PHY 113 A General Physics I
9-9:50 AM MWF Olin 101
Plan for Lecture 25:
Review: Chapters 10-13, 15
1. Advice on how to prepare for exam
2. Review of rotational motion, angular
momentum, static equilibrium, simple
harmonic motion, universal
gravitational force law
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PHY 113 A Fall 2012 -- Lecture 25
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PHY 113 A Fall 2012 -- Lecture 25
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Format of Friday’s exam
What to bring:
1. Clear, calm head
2. Equation sheet (turn in with exam)
3. Scientific calculator
4. Pencil or pen
(Note: labtops, cellphones, and other electronic equipment
must be off or in sleep mode.)
Timing:
May begin as early as 8 AM; must end ≤ 9:50 AM
Probable exam format
 4-5 problems similar to homework and class examples;
focus on Chapters 10-13 & 15 of your text.
 Full credit awarded on basis of analysis steps as well as
final answer
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Examples of what to include on equation sheet
Given information
Suitable for equation sheet
on exam
Universal or common
constants (such as g, G,
ME, MS, RE…)
Basic equations from material from earlier
Chapters: Newton’s laws, energy, momentum,
center of mass
Particular constants (such Simple derivative and integral relationships,
as k,m,I… )
including trigonometric functions
Unit conversion factors
such as Hz and rad/s
Definition of moment of inertia, torque, angular
momentum, rotational kinetic energy
Newton’s law for rotational motion; combination
of rotational and center of mass motion
Equations describing simple harmonic motion
and driven harmonic motion
Newton’s universal gravitation force law and
corresponding gravitational potential energy
Gravitational stable circular orbits
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Possible extra review session on Thursday:
iclicker question:
Which of the following possible times would work
with your schedules (vote for one)?
A. 2 PM
B. 3 PM
C. 4 PM
D. Prefer to meet individually or in small groups
in my office (Olin 300).
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Rotations:
Angular variables
angular “displacement”  q(t)
dθ
ω
(t)

angular “velocity” 
dt dω
angular “acceleration”  α(t) 
dt
s
“natural” angular unit = radian
Relation to linear variables:
sq = r (qf-qi)
vq = r w
aq = r a
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Object rotating with constant angular velocity (a = 0)
w
R
vi=riw
mi
v=Rw
v=0
Kinetic energy associated with rotation:
K   12mi vi2   12mi ri2ω2 
i
i
1
2
I
ω
;
2
where : I   mi ri2 “moment of inertia”
i
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Moment of inertia:
I   mi ri
2
i
I  2Ma
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2
I  2Ma  2mb
2
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Total kinetic energy of
rolling object :
K total  K rolling  K CM
1 2 1
2
 Iw  MvCM
2
2
Note that :
dq
w
dt
ds
dq
R
 Rw  vCM
dt
dt
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K total  K rolling  K CM
1 I
1
2
2
Rw   MvCM

2
2R
2
1 I
 2
  2  M vCM
2 R

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How to make objects rotate.
q
r
r sin q
Define torque:
t=rxF
q
t = rF sin q
F
F  ma
r  F  τ  r  ma  Iα
Note: We will define and use the “vector cross product”
next time. For now, we focus on the fact that the direction
of the torque determines the direction of rotation.
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Vector cross product; right hand rule
C  AB
C  A B sin q
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
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Note that rolling motion is caused by the torque of friction:
Newton’s law for torque:
τ total  I
dω
 Iα
dt
F  f s  MaCM
F
fs
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f s R  Ia  IaCM / R
 aCM


1


fs  F
2
 1  MR / I 
For a solid cylinder, I  12 MR 2

fs R2

I

PHY 113 A Fall 2012 -- Lecture 25
 f s  13 F
12
Torque and Newton’s second law:
F  ma
dv
d mv  d
r  F  τ  r  ma  r  m
 r
 r  p 
dt
dt
dt
d
r  F  τ  r  p 
dt
Define :
L  rp
Note that : if τ  0
Then :
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dL
0
dt
L  (constant)
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Example of conservation of angular momentum:
Lbf  Lwheelf  Lbi  Lwheeli
Lbf  Lwheel  0  Lwheel
Lbf  2 Lwheel
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Another example of conservation of angular momentum
w1
m
w2
m
m
d1
d2
d1
m
d2
I2=2md22
I1=2md12
I1w1=I2w2 w2=w1 I1/I2
 w2=w1 (d1/d2)2
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Conditions for stable equilibrium
Balance of force :
F  0
i
i
Balance of torque :
τ
i
0
i
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Example:
*
X
*
Forces :
Torques :
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n  M D g  mc g  mP g  0
M D g (1m)  mc gx  0
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Simple harmonic motion:
Newton' s law for mass - spring system :
d 2x
ma  m 2  F  kx
dt
d 2x
k
 x
2
dt
m
Guess that solution for x(t ) has the form :
x(t )  A cos(wt   ) where A and  and w are unknown constants
Condition that guess satisfies the equation :
d 2 A cos(ωt  φ)
k
2
 ω Acos(ωt  φ)   Acos(ωt  φ)
2
dt
m
k
2
ω 
(determines w - - " natural frequency")
m
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Summary -Simple harmonic motion:
d 2x
F   kx  m 2
dt
d 2x
k
 x
2
m
dt
Conveniently
evaluated in
radians
k
x(t )  Acos(ωt  φ); ω 
m
Constants
Note that:
dx
  Aω sin( ωt  φ)
dt
dv
a (t ) 
  Aω 2 cos( ωt  φ)
dt
v(t ) 
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Energy associated with simple harmonic motion
Energy :
1 2 1 2
E  mv  kx
2
2
1
1 2
2
2
2 2
E  mw A sin(wt   )   kA cos(wt   ) 
2
2
k
2
But w 
m
1 2
1 2
2
2
 E  kA sin(wt   )   cos(wt   )   kA
2
2

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
PHY 113 A Fall 2012 -- Lecture 25
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Simple harmonic motion for a pendulum:
L
Q
d 2Q
τ  mgL sin Q   Iα  -I 2
dt
d 2Q
mgL
g
2


sin
Q


sin
Q
(since
I

mL
)
2
I
L
dt
Approximation for small Q:
sin Q  Q
d 2Q
g
 2  Q
L
dt
Solution :
g
Q(t )  A cos( ωt  φ); ω 
L
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The notion of resonance:
Suppose F=-kx+F0 sin(Wt)
According to Newton:
d 2x
 kx  F0 sin( Wt )  m 2
dt
Differenti al equation (" inhomogene ous" ) :
F0
d 2x
k
  x  sin( Wt )
2
dt
m
m
Solution :
F0 / m
F0 / m
x(t ) 
sin( Wt )  2
sin( Wt )
2
2
k /mW
ω W
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Universal law of gravitation
 Newton (with help from Galileo, Kepler, etc.) 1687
Gm1m2rˆ12
F12 
r122
G  6.674 10 11
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PHY 113 A Fall 2012 -- Lecture 25
N  m2
kg 2
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Review:
Gravitational force of the Earth
RE
m
GM E m
F
RE2
GM E 6.67 10 11  5.98 10 24
2
2
g 2 
m/s

9
.
8
m/s
RE
(6.37 106 ) 2
Note: Earth’s gravity acts as a point mass located at the
Earth’s center.
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Example:
Satellite in circular Earth orbit
For m  M E
r  RE  h

v
2r / T 
GM E m
 2 
m m
 m
 r
2
r
r
T
r


2
2
T  2
2
RE  h 3
GM E
If h  35.83 106 m
T  8.53 10 4 s  1 day
(geosynchronous)
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Gravitational potential energy
r
U gravity (r )    F  dr
rref
Gm1m2rˆ
F
r2
 Gm1m2
Gm1m2
U gravity (r )   
dr '  
2
r'
r

r
Example:
GM E m
U gravity (r  RE  h)  
RE  h
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