PHYS 1443 – Section 003
Lecture #23
Wednesday, Dec. 4, 2002
Dr. Jaehoon Yu
Review of Chapters 11 - 15
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Announcements
• Final Term Exam
– Monday, Dec. 9, between 12:00pm – 1:30pm for 1.5
hours in the class room
– Covers chapters 11 – 15
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Similar Quantity
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Wednesday, Dec. 4, 2002
Linear
Mass
Rotational
Moment of Inertia
M
Distance
I   r 2 dm
L
v
dr
dt
a
dv
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Angle  (Radian)

d
dt
 
d
dt
Torque
Work
  I
W 
P  F v
P  
p  mv
L  I
K 
1
mv 2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Rotational
f

d 
i
KR 
1
I 2
2
3
Work, Power, and Energy in Rotation
F
f
d r
ds
O
Let’s consider a motion of a rigid body with a single external
force F exerting on the point P, moving the object by ds.
The work done by the force F as the object rotates
through the infinitesimal distance ds=rd is
dW  F  d s  F sin f rd
What is Fsinf?
What is the work done by
radial component Fcosf?
Since the magnitude of torque is rFsinf,
The rate of work, or power becomes
The tangential component of force F.
Zero, because it is perpendicular to the
displacement.
dW  d
P 
The rotational work done by an external force
equals the change in rotational energy.
dW d
 

dt
dt
 d   d  d 
  I


dt
d

dt

 


  I  I 
The work put in by the external force then
dW   d  Id
f
Wednesday, Dec. 4, 2002
How was the power
defined in linear motion?
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
W  

f
 d   Id 

1 2 1 2
I f  I i
4 2
2
Rolling Motion of a Rigid Body
What is a rolling motion?
A more generalized case of a motion where the
rotational axis moves together with the object
A rotational motion about the moving axis
To simplify the discussion, let’s
make a few assumptions
1.
2.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
The object rolls on a flat surface
Let’s consider a cylinder rolling without slipping on a flat surface
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
s  R
R  s
s=R
Wednesday, Dec. 4, 2002
Thus the linear
speed of the CM is
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
vCM 
ds
d
R
 R
dt
dt
Condition for “Pure Rolling”
5
Total Kinetic Energy of a Rolling Body
What do you think the total kinetic
energy of the rolling cylinder is?
P’
CM
Since it is a rotational motion about the point
P, we can writ the total kinetic energy
1
K  I P 2
2
2vCM
vCM
Where, IP, is the moment of
inertia about the point P.
Using the parallel axis theorem, we can rewrite

P

1
1
1
1
2
2
2
2
2 2
K  I P  I CM  MR   I CM   MR 
2
2
2
2
1
1
Since vCM=R, the above
2
2
K  I CM   MvCM
relationship can be rewritten as
2
2
What does this equation mean?
Rotational kinetic
energy about the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
6
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down a hill without slipping.
R

h
1
1 2 2
2
K  I CM   MR 
2
2

vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
2
1
 vCM  1
2
 I CM 
  MvCM
2
 R  2
1  I CM
 2
  2  M vCM
2 R

Since the kinetic energy at the bottom of the hill must
be equal to the potential energy at the top of the hill
1  I CM
 2


M
 2
vCM  Mgh
K
2 R

2 gh
2
1

I
/
MR
CM
PHYS 1443-003, Fall 2002
vCM 
Wednesday, Dec. 4, 2002
Dr. Jaehoon Yu
7
Example 11.1
For solid sphere as shown in the figure, calculate the linear speed of the CM at the
bottom of the hill and the magnitude of linear acceleration of the CM.
What must we know first?
R

h
I CM   r 2 dm 

vCM 
2
vCM

The linear acceleration
of the CM is
Wednesday, Dec. 4, 2002
2 gh
2 
1  I CM / MR
10
gx sin 
7
aCM
2
MR 2
5
Thus using the formula in the previous slide
vCM
Since h=xsin,
one obtains
The moment of inertia the
sphere with respect to the CM!!
2 gh

1 2 / 5
Using kinematic
relationship
2
vCM
5

 g sin 
2x 7
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10
gh
7
2
vCM
 2aCM x
What do you see?
Linear acceleration of a sphere does
not depend on anything but g and .
8
Example 11.2
For solid sphere as shown in the figure, calculate the linear speed of the CM at the
bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem
using Newton’s second law, the dynamic method.
What are the forces involved in this motion?
Gravitational Force, Frictional Force, Normal Force
Newton’s second law applied to the CM gives
n
f
M
h
Mg
F
F

x
 Mg sin   f  MaCM
y
 n  Mg cos  0
Since the forces Mg and n go through the CM, their moment arm is 0

and do not contribute to torque, while the static friction f causes torque CM
We know that
I CM 
2
MR 2
5
aCM  R
We
obtain
2
MR 2
I CM 
2
f 
 aCM   MaCM
5
R  R  R  5


Substituting f in
dynamic equations
Wednesday, Dec. 4, 2002
 fR  I CM 
7
Mg sin   MaCM
5
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
aCM 
5
g sin 
7
9
Torque and Vector Product
z
O
r
rxF
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
p
The disk will start rotating counter clockwise about the Z axis
y
The magnitude of torque given to the disk by the force F is

  Fr sin f
F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
What is the direction?
The direction of the torque follows the right-hand rule!!
The above quantity is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Wednesday, Dec. 4, 2002
  rF
C  A B
C  A  B  A B sin 
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
C  A  B  A B cos
Result? A scalar
10
Angular Momentum of a Particle
If you grab onto a pole while running, your body will rotate about the pole, gaining
angular momentum. We’ve used linear momentum to solve physical problems
with linear motions, angular momentum will do the same for rotational motions.
z
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
L=rxp
The instantaneous angular momentum
L  r p
L of this particle relative to origin O is
O
y
m
r
What is the unit and dimension of angular momentum? kg m2 / s 2
f p
Note that L depends on origin O. Why? Because r changes
x
What else do you learn? The direction of L is +z
Since p is mv, the magnitude of L becomes L  mvr sin f
What do you learn from this?
The point O has
to be inertial.
Wednesday, Dec. 4, 2002
If the direction of linear velocity points to the origin of
rotation, the particle does not have any angular momentum.
If the linear velocity is perpendicular to position vector, the
particle
moves exactly the same way as a point on a11rim.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Angular Momentum and Torque
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
F 
dp
dt
Total external forces exerting on a particle is the same as the change of its linear momentum.
The same analogy works in rotational motion between torque and angular momentum.


Net torque acting on a particle is
F  r d p
  r
dt
z
d r p
dr
dp
dp
dL


 pr
 0r
L=rxp
dt
dt
dt
dt
dt

O
x
r
y
m
f

Why does this work?
p
Thus the torque-angular
momentum relationship
Because v is parallel to
the linear momentum
 
dL
dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
12
Example 11.4
A particle of mass m is moving in the xy plane in a circular path of radius r and linear
velocity v about the origin O. Find the magnitude and direction of angular momentum with
respect to O.
Using the definition of angular momentum
y
v
L  r  p  r  mv  mr  v
r
O
x
Since both the vectors, r and v, are on x-y plane and
using right-hand rule, the direction of the angular
momentum vector is +z (coming out of the screen)
mrrvv  mrv sin f  mrv sin 90  mrv
The magnitude of the angular momentum is LL  m
So the angular momentum vector can be expressed as
L  mrvk
Find the angular momentum in terms of angular velocity .
Using the relationship between linear and angular speed
2
2
L  mrvk  mr  k  mr   I 
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
13
Angular Momentum of a Rotating Rigid Body
z
Let’s consider a rigid body rotating about a fixed axis
Each particle of the object rotates in the xy plane about the zaxis at the same angular speed, 
L=rxp
O
y
m
r
f
x
Magnitude of the angular momentum of a particle of mass mi
about origin O is miviri
2
Li  mi ri vi  mi ri 
p
Summing over all particle’s angular momentum about z axis
Lz   Li   mi ri 2 
i
i
Since I is constant for a rigid body
Thus the torque-angular momentum
relationship becomes
What do
you see?
Lz 
ext
2
i i
i
dL z
d
 I
dt
dt

 m r   I
 I
 is angular
acceleration
dLz

 I
dt
Thus the net external torque acting on a rigid body rotating about a fixed axis is equal to the moment
of inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
14
Example 11.6
A rigid rod of mass M and length l pivoted without friction at its center. Two particles of mass
m1 and m2 are connected to its ends. The combination rotates in a vertical plane with an
angular speed of . Find an expression for the magnitude of the angular momentum.
y
m2
l

O
m1
m1 g
m2 g
x
The moment of inertia of this system is
1
1
1
2
2

I

I

I

Ml

m
l

m2l 2
I
rod
m1
m2
1
12
4
4
2
l2  1


l
1

  M  m1  m2  L  I 
 M  m1  m2 
4 3
4 3


Find an expression for the magnitude of the angular acceleration of the
system when the rod makes an angle  with the horizon.
If m1 = m2, no angular
momentum because net
torque is 0.
If /p/2, at equilibrium
so no angular momentum.
Wednesday, Dec. 4, 2002
First compute net
external torque
   m1 g
l
cos
2
 ext     2

 2  m2 g
gl cos  m1  m2 
2
1
m1  m1 gl cos 
2



Thus 
ext


2
l
1


I
 M  m1  m2 
becomesPHYS 1443-003, Fall
2002 4  3

Dr. Jaehoon Yu
l
cos
2
2m1  m1  cos 
g /l
1

 M  m1  m2 
3
15 
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
Linear momentum is conserved when the net external force is 0.  F  0 
dp
dt
p  const
By the same token, the angular momentum of a system
is constant in both magnitude and direction, if the
resultant external torque acting on the system is 0.
What does this mean?
 ext 
dL
0
dt
L  const
Angular momentum of the system before and
after a certain change is the same.
Li  L f  constant
Three important conservation laws
for isolated system that does not get
affected by external forces
Wednesday, Dec. 4, 2002
 K i  U i  K f  U f Mechanical Energy


Linear Momentum
 pi  p f


Angular Momentum
Li  L f
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
16
Example 11.8
A star rotates with a period of 30days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses
into a neutron start of radius 3.0km. Determine the period of rotation of the neutron star.
What is your guess about the answer?
Let’s make some assumptions:
Using angular momentum
conservation
The period will be significantly shorter,
because its radius got smaller.
1. There is no torque acting on it
2. The shape remains spherical
3. Its mass remains constant
Li  L f
I i  I f  f
The angular speed of the star with the period T is
Thus

Tf 
I i
mri 2 2p


f
If
mrf2 Ti
2p
f
 r f2
 2
r
 i
Wednesday, Dec. 4, 2002
2p

T
2

3
.
0


6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
17
Conditions for Equilibrium
What do you think does the term “An object is at its equilibrium” mean?
The object is either at rest (Static Equilibrium) or its center of mass
is moving with a constant velocity (Dynamic Equilibrium).
When do you think an object is at its equilibrium?
Translational Equilibrium: Equilibrium in linear motion
Is this it?
The above condition is sufficient for a point-like particle to be at its static
equilibrium. However for object with size this is not sufficient. One more
condition is needed. What is it?
Let’s consider two forces equal magnitude but opposite direction acting
on a rigid object as shown in the figure. What do you think will happen?
F
d
d
F 0
CM
-F

The object will rotate about the CM. The net torque
 0
acting on the object about any axis must be 0.
For an object to be at its static equilibrium, the object should not
have linear or angular speed. v
0  0
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
CM
18
More on Conditions for Equilibrium
To simplify the problems, we will only deal with forces acting on x-y plane, giving torque
only along z-axis. What do you think the conditions for equilibrium be in this case?
The six possible equations from the two vector equations turns to three equations.
 F  0 F
F
x
0
y
0
  0 
0
z
What happens if there are many forces exerting on the object?
If an object is at its translational static equilibrium, and if the
net torque acting on the object is 0 about one axis, the net
torque must be 0 about any arbitrary axis.
r’
r5 O O’
Net Force exerting on the object  F  F 1  F 2  F 3      0
Net torque about O

O
 r1  F 1  r 2  F 2  r 3  F 3       r i  F i
Position of force Fi about O’
Net torque about O’

Wednesday, Dec. 4, 2002
O'
 
'
i
r  r i  r'


0
 r '1F 1  r '2  F 2      r1  r '  F 1  r 2  r '  F 2       r i  F i  r ' F i

PHYS 1443-003, Fall 2002 O '
Dr. Jaehoon Yu
  r i  F i  r '0   O
 019
Example 12.1
A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N,
respectively. If the support (or fulcrum) is under the center of gravity of the board and the
father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board
by the support?
1m
F
MFg
x
n
MBg
Since there is no linear motion, this system
is in its translational equilibrium
D
F
F
MFg
x
 MBg  MF g  MDg n
0
n  40.0  800  350  1190N
y
Therefore the magnitude of the normal force
0
Determine where the child should sit to balance the system.
The net torque about the fulcrum
by the three forces are
Therefore to balance the system
the daughter must sit
Wednesday, Dec. 4, 2002
  M B g  0  M F g 1.00  M D g  x  0
x

MFg
800
1.00m 
1.00m  2.29m
MDg
350
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
20
Example 12.1 Continued
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg

x
n
x/2
D
MFg
MBg
The net torque about the axis of
rotation by all the forces are
 M B g  x / 2  M F g  1.00  x / 2  n x / 2  M D g  x / 2  0
n  MBg  MF g  MDg
  M B g  x / 2  M F g  1.00  x / 2
 M B g  M F g  M D g  x / 2  M D g  x / 2
Since the normal force is
The net torque can
be rewritten
 M F g 1.00  M D g  x  0
Therefore
x
MFg
800

1.00m 
1.00m  2.29m
MDg
350
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
21
Example 12.2
A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps
muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find
the upward force exerted by the biceps on the forearm and the downward force exerted
by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.
FB
Since the system is in equilibrium, from
the translational equilibrium condition
F  0
O
l
mg
 F  F  F  mg  0
F
From the rotational equilibrium condition   F  0  F  d  mg  l  0
d
x
U
y
B
U
U
B
FB  d  mg  l
mg  l 50.0  35.0

 583N
FB 
3.00
d
Force exerted by the upper arm is
FU  FB  mg  583  50.0  533N
Thus, the force exerted by
the biceps muscle is
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
22
Example 12.3
A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by
a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the
horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as
the magnitude and direction of the force exerted by the wall on the beam.
R

53.0o
200N
600N
2m
53.0o
FBD
Rsin
Rcos
8m
From the rotational equilibrium
Using the
translational
equilibrium
T
First the translational equilibrium,
using components
F
x
Tsin53
Tcos53
F
y
 R sin   T sin 53.0  600N  200N  0
  T sin 53.0  8.00  600N  2.00  200N  4.00m  0

T  313N
And the magnitude of R is
R cos  T cos 53.0
R sin   T sin 53.0  600 N  200 N
 800  313  sin 53.0 
  71.7
  tan 

 313 cos 53.0

Wednesday, Dec. 4, 2002
 R cos  T cos 53.0  0
1
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
T cos 53.0 313  cos 53.0
R

 582 N

cos 71.1
cos
23
Example 12.4
A uniform ladder of length l and weight mg=50 N rests against a smooth, vertical wall. If
the coefficient of static friction between the ladder and the ground is ms=0.40, find the
minimum angle min at which the ladder does not slip.
First the translational equilibrium,
using components
P
l
FBD

n
O
f
x
 f P 0
y
 mg  n  0
n  mg  50 N
Thus, the normal force is
The maximum static friction force
just before slipping is, therefore,
From the rotational equilibrium
F
F
mg
f smax  m s n  0.4  50N  20N  P

O
l
 mg cos min  Pl sin  min  0
2
 mg 
1  50 N 

  tan 
  51
 40 N 
 2P 
 min  tan 1 
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
24
How did we solve equilibrium problems?
1.
2.
3.
4.
5.
6.
Identify all the forces and their directions and locations
Draw a free-body diagram with forces indicated on it
Write down vector force equation for each x and y
component with proper signs
Select a rotational axis for torque calculations  Selecting
the axis such that the torque of one of the unknown forces
become 0.
Write down torque equation with proper signs
Solve the equations for unknown quantities
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
25
Elastic Properties of Solids
We have been assuming that the objects do not change their
shapes when external forces are exerting on it. It this realistic?
No. In reality, the objects get deformed as external forces act on it,
though the internal forces resist the deformation as it takes place.
Deformation of solids can be understood in terms of Stress and Strain
Stress: A quantity proportional to the force causing deformation.
Strain: Measure of degree of deformation
It is empirically known that for small stresses, strain is proportional to stress
The constants of proportionality are called Elastic Modulus Elastic Modulus 
Three types of
Elastic Modulus
Wednesday, Dec. 4, 2002
1.
2.
3.
stress
strain
Young’s modulus: Measure of the elasticity in length
Shear modulus: Measure of the elasticity in plane
Bulk modulus: Measure of the elasticity in volume
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
26
Young’s Modulus
Let’s consider a long bar with cross sectional area A and initial length Li.
Li
Fex
After the stretch
F
Tensile Stress  ex
A
Young’s Modulus is defined as
Fex
Fex=Fin
A:cross sectional area
Tensile stress
Lf=Li+DL
Tensile strain
Tensile Strain 
F
Y
ex
Tensile Stress
A


Tensile Strain DL L
i
DL
Li
Used to characterize a rod
or wire stressed under
tension or compression
What is the unit of Young’s Modulus?
Force per unit area
1. For fixed external force, the change in length is
Experimental
proportional to the original length
Observations
2. The necessary force to produce a given strain is
proportional to the cross sectional area
Elastic limit: Maximum stress that can be applied to the substance
before it becomes permanently
deformed
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
27
Simple Harmonic Motion
What do you think a harmonic motion is?
Motion that occurs by the force that depends on displacement, and the
force is always directed toward the system’s equilibrium position.
What is a system that has this kind of character? A system consists of a mass and a spring
When a spring is stretched from its equilibrium position
by a length x, the force acting on the mass is
F  kx
It’s negative, because the force resists against the change of
length, directed toward the equilibrium position.
From Newton’s second law
F  ma  kx
we obtain
a

k
x
m
Condition for simple
k
This is a second order differential equation that can d 2 x


x
harmonic motion
be solved but it is beyond the scope of this class.
m
dt 2
Acceleration is proportional to displacement from the equilibrium
What do you observe
Acceleration is opposite direction to displacement
from this equation?
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
28
Equation of Simple Harmonic Motion
d 2x
k


x
2
dt
m
The solution for the 2nd order differential equation
x  Acost  f 
Amplitude
Phase
Angular
Frequency
Phase
constant
Generalized
expression of a simple
harmonic motion
Let’s think about the meaning of this equation of motion
What happens when t=0 and f=0?
What is f if x is not A at t=0?
x  A cos0  0  A
x  A cosf   x'
f  cos 1 x'
What are the maximum/minimum possible values of x?
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
A/-A
An oscillation is fully
characterized by its:
•Amplitude
•Period or frequency
•Phase constant
29
More on Equation of Simple Harmonic Motion
What is the time for full
cycle of oscillation?
The period
Since after a full cycle the position must be the same
x  Acos t  T   f   A cost  2p  f 
T

How many full cycles of oscillation
does this undergo per unit time?
2p

One of the properties of an oscillatory motion
f
1 
 
T 2p
Frequency
Let’s now think about the object’s speed and acceleration.
What is the unit?
1/s=Hz
x  Acost  f 
dx
 Asin t  f  Max speed v
max  A
dt
Max acceleration
dv
2
2
Acceleration at any given time a    A cos  t  f    x a   2 A
dt
max
What do we learn
Acceleration is reverse direction to displacement
about acceleration?
Acceleration and speed are p/2 off phase:
When v is maximum, a is at its minimum
Speed at any given time
Wednesday, Dec. 4, 2002
v

PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
30
Simple Block-Spring System
A block attached at the end of a spring on a frictionless surface experiences
acceleration when the spring is displaced from an equilibrium position.
a
k
x
m
This becomes a second
order differential equation
If we  2  k
d 2x
k
  x denote
2
m
dt
m
2
d
x
2
The resulting differential equation becomes



x
2
dt
Since this satisfies condition for simple
x  A cost  f 
harmonic motion, we can take the solution
Does this solution satisfy the differential equation?
Let’s take derivatives with respect to time dx  A d cost  f    sin t  f 
Now the second order derivative becomes
dt
dt
d
d 2x
sin t  f    2  cost  f    2 x




2
dt
dt
Whenever the force acting on a particle is linearly proportional to the displacement from some
equilibrium position and is in the opposite direction, the particle moves in simple harmonic motion.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
31
More Simple Block-Spring System
How do the period and frequency of this harmonic motion look?
Since the angular frequency  is
The period, T, becomes
So the frequency is
Special case #1
x   cost
f
T



2p


 2p
1

1


T
2p
2p
k
m
m
k
k
m
What can we learn from these?
•Frequency and period do not
depend on amplitude
•Period is inversely proportional
to spring constant and
proportional to mass
Let’s consider that the spring is stretched to distance A and the block is let
go from rest, giving 0 initial speed; xi=A, vi=0,
2
dx
d
x
v
  sin t a  2   2  cos t ai   2   kA/ m
dt
dt
This equation of motion satisfies all the conditions. So it is the solution for this motion.
Suppose block is given non-zero initial velocity vi to positive x at the
instant it is at the equilibrium, xi=0
Is this a good
p
vi 

p
1 
1
  tan     
 tan  
x   cos t    A sin t  solution?
x 
2
2
Special case #2
f

i

Wednesday, Dec. 4, 2002


PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
32
Example 13.3
A block with a mass of 200g is connected to a light spring for which the force constant is
5.00 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced
5.00 cm from equilibrium and released from reset. Find the period of its motion.
From the Hook’s law, we obtain

X=0
X=0.05
k

m

5.00
 5.00s 1
0.20
As we know, period does not depend on the
amplitude or phase constant of the oscillation,
therefore the period, T, is simply
T

2p


2p
 1.26 s
5.00
Determine the maximum speed of the block.
From the general expression of the
simple harmonic motion, the speed is
Wednesday, Dec. 4, 2002
dx
 Asin t  f 
dt
 A  5.00  0.05  0.25m/ s
vmax 
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
33
Energy of the Simple Harmonic Oscillator
How do you think the mechanical energy of the harmonic oscillator look without friction?
Kinetic energy of a
1
1
2

mv

m 2  2 sin 2 t  f 
KE
harmonic oscillator is
2
2
The elastic potential energy stored in the spring
PE

1 2 1
kx  k 2 cos 2 t  f 
2
2
Therefore the total
1
2 2
2
2
2
mechanical energy of the E  KE  PE  m  sin t  f   k cos t  f 
2
harmonic oscillator is




1
1 2
2
2
2
2
k







k

sin

t

f

k

cos

t

f

kA
Since
m E  KE  PE
2
2
Total mechanical energy of a simple harmonic oscillator is a constant of
a motion and is proportional to the square of the amplitude
Maximum KE
is when PE=0
KEmax 
One can obtain speed
Wednesday, Dec. 4, 2002
1
1
1
1
2
mvmax
 m 2  2 sin 2 t  f   m 2  2  k 2
2
2
2
2
E  KE  PE 
v 

KE/PE
1
1
1
mv 2  kx2  k 2
2
2
2

PHYS 1443-003, Fall 2002
2
k m A
 x 2 Yu
 A2  x 2
Dr. Jaehoon
-A
A
E=KE+PE=kA2/2
x
34
Example 13.4
A 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a
horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of
the cube if the amplitude of the motion is 3.00 cm.
k  20.0N / m
A  3.00cm  0.03m
1 2 1
2
3
 KE  PE  kA  20.0 0.03  9.00 10 J
2
2
From the problem statement, A and k are
The total energy of
the cube is
E
E
1
1 2
 kA2
mvmax
2
2
Maximum speed occurs when kinetic
energy is the same as the total energy
vmax  A k  0.03 20.0  0.190m / s
m
0.500
KEmax 
b) What is the velocity of the cube when the displacement is 2.00 cm.
velocity at any given
displacement is
v




k m A2  x 2  20.0  0.032  0.02 2 / 0.500  0.141m / s
c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.
1 2 1
Potential
2
3
Kinetic
KE  mv  0.500  0.141  4.97 10 J
2
2
energy, PE
energy, KE
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
PE 
1 2 1
2
kx  20.0  0.02  4.00 10 3 J
2
2
35
The Pendulum
A simple pendulum also performs periodic motion.
The net force exerted on the bob is
 Fr  T  mg cos A  0
L

T
m
s
2
d
s

ma
F


mg
sin


m
 t
A
dt 2
mg
Since the arc length, s, is s  L A
2
d 2s
d 2   g sin 
d
   g sin 

L
results
2
2
2
dt
dt
dt
L
Again became a second degree differential equation,
satisfying conditions for simple harmonic motion
g
d 2
g
2









If  is very small, sin~
giving
angular
frequency
2
L
dt
The period for this motion is
Wednesday, Dec. 4, 2002
T 
2p

L
 2p
L
g
The period only depends on the
length of the string and the
gravitational acceleration
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
36
Example 13.5
Christian Huygens (1629-1695), the greatest clock maker in history, suggested that an
international unit of length could be defined as the length of a simple pendulum having a
period of exactly 1s. How much shorter would out length unit be had this suggestion
been followed?
Since the period of a simple
pendulum motion is
T

2p

 2p
L
g
The length of the pendulum
in terms of T is
T 2g
L
4p 2
Thus the length of the
pendulum when T=1s is
T 2 g 1 9.8
L

 0.248m
2
2
4p
4p
Therefore the difference in
length with respect to the
current definition of 1m is
DL  1  L  1  0.248  0.752m
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
37
Physical Pendulum
Physical pendulum is an object that oscillates about a fixed
axis which does not go through the object’s center of mass.
O

Consider a rigid body pivoted at a point O that is a distance d from the CM.
d
dsin
CM
The magnitude of the net torque provided by the gravity is
  mgd sin 
mg
Then
d 2  mgd sin 
  I  I dt 2
Therefore, one can rewrite
mgd
 mgd 
d 2




sin


   2
2
I
 I 
dt
Thus, the angular frequency  is

And the period for this motion is T 
Wednesday, Dec. 4, 2002
2p

mgd
I
 2p
I
mgd
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
By measuring the period of
physical pendulum, one can
measure moment of inertia.
Does this work for
simple pendulum?
38
Example 13.6
A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical
plane. Find the period of oscillation if the amplitude of the motion is small.
O
Pivot
L
CM
1
Moment of inertia of a uniform rod,
I  ML2
rotating about the axis at one end is
3
The distance d from the pivot to the CM is L/2,
therefore the period of this physical pendulum is
Mg
T

2p

 2p
I
2ML2
 2p
 2p
Mgd
3MgL
2L
3g
Calculate the period of a meter stick that is pivot about one end and is oscillating in a
vertical plane.
Since L=1m,
the period is
T  2p
Wednesday, Dec. 4, 2002
2L
 2p
3g
2
 1.64 s
3  9.8
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
So the
f   0.61s 1
frequency is
T
39
Simple Harmonic and Uniform Circular Motions
Uniform circular motion can be understood as a
superposition of two simple harmonic motions in x and y axis.
y
y
A
O
P
y
f
x
P
y

A

O x Q
y

A a

O
Q
A
O vx Q
x
t=t
t=0
=t+f
When the particle rotates at a uniform angular
speed , x and y coordinate position become
Since the linear velocity in a uniform circular
motion is A, the velocity components are
Since the radial acceleration in a uniform circular
motion is v2/A=2, the components are
Wednesday, Dec. 4, 2002
v
P
x
ax P
x  A cos  Acost  f 
y  A sin   Asin t  f 
vx  v sin    A sin t  f 
v y  v cos  A cost  f 
a x  a cos   A 2 cost  f 
a y  a sin    A 2 sin t  f 
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
40
x
Example 13.7
A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular
speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to
the right. A) Determine the x coordinate as a function of time.
Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the
angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is
x  A cos  3.00mcos8.00t  f 
Since x=2.00, when t=0
 2.00 

  48.2
 3.00 
1
2.00  3.00mcos f f  cos 
However, since the particle was
moving to the right f=-48.2o,

x  3.00m cos 8.00t  48.2

Find the x components of the particle’s velocity and acceleration at any time t.
Using the
displcement
Likewise,
from velocity


 

dx
vx 
 3.00  8.00sin 8.00t  48.2   24.0m / s sin 8.00t  48.2
dt
dv
a x    24.0  8.00cos8.00t  48.2   192m / s 2 cos 8.00t  48.2
dt
Wednesday, Dec. 4, 2002

PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
41
Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making
observations for a long time. But the data people collected have not been
explained until Newton has discovered the law of gravitation.
Every particle in the Universe attracts every other particle 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
principle mathematically?
G is the universal gravitational
constant, and its value is
Fg 
m1m2
r122
With G
G  6.673 10
11
Fg  G
Unit?
m1m2
r122
N  m 2 / kg 2
This constant is not given by the theory but must be measured by experiment.
This form of forces is known as an inverse-square law, because the magnitude of the
force is inversely proportional to the square of the distances between the objects.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
42
More on Law of Universal Gravitation
Consider two particles exerting gravitational forces to each other.
m1
r̂12
F21
r
m2
Two objects exert gravitational force on each other
following Newton’s 3rd law.
F12
Taking r̂12 as the unit vector, we can
write the force m2 experiences as
What do you think the
negative sign mean?
F 12
m1m2
 G 2 r̂12
r
It means that the force exerted on the particle 2 by
particle 1 is attractive force, pulling #2 toward #1.
Gravitational force is a field force: Forces act on object without physical contact
between the objects at all times, independent of medium between them.
How do you think the
The gravitational force exerted by a finite size,
gravitational force on the
spherically symmetric mass distribution on a particle
outside the distribution is the same as if the entire mass surface of the earth look?
of the distributions was concentrated at the center.
M Em
Fg  G 2
RE
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
43
Dr. Jaehoon Yu
Free Fall Acceleration & Gravitational Force
Weight of an object with mass m is
mg. Using the force exerting on a
particle of mass m on the surface of
the Earth, one can get
What would the gravitational
acceleration be if the object is at
an altitude h above the surface of
the Earth?
mg
g
M Em
RE2
ME
G
RE2
G
M Em  G M Em
Fg  mg '  G
2
2


R

h
r
E
ME
g'  G
RE  h 2
What do these tell us about the gravitational acceleration?
•The gravitational acceleration is independent of the mass of the object
•The gravitational acceleration decreases as the altitude increases
•If the distance from the surface of the Earth gets infinitely large, the weight of the
object approaches 0.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
44
Example 14.2
The international space station is designed to operate at an altitude of 350km. When
completed, it will have a weight (measured on the surface of the Earth) of 4.22x106N.
What is its weight when in its orbit?
The total weight of the station on the surface of the Earth is
FGE  mg
ME
M Em
6
G
2  4.22  10 N
RE
Since the orbit is at 350km above the surface of the Earth,
the gravitational force at that height is
FO
M Em
RE2
FGE
 mg'  G R  h 2 
2
RE  h 
E
Therefore the weight in the orbit is
FO


2
RE2
6.37 106

FGE 
2
RE  h 
6.37 106  3.50 105

Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
6
6

4
.
22

10

3
.
80

10
N
2

45
Example 14.3
Using the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
g
So the mass of the Earth is
Therefore the density of the
Earth is
G
ME
11 M E

6
.
67

10
2
2
RE
RE
2
R g
ME  E
G
2


ME
VE
RE g
3g
G


4pGRE
4p
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4p  6.67 10 11  6.37 106
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
46
Kepler’s Laws & Ellipse
a
b
F1
c
F2
Ellipses have two different axis, major (long) and
minor (short) axis, and two focal points, F1 & F2
a is the length of a semi-major axis
b is the length of a semi-minor axis
Kepler lived in Germany and discovered the law’s governing planets’
movement some 70 years before Newton, by analyzing data.
•All planets move in elliptical orbits with the Sun at one focal point.
•The radius vector drawn from the Sun to a planet sweeps out equal
area in equal time intervals. (Angular momentum conservation)
•The square of the orbital period of any planet is proportional to the
cube of the semi-major axis of the elliptical orbit.
Newton’s laws explain the cause of the above laws. Kepler’s third law is
the direct consequence of law of gravitation being inverse square law.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
47
Kepler’s Third Law
It is crucial to show that Keper’s third law can be predicted from the
inverse square law for circular orbits.
v
r
Since the gravitational force exerted by the Sun is radially
directed toward the Sun to keep the planet circle, we can
apply Newton’s second law
GM s M P M p v 2

r2
r
Ms
2pr
Since the orbital speed, v, of the planet with period T is v  T
2
GM s M P
M p 2pr / T 
The above can be written

2
r
Solving for T one
can obtain
T
2
2  4p r 3  K r 3 and


s
 GM s 
 4p 2
K s   GM
s

r

  2.97 10 19 s 2 / m3

This is Keper’s third law. It’s also valid for ellipse for r being the length of the
semi-major axis. The constant Ks is independent of mass of the planet.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
48
Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass Mp moving around the Sun in an elliptical orbit.
D
S
C
r
Since the gravitational force acting on the planet is
A always toward radial direction, it is a central force
dr
B Therefore the torque acting on the planet by this
force is always 0.
  r  F  r  Frˆ  0
Since torque is the time rate change of angular
momentum L, the angular momentum is constant.


dL
0
dt
L  const
Because the gravitational force exerted on a
 r  p  r  M p v  M p r  v  const
planet by the Sun results in no torque, the
angular momentum L of the planet is constant.
Since the area swept by the
L
1
1
dA  L

dt
 const
dA  r  d r  r  vdt
motion of the planet is
2M p
2M p
2
2
dt
L
This is Keper’s second law which states that the radius vector from the Sun
to a planet sweeps our equal areas in equal time intervals.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
49
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?
Because 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.
m
Fg
rf
m
ri
RE
Wednesday, Dec. 4, 2002
Fg
The path can be looked at as consisting of
many tangential and radial motions.
Tangential motions do not contribute to work!!!
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
50
More on The Gravitational Potential Energy
Since the gravitational force is a radial force, it only performed work while the
path was radial direction only. Therefore, the work performed by the gravitational
force that depends on the position becomes
dW  F  d r  F r dr    
 W  r F r dr
i
rf
For the whole path
Potential energy is the negative
change of work in the path
DU  U f  U i   r F r dr
rf
i
Since the Earth’s gravitational force is
So the potential energy
function becomes
F r   
U f Ui  
rf
ri
GM E m
r2
 1 1
GM E m
dr  GM E m   
2
r
 rf ri 
Since potential energy only matters for differences, by taking the
infinite distance as the initial point of the potential energy, we get
For any two
particles?
Gm1m2
U 
r
Wednesday, Dec. 4, 2002
The energy needed
to take the particles
infinitely apart.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
For many
particles?
U 
GM E m
r
U  U i, j
i, j
51
Example 14.6
A particle of mass m is displaced through a small vertical distance Dy near the Earth’s
surface. Show that in this situation the general expression for the change in gravitational
potential energy is reduced to the DU=mgDy.
Taking the general expression of
gravitational potential energy
The above
formula becomes
DU  GM
Since the situation is close to
the surface of the Earth
Therefore, DU becomes
DU
E

r
m
 ri 
rf ri
Dy
 GM E m
rf ri
ri  RE and rf  RE
DU  GM E m
GM E
Since on the surface of the
g
RE2
Earth the gravitational field is
Wednesday, Dec. 4, 2002
f
 1 1
 GM E m  
r

r
f
i


Dy
RE2
The potential
energy becomes
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
DU  mgDy
52
Energy in Planetary and Satellite Motions
v
r
M
Consider an object of mass m moving at a speed
v near a massive object of mass M (M>>m).
What’s the
total energy?
1 2 GMm
E  K  U  mv 
2
r
Systems like the Sun and the Earth or the Earth and the Moon whose
motions are contained within a closed orbit is called Bound Systems.
For a system to be bound, the total energy must be negative.
Assuming a circular orbit, in order for the object to be kept in
2
GM E m
v
the orbit the gravitational force must provide the radial
 ma  m
2
r
r
acceleration. Therefore from Newton’s second law of motion
The kinetic energy for this system is
Therefore the total
mechanical energy
of the system is
Wednesday, Dec. 4, 2002
1
GM E m
mv 2 
2r
2
GMm
E  K U  
2r
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Since the gravitational
force is conservative, the
total mechanical energy of
the system is conserved.
53
Example 14.7
The space shuttle releases a 470kg communication satellite while in an orbit that is
280km above the surface of the Earth. A rocket engine on the satellite boosts it into a
geosynchronous orbit, which is an orbit in which the satellite stays directly over a single
location on the Earth, How much energy did the engine have to provide?
What is the radius of the geosynchronous orbit?
From Kepler’s
3rd
law
Therefore the
geosynchronous radius is
Because the initial position
before the boost is 280km
The total energy needed to
boost the satellite at the
geosynchronous radius is the
difference of the total energy
before and after the boost
Wednesday, Dec. 4, 2002
T K r
2
rGS  3
3
E GS
T  1day  8.64 10 4 s
Where KE is


4p 2
K E  GM  9.89 1014 s 2 / m3
E


4 2
4 2
T2
8
.
64

10
8
.
64

10
3
3
 4.23 107 m

14

14
KE
9.89 10
9.89 10
ri  RE  2.80 105 m  6.65  10 6 m
DE  
GM E ms
2
 1 1

 
r
 GS ri 
6.67 1011  5.98 1024  470 
1
1

10



1
.
19

10
J


7
6
2
 4.23 10 6.65 10 
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
54
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.
Because the total
energy is conserved
ME
E
 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
GM E m
GM E m
mvi2 

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.
Wednesday, Dec. 4, 2002
How does this depend
on the mass of the
escaping object?
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Independent of
the mass of the
escaping55object
Fluid and Pressure
What are the three states of matter?
Solid, Liquid, and Gas
By the time it takes for a particular substance to
How do you distinguish them?
change its shape in reaction to external forces.
A collection of molecules that are randomly arranged and loosely
What is a fluid? bound by forces between them or by the external container.
We will first learn about mechanics of fluid at rest, fluid statics.
In what way do you think fluid exerts stress on the object submerged in it?
Fluid cannot exert shearing or tensile stress. Thus, the only force the fluid exerts
on an object immersed in it is the forces perpendicular to the surfaces of the object.
F
This force by the fluid on an object usually is expressed in the form of
P
A
the force on a unit area at the given depth, the pressure, defined as
Expression of pressure for an
dF Note that pressure is a scalar quantity because it’s
P

infinitesimal area dA by the force dF is
dA the magnitude of the force on a surface area A.
What is the unit and
Unit:N/m2
Special SI unit for
2
1
Pa

1
N
/
m
dimension of pressure?
pressure is Pascal
Dim.: [M][L-1][T-2]
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
56
Example 15.1
The mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep.
a) Find the weight of the water in the mattress.
The volume density of water at the normal condition (0oC and 1 atm) is
1000kg/m3. So the total mass of the water in the mattress is
m  W VM  1000  2.00  2.00  0.300  1.20 103 kg
Therefore the weight of the water in the mattress is
W  mg  1.20 103  9.8  1.18 10 4 N
b) Find the pressure exerted by the water on the floor when the bed rests in its
normal position, assuming the entire lower surface of the mattress makes contact
with the floor.
Since the surface area of the
mattress is 4.00 m2, the
pressure exerted on the floor is
Wednesday, Dec. 4, 2002
F mg 1.18 10 4
3




2
.
95

10
P A A
4.00
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
57
Variation of Pressure and Depth
Water pressure increases as a function of depth, and the air pressure
decreases as a function of altitude. Why?
It seems that the pressure has a lot to do with the total mass of
the fluid above the object that puts weight on the object.
P0A
h
Mg
PA
Let’s consider a liquid contained in a cylinder with height h and
cross sectional area A immersed in a fluid of density  at rest, as
shown in the figure, and the system is in its equilibrium.
If the liquid in the cylinder is the same substance as the fluid,
the mass of the liquid in the cylinder is
M  V  Ah
Since the system is in its equilibrium
Therefore, we obtain P  P0  gh
Atmospheric pressure P0 is
1.00atm  1.013 105 Pa
Wednesday, Dec. 4, 2002
PA  P0 A  Mg  PA  P0 A  Ahg  0
The pressure at the depth h below the surface of a fluid
open to the atmosphere is greater than atmospheric
pressure by gh.
What else can you learn from this?
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
58
Pascal’s Law and Hydraulics
A change in the pressure applied to a fluid is transmitted undiminished
to every point of the fluid and to the walls of the container.
P  P0  gh
What happens if P0is changed?
The resultant pressure P at any given depth h increases as much as the change in P0.
This is the principle behind hydraulic pressure. How?
d1
F1
A1
A2
F2
Since the pressure change caused by the
d2 the force F1 applied on to the area A1 is
transmitted to the F2 on an area A2.
P

F1 F2

A1 A2
In other words, the force get multiplied by
A2
Therefore, the resultant force F2 is F2  A F1 the ratio of the areas A2/A1 is transmitted
1
to the F2 on an area.
No, the actual displaced volume of the
This seems to violate some kind
d1

F1
F
2
of conservation law, doesn’t it?
fluid is the same. And the work done
d2
by the forces are still the same.
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
59
Example 15.4
Water is filled to a height H behind a dam of width w. Determine the resultant
force exerted by the water on the dam.
H
Since the water pressure varies as a function of depth, we
will have to do some calculus to figure out the total force.
h
dy
y
The pressure at the depth h is
P  gh  g H  y 
The infinitesimal force dF exerting on a small strip of dam dy is
dF  PdA g H  y wdy
Therefore the total force exerted by the water on the dam is
yH
F




g
H

y
wdy
 g  Hy 

y 0
Wednesday, Dec. 4, 2002

PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
yH
1
1 2
 gH 2
y 
2  y 0 2
60
Absolute and Relative Pressure
How can one measure pressure?
P0
P
h
One can measure pressure using an open-tube manometer,
where one end is connected to the system with unknown
pressure P and the other open to air with pressure P0.
The measured pressure of the system is
P  P0  gh
This is called the absolute pressure, because it is the
actual value of the system’s pressure.
In many cases we measure pressure difference with respect to
P  P0 
atmospheric pressure due to changes in P0 depending on the
environment. This is called gauge or relative pressure.
gh
The common barometer which consists of a mercury column with one end closed at vacuum
and the other open to the atmosphere was invented by Evangelista Torricelli.
Since the closed end is at vacuum, it
does not exert any force. 1 atm is
Wednesday, Dec. 4, 2002
P0  gh  (13.595 103 kg / m3 )(9.80665m / s 2 )(0.7600m)
 1.013 105 Pa  1atm
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
61
Buoyant Forces and Archimedes’ Principle
Why is it so hard to put a beach ball under water while a piece of small
steel sinks in the water?
The water exerts force on an object immersed in the water.
This force is called Buoyant force.
How does the
The magnitude of the buoyant force always equals the weight of
Buoyant force work? the fluid in the volume displaced by the submerged object.
This is called, Archimedes’ principle. What does this mean?
Let‘s consider a cube whose height is h and is filled with fluid and at its
equilibrium. Then the weight Mg is balanced by the buoyant force B.
the pressure at the bottom of the
B  Fg  Mg And
cube is larger than the top by gh.
h
Mg B
Therefore, DP  B / A  gh
Where Mg is the
B  DPA  ghA  Vg
weight of the fluid.
B  Fg  Vg  Mg
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
62
More Archimedes’ Principle
Let’s consider buoyant forces in two special cases.
Case 1: Totally submerged object Let’s consider an object of mass M, with density 0, is
immersed in the fluid with density f .
The magnitude of the buoyant force is
B   f Vg
The weight of the object is Fg  Mg   0Vg
h
Mg B


Therefore total force of the system is F  B  Fg   f   0 Vg
What does this tell you?
Wednesday, Dec. 4, 2002
The total force applies to different directions, depending on the
difference of the density between the object and the fluid.
1. If the density of the object is smaller than the density of
the fluid, the buoyant force will push the object up to the
surface.
2. If the density of the object is larger that the fluid’s, the
object will sink to the bottom of the fluid.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
63
More Archimedes’ Principle
Case 2: Floating object
h
Mg B
Let’s consider an object of mass M, with density 0, is in
static equilibrium floating on the surface of the fluid with
density f , and the volume submerged in the fluid is Vf.
The magnitude of the buoyant force is
The weight of the object is
Therefore total force of the system is
Since the system is in static equilibrium
B   fVf g
Fg  Mg   0V0 g
F  B  Fg   f V f g   0V0 g
0
 f V f g  0V0 g
0 V f

f
V0
What does this tell you?
Wednesday, Dec. 4, 2002
Since the object is floating its density is always smaller than
that of the fluid.
The ratio of the densities between the fluid and the object
determines the submerged volume under the surface.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
64
Example 15.5
Archimedes was asked to determine the purity of the gold used in the crown.
The legend says that he solved this problem by weighing the crown in air and
in water. Suppose the scale read 7.84N in air and 6.86N in water. What
should he have to tell the king about the purity of the gold in the crown?
In the air the tension exerted by the scale on
the object is the weight of the crown
Tair  mg  7.84 N
Twater  mg  B  6.86 N
In the water the tension exerted
by the scale on the object is
B  Tair  Twater  0.98N
Therefore the buoyant force B is
Since the buoyant force B is
B   wVw g   wVc g  0.98N
The volume of the displaced
0.98 N
0.98
4
3

V

V


1
.
0

10
m
c
w
water by the crown is
 w g 1000  9.8
Therefore the density of
mc mc g 7.84
7.84
3
3





8
.
3

10
kg
/
m


4
the crown is
c V
Vc g Vc g 1.0 10  9.8
c
3kg/m1443-003,
3, this crown
Wednesday,
Dec. 4, of
2002
Fall 2002
Since
the density
pure gold is 19.3x10PHYS
is either not made of pure gold or hollow. 65
Dr. Jaehoon Yu
Example 15.6
What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Fgi  iVi g
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
B   wVw g
Since the whole system is at its
static equilibrium, we obtain
Therefore the fraction of the
volume of the iceberg
submerged under the surface of
the sea water is
iVi g   wVw g
Vw  i 917kg / m3


 0.890
3
Vi  w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
66
Congratulations!!!!
You all have done very well!!!
•Good luck with your exams!!!
Happy Holidays!!
Enjoy the winter break!!!
Wednesday, Dec. 4, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
67