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Chapter 9 Rotational Dynamics
9-1 Torque
Torque and rotation
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Chapter 9 Rotational Dynamics
Torque as a vector
 
  rF

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Chapter 9 Rotational Dynamics
The torque about z axis:
 
 i  Ri  Fi



 Ri  ( Fit  Fin )

 iz  Ri Fit sin   Fit ri
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Chapter 9 Rotational Dynamics
The magnitude of τ is rFsinθ, and its direction is
determined by right-hand rule.
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Chapter 9 Rotational Dynamics
9-2 Rotational inertia and Newton’s second law
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Chapter 9 Rotational Dynamics
1. Rotational inertia of a single particle
F sin  ma T  mr
F
y
Fsinθ
θ
  rF sin  mr   I
2
m
r
o
τ
x
I= mr2 is the rotational inertia
of the particle.
The rotational inertia depends on the mass of the particle
and on the perpendicular distance between the particle and
the axis of rotation.
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Chapter 9 Rotational Dynamics
2. Newton’s second law for rotation
For rotational motion of a single particle, we obtain:
τ=Iα
Let us consider a more complicated rigid system consisting
of many particles.
 

Fi  fi  mi ai
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Chapter 9 Rotational Dynamics
The tangential component formula of the equation is:
FiT + fiT = mi aiT=miriα
FiTri + fiTri = miri2α
2
F
r

f
r

m
r
 iT i  iT i  i i 
F
r   mi ri  ;
2
iT i
 ext  I
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I   mi ri2
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Chapter 9 Rotational Dynamics
 ext  I
This is the rotational form of Newton’s law, called
rotational theorem of rigid body about fixed axis
Here the torque τext, the rotational inertia I and the angular
acceleration α are all about the z axis.
Note that : τext is the sum of the torques due to all the
external forces, is not the torque of the sum of all the
external forces.
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Chapter 9 Rotational Dynamics
3. The parallel-axis theorem
Rotation about an arbitrary axis y
  2
I y   mi r i   mi ( ri  h )
2
 
  mi ri  h  mi  2 mi ri  h
2
2
 I cm  Mh 2

rc 
I  ICM  mh
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2

 mi ri
m
0
i
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Chapter 9 Rotational Dynamics
Rotation about an arbitrary axis z
I z   mi r i
2
  mi ( x i  y i2 )
2
  mi x i2   mi y i2
 Iy  Ix
I z = Iy + Ix
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Chapter 9 Rotational Dynamics
9-3 Rotational inertia of solid bodies
If the body is one it as a continuous distribution of matter,
we can imagine it divided into a larger number of small
mass elements δm,
I  lim
 mn  0
r δ m
2
n
n
  r 2 dm
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Chapter 9 Rotational Dynamics
dm   dV
I   r  dV
2
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dm   dS
dm   dl
I   r  dS I   r  d l
2
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Chapter 9 Rotational Dynamics
Example: A thin rod with mass M, it is uniform density ,
and length L rotate about z axis through the center of
mass.
Ic  
L
2
L
2
L
x dm   L2 x 2 dx
2
2
M
 x
dx
L
M1 L 3
L 3

[( )  (  ) ]
L 3 2
2
1

ML2
12
L
2
L
2
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2
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Chapter 9 Rotational Dynamics
z
If the axis through the one
end of the rod in parallel
with axis through the center
of mass.
dm
x
dx
I   x dm   x  dx  
L
0
2
L
0
2
L
0
M
x
dx
L
2
M 1 3 1

L  ML2
L 3
3
We can see: I=Ic+M(L/2)2
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Chapter 9 Rotational Dynamics
Example: A merry-go-round with radius R, and mass M
rotate about z axis perpendicular to the plate.
R
I   r dm  
2
0
R
0
M
r
2rdr
2
R
2
2 M R4 1
 2
 MR2
R 4
2
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Chapter 9 Rotational Dynamics
Example: A spherical shell with the axis across the
center
2
r
 dm
I

 R sinθ 
π
0
2
M
2R sinRdθ
2
4R
2
 MR 2
3
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Chapter 9 Rotational Dynamics
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Chapter 9 Rotational Dynamics
9-4 Torque due to gravity
Each particle in the body, such as mass mn, experience a
gravitational force mng
1


r 
mr
M
cm

rcg 
i
i

 mi gri
m g
i
The center of gravity coincides
with the center of mass, if we
consider the gravitational field
is uniform.
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Chapter 9 Rotational Dynamics
The force on the entire body due to
gravity is:



F   mi g  Mg


 

   ri  mi g  (  mi ri )  g


 Mrcm  g

  
  r  Mg  r W

cm
cm
The toque acted on a body due to gravity about any
point is equal to the toque of total gravity acted at
the CM about this point.
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Chapter 9 Rotational Dynamics
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Chapter 9 Rotational Dynamics
How determine the position of center of gravity in a
extended object?
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Chapter 9 Rotational Dynamics
9-5 Equilibrium applications of Newton’s law
for rotation
For any body :

F  0,
ext
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
 0
ext
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Chapter 9 Rotational Dynamics
Example: Fine the acceleration of the blocks(m1>m2),the
tension in the rope and the angular acceleration of the
pulley.
T

T1
T2
m,R
T1
T2
mg
m1
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a1
m1 g
a2
m2g
m2
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Chapter 9 Rotational Dynamics
T
T1

T2
mg
T1
a1
m1 g  T1  m1a1
T2  m2 g  m2 a2
(T1  T2 ) R  I
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T2
m1g
a2
m2g
a1  a2  R
1
2
I  mR
2
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Chapter 9 Rotational Dynamics
m1  m2
m1  m2
g
a
g

m1  m2  m
m1  m2  m R
2
2
2m 2  m
2m1  m
2 mg T 
2 m g
T1 
1
2
2
m
m
m1  m2 
m1  m2 
2
2
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Chapter 9 Rotational Dynamics
Exercises
19
38
39
Problems
13
14
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Chapter 9 Rotational Dynamics
9-6 Nonequilibrium applications of Newton’s
law for rotation
F
x
 max
 ext ,z  I z
Analysis sample problem 9-10,detail please see
page192!
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Chapter 9 Rotational Dynamics
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Chapter 9 Rotational Dynamics
Example: A rod, mass m and length l, can rotate about its
axis perpendicular to the page surface through the point o.
The rod rotates from the rest when it is horizontal to the
angle θ. Find (1) α; (2) ω; (3)acm; (4) Nx,Ny.
Solution:
l
(1) mg cos  I
2
1 2
I  ml
3
3g
  cos
2l
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Chapter 9 Rotational Dynamics
d d 3 g
(2)  


cos 
dt
d
2l

 3g
0 d  0 2l cos d
3 g sin 

l
(3) acT
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l 3g
3g
2 l
 
cos , acN  

sin
2 4
2 2
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Chapter 9 Rotational Dynamics
(4) motion of CM
N x sin  N y cos  mg cos   macT
N x cos   N y sin  mg sin  macN
9
N x  mg sin cos
4
3 2
3
N y  mg (1  sin   cos 2  )
2
4
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Chapter 9 Rotational Dynamics
Discussion:
(1) If θ=0
3g
3g
  ;  0  0; acT  ; a cN  0;
2l
4
1
N x  0; N y  mg;
4
(2) If θ=π/2,
3g
3g
  0;  0 
; acT  0; acN 
;
l
2
5
N x  0; N y  mg
2
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Chapter 9 Rotational Dynamics
9-7 Combined rotational and translational motion
For arbitrary motion of an
object we can consider the
motion to be a
combination of two
motion.
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Chapter 9 Rotational Dynamics
the center of mass(CM) of the rigid body:
translational motion
any other point of the rigid body:
rotational motion about the CM
Now we restrict our discussion of this combined motion to
the cases satisfying two conditions: (1) the axis of rotation
passes through the center of mass, and (2) the axis always
has the same direction in space. We call this motion as plane
parallel motion.
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Chapter 9 Rotational Dynamics
The motion of CM is two-dimensional motion. We can
apply Newton’s second law:


F  macm
For the motion of other particles we can discuss them in
CM reference frame, in which they rotate about fixed axis it
through CM. We can apply rotate theorem:
τext=Iα
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Chapter 9 Rotational Dynamics
1. Kinematics
For a every point of the rigid body it moves in plane
motion, we have:
 
 
 
vi  vcm    ric  vc  vc
vc is the velocity of CM, and v'c is the velocity
of any point respect to CM
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Chapter 9 Rotational Dynamics
2. Rolling without slipping
There is not relative motion at the point of contact.
v p  vcm  R  0
vcm  R
This is the condition of rolling without slipping.
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Chapter 9 Rotational Dynamics
xC  R
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vC  R
aC  R
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Chapter 9 Rotational Dynamics
3. Instantaneous axis
 
 
v  vC    r
   
  R r
  
   (R  r )
— The axis is changing
— Pure rotation about P
— The same angular velocity about different axis
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Chapter 9 Rotational Dynamics
Instantaneous axis
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Chapter 9 Rotational Dynamics
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Chapter 9 Rotational Dynamics
4. The frictional force in rolling without slipping.
Generally the frictional force is necessary for rolling
without slipping.
Analysis sample problem 9-11,detail please see
page195!
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Chapter 9 Rotational Dynamics
Example: A solid cylinder of mass m and radius R rolls
without slipping down an inclined plane with angle θ.
Find the acm and f.
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Chapter 9 Rotational Dynamics
Solution: translational motion of CM
mg sin   f  macm
N-mgcos  0
the rotational motion of the cylinder about the axis it
through CM
1
2
fR  I  mR 
2
and
acm  R
We can obtain:
acm
2
 g sin ;
3
2g

sin ;
3R
1
f  mg sin
3
The result means that the rolling without slipping needs f.
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Chapter 9 Rotational Dynamics
The maximum friction is μN=μmgcosθ,
1
1
f  mg sin   N  mg cos ,    tan  .
3
3
If μ<(1/3)tanθ, the cylinder may be slipping.If μ=0, the
cylinder is complete slipping (no rotate).
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Chapter 9 Rotational Dynamics
5. The direction of frictional force in rolling without
slipping
Example:The cylinder with mass m and radius R is
twined round by rope. A force F act on it. The cylinder
rolls without slipping from the rest. Find the direction of
the frictional force.
Solution: if there is no frictional
F
force, we obtain:
F  macm
F
acm 
m
the direction of acm :
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P
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Chapter 9 Rotational Dynamics
1
FR  mR 2
2
2F
R 
m
the direction of  : 


 
 
ap  acm    R; the direction of   R :
 
  R  R  acm

The direction of ap :
The direction of f :
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Chapter 9 Rotational Dynamics
Example: If the force F act on the point R/4 from the
center, we can see:
F
P
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R 1
F
2
F  mR  ; R 
;
4 2
2m
 
   R  R  acm ;  a p :, f :
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Chapter 9 Rotational Dynamics
Analysis sample problem 9-12,detail please see
page195-196!
Discussion: In the time interval, the velocity of CM is
increased and the angular velocity of the cylinder about
CM is decreased. When the velocity vcm and angular
velocity ωf are satisfactory to the formula vcm = ωf R, the
rotation without slipping is beginning.
Analysis sample problem 9-13,detail please see
page196!
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Chapter 9 Rotational Dynamics
Exercises
42
Problems
16
21
22
25
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