Transcript Monday, Oct. 7, 2002

PHYS 1443 – Section 003
Lecture #7
Monday, Oct. 7, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Numerical Modeling in Particle Dynamics (Euler Method)
Work & Scalar product of vectors
Kinectic Energy
Power
Potential Energy
•
Gravitational Potential Energy
•
Elastic Potential Energy
Conservative Forces and Mechanical Energy Conservation
Today’s homework is homework #8, due 12:00pm, next Monday!!
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Announcements
• Term Exam
– Grading is completed
• Maximum Score: 85
• Numerical Average: 51.7  Very good!!!
• One person missed the exam without a prior approval
– Can look at your exam after the class
– All scores are relative based on the curve
• One worst after the adjustment will be dropped
– Exam constitutes only 50% of the total
• Do your homework well
• Come to the class and do well with quizzes
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
First Term Exam Distributions
Gaussian Mean: 51
This is what I will use to
adjust term exam scores
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
3
Resistive Force Proportional to Speed
Since the resistive force is proportional to speed, we can write R=bv
Let’s consider that a ball of mass m is falling through a liquid.
R
v
m
F  F
mg
g
F
R
x
 Fy  mg  bv  ma  m
dv
dt
0
dv
b
g v
dt
m
This equation also tells you that
dv
b
 g  v  g , when v  0
dt
m
The above equation also tells us that as time goes on the speed
increases and the acceleration decreases, eventually reaching 0.
What does this mean?
An object moving in a viscous medium will obtain speed to a certain speed (terminal speed)
and then maintain the same speed without any more acceleration.
What is the
terminal speed
in above case?
How do the speed
and acceleration
depend on time?
dv
b
mg
 g  v  0; vt 
dt
m
b
Monday, Oct. 7, 2002
mg 
bt
1  e m ; v  0 when t  0;

b 
dv mg b bt m
t
a

e
 ge t ; a  g when t  0;
dt
b m
dv mg b  t t mg b 
b
t

e

1  1  e t   g  v

dt
b m
b m
m
v
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
The time needed to
reach 63.2% of the
terminal speed is
defined as the time
constant, tm/b.
4
Example 6.11
A small ball of mass 2.00g is released from rest in a large vessel filled with oil, where
it experiences a resistive force proportional to its speed. The ball reaches a terminal
speed of 5.00 cm/s. Determine the time constant t and the time it takes the ball to
reach 90% of its terminal speed.
vt 
R
v
m
Determine the
time constant t.
mg
Determine the time it takes
the ball to reach 90% of its
terminal speed.
mg
b
mg 2.00 10 3 kg  9.80m / s 2
b 

 0.392kg / s
vt
5.00 10  2 m / s
m 2.00 103 kg
t 
 5.10 103 s
b
0.392kg / s
v
mg 
t
t
1

e

b 
  v 1  e  t t 
 t

t
0.9vt  vt 1  e t 


1  e  t t   0.9; e  t t  0.1


t  t  ln 0.1  2.30t  2.30  5.10 103  11.7ms
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
5
Numerical Modeling in Particle Dynamics
•
The method we have been using to solve for particle dynamics is called
Analytical Method  Solve motions using differential equations
–
–
–
–
•
Use Newton’s second law for net force in the motion
Use net force to determine acceleration, a=SF/m
Use the acceleration to determine velocity, dv/dt=a
Use the velocity to determine position, dx/dt=v
Some motions are too complicated to solve analytically  Numeric method or
Euler method to describe the motion
– Differential equations are divided in small increments of time or position
– Acceleration is determined from net force:
a ( x, v, t ) 
 F ( x, v, t )
m
– Determine velocity using a(x,v,t):
v( x, t )  v( x, t  t )  a( x, v, t )t
– Determine position using a and v:
xt   x(t  t )  v( x, t )t
Compute the quantities at every small increments of time t and plot position,
velocity, or acceleration as a function of time to describe the motion.
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
6
Work Done by a Constant Force
Work in physics is done only when a sum of forces
exerted on an object made a motion to the object.
F
M
y
q
FN
M
Free Body
Diagram
F
q
d
x
FG  M g
Which force did the work? Force F
How much work did it do? W   F  d Fd cosq
What does this mean?
Monday, Oct. 7, 2002
Unit?
N m
 J (for Joule)
Physical work is done only by the component of of
the force along the movement of the object.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Work is energy transfer!!
7
Example 7.1
A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at
an angle of 30.0o with East. Calculate the work done by the force on the vacuum
cleaner as the vacuum cleaner is displaced by 3.00m to East.
F
M
30o
M
W   F  d  F d cosq
W  50.0  3.00  cos 30  130 J
d
Does work depend on mass of the object being worked on?
Why don’t I see the mass
term in the work at all then?
Monday, Oct. 7, 2002
Yes
It is reflected in the force. If the object has smaller
mass, its would take less force to move it the same
distance as the heavier object. So it would take less
work. Which makes perfect sense, doesn’t it?
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
8
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
angle between them
A  B  A B cos q
• Operation is commutative A  B  A B cosq  B A cosq  B  A


• Operation follows distribution
A B  C  A B  AC
law of multiplication
• Scalar products of Unit Vectors i  i  j j  k  k  1 i  j  j k  k  i  0
 




 


 
• How does scalar product look in terms of components?



A  Ax i  Ay j  Az k


B  Bx i  By j  Bz k




 
  

A  B   Ax i  Ay j  Az k    Bx i  By j  Bz k 

 

A  B  Ax Bx  Ay B y  Az Bz
Monday, Oct. 7, 2002




C  Cx i  C y j  Cz k
 
 
 


A  B   Ax Bx i  i  Ay B y j j  Az Bz k  k 


Angle between
the vectors
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
cos q 
Ax Bx  Ay B y  Az Bz
AB
9
Example 7.3
A particle moving in the xy plane undergoes a displacement d=(2.0i+3.0j)m as a
constant force F=(5.0i+2.0j) N acts on the particle.
a) Calculate the magnitude of the displacement
and that of the force.
Y
d
F
X
d  d x2  d y2  2.0  3.0  3.6m
2
F  Fx2  Fy2 
2
5.02  2.02  5.4 N
b) Calculate the work done by the force F.
W  F d


 
  
  2.0  5.0 i  i  3.0  2.0 j j  10  6  16( J )
  2 . 0 i  3 . 0 j    5 . 0 i  2 .0 j 

 

Can you do this using the magnitudes and the angle between d and F?
W  F  d  F d cosq
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
Work Done by Varying Force
• If the force depends on position of particle during the motion
– one must consider work segment in small segment of the position where
the force can be considered constant
W  Fx  x
– Then add them all up throughout the entire motion (xi xf)
xf
W   Fx  x
xi
In the limit where x0
xf
lim
x 0
 Fx  x   Fx dx  W
xf
xi
xi
– If more than one force is acting, the net work is done by the net force
W (net )  
xf
xi
 F dx
ix
One of the forces depends on position is force by a spring
The work done by the spring force is
W 
0
 xmax
Monday, Oct. 7, 2002
Fs dx  
0
 xmax
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Fs  kx
 kxdx  1 kx2
2
11
Kinetic Energy and Work-Kinetic Energy Theorem
• Some problems are hard to solve using Newton’s second law
– If forces exerting on the object during the motion are so complicated
– Relate the work done on the object by the net force to the change of the
speed of the object
M
SF
vi
M
vf
Work
The work on the object by the net force SF is
W 
d
Displacement
Suppose net force SF was exerted on an object for
displacement d to increase its speed from vi to vf.
d 
 F  d  mad cos 0  mad
1
v f  vi  t
2
Acceleration
  v f  vi  1
1
1
 v f  vi  t  mv 2f  mvi2
W  mad  m
2
2
  t  2
a
v f  vi
t
Kinetic KE  1 mv2
2
Energy
1 2 1 2
W

mv f  mvi  KE f  KEi  KE The work done by the net force caused
Work
change of object’s kinetic energy.
2
2
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
12
Example 7.7
A 6.0kg block initially at rest is pulled to East along a horizontal, frictionless surface by a
constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m.
M
F
M
vi=0
vf
Work done by the force F is
W  F  d  F d cosq  12  3.0 cos 0  36J 
d
From the work-kinetic energy theorem, we know
Since initial speed is 0, the above equation becomes
Solving the equation for vf, we obtain
Monday, Oct. 7, 2002
1
1
W  mv 2f  mvi2
2
2
1 2
W  mv f
2
2W
2  36
vf 

 3.5m / s
m
6.0
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
13
Work and Energy Involving Kinetic Friction
• Some How do you think the work looks like if there is
friction?
– Why doesn’t static friction matter?
Ffr
M
M
vi
vf
d
Friction force Ffr works on the object to slow down
The work on the object by the friction Ffr is
W fr  F fr d cos180   F fr d
KE   F fr d
The final kinetic energy of an object with initial kinetic energy, friction
force and other source of work is
KE f  KEi  W Ffr d
t=0, KEi
Monday, Oct. 7, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Friction
Engine work
t=T, KEf
14
Example 7.8
A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient
of kinetic friction mk=0.15 by a constant horizontal force of 12N. Find the speed of the
block after it has moved 3.0m.
Fk
M
F
vi=0
M
WF  F  d  F d cosq  12  3.0 cos 0  36J 
vf
d=3.0m
Work done by friction Fk is
Thus the total work is
Work done by the force F is
Wk  Fk  d  m k mg d cosq
 0.15  6.0  9.8  3.0 cos180  26J 
W  WF  Wk  36  26  10( J )
Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain
1
W  WF  Wk  mv 2f
2
Monday, Oct. 7, 2002
Solving the equation
for vf, we obtain
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
vf 
2W
2 10

 1.8m / s
m
6.0
15