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Chapter 9-10 rotation of rigid bodies
Rotational motion
dynamics
kinematics
Instantaneous
effect
1. model
2.Angular
quantity
theorem
of
rotation
Space effect
1.work
2.kinetic and
potential
energy
3.conservation
of energy
Time effect
1.Theorem of
angular
momentum
2.conservatio
n of angular
momentum
Chapter 9-10 rotation of rigid bodies
Key terms:
rigid body
rotation
translation
radianrad
angular velocity
angular acceleration
angular speed
moment of inertia
rotational kinetic energy
parallel-axis theorem
Chapter 9-10 rotation of rigid bodies
Key terms:
torque
line of action
lever arm
rotational analog of Newton’s second law
combined translation and rotation
angular momentum
conservation of angular momentum
Types of Motion
Translation
The linear position of a body changes with
respect to a fixed frame, but its angular orientation
remains unchanged.
Rotation
The angular orientation of a body changes about
a fixed frame of reference, but its linear position
remains fixed.
Complex
Simultaneous combination of translation and
rotation motions
1. Ideal Model-----Rigid Body
we neglect the deformations and assume that the
body has a perfectly definite and unchanging shape
and size.
Definite shape and definite size.
The distance between any two points on a rigid body
remain unchanging.
A rigid body can be considered as a special
collection of particles with mass continuously
distributed.
A rigid body in general can have rotational as well as
translational motion.
2. Angular Quantities
let’s think first about a rigid body that rotates
about a fixed axis. (The axis passes through point O
and is perpendicular to the x-y plane)
y
P
r
o
x
We can use the linear quantities
( r, v, a) as well as the angular
quantities (  ,  ,  ) to describe
the motion of the rigid body.

2.1 Angular Coordinate
s
rad    2   1
r
2.2 Angular Velocity
d
  

rad / s
v



r
dt
Counterclockwise(+)
2.3 Angular Acceleration
y
d d 2

 2 rad / s 2
dt dt
P
s
r
If  is a constant (=c):

   0  t
o
x

   0 t  1 / 2t 2
2
2
2     0
2.4 Relating Linear and Angular Kinematics
We can use linear quantities ( r, v, a) or angular
quantities (  ,  ,  ) to describe the rotational motion.
The relationships between them are:
  r
s  r

an 

2
r
 r
2
d
at 
 r
dt
y
v
r

P
s
o
x

The linear quantities of
any point on the rigid body
are different, while angular
quantities are the same.
3. Theorem of Rotation
3.1 torque (/moment) of a force
  
  r F
force times lever-arm, tends to produce rotation.

  rF sin   Fd
z

Vector: right-hand rule.
o
x
d

r
 y
F
mi

    
discussion : F  r , F // r , F // axis
Net torque is the vector sum of individual torques:
   
 
 i  r1  F1  r1  F2     ri  Fi

i
i
Example: a force on particle
 

F  ( xi  yj )
find the torque on particle if the fixed axis
is in original point.
Solution:
  
 
(1)   r  F  r  (  r )  0

 
  
 
( 2)   r  F  ( xi  yj )  (  xi  yj )  0

3.2 Rotation Analog of Newton’s Second Law
  I
Proof: A rigid body can be looked as a special collection
of particles.
  
 i  ri  Fi  ri 

The net torque on mi to point O:
z
 o
f in

ri

Fex

j( i j )
The net torque on the rigid body to
point O:
 

 i   ri  Fi   ri 

i
mi

f ji
i

 ri 
i
i


f ji  0
j( i j )


f ji
j( i j )
 

 i   ri  Fi   ri 

i
i
i


f ji
j( i j )
 

 i   ri  Fi   ri Fi ,tan
i
i
i
  ri ( mi ai ,tan )   ri mi ( ri )
i
z
o

 ri 
i

j(i j )

f ji  0
Fi ,tan  mi ai ,tan
ai ,tan  ri
i
  (  ri2 mi ) “”s are the same for every part

ri
i
of the rigid body.

Fex
I   ri2 mi
i
mi
  I
Rotational analog of Newton’s
law for a rigid body.
  I
Caution:
1)

2) 
3)
Means external torque, the torque offered
by a pair of internal force equals to zero
The angular acceleration of rigid body
with fixed axis
  0   0
The rigid body keep rest or rotate in uniform
angular velocity
4) I
moment of inertial, measurement of rotational
inertial
1
Given  , I 

3.3 Calculation of Moment of Inertial
I   mi ri2
i
1) A few point masses: I 
2
m
r
 ii
i
2) Continuous distribution of mass:
I   r dm
2
Moment of inertial depends on the distribution of
mass in the system.
Moment of inertia depends on the location and
orientation of the axis.
Example: An engineer is designing a one-piece machine
part consisting of three heavy connectors linked by
light molded struts. The connectors can be considered
as massive particles connected by massless rods.
a) What is the moment of inertia of this body about an
axis through point A, perpendicular to the plane of the
diagram?
b) What is the moment of inertia about an axis coinciding
with rod BC?
(see page 277)
B
mB=0.10kg
Solution:
0.50m
I   mi ri2
i
0.30m
A
mA=0.30kg
mC=0.20kg
0.40m
C
Solution:
a) I a 
2
2
2
m
r

(
0
.
10
kg
)(
0
.
50
m
)

(
0
.
20
kg
)(
0
.
40
m
)
 ii
i
 0.057 kg .m 2
a)
b) I b 
2
2
2

(
0
.
30
kg
)(
0
.
40
m
)
m
r

0
.
048
kg
.
m
 ii
i
B
mB=0.10kg
0.50m
0.30m
A
mA=0.30kg
mC=0.20kg
0.40m
C
Example: A slender uniform rod with mass M and
length L. It might be a baton held by a twirler in a
marching band. Compute its moment of inertia about
an axis through center of mass point O.
Solution: I 
2
r
 dm
I   r dm
2
  x dm
2
M

L
L
2
L

2

2
x dx
1
2

ML
12
C
o
dm
x dx
x
Uniform thin loop (M ,R):
I m c   R 2 dm  MR 2
R
Uniform thin disc(M, R)

R
M
Ic  r
2 2rdr
0 R
2
1
 MR 2
2
r dr
2) Parallel-Axis Theorem: (p281)
I p  I cm  Md
2
p
Icm: about axis through CM
M: total mass
d : distance between CM and
point p.
Icm
Ip
d
O
CM
I p   r 2dm   [( x  a )2  ( y  b)2 ]dm
I p   ( x 2  y 2 )dm  2a  xdm  2b ydm   (a 2  b2 )dm
xc
xdm


0
M
yc
ydm


o
I p  I cm  Md
M
2
Example: If the rod rotates about the axis through
one end, Use parallel-axis theorem:
I  I cm  Md 2
1
1
2

ML  ML2
12
4
1
 ML2
3
o
3.3 Application of rotational analog of Newton’s second law
  I
Caution:
1) The left is vector sum of external torque about a fixed
axis, and the direction is parallel to axis.
2) “ I ” is the Moment of inertial
3) “” means the angular acceleration to fixed axis,
same direction with torque
Example: In judo, a weaker and smaller fighter who
understands physics can defeat a stronger and larger
fighter who doesn’t. The fact is demonstrated by the basic
‘ hip throw ’, in which a fighter rotates his opponent
around his hip.
Solution:
Fd1  I
F d1  mgd2  I
FF
Example: The massive shield door at
a neutron test facility at Lawrence
Livermore Laboratory is the world’s
heaviest hinged door M=44000kg,
I=8.7*104 kgm2, and face width 2.4m
find the smallest force to move the
door from rest through an angle of
900 in 30s.
Example: We wrap a light, flexible cable around a solid
cylinder with mass M and radius R. The cylinder rotates
with negligible friction about a stationary horizontal
axis. We tie the free end of the cable to an object of
mass m and release the object with no initial velocity.
Find the acceleration of the object of mass m and the
angular acceleration of the cylinder
N
R
For m:
O
M
T
m
Mg
T
y
h
Solution:
mg
mg  T  ma
(1)
For M: TR  ( 1 MR 2 )
2
a  R
(3)
(2)
Example: An Atwood’s machine. There is no slipping
between the cord and the surface of the wheel. Find
the linear accelerations of blocks A and B, the
angular acceleration of the wheel C, and the tension
in each side of the cord
a) If weight of cylinder is negligible ;
b) If weight of cylinder is not negligible.
T
Solution:
a) The weight of cylinder is negligible …
T’2
R
C
O
T’1
b)If weight of cylinder is not negligible:
For m1: m1 g  T1  m1a1 y
For m2: m2 g  T2  m2 a 2 y
(1)
(2)
T1
T2
mB
m2g
mA
m1g
y
For m1:
For m2:
m1 g  T1  m1a1 y
m2 g  T2  m2 a2 y
1
For M: T R  T R  ( MR 2 )
2
'
2
'
1
(1)
(2)
(3)
a1 y  a2 y ( 4 ) a1 y  R ( 5 )
T
O
From (1),(2),(3),(4),(5) :
m1  m 2
a1 y  a 2 y 
g
m1  m 2  M 2
m1 ( 2 m 2  M 2 )
m2 ( 2m 1  M 2 )
T1 
g , T2 
g
m1  m 2  M 2
m1  m 2  M 2
T’2
R
C
T’1
T1
T2
m2
m2g
m1
m1g
y

Solution:
 mg
L
cos   I
2
L
d
d
 mg cos  I
I

2
dt
d


L
  mg 2 cosd  0 Id
2
L
1
mg (1  sin )  I 2
2
2
Example: A grindstone 1.0 m in diameter, of mass 50 kg,
is rotating at 900 rev.min-1. A tool is pressed against the
rim with a normal force of 200N, and the grindstone
comes to rest in 10s. Find the coefficient of friction
between the tool and the grindstone. Neglect friction in
the bearings.
Solution:
  I ,   0  t
  f k R , f k  N k
 0  900  2 / 60  30 ( rad / s )
 t  0 , t  10 s
  (  t   0 ) / t  3 ( rad / s 2 )
1
I  MR 2
2
 k  o .59
Example: A wheel rotates about a fixed axis from rest. A
constant torque,=20N.m, acts on it for 10 seconds.
=100rev/min at t=10s. Then it takes 100 seconds for
the wheel to come to stop again. Find the moment of
inertial of the wheel with respect to the axis.
Solution:  =I, = o+ t
acted by external torque:
20=I r=I
20= 1/t
(for:
o=0)
o=0)
1， 
1，
1=1/t
1 (for:
When release external torque:
-  r=I2 , 2=- /t2
(1)
(2)
t1=10s , t2=100s , =(100×2)/60=10.5rad/s,
From (1)、(2):
I=17.3kg.m2 。
4. Energy in Rotational Motion
4.1 Rotational Kinetic Energy
The kinetic energy of mi :
1
1
2
K i  m i v i  m i ri2 2
2
2
The rigid body :

vi


O
r
P
mi
 Ki  
i
i
1
1
mi v i2  (  mi ri2 ) 2
2
2 i
I   mi ri2
i
vi  ri i
1
K  I 2
2
Example: Each of the helicopter rotor
blades is 5.2m long, m=240kg. The rotor
is rotating at 350rev/min.
Find 1) the rotational inertial of the rotor
assemble about the axis of rotation.
2)what is the total kinetic energy of
rotation.
4.2 Gravitational Potential Energy
The gravitational potential energy is the same as
though all the mass were concentrated at the center
of mass of the rigid body.
U   mi yi g  Mgy cm
i
m y
i
i
i
 My cm
4.3 Work and Power in Rotational Motion
Work of the torque
Suppose a force F acts on the rigid body. The rigid
body rotates through an infinitesimal angle d about a
fixed axis. The work dW done by the force while point P
moves distance ds is:
 
dW  F  ds  Fr sind  d
2
 d
r
P

ds F

W   d
1
work done by a torque
2
W   d
1
d
d
W   d   I
1
1
dt
2
1
1
2
  Id  I 2  I 12
1
2
2
2
2
  I
d
 
CAUTION:
 a work done by the forces
doesn’t equal to the work
done by the torques.
  1
1
For a particle: W   F  dr  m 2 2  m1 2
2
2
dt
 Compare with the work-energy theorem for
collection of particles:
Wtot
1
1
2
2
 I 2  I 1
2
2
Wtot
1
1
2
2
  mv 2   mv 1
2
2
Power of a torque:
dW
d
P

 
dt
dt
 
p  F v
4.4 Conservation of Mechanical Energy
If the work done by all forces other than
conservative force equals zero. The total mechanical
energy of the rigid body is conservative.
Wext  Winnon  0
E C
Example: We wrap a light, flexible cable around a solid
cylinder with mass M and radius R. The cylinder
rotates with negligible friction about a stationary
horizontal axis. We tie the free end of the cable to an
object of mass m and release the object with no initial
velocity. Find the speed of the falling object and the
cylinder just as the object strikes the floor.
See page 280, example 9-9.
N
R
No work is done by friction, so
O
M
T
m
Mg
T
y
h
Solution:
mg
1
v 2
2 1 1
2
mgh  mv  ( MR )( )
R
2 2
2

Solution:
L
L
1 2
mg  mg sin  I
2
2
2
1
I  ml 2
3
We get

( 2)
(1)
Example: A uniform spherical shell of mass M and
radius R rotates about a vertical axis on frictionless
bearings. A massless cord passes around the equator
of the shell, over a pulley .
Find the speed of the object when it falls a distance h
from rest.
E  mgh
(1)
(2)
v
s 
R
v
p 
r
From the conservation law of energy
Example: the given is as follows,find the velocity of the
object and the acceleration of M。
Solution:
1)
1
1
1
Mgh  Mv 2  I 2  kh2
2
2
2
1
I  mr 2，v  r
2
2 Mgh  kh2
v
1
M m
2
k
M
h
Example: the given is as follows,find the velocity of the
object and the acceleration of M。
Solution:
2)
1
1 2 1 2
2
Mgh  Mv  I  kh
2
2
2
k
Make a derivation
mgv  mva  I   khv
a
mg  ma  I 2  kh
R
mg  kh
We get:
a
I
m 2
R
M
h
Alternative solution:
mg  T1  ma (1)
(T1  T2 ) R  I (2)
T2  kh
( 3)
a  R
(4)
We get:
k
M
h
mg  kh
a
I
m 2
R
Example: The power output of an automobile engine is
advertised to be 200 hp at 6000 rpm. What is the
corresponding torque?
Solution:
1 HP = 746 W
746W
P  200hp  200hp(
)  1.49  105W
1hp
  6000rev / min  628rad / s
1.49  10 N  m / s
 
237 N  m

628rad / s
P
5
5. Angular Momentum
Coordinate(x)
---
Angular Coordinate()
Distance(s)
---
Angular Displacement()
Velocity(v)
---
Angular Velocity()
Acceleration(a) ---
Angular Acceleration()
Momentum(P) ---
?

M, L



P   mi v i  0
i
'
v ?

v
m
5.1 Angular Momentum of a particle
   

L  r  p  r  mv Unit: kg m2/s
 z
L

r
o
x
d
m
y 
 mv
Magnitude:
L  rmv sin  mvd
Direction: determined by right-hand
rule for vector products.
z

o
x
d

r
m
Compare with torque
y
F
 
  r F

Caution: The Value of L and 
both depends on the choice of
origin O.
 z
L

r
o
x
   

L  r  p  r  mv
d
m
y 
 mv

mv
Angular momentum of collection of
particles:


 


L   Li   ri  pi   ri  mv i
i
i
O
R
O
i
P
d
mv
To point O: L=mvd
L=mvR
To point P: L=0
5.2 Angular Momentum of a rigid body
The angular momentum of mi to point O:
v i  ri
Li  ri mi vi  mi ri2
The angular momentum of the rigid
body to the fixed axis (z-axis):
z

Lz   Li  (  mi ri2 )
i

r
O i
x
mi

vi
y
I   mi ri2
i
i


Lz  I z
CAUTION: Giving angular
momentum you must always
specify about axis
5.3 The Theorem of Angular Momentum
   

for one single particle: L  r  p  r  mv






 dp
dL d ( r  p )  dp dr 

F
,
r   p
dt
dt
dt
dt dt


dr 
 v,
dL   
dt
 r F 


dt
v  mv  0
The rate of change of angular momentum of a
particle equals the torque of the net Force acting on It.
t2
t1


L2

 dt   dL  L2  L1
L1
for collection of particles:


i
i


d (  Li )
i
 

 i   ri  Fi   ri 

i
i

 ri 
dt

 dL
 
dt
i


f ji  o
i
j( i j )
 
 i   ri  Fi

i
i
for rigid body:
dLz
d ( I z )
 I z
z 

dt
dt

t2
t1
L2
dt   d ( I )  I 2 2  I11
L1
Lz  I z


f ji
j( i j )

L2

L1
  

t1 dt   d L  L2  L1
 
t2 
P2 
 Fdt   dp  P2  P1
t2
t1

 dL
 
dt
P1
If the torque of all the external forces equals 0:


   i  0 ,
 
L2  L1
i
------conservation of the angular momentum for
collection of particles or one single particle.
Compare with conservation of momentum:


F   Fi  0 ,
i


   i  0 ,
i
 
P2  P1
 
L2  L1
Example: the disk rotate on the table, angular velocity 
and friction coefficient is  , how much time it will take
for the disk to stop its motion
Solution :
Theorem of angular momentum

dt

I


I

2
1

d  dmgr  2r gdr
2
R
2
   d   RMg
3
0
  t   I
I 3 R 
t  


4 g
r dr
5.4 Conservation of Angular Momentum
When the net external torque acting on a system is
zero, the total angular momentum of the system is
 
constant.

 ex  0 , L2  L1 , I11  I 2 2
Example: The position of a object with mass m is given
by r=acost i+bsint j (a, b, and  are all constant),
find the angular moment of the object and the net
torque acting on it.
 
Solution: L  r  mv

r  a cos tî  b sin tĵ
 dr
v
 a sin tî  b cos tĵ
dt
î
ĵ
k̂
L
a cos t b sint 0
 ma sint mb sint 0

 mabk
  

  r  F  r  ma
2
 d r
2
a
  r
2
dt



2
  r  (  r )  0
Example: A small block on a frictionless horizontal
surface has a mass of m. it is attached to a massless
cord passing through a hole in the surface. The block
is originally revolving at a distance of r from the hole
with an angular speed 0. Pull the cord from below
slowly, shorten the radius of the circle to r/2. How
much work was done in pulling the cord?
See page 322 10-33, 10-76, 10-87

Solution:  net to O is zero
conservation of L : L1  L2
mr 2 0  m( r 2 )2 
0
m
r
o
  40
1
r 2 2 1
3
2
2
W  m( )   mr  0  mr 2 0 2
2
2
2
2
F
Example: see page 314 example 10-16
Solution: Consider the door and bullet as a system, net
torque about the axis through O equals zero,
so angular momentum is conserved.
l
initial L : L1  mv
2
final L : L2  ( I bul  I rod )
1
l 2
I bul  m( ) , I rod  Ml 2 ,
3
2
0.50
conservation of momentum
m 1.00
L1  L2
m

v  400 m / s
so :
  0.40 rad / s
Example: see page 327 10-75
Solution:
Consider A and B as a system, when remove the
accelerating torque from A, angular momentum is
conserved.
I A 0  ( I A  I B )
The thermal energy equals the lose of the total
rotational kinetic energy:
1
Q  I A 02  1 ( I A  I B ) 2
2
2
S
S
A
C
B
1
Q  2400 J , I A  I B
3
so :
1
K A0  I A 02  3200 J
2
Example：a block with spring was in a frictionless
surface, k=100N/m, a spring was fixed in o point，m=1kg,
the natural length of the spring l0=0.2m, the velocity of
the block is v0=5m/s, perpendicular to the spring。When
the spring rotate 900, the length of l=0.5m find the
velocity of the block.
Solution:
mv0l0  mvl sin
1
1 2 1
2
mv 0  mv  k ( l  l0 )2
2
2
2
We get：v =4m/s， =300
l
O
l0
d
v0
v

Example: a rocket’s mass is m, velocity is v0, oc=3R。
Find the angle between v and v0。(the earth mass is M、
radius=R)
Solution:
1
Mm 1 2
Mm
2
mv 0  G
 mv  G
Conservation of the energy
2
R
2
3R
mv 0 R  mv  3 R sin
Conservation of the angular
momentum
We get：
2
A
m
O
M
R
Rv 0
sin  
2
3( 3 Rv 0  4GM )
v0
3R
O’
C

v
Example: A man of mass m runs around the edge of a
horizontal turntable that is mounted on a frictionless
vertical axis through its center. The velocity of the man,
relative to the disk, is v. The turntable is rotating in the
opposite direction with an angular velocity of , relative
to the earth. The radius of the turntable is R, and its
mass is M. Find the final angular velocity of the system if
the man comes to rest, relative to the turntable.
Solution: Consider the man and turntable as a
system, angular momentum is conserved.
'
Rm  I disc  ( I disc  I man )
The equation is wrong, because
conservation of L is only valid in
inertial reference frame

R




 ME   MD  D E
 ME
v
 
R
'

(
I

I
)


I

I man  ME
disc
man
disc
I man
1
 mR , I disc  MR 2 ,
2
2
mv / R  m  M / 2
 
mM / 2

'
R

Summary:
Instantaneous Effect
 Fx  ma x

 
 F  ma  Fy  ma y
(One particle) 
  Fz  ma z
  I
(Rigid body)
I   m i ri
i
I   r 2 dm

v2

 Fn  ma n , ( a n  R )

  Ft  ma t , ( a t  dv )

dt

2
I p  I cm  Md
2
Accumulation Effect of Space
  1
1
2
2
(One particle) W 
F

d
r
 mv 2  mv 1
tot
a 
21
2
1
2
(particles) Wex  Win K 2  K 1   mi v i 2   mi v i21
i 2
i 2
Wcon  ( U 2  U 1 )
b
Wex  Win ,non  ( K 2  U 2 )  ( K 1  U 1 )  E2  E1
if Wex  Win ,non  0 , E2  E1
(Rigid body)
1 2 1 2

d

  I 2  I 1
a
Wtot  
b
2
1 2
E  mgy cm  I
2
2
时间积累
 


t2


(One particle) J    F dt  P2  P1  mv2  mv1
t1


 
t2


(particles) J    Fdt  P2  P1   mi v i 2   mi v i 1
t1
i
i



if  F  0 , P2  P1


dLz
 dL
 dL
z 
 
 


dt
dt
dt
(One particle)
(particles)
(Rigid body)


t2
 

矩冲量= t   dt  L2  L1   Li 2   Li 1
1
i
i



if   0 , L2  L1 or I 1 1 I 2 2
6* Rigid-body Rotation about a Moving Axis
6.1 Combined translation and rotation:
Energy Relations
For mi:
 
'
v i  vcm  v i
1

'

'
K i  mi ( vcm  v i )  ( vcm  v i )
2
1
 ' ' 2
2
 mi ( vcm  2vcm  v i  v i )
2

cm

v cm
ri
mi
'
vi

vi

v cm
1


2
For mi: K i  m i ( v cm  2v cm  v i'  v'i 2 )
2
2
For body: K   K i   ( 1 m i vcm
)
2
1
 '
'2
m
v

v

(
m
v
 i cm i  2 i i )
 ' 
'
 mi vcm  vi  vcm  (  mi vi )  0
1
1
2
K   ( m i v cm )   ( m i v'i 2 )
2
2
1
1
2
 Mv cm
 I cm 2
2
2
2. Rolling without slipping
vcm  R
1
1
2
K  Mv 1  I 12
2
2
1
 0  ( I cm  MR 2 )2
2
1
1
2
2
 I cm   Mv cm
2
2


F

M
a
Center of mass:  i
cm
Rotation about
axis through CM:

   Icm
3
'
 v3
 v
v 2 cm
'
v2

2 
v cm
R

v cm
'
 1 
v1  vcm v cm

v1  0

v3

4 v cm
' v
v4 4
Example13: See page 303 example 10-6.
Use two ways to find the speed vcm of solid cylinder
after it has dropped a distance h.
1
1
2
Mv cm  I cm 2
2
2
Solution2: Mg  T  Ma
(1)
cm
Solution1: Mgh 
TR  I cm
(2)
vcm  R
I cm 
From (1),(2): a  2 g ,T  1 mg
cm
3
3
v
2
cm
 2acm h  vcm 
4
gh
3
1
MR 2
2
h
T
Mg
7* Gyroscopes and precession
p

dL
L  I

r
O
 mg


 r  mg
dL   mg dt
d



mg
dL  L sind
 mg
d
p 

dt I sin 