R-1
Rotational Motion
We are going to consider the motion of a rigid body about a fixed axis of rotation.
(rads) 
The angle of rotation is measured in radians:

r
s
s
(dimensionless)
r
Notice that for a given angle , the ratio s/r is
independent of the size of the circle.
s

r
Example: How many radians in 180o? Circumference C = 2 r
r
s
 
=
  rads  rads = 180o, 1 rad = 57.3o
r
r
s=r

r
Angle of a rigid object is measured relative to some reference orientation, just like 1D position
x is measured relative to some reference position (the origin).
Angle  is the "rotational position".
Like position x in 1D, rotational
position  has a sign convention.
Positive angles are CCW (counterclockwise).
Definition of angular velocity:  
units   =
x

x+
x


0
d
,
dt
=

(rad/s)
t
(like v 
dx
,
dt
v 
x
)
t
rad
s
In 1D, velocity v has a sign (+ or –) depending on direction. Likewise, for fixed-axis rotation, 
has a sign convention, depending on the sense of rotation.

v:
(+)
(–)
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(+)
(–)
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R-2
More generally, when axis not fixed, we define vector angular velocity 
with direction = the direction of the axis + "right hand rule". Curl fingers of
right hand around rotation, thumb points in direction of vector.
For rotational motion, there is a relation between tangential velocity v
(velocity along the rim) and angular velocity .
s
 
 s = r  ,
r
r 
s
v = r
v 
=
= r
t
t
Definition of angular acceleration :  
( like a 
dv
,
dt
a 
v
)
t
d
,
dt
Units:    =
 

r

s in
time t


(rad/s 2 )
t
rad
s2
 = rate at which  is changing.
 = constant   = 0  speed v along rim = constant = r 

Equations for constant :
Recall from Chapter 2: We defined v =
and then showed that, if a = constant,
Now, in Chapter 9, we define  =
So, if  = constant,
dx
dv
, a=
,
dt
dt
v = v0  a t

2
1
x  x 0  v0 t  2 a t
 2
2
 v  v 0  2 a ( x  x 0 )
d
d
.
, =
dt
dt
 = 0   t

2
1
  0  0 t  2  t
 2
2
  0  2  (   0 )
Same equations, just different symbols.
Example: Fast spinning wheel with 0 = 50 rad/s ( 0 = 2f  f  8 rev/s ). Apply brake and
wheel slows at  = 10 rad/s. How many revolutions before the wheel stops?
2/6/2016
©University of Colorado at Boulder
R-3
Use 2  02  2   , final = 0  0  02  2      
125 rad 
02
502

 125 rad
2
2( 10)
1 rev
 19.9 rev
2 rad
Definition of tangential acceleration atan = rate at which speed v along rim is changing
d( r )
dv
d
a t an 
=
 r

atan = r 
dt
dt
dt
v2
atan is different than the radial or centripetal acceleration a r 
r
ar is due to change in direction of velocity v
atan is due to change in magnitude of velocity, speed v
atan and ar are the tangential and radial components of the
acceleration vector a.
a
atan
ar
|a|  a 
a tan 2  a r 2
Angular velocity  also sometimes called angular frequency.
Difference between angular velocity  and frequency f:
 
# radians
,
sec
f 
# revolutions
sec
T = period = time for one complete revolution (or cycle or rev) 
 
2 rad
2

,
T
T
f 
1 rev
1

T
T

  2f
Units of frequency f = rev/s = hertz (Hz) . Units of angular velocity = rad /s = s-1
Example: An old vinyl record disk with radius r = 6 in = 15.2 cm is spinning at 33.3 rpm
(revolutions per minute).
 What is the period T?
33. 3 rev
1min

33.3 rev
60 s

60 s
33.3 rev

( 60 / 33.3) s
1rev
1.80 s/rev
 period T = 1.80 s
2/6/2016
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R-4
 What is the frequency f ?
f = 1 / T = 1 rev / (1.80 s) = 0.555 Hz
 What is the angular velocity  ?   2  f  2  ( 0.555 s 1 )
3.49 rad / s
 What is the speed v of a bug hanging on to the rim of the disk?
v = r  = (15.2 cm)(3.49 s-1) = 53.0 cm/s
 What is the angular acceleration  of the bug?  = 0 , since  = constant
 What is the magnitude of the acceleration of the bug? The acceleration has only a radial
component ar , since the tangential acceleration atan = r  = 0.
( 0.530 m/s)2
v2
a = ar 

 1.84 m/s 2 (about 0.2 g's)
r
0.152 m
For every quantity in linear (1D translational) motion, there is corresponding quantity in
rotational motion:
Translation 
x
Rotation


v
dx
dt

=
d
dt
a
dv
dt

=
d
dt
F

(?)
M

(?)
F = Ma

(?) = (?) 

KE = (1/2) (?) 2
KE = (1/2) m v2
The rotational analogue of force is torque.
Force F causes acceleration a
F = F sin

Torque causes angular acceleration 
The torque (pronounced "tork") is a kind of
F
axis
r
  r  F  r Fsin 
  r F  m N
2/6/2016

F||
"rotational force".
magnitude of torque:
F
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R-5
r = "lever arm" = distance from axis to point of application of force
F = component of force perpendicular to lever arm
Example: Wheel on a fixed axis:
Notice that only the perpendicular component of the force F will rotate the wheel. The
component of the force parallel to the lever arm (F||) has no effect on the rotation of the wheel.
If you want to easily rotate an object about an axis, you want a large lever arm r and a large
perpendicular force F:
best
better
bad
axis
no good!
(F = 0)
no good!
(r = 0)
Example: Pull on a door handle a distance r = 0.8 m from the hinge with a force of magnitude F
= 20 N at an angle  = 30o from the plane of the door, like so:
F
F
hinge
 = r F = r F sin  =

(0.8 m)(20 N)(sin 30o) = 8.0 mN
r
Another example:
a Pulley
 = r F
r
F
For fixed axis, torque has a sign (+ or –) :
Positive torque causes counter-clockwise CCW rotation.
Negative torque causes clockwise (CW) rotation.
+
–
If several torques are applied, the net torque causes angular acceleration: net 

 
Aside: Torque, like force, is a vector quantity. Torque has a direction.
Definition of vector torque :   r  F = cross product of r and F: "r cross F"
Vector Math interlude: The cross-product of two vectors is a third vector A  B  C defined
like this: The magnitude of A  B is A B sin . The direction of A  B is the direction
perpendicular to the plane defined by the vectors A and B plus right-hand-rule. (Curl fingers
from first vector A to second vector B, thumb points in direction of A  B
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©University of Colorado at Boulder
R-6
AB
A

B
To see the relation between torque  and angular acceleration  , consider a mass m at the end of
light rod of length r, pivoting on an axis like so:
m
r
axis
Apply a force F to the mass, keeping the force perpendicular to the lever arm r.
F
acceleration atan = r 

Apply Fnet = m a, along the tangential direction:
F = m atan = m r 
F
axis
Multiply both sides by r ( to get torque in the game ): r F = (m r 2) 

Define "moment of inertia" = I = m r 2


 = I
( like F = m  a )
Can generalize definition of I:
Definition of moment of inertia of an extended object
about an axis of rotation:
I 
m r
2
i i
 m1 r12  m2 r2 2  ...
i
axis
ri
mi
Examples:
 2 small masses on rods of length r:
r
r
m
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axis
m
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R-7
I = 2 m r2
mi
 A hoop of total mass
center, has Ihoop = M R2
M, radius R, with axis through the
R
I 
m r
2
i i
i


   mi  R 2  M R 2
 i

(since ri = R for all i )
I  m1r12  m2 r2 2  m3r32 
In detail:
 ( m1  m2  m3 
 m1R 2  m 2R 2  m3R 2 
)R 2  MR 2
mass M
 A solid disk of mass M, radius R, with axis through the center:
Idisk = (1/2) MR2 (need to do integral to prove this)
R
See Appendix for I’s of various shapes.
Moment of inertia I is "rotational mass".
Big I 
( like Big M  hard to get moving )
hard to get rotating
If I is big, need a big torque  to produce angular acceleration according to
 net = I  
( like Fnet = m a )
Example: Apply a force F to a pulley consisting of solid disk of radius R, mass M.  = ?
  I
R
RF 
F
2/6/2016

1
2

MR 2 

 
2F
MR
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R-8
Parallel Axis Theorem
Relates Icm (axis through center-of-mass) to I w.r.t. some other axis:
I = Icm + M d2
(See proof in text.)
Example: Rod of length L, mass M
1
ICM 
M R 2 , d = L/2 
12
axis here
(I)
1
1
1
M L2  M L2  M L2
12
4
3
Iend axis  ICM  M d 2 
rod mass M
length L
d
axis here
( Icm )
Rotational Kinetic Energy
How much KE in a rotating object? Answer: KE rot 
KE tot 
Proof:
(
1
2
1
2
I 2
( like KE trans 
1
2
m v2 )
mi vi 2 )
i
v  r,
KE 
vi   ri
 ( 12 mi ri 2 ) 
axis
1
2
i

2 2
  mi ri   
 i

1
2
ri
mi
I 2
How much KE in a rolling wheel?
The formula v = r  is true for a wheel spinning about a fixed
axis, where v is speed of points on rim. A similar formulas vCM

= r works for a wheel rolling on the ground. Two very
v
different situations, different v’s: v = speed of rim vs. vcm =
speed of axis. But v = r  true for both.
v


v
axis stationary: v = r
2/6/2016
vcm = center-of-mass
velocity =  r
point touching ground
instantaneously at rest
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R-9
To see why same formula works for both, look at situation from the bicyclist's point of view:

v
axis stationary,
ground moving
v
Rolling KE: Rolling wheel is
simultaneously translating and
KEtot
KE tot 
1
2
M v2 
1
2
+
=
rotating:
I 2
=
KEtrans
+
KErot
(v = Vcm )
(See proof in text.)
Conservation of energy problem with rolling motion:
A sphere, a hoop, and a cylinder, each with mass M and radius R, all start from rest at the top of
an inclined plane and roll down to the bottom. Which object reaches the bottom first?
M
R
(vi = 0)
h
vf = v =?

Apply Conservation of Energy to determine vfinal. Largest vfinal will be the winner.
KE i
0
 PE i

 Mgh 

KE f
1
2
Mv 
2
KE trans
1
2
PE f
I  0
2
KE rot
Value of moment of inertia I depends on the shape of the rolling thing:
Idisk = (1/2)M R2 , Ihoop = M R2 , Isphere = (2 /5)M R2 (Computing coefficient requires integral.)
Let's consider a disk, with I = (1/2)MR2. For the disk, the rotational KE is
2/6/2016
©University of Colorado at Boulder
R-10
2
1
2
I 2 
1
2
v
( 12 M R 2 )   
R
 Mgh 
gh 
3
4
1
2
M v2 
v2 ,
1
4
1
4
M v2
[used   v / r]
M v 2  ( 12  14 ) M v 2 
v 
4
3
3
4
M v2
g h  1.16 g h
Notice that final speed does not depend on M or R.
Let's compare to final speed of a mass M, sliding down the ramp (no rolling, no friction).
M
Mgh 
1
2
v =
2 g h  1.4 g h
M v2
(M's cancel)
(vi = 0)
h

vf =
v =?
Sliding mass goes faster than rolling disk.
Why?
As the mass descends, PE is converted into KE. With a rolling object, KEtot = KEtrans + KErot , so
some of the PE is converted into KErot and less energy is left over for KEtrans. A smaller KEtrans
means slower speed (since KEtrans = (1/2) M v2 ). So rolling object goes slower than sliding
object, because with rolling object some of the energy gets "tied up" in rotation, and less is
available for translation.
Comparing rolling objects: Ihoop > Idisk > Isphere  Hoop has biggest KErot = (1/2) I 2, 
hoop ends up with smallest KEtrans  hoop rolls down slowest, sphere rolls down fastest.
Another conservation of rotational energy problem:
Rod of mass M, length L, one end stationary on ground, starts from rest at angle  and falls.
What is speed v of end of stick, when stick hits ground?
I
L
Plan: Use conservation of energy to get ,
CM

fixed axis
2/6/2016
1
M L2 (axis at end)
3
hCM
v=?
then v =  r =  L
Ei  E f

M g h CM 
1
2
I 2
©University of Colorado at Boulder
R-11
Important point:
PEgrav = Mgh where h = height of center-of-mass, independent of the orientation of the stick.
Proof: PE grav 
m gh
i
i
i
 g  mi h i  g M YCM  M g h CM
i
( Have used definition of center-ofmass: MYCM 
m
i
yi )
same hCM , same PEgrav
i
Back to the problem: M g h CM 
M g L sin  
1
2
1 1
2 3
1
2
I 2 ,
h  12 Lsin  , I 
M L   g sin  
2
2
1
3
1
M L2
3
L
2
v2
v2
Use   v / r  v / L to get: 3g sin   L 2 
 v
L
L
L/2

3g Lsin  Done.
The tip of the stick starts at height htip = L sin, but its final speed v
is faster than the speed of an object that falls from that height h [
1
2
m v2  m g h  v 
2g h ]. The tip of the stick falls faster
than it would in free-fall, because the central part of the rod pulls it
down. This is why tall chimneys always break apart when toppled:
:
Let's Review: Translation
x
Rotation


v
dx
dt

=
d
dt
a
dv
dt

=
d
dt
F


M

I

 =I

KE = (1/2) I 2
F = Ma
KE = (1/2) m v2
2/6/2016

©University of Colorado at Boulder
h
R-12
Appendix:
Moments of Inertia for some shapes:
R
Hoop
Disk
I = M R2
I = (1/2) M R2
Solid sphere
Thin spherical shell
I = (2/5) M R2
I = (2/3) M R2
L
L
2/6/2016
Thin rod, axis thru center
Thin rod, axis thru end
I = (1/12) M L2
I = (1/3) M L2
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