Transcript Ch 12 Rotation of a Rigid body

講者： 許永昌 老師
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Contents
 Rigid body
 Center of mass: rCM
 Rotational Energy
 Moment of inertia: I
 Mathematics: cross product
 Torque
 Properties
 Applications



Rotation about a fixed axis
Static Equilibrium
Rolling Motion
 Angular Momentum
 Examples
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Rigid Body (請預讀P340~P345)
 A rigid body is an extended object whose
R̂ 
as it moves.
 E.g.
 1st point: r .
 2nd point:
.
R̂
 3rd point:  .
 Others: .
z
rCM
x
y
 We just need 6 parameters
to describe a rigid body:
 Translational: rCM.
 Rotational: axis R̂ and angle
z
R̂

r
.
x
y
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Rotational Motion
 Top view from the tip of the axis of rotation:
 Right hand rule.
ˆ.
 Angular velocity:    R
R̂ =?
 Exercise:
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Center of Mass

rCM 
1
M
m r .
i i
i
 Benefits:
 MvCM  P   mi vi
 MaCM  Fnet   mi ai
i
 MrCM 
M r
i
i CM ,i
i
 Exercise:
 Find the center of mass of this object:
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Homework
 Student Workbook:
 P12-1~P12-2
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Rotational Energy & Moment of
inertia (請預讀P346~P350)
K micro  K rot
1
1
1
2
2
2
  mi v 'i   mi raxis to i   I CM  2 .
2
i 2
i 2
2
I   mi raxis
to i
i

I   mi xi2  yi2
Rˆ  zˆ

i
K
1
1
2
MvCM
 I CM  2  K vibrate
2
2
KCM
K rot
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Parallel-axis Theorem (I=ICM+MR2)
 Total kinetic energies of these two cases:
R

'
K=?
K=?

When    ',
we get
I=ICM+MR2.
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Exercises
 ICM of
 A cylinder.
 A thin rod.
 A sphere.
 Example 12.5
 A 1.0-m-long, 200 g rod is hinged at one end and
connected to a wall. It is held out horizontally, then
released. What is the speed of the tip of the rod as it hits
the wall?
v=?
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Homework
 Student Workbook
 12.7
 12.8
 12.12
 12.13
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Cross Product (請預讀P368~P369)
 Geometrical definition:

A  B  C , where
(1) AB sin   C ,

(2) C is perpendicular to A and B, i.e. A Cˆ  B Cˆ  0,
(3) Its direction is given by the right-hand rule.

A
B
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Cross Product
 Properties:
 Anti-commutative: AB=-BA.
 Distributive: A(aB+bC)=aAB+ bAC.
 Based on the right-hand rule, we get:
ˆ  yˆ  zˆ, yˆ  zˆ  xˆ, and zˆ  xˆ  yˆ.
 x
 Exercise:
 A  B  ? xˆ  ? yˆ  ? zˆ.

If


A=(1, 0, 0),
B=(1, 1 , 0).
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Torque (請預讀P370, P351~P356)
 Think of that, since
 Kinetic energy:
1 2
1
mv  I  2
2
2
translational
 Can we define
rotational
Fnet  MaCM   net  Ia ?
cause
result
cause
translational
2
 If so, Ia   mi raxis
to ia
i
rotational

r
axis to i
raxis to i is fixed
for a rigid body.
i
  net on i
i
mi at ,i   raxis to i Fnet on i , t
i
 i  ri  Fi
raxis to i
Ft
result
Fnet on i
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Torque
 Definition:
 The torque contributed by Fi exerted on particle i is
 i  ri  Fi .


It is dependent of the origin we chose.
 i = riFisin=diFi=riFi,t.
顯然，在討論旋轉時，particle model 並不適用。
Particle model 適用於合力與動量的部分，而旋轉
 di: (力臂)
則必須知道受力點與origin的相對位置。
 Moment arm.
 Lever arm.
di
ri
Fi,t
Fnet on i
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Fext on i
Torque
FG ,i
:
1.
 net on system   ext on system
1.
FG
Based on Newton’s 3rd Law of motion and assume that Fi on j // rij
Gravitational torque:  G  rCM  FG .
2.
1.
Owing to g=constant.
When Fnet=0, net is independent of the choose of the
origin.
3.
1.
 net   ri  Fi   ri  Fi - r0   Fi
i
i
i
if Fnet  0
   ri - r0   Fi , origin is changed to r0 .
i
4.  net , ref II   net - rCM   Maref II 
1.
Fictitious force = -mia.
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Action (torque)
 Purpose:
 Understand how to find the
 Get the feeling of =I = F
.
 Objects:
 A coat hanger (lever)
 Actions:
1. Find its center of mass.
2. Exert different forces on this lever.
3. Which point you chose as the origin.
.
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 Is it possible for us to explain the equation r1F1=r2F2 for
a balance by Newton’s 2nd and 3rd Law directly?
F1
r1
r2
F2
 Hint: acceleration constraint &
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Homework
 Student Workbook:
 12.15,
 12.17,
 12.20,
 12.22,
 12.24
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Applications (請預讀P357~P367)
Rotation about a
1.
1.
.
Rotation  Tangential (linear) motion
2. Static Equilibrium.
3. Rolling Motion.
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Rotation about a fixed axis (請預讀
P357~P359)
 E.g. rope and pulley.
 Kinematics:

Translational
velocity
acceleration
vtˆ
atˆ
Rotational
vt=r
at=ra.
r
ˆ
 Dynamics:

  r F .

 Rˆ  Ia
 This is not a general form.
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Exercise
 The acceleration of box m1.
 Frictionless.
 Ignore drag.
 Massless string.
 The rope turns on the pulley without slipping.
m1
I
m2
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Static Equilibrium (請預讀P360~P363)
 The condition for a rigid body to be in static
equilibrium is both
 Fnet  0,

 net  0.
 aCM  0.
 a  0  For a rigid body  .
 Exercise: problem 12.62 (P381)
 A 3.0-m-long rigid beam with a mass of 100 kg is
supported at each end. An 80 kg student stands 2.0 m
from support I. How much upward force does each
support exert on the beam?
1
2
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Rolling Motion (請預讀P364~P367)
 Condition: vCM=R.
 Reason:

 aCM=Ra.
The contact point P, which is instantaneously
2R
.
R
R
P
Inertial reference frame
P
Viewed from its center of mass
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Exercise
 The acceleration of this cylinder.
 I= ½MR2.
 No slipping.

Friction?

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Homework
 Student Workbook
 12.25,
 12.29,
 12.30
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Angular Momentum (請預讀P371~P375)

 Angular Momentum:
 Li  ri  pi .
 Properties:
1.
 net on i
dLi

dt
Li
dLi d
dr
dp
  ri  pi   i  pi +ri  i  ri  Fnet on i
dt dt
dt
dt
ri
=0, pi  mi vi
  net on i
1 vi    ri
 2  Ltot   ri   mi vi    mi ri 2 -  mi ri ri
2. Ltot  I 
1.
For a rigid body
2.
I is a tensor in general.
i
i

 
 
A BC  B A C -C A B
i.e. Ltot , s

s x, y ,z


   m r 
i

I stt .
t  x, y , z
 I xx

Iˆ   I yx
 I zx

I xy
I yy
I zy
2
i i
t  x, y , z

i
st


- mi si ti  t


yi2  zi2



I xz
i
 
I yz    - yi xi

i
I zz  
  zi xi
i

- xi yi
i
x z
- z y
2
i
2
i
i
i
i
i

- xi zi 
i

- yi zi  &

i
2
2
x

y
i i i 
 Lx 
 x 
  ˆ 
 Ly   I   y 
L 
 
 z
 z
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Exercises
 The change of the diver’s angular velocity when he
extend his legs and arms.
ICM2
ICM1
 Rotating bike’s wheel and rotating coins.
 Questions:

   constant ?



The motion of this wheel or coin.
Frictionless?
Which point you chose to be as the origin, and why?
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Homework
 Student Workbook
 P12-11~P12-12
 Textbook
 12.20
 12.51
 12.54
 12.65
 請製作卡片
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Summary
Kinematics
Single particle
Many particles
Rotation
Position
ri
rCMSimiri/M
=s/r
Mass (moment of
mi
MSimi
ISimiri2.
Velocity
vi=dri/dt
vCM =drCM/dt
=d/dt
Momentum
pi=mivi
PCM=MvCM
Li=ripi
Ltot=I (*)
Acceleration
ai=dvi/dt
aCM =dvCM/dt
a=d/dt
Fnet on i
Fnet on system =Fext
inertia for rotation)
New
Dynamics
New
Force (torque)
 net on i  ri  Fi
 net on system   ext
 ext  Ia *
Newton’s 2nd
Law
Fnet on i  mi ai 
Kinetic Energy
Ki= ½mivi2.
dpi
dt
Fnet  MaCM 
dPtot
dt
 net 
dLtot
dt
Ktot= ½MvCM2+Kmicro. Krot= ½ I2.
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