Physics 121
Mechanics
Lecture notes are posted on
www.physics.byu.edu/faculty/chesnel/physics121.aspx
InstructorKarine Chesnel
April 14, 2009
Final Review
Have you evaluated the class on Route Y?
YES! Click!
Do not forget to evaluate this classonline until April 15th
Thank you!
Quiz # 40
Final review 04/14/09
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 Average
First test 83 /100Second test 68/100Third test 74/100
Test average 75/100 (30%)Final (20%)Homework 85/100 (25%)Labs 96/100 (15%)quizzes 113/100 (10%)
Prepare well for the FINAL test!Try to increase your average test score
Class statistics
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. 2 Motion in one dimension• Position, velocity, acceleration• Case of constant velocity• Case of constant acceleration• Free falling motion
Ch. 4 Motion in two dimensions• Position, velocity, acceleration• Case of constant acceleration• Projectile motion• Circular motion (uniform & non-uniform)• Tangential and radial acceleration
Ch. 3 Vectors• Coordinate systems• Algebra
Ch. 1 Physics & measurments• Standard units• Dimensional analysis
Kinematics
The displacement is the difference between two positions
x1
x2
x= x2 – x1
Average accelerationt
Vaavg
(Instantaneous) acceleration dt
dV
t
Va
ttx
0
, lim
t
xVavg
Average velocity
(Instantaneous) velocity dt
dx
t
xV
ttx
0
, lim
The speed is the amplitude of the velocity
Final review 04/14/09
Motion under constant velocity
V = constant
x(t)
tt
x
0
x0
= Slope x/t
Position at t=0
x(t) = x0 + v.t
Final review 04/14/09
Motion under constant acceleration
a = constant
a(t)
t0
v(t) = v0 + a.t
v(t)
tt
v
0v0
V(t) is linear
x(t)
t0
x0
x(t) = x0 +v0.t+ ½ a.t2
x(t) is parabolic
Application: Free Fall a = g
Final review 04/14/09
Vectors components
Cartesian to polar conversion in 2D
x
y
yx
tan
22
)sin(
)cos(
y
x
x
y
(0,0)
A=(x,y)
x
y
Final review 04/14/09
Projectile Motion
r(t) = r0 + V0t + ½ g t2
r0
V0
y
x(0,0)
a = g = (0, -g)
V2
V1
V3
x(t) = x0 + V0,x t
y(t) = y0 + V0,y t - ½ gt2
v(t) = v0 + gt
Uniform velocity along x
Free falling along y
vertical
horizontal
Final review 04/14/09
Projectile
V0
y
x(0,0)
a = g
Vmax
Performances
H
Hits the ground
R
R = V02
sin (2 /g
The particle is projected with a speed V0 at angle
The maximal height is
The horizontal range is
H = (V0 sin )2 / 2g
ReviewCh.4 Motion in two dimensions 1/27/09Final review 04/14/09
Uniform Circular Motion
Angular speed is constant:
= .t
Velocity:
Vx = - R0 sin (tVy = R0 cos (t
Position
x = R0 cos (ty = R0 sin (t
Acceleration
ax = - R0 cos (tay = - R0 sin (t
a = - r
|V| = R0
The acceleration is centripetal. Its magnitude is
|a| = R0
(0,0) x
y
R0
V
a
Final review 04/14/09
Tangential and radialacceleration
General case
V1
V2
V3
a
a a
V is tangential to the trajectory
The components of the acceleration in the Frenet frame are:
• TangentialThe sign tells if the particlespeeds up or slows downat= dV/dt
• centripetalAlways directed toward the center of curvatureR radius of curvature
ac= V2/R
Final review 04/14/09
Generalization
positionr (t)
velocity
First derivative
dt
rdV
acceleration
Second derivative
2
2
dt
rd
dt
Vda
First integration
t
dtaV0
.
Second integration
t tt
dttadttVtr0 0
2
0
).().()(
a(t)
Final review 04/14/09
Quiz # 41Two 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 Which ball will hit the floor first?
A The blue ballB The yellow ballC Both of them
V0,yV0
x(t) = x0 + V0,x t
y(t) = y0 + V0,y t - ½ gt2
For the yellow ball the motion will be given by
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!
y
H
Final review 04/14/09
Part II
Laws of motion
Ch. 6 Newton’s laws applications• Circular Motion• Drag forces and viscosity• Friction• Fictitious forces
Ch. 8 Conservation of Energy• Mechanical energy• Conservation of energy
Ch. 7 Work and energy• Work• Kinetic energy• Potential energy• Work- kinetic energy theorem
Ch. 5 The Laws of Motion• Newton’s first law• Newton’s second law• Newton’s third law
Summary ofthe Laws of Motions
First Law: Principle of InertiaIn a inertial frame,
an isolated system remains at constant velocity or at rest
Second Law: Forces and motionIn an inertial frame
the acceleration of a systemis equal to the sum of
all external forcesdivided by the system mass
Fam
m
Fa
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 < fs,max= s N
s is called the coefficient of static friction
• Static regime
fk= k N
k is called the coefficient of kinetic friction
• Kinetic regime
F
f
Static regime Kinetic regime
F
f
maxf
kf
Two regimes
Final review 04/14/09
amF
Work and kinetic energy
B
A
B
ABA rdamrdFW
..
Defining the kinetic energy
2
2
1mVK
Using Newton’s second law
Work- Kinetic energy theorem
ABBA KKKW
Final review 04/14/09
Mechanical energy
nccons WWK
ncmech WE
ncWUK
ncWUK
ncWUK )(
UKEmech
We define the mechanical energy Emech
as the sum of kinetic and potential energies
Sometimes, the work of non conservatives forces (friction, collision)is transformed into internal energy
intEWnc 0int EEmechthus
Final review 04/14/09
Closed System with conservative forces only
0 ncmech WE
Fcons
cstEmech
cstUK
There are no non-conservative forces working
The mechanical energy is constant
iiff UKUK
The mechanical energy is conserved between initial and final points
Final review 04/14/09
Examples ofPotential energy
2
2
1kxU k x
L0
Spring
Gravity
g
mghU g h
r
mMGU g
Gravitational fieldm
r
M
Final review 04/14/09
Quiz # 42
`F
mg
N
F
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 Larger than the weight of the boxB Same than the weight of the boxC Same than the friction forceD Larger than the friction forceE None of the answers
amF
According to Newton’s law
Here the velocity is constant, so 0
F
fF
Thus and on the other hand
mgmgNf kk
This is true both vertically
and horizontally
gmN
fF
f
Final review 04/14/09
Part III
Laws of motion for Solids
Ch. 10 Rotation of solid• Rotational kinematics• Rotational and translational quantities• Rolling motion• Torque
Ch. 12 Static equilibrium and elasticity• Rigid object in equilibrium• Elastic properties of solid
Ch. 11 Angular momentum• Angular momentum• Newton’s law for rotation• Conservation of angular momentum• Precession motion
Ch. 9 Linear Momentum & collision• Linear momentum• Impulse• Collisions 1D and 2D
Ch. 13 Universal gravitation• Newton’s law of Universal gravitation• Gravitational Field & potential energy • Kepler’s laws and motion of Planets
Linear Momentum & Impulse
• The Newton’s second law can now be written as F
dt
pd
Vmp
• The linear momentum of a particle is the product of its mass by its velocity
Units: kg.m/s or N.s
cstp
• For an isolated system 0
dt
pd
• The impulse is the integral of the net force, during an abrupt interaction in a short time
f
idtFI
Ip
• According to Newton’s 2nd law:
Final review 04/14/09
Collisions
iiff pppp ,2,1,2,1
1. Conservation of linear momentum
2,22
2,11
2,22
2,11 2
1
2
1
2
1
2
1iiff VmVmVmVm
2. Conservation of kinetic energy
(2)
Elastic collision
iiff VmVmVmVm ,22,11,22,11
(1)
V1,i
V2,i
V1,
f
V2,f
Inelastic collision: change in kinetic energyPerfectly inelastic collision: the particles stick together
Final review 04/14/09
Collisions 1D
if Vmm
mmV ,1
21
21,1
if V
mm
mV ,1
21
1,2
2
• If one of the objects is initially at rest:
iif Vmm
mV
mm
mmV ,2
21
2,1
21
21,1
2
• Combining (1) and (2), we get expression for final speeds:
iif Vmm
mV
mm
mmV ,1
21
1,2
21
12,2
2
V1,i
V1,f V2,f
V2,i
Collisions 2D
• 3 equations
• 4 unknow parametersV1,i
x
y V1,f
V2,f
Final review 04/14/09
Solid characteristics
M OC dmr
• The center of mass is defined as:
C
O
dm
rdVdm
Ctot VMp
FaM C
• The moment of inertia of the solid about one axis:
dmrI 2
I
2' MDII
I’
D
Final review 04/14/09
Rotational kinematics
dt
d
dt
d
• Solid’s rotation
Angular position
Angular speed
Angular acceleration
RVt
• Linear/angular relationship
Velocity
Acceleration
• Tangential
• Centripetal
Rat 2RaC
For any point in the solid
• Rotational kinetic energy2.
2
1 IK
Final review 04/14/09
Motion of rolling solid
P
C
R
Non- sliding situation
• The kinetic energy of the solid is given by the sum of the translational and rotational components:
Ksolid = Kc + Krot
22
2
1
2
1 IVMK Csolid
222
2
1
2
1 IMRK solid
22 )(2
1 IMRK solid
cstKU sol If all the forces are conservative:
Final review 04/14/09
Torque & angular momentum
Fr
We define the torque
F
The angular momentum is defined as
dt
Ld
Deriving Newton’s second law in rotation
angular momentum Linear momentum
prL
sinrF
When a force is inducing the rotation of a solid about a specific axis:
For an object in pure rotation
Inet IL
Final review 04/14/09
Solving a problem
Static equilibrium
• Define the system
• Locate the center of mass (where gravity is applied)
• Identify and list all the forces
0
F• Apply the equality
• 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.
0
• Apply the equality
Final review 04/14/09
Gravitational laws
rMg ur
mMGrgmF
2
)(
Any object placed in that field experiences a gravitational force
Any material object is producing a gravitational field
rM ur
MGrg
2
)( M
r
ur
m
Fg
The gravitational field created by a spherical object is centripetal
(field line is directed toward the center)
The gravitational potential energy is
r
mMGU g
Ug
0 r
Final review 04/14/09
Kepler’s Laws
“The orbit of each planet in the solar system is an ellipse with the Sun as one focus ”
First Law
0LcstL
“The line joining a planet to the sun sweeps out equal areas during equal time intervals as the planet travels along its orbit.”
Second Law
cstmL
dtdA
20
“The square or the orbital period of any planet is proportional to the cube of the semimajor axis of the orbit”
Third Law
32
2 4R
GMT
S
Final review 04/14/09
Satellite Motion
Tsat =TEarth
= 1 day
Geostationary orbit
T
RV GSGS
2
• Satellite speed
(1)GS
EGS R
MGV 2
From Newton’s law
(2)
Escape speed
R
GMVesc
2
GS
E
R
mMGE
2The mechanical energy of
The satellite on orbit is
Final review 04/14/09
Quiz # 43
A planet has a mass twice the mass of the Earthand a diameter 0.7 times the Earth diameter.What would be the weight, in Newtons, of a 82kg personstanding at the surface of this planet?
A 334 NB 803 NC 3280ND 483 NE 4830 N
2p
ppp
R
mMGmgF
The weight of this person at the surface of this planet is
Compared to the weight on earth2
p
E
E
pEp R
R
M
MmgF
9.8 m/s2
Gravitational field on Earth
2
7.0
128.982
N3280
(Equivalent “mass’ on earth: 334kg!)
Final review 04/14/09
Part IV
Oscillatory motion (ch15)
• Harmonic equation and solutions
• Energy of harmonic oscillator
• Spring motion • Pendulum motions
• Damped oscillation
• Forced oscillation
x
Harmonic motionSpring
Equation ofthe motion is 0
2
2
xm
k
dt
xd Harmonicequation
A general solution to this harmonic equation is
)cos()( tAtx
Amplitudeangular
frequency
Phase constant(phase at t=0)
m
k
with
Unit = rad/s
frequency
m
kf
21
Unit = Hz
period
k
mT 2
Unit = s
F
Final review 04/14/09
Position, velocity and acceleration
)cos()( tAtx
)sin()( tAdt
dxtV
)cos()( 2 tAdt
dVta
Position, velocity and acceleration are all sinusoidal functions
Tx(t)
t
Position
Velocity
Acceleration
t
V(t)
Phasequadrature
t
a(t)
Opposite phase
Final review 04/14/09
Energy
cstkAUKEmech 2
2
1
0 t
U
K
0 x
U
K
Final review 04/14/09
The pendulum(punctual)
L
m
0The equation for motion is
02
2
L
g
dt
d
The solution for the angle position is
)cos()( max tt
L
gwith
The oscillation frequency does not depend on the mass m
The period of the oscillation is
g
LT
22
Final review 04/14/09
Torsional pendulum
A torsional pendulum uses the torqueinduced by torsion to oscillate
0 (rest)
For an angular displacement , the torque is k
The equation for the motion is then 2
2
dt
dI
dt
dLk
02
2
I
k
dt
d
A general solution is:
)cos()( max ttI
kwith
L
dt
Ld
According to Newton’s second law
Final review 04/14/09
Damped oscillations
If the oscillator is moving in a resistive medium:friction, viscosity…. The oscillation will be damped.
L0
x0
FA
Compressed
The expression of the spring force is xk ukxF
Applying the Newton’s second law
dg FFam
dt
dxbkx
dt
xdm
2
2Equation ofthe motion
The expression of resistive force is VbFd
Damping
coefficient
Final review 04/14/09
Damped oscillations
)cos(2
exp)(
tt
m
bAtx
with 2
20 2
m
b
The general solution for the motion is
T
x(t)
t
Underdamped
02
m
b
x(t)
t
overdamped
02
m
b
TCritical
02
m
b
x(t)
t
0
Final review 04/14/09
Forced oscillations
A general solution to this equation is
)cos()( tAtx
Amplitude Frequencyforced
Phase constant(phase at t=0)
kxdt
dxbtF
dt
xdm sin02
2
An external force is applied to the system, forcing the oscillation to a frequency
If any, resistive force is VbFd
xk ukxF
The spring force is
The external force is tFFosc sin0
dgosc FFFma
Final review 04/14/09
Forced oscillations
A general solution to a forced oscillation motion
)cos()( tAtx
with
2
220
2
0
mb
mF
A
m
k0and
Resonance
In absence of resistive forces, the amplitude of the oscillation is amplified to infinity when the force frequency
approaches the proper frequency 0
Low damping b
No damping b=0
Large damping b
Final review 04/14/09
GOOD LUCK With the FINALS !
To contact me:
Karine CHESNEL
Office: N319 ESC
801 – 422 – 5687
Final review 04/14/09
Top Related