#### Transcript Document

```PHY 113 C General Physics I
11 AM-12:15 PM MWF Olin 101
Plan for Lecture 9:
1. Review (Chapters 1-8)
3. Example problems
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
1
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
2
iclicker question
What is the best way to prepare for Thursday’s exam?
A. Read Lecture Notes and also reread Chapters 18 in Serway and Jewett.
B. Prepare equation sheet.
C. Solve problems from previous exams.
D. Solve homework assignments (both graded and
E. All the above
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
3
iclicker question
Have you (yet) accessed the online class
lecture notes from previous classes?
A. yes
B. no
iclicker question
Have you (yet) accessed the passed Webassign
Assignments (with or without the answer key)?
A. yes
B. no
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
4
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
5
Previous exam access -- continued
Overlap with
2013 schedule
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
6
Comments on preparation for next Thursday’s exam –
continued
What you should bring to the exam (in addition to your
well-rested brain):
 A pencil or pen
 An 8.5”x11” sheet of paper with your favorite
equations (to be turned in together with the exam)
What you should NOT use during the exam
 Electronic devices (cell phone, laptop, etc.)
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
7
F  ma
dv
a
dt
dr
v
dt
Problem solving skills
Math skills
Equation Sheet
1. Keep basic concepts and equations at the top of your head.
2. Practice problem solving and math skills
3. Develop an equation sheet that you can consult.
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
8
iclicker exercise
Does the previous slide annoy you?
A. yes
B. no
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
9
Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle(s) apply
3. Write down the appropriate equations using the variables
defined in step 1.
4. Check whether you have the correct amount of
information to solve the problem (same number of
knowns and unknowns).
5. Solve the equations.
magnitude, etc.).
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
10
Likely exam format (example from previous exam)
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
11
Likely exam format (example from previous exam)
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
12
Review of slides from previous lectures
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
13
Mathematics Review -- Appendix B Serwey & Jewett
ax 2  bx  c  0
c
 b  b 2  4ac
x
2a
a
b
Differential calculus :
d n
at  ant n 1
dt
d t
e   e t
dt
d
sin( t )   cos( t )
dt
9/24/2013
q
Integral calculus :



Trigonometry :
b
cos q 
c
a
sin q 
c
a
tan q 
b
n 1
at
at n dt 
n 1
1
et dt  et

sin( t )dt  
PHY 113 C Fall 2013-- Lecture 9
1

cos( t )
14
One dimensional motion -Summary of relationships
dx
v(t ) 
dt
dv
a (t ) 
dt
9/24/2013

t
x(t )   v(t ' )dt '
t0
t

v(t )   a (t ' )dt '
t0
PHY 113 C Fall 2013-- Lecture 9
15
Special relationships between t,x,v,a for constant a:
General relationship :
dx
v(t ) 
dt
dv
a (t ) 
dt
t

x(t )   v(t ' )dt '
t0
t

v(t )   a (t ' )dt '
t0
Special relationship :
v(t )  v0   at  v0  at
1 2
1 2
x(t )  x0   v0 t  at  x0  v0t  at
2
2
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
16
Introduction
of vectors
a+b
b
a
Vector subtraction:
a
a–b
-b
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
17
Treatment of vectors in
component form
by
b  bx xˆ  by yˆ
ab  c
bx
ay
a  ax xˆ  ay yˆ
ax
For a  a x xˆ  a y yˆ and b  bx xˆ  by yˆ
a  b  a x  bx xˆ  a y  by yˆ  c
c  c x2  c y2
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
18
Vectors relevant to motion in two dimenstions
Displacement: r(t) = x(t) i + y(t) j
Velocity: v(t) = vx(t) i + vy(t) j
vx 
Acceleration: a(t) = ax(t) i + ay(t) j
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
dx
dt
dv
ax  x
dt
vy 
dy
dt
ay 
dv y
dt
19
Visualization of the position vector r(t) of a particle
r(t1)
9/24/2013
r(t2)
PHY 113 C Fall 2013-- Lecture 9
20
Visualization of the velocity vector v(t) of a particle
dr
r (t 2 )  r (t1 )
vt  
 lim
dt t2 t1 0 t 2  t1
v(t)
r(t2)
r(t1)
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
21
Visualization of the acceleration vector a(t) of a particle
dv
v (t 2 )  v (t1 )
at  
 lim
dt t2 t1 0 t 2  t1
v(t2)
v(t1)
r(t2)
r(t1)
9/24/2013
a(t1)
PHY 113 C Fall 2013-- Lecture 9
22
Projectile motion (near earth’s surface)
j
vertical direction (up)
r (t )  x(t )ˆi  y (t )ˆj
v (t )  v (t )ˆi  v (t )ˆj
x
y
a(t )   gˆj
g = 9.8 m/s2
i
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
horizontal
direction
23
Projectile motion (near earth’s surface)
r (t )  x(t )ˆi  y (t )ˆj
dr
v (t ) 
 v x (t )ˆi  v y (t )ˆj
dt
dv
a(t ) 
  gˆj
dt
 vt   v  gtˆj note that vt  0   v
i
i
1 2ˆ
 r t   ri  v i t  gt j note that r t  0   ri
2
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
24
Projectile motion (near earth’s surface)
Trajectory equation in vector form:
r t   ri  v i t  gt ˆj
1
2
2
v (t )  v i  gtˆj
Trajectory equation in component form:
xt   xi  v xi t
v x (t )  v xi
2
1


y t  yi  v yi t  2 gt
v y (t )  v yi  gt
Aside: The equations for position and velocity written
in this way are call “parametric” equations. They are
related to each other through the time parameter.
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
25
Diagram of various trajectories reaching the same
height h=1 m:
y
q
x
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
26
Projectile motion (near earth’s surface)
Trajectory equation in component form:
xt   xi  v xi t  xi  vi cos q i t
y t   yi  v yi t  gt  yi  vi sin q i t  gt
v x (t )  v xi  vi cos q i
1
2
2
1
2
2
v y (t )  v yi  gt  vi sin q i  gt
Trajectory path y(x); eliminating t from the equations:
x  xi
t
vi cos q i


x  xi
1  x  xi 
y  x   yi  vi sin q i
 2 g

vi cos q i
v
cos
q
i 
 i
 x  xi 

y  x   yi  tan q i  x  xi   g 
vi cos q i 

9/24/2013
PHY 113 C Fall 2013-- Lecture 9
2
1
2
27
2
Isaac Newton, English physicist and mathematician
(1642—1727)
1. In the absence of a
net force, an object
remains at constant
velocity or at rest.
2. In the presence of a
net force F, the
motion of an object of
mass m is described
by the form F=ma.
3. F12 =– F21.
http://www.newton.ac.uk/newton.html
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
28
Newton’s second law
F=ma
Types of forces:
Fundamental
Approximate
Gravitational
F=-mg j
Empirical
Friction
Electrical
Support
Magnetic
Elastic
Elementary
particles
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
29
Example of two dimensional motion on a frictionless
horizontal surface
F1  F2  ma
F1  F2
a
m
m
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
30
Example – support forces
Fg  mgyˆ
Fsupport  Fapplied
Fsupport acts in direction  to surface
(in direction of surface “normal”)
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
31
Example of forces in equilibrium
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
32
Example: 2-dimensional forces
A car of mass m is on an icy (frictionless)
driveway, inclined at an angle t as shown.
Determine its acceleration.
Conveniently tilted
coordinate system:
Along y : n  mg cos q  0
Along x : mg sin q  ma x
a x  g sin q
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
33
Another example:
Note: we are using a tilted coordinate frame
i
vf=0
vi
mg
Along incline : x(t )  xi  vi t  12 g sin q t 2
v(t )  vi  g sin q t
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
34
Friction forces
The term “friction” is used to describe the category of forces
that oppose motion. One example is surface friction which
acts on two touching solid objects. Another example is air
friction. There are several reasonable models to quantify
these phenomena.
 Fapplied
Surface friction: f  
Normal force between
 N
surfaces
Material-dependent
coefficient
at low speed
 Kv
D
Air friction:
2
at high speed
 K v
K and K’ are materials and
shape dependent constants
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
35
surface friction force
Models of surface friction forces
fs,max=sn
(applied force)
Coefficients s , k depend on the surfaces; usually, s > k
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
36
Consider a stationary block on an incline:
n  mg cos q  0  n  mg cos q
n
f  mg sin q  0  f  mg sin q
If f  f S ,max   S n   S mg cos q
Then  S mg cos q  mg sin q
 S  tan q
q mg cos q
mg
q
9/24/2013
mg sin q
PHY 113 C Fall 2013-- Lecture 9
37
Consider a block sliding down an inclined surface;
constant velocity case
n  mg cos q  0  n  mg cos q
n
f  mg sin q  0  f  mg sin q
If f   K n   K mg cos q
Then  K mg cos q  mg sin q
f=kn
 K  tan q
mg
q
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
38
Summary
 K  tan q when block moves at constant velocity
 S  tan q when block is just about to slip
n
q
mg cos q
mg
q
9/24/2013
mg sin q
PHY 113 C Fall 2013-- Lecture 9
39
Uniform circular motion and Newton’s second law
r
F  ma
v2
a c   rˆ
r
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
40
F
Definition of work:
dr
ri
rj
rf
Wi f   F  dr
ri
Units of work :
Work  Newtons meters  Joules
1 J  0.239 cal
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
41
Example:
rf
xf
ri
xi
Wi  f   F  dr   Fx dx  (5 N )(4m)  12 5 N 2m   25 J
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
42
Work and potential energy
rf
Definit ionof work :
Wi  f   F  dr
ri
Definition of potential energy :
r
U r   Wref r    F  dr
rref
Note: It is assumed that F is conservative
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
43
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
9/24/2013
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
PHY 113 C Fall 2013-- Lecture 9
44
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
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
45
Example problem from Webassign #8
A baseball outfielder throws a 0.150-kg baseball at a speed of
37.2 m/s and an initial angle of 31.0°. What is the kinetic energy
of the baseball at the highest point of its trajectory?
vf
yf
vi
qi
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
46
Example problem from Webassign #8
f
h
h
The coefficient of friction between the
block of mass m1 = 3.00 kg and the surface
in the figure below is μk = 0.440. The system
starts from rest. What is the speed of the
ball of mass m2 = 5.00 kg when it has fallen
a distance h = 1.85 m?
dissipative
K f  U f  K i  U i  Wi 
f
dissipative
Wi 
  fh    k m1 gh
f
dissipative
K f  K i  U i  U f   Wi 
f
1
m1  m2 v 2f  0  m2 gh   k m1 gh
2
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
47
Example problem from Webassign #8
A block of mass m = 3.40 kg is
released from rest from point A and
slides on the frictionless track
shown in the figure below. (Let ha =
6.70 m.)
.
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
48
Example problem
from 2012 Exam #2:
conservative
Wi 
f
 U f  U i 
 U x f   U  xi 
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
49
Example problem from Webassign #5
 T1 cos q1  T2 cos q 2  0
T1 sin q1  T2 sin q 2  T3  0
T3  m1 g  0
9/24/2013
PHY 113 C Fall 2013-- Lecture 9
50
```