Comprehensive Review
Comprehensive Review
a) Exam information
b) What kind of questions?
c) Review
Mechanics Lecture 4, Slide 1
Final Exam


In-class only!!!
Review the following material







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homework problems.
video pre-lectures/textbook.
Lecture slides
Unit Main Points
Multiple choice…but show your work and justification.
Mostly Calculations…”step by step”
Some Conceptual questions…like checkpoint problems.
Bring calculators and up to ten sheets of notes.

It is best to prepare your own hand-written notes!

Derived Equations…(e.g. projectile motion)
Mechanics Lecture 8, Slide 2
General Topics
Kinematics
 Description of Motion
Force
 Dynamics-how objects change velocity
Energy
 Kinetic and Potential
Conservation Laws
 Momentum and Energy
Collisions
 Elastic and In-elastic
Rotations
 Torque/ Angular Momentum/Statics
Mechanics Lecture 8, Slide 3
Problem Solving Techniques
Visualize/Diagram
 “Sketch” problem
 Identify variables, input and what we are trying to solve
 Free-body diagrams
Express in Mathematical Equations




Scalars-1d
Vectors-2d,3d Break into components
System of n-equations with n-unknowns
Use Mathematical tools to solve:
 Quadratic Equation
 Vector operations
 Trigonometry
Conceptual Understanding
 Does answer make sense?
Mechanics Lecture 8, Slide 4
Potential Problem Topics
Projectile Motion
Center of Mass
Relative Motion - 2d
Conservation of
Uniform Circular Motion Momentum
Forces
Collisions






Weight (near earth)
Gravitational (satellite)
Springs
Normal Force
Tension
Friction
Free-Body Diagrams
Work-Kinetic Energy
Potential Energy
 In-elastic
 Elastic
Rotations





Kinematics
Dynamics
Statics
Moment of Inertia
Torque
 Angular Momentum
Mechanics Lecture 8, Slide 5
Relevant Formulae
Mechanics Review 2 , Slide 6
Relevant Formulae
Mechanics Review 2 , Slide 7
Kinematics
Mechanics Lecture 8, Slide 8
Hyperphysics
Motion
Displacement vs timet
Velocity vs timet
Acceleration vs timet
Mechanics Lecture 1, Slide 9
Hyperphysics
Motion
Mechanics Lecture 1, Slide 10
1d-Kinematic Equations for constant acceleration
a (t )  a0
v(t )  a0t  v0
1 2
x(t )  a0t  v0t  x0
2
2
2
(v(t ))  v0 )  2a ( x(t )  x0 )
 Basic Equations to be used
for 1d – kinematic problems.
 Need to apply to each object
separately sometimes with
time offset
 When acceleration changes
from one constant value to
another say a=0 The
problem needs to be broken
down into segments
Mechanics Lecture 1, Slide 11
Ballistic Projectile Motion Quantities
Initial velocity
speed,angle
Maximum Height of trajectory, h=ymax
“Hang Time”
Time of Flight, tf
Range of trajectory, D
Height of trajectory at arbitrary x,t
Mechanics Lecture 2, Slide 12
Derived Projectile Trajectory Equations
Maximum height
v02 sin 2 
h  y0 
2g
Time of Flight (“Hang Time”)
tf 
2v0 y
g

2v0 sin 
g
Range of trajectory
v02 sin 2
D
g
Height of trajectory as f(t) , y(t)
y (t )  y0  v0 y t 
1 2
gt
2
Height of trajectory as f(x), y(x)
 x  1  x 
  g 

y ( x )  v0 sin  
v
cos

2
v
cos

 0

 0

2
Mechanics Lecture 1, Slide 13
Relative Motion in 2 Dimensions
Direction w.r.t shoreline
Speed relative to shore
Mechanics Lecture 3, Slide 14
Uniform Circular Motion
Mechanics Lecture 8, Slide 15
Uniform Circular Motion
Constant speed in circular
path
v2
ac 
R
Acceleration directed toward
center of circle
What is the magnitude of
acceleration?
Proportional to:
1. Speed v
= R
1. time rate of change
of angle or angular
d 2
velocity

dt

T
 2f
Mechanics Lecture 3, Slide 16
Dynamics
Mechanics Lecture 8, Slide 17
Inventory of Forces
 Weight
 Normal Force
 Tension
 Gravitational
 Springs
 …Friction
Mechanics Lecture 5, Slide 18
Mechanics Lecture 5, Slide 19
http://hyperphysics.phy-astr.gsu.edu/hbase/N2st.html#c1
Mechanics Lecture 5, Slide 20
Mechanics Lecture 5, Slide 21
Mechanics Lecture 5, Slide 22
m1m2
mmars msatellite
F G 2 G 2
r
rmarscenter satellite
Mechanics Lecture 5, Slide 23
v  ac r 
m1m2
Fgrav  G 2
r
v 2 Fgrav
ac 

R
m
v
Fgrav
v
Fgrav
msat
msat
Fgrav
msat
r
mmars msat r
r G
r2
msat
mmars
r G
r
Be careful what value you use for r !!! Should be
distance between centers of mass of the two
objects
Mechanics Lecture 5, Slide 24
1) FBD
m2
N
f
m2
T
g
T
m2g
m1
m1
m1g
Mechanics Lecture 6, Slide 25
1) FBD
2) SF=ma
m2
N
T
m2
f
g
T
m2g
N = m2g
T – m m2g = m2a
m1g – T = m1a
m1
m1
m1g
add
m1g – m m2g = m1a + m2a
a=
m1g – m m2g
m 1 + m2
Mechanics Lecture 6, Slide 26
1) FBD
2) SF=ma
m2
N
f
m2
T
g
T
m1
m1
m2g
m1g
a=
m1g – m m2g
m1 + m2
m1g – T = m1a
T = m1g – m1a
T is smaller when a is bigger
Mechanics Lecture 6, Slide 27
Work-Kinetic Energy Theorem
The work done by force F as it acts on an object that
moves between positions r1 and r2 is equal to the
change in the object’s kinetic energy:
But again…!!!

r2
W  K
 
W   F dl

r1
1 2
K  mv
2
Mechanics Lecture 7, Slide 28
Energy Conservation Problems in general
For systems with only conservative forces acting
Emechanical  0
Emechanical is a constant
Emechanical  Ki  Ui  K f  U f  K (t )  U (t )
Mechanics Lecture 8, Slide 29
Determining Motion
Force

Unbalanced Forces  acceleration
(otherwise objects velocity is constant)
Energy

Total Energy  Motion, Location


F  a
 F
a
m


F12  F21

Emechanical  K  U
Emechanical  Wnonconservative

rf

Determine Net Force acting on
object
Work

Wnet


  Fnet  dl  K

Conservative forces r0
U  Wnet  K
 Emechanical  U  K  0

Motion from Energy conservation
Emechanical, final  K f  U f

Use kinematic equations to
determine resulting motion
v f  at  v0 ;...
Emechanical, final  K i  U i  Emechanical
K f  K i  U i  Emechanical  U f
vf 
2
Ki  U i  Emechanical  U f 
m
Mechanics Lecture 8, Slide 30
Friction
“It is what it has to be.”
Mechanics Lecture 8, Slide 31
Block
1 2
2x
at  a  2
2
t
F
m g sin   f k
a  net 
m
m
2x 

f k  m g sin   m a  m g sin   2 
t 

x 
Mechanics Lecture 5, Slide 32
Work & Kinetic Energy
Mechanics Lecture 8, Slide 33
Example Problem
Wtension
 
  T  dl  Tx
Wnet  Wtension  W friction  K
W friction  Wtension  K

1
K  m v 2f  v02
2


1
W friction  Wtension  m v 2f  v02
2

Mechanics Lecture 8, Slide 34
Potential Energy
Mechanics Lecture 8, Slide 35
Example Problems
Emechanical  W friction
W friction   m k m gx
Emechanical, final  Emechanical,initial  W friction
m gh  m ghi  m k m gx  m ghi  m k x 
h  hi  m k x 
Mechanics Lecture 8, Slide 36
Example: Pendulum
vv 22gh
gh
hh
Conserve Energy from initial to final position.
1 2
mgh  mv
2
v  2gh
Mechanics Review 2 , Slide 37
Gravitational Potential Problems
 
r  rE

rE
 
r  rM
 conservation of mechanical energy
can be used to “easily” solve
problems.
Emechanical  K  U 

r

rM
 Add potential energy from each
source.
GM E m
U Earth (rE )  
rE
U Moon (rM )  
1
mv (h) 2  U (h) gravity
2
 Define coordinates: where is
U=0?
U (r )  
GM E m
 0 as r  
r
GM M m
rM

GM E m GM M m
U total (r )       
r  rE
r  rM
Mechanics Lecture 8, Slide 38
Collisions
Center of Mass
 Conservation of
Momentum


Inelastic collisions
Multiple particles, Solid Objects
Isolated system, No external force
Fext 
dPtot
 0  Ptot  0
dt
Non-conservative internal force
K tot  0
Ptot  0

Elastic Collisions
Conservative internal force
Ktot  0
Ptot  0

Impulse and Reference
Frames
Individual Particle changes
momentum due to Force acting over a
given duration
Favg = P/t
Mechanics Lecture 8, Slide 39
Systems of Particles
Mechanics Lecture 8, Slide 40
Example Problem
m x

m
i i
xcm
i
i

1 0  3  0.5  2 1
 0.58
1 3  2
i
m v
0
m
i i
vcm
i
i

m1v1  m2 v2
 0  m1v1  m2 v2
m1  m2
i
2
m 
 m1v1
2
v2 
 v2   1  v12
m2
 m2 
1
m v2
K1 2 1 1
m1v12
m1
m



 2
2
2
K2 1 m v2
m1
 m1  2
 m1 
2 2




m
v
m
2
2
2
 1

 m2 
 m2 
Mechanics Lecture 8, Slide 41
Collisions
Mechanics Lecture 8, Slide 42
Example Problem


v2, f  v1, f

ptotal  0



ptotal,i  m1v1,i  m2 v2,i



ptotal, f  m1v1, f  m2 v2, f




m1v1,i  m2 v2,i  m1v1, f  m2 v2, f

1
m1v1,i  m2v2,i  m1v1, f 
v2 , f 
m2

1
m1v1,i  m2v2,i  m1v2, f 
v2 , f 
m2
(1 

v2 , f
Kf 0
K  3m v2
m1 
1
m1v1,i  m2v2,i 
) v2 , f 
m2
m2



m1v1,i  m2 v2,i  m2v  2m v

 v1, f 

0
m1
1
m2 (1  )
2m(1  )
m2
2
Ki 
1
1
m(2v) 2  2m(v) 2  3m v2
2
2
Mechanics Lecture 8, Slide 43
Example Problem :
vCM 
1
m1v1  m2v2   m1v1
m1  m2 
m1  m2 


m1

v1*,i  v1,i  vCM  v1,i 1 


m

m
1
2 



m1

v1*, f  v1*,i  v1,i 1 


m

m
1
2 

Mechanics Review 2 , Slide 44
Impulse
Mechanics Lecture 8, Slide 45
|Favg | = |P | /t = 2mv cos /t
Mechanics Lecture 13, Slide 46
Rotations

Rotational Kinematics

Moment of Inertia

Torque
Force applied at a lever arm
resulting in angular acceleration

Rotational Dynamics
Newton’s 2nd law for rotations

Rotational Statics
How to ensure stability

Angular Momentum
Vector Quantity describing
object(s) rotation about an axis
Description of motion about a
center of mass
Resistance to changes in angular
velocity
Mechanics Lecture 8, Slide 47
Rotational Kinematics
Mechanics Lecture 8, Slide 48
Rotational Dynamics
Mechanics Lecture 8, Slide 49
Example Problem
1
MR 2
2
1
L
1
I rod ,end   ML2  M ( ) 2  ML2
12
2
3
I hoop  MR 2
I disk 


I
Mechanics Lecture 8, Slide 50
Example Problem
Mechanics Lecture 8, Slide 51
Work & Energy (rotations)
Mechanics Lecture 8, Slide 52
Example Problem
N  Mg cos
Ma  Mg sin   F f
2
MR 2  F f R
5
a

R
2
Ma  F f
5
2
Ma  Mg sin   Ma
5
2
a(1  )  g sin 
5
5
a  g sin 
7
2
Ma
5
2 5
2
F f  M g sin   Mg sin 
5 7
7
Ff 
Mechanics Lecture 8, Slide 53
Statics
Mechanics Lecture 8, Slide 54
Statics Problems
Mechanics Lecture 18, Slide 55
Example Problem
Mechanics Lecture 8, Slide 56
Angular Momentum
Mechanics Lecture 8, Slide 57
Example Problem
Mechanics Lecture 8, Slide 58
Relevant Formulae
Mechanics Review 2 , Slide 59
Relevant Formulae
Mechanics Review 2 , Slide 60