Transcript v - Mr. Lee at Hamilton High School

Physics Chapter 6
Motion in Two
Dimensions
1
Physics



2
Turn in Chapter 5 Homework & Worksheet &
Lab
Lecture
Q&A
1-D Review



Position: d (unit: m)
Displacement: d = df – di
Velocity:
v
–

d
t
(unit: m/s)
Instantaneous velocity = slope of Position-Time graph
Acceleration:
a
–
3
(unit: m)
v
t
(unit: m/s2)
Instantaneous acceleration = slope of Velocity-Time graph
1D Review (2)

The 3 Great Equations of constant
acceleration motion:
 v f  vi  at f


1 2
 d f  di  vi t f  at f
2

 2
v f  vi 2  2a d
4
 v  vi  at

1 2

  d  di  vi t  at
2

 v 2  v 2  2a d
i
Projectile Motion: 2-D

Choose coordinate so that the motion is confined in
one plane  2-D motion
y
1D
d
v
vi
a
5
2D
x
x
vx
vix or vxi
ax
y
y
vy
viy or vyi
ay
x
Package Dropped From Airplane
6
Horizontal Direction: velocity is _______
constant
Vertical Direction: velocity is __________
increasing
Snapshots
7
Ball projected straight up from
truck
8
Snapshots
9
Projectile Motion: Horizontal and
Vertical Displacements
10
Velocity Components of Projectile
11
Independence of Motion

From observation:
–

Connection:
–
–
12
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 these two motions are independent, they
are connected to each other by time.
Initial Velocity

Initial velocity: vi at angle i with the horizontal
Set up coordinate:


Horizontal:
Vertical:
x direction
y direction
Decompose vi into horizontal
and vertical direction:
 vix  vi cos i

 viy  vi sin i
13
y
viy
vi
i
vix
x
Projectile Motion Breakdown


Projectile: Object launched into the air
Horizontal:
Fnet
 0  Constant velocity
Fnet  0  a 
m
 x  vix t

 vx  vix
14
Projectile Motion Breakdown (2)
Similarly,
 Vertical: Constant acceleration (ay = g,
downward)


15
ay = g if downward is defined as +y direction
ay = -g if upward is defined as +y direction

1 2
 y  yi  viy t  a y t
2

 v y  viy  a y t

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.

16

vy is upward when the projectile is going up and
vy is downward when coming down
Symmetry of Trajectory (2)
 Velocity at any moment is tangent to the actual path.
 Velocity is horizontal at the top of the trajectory.
vy
v
vx
vx
17
vy
v
y
Highest Point of Trajectory


Minimum speed at top, but  0
 vy = 0
 vx = vix = vi cos i  vmin  vi cos i
Maximum Height is
v y  viy  gt  0  t 
18
viy
g
vi 2 sin 2 i
H
2g

vi sin i
g
1 2
 v sin i
y  viy t  gt   vi sin i   i
2
 g
 1  vi sin i 
 g

2
g



2
vi 2 sin 2 i

2g
x
y
Horizontal Range
x
2vi sin i
1 2
y   vi sin i  t  gt  0  t  0 or
g
2
2vi sin i
vi 2

sin  2i 
x   vi cos i  t   vi cos i 
g
g
vi 2
 R
sin  2i 
g



19
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.
y
Equation of Path
x  xi   vi cos i   t  t 
y  yi   vi sin i   t 




20
1 2
gt
2
x
vi cos i
x





y   tan i  x 
gx 2
2  vi cos i 
2
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.
Example: 150-1
A 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?
Set up coordinates as to the right. Then
ay = g, yi = 0, xi = 0, vix = 5m/s, viy = 0
0 vi = 5m/s
y = 78.4m
a) t = ?
In y-direction:
y  yi  viy t 
1 2 1 2
ayt  a y t
2
2
2y

ay
t 
b) In x-direction:
m

v
t

5.0
x  xi  vixt  ix 
   4.0s   20.m
s

0
21
2  78.4m
 4.0s
9.8m / s 2
y
x
0 vi = 5m/s
x
Continues …
vx

c) Horizontal:
m
vx  vix  5.0
s
y
Vertical:
m
m
v y  viy  ay t  gt  9.8 2  4.0s  39.
s
s
Speed:
v
vx  v y
Direction:
22
2
2
m 2
m 2
m
 (5.0 )  (39. )  39.
s
s
s
m
v
1 y
s  83o , below the horizontal
 tan 1
  tan
m
vx
5.0
s
39
vy v
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  90.0o  50.0o  40.0o , vi  11.0m / s, xi  0, yi  0,
ax  0, a y   g , v y ,max  0
y
ymax  ?
m
m

viy  vi sin  i   11.0  sin  40.0o   7.07
s
s

v y  viy  a y t  t 
ymax  yi  viy t 
v y  viy
ay

x
0  7.07 m / s
 0.721s
2
9.8m / s
1 2 
m
1
m
2
a y t   7.07   0.721s    9.8 2   0.721s   2.55m
2
s
2
s 

2
23
m

Or Max. Height Eqn.
11.0
sin 2 40.0o
2
2


v sin i 
s
H i

 2.55m
m
2g
2  9.8 2
s
Practice: 152-10
A 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?
xi  0, yi  0, y  28m, a x  0, a y  g ,
vi  15.0m / s,i  20.0o , x  ?
0
vix  vi cos i  15.0m / s  cos  20.0o   14.1m / s
viy  vi sin  i  15.0m / s  sin  20.0o   5.13m / s
y  yi  vi y t 
1 2
ayt
2
y
m  9.8m / s 2 2

 28m   5.13  t 
t  t  1.92 s or 2.97 s
s
2

24
m

x  xi  vix t   14.1  1.92s   27.1m
s

x
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
25
Velocity: changing
changing
Centripetal Acceleration
v2
a
r


26
, 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.
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
27
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
v
1
 f 
for uniform circular motion
f 
2 r
T
Unit:  f  
28
1
1
  Hertz  Hz
T  s
Centripetal Force
v2
Fc  mac  m
r


29
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)
Example: 156-12
A 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)?
m
a )v  8.8 , r  25m, ac  ?
s
8.8m / s 
v
m

 3.1 2 , towards the center
ac 
r
25m
s
2
2
b)
The friction the ground giving to the shoes provides the
centripetal force.
30
Practice: 156-13
A car racing on a flat track travels at 22 m/s around a curve with a 56m 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?
m
a)v  22 , r  56m, ac  ?
s
2
2
22m / s 

v
m
ac 

 8.6 2 , towards the center
r
56m
s
b) The friction provides the centripetal force, so f  mac , and   ?
N  W  mg
f  N
31
m
8.6 2
mac ac
f
s  0.88


  
N
mg
g 9.8 m
s2
N
f
W
v2
Fc  m
r
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.
32
•
Insufficient Centripetal force.
+
2
Relative Velocity
VA,B



Velocity of A relative to B
Velocity of A measured by B
Velocity of A at reference frame B
m
s
10
12
m
m
m
 2  10
s
s
s
v p , g  v p ,t  vt , g
vA,C  v A, B  v B ,C



33
m
s
1D or 2D
1D: Define positive direction
2D: vector addition (head-tail or parallelogram)
Example:159-22
You 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,
then vb,s = 2.0m/s, Vy,b = 4.0 m/s, Vy,s =?
v y , s  v y ,b  vb , s
34
m
m
m
 4.0  2.0  6.0
s
s
s
Example: 167-71
A 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?
y
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 = ?, = ?
va,g
vb,a

vb , g 
  tan
35
vb,a  va , g 
2
1
vb ,a
va , g
2
2
2
m
m
 m 
15

6.5

16

 

s
s
 s 
m
s  67 o
 tan 1
m
6.5
s
15
vb,g
x
Practice:159-24
A 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,
then vb,w = 2.5 m/s, Vb,s = 0.5 m/s, Vw,s =?
vb , s  vb , w  vw, s
 vw , s 
Or
vb , s  vb , w
m
m
m
 0.5  2.5  2.0
s
s
s
vw, s  vw,b  vb , s
 vb , w  vb , s  2.5
36
m
m
m
 0.5  2.0
s
s
s
against