Chapter 11A – Angular Motion
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
©
2007
WIND TURBINES such
as these can generate
significant energy in a
way that is environmentally friendly and
renewable. The
concepts of rotational
acceleration, angular
velocity, angular
displacement, rotational
inertia, and other topics
discussed in this
chapter are useful in
describing the operation
of wind turbines.
Objectives: After completing this
module, you should be able to:
• Define and apply concepts of angular
displacement, velocity, and acceleration.
• Draw analogies relating rotational-motion
parameters (, , ) to linear (x, v, a)
and solve rotational problems.
• Write and apply relationships between
linear and angular parameters.
Objectives: (Continued)
• Define moment of inertia and apply it for
several regular objects in rotation.
• Apply the following concepts to rotation:
1. Rotational work, energy, and power
2. Rotational kinetic energy and
momentum
3. Conservation of angular momentum
Rotational Displacement, 
Consider a disk that rotates from A to B:
B

Angular displacement :
A
Measured in revolutions,
degrees, or radians.
1 rev = 360 0 = 2 rad
The
The best
best measure
measure for
for rotation
rotation of
of
rigid
.
rigid bodies
bodies isis the
the radian
radian.
Definition of the Radian
One radian is the angle  subtended at
the center of a circle by an arc length s
equal to the radius R of the circle.
s
s

R
1 rad =
R
R
= 57.30
Example 1: A rope is wrapped many times
around a drum of radius 50 cm. How many
revolutions of the drum are required to
raise a bucket to a height of 20 m?
s
20 m
 
R 0.50 m
 = 40 rad
R
Now, 1 rev = 2 rad
 1 rev 
   40 rad  

 2 rad 
 == 6.37
6.37 rev
rev
h = 20 m
Example 2: A bicycle tire has a radius of
25 cm. If the wheel makes 400 rev, how
far will the bike have traveled?
 2 rad 
   400 rev  

 1 rev 
 = 2513 rad
s = R = 2513 rad (0.25 m)
ss == 628
628 m
m
Angular Velocity
Angular velocity, is the rate of change in
angular displacement. (radians per second.)


t
Angular velocity in rad/s.
Angular velocity can also be given as the
frequency of revolution, f (rev/s or rpm):

 
ff Angular
Angular frequency
frequency ff (rev/s).
(rev/s).
Example 3: A rope is wrapped many times
around a drum of radius 20 cm. What is
the angular velocity of the drum if it lifts the
bucket to 10 m in 5 s?
s
10 m
 
R 0.20 m
 = 50 rad
R

50 rad


t
5s
h = 10 m

 == 10.0
10.0 rad/s
rad/s
Example 4: In the previous example, what
is the frequency of revolution for the drum?
Recall that  = 10.0 rad/s.

  2 f or f 
2
10.0 rad/s
 1.59 rev/s
f 
2 rad/rev
R
Or, since 60 s = 1 min:
rev  60 s 
rev
f  1.59

  95.5
s  1 min 
min
ff == 95.5
95.5 rpm
rpm
h = 10 m
Angular Acceleration
Angular acceleration is the rate of change in
angular velocity. (Radians per sec per sec.)


t
Angular acceleration (rad/s 2 )
The angular acceleration can also be found
from the change in frequency, as follows:
2 (f )

t
Since
  2 f
Example 5: The block is lifted from rest
until the angular velocity of the drum is
16 rad/s after a time of 4 s. What is the
average angular acceleration?

 f  o
t
0
or

f
R
t
16 rad/s
rad

 4.00 2
4s
s
h = 20 m
22

=
4.00
rad/s
 = 4.00 rad/s
Angular and Linear Speed
From the definition of angular displacement:
s = R Linear vs. angular displacement
s    R   
v


t  t   t

R

v = R
Linear
Linear speed
speed == angular
angular speed
speed xx radius
radius
Angular and Linear Acceleration:
From the velocity relationship we have:
v = R Linear vs. angular velocity
v  v  R   v 
v

  R
t  t   t 
a = R
Linear
Linear accel.
accel. == angular
angular accel.
accel. xx radius
radius
Examples:
R1
Consider flat rotating disk:
B
 = 0; f = 20 rad/s
t=4s
What is final linear speed
at points A and B?
A
R2
R1 = 20 cm
R2 = 40 cm
vAf = Af R1 = (20 rad/s)(0.2 m);
vAf = 4 m/s
vAf = Bf R1 = (20 rad/s)(0.4 m);
vBf = 8 m/s
Acceleration Example
Consider flat rotating disk:
R1
A
B
 = 0; f = 20 rad/s
t=4s
What is the average angular
and linear acceleration at B?

 f  0
t
20 rad/s

4s
a = R = (5 rad/s2)(0.4 m)
R2
R1 = 20 cm
R2 = 40 cm
22

=
5.00
rad/s
= 5.00 rad/s
aa == 2.00
2.00 m/s
m/s22
Angular vs. Linear Parameters
Recall the definition of linear
acceleration a from kinematics.
a
v f  v0
t
But, a = R and v = R, so that we may write:
a
v f  v0
t
becomes
R 
R f  R 0
t
 f  0
Angular acceleration is the time
rate of change in angular velocity.  
t
A Comparison: Linear vs. Angular
 v0  v f
s  vt  
 2

t

 0   f
  t  
2

v f  vo  at
 f  o   t
s  v0t  at
1
2
2
s  v f t  at
1
2
2
2as  v  v
2
f
2
0

t

   0t   t
1
2
2
   f t  t
2
2    
2
0
1
2
2
f
Linear Example: A car traveling initially
at 20 m/s comes to a stop in a distance
of 100 m. What was the acceleration?
100 m
Select Equation:
2as  v 2f  v02
a=
0 - v o2
2s
vo = 20 m/s vf = 0 m/s
-(20 m/s)2
=
2(100 m)
22
aa == -2.00
m/s
-2.00 m/s
Angular analogy: A disk (R = 50 cm),
rotating at 600 rev/min comes to a stop
after making 50 rev. What is the
acceleration?
Select Equation:
2
2
2   f   0
o = 600 rpm
R
f = 0 rpm
 = 50 rev
rev  2 rad   1 min 
600


  62.8 rad/s
min  1 rev   60 s 
=
0 - o2
2
-(62.8 rad/s)2
=
2(314 rad)
50 rev = 314 rad
22
 == -6.29
m/s
-6.29 m/s
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction of rotation.
 List givens and state what is to be found.
Given: ____, _____, _____ (,,f,,t)
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
Example 6: A drum is rotating clockwise
initially at 100 rpm and undergoes a constant
counterclockwise acceleration of 3 rad/s2 for
2 s. What is the angular displacement?
Given: o = -100 rpm; t = 2 s
 = +2 rad/s2
rev  1 min   2 rad 
100


  10.5 rad/s
min  60 s   1 rev 

   ot   t  (10.5)(2)  (3)(2)
1
2
2
 = -20.9 rad + 6 rad
1
2
R
2
 == -14.9
-14.9 rad
rad
Net displacement is clockwise (-)
Summary of Formulas for Rotation
 v0  v f
s  vt  
 2

t

 0   f
  t  
2

v f  vo  at
 f  o   t
s  v0t  at
1
2
2
s  v f t  at
1
2
2
2as  v  v
2
f
2
0

t

   0t   t
1
2
2
   f t  t
2
2    
2
0
1
2
2
f
CONCLUSION: Chapter 11A
Angular Motion