Transcript phys1441-summer14-061914

PHYS 1441 – Section 001
Lecture #10
Thursday, June 19, 2014
Dr. Jaehoon Yu
•
•
•
•
•
•
Uniform Circular Motion
Centripetal Acceleration
Unbanked and Banked highways
Newton’s Law of Universal Gravitation
Weightlessness
Work done by a constant force
Today’s homework is homework #6, due 11pm, Tuesday, June 24!!
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
• Quiz 3
Announcements
– Beginning of the class Monday, June 23
– Covers CH 4.7 to what we finish today
– Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!
– You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam  no
solutions, derivations or definitions!
• No additional formulae or values of constants will be provided!
• Mid-term result
– Class average: 67/97
• Equivalent to 69.1/100
• Previous exam: 61.8/100
– Top score: 89/97
• Mid-term grade discussion bottom half of the class
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
Reminder: Special Project #3
• Using the fact that g=9.80m/s2 on the Earth’s
surface, find the average density of the Earth.
– Use the following information only but without
computing the volume explicitly
-11
2
2
• The gravitational constant G = 6.67 ´ 10 N × m kg
• The radius of the Earth
RE = 6.37 ´ 103 km
• 20 point extra credit
• Due: Monday, June 23
• You must show your OWN, detailed work to
obtain any credit!! Much more than in this
lecture note!
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
3
Definition of the Uniform Circular Motion
Uniform circular motion is the motion of an object
traveling at a constant speed on a circular path.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
4
Speed of a uniform circular motion?
Let T be the period of this motion, the time it takes for the object
to travel once around the complete circle whose radius is r.
r
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
distance
v
time
2p r

T
5
Ex. : A Tire-Balancing Machine
The wheel of a car has a radius of 0.29m and is being rotated at
830 revolutions per minute on a tire-balancing machine.
Determine the speed at which the outer edge of the wheel is
moving.
1
-3
= 1.2 ´ 10 min revolution
830 revolutions min
T  1.2 ´ 10-3 min = 0.072 s
(
)
2p r 2p 0.29 m
v
=
= 25m s
T
0.072 s
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
6
Newton’s Second Law & Centripetal Force
The centripetal * acceleration is always perpendicular to the
velocity vector, v, and points to the center of the axis (radial
direction) in a uniform circular motion.
2
v
ac =
r
Are there forces in this motion? If so, what do they do?
The force that causes the centripetal acceleration
acts toward the center of the circular path and
causes the change in the direction of the velocity
c
vector. This force is called the centripetal force.
What do you think will happen to the ball if the string that
holds the ball breaks?
v
å F = mac = m r
2
The external force no longer exist. Therefore, based on Newton’s
1st law, the ball will continue its motion without changing its velocity
and will fly away following the tangential direction to the circle.
*Mirriam Webster: Proceeding or acting in the direction toward the center or axis
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
7
Ex. Effect of Radius on Centripetal Acceleration
The bobsled track at the 1994 Olympics in Lillehammer, Norway, contain turns with
radii of 33m and 23m. Find the centripetal acceleration at each turn for a speed of
34m/s, a speed that was achieved in the two–man event. Express answers as
multiples of g=9.8m/s2.
Centripetal acceleration:
R=33m
2
m
v

r
ar=33m =
R=24m
ar=24m =
Thursday, June 19, 2014
v2
ar 
r
(34)2
33
( )
34
24
2
= 35m s2 = 3.6g
= 48m s = 4.9g
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
8
Example 5.1: Uniform Circular Motion
A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is
moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N,
what is the maximum speed the ball can attain before the cord breaks?
Centripetal
acceleration:
When does the
string break?
v2
ar 
r
v2
 Fr  mar  m r  T
when the required centripetal force is greater than the sustainable tension.
v2
m
 T
r
v  Tr  50.0 1.5  12.2  m / s 
m
0.500
Calculate the tension of the cord
when speed of the ball is 5.00m/s.
Thursday, June 19, 2014
v2
 5.00   8.33 N
 0.500 
T m
 
r
1.5
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
9
Unbanked Curve and Centripetal Force
On an unbanked curve, the static frictional force provides
the centripetal force.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
10
Banked Curves
On a frictionless banked curve, the centripetal force is the
horizontal component of the normal force. The vertical
component of the normal force balances the car’s weight.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
11
Ex. The Daytona 500
The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona
International Speedway in Daytona, Florida. The turns in this oval track have a
maximum radius (at the top) of r=316m and are banked steeply, with θ=31o. Suppose
these maximum radius turns were frictionless. At what speed would the cars have to
travel around them?
x comp.
y comp.
y
v2
 Fx  FN sin   m r  0
 Fy  FN cos  mg  0
x
mv 2
v2
tan  

mgr gr
v 2  gr tan 
v
gr tan  
 
9.8  316 tan 31  43m s  96mi hr
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
12
Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making
observations for a long~ time. The data people collected, however,
have not been explained until Newton has discovered the law of
gravitation.
Every object in the Universe attracts every other object with a force that
is directly proportional to the product of their masses and inversely
proportional to the square of the distance between them.
How would you write this
law mathematically?
G is the universal gravitational
constant, and its value is
Fg 
m1 m2
r
2
12
With G
G = 6.673 ´10
m1m2
Fg  G
r122
-11 Unit?
N  m2 / kg 2
This constant is not given by the theory but must be measured by experiments.
This form of forces is known as the inverse-square law, because the magnitude of the
force is inversely proportional to the square of the distances between the objects.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
13
Ex. Gravitational Attraction
What is the magnitude of the
gravitational force that acts on each
particle in the figure, assuming
m1=12kg, m2=25kg, and r=1.2m?
m1m2
F G 2
r
  6.67 10
 1.4 10
Thursday, June 19, 2014
8
11
N  m kg
2
2
12 kg  25 kg 

2
1.2 m 
N
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
14
Why does the Moon orbit the Earth?
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
15
Gravitational Force and Weight
Gravitational Force, Fg
The attractive force exerted
on an object by the Earth
Weight of an object with mass M is W 
What is the SI unit of weight?
Mg
N
Since weight depends on the magnitude of gravitational
acceleration, g, it varies depending on geographical location.
By measuring the forces one can determine masses. This is
why you can measure mass using the spring scale.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
16
Gravitational Acceleration
M Em
W G 2
r
W  mg
M Em
mg  G 2
r
g  G ME
2
r
Gravitational acceleration at
distance r from the center
of the earth!
What is the SI unit of g?
Thursday, June 19, 2014
m/s2
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
17
Magnitude of the gravitational acceleration
on the surface of the Earth
Gravitational force on
the surface of the earth:
ME
gG 2
RE
(
= 6.67 ´10
 9.80 m s
Thursday, June 19, 2014
M Em
M Em
FG  G 2  G 2
r
RE
 mg
G = 6.67 ´10-11 N × m 2 kg 2
M E = 5.98 ´1024 kg; RE = 6.38 ´106 m
5.98 ´10 kg )
(
kg )
(6.38 ´10 m)
24
-11
N×m
2
2
2
6
2
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
18
Example for Universal Gravitation
Using the fact that g=9.80m/s2 on the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
Fg
G
M Em
RE2
 mg
Solving for g
Solving for ME
Therefore the
density of the
Earth is
g
ME
M
 G 2  6.67 1011 E2
RE
RE
RE 2 g
ME 
G
2

ME

VE
RE g
3g
G


4GRE
4
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67 10 11  6.37 106
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
19
Satellite in Circular Orbits
There is only one speed that a satellite can have if the
satellite is to remain in an orbit with a fixed radius.
What acts as the centripetal force?
The gravitational force of the earth
pulling the satellite!
2
v
mM E
Fc  G 2  m
r
r
GM E
v 
r
2
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
GM
E
v
r
20
Ex. Orbital Speed of the Hubble Space Telescope
Determine the speed of the Hubble Space Telescope
orbiting at a height of 598 km above the earth’s surface.
v  GM E
r

 6.67 10
11
2
5.98 10 kg 
24
6.38 10 m  598 10 m
6
 7.56 103 m s
Thursday, June 19, 2014
N  m kg
2
3
16900mi h 
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
21
Period of a Satellite in an Orbit
Speed of a satellite
GM E  2 r 


r
 T 
2
GM E 2 r
v

r
T
2 3
2  r

2
Square either side
T 
and solve for T2
GM E
2 r
T
GM E
32
Period of a satellite
Kepler’s 3rd Law
This is applicable to any satellite or even for planets and moons.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
22
Geo-synchronous Satellites
Global Positioning System (GPS)
Satellite TV
What period should these
satellites have?
The same as the earth!! 24 hours
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
23
Ex. Apparent Weightlessness and Free Fall
0
0
In each case, what is the weight recorded by the scale?
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
24
Ex. Artificial Gravity
At what speed must the surface of the space station move so
that the astronaut experiences a push on his feet equal to his
weight on earth? The radius is 1700 m.
v
Fc = m = mg
r
2
v = rg
=
(1700 m )(9.80 m s )
2
= 130 m s
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
25
Motion in Resistive Forces
Medium can exert resistive forces on an object moving through it due
to viscosity or other types frictional properties of the medium.
Some examples?
Air resistance, viscous force of liquid, etc
These forces are exerted on moving objects in opposite direction of
the movement.
These forces are proportional to such factors as speed. They almost
always increase with increasing speed.
Two different cases of proportionality:
1. Forces linearly proportional to speed:
Slowly moving or very small objects
2. Forces proportional to square of speed:
Large objects w/ reasonable speed
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
26
Work Done by a Constant Force
A meaningful work in physics is done only when the net
forces exerted on an object changes the energy of the object.
F
M
y

Free Body
Diagram
M

d
x
Fg  M g
Which force did the work?
How much work did it do?
Force
Why?
W
Fd cos
What kind? Scalar
Unit? N  m
 J (for Joule)
Physically meaningful work is done only by the component
What does this mean? of the force along the movement of the object.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
Work is an energy transfer!!
27
Let’s think about the meaning of work!
• A person is holding a grocery bag and
walking at a constant velocity.
• Are his hands doing any work ON the bag?
– No
– Why not?
– Because the force hands exert on the bag, Fp,
is perpendicular to the displacement!!
– This means that hands are not adding any
energy to the bag.
• So what does this mean?
– In order for a force to perform any meaningful
work, the energy of the object the force exerts
on must change due to that force!!
• What happened to the person?
– He spends his energy just to keep the bag up
but did not perform any work on the bag.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
28
Work done by a constant force
s
W  F ×s
= F cosq s
(
Thursday, June 19, 2014
)
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
cos 0  1
cos 90  0
cos180  1
29
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
angle between them
• Operation is commutative
• Operation follows the distribution
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?
 
 
 


  Ax Bx i  i  Ay By j j  Az Bz k  k  cross terms


Ax Bx  Ay By  Az Bz
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
=0
30
Example of Work by Scalar Product
A particle moving on 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
d x2  d y2 
X
2.02  3.02  3.6m
Fx2  Fy2  5.0  2.0  5.4 N
2
2
b) Calculate the work done by the force F.
W


 
  
  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
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
31
Ex. Pulling A Suitcase-on-Wheel
Find the work done by a 45.0N force in pulling the suitcase in the
figure at an angle 50.0o for a distance s=75.0m.
W
(
)
= 45.0 × cos50 × 75.0 = 2170J
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?
Thursday, June 19, 2014
Yes
It is reflected in the force. If an object has smaller
mass, it would take less force to move it at the same
acceleration than a heavier object. So it would take
less work. Which makes perfect sense, doesn’t it?
PHYS 1441-001, Summer 2014
32
Dr. Jaehoon Yu
Ex. 6.1 Work done on a crate
A person pulls a 50kg crate 40m along a horizontal floor by a constant force Fp=100N, which
acts at a 37o angle as shown in the figure. The floor is rough and exerts a friction force
Ffr=50N. Determine (a) the work done by each force and (b) the net work done on the crate.
What are the forces exerting on the crate?
Fp
Ffr
FG=-mg
FN=+mg
Which force performs the work on the crate?
Fp
(
Ffr
)
Work done on the crate by Fp:
WG = FG × x = -mg cos -90 × x = 0J
WN = FN × x = mg cos90 × x = 100 × cos90 × 40 = 0J
Wp = F p × x = F p cos37 × x = 100 × cos37 × 40 = 3200J
Work done on the crate by Ffr:
W fr = F fr × x = F fr cos180 × x = 50 × cos180 × 40 = -2000J
Work done on the crate by FG
Work done on the crate by FN
( )
So the net work on the crate Wnet =WN +WG +W p +W fr =0 + 0 + 3200 - 2000 = 1200 J
This is the same as
Thursday, June 19, 2014
Wnet =
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
33
Ex. Bench Pressing and The
Concept of Negative Work
A weight lifter is bench-pressing a barbell whose weight is
710N a distance of 0.65m above his chest. Then he lowers it
the same distance. The weight is raised and lowered at a
constant velocity. Determine the work in the two cases.
What is the angle between the force and the displacement?
W
 F cos0  s 
Fs
( )
= 710 × 0.65 = +460 J
  s   Fs
W   F cos180
( )
= -710 × 0.65 = -460 J
What does the negative work mean? The gravitational force does the
work on the weight lifter!
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
34
Ex. Accelerating a Crate
The truck is accelerating at a rate of +1.50
m/s2. The mass of the crate is 120-kg and
it does not slip. The magnitude of the
displacement is 65 m. What is the total
work done on the crate by all of the forces
acting on it?
What are the forces acting in this motion?
Gravitational force on the crate,
weight, W or Fg
Normal force force on the crate, FN
Static frictional force on the crate, fs
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
35
Ex. Continued…
Lets figure out what the work done by
each of the forces in this motion is.
Work done by the gravitational force on the crate, W or Fg
W


Fg cos  90o  s  0
Work done by Normal force force on the crate, FN
W


FN cos  90o  s  0
Work done by the static frictional force on the crate, fs
2
120
kg
1.5m
s
ma

fs 


  180N
(
)
(
)
W  f s  s  éë 180N cos0 ùû 65 m = 1.2 ´ 104 J
Which force did the work? Static frictional force on the crate, fs
How?
By holding on to the crate so that it moves with the truck!
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
36
Kinetic Energy and Work-Kinetic Energy Theorem
• Some problems are hard to solve using Newton’s second law
– If forces exerting on an object during the motion are complicated
– Relate the work done on the object by the net force to the change of the
speed of the object
M
ΣF
M
Suppose net force ΣF was exerted on an object for
displacement d to increase its speed from vi to vf.
The work on the object by the net force ΣF is
W
 ma cos 0 s   ma  s
s
v 2f  v02
Using the kinematic 2as  v 2  v 2
as 
2
f
0
equation of motion
Kinetic
1 2
1 2 1 2
1
2
2
KE  mv
Work W   ma  s  2 m  v f  v0   2 mv f  2 mv0
Energy
2
vi
vf
1
2
1
2
Work W  mv 2f  mvi2  KE f  KEi  KE
Thursday, June 19, 2014
Work done by the net force causes
change in the object’s kinetic energy.
PHYS 1441-001, Summer 2014Work-Kinetic Energy
Dr. Jaehoon Yu
37
Theorem
Work-Kinetic Energy Theorem
When a net external force by the jet engine does work on an
object, the kinetic energy of the object changes according to
W  KEf - KE o = mv - mv
1
2
Thursday, June 19, 2014
2
f
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
2
2
o
38
Ex. Deep Space 1
The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If
a 56.0-mN force acts on the probe parallel through a displacement of
2.42×109m, what is its final speed?
(
é
ë
)
2
2
1
1
ù
F
cos
q
s
=
mvf - 2 mvo
å
û
2
v f = vo2 + 2
(å F cosq ) s
Thursday, June 19, 2014
m
=
Solve for vf
( 275m s)2 + 2 ( 5.60 ´10-2 N) cos0 ( 2.42 ´109 m)
v f = 805m s
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
39
474
Ex. Satellite Motion and Work By the Gravity
A satellite is moving about the earth in a
circular orbit and an elliptical orbit. For
these two orbits, determine whether the
kinetic energy of the satellite changes
during the motion.
For a circular orbit No change! Why not?
Gravitational force is the only external
force but it is perpendicular to the
displacement. So no work.
For an elliptical orbit Changes! Why?
Gravitational force is the only external
force but its angle with respect to the
displacement varies. So it performs work.
Thursday, June 19, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
40