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Chapter 4 Rotation of rigid body
§4.1 The rotation of a rigid body about a fixed
axis
§4.2 Torque , the law of rotation , moment of
inertia
§4.3 Angular momentum the law of angular
momentum conservation
§4.4 work done by a torque, the theorem of
kinetic energy of a rigid body rotating about a
fixed axis
Summary
§4.1 The rotation of a rigid body about
a fixed axis
1. the angular velocity and angular acceleration of a
rotating rigid body
2. Formulation of fixed axis rotation with constant
angular acceleration
3. The relationship between angular quantities and
linear quantities
Rigid body：under external forces if the shape
and size of an object do not change ．(the distance
between any two arbitrary points in the object is a
constant)
explain：⑴ Rigid body is an idealized object
⑵ The model of the rigid body is introduced
for trying to simplify the study of the motion.
The motion of a rigid body：the translation
rotation．
1
the angular velocity and angular acceleration of a
rotating rigid body
1. Angular coordinate
Rigid body
Equation motion of
rotating rigid body
about a fixed axis
Rotation clockwise
rotation counter
clockwise
An arbitray
point in
(t+△t)
(t) Reference
q>0
q<0
direction
Reference plane which is perpendicular to the Oz
2. Angular displacement
axis）
Axis
q  q (t  t )  q (t )
3. Angular velocity

q dq direction of  : determined by right
  lim

t 0
hand rule
t
dt
1
the angular velocity and angular acceleration of a
rotating rigid body
The rotation direction can be expressed by the positive or negative
of the angular velocity, when the rigid body rotating about a fixed
z
z

axis



 <0
 >0
(i) When rotating counter clockwise,  > 0;
(ii) When rotating clockwise,  < 0;
4. angular acceleration
When  > 0





d

dt

(i) When >0, the rigid body rotates with acceleration;.
(ii) When <0, the rigid body rotates with deceleration.
The Features of Rotation about a Fixed Axis:
(1) The location and direction of rotation axis are
fixed relative to a inertial reference frame.
( 2 ) Every point of body moves in a circle whose
center lies on rotation axis and radius different.
(3) Every point moves through the same angle


during a particular time interval；q ,  , 
(4) Each rotation planeis perpendicular to
rotation axis.．
2
Formula of fixed axis rotation with constant
angular acceleration
A rotation with variable angular velocity is
called fixed axis rotation with constant angular
acceleration ．
Straight line motion with constant
linear acceleration
Fixed axis rotation constant angular
acceleration
   0  t
v  v 0  at
2
2
1
1
x  x0  v0t  2 at q  q 0  0t  2 t
2
2
2
2
v  v 0  2a( x  x0 )    0  2 (q  q 0 )
3 The relationship between angular quantities
and linear quantities
dq
ω
dt
2
dω d q

 2
dt
dt


v  rωet
a t  r
an  rω
2




an a
r P

at


2
a  ret  rω en

et

v
Example 1 In a mini-motor rotates in a high
speed, in which a cylindrical rotor rotates about an
axis that is perpendicular to the cross-section and
passed through is center. Initially, the angular speed of
the mini-motor is zero then it up in a time relationship
of   m (1  e t / ) , where m  540 r  s 1，．
  2.0 s　(1) what is the rotate speed of the mini-motor at t =6 s .
(2)how many turns it has made in the time interval of
at t =6 s ．(3)what is the discipline of the angular
acceleration varying with respect of time．
Solution: (1) substitute t=6 s to ω  m (1  e t / )
ω  0.95ωm  513 r  s
1
(2) The turns the motor has made in the time
interval of t=6 s is
1 6
1 6
t / 
N
ω
d
t

ω
(
1

e
)dt
m


2π 0
2π 0
 2.2110 r
3
(3) The angular acceleration of the motor is
d m t /
t / 2
2


e  540 πe rad  s
dt

Example 2 An electric motor rotates with a high
speed, in which a cylindrical rotor can rotate about the
axis going through its center and perpendicular to the
cross – section area of the rotator. Initially, the angular
velocity is ω0  0 . After 300 s the speed reached
18000 r/min. it is known that the angular acceleration a
of the rotation is proportional to time. How many
revolutions has the turned in this time interval？
d
Solution :   ct ，  ct ，
d
t
integrating both side

t
0 d  c0 tdt
1 2
得   ct
2
1 2
  ct
2
at t =300 s
  18 000 r  min
1
 600π rad  s
2 2  600 π π
3
c 2 

rad

s
2
t
300
75
1 2
π 2
  ct 
t
2
150
1
dq
π 2
from  

t
dt 150
q
π t 2
We have  dq 
t dt

0
150 0
π 3
q
t rad
450
The rotator has made the following revolutions
in 300 s
q
π
N
2π

(300)  3 10
3
2π  450
4
§4.2 Torque, the law of rotation,
moment of inertia
1.
2.
3.
4.
Torque
The law of rotation
The power of torque
The theorem of parallel axes
1
Torque
Describe the effect of force on the rotation of a rigid body．
d  r sin q

the torque of force F
Right hand rule
Is the arm of force
with respect to
axis z is

M
M  Fr sin q  Fd
  
M  r F

F
z
O

r

 Fi  0,

F

 Mi  0

F
d

 Fi  0,
i
q
P
When the combined force is zero, their combined torque
may not be zero

F
*

F

 Mi  0
i
discussion

F is not in the rotation plane ,
（1）If the force
it can
be decomposed into two components with parallel and
normal to the rotation axis.
  
F  Fz  F

where the torque of Fz
is zero, so the torque

of F is
  
M z k  r  F
M z  rF sin q
z

k
O

F

Fz

r
q

F
（2） External torque and total external torque
the torque of the external
forces drives the plate to rotate
M1
F2
F 1
F 2
j2
r2
P2
O
r1
d2 d1
P1
F1
j1
torque
Magnitude
方向
M2
Magnitude
M1 = r1 × F1
M1 = r1 F1 sin j1
= F1 d 1 = F 1 r1
M
M 2 = r 2 × F2
M 2 = r 2F 2 sin
F j2
= F2 d 2 = F 2 r2
= M1 + M 2
r
Magnitude M = F1 d 1 F2 d 2 = F 1 r1 叉乘右螺旋
F 2 r2
Total internal torque M
The total external torque is equal to the algebraic sum of these
external torque 



M  M1  M 2  M 3  
（3）The combined torque generated by the internal forces
between the mass points in a rigid body is 0, i.e.

M ij
M内 
M ij  0

j 
rj
i, j
F
ji
O r i F
ij
z 
d i


M ji
F


M ij  M ji
(4) The torque is zero, when the force
acting on the axis
O

M 0
2
M

(

m
r
 ej  j j )α
j
z
Moment of inertia
J   m r
2
j j
j
J   r dm
2
O

r
j

Fej
m j

Fij
The law of rotation M  J
when the rigid body rotates about a fixed axis, the
angular acceleration is proportional to the combined external
torque that the rigid body is subject to, and it is inversely
proportional to the moment of inertia of the rigid body.
The law of rotation
M  J
discussion
M  0， ω
M
（2）  
J
（1）
（3）
is a constant
d
M  J  J
dt
Exercises
p.144 / 4- 6,
7, 9
§4.3 Angular momentum, the law of
angular momentum conservation
1. The theorem of angular momentum and the
law of the conservation of the angular
momentum of mass points
2. The theorem of angular monentum and the
priciple of conservation of angular momentum
of a rigid body rotating
§4.3 Angular momentum, the law of
angular momentum conservation
The accumulation effect of forces over time：
Impulse 、Momentum、 the theorem of momentum．
The accumulation effect of Torque over time：
Impulse torque、 Angular momentum 、 the theorem of
angular momentum of a rigid body rotating about a fixed axis．
1 The theorem of angular momentum and the law of the
conservation of the angular momentum of mass points


2
p  mv，Ek  mv 2
Description for the motion of
a mass point
Can we use the momentum to descript the
motion of a rigid body ?


  0, p  0


  0, p  0

pi



pj
“momentum ” is not good physical q
the rotation of the rigid body .
to describe to
1
mM m 1
mM m
2
2
mv A  G
 mvB  G
2
Rh 2
R
mM
mM
v  v  2G
 2G
Rh
R
2
A
2
B
v A  1 615 m  s
so
1
v  ( v  v )
and (m)u
2
A
2 12
0
 100 m  s
1
 mv m  mv u  120 kg
Exercises
p.144 / 4- 13,
19, 28
§4.4 Work Done by a Torque
1. Work done by a torque
2. The power of a torque
3. The kinetic energy of rotation
4. The theorem of kinetic energy of a
rigid body rotating about a fixed axis
The accumulation effect of forces over space：
Work 、Kinetic energy、 the theorem of Kinetic energy ．
The accumulation effect of Torque over space：
Work done by a torque、the kinetic energy of rotation、
the theorem of Kinetic energy of a rigid body rotating about a fixed
axis．
1 Work done by a Torque
 
dW  F  dr  Ft ds
 Ft rdq
dW  Mdq
Work done
by a torque:
2
q2
W   Mdq
The power of a
torque
q1
dW
dq
P
M
 M
dt
dt
 
Comparing to W  F  dr

 
P  F v
3
The kinetic energy of rotation
For a mass element
the line sped is
The kinetic energy of the mass element is
The kinetic energy of the entire rigid body is
∑
∑
Moment of inertia J
We have
J
4 The theorem of kinetic energy of a rigid
body rotating about a fixed axis
The theorem of Kinetic
energy of a mass point:
The theorem of Kinetic energy of a rigid body rotating:
From the element of work
done by the external torque
The law of rotation
so
The work done by the
combined external torque
The increment of the rotation
kinetic energy of the rigid body
Example 1
The turnplate of a gramophone rotates in
an angular velocity ω about the axis which goes through
the center of the plate. After a record being put on it, the
record will rotate will rotate with the turnplate under the
action of friction force. Assume the radius of the plate is R
and the mass is m，the friction factor is  .(1)what is the
magnitude of the torque of the friction force； (2)how
long does the record need when its angular velocity
reachesω ；(3)what is the work done by the drive force
of the turn plate in this period of time ？
Solution (1) as shown in

df
figure, an element area, ds = drdl，
the friction force of which the
element subjected is
df 
mg
πR
2
o
drdl
R
The torque of the friction force to the
point O on the rotating axis of the
trunplate is
rdf 
mg
πR
2
rdrdl
r
dl
dr

df
for the cirque with a width
dr, the torque of friction force
that the turnplate subject is
mg
dM 
rdr (2πr )
2
πR
o
R
2mg 2

r dr
2
R
2mg R 2
2
M
r dr  Rmg
2

0
R
3
r
dl
dr
 is (the
(2) According to the theorem of angular momentum
moment of inertia of record is J=mR2/2)
M 4g


J
3R
Based on   0  t
3R
t
4g
（ the record makes a
rotation motion with a
uniform acceleration ）
(3) From  2   2  2q
0
the time interval of 0 to t is
the angle that the record turns in
The word done by the drive
force of the turnplate in this
time interval is
1
2 2
W  Mq  mR 
4
3 2 R
q
8g
Example 2
A rod of length is l and
mass m’ can rotate about pivot O freely, a
bullet with mass m and speed v is shot into
the point with distance a away from the
pivot, rendering a 30o . Deflection angle
of the rod with respect to the vertical axis,
what is the initial speed of the?
Solution
o
a

v
m
30

'
Take the bullet and the rod as a system, the angular
momentum of the system should be conserved
1 2
3
m
v
a
2
mva  ( ml  ma )  
3
m'l 2  3ma 2
After the bullet gets into the rod, taking the
bullet, the rod and the earth as a system, the
mechanical momentum of the system
should be conserved ，E =constant．
1 1 2
2
2

( m l  ma ) 
2 3
o
a

v
m
30 
'
l
o

mga(1  cos 30 )  m g (1  cos 30 )
2
o
2
2


v  g (2  3 )(m l  2ma)(m l  3ma ) 6 ma
Exercises
p.147 / 4- 30,
31, 36
The end