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Dynamics of Rotational Motion
The main problem of dynamics: How
 a netforce affects
(i) translational (linear) motion  F  ma (Newtons’ 2nd law)
(ii) rotational motion ???

m
(iii) combination of translational
F
and rotational motions ???
αz

a
Lever arm l is the distance between the
line of action and the axis of rotation,
measured on a line that is
to both.

Axis of
rotation
  
  rF
Definition of torque:
τz > 0 if the force acts counterclockwise
τz < 0 if the force acts clockwise
Units: [ τ ] = newton·meter = N·m
A torque applied to a door
Newton’s Second Law for Rotation about a Fixed Axis
Only Ftan contributes to the torque τz .
(i) One particle moving on a circle: Ftan=matan and atan= rαz
rFtan= mr2 αz
τz
τ z = I αz
I
(ii) Rigid body (composed of many particles m1, m2, …)

2
i  iz   i m i ri   z   z  I z
Example 10.3: a ,T ,T ?
1x
1
Pulley: T2R-T1R=Iαz
T2-T1=(I/R2)a1x
Glider: T1=m1a1x
Object: m2g-T2=m2a1x
Only
external
torques
(forces)
count !
a1 x 
m2 g
m1  m 2  I / R 2
T1 
m1m 2 g
m1  m 2  I / R 2
T2 
(m1  M )m 2 g
m1  m 2  I / R 2
2
a1x
a1x=a2y=Rαz
a2y
y
Work-Energy Theorem and Power in Rotational Motion
Rotational work:
dW R  Ftan ds  Ftan Rd   z d 
2
W R    z d
1
For  z  const  WR   z ( 2  1 )   z 
Work-Energy Theorem for Rigid-Body Rotation:
2
1 2 1 2
W R   I  z d  z  I  2  I  1  K R 2  K R1
2
2
1
d z d
 I z d z
Proof:  z d  I
dt
Power for rotational work or energy change:
dW R
d
PR 
z

dt
dt
PR   z z
( analo g of
 
P  F v)
Rigid-Body Rotation about a Moving Axis
General Theorem: Motion of a rigid body is always
a combination of translation of the center of
mass and rotation about the center of mass.
Energy:
Proof:
1
1
2
2
K  Mv cm  I cm 
2
2
2  2  
1 2
1
K   m i v i   m i (v cm  v 'i 2v cm  v 'i ) 
i 2
i 2
2 2
1
2 1
   m i v cm   m i r 'i 
2 i
2 i



since  mi v i ' 

 drcm '
d
m
r
0

i i '
dt i
dt
Rolling without
slipping: vcm= Rω,
ax = R αz

General Work-Energy Theorem:
E – E0 = Wnc ,
E=K+U
Rolling Motion
Sliding and deformation of a tire also cause rolling friction.
Rolling Friction
Combined Translation and Rotation: Dynamics


 Fi  Macm
i
and

iz
 I cm z
i
Note: The last equation is valid only if the axis
through the center of mass is an axis of
symmetry and does not change direction.
Exam Example 24: Yo-Yo has Icm=MR2/2 and
rolls down with ay=Rαz (examples 10.4, 10.6; problems 10.20, 10.67)
Find: (a) ay, (b) vcm, (c) T
Mg-T=May
τz=TR=Icmαz
y
ay
ay=2g/3 , T=Mg/3
4 gy
v cm  2ay 
3
Exam Example 25: Race of Rolling Bodies (examples 10.5, 10.7; problem 10.22,
y
Data: Icm=cMR2, h, β problem 10.27)
Find: v, a, t, and min μs
FN
preventing from slipping
β
 
v a
Solution 1: Conservation of Energy Solution 2: Dynamics
K1  U1  K 2  U 2 , K1  0, U 2  0  (Newton’s 2nd law) and
rolling kinematics a=Rαz
fs
1
1
1
Mv 2  I 2  (1  c ) Mv 2
2
2
2
I  cMR 2 and   v / R
Mgh 
for
x
2 gh
v
1 c
v2=2ax
g sin 
 F x  Mg sin   f s  Ma  a  1  c
g sin    z  f s R  I cm z  cMRa  f s  cMa
a
1  c v  2 ax  2 g sin  h  2 gh
1  c sin 
1 c
x 2x
1
t 

v
v
sin 
2 h (1  c )
g
fs
c
c Mg sin  c tan 
f s  Mg sin   Ma 
Mg sin   min  s 


1 c
F N 1  c Mg cos 
1 c
Angular
   Momentum


(i) One particle: L  r  p  r  mv  L  mvr sin 




 dL   
dL  dr
dv  
 
   mv    r  m   r  ma 
 r  F 
dt  dt
dt 
dt
 
Impulse-Momentum Theorem for Rotation




dL
 
(ii) Any System of Particles: L   Li 
dt
(nonrigid or rigid bodies)
i
It is Newton’s 2nd law
for arbitrary rotation.
 
m (v  v )  0

Note: Only external torques count
since

i nt er na l
 0.
Unbalanced wheel: torque of friction in bearings.
(iii) Rigid body rotating
around a symmetry axis:
 
Lz   Li   mi ri  z  I z  L  I
2
dLz
d z
I  const and
I
 I z   z  I z
dt
dt
Principle of Conservation of Angular Momentum
Total angular momentum of a system is constant (conserved),
if the
acting on the system is zero:
 net external torque



dL
dL
  
 0  L  const
dt
dt
if

  0
For a body rotating around a symmetry axis:
I1ω1z = I2ω2z
Example: Angular acceleration due to sudden
decrease of the moment of inertia
I0
 f   0   0 si nce I 0  I f
If
Origin of Principles of Conservation
ω0 < ωf
There are only three general principles of conservation
(of energy, momentum, and angular momentum) and
they are consequences of the symmetry of space-time
(homogeneity of time and space and isotropy of space).
Hinged board (faster than free fall)
2
mgh

mv
/ 2  tball  2h / v  (2L / g ) sin  0
m Ball:
h=L sinα

Mg
In[28]:=
f s_ : NIntegrate 1 Sqrt 1 x x
tcup
Plot f s , 1 , s, 0, 1
s x
Board: I=(1/3)ML2
h0  h I 2 ML2 (d / dt ) 2
Mg


2
2
3
2
d
3g


(sin  0  sin  )
dt
L
6 s , x, 0, s ;
tcup 
1.2
tball
tcup
1.1
tball
Out[29]=
1.0
0.2
2/3
0.4
0.6
0.8
1.0
Critical  0  500
sin  0

L
3g
0

0
d
sin  0  sin 
1
6 sin  0
sin 0

0
dx
(1  x 2 )(sin  0  x)
Gyroscopes and Precession
Precession is a circular motion of the axis
due to spin motion of the flywheel about axis
Period of earth’s precession is 26,000 years.
 
Dynamics of precession: d L   dt

L0  0 (  0)

 n
Fc ac

L0  0 (  )
Precession angular speed:

dL
 z mgr
d

  

dt
I
L dt Lz
Circular motion of the
center of mass requires
a centripetal force
Fc = M Ω2 r
supplied by the pivot.
Nutation is an up-and-down

w
wobble of flywheel axis
that’s superimposed on the
precession motion if Ω ≥ ω.
Analogy between Rotational and Translational Motions
Physical Concept
Rotational
θ
ω
α
Torque τ
Moment of
inertia I = Σmr2
Στ = I α
Newton’s second law
τθ
Work
Kinetic Energy
(1/2) Iω2
Momentum
L=Iω
Displacement
Velocity
Acceleration
Cause of acceleration
Inertia
Translational
s
v
a
Force F
Mass m
ΣF = ma
Fs
(1/2) mv2
p = mv