
t 
2

1
g
2
Physics Chapter 3
28
Example – the policeman
when he hits the roof,
y  3 m
t 
2
3m
m

  4 .9 2 
s 

t  0 . 61 s
2
2
t  0 . 78 s
Physics Chapter 3
29
Example – the policeman
 m
x  5
t
 s 
 m
x  5
 0 . 78 s 
 s 
x  3 .9 m
• No, he does not clear the gap
Physics Chapter 3
30
Example 2
•
A rookie quarterback throws a football
with an initial upward velocity
component of 16.0 m/s and a horizontal
component of 20.0 m/s.
a) How much time is required for the
football to reach the highest point of the
trajectory?
Physics Chapter 3
31
Example 2 a)
• At the highest point, vy must be zero.
• Before that, it was moving up, and after
that it is moving down, so it must
momentarily stop at that point.
v y  v 0 y  gt
vy 0
v 0 y  16 . 0
Physics Chapter 3
m
s
32
Example 2 a)
m

0  16 . 0
  9 .8 2 t
s 
s 
m
16 . 0
t 
m
s
m

 9 .8 2 
s 

t  1 .6 s
Physics Chapter 3
33
Example 2 b)
• b)How high is this point?
y  y0  v0 yt 
y0  0
1
gt
2
2
v 0 y  16 . 0
m
s
t  1 .6 s
Physics Chapter 3
34
Example 2 b)
m
1
m

2
y   16
 1 . 6 s    9 . 8 2  1 . 6 s 
s 
2
s 

y  25 . 6 m  12 . 5 m
y  13.1 m
Physics Chapter 3
35
Example 2 c)
•
c) How much time is required for the ball
to return to its original height?
y  y0  v0 yt 
1
gt
2
2
y0
y0  0
v 0 y  16 . 0
m
s
Physics Chapter 3
36
Example 2 c)
0  v0 y t 
1
2
1
gt
2
2
gt  v 0 y t  0
2
1

t  gt  v 0 y   0
2

Physics Chapter 3
37
Example 2 c)
1

or  gt  v 0 y   0
2

t  0
1
2
gt  v 0 y
t 
2v0 y
g
Physics Chapter 3
38
Example 2 c)
m

2  16

s 

t
m
9 .8 2
s
t  3 .2 s
Physics Chapter 3
39
Example 2 d)
•
d) How does the answer to part c)
compare to the answer to part a)?
•
It is double. When something is thrown
upward, it always takes the same amount
of time to go up as it does to come down.
So the total flight time (back to the
original height) is always twice the time
to the highest point.
•
Physics Chapter 3
40
Example 2 e)
•
e) How far has the football traveled
horizontally?
x  x0  v xt
x0  0
v x  20
m
s
t  3 .2 s
Physics Chapter 3
41
Example 2 e)
m

x   20
 3 . 2 s 
s 

x  64 m
Physics Chapter 3
42
Normal acceleration
• Normal means perpendicular
• The component of acceleration that is
perpendicular to the path
• Shows change in direction of velocity
a  or a norm
Physics Chapter 3
43
Tangential acceleration
• Parallel to the path
• Shows change in magnitude of velocity
(change in speed)
a
or a tan
Physics Chapter 3
44
Uniform Circular Motion
• When an object moves around a circle with
constant speed, the acceleration vector
points towards the center of the circle.
• The velocity is tangent to the circle. In
order for the speed to remain constant, the
acceleration must be towards the center of
the circle.
• We call this centripetal acceleration.
Physics Chapter 3
45
Uniform circular motion
• See Fig. 3-11 on page 63
v

v1
a av 
s
v 
R
v
t
v1
s
R

v1  s
a
Rt
v
R
a 
v
lim
t0
s
t
2
R
Physics Chapter 3
46
Uniform circular motion
• The period of any motion is the time it takes to
make one complete cycle. It is denoted with the
letter T.
• In this case it’s the time to go around the circle
once.
v av 
d
v
t
2 R
T
4 R
2
a 
T
2
Physics Chapter 3
47
Example
• The radius of the earth’s orbit around the
sun is 1.49 x 1011 m, and the earth travels
around the sun in 365.25 days.
• What is the earth’s orbital speed in m/s?
• What is the acceleration of the earth
toward the sun in m/s2?
Physics Chapter 3
48
Download