Download V 0 - BYU Physics and Astronomy

Physics 121
Mechanics
Instructor
Karine Chesnel
April 14, 2009
Final Review
Lecture notes are posted on
www.physics.byu.edu/faculty/chesnel/physics121.aspx
Final review
04/14/09
Quiz # 40
Have you evaluated the class on Route Y?
YES!
Click!
Do not forget to evaluate this class
online until April 15th
Thank you!
Physics 121 Winter 2009
* Today *
last lecture
Last assignment:
• Online homework 24: (70pts)
until midnight today!
http://gardner.byu.edu/121w2/homework.html
For any question after this class:
Karine Chesnel
N319 ESC
801-422-5687
[email protected]
I will be available until Friday April 24th
Physics 121 Winter 2009 - section 2
Class statistics
Class Average
First test
Second test
Third test
83 /100
68/100
74/100
Test average
Final
Homework
Labs
quizzes
75/100
85/100
96/100
113/100
(30%)
(20%)
(25%)
(15%)
(10%)
Prepare well for the FINAL test!
Try to increase your average test score
Final exam
• Friday April 17 through Wednesday April 22
• At the testing center : 8 am – 9 pm (Mo- Fri)
10 am – 4 pm (Sat)
• Closed Book
• Only bring: - Phys 121 (Chesnel)
Memorization sheets (5pages)
- Math reference sheet
- Pen / pencil
- Calculator
- your CID
• No time limit
• Multiple Choice Questions: 30 questions
Final Review
ch 1 – ch 15
Part I: Kinematics
Ch. 1- 4
Part II: Laws of Motion
(material points)
Ch. 5- 8
Part III: Laws of Motion
(Solids rotation)
Ch. 9 - 13
Part IV: Oscillatory motion
Ch. 15
Part I
Kinematics
Ch. 1 Physics & measurments
• Standard units
• Dimensional analysis
Ch. 2 Motion in one dimension
• Position, velocity, acceleration
• Case of constant velocity
• Case of constant acceleration
• Free falling motion
Ch. 3 Vectors
• Coordinate systems
• Algebra
Ch. 4 Motion in two dimensions
• Position, velocity, acceleration
• Case of constant acceleration
• Projectile motion
• Circular motion (uniform & non-uniform)
• Tangential and radial acceleration
Final review
04/14/09
Kinematics
The displacement is the difference between two positions
x1
Dx= x2 – x1
x2
Average velocity
(Instantaneous)
velocity
Vavg
Dx

Dt
V x ,t
 Dx  dx
 lim   
Dt  0 Dt
  dt
The speed is the amplitude of the velocity
Average acceleration
(Instantaneous)
acceleration
aavg
DV

Dt
a x ,t
 DV
 lim 
Dt 0 Dt

 dV

 dt
Final review
04/14/09
Motion under constant velocity
V = constant
= Slope Dx/Dt
x(t)
Dx
x0
0
Dt
x(t) = x0 + v.t
Position at t=0
t
Final review
04/14/09
Motion under constant acceleration
a(t)
a = constant
0
t
v(t)
V(t) is linear
Dv
v(t) = v0 + a.t
v0
0
Dt
x(t)
t
x(t) is parabolic
x(t) = x0 +v0.t+ ½ a.t2
x0
t
0
Application: Free Fall
a=g
Final review
04/14/09
Vectors components
Cartesian to polar conversion in 2D
y
A=(x,y)
y


(0,0)
x
x
x   cos( )
y   sin(  )
  x2  y2
y
tan  
x
Final review
04/14/09
Projectile Motion
vertical
y
V1
V0
r0
(0,0)
a = g = (0, -g)
V2
V3
horizontal x
v(t) = v0 + gt
r(t) = r0 + V0t + ½ g t2
x(t) = x0 + V0,x t
y(t) = y0 + V0,y t - ½ gt2
Uniform velocity
along x
Free falling
along y
Final
review in two dimensions
Ch.4 Motion
Review
04/14/09
1/27/09
Projectile
Performances
Vmax
y
a=g
H
V0
(0,0)

Hits the ground
R
x
The particle is projected with a speed V0 at angle 
The maximal height is
H = (V0 sin )2 / 2g
The horizontal range is
R = V02 sin (2) /g
Final review
04/14/09
Uniform Circular Motion
y
V
R0

Angular speed is constant:
 = w.t
a
(0,0)
x
Position
x = R0 cos (wt)
y = R0 sin (wt)
Velocity:
Vx = - R0 w sin (wt)
Vy = R0 w cos (wt)
Acceleration
ax = - R0 w2 cos (wt)
ay = - R0 w2 sin (wt)
|V| = R0 w
a = - w2 r
The acceleration is
centripetal.
Its magnitude is
|a| = R0 w2
Final review
04/14/09
Tangential and radial
acceleration
General case
V3
V1
a
a
a
V2
V is tangential to the trajectory
The components of the acceleration in the Frenet frame are:
• Tangential
at= dV/dt
The sign tells if the particle
speeds up or slows down
• centripetal
ac=
V2/R
Always directed toward
the center of curvature
R radius of curvature
Final review
04/14/09
Generalization
t t



2
r (t )   V (t ).dt    a (t ).dt
t
0
r (t)
0 0
position
First derivative
Second
integration
 dr
V 
dt
 t 
V   a.dt
velocity
0
Second
derivative
First integration
acceleration

2
 dV d r
a
 2
dt
dt
a(t)
Final review
04/14/09
Quiz # 41
Two racquet balls are thrown in the air at the same time
from the same height H. One ball (yellow) is thrown at some angle ,
with a vertical velocity V0,y, and horizontal velocity V0,x.
The other ball (blue) is thrown vertically
with the same vertical velocity V0,y
y
Which ball will hit the floor first?
V0,y
V0
H
A
B
C
The blue ball
The yellow ball
Both of them
For the yellow ball the motion will be given by
x(t) = x0 + V0,x t
y(t) = y0 + V0,y t - ½ gt2
For the blue ball the motion will be given by
y(t) = y0 + V0,y t - ½ gt2
Both balls touch the ground at the same time!
Part II
Laws of motion
Ch. 5 The Laws of Motion
• Newton’s first law
• Newton’s second law
• Newton’s third law
Ch. 6 Newton’s laws applications
• Circular Motion
• Drag forces and viscosity
• Friction
• Fictitious forces
Ch. 7 Work and energy
• Work
• Kinetic energy
• Potential energy
• Work- kinetic energy theorem
Ch. 8 Conservation of Energy
• Mechanical energy
• Conservation of energy
Final review
04/14/09
Summary of
the Laws of Motions
First Law: Principle of Inertia
In a inertial frame,
an isolated system remains
at constant velocity or at rest
Second Law: Forces and motion
In an inertial frame
the acceleration of a system
is equal to the sum of
all external forces
divided by the system mass

a

F
m


ma   F
Third Law: Action and reaction
If two objects interact,
the force exerted by object 1 on object 2
is equal in magnitude and opposite in direction
to the force exerted by object 2 on object 1.
F1
F2
Final review
04/14/09
Forces of Friction
f
Two regimes

f
Static regime
F
Kinetic regime
f max
fk

F
• Static regime
F < fs,max= ms N
ms is called the coefficient of static friction
• Kinetic regime
fk= mk N
mk is called the coefficient of kinetic friction
Final review
04/14/09
Work and kinetic energy
Using Newton’s second law


 F  ma
WAB  
B
A
 
B
 
F .dr   ma.dr
A
WAB  DK  K B  K A
Work- Kinetic energy theorem
Defining the
kinetic energy
1
K  mV 2
2
Final review
04/14/09
Mechanical energy
DK  Wcons  Wnc
DK  DU  Wnc
DK  DU  Wnc
D( K  U )  Wnc
We define the mechanical energy Emech
as the sum of kinetic and potential energies
Emech  K  U
DEmech  Wnc
Sometimes, the work of non conservatives forces (friction, collision)
is transformed into internal energy
Wnc  DEint
thus
DEmech  DEint  0
Final review
04/14/09
Closed System with
conservative forces only
Fcons
There are no non-conservative forces working
DEmech  Wnc  0
The mechanical energy is constant
Emech  cst
K  U  cst
The mechanical energy is conserved
between initial and final points
K f  U f  Ki  U i
Final review
04/14/09
Examples of
Potential energy
Spring
x
1 2
U k  kx
2
L0
Gravity
U g  mgh
h
g
Gravitational field
m
r
M
mM
U g  G
r
Final review
04/14/09
Quiz # 42
This man is pushing this box of 85kg on the carpet
at a constant speed.
How is the magnitude of the force he needs to apply?
A
B
C
D
E
F
N
`
f
mg
Larger than the weight of the box
Same than the weight of the box
Same than the friction force
Larger than the friction force
None of the answers
F


According to Newton’s law
 F  ma
 
Here the velocity is constant, so
F  0


This is true both vertically
N   mg


and horizontally
F f
and on the other hand


Thus

F  f
f  k N  k mg  mg
Part III
Laws of motion for Solids
Ch. 9 Linear Momentum & collision
• Linear momentum
• Impulse
• Collisions 1D and 2D
Ch. 10 Rotation of solid
• Rotational kinematics
• Rotational and translational quantities
• Rolling motion
• Torque
Ch. 11 Angular momentum
• Angular momentum
• Newton’s law for rotation
• Conservation of angular momentum
• Precession motion
Ch. 12 Static equilibrium and elasticity
• Rigid object in equilibrium
• Elastic properties of solid
Ch. 13 Universal gravitation
• Newton’s law of Universal gravitation
• Gravitational Field & potential energy
• Kepler’s laws and motion of Planets
Final review
04/14/09
Linear Momentum & Impulse
• The linear momentum of a particle is the product of its mass by its velocity


p  mV
• The Newton’s second law
can now be written as
• For an isolated system
Units: kg.m/s or N.s


dp
F
dt
 
dp
0
dt

p  cst
• The impulse is the integral of the net force, during
an abrupt interaction
in a short time


f
I    Fdt
i
• According to Newton’s 2nd law:
 
Dp  I
Final review
04/14/09
Collisions
V1,i
V1
,f
V2,f
V2,i
Elastic collision
1. Conservation of linear momentum




p1, f  p2, f  p1,i  p2,i




m1V1, f  m2V2, f  m1V1,i  m2V2,i
(1)
2. Conservation of kinetic energy
1  2 1  2 1  2 1  2
m1V1, f  m2V2, f  m1V1,i  m2V2,i
2
2
2
2
Inelastic collision: change in kinetic energy
Perfectly inelastic collision: the particles stick together
(2)
Final review
04/14/09
Collisions 1D
• Combining (1) and (2), we get expression for final speeds:
V1,i
V2,i
V1,f
V2,f
V1, f
m1  m2
2m2

V1,i 
V2,i
m1  m2
m1  m2
V2, f
m2  m1
2m1

V2,i 
V1,i
m1  m2
m1  m2
• If one of the objects is initially at rest:
V1, f
m1  m2

V1,i
m1  m2
V2, f
2m1

V1,i
m1  m2
Collisions 2D
y
• 3 equations
V1,f
V1,i
1
2
V2,f
• 4 unknow parameters
x
Final review
04/14/09
Solid characteristics
• The center of mass is defined as:

M OC   r dm
dm  dV


ptot  MVC
C
dm
r
O

 
MaC  F
• The moment of inertia of the solid about one axis:
I
I   r dm
I’
2
D
I '  I  MD
2
Final review
04/14/09
Rotational kinematics
• Solid’s rotation
Angular position 
Angular speed w
d
w
dt
Angular acceleration 

dw
dt
• Linear/angular relationship
For any point in the solid
Velocity
Acceleration
Vt  Rw
• Tangential
at  R
• Centripetal
aC  Rw 2
• Rotational kinetic energy
1
K  I .w 2
2
Final review
04/14/09
Motion of rolling solid
w
C
R
P
• The kinetic energy of the solid is given by the sum
of the translational and rotational components:
Ksolid = Kc + Krot
Non- sliding
situation
K solid
1 2 1 2
 MVC  Iw
2
2
K solid
1
1 2
2 2
 MR w  Iw
2
2
K solid
1
 ( MR 2  I )w 2
2
If all the forces are conservative:
U  K sol  cst
Final review
04/14/09
Torque & angular momentum
When a force is inducing the rotation
of a solid about a specific axis:
We define the torque
F
 
  r F




  rF sin 
The angular momentum is defined as
  
Lrp
angular
momentum
Linear momentum
Deriving Newton’s second law in rotation

 dL
  dt

 net  I

L  Iw
For an object in pure rotation
Final review
04/14/09
Solving a problem
Static equilibrium
• Define the system
• Identify and list all the forces
• Apply the equality
 
F  0
• Locate the center of mass (where gravity is applied)
• Choose a convenient point to calculate the torque
(you may choose the point at which most
of the forces are applied, so their torque is zero)
• List all the torques applied on the same point.
• Apply the equality

  0

Final review
04/14/09
Gravitational laws
Any material object is producing a gravitational field
M
 
M 
g M (r )  G 2 ur
r
ur
r
Fg
m
Any object placed in that field experiences a gravitational force

 
mM 
Fg  mg M (r )  G 2 ur
r
The gravitational field created
by a spherical object is centripetal
(field line is directed toward the center)
Ug
0
The gravitational potential energy is
r
mM
U g  G
r
Final review
04/14/09
Kepler’s Laws
First Law
“The orbit of each planet in the solar system is an ellipse
with the Sun as one focus ”


L  cst  L0
Second Law
“The line joining a planet to the sun sweeps out equal areas
during equal time intervals as the planet travels along its orbit.”
dA  L0 
cst
dt 2 m
Third Law
“The square or the orbital period of any planet is proportional
to the cube of the semimajor axis of the orbit”
2

4

T 2  
 GM S
 3
 R

Final review
04/14/09
Satellite Motion
Geostationary
orbit
Tsat =TEarth
= 1 day
From Newton’s law
• Satellite speed
VGS
2RGS

T
(1)
The mechanical energy of
The satellite on orbit is
VGS
2
ME
G
RGS
(2)
mM E
E  G
2 RGS
Escape speed
Vesc
2GM

R
Final review
04/14/09
Quiz # 43
A planet has a mass twice the mass of the Earth
and a diameter 0.7 times the Earth diameter.
What would be the weight, in Newtons, of a 82kg person
standing at the surface of this planet?
A
B
C
D
E
334 N
803 N
3280N
483 N
4830 N
The weight of this person at the surface of this planet is
Fp  mg p  G
mM p
Rp
2
Compared to the weight on earth
M p  RE 


Fp  mg E
M E  R p 
2
 1 
 82  9.8  2  

 0.7 
2
 3280 N
9.8 m/s2
Gravitational field on Earth
(Equivalent “mass’
on earth: 334kg!)
Part IV
Oscillatory motion
(ch15)
• Harmonic equation and solutions
• Energy of harmonic oscillator
• Spring motion
• Pendulum motions
• Damped oscillation
• Forced oscillation
Final review
04/14/09
Harmonic motion
Spring
F
x
d 2x k
 x0
2
dt
m
Equation of
the motion is
Harmonic
equation
A general solution to this harmonic equation is
x(t )  A cos(wt   )
Amplitude
angular
frequency
Phase constant
(phase at t=0)
with
k
w
m
Unit = rad/s
frequency
1
f 
2
Unit = Hz
k
m
period
m
T  2
k
Unit = s
Final review
04/14/09
Position, velocity and acceleration
Position
Velocity
x(t )  A cos(wt   )
dx
V (t ) 
  Aw sin( wt   )
dt
dV
Acceleration a (t ) 
  Aw 2 cos(wt   )
dt
Position, velocity and acceleration are all sinusoidal functions
T
x(t)
t
V(t)
Phase
quadrature
a(t)
t
Opposite
phase
t
Final review
04/14/09
Energy
Emech
1 2
 K  U  kA  cst
2
U
K
0
t
U
K
0
x
Final review
04/14/09
The pendulum
(punctual)
0
The equation for motion is
d 2 g
  0
2
dt
L

L
The solution for the angle position is
 (t )   max cos(wt   )
m
with
w
The oscillation frequency
does not depend on the mass m
The period of the oscillation is
T
2
w
 2
L
g
g
L
Final review
04/14/09
Torsional pendulum
A torsional pendulum uses the torque
induced by torsion to oscillate
w
For an angular displacement ,
the torque is
  k
L
According to
Newton’s second law

0 (rest)

The equation for
the motion is then

 dL
 
dt
dL
d 2
 k 
I 2
dt
dt
d 2 k
  0
2
dt
I
A general solution is:
 (t )   max cos(wt   )
with
k
w
I
Final review
04/14/09
Damped oscillations
If the oscillator is moving in a resistive medium:
friction, viscosity…. The oscillation will be damped.
L0
F
A
0
Compressed
x
The expression of the spring force is
The expression of resistive force is


Fk  kxux


Fd  bV
Applying the Newton’s second law

 
ma  Fg  Fd
d 2x
dx
m 2  kx  b
dt
dt
Damping
coefficient
Equation of
the motion
Final review
04/14/09
Damped oscillations
The general solution for the motion is
 b 
x(t )  A exp  
t  cos(wt   )
 2m 
with
 b 
w  w02  

 2m 
2
x(t)
Underdamped
b
 w0
2m
Critical
t
x(t)
b
 w0
2m
T
w 0
T 
t
overdamped
x(t)
b
 w0
2m
t
Final review
04/14/09
Forced oscillations
An external force is applied to the system,
forcing the oscillation to a frequency w
The external force is
The spring force is
If any, resistive force is


Fosc  F0 sin wt )


Fk  kxux


Fd  bV
ma  Fosc  Fg  Fd
d 2x
dx
m 2  F0 sin wt )  b  kx
dt
dt
A general solution to this equation is
x(t )  A cos(wt   )
Amplitude
Frequency
forced
Phase constant
(phase at t=0)
Final review
04/14/09
Forced oscillations
A general solution to a forced oscillation motion
x(t )  A cos(wt   )
F0
with
A
w
2
 w0
m
)
2 2
 bw 


m


2
and
w0 
k
m
Resonance
A
No damping b=0
Low damping b
Large damping b
w0
w
In absence of resistive forces, the amplitude of the oscillation is
amplified to infinity when the force frequency w
approaches the proper frequency w0
Final review
04/14/09
GOOD LUCK
With the FINALS !
To contact me:
Karine CHESNEL
[email protected]
Office: N319 ESC
801 – 422 – 5687