PHY 113 C General Physics I
11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 26:
1. Comments on preparing for Final Exam
2. Comprehensive review – Part II
3. Course assessment
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PHY 113 C Fall 2013 -- Lecture 26
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PHY 113 C Fall 2013 -- Lecture 26
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Final exam schedule for PHY 113 C
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PHY 113 C Fall 2013 -- Lecture 26
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Comments on Final Exam
 It will be comprehensive (covering material
from Chapters 1-22)
 It is scheduled for 9 AM Dec. 12th in Olin 101
 In class format only; no time pressure
 May bring 4 equation sheets
 Format will be similar to previous exams; may
see problems similar to those on previous
exams
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PHY 113 C Fall 2013 -- Lecture 26
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General advice on how to prepare for Final Exam
 Review fundamental concepts and their
corresponding equations
 Develop equation sheets that help you solve
example problems on all of the material. (You can
assume that empirical constants and parameters
will be given to you; they need not take up space on
your equation sheet.)
 Practice problem solving techniques.
 If you find mysteries, unanswered questions, etc.,
please contact me.
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PHY 113 C Fall 2013 -- Lecture 26
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Review of some basic concepts
Vectors
 Keep track of 2 or
more components
(or magnitude and
direction)
 Examples
 Position vector
 Velocity
 Acceleration
 Force
 Momentum
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Scalars
 Single (signed)
quantity
 Examples
 Time
 Energy
 Kinetic energy
 Work
 Potential energy
 Pressure
 Temperature
 Mass
 Density
 Volume
PHY 113 C Fall 2013 -- Lecture 26
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Review of some basic concepts
Newton’s second law
For single point particle (or center of mass of extended system)
ma  F
dv
d 2r
m
m 2 F
dt
dt
d mv  dp

F
dt
dt
For system of particles
m a  F
i i
i
i
i
dp i
i dt  i Fi
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PHY 113 C Fall 2013 -- Lecture 26
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Review of some basic concepts
Newton’s second law for angular motion
dp
F
dt
dp d r  p  dL
r


 rF  τ
dt
dt
dt
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PHY 113 C Fall 2013 -- Lecture 26
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Review of energy concepts:
rf
Definition of work :
Wi  f   F  dr
ri
Definition of kinetic energy :
1 2
K  mv
2
Work - kinetic energy theorem :
f
total
i f
W
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1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
PHY 113 C Fall 2013 -- Lecture 26
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Summary of work, potential energy, kinetic energy
relationships
Work - kinetic energy theorem :
f
total
i f
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
total
i f
W
W
W
 U
total
i f
W
conservative
i f
f
W
 U  W
i
dissipative
i f
dissipative
i f
 U r f   U ri   W
dissipative
i f
 K f  Ki
Rearranging : K f  U f  K i  U i  W
dissipative
i f
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PHY 113 C Fall 2013 -- Lecture 26
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Extension of concepts of energy conservation to
extended objects
Kinetic energy
K total  K center of mass  K rotation
K f  U f  Ki  U i  W
dissipative
i f
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PHY 113 C Fall 2013 -- Lecture 26
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kinetic energy
energy of
Total kinetic
rolling
rolling object :
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 26
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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?
B
A
C
I A  MR 2
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 26
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iclicker exercise:
In previous example which of the equations on your
equation sheet would be most useful?
A. K i  U i  K f  U f
1
1 2
2
B. K total  MvCM  I ; For rolling vCM  R
2
2
C. A & B
D. vCM
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2 gh

1  I / MR 2


PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- (question from Exam 2)
Gm1m2
F12 
rˆ
2 12
R1  R2 
U 
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τ0
Gm1m2
R1  R2 
PHY 113 C Fall 2013 -- Lecture 26
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Comment on circular motion -- uniform circular motion
If vi  v f  v, then the acceleration
in the radial direction and the
centripetal acceleration is :
v2
a c   rˆ
r
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PHY 113 C Fall 2013 -- Lecture 26
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Comment on circular motion -- uniform circular motion
r
v2
a c   rˆ
r
 2  ˆ
a c  
 rr
 T 
2
a c  2f  rrˆ
2
In terms of time period T for one cycle:
2r
v
T
In terms of the frequency f of complete cycles:
1
f  ;
T
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v  2πfr
PHY 113 C Fall 2013 -- Lecture 26
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Comment on circular motion -- uniform circular motion –
effects on gravitationally attractive bodies
F12 
Gm1m2
rˆ
2 12
R1  R2 
Gm1m2 ˆ
ˆ
m1a1R1  
R1
2
R1  R2 
v12
Gm1m2
 m1

R1
R1  R2 2
 2

 T1
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2

Gm2
 R1 
2


R

R

1
2
PHY 113 C Fall 2013 -- Lecture 26
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Comment on circular motion -- non-uniform circular motion
v2
a c   rˆ
r
r
 2  ˆ
a c  
 rr
 T 
a
2
a c  2f  rrˆ
2
ac
At each
instant of
time
Note that if speed v is not constant, then there will also
be a tangential component of acceleration:
dv ˆ
a  θ
dt
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PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- (question from Exam 1)
a. Neglecting any possible dissipative
forces acting on this system, determine
the magnitude of the velocity of the ball
vf as it is caught by the person at the
coordinates (xf,yf).
b. What is the angle f?
c. Determine the net work of gravity on the
ball at it moves from the initial to final
positions in its trajectory: .
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PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- (question from Exam 1)
a. Neglecting any possible dissipative
forces acting on this system, determine
the magnitude of the velocity of the ball
vf as it is caught by the person at the
coordinates (xf,yf).
b. What is the angle f?
c. Determine the net work of gravity on the
ball at it moves from the initial to final
positions in its trajectory: .
Solution using conservation of energy :
Ki  U i  K f  U f
1 2
1 2
mvi  0  mv f  mgy f  Solve for v f for (a)
2
2
Note that horizontal velocity is constant :
vi cos  i  v f cos  f
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 Solve for  f for (b)
(c) Work by gravity : W  mg ( y f  yi )
PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- force diagrams
1
2
F2
F1
mg
m
For system in equilibrium :
F  0
i
i
 F1 cos 1  F2 cos  2  0
F1 sin 1  F2 sin  2  mg  0
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PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- pendulum

T- mg cos   0
F=ma 
r
mg sin   ma
T
t=I a  r mg sin  = mr2 a  mra
mg(-j)
Alternatively :
dL
τ
dt


2
d mr 2
2 d 
r mg sin   
 mr
dt
dt 2
d 2
g
g
Pendulum equations :


sin




2
dt
r
r
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PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- driven Harmonic oscillator
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From your questions -- driven Harmonic oscillator
ma  Ftotal
d 2x
m 2  kx  F0 sin t 
dt
General solution :
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 k
 F0 / m
x(t )  A cos
t    
sin t 
 m
 k  2
m
PHY 113 C Fall 2013 -- Lecture 26
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Similar problem from webassign:
Damping is negligible for a 0.165-kg object
hanging from a light, 6.30-N/m spring. A
sinusoidal force with an amplitude of 1.70 N
drives the system. At what frequency will the
force make the object vibrate with an amplitude
of 0.600m?
 k
 F0 / m
x(t )  A cos
t    
sin t 
 m
 k  2
m
F0 / m
In
this
case
:
 0.6
(usually neglected)
k
 2
m
F0
k
2
  
m 0 .6 m
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PHY 113 C Fall 2013 -- Lecture 26
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Examples of two-dimensional collision; balls moving on a
frictionless surface
Suppose : m1  m2  0.06kg , v1i  2m / s,
v2 f  1m / s,
  20o
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
v1 f sin   v2 f sin 
 1m / s sin 20o  0.342m / s
v1 f cos   v1i  v2 f cos 


 2m / s   1m / s  cos 20o  1.060m / s
0.342
   17.88o
1.060
0.342m / s 1.060m / s
v1 f 

 1.11m / s
o
o
sin 17.88
cos17.88
tan  
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PHY 113 C Fall 2013 -- Lecture 26
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Examples of two-dimensional collision; balls moving on a
frictionless surface – energy conservation?
Note: In these collision analyses, we
are neglecting forces and potential
energy
iclicker question
Why?
A. We are cheating physics
B. We are applying the laws of
physics correctly
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PHY 113 C Fall 2013 -- Lecture 26
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Examples of two-dimensional collision; balls moving on a
frictionless surface – energy conservation?
Assuming that we applying the laws of
physics correctly – we can ask the
question – Is (kinetic) energy
conserved?
1
m1v12i  0
2
1
1
2
K f  m1v1 f  m2 v22 f
2
2
If K i  K f  Energy is conserved
Ki 
If
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Ki  K f
 Energy added or lost in process
PHY 113 C Fall 2013 -- Lecture 26
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From your questions -- conservation of angular momentum
L   ri  mi v i
L  I
i
dL
τ
dt
If τ  0, then L is constant
1
m
2
m
m
d1
d2
d1
I1=2md1
2
m
d2
I2=2md22
I11=I22 2=1 I1/I2
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PHY 113 C Fall 2013 -- Lecture 26
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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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PHY 113 C Fall 2013 -- Lecture 26
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An example of the application of torque on a rigid object:
A horizontal 800 N merry-go-round is a solid disc of radius
1.50 m and is started from rest by a constant horizontal force
of 50 N applied tangentially to the cylinder. Find the kinetic
energy of solid cylinder after 3 s.
F
R
K = ½ I 2
t Ia
In this case I = ½ m R2
FR  Ia
  at 
FR
t
I
I
and
  i  at = at
t = FR
1 mg 2
R
2 g
1 2 1  FR 
1 F 2t 2
F 2t 2
2 50 N 
2
K  I  I 
t 

g

9
.
8
m/s
(
3
s
)
 275.625 J
2
2
2  I 
2 I/R
mg
800 N
2
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2
PHY 113 C Fall 2013 -- Lecture 26
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Webassign questions on fluids (Assignment #17)
A hypodermic syringe contains a medicine with the density of
water (see figure below). The barrel of the syringe has a crosssectional area A = 2.40 10-5 m2, and the needle has a crosssectional area a = 1.00 10-8 m2. In the absence of a force on
the plunger, the pressure everywhere is 1.00 atm. A force of
magnitude 2.65 N acts on the plunger, making medicine squirt
horizontally from the needle. Determine the speed of the
medicine as it leaves the needle's tip.
v  v2 
1
2
v12  gy1  P1  12 v22  gy2  P2
2F
  a 2 
A1    
  A 


In this case : y1  y2 ; P1  P2  F / A; av2  Av1
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PHY 113 C Fall 2013 -- Lecture 26
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Send email or come to see me if you have
further questions.
THANKS!
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PHY 113 C Fall 2013 -- Lecture 26
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