  2 f
Chapter 8:
Rotational Motion
  I
Sec 8-1
“Please – Make it Stop”
Rigid Body – the distance between any two particles remains
constant.
In general the motion of a rigid body consists of translation of the
center of mass plus rotation about the center of mass (CM).
Rotation about
CM
The Actual Path
of the object’s
center of mass
2
Angular Movement of a Rigid Body
Consider a particle P at position PI
moving along a circular arc S

S
Pi
Linear Speed of P is
v

Reference Line
s
t
 
ArcLength s

radius
r
Rotating Wheel “system” made up
of many particles P.

All particles in the wheel “system” will sweep out the same angle 
in a given amount of time.

Therefore, the rate of change of the angle  with respect to time,
/t, is the same for all particles on the wheel.
Multiply both sides of above equation by (1/t)

s

t t r

v
r
v  r
3
Units for “omega” :

 radians rad 1



t
sec
s
s
The Angular Acceleration,  (alpha) is


t
rad
units for alpha are
s2
Relating Angular Acceleration to Linear Acceleration
v  r
atan
v

t
atan 
( r )
t
atan   r
 Remember !!!!!!!!
2
v



atotal  a tan  a rad   r (tan) 
(rad)
r

a total   r (tan)   2 r (rad)
4
Sec 8-2
CONSTANT ANGULAR ACCELERATION EQUATIONS
RECALL:
THEN:
a  r
v  r ,
s  r ,
v f  vi  at
 f r   i r  rt
 f   i  t
1
s f  si  vi t  at 2
2
1
 f r   i r   i rt  rt 2
2
1
 f   i   i t  t 2
2
v 2f  vi2  2as
 2f r 2   i2 r 2  2r  f r   i r 
 2f   i2  2  f   i 
OTHERS:

 f  i 

2


t
5


Direction for the  and  vectors
Any linear direction in the plane of the wheel does not uniquely define the
rotation of the wheel because of symmetry
Right-Hand Rule: curl the fingers of your
right hand in the direction of the rotation,
and thumb will point in the direction of 


 and  Vectors

   ẑ

   ẑ

   ẑ
And Speeding Up
6
Angular Velocity and Frequency
Period, T = time to complete 1 revolution
Frequency, f = # of revolutions per second.
f 
#revs
sec
i.e. 3 rev/sec = 3 Hertz (Hz)
1Hz 
1
sec
Period and Frequency are the reciprocal of each other!
Example:
A frequency of 1 rev/sec equals a period of 1 sec. (or sec/rev)
A frequency of 2 rev/sec equals a period of 0.5 sec. (or sec/rev)
1
f
Therefore,
T
Since
1 rev = 2 radians
OR
 rev  2 rad rad
So, f 


rev
sec
 sec 
Therefore,
0r
f 
1
T
(Wow! this is  )
f (2 )  
  2 f
Angular Velocity,  , is also called Angular Frequency.
7
EXAMPLE PROBLEMS
A particle moves in a circle of radius 100 m with a constant velocity of 20 m/s. (a) What is the
angular velocity in radians per second about the center of the circle? (b) How many revolutions
does it make in 30 s? Answer: 0.2 rad/s, 0.955 rev
A 30 cm diameter music record (vinyl) rotates at 33 1/3 rev/min. (a) What is its angular velocity in
radians per second? (b) Find the speed and linear acceleration of a point on the rim of the record.
a.) 3.49 rad/s b.) 52.35 cm/s c.) a total = 0 tan + 182 cm/s2 rad
A turntable rotating at 33 1/3 rev/min is shut off. It brakes with a constant angular acceleration and
comes to rest in 2.0 min. (a) Find the angular acceleration. (b) What is the average angular velocity
of the turntable? (c) How many revolutions does it make before stopping?
a.) -16.7 rev/min2 b.) 16.63 rev/min c.) 33.26 rev
8
Comparison of Linear Motion and Rotational Motion
Linear
Motion
Rotational Motion
Displacement
x
Velocity
v
dx
dt
Acceleration
a
dv
dt
Constant
Acceleration
Equations
x  xo  vo t  12 at 2
Mass
Momentum
Force
Power
Newton’s 2nd
Law

d 2x
dt 2
Angular Displacement

Angular Velocity
  ddt
Angular Acceleration

Constant Angular
Acceleration Equations
   o   o t  12 t 2
d
dt

d 2
dt 2
v  vo  at
   o  t
v 2  vo2  2ax
 2   o2  2
v av  12 (vo  v)
 av  12 ( o   )
m


p  mv

F
 
P  F v

dp

 F   ma
dt
Moment of Inertia
I
Angular Momentum


L  I

Torque
Power
Newton’s 2nd Law

 
P   

dL

   I
dt
9
8-3 Rolling Motion
Ffriction
Ffriction
How can friction point in opposite directions?
Rolling Without Slipping – Velocity and Acceleration
A
O
vcm
s
O vcm
A
s
s  r
s = distance x the cm of the wheel moved

vcm 
x ( R )


R
t
t
t
vcm   R
Conditions for Rolling
Without Slipping
Then
acm   R
10
Rotation and Translation of Points on the Wheel
While Rolling Without Slipping
O
O
Spinning and
Translating
Translating
Spinning
vtop of
O
wheel
 2 vcm
vcenter of
wheel
 vcm
vbottom  0
Two Ways to achieve Rolling without Slipping
I.
Chain/Gears Spin Wheel – Static Friction from Ground
Translates Wheel
The gears spin the wheel.
When the wheel hits the
ground it lurches forward Translation
Fon wheel
by ground
II.
Push on Axel Translates Wheel – Static Friction from
Ground allows wheel to Rotate
Fpush
Fon wheel
by ground
Fpush translates the wheel.
Static friction at the ground
lets the wheel fall over itself
– Rotation
11
8-4 TORQUE
Archimedes (287-312 BC) – examined the turning “effect” of forces in his
analysis of levers.
Pivot Point
Boulder
FPUSH
By Definition:
  (lever arm distance) ( Force )
Greek Letter “Tau” is
the symbol for torque.
Lever Arm = The perpendicular distance from the pivot point to
the line of action of the force.
Often the distance along the stick itself is the lever arm distance BUT NOT ALWAYS, only when Fpush is tangent to arc of
rotation.
12
d  lever arm
Line-of-action
of force


FPULL
O
r
Lever Pole
Radial vector from pivot point
r
  
  r  F  r F sin  r , F nˆ
LEVER ARM

FORCE TANGENT
  (r sin )F
  (F sin )r
  rF
 F r
WHERE
r  r sin(r ,F )
WHERE
Ftan  F sin(r ,F )
( r ,F )
( r ,F )
tan
FPULL
13
EXAMPLE: A 30 N and a 15 N force are applied to the rod
shown below. Determine the net torque and direction of rotation.
1m
40o
2m
F2=15 N
F1=30 N
40o
30o
14
8-5 Torque and
Moment of Inertia
F
A force, F, is acting on
a wheel
Fradial, does not effect
the rotation of the
wheel. The torque
exerted byF is

Fradial
Ftangent
m
r

  r  F nˆ
BY DEFINITION
The magnitude of the
torque, , “tau” is equal to
the product of Ftangent , and the magnitude of the radius vector.
 F
tan
SINCE:
F = ma
r
A
and a   r

 mr 2
B
Now if we sum all of the particles, m, on the wheel.

 mr 2
C
The sum, mr 2 , is a property of the wheel called the moment of inertia, I.
The moment of inertia depends on the distribution of the mass relative to
the axis of rotation.
I   mr 2
D
15
From


C and
 mr 2
 I
D
I   mr 2
Newton’s 2nd
Law for rotation
How to “think” about Torque
1. Torque must be specified about a pivot point
2. Torque is a product quantity made up of distance and force.
3. Torque causes angular acceleration, , in the same way that
forces cause linear accelerations.
4. The Moment of Inertia, I, is a measure of resistance to rotation
analogous to mass as a measure of inertia for linear motion.
5. We can’t “see” or “touch” torque, in the same sense that we
can’t “see” or “touch” forces – We see the effects of both.
6. Equilibrium means
 F  0 and   0
Object at rest totally
or translating and/or rotating with constant v or 
16
8-6 Solving Problems in Rotational Dymanics
Example: A uniform cylinder of mass M and radius R is held by a hand that is
accelerated upward so that the center of mass of the cylinder does not move. Find (a)
the tension in the string, (b) the angular acceleration of the cylinder, and (c) the
acceleration of the hand
Answers: (a) Mg (b) 2g/R (c) 2g
17
Example: A body of mass 1.2 kg is tied to a
light string wound around a 2.5 kg wheel of
radius 0.2 m. The wheel bearing is frictionless.
Find the tension is the string, the acceleration
of the block, and its speed after it has fallen a
distance h = 0.25 m from rest. T=6N, a = -4.8 m/s2 ,
s

R
v = 1.55 m/s
M
M
18
8-7 Rotational Kinetic Energy
Consider of piece of mass rotating
a central point
about
m
r
The Kinetic energy of the mass is
v
1
KE  m v 2
2
Consider a wheel made up of many such masses, then
1
1
1
KE   mi vi 2   mi (ri ) 2  ( mi ri 2 ) 2
2 i
i 2
i 2
KE rotation
1 2
 I
2
 Remember that the wheel may also be translating

Vcm
Then
1
1
2
KE total  KE trans  KE rotation  MVcm
 I 2
2
2
19
Rolling Without Slipping – Different Shapes
Moment of Inertia Considerations (Same Mass, Same
Radius)
Hoop
Disk
Sphere
1
I cm  M R 2
2
Disk
I cm
 M R2
I cm
Hoop

Sphere
2
MR 2
5
KEtotal  KErot  KEtrans
KEtotal
Hoop
2
1 Vcm
1
2
 I 2  M Vcm
2 R
2
Disk
Sphere
V2 1
1
2
( MR 2 ) cm
 MVcm
2
2
2
R
2
1 1
1
2 Vcm
2
( 2 MR ) 2  MVcm
2
2
R
2
1 2
1
2 Vcm
2
( 5 MR ) 2  MVcm
2
2
R
1
1
2
2
MVcm
 MVcm
2
2
1
1
2
2
MVcm
 MVcm
4
2
1
1
2
2
MVcm
 MVcm
5
2
50 %
KErot
50 %
KEtrans
33 %
KErot
67 %
KEtrans
29 %
KErot
71 %
KEtrans
20
If started from rest at the top of a hill, which will move the fastest
at the bottom of a hill?
KEi  PEi  KE f  PE f
Hoop
Disk
Sphere
2
mgh  MVcm
2
mgh  34 MVcm
2
7
mgh  10
MVcm
PEi  KE f
PEi  KE f
PEi  KE f
Example: A hoop of radius 0.5 m and mass 0.8 kg is rolling without slipping at a
o
speed of 20 m/s toward an incline of slope 30 . How far along the incline will the
hoop roll? Assuming it rolls with out slipping. Answer 81.5 m
30
o
21
Example: Four particles of mass m are connected by massless rods to form a
rectangle of sides 2a and 2b, as shown below. The system rotates about an axis in the
plane of the figure through the center. Find the moment of inertia about this axis and
the kinetic energy of rotation if the angular velocity is 
m
m
m
m
2b
2a
Example: Find the moment of inertia and Kinetic Energy of rotation of the system
shown below.
m
m
2b
m
m
2a
22
8-8 Angular Momentum and Its Conservation
Definition of Angular Momentum
Recall
Now

F  ma
m
v
t
I

t
 ( I )
t
 (mv )

t

p

t
 
L  I
p  mv
is Linear Momentum
is Angular Momentum
dL
 
dt
Since

L
t
with
with
If
 I
0
then
Li  L f
and
I ii  I f  f
23
Torque and Angular Momentum – Formal Definitions



 r F
Torque analogous to Force
  
l rp
Angular momentum analogous to Linear Momentum


p  mv

r
r,p
r
m
 An object moving in a straight line has an angular momentum about
the pivot point O.
24
Newtons 2nd Law for Torques in momentum form
  
l rp

d l
dt

d
dt
 
m(r  v )



dl


 m (r  ddtv )  ( ddtr  v )
dt



dl
 
 
 m(r  a )  (v  v )
dt

dl


  
 (r  ma )  r  F  
dt


d
l

 
dt
Conservation of Angular Momentum
If the resultant external torque acting on a system is zero, the total
angular momentum of the system is constant.

dL
 
dt


dL
0
dt


L  constant
25
Conservation of Angular Momentum
Example: A playground merry-go-round is at rest, pivoted about a frictionless axis.
A ferocious rabbit runs along a path tangential to the rim with initial speed v i and
jumps onto the merry-go-round. What is the angular velocity of the merry-go-round
and the rabbit?
r
v
26
Example: A disk with a moment of inertia I1 is rotating with angular velocity 1
about a frictionless shaft. It drops onto another disk with moment of inertia I2
initially at rest. Because of surface friction, the two disks eventually attain a common
angular velocity. What is it?
I1
I1
I2
I2
27
Appendix I
Continuous Mass Distributions
 For a continuous body replace I   mi ri2 with
i
I   r 2 dm
Example: Calculate the Moment of Inertia of a uniform stick about an axis
perpendicular to the stick through one end
dm
x
dx
28
Appendix II - Some Rotational Inertias