# Phys110_Lecture18Rotational Kinematics

Document Sample

```					Physic 110          Lecture 18
from Chapter 7 Sections 1 to 3

Rotational Kinematics
Homework Assignment 18:

Problems:
Chapter 7,   Problem   2 on page 218
Chapter 7,   Problem   4 on page 219
Chapter 7,   Problem   6 on page 219
Chapter 7,   Problem   10 on page 219
   The radian is a
unit of angular
measure
   The radian can be
defined as the arc
length s along a
circle divided by
s
    
r
   Comparing degrees and radians

1 revolution  360  2 rad
   Converting from degrees to

 [rad]        [deg rees]
180
Example 1:
A bicycle with 68 cm diameter tires
travels 2.0 km. How many
revolutions do the wheels make?
Example 1:
A bicycle with 68 cm diameter tires
travels 2.0 km. How many
revolutions do the wheels make?
r =34 cm = 0.34 m
x = s = 2 km = 2000 m

s 2000 m                1 rev
           5880 rad          936 rev
r  0.34 m              2 rad
Angular Displacement
   Axis of rotation is
the center of the
disk
   Need a fixed
reference line
   During time t, the
reference line
moves through
angle θ
Angular Displacement:
the angle the object rotates through

   f   i

where
i    is the initial angular position

 f is the final angular position
Average Angular Speed
    The average
angular speed, ω,
of a rotating rigid
object is the ratio
of the angular
displacement to
the time interval
f  i    
 av            
tf  ti   t
Angular Speed, cont.
   The instantaneous angular speed is
defined as the limit of the average
speed as the time interval approaches
zero
   Units of angular speed are radians/sec
   Speed will be positive if θ is increasing
(counterclockwise)
   Speed will be negative if θ is decreasing
(clockwise)
Example 2
Find the angular speed of the Earth
around the sun in radians per
second and degrees per day.
Example 2
Earth makes 1 rev around sun in 1 year.

   1rev 360 deg     1year
                              0.986o / day
t 1year   1rev     365.25days
deg   2 rad   1day   1hr
 0.986                         2x107 rad / s
day 360 deg 24hr 3600s
Average Angular Acceleration
   the ratio of the change in the angular
speed to the time it takes

f  i        
 av                 
tf  ti       t
Units of angular acceleration: rad/s2
Analogies Between Linear
and Rotational Motion
Uniform motion: ω = constant

Linear:                     Rotational:

x f  xi  v t            f  o   t
where θi and θf are the angular displacements
and ω is the angular velocity
Analogies Between Linear
and Rotational Motion
Uniform accelerated motion: α= constant

Linear:                     Rotational:
v f  vo  a t               f  o   t

x f  xo   vo  v f  t    f   o   o   f  t
1                            1
2                            2
1 2                         1 2
x f  xo  vot  a t         f   o  o t   t
2                           2
v 2  vo  2a( x f  xo )
f
2
 2  o2  2 ( f   o )
f
Example 3:
A tire on a balancing machine starts from rest
and turns through 4.7 revolutions in 1.2 s
before reaching its final speed.

Assuming the acceleration was constant, find
a) the angular velocity at the end of this time
b) the angular acceleration
Example 3:
θo = 0.0              θf = 4.7 rev = 29.5 rad
t = 1.2 s             ωo = 0.0 rad/s
ωf = ?                 α=?
For uniform angular acceleration:
 f   o  o   f  t
1
2
2  f   o 
 f  o 
t
2  29.5rad  0 
f  0                         49.2rad / s
1.2s
Example 3:
θo = 0.0          θf = 4.7 rev = 29.5 rad
t = 1.2 s         ωo = 0.0 rad/s
ωf =49.2 rad/s     α=?
For uniform angular acceleration:
 f  o   t
 f  o

t
49.2rad / s  0
                  41rad / s 2

1.2s
Relationship Between Angular
and Linear Quantities
Consider a ball moving            arc length
s r

along an arc. v at
velocity along arc
an               v  r
r             s
acceleration along arc
ω
θ
at   r
α
acceleration normal to arc
2
v
an   r 
2

r
Example 4:
A 7.60 m diameter helicopter rotor rotates at a
constant speed of 450 rev/min.
What is the speed of the tip?
What is the acceleration of the tip?

r = 7.6/2 = 3.8 m    ω = 450 rev/min = 47.1 rad/s

v  r
 47.1rad / s  3.8m  179 m / s
Example 4:
A 7.60 m diameter helicopter rotor rotates at a
constant speed of 450 rev/min.
What is the speed of the tip?
What is the acceleration of the tip? (α = 0 rad/s2)

at   r                     an   r   2

 0rad / s 2  3.8m             47.1rad / s  (3.8m)   2

 0m / s   2
 680m / s   2

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 8 posted: 7/11/2011 language: English pages: 21