Download + Rotational motion about its CM

Chapter 10
Rotational Motion
About a Fixed Axis
Rigid body
A body with a definite shape that doesn’t change
1. Vibrating or deforming can be ignored
2. Distance between particles does not change
Motion of a rigid body can be expressed as
Translational motion of its CM
+ Rotational motion about its CM
Pure rotational motion: all move in circles
centers of these circles all lie on a line
Axis of rotation
Fixed axis
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Angular quantities
Angular coordinate / Angle   s / R
Angular velocity
y v
v
d 1 ds



R
R dt
R
dt

o
Angular acceleration
d  1 dv atan



R dt
R
dt
Radial acceleration aR  v 2 / R   2 R

1
Frequency f 
,
Period T 
2
f
s
x
3
Hard drive
Example1: The platter of hard disk rotates 5400rpm. a) Angular
velocity; b) Speed of reading head located 3cm from the axis;
c) Acceleration at that point; d) How many bits the writing head
writes per second at that point if 1 bit needs 0.5µm length.
Solution: a) f  5400 rev / min  90 rev / s  90Hz
angular velocity   2 f  565 rad/s
b) speed v  R  17 m/s
2
2
c) acceleration aR   R  9580m/s
d)
17
7
N

3.4

10
bit/s
7
5 10
 4 MB/s
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Useful Similarities
Notice the similarities in different motion
d
d
dx
dv

, 
v
, a
dt
dt
dt
dt
Uniformly accelerated rotational motion
v  v o  at
1 2
 x  v0 t  at
2
v 2  v 02  2 a  x
  0   t
1 2
   0t   t
2
 2   02  2  
Analogous thinking is very helpful
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Vector nature of angular quantities
Angular velocity and acceleration → vectors

d

dt
Points along the axis,
follows the right-hand rule
v  r  v    r
How does
 change?
Angular
acceleration
d

dt
pulley
spinning top
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Rotational dynamics
What causes acceleration of rotational motion?
Force: magnitude, direction and point of action
Push a door
Archimedes’ Lever
“Give me a fulcrum, and I shall
move the world. ” —— Archimedes
Lever arm: the perpendicular distance from the
axis to the line along which the force acts
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Torque about fixed axis
The effect of force → angular acceleration
Torque = force × lever arm
  R  F  R F s in   F tan R
R
R
R⊥: lever arm or moment arm

F
Net torque causes acceleration of rotational motion


positive rotational direction
Balance of rigid body
Play on a seesaw
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Torque and balance
Example2: A 15kg mass locates 20cm from the axis
of massless lever. Determine: a) torque on the lever;
b) Force required respectively to balance the lever.
Solution: a)   R  F  3 0 N  m
b) F1  2 5 c m  3 0 N  m
20cm
5cm
25cm
30
 F1  1 2 0 N
15cm
F2  5 c m  3 0 N  m  F2  6 0 0 N
F 3  1 5 c m  s in 3 0   3 0 N  m
F3
F4
150N
F2
F1
 F3  4 0 0 N
What about F4 ?
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Rotational theorem
A particle mi in rigid body
fi
F i ta n  f i ta n  m i a ta n  m i R i
Fi
Ri
mi
 F i ta n R i  f i ta n R i  m i R i 2 


F i tan R i   f i tan R i 
   ext

Net external torque
  I
in
0

m i Ri2  
I
Rotational inertia
or moment of inertia
Rotational theorem about fixed axis
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Properties of Rotational theorem
1) Only external torques are effective
Sum of the internal torques is 0 from N-3
2) Rotational equivalent of Newton’s second law.
  I
F  ma
3) Analogy of rotational and translational motion
change in motion
cause of the change
inertia of motion


I
a
F
m
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Determining rotational inertia
I

I   R 2 dm
m i R i2
several particles
continuous object
How mass is distributed with respect to the axis
Example3: Rotational inertia of 3 particles fixed
on a massless rod about a, b, c axis.
Solution:
I a  2 m  l  3m   2l   14ml 2
2
2
I b  m  l  3m  l  4ml 2
2
2
I c  2 m  l  m   2l   6ml
2
2
2
l
m
a
l
2m
b
3m
c
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Uniform thin rod
Example4: Rotational inertia of uniform thin rod
with mass m and length l. a) Through center;
Solution: Choose a segment dx
l
2
l

2
I 
m
1 2
2
x dx  ml
l
12
C
dm
o
x dx
x
b) Through end
1 2
m
I   x dx  ml
0
3
l
l
2
o
Typical result, should be memorized
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Uniform circles
Example5: Uniform thin hoop (mass m, radius R)
Solution: Choose a segment dm
I   R dm  m R
2
2
R
dm
Example6: Uniform disk/cylinder (m, R)
Solution: Choose a hoop dm
I
R
0
r dr
m
1
2

mR
r
2 2  rd r
R
2
2
Homework: Uniform sphere (m, R) (P246)
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Parallel-axis theorem
If I is the rotational inertia about any axis, and IC
is the rotational inertia about an axis through the
CM, and parallel to the first but a distance l, then
I  IC  M l
2
I
 r dm   ( r  r ) dm
  ( r   l )( r   l ) dm
  ( r   2 l  r   l ) dm
 I  M l  2 l   r dm
Proof: I 
l
2
2
o
l
r
C
r
M
2
2
c
Ic
 M rC  0
IC is always less than other I of parallel axes
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Perpendicular-axis theorem
The sum of the rotational inertia of a plane about
any two perpendicular axes in the plane
is equal to the rotational inertia about an axes
through the point of intersection ⊥ the plane.
z
Iz  Ix  Iy
1) Only for plane figures or
2-dimensional bodies
y
x
2) x ⊥ y ⊥ z and intersect at one point
3) Try to prove it by yourself
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Application of two theorems
Example7: Rotational inertia of uniform thin disk
about the line of diameter
Solution: You can choose a dm
or apply the ⊥-axis theorem
1
4
I z  I x  I y  2 I x  I  mR 2
dm
Example8: Thin hoop about a tangent line
Solution: Perpendicular-axis theorem
mR 2  mR 2 / 2
Parallel-axis theorem mR 2 / 2  3mR 2 / 2
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Massive Pulley
  I
Example9: A box is hanging on a pulley by massless
rope, then the system starts to move without slip,
determine the angular acceleration and tension.

Solution: T = mg ?
Free-body diagram
M

R
Rotational theorem:   TR  I
Newton’s second law: mg  T  ma
No slip motion:
 
a  R
2mg
Mmg
, T
2m  M
 2m  M  R
1
( I  MR 2 )
2
T
m
mg
Two boxes?
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Rotating rod
Example10: A uniform rod of mass M and length l
can pivot freely about axis o, released horizontally.
Determine: a) α; b) aC; c) force acted by the axis.
Solution: a) Gravity → torque
 3g
l
  Mg    
I
2l
2
b) Acceleration of CM
l
3
aC    g
2
4
F
o
C
Mg
a of the end? when   0?
3
1
c) M a C  M g  F  F  M g  M g  M g
4
4
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Angular momentum
Linear momentum p = mv
Angular momentum L  I 
rotational analog
Rotational theorem can be written in terms of L
d  d  I   dL


  I  I
dt
dt
dt
dL

dt
The rate of change in angular momentum of a
rigid body is equal to the net torque applied on it.
Comparing with Newton’s second law
dp
F  ma  F 
dt
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Conservation of angular momentum
dL
Torque & angular momentum   I  
dt
1) It is valid even if I changes
2) Valid for a fixed axis or axis through its CM
Law of conservation of angular momentum:
The total angular momentum of a rotating body
remains constant if the net external torque is zero.
1) It holds for inertial frames or frame of CM
2) One of fundamental laws of conservation
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Examples in sports
Figure skating
Diving
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Helicopter
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Falling cat
A falling cat can adjust his posture
to avoid injury, how to make it?
Angular momentum is conserved
1) Bend his body
2) Rotate his upper part to proper
position about blue axis, meanwhile
the lower part rotates a less angle
3) Rotate his lower part to proper
position about the red axis
4) Get the work done
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Rotating disk
Example11: A disk is rotating about its center axis,
and two identical bullets hit into it symmetrically.
The angular velocity of system will ________
A. increase; B. decrease; C. remain constant
Solution: Total angular momentum of the system is
conserved
L  I   c o n sta n t

.o
Total rotational inertia increases
So angular velocity will decrease
What if two opposite forces act on the disk?
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Man on rotating platform
Example12: A platform is rotating about its center
axis, and a man standing on it (treat as a particle)
starts to move. How does  change if he goes: a) to
point o; b) along the edge with relative speed v.
Solution: a) Conservation of angular momentum
2
m
I

mR
2
(I  m R )  I      

I

b) choose a positive direction
(I  m R )  (I  m R )   m v R
mvR
    
I  mR 2
2
2
I o
R
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Hits on a rod
Example13: A bullet hits into a hanged uniform rod,
determine the angular velocity after collision.
Solution: Conservation of total L
1
m v r  [ M l 2  mr 2 ] 
3
3 mvr
 2
M l  3 mr 2
o
r
mv
A
M, l
Notice: Momentum is not conserved in general!
Require: Force acted by axis remains constant
Only if the bullet hits on position r = 2l /3
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Rotational kinetic energy
Kinetic energy in translational motion
1 2
E k  mv
2
1 2
Ek  I
2
rotational motion
Total Ek is the sum of Ek of all particles
Ek  
1
1
1
1 2
2
2 2
2
2
m i vi   m i Ri     m i R i    I 
2
2
2
2
Work done on a rotating body:

W   F  dl   Ftan dl   Ftan Rd   d 
Power
dW
d
P

 
dt
dt
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Energy in rotational motion
Rotational theorem:
d
d  d
d
  I  I
I
 I
dt
d  dt
d
Work-energy principle:
2
2
1 2 1 2
W    d   I  d   I  2  I 1
1
1
2
2
Comparing with W   F  dl  1 mv 22  1 mv12
2
2
Total mechanical energy is conserved in rotational
motion if only conservative forces do work.
Potential energy of gravity: U  mghC
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Rotating rod
Example14: A uniform rod (M, l) can pivot freely
about axis o (Ao=l/3), and it is released horizontally.
Determine ω at the vertical position.
Solution: Distance oC=l/6
A
Rotational inertia
2
1
l
I  I C  M    Ml 2  M
 6  12
o
.
.
C
2
l 1 2
   Ml
6 9
Conservation of mechanical energy
l 1 2
Mg  I     3 g / l
6 2
Any position?
α=?
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General motion
For general motion of a rigid body
Translational motion of its CM:
Fnet  maC
+ Rotational motion about its CM:
 C  IC
The total kinetic energy
1 2 1
E k  E kC  E kR  m v C  I C  2 (Proved in P259)
2
2
Examples:
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Rolling motion
A typical motion: rolling without slipping
To make sure v P  0
Relationship between translational
and rotational motion

vC
vC  R
Valid only if no slipping
Motion of pulley, tire, …
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Rolling down an incline
Example15: A uniform cylinder (m, R) rolls down
an incline without slipping, determine its speed if
the CM moves a vertical height H.
Solution: Conservation of energy
1 2 1
mgH  mv C  I C  2
2
2

where I C  1 mR 2 , v C  R 
2
 vC  2 gH / 3
Comparing with sliding:
Which is faster?
vC  2 gH
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Dynamics in rolling
Forces acting on the cylinder
Translational motion of CM
m g sin   F  m a C

Rotational motion about CM
1
 C  IC  FR  mR 2 (aC  R )
2
We can obtain: aC = 2gsin /3， F = mgsin /3
Static friction causes the rolling motion
It also rearranges the kinetic energy
34
Challenging question
A stick (M, l) stands vertically on a frictionless table,
then it falls down. Describe the motion of its CM,
and of each end.
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