Download rotation


r1
When the axis of rotation is fixed, all particles
move in a circle.
Because the object is rigid, they move through
the same angular displacement in the same
time period.

r2
x
r

Radians only!
x  r
v  r
at  r
d

dt
d d 2

 2
dt
dt
Note that as long as there is rotation there will be a radial or centripetal acceleration
given by
v2
acp 
  2R
R
Linear 1-D vs Fixed Axis Rotation
1-D
linear
motion
Rotation: fixed
axis
∆x
∆
a


m
I
v
K
1 2
mv
2
F
F=ma
p=mv

W  Fdx
P=Fv
 mr
K

2
1 2
I
2
  I
L  I
W   d
P  
Constant Angular Acceleration Kinematics
The equations for 1-D motion with constant acceleration are a result the
dv
definitions of the quantities; because a  dt it immediately follows that
v  v0  at if acceleration is constant.
Since the angular variables θ, ω, α are related to each other in exactly the same
way as x, v, and a are, it follows that they will obey analogous kinematic equations:
1 2
   0t  t
2
 2  02  2
   0  t
1
     0 t
2
Exercise: A disc initially rotating at 40 rad/s is slowed to 10 rad/s in 5 s. Find
the angular acceleration and the angle through which it turns during this time.
   0  t
10  40   5
rad
  6 2
s
102  402  2 6
  125 rad

r1

r2
K
Ki 

I
Ki 

m r
K rot
1
1
mi vi2  mi ri2 2
2
2
1
1
2
mi vi 
2
2
2
i i
1 2
 I
2
 m r 
2
i i
Moment of inertia about the
axis of rotation
2
Moment of Inertia Examples
R
m1
I L  m1 0  m2 R   m2 R
2
m2
2
I R  m1  R   m2 0  m1R 2
2
L
C
R
2
2
2
R
 R
 R
I C  m1     m2    m1  m2 
4
 2
2
 2,1
2
2,2
Find I origin and I 2,2
2
1
4
4,1
2
Continuous Matter Distribution
m r
i i
2


2
r dm
body
Example: Uniform rod, length L, mass M. Find ILeft
x
dx
1) Find contribution of arbitrary little piece dm.
dI left
M  2
  dx  x
 L 
2) Sum up (integrate over correct limits) all the little pieces.
I left 

L
dI left 
body

0
M 2
1
x dx  ML2
L
3
Uniform disc, radius R, mass M. Find IC
1) Find contribution of arbitrary little piece dm.
dI  dm r
r
2
All the mass in ring is same r away.
But what is dm?
The area in the ring is an infinitesimal:
dr

Let σ designate mass per area:
dm  dA 
Then
M
R
2
2 rdr
dA  2r dr
M
R 2
2) Sum up (integrate over correct limits) all the little pieces.
I

R
 
0
0
R

4
R
1
3
2
dI   dAr   2 r dr  
 MR 2
2
2
body

Parallel Axis Theorem

r2

r1
A
IA 


r1

r2
CM
mr 2
rel A
I CM 

mr 2
rel CM
R
CM
I A  I CM  MR
2
A
Total mass
Example: Find ICM for a uniform stick of length L, mass M.
L
2
 L
I end  I CM  M  
2
2
1
1
 L
2
2
ML  I CM  M    I CM  ML
3
12
2
Exercise: Find I about a point P at the rim of a uniform disc of mass M, radius R.
R
P
Torque
From Newton’s 2nd law we know that forces cause accelerations. We might ask
what particular quantity, obviously related to force, will cause angular
accelerations.
Consider a 10 N force applied to a rod pivoted about the left end. We can apply
the force in a variety of ways, not all causing the same angular acceleration:
From the examples it should be clear that it is a combination of the force
applied, direction of application, and distance from axis that need to be
accounted for in determining if a force can cause angular acceleration.
The previous discussion is consistent with focusing on the following quantity:


F
  Feff R  F sin R
R

Reff or lever arm
  F sin  R  F R sin  
  FReff
Line of action of F
The greater the torque associated with a force, the more efficient it is in creating
angular acceleration.
Fi t

Fi

ri
t
Fi  mi ai
t
t
2
ri Fi  mi ri ai  mi ri 
2
i 
mi ri 
t
Torque on piece
i


net  I

Refers to a particular
axis
Example: Find the acceleration of the mass, the
tension in the rope and the speed of the mass having
fallen H.
I
R
  RT
T
T
m
H
net  I
  RT  I
a
RT  I
R
mg
mg  T  ma
a
m
I
m 2
R
g
I
R
m
H
1 2 1 2
mgH  mv  I
2
2
2
1 2 1 v
mgH  mv  I  
2
2  R
v
2mgH
I
m 2
R
Rolling
When an object rolls, the point of contact is instantaneous at rest. It can be
thought of as an instantaneous axis of rotation, and relative to this point we
have pure rotation at that instant.
At this instant, each point shown will have a different speed, but
is the same for each point on the object.
vi

ri
2v
v  R
v  vCM
v
=
rolling condition
v
v
v
v
+
v
Conceptually, we can think of rolling as pure translation at vCM coupled with a
simultaneous rotation about CM with   vCM
R
2v
v
K rot
K rot
v
v
K rot


1 2 1
2
 I  Mv
2
2
v
+
v
1
2
 Mv
2
1
2
K rot  I cp
2
1
 I  MR 2  2
2
v

1 2
I
2
We can use energy concepts coupled with kinematics analyze rolling dynamics.
I   MR
2
v  R
v
H

1
1 2
2 gH
2
2
MgH  Mv  I  v 
2
2
1 
H
g sin 
v  0  2a
a
sin 
1 
2
N
fs

g sin 
Mg sin   f s  Ma  M
1 
  

f s  Mg sin  
 1  
Exercise: Find and interpret the work done by fs.
Exercise: At what angle will slipping start?
Mg
Torque
A more systematic approach to rotation involves relating the torques associated
with forces to the angular accelerations they produce. This can be more
complicated if the axis is not fixed, but we shall see that the rolling case is
especially simple in this approach.
First generalize torque:

F

r
 
  r F
  rF sin 


Right hand rule, RHR
Note again that torques are calculated relative to a specific origin and will
change if we change the origin.
Where could I place an origin above so that F would exert zero torque?
Angular Momentum

r
  
l rp


p
l  rp sin   r sin   p
l  pr
r
1) a vector perpendicular to both
RHR

r
and

p
2) magnitude equals linear momentum times “closest approach distance”.
3) value clearly depends choice of origin.
What kind of motion might have constant angular momentum?
Why Angular Momentum?
  
l rp



dl dr   dp

 pr
dt dt
dt


  
dl
 mv  v   r  netF
dt


dl
 net
dt
Rotational analog of 2nd law
For a system of particles we can define the total angular momentum:

L


li
Applying the second and third law we get::


dL
 net ext
dt
Dynamics of rolling and slipping
We will assume that the axis along which the angular momentum points does not
change direction. A baseball curve ball is too hard for us to deal with.
Just as we could break the kinetic energy into two parts, so too can we break
down the angular momentum:
L  Lof CM  LaboutCM

Treat CM as point particle

v

Pure rotation around CM axis

v


There are two “natural” origins to choose for calculating torques and angular
momentum: the center of mass CM and the contact point CP.
Let us look at the “L” analysis in each case.


v
CP



L  mvR  I
CM

v

The second term will
always be I



L  0  I

Note the simplicity
Example:
N
fs

L  I   Rf s
Mg
dL
   I  Rf s
dt
C`M
a

mg sin   f s  ma
R
I
mg sin 
2
a
R
f s  mg sin  
I
I
m 2
m 2
R
R
T
Yo-yo
R
CM
a

R
a
mg
I
m 2
R
mg
mg  T  ma
L  I   RT
dL
   I  RT
dt
Conservation of Angular Momentum


dL
 net ext
dt
Even if there are external forces present, if
they exert no torque about a given axis, then L
will be conserved.
Note that if L is conserved about one axis, it
need not be conserved about any other axis.
Disc of mass M rotating with ωo . Bog of mass
m lands at center and walks out to rim. Find
the final ω
R
L0  L f   f 
1
2
L0   MR  0
2

1
2
2
L f   MR  mR  f
2

1
MR 2
2
1
MR 2  mR 2
2
0
Frictionless table viewed from above. Stick pivoted at one end.
Pivot exerts force but no torque about pivot. Linear momentum not conserved,
but angular momentum about pivot will be.
L, M
v
L f   m L  I p
3
L0  mvL
v
m
Was this elastic?
v

3

4
mvL
m v
3
L0  L f   
4
1
M
L
2
ML
3
L, M
L0  mv
L
2
v
m
Remove pivot
V
v

3
2
mvL
m v
3
L0  L f   
8
1
M L
ML2
12
v
Po  Pf  mv  m  MV
3
4 m
V 
v
3M
vL
L f  m
 I
32
  CM
L, M
Axis at bottom
V
L0  0
v
m
v

3
L
L f   MV  I
2
  CM
L
V
2
L0  L f   
6
1
L
ML2
12
v
Po  Pf  mv  m  MV
3
MV
4 m
m v
V 
v  8
3M
M L
as before