Download a 2 - BYU Physics and Astronomy

Physics 121
Newtonian Mechanics
Instructor: Karine Chesnel
Feb 26th , 2009
Review for Exam 2
Class website:
www.physics.byu.edu/faculty/chesnel/physics121.aspx
Mid-term exam 2
• Fri Feb 27 through Tuesday Mar 3
• At the testing center : 8 am – 9 pm
• Closed Book and closed Notes
• Only bring: - Pen / pencil
- Calculator
- Math reference sheet
- dictionary (international)
- your CID
• No time limit (typically 3 hours)
Midterm exam 2
Review: ch 5 – ch 8
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
Ch.5 Laws of motion
2/26/09
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
Ch.5 Laws of motion
2/26/09
Review of basic forces
• The weight
Object
of mass m
Fg
• Normal reaction
When two objects are in contact
N


Fg  mg
-N
• Force of tension
The reaction exerted
by the support on the object
is NORMAL to the surface
• Spring force
T
F
0
x
F=-kxî
The tension exerted by a rope
on the object is ALONG
the direction of the rope
The spring force tends
to bring the object back to rest
Ch.5 Friction
2/26/09
Forces of Friction
Two regimes
R
f
`
F
If one applies a force F
at which point
the system starts
to move?
mg
• If F is smaller than a maximum value fmax
then the system does not move
Static
regime
Dynamic
regime
• If F is larger than the maximum value fmax
the system starts to move and the friction is constant

f
Static regime
Kinetic regime
f max
fk

F
Ch.5 Friction
2/26/09
Friction
Summary of two regimes

f
Static
regime
Kinetic regime
f max
fk

F
In the static regime, the magnitude of the friction
is equal to the force pushing the object


fs  F
When the system is on the verge to move

f s , max   s N
Static coefficient
of friction
Once the system is moving

f k  k N
Kinetic coefficient
of friction
Ch.6 Special applications of Newton’s law
2/26/09
Resistive forces
Low speed regime
The equation of the motion is given by
dV
bV
g
dt
m
F
V
V (t )  V f (1  et /t )
Where t = m/b the time constant
and Vf = mg/b is the terminal speed
V(t)
Vf
(1-1/e)Vf
0
t
t
The motion starts at t = 0 with no initial speed
The speed increases to reach the limit Vf
When t = t the speed value is V = (1-1/e) Vf ~ 0.63 Vf
Ch.5&6 Laws of motion
2/26/09
General method
To solve a given problem:
m
1. Define system
Define the object you will consider
and identify its mass m
2. List the forces
List all the forces applied on the system, for example:
- the weight mg
- the normal reaction of a support N
- a force of tension T
- a force of friction f … etc
3. Apply Newton’s law


ma   F
4. Define a frame and project
• Define a frame of work that suits with the situation:
either Cartesian coordinates (x, y) or polar coordinates (r, q)
• Project the Newton’s law along each axis separately.
•Be careful with the SIGN!!
Ch.5 Laws of motion
2/26/09
Pitfalls to avoid


ma   F
• Newton’s second law
This is an ABSOLUTE equation (vectors)
. Projection along specific axis
The projection is not an absolute equation:
the sign depends on your choice of axis orientation.
Be CONSISTENT with your choice of axis!
Axis choice
Example
z
T
T
T
m
m
mg

 
ma  mg  T
Vectorial
equation
m
mg
Choice 1
ma  T  mg
mg
z
Choice 2
ma  mg  T
Ch.5&6 Laws of motion
2/26/09
Pitfalls to avoid
• Newton’s second law


ma   F
This is a VECTORIAL equation
• What if forces are in different directions?
Be careful: do not mix forces in different directions!!
Examples
q
q
R
T
H
m
mg
q
mg
Take into account the direction,
possibly by using inclination angles (q).
Project Newton’s law along each axis separately
Ch.6 Motion
2/26/09
Tangential and radial
acceleration
General case
V3
V1
a
a
a
V2
V is tangential to the trajectory
• Tangential acceleration
The sign tells if the particle
speeds up or slows down
at= dV/dt
• Centripetal acceleration
ac=
Rw2
=
V2/R
The centripetal acceleration
is toward the center of curvature
Ch.5&6 Newton’s law
2/26/09
Problem (Attwood machine)
z
Two objects of different mass
are suspended at each end of a string
with a frictionless pulley
Will the system move?
If so, in which direction
and with what acceleration?
T2
T1
m2
Let’s apply Newton’s Law
on each object:


ma   F
m2g
m1
m1g
Object 1:
m1a1 = T1 + m1g
Object 2:
m2a2 = T2 + m2g
Let’s project these equations along z axis
m1a1 = T1 - m1g
m2a2 = T2 - m2g
So
T1 = m1a1 + m1g
T2 = m2a2 + m2g
Knowing that T1 = T2 and that a2= - a1 we get
m1a1 + m1g = -m2a1 + m2g
(m2 - m1) g = (m2 + m1) a1
m2  m1
a1 
g
m2  m1
Ch.5 Laws of motion
2/26/09
Problem (Attwood machine)
z
T2
T1
m2
m2g
m1
m1g
Two objects of different mass
are suspended at each end of a string
with a frictionless pulley
Will the system move?
If so, in which direction
and with what acceleration?
We have T1 = T2
and a2 = - a1
m2  m1
a1 
g
m2  m1
• If m2 > m1 : then a1 > 0
the red sphere moves down and green cube moves up
• If m2 < m1 : then a1 < 0
the red sphere moves up and green cube moves down
Ch.7 Work and energy
2/26/09
Work of a force
A particle moves under the action of a force F
from initial point A to final point B
B
F
The total work done
by the force F on the particle
A
from point A to point B is
dr
B
B
A
A
WAB   dW  
 
F .dr
F
dr
A
B
If at any time along the path,
the force F is perpendicular
to the displacement, then:
WAB  
B
A
 
F .dr  0
F q
A
B
dr
If the force is constant and
working along a straight line
WAB

 F d AB cos q
Ch.7 Work and energy
2/26/09
Conservative force
A force is conservative when:
its work does not depend on the path.
The force conserves the energy
A
• Path independence
 
Wloop   F .dr  0
The work done by a conservative force
on a closed path is zero
• Examples of conservative forces:
- Gravity
- Elastic force
- Gravitational field
- Electric force
- Magnetic force
- any constant force
For conservative forces,
we can express the work in terms of
potential energy
Ch.7 Work and energy
2/26/09
Gravity
potential energy
WAB  mg.H  mg ( z A  z B )
WAB  mgz A  mgzB
WAB  (mgzB  mgz A )
We can express the work of the weight
as a variation of a potential function Ep
WA B  E p
E p  mgz
A
mg
H
B
Ch.7 Work and energy
2/26/09
Elastic potential energy
L0
F
A
x
0
F
B
W A B
W A B
x
k 2
 ( x A  xB2 )
2
k 2 k 2
 ( x B  x A )
2
2
WA B   E p
We can express the work of a spring force
as a variation of the elastic potential Ep
1 2
E p  kx
2
Ch.7 Work and energy
2/26/09
Work and kinetic energy
Using Newton’s second law


 F  ma
WAB  
B
A
 
B
 
F .dr   ma.dr
A
WAB  K  K B  K A
Work- Kinetic energy theorem
Defining the
kinetic energy
1
K  mV 2
2
Ch.7 Work and energy
2/26/09
Mechanical energy
K  Wcons  Wnc
K  E p  Wnc
K  E p  Wnc
( K  E p )  Wnc
We define the mechanical energy Emech
as the sum of kinetic and potential energies
Emech  K  E p
Emech  Wnc
Ch.7 Work and energy
2/26/09
Closed System with
conservative forces only
Fcons
There are no non-conservative forces working
Emech  Wnc  0
The mechanical energy is constant
Emech  cst
K  E p  cst
The mechanical energy is conserved
between initial and final points
K f  E p , f  K i  E p ,i
Next Class
Tuesday March 3rd
Read
Textbook:
Chapter 9
Homework assignment:
Today Feb 26th 7pm
Problems 8: 5-7
Good luck on
You exam!!