Chapter4-F08.ppt 1911KB Jun 23 2011 10:24:02 AM

Chapter 4:Kinematics in Two
Dimensions
1.Two-Dimension Kinematics
2.Projectile Motion
3.Relative Motion
4.Uniform Circular Motion
5.Velocity and Acceleration in Uniform Circular
Motion
6.Nonuniform Circular Motion

Stop to think 4.1
P 93
Stop to think 4.2
P 97
Stop to think 4.3
P 102
Stop to think 4.4
P 107
Stop to think 4.5
P 110
Stop

to think

Example
4.3 4.6
P97 P 113
 Example 4.4
P98
 Example 4.5
P100
 Example 4.6
P101
 Example 4.9
P106
 Example 4.13 P110
 Example 4.15 P114

Position and Velocity
v
v
v

r  xi  yj

v v
 x1i  y1 j
v drv dx v dy v
V
 i
j
dt dt
dt

Instantaneous velocity

The Instantaneous velocity vector
is tangent to the trajectory.
The direction of the velocity is
to the curve.

Don’t confuse these two
graphs

ds
Vs 
dt

dx 2 dy 2
V  ( ) ( )
dt
dt

Acceleration

v
V
v
aavg 
t
v
v dV
a

dt

The instantaneous acceleration
can be
decomposed into parallel and
perpendicular components

Stop to think:
This acceleration will cause the
particle to:
a. Speed up and curve upward
b. Speed up and curve downward
c. Slow down and curve upward
d. Slow down and curve downward
e. Move to the right and down
f. Reverse direction

Projectile Motion
object moves in two dimensions under the
gravitational force.

B

ax  0
ay   g
A
1. What is the accelerations at position A and B?
2. What is the velocities at position A and B?

A projectile launched horizontally falls in
the same time as projectile that is released
from rest

Plot of projectile motion in txy
20
18
16
14
12
10
8

6
4
2
0
0
1
2

40

3

20
4

0

60

80

140
120
100

Launch angle

Vix  Vi cos 
Viy  Vi sin 
x  Vixt
y  Viyt  1/ 2 g (t )

2

Ex. A ball thrown horizontally at
velocity Vi , travels a horizontal
distance of R m before hitting the
ground. From what height was the
ball thrown?
(1) Since ball is thrown horizontally, Vi =Vx

There is no acceleration at x direction.
ie. R = Vxt, t = R/Vx
(2) Viy=0,

h = -1/2gt2

Problem 50

Vox  6cos(15o )m / s
Voy  6sin( 15o ) m / s

y  3  Voyt  1/ 2 gt 2
4.9t 2  1.55t  3  0
Solve a quadratic equation to get
t

d  Vox * t

The maximum height and distance
of fly ball



For projectile motion, always
remember:

ax  0, ay   g

2
i

2

v sin  i
h
2g

2
i

v sin 2 i

R
g

Trajectories of a projectile launched at
different angles with the same speed

Relative Motion


Relative position

v v v
r  r ' R
Relative velocity

v
v v
Vab  Vac  Vcb

Uniform Circular Motion





Period

1 circumference
T
speed

Angular Position

s
  (radians)
r

2 r
 full circle=
 2 rad
r
360o
1 rad 
 57.3o
2

2 r
T
V

Angular Velocity


Average angular velocity =∆θ/∆t



Instantaneous angular velocity



d

dt

The angular velocity is constant during
uniform circular motion

  t

2

T

An old-fashioned single-play vinyl record rotates
30.0 rpm . What are (a) the angular velocity in
rad/s and (b) the period of the motion?


rpm: revolution per
minute.
1 rpm = 2π/60 (rad)/s

2

T

2
T


Velocity and acceleration in
uniform circular motion

Velocity in uniform circular
The magnitude of velocity is a
motion
constant

Vt =r dθ/dt =ωr

Has only a tangential
Component

Centripetal acceleration

The magnitude of centripetal
acceleration

P184

2

V
2
ar 
 r
r
Towards center of circle

Velocity and acceleration in Uniform
Circular Motion
The velocity has only a tangential component Vt

ds
d
Vt 
r
  r (with  in rad/s)
dt
dt
2

V
a
(toward center of ciecle)
r

Nonuniform Circular Motion
dV
at 
dt

Change the speed

d
V   r with  =
dt

Here α is angular acceleration

at   r

 f   i  t if  is constant

Rotational kinematics


For constant angular acceleration

 f  i  t

 f   i   it  1/ 2 (t )
i   f
 f  i 
t
2

 f   i  2
2

2

2