PHYS 1443 – Section 003
Lecture #15
Monday, Oct. 18, 2004
Dr. Jaehoon Yu
Gravitational Potential Energy
1.
•
2.
3.
4.
5.
6.
Escape Speed
Power
Linear Momentum
Linear Momentum and Forces
Conservation of Momentum
Impulse
2nd Term Exam Monday, Nov. 1!!
Monday, Oct. 18, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
Announcements
• Remember the 2nd term exam, Monday, Nov. 1, two
weeks from today
– Covers Chapter 6 – wherever we get to (~Chapter 11?).
– No make-up exams
• Miss an exam without pre-approval or a good reason: Your
grade is F.
– Mixture of multiple choice and free style problems
Monday, Oct. 18, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2
The Gravitational Potential Energy
What is the potential energy of an object at the
height y from the surface of the Earth?
U  mgy
Do you think this would work in general cases?
No, it would not.
Why not?
Because this formula is only valid for the case where the gravitational force
is constant, near the surface of the Earth and the generalized gravitational
force is inversely proportional to the square of the distance.
OK. Then how would we generalize the potential energy in the gravitational field?
m
Fg
rf
m
ri
RE
Monday, Oct. 18, 2004
Fg
Since the gravitational force is a central force, and a
central force is a conservative force, the work done by
the gravitational force is independent of the path.
The path can be considered as consisting of
many tangential and radial motions.
Tangential motions do not contribute to work!!!
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
3
More on The Gravitational Potential Energy
Since the gravitational force is a radial force, it performs work only when the path
is radial direction. Therefore, the work performed by the gravitational force that
depends on the position becomes
dW  F  d r  F r dr
Potential energy is the negative
change of work in the path
For the whole path
i
U  U f  U i   r F r dr
rf
i
Since the Earth’s gravitational force is F  r   
So the potential energy
function becomes
W  r F r dr
U f Ui  
rf
ri
GM E m
r2
 1 1
GM E m
dr  GM E m   
2
r
 rf ri 
Since only the difference of potential energy matters, by taking the
infinite distance as the initial point of the potential energy, we obtain
For any two
particles?
Gm1m2
U 
r
Monday, Oct. 18, 2004
rf
The energy needed
to take the particles
infinitely apart.
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
For many
particles?
U 
GM E m
r
U  U i, j
i, j
4
Example of Gravitational Potential Energy
A particle of mass m is displaced through a small vertical distance y near the Earth’s
surface. Show that in this situation the general expression for the change in gravitational
potential energy is reduced to the U=mgy.
Taking the general expression of
gravitational potential energy
The above equation
becomes
U  GM
Since the situation is close to
the surface of the Earth
Therefore, U becomes
U
E

r
m
 ri 
rf ri
y
 GM E m
rf ri
ri  RE and rf  RE
U  GM E m
GM E
Since on the surface of the
g
RE2
Earth the gravitational field is
Monday, Oct. 18, 2004
f
 1 1
 GM E m  
r

r
f
i


y
RE2
The potential
energy becomes
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
U  mgy
5
Escape Speed
vf=0 at h=rmax
m
h
vi
RE
Consider an object of mass m is projected vertically from the surface of
the Earth with an initial speed vi and eventually comes to stop vf=0 at
the distance rmax.
Since the total mechanical
energy is conserved
ME
ME  K  U 
Solving the above equation
for vi, one obtains
vi
Therefore if the initial speed vi is known, one can use
this formula to compute the final height h of the object.
In order for the object to escape
Earth’s gravitational field completely, vesc 
the initial speed needs to be

1 2 GM E m
GM E m
mvi 

2
RE
rmax
 1
1
2GM E 

 RE rmax
hr
2GM E

RE
max



vi2 RE2
 RE 
2GM E  vi2 RE
2  6.67 10 11  5.98 10 24
6.37 106
 1.12 10 4 m / s  11.2km / s
This is called the escape speed. This formula is
valid for any planet or large mass objects.
Monday, Oct. 18, 2004
How does this depend
on the mass of the
escaping object?
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Independent of
the mass of the
escaping6object
Power
• Rate at which the work is done or the energy is transferred
– What is the difference for the same car with two different engines (4
cylinder and 8 cylinder) climbing the same hill?
–  8 cylinder car climbs up faster
Is the total amount of work done by the engines different? NO
Then what is different?
Average power
The rate at which the same amount of work
performed is higher for 8 cylinders than 4.
W
P
t
Instantaneous power
r
u
r
s
W dW
F 


 lim

P  lim
t 0

t
t  0  t
dt


Unit? J / s  Watts 1HP  746Watts


ur r
F v 
ur r
 F v cos
What do power companies sell? 1kWH  1000Watts  3600s  3.6  106 J
Monday, Oct. 18, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Energy
7
Energy Loss in Automobile
Automobile uses only 13% of its fuel to propel the vehicle.
67% in the engine:
Why?
•
•
•
16% in friction in mechanical parts
Incomplete burning
Heat
Sound
4% in operating other crucial parts
such as oil and fuel pumps, etc
13% used for balancing energy loss related to moving vehicle, like air
resistance and road friction to tire, etc
Two frictional forces involved in moving vehicles
Coefficient of Rolling Friction; m=0.016
Air Drag
fa 
mcar  1450kg Weight  mg  14200 N
m n  m mg  227 N
1
1
DAv 2   0.5 1.293  2v 2  0.647v 2
2
2
Total power to keep speed v=26.8m/s=60mi/h
Power to overcome each component of resistance
Monday, Oct. 18, 2004
Total Resistance
ft  f r  f a
P  ft v   691N   26.8  18.5kW
Pr  f r v  227   26.8  6.08kW
PHYS 1443-003, FallP2004
a  fav
Dr. Jaehoon Yu
 464.7 26.8  12.5kW8
Linear Momentum
The principle of energy conservation can be used to solve problems
that are harder to solve just using Newton’s laws. It is used to
describe motion of an object or a system of objects.
A new concept of linear momentum can also be used to solve physical
problems, especially the problems involving collisions of objects.
Linear momentum of an object whose mass is m
and is moving at a velocity of v is defined as
What can you tell from this
definition about momentum?
What else can use see from the
definition? Do you see force?
Monday, Oct. 18, 2004
1.
2.
3.
4.
p  mv
Momentum is a vector quantity.
The heavier the object the higher the momentum
The higher the velocity the higher the momentum
Its unit is kg.m/s
The change of momentum in a given time interval


r r
ur
r
r
r
ur
r
m
v

v
0
 p mv  mv 0
v


ma   F
 m
t
t
t
t
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
9