PHY 113 C General Physics I
11 AM – 12:15 PM TR Olin 101
Plan for Lecture 17:
Review of Chapters 9-13, 15-16
1. Comment on exam and advice for
preparation
2. Review
3. Example problems
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PHY 113 C Fall 2013 -- Lecture 17
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Webassign questions – Assignment #15
Consider the sinusoidal wave of the figure below with the wave
function y = 0.150 cos(15.7x − 50.3t)
where x and y are in meters and t is in seconds. At a certain
instant, let point A be at the origin and point B be the closest
point to A along the x axis where the wave is 43.0° out of phase
with A. What is the coordinate of B?
  
43 
 15.7 x
o 
 180 
o
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Webassign questions – Assignment #15
A transverse wave on a string is described by the following
wave function. y = 0.115 sin ((π/9)x + 5πt)
where x and y are in meters and t is in seconds.
(a) Determine the transverse speed at t = 0.150 s for an
element of the string located at x = 1.50 m.
y ( x, t )
  

 0.115  cos x  5t 
t
9 9

(b) Determine the transverse acceleration at t = 0.150 s for an
element of the string located at x = 1.50 m.
 y ( x, t )
 


 0.115  sin  x  5t 
2
t
9
9

2
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PHY 113 C Fall 2013 -- Lecture 17
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Webassign questions – Assignment #15
A sinusoidal wave in a rope is described by the wave function
y = 0.20 sin (0.69πx + 20πt)  y0 sin kx  t
where x and y are in meters and t is in seconds. The rope has
a linear mass density of 0.230 kg/m. The tension in the rope is
provided by an arrangement like the one illustrated in the figure
below. What is the mass of the suspended object?



20
T
mg
c 


k 0.69


T
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mg
PHY 113 C Fall 2013 -- Lecture 17
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Comment about exam on Tuesday 10/29/2013
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iclicker question
What is the purpose of exams?
A. Pure pain and suffering for all involved.
B. To measure what has been learned.
C. To help students learn the material.
D. Other.
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Advice on how to prepare for the exam
Review lecture notes and
Prepare equation sheet
Work practice problems
text chapters 9-13,15-16
Topics covered
Linear momentum
Rotational motion and angular momentum
Gravitational force and circular orbits
Static equilibrium
Simple harmonic motion
Wave motion
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What to bring to exam:
Clear head
Calculator
Equation sheet
Pencil or pen
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iclicker question:
Have you looked at last year’s exams?
A. Yes
B. No
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Linear momentum
 What is it?
 When is it “conserved”?
 Conservation of momentum in analysis of collisions
 Notion of center of mass
Define " linear momentum": mv  p
Units of linear momentum : kg  m/s  N  s
dv d mv  dp
F  ma  m


dt
dt
dt
Fdt  dp
f
f
i
i
Impulse : I   Fdt   dp  p f  p i
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Linear momentum -- continued
Physics of composite systems
Newton' s second law :
dv i
d mi v i  d 

i Fi  i miai  i mi dt i dt  dt  i pi 
Note that if
 F  0,
i
then :
i
d 

  pi   0
dt  i

  p i  (constant)
i
  p i initial 
i
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p
i final
i
PHY 113 C Fall 2013 -- Lecture 17
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Example – completely inelastic collision; balls moving
in one dimension on a frictionless surface
p
i initial
i

p
i final
i
m1 v1i  m2 v 2i  m1  m2 v f
m1 v1i  m2 v 2i
vf 
m1  m2
For
m1  0.3kg , m2  0.5kg
v  2m / sˆi , v  1m / sˆi
1i
vf
2i

0.32 ˆi  0.5 1ˆi

m/s
0.3  0.5
 0.125 m/s ˆi
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Examples of two-dimensional collision; balls moving on a
frictionless surface
p
i
i initial

p
i final
i
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
Knowns : m1 , m2 , v1i
Unknowns : v1 f , v2 f ,  , 
Need 2 more equations - -
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The notion of the center of mass and the physics of
composite systems
Newton' s second law :
d 2ri
d 2 mi ri 
i Fi  i miai  i mi dt 2 i dt 2
rCM 
Define :
F  F
i
i
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total
 m r 
i i
i
M
M   mi 
i
2
d rCM
M
dt 2
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Finding the center of mass
rCM 
 m r 
i i
i
M
M   mi 
i
In this example : m1  m2  1kg ; m3  2kg
m1 x1ˆi  m2 x2 ˆi  m3 y3ˆj
rCM 
m1  m2  m3
(1)(1m)ˆi  12m ˆi  2 2m ˆj
rCM 
4
 0.75mˆi  1.00mˆj
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Rotational motion and angular momentum
 Angular variables
 Newton’s law for angular motion
 Rotational energy
 Moment of inertia
 Angular momentum

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PHY 113 C Fall 2013 -- Lecture 17
d

dt
d

dt
17
Review of rotational energy associated with a rigid body
Rotational energy :
1
1
2
2
K rot   mi vi   mi ri 
2 i
2 i
1
1 2
2 2
  mi ri   I
2 i
2
where I   mi ri
2
i
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Moment of inertia:
I   mi ri
2
i
I  2Ma
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2
I  2Ma  2mb
2
PHY 113 C Fall 2013 -- Lecture 17
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19
kinetic energy
energy of
Total kinetic
rolling object :
rolling
K total
total  K rolling
rolling  K CM
CM
CM
1 2 1
2
 I  MvCM
2
2
Note that :
d

dt
ds
d
R
 R  vCM
dt
dt
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K total
total  K rolling
rolling  K CM
1 I
1
2
2
R   MvCM

2
2R
2
1 I
 2
  2  M vCM
2R

PHY 113 C Fall 2013 -- Lecture 17
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iclicker exercise:
Three round balls, each having a mass M and
radius R, start from rest at the top of the incline.
After they are released, they roll without slipping
down the incline. Which ball will reach the bottom
first?
I  MR 2
A
B
A
C
1
I B  MR 2  0.5MR 2
2
2
I C  MR 2  0.4 MR 2
5
Ki  U i  K f  U f
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0  Mgh 
1 
I  2
M 1 
v 0
2  CM
2  MR 
 vCM 
2 gh
1  I / MR 2
PHY 113 C Fall 2013 -- Lecture 17


21
How can you make objects rotate?

r sin 
Define torque:
t=rxF
r

t = rF sin 
F
F  ma
r  F  τ  r  ma  Iα
Newton' s law for rotational motion :
τ  Iα
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Example form Webassign #11
t3
X
iclicker exercise
When the pivot point
is O, which torque is
zero?
A. t1?
B. t2?
C. t3?
t2
t1
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Vector cross product; right hand rule
C  AB
C  A B sin 
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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From Newton’s second law – continued –
conservation of angular momentum:
d
r  F  τ  r  p 
dt
Define :
L  rp
dL
If τ  0
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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Summary – conservation laws we have studied so far
Conserved quantity
Linear momentum p
Angular momentum L
Mechanical energy E
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Necessary condition
Fnet = 0
tnet = 0
No dissipative forces
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Fundamental gravitational force law and planetary motion
 Newton’s gravitational force law
 Gravity at Earth’s surface
 Circular orbits of gravitational bodies
 Energy associated with gravitation and orbital motion
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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 C Fall 2013 -- Lecture 17
N  m2
kg 2
29
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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Stable circular orbit of two gravitationally attracted
objects (such as the moon and the Earth)
Newton' s law for Moon due to Earth :
FM  M M a M
REM
v2
GM E
a
 2
REM
REM
F
v
a
v  ωREM 
2π
REM
T
3
REM
T  2π
GM E
(3.84 108 ) 3
 2π
6.67 10 11  5.98 10 24
 2367353.953 s  27.4 days
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Circular orbital motion about center of mass
v2
R2
R1
m1
v1
CM
v12
Gm1m2
v22
m1 
 m2
2
R1 R1  R2 
R2
m2
m1 R1  m2 R2
2
 2R1  1
 2
v

m1  m1 
 m1 R1 
R1
 T1  R1
 T1
2
1
T1  T2  2



R1  R2 
G m1  m2 
3
Note that if m2  m1 then R2  R1
T1  T2  2
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R1  R2 3  2
G m1  m2 
R13
Gm2
PHY 113 C Fall 2013 -- Lecture 17
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2
dL
0
dt
 L  (const)
τ
v2
R2
L1=m1v1R1
R1
L2=m2v2R2
L = L1 + L2
L1 L2

R1 R2
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m2
m1
v1
Note: More generally,
stable orbits can be
elliptical.
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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 mS
U gravity (r  RE  h)  
RE  h
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Analysis of static equilibrium
Meanwhile – back on the surface of the Earth:
Conditions for stable equilibrium
Balance of force :
F  0
i
i
Balance of torque :
τ
i
0
i
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Torques :
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 Fg1 (2m)  mg ( RCM )  0
PHY 113 C Fall 2013 -- Lecture 17
36
t 0
T

 mgx  Mg  T sin   0
2
mgx /   Mg / 2
T
sin 
For   53o   8m x  2m
mg  600 N Mg  200 N
T  313 N
*
X*
x
Mg

mg
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/2
PHY 113 C Fall 2013 -- Lecture 17
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Some practice problems
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From webassign:
A 100-kg merry-go-round in the shape of a uniform,
solid, horizontal disk of radius 1.50 m is set in motion by
wrapping a rope about the rim of the disk and pulling on
the rope. What constant force would have to be exerted
on the rope to bring the merry-go-round from rest to an
angular speed of 0.800 rev/s in 2.00 s? (State the
magnitude of the force.)
view from top:
F
R
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τ  r  F  Iα
1
I  MR 2
2
PHY 113 C Fall 2013 -- Lecture 17
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From webassign:
A 10.3-kg monkey climbs a uniform ladder with
weight w = 1.24 102 N and length L = 3.35 m as shown in the
figure below. The ladder rests against the wall and makes an
angle of θ = 60.0° with the ground. The upper and lower ends
of the ladder rest on frictionless surfaces. The lower end is
connected to the wall by a horizontal rope that is frayed and
can support a maximum tension of only 80.0 N.
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