PHYS 1441 – Section 004
Lecture #25
Wednesday, May 5, 2004
Dr. Jaehoon Yu
Review of Chapters 8 - 11
Final Exam at 11am – 12:30pm, Next Monday, May. 10 in SH101!
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
1
Fundamentals on Rotation
Linear motions can be described as the motion of the center of
mass with all the mass of the object concentrated on it.
Is this still true for
rotational motions?
No, because different parts of the object have
different linear velocities and accelerations.
Consider a motion of a rigid body – an object that
does not change its shape – rotating about the axis
protruding out of the slide.
The arc length, or sergita, is l  R
Therefore the angle, , is 
the angle is in radian.

l
R
. And the unit of
One radian is the angle swept by an arc length equal to the radius of the arc.
Since the circumference of a circle is 2pr,
360  2pr / r  2p
The relationship between radian and degrees is 1 rad  360  / 2p  180 / p
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
 180o 3.14  57.3o
2
Example 8-1
A particular bird’s eyes can just distinguish objects that subtend an angle no
smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small
an object can the bird just distinguish when flying at a height of 100m?
(a) One radian is 360o/2p. Thus


4
3  10 rad  3 10 rad 
4
 360
o

2p rad  0.017
(b) Since l=r and for small angle
arc length is approximately the
same as the chord length.
l  r 
4
100m  3 10 rad 
2
Wednesday, May 5, 2004
3 10 m  3cm
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
3
o
Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how

 i



f
would you define the angular displacement?
 f  i
How about the average angular speed?
Unit? rad/s

And the instantaneous angular speed?
Unit? rad/s
  lim
By the same token, the average angular
acceleration
Unit? rad/s2
And the instantaneous angular
acceleration? Unit? rad/s2

t f  ti
t 0
t f  ti
t 0

t
 d

t
dt
 f  i
  lim


f
i

t
 d

dt
t
When rotating about a fixed axis, every particle on a rigid object rotates through
the same angle and has the same angular speed and angular acceleration.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
Rotational Kinematics
The first type of motion we have learned in linear kinematics was
under a constant acceleration. We will learn about the rotational
motion under constant angular acceleration about a fixed rotational
axis, because these are the simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant
angular acceleration:
Angular displacement under
constant angular acceleration:
One can also obtain
Wednesday, May 5, 2004
 f i  t
1 2
 f   i  i t   t
2
    2  f  i 
2
f
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
i
5
Example for Rotational Kinematics
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If
the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what
angle does the wheel rotate in 2.00s?
Using the angular displacement formula in the previous slide, one gets
 f i
1 2
 t   t
2
1
2
 2.00  2.00  3.50  2.00
2
11.0

rev.  1.75rev.
2p
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
 11.0rad
6
Example for Rotational Kinematics cnt’d
What is the angular speed at t=2.00s?
Using the angular speed and acceleration relationship
 f   i  t
 2.00  3.50  2.00  9.00rad / s
Find the angle through which the wheel rotates between t=2.00
s and t=3.00 s.
Using the angular kinematic formula
f
1 2
  i   t  t
 11.0rad 2
1
At t=2.00s t 2  2.00  2.00  2 3.50  2.00
1
2


2.00  3.00 3.50   3.00   21.8rad
t 3
At t=3.00s
2
Angular
displacement
Wednesday, May 5, 2004
      2
10.8
rev.  1.72rev.
 10.8rad 
2p
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
7
Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates
about a fixed axis of rotation?
Every particle (or masslet) in the object moves in a
circle centered at the axis of rotation.
When a point rotates, it has both the linear and angular motion
components in its motion.
The
direction
What is the linear component of the motion you see?
Linear velocity along the tangential direction.
of 
follows a
right-hand
rule.
How do we related this linear component of the motion
with angular component?
l


The arc-length is l  R So the tangential speed v is v    r   r
t t
t
 r
What does this relationship tell you about Although every particle in the object has the same
the tangential speed of the points in the angular speed, its tangential speed differs
object and their angular speed?:
proportional to its distance from the axis of rotation.
Wednesday, May 5, 2004
The farther
PHYS 1441-004,
Springaway
2004 the particle is from the center of
Dr. rotation,
Jaehoon Yu
the higher the tangential speed.
8
Example 8-3
(a) What is the linear speed of a child seated 1.2m from the center of
a steadily rotating merry-go-around that makes one complete
revolution in 4.0s? (b) What is her total linear acceleration?
First, figure out what the angular
speed of the merry-go-around is.
Using the formula for linear speed
1rev 2p rad


 1.6rad / s
4.0 s
4.0 s
v  r  1.2m 1.6rad / s  1.9m / s
Since the angular speed is constant, there is no angular acceleration.
Tangential acceleration is
Radial acceleration is
Thus the total
acceleration is
Wednesday, May 5, 2004
at  r  1.2m  0rad / s 2  0m / s 2
2
2
ar  r  1.2m  1.6rad / s   3.1m / s 2
a  a  a  0  3.1  3.1m / s 2
2
t
2
r
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
9
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, May 5, 2004
Thus the linear
speed of the CM is
vCM 
PHYS 1441-004, Spring 2004Condition
Dr. Jaehoon Yu
ds
d
R
 R
dt
dt
for “Pure Rolling”
10
More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is
P’
CM
aCM
vCM


R
 R
t
t
2vCM As we learned in the rotational motion, all points in a rigid body
moves at the same angular speed but at a different linear speed.
vCM
P
At any given time the point that comes to P has 0 linear
speed while the point at P’ has twice the speed of CM
CM is moving at the same speed at all times.
Why??
A rolling motion can be interpreted as the sum of Translation and Rotation
P’
CM
P
vCM
P’
vCM
CM
v=0
vCM
Wednesday, May 5, 2004
+
v=R
v=R
2vCM
P’
=
P
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
CM
vCM
P
11
Torque
Torque is the tendency of a force to rotate an object about an axis.
Torque, t, is a vector quantity.
F
f
r
P
Line of
Action
d2
d
Moment
arm
F2
Consider an object pivoting about the point P
by the force F being exerted at a distance r.
The line that extends out of the tail of the force
vector is called the line of action.
The perpendicular distance from the pivoting point
P to the line of action is called Moment arm.
Magnitude of torque is defined as the product of the force
exerted on the object to rotate it and the moment arm.
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force vectorially. The convention for sign of the torque is positive if
rotation is in counter-clockwise and negative if clockwise.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
t  rF sin f  Fd
t  t
1
t 2
 F1d1  F2 d2
12
Example for Torque
A one piece cylinder is shaped as in the figure with core section protruding from the larger
drum. The cylinder is free to rotate around the central axis shown in the picture. A rope
wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder,
and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A)
What is the net torque acting on the cylinder about the rotation axis?
R1
F1
The torque due to F1
t1  R1F1 and due to F2
So the total torque acting on
the system by the forces is
R2
t  t
1
t 2  R2 F2
 t 2  R1F1  R2 F2
F2
Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque
about the rotation axis and which way does the cylinder rotate from the rest?
Using the
above result
Wednesday, May 5, 2004
t  R F  R F
1 1
2 2
 5.0 1.0 15.0  0.50  2.5N  m
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
The cylinder rotates in
counter-clockwise.
13
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating on a circle.
What forces do you see in this motion?
m
The tangential force Ft and radial force Fr
Ft  mat  mr
The tangential force Ft is
The torque due to tangential force Ft is
What do you see from the above relationship?
What does this mean?
t  Ft r  mat r  mr 2  I
t  I
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2nd law of motion in rotation.
How about a rigid object?
The external tangential force dFt is dFt  dmat  dmr
dFt
dt dFt r  r 2 dm
The torque due to tangential force Ft is
dm
The total torque is t    r 2 dm I
r
Contribution from radial force is 0, because its
What is the contribution due
line of action passes through the pivoting
O
to
radial
force
and
why?
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004point, making the moment arm 0.
14
Dr. Jaehoon Yu
Moment of Inertia
Measure of resistance of an object to
changes in its rotational motion.
Equivalent to mass in linear motion.
Rotational Inertia:
For a group
of particles
I   mi ri
i
What are the dimension and
unit of Moment of Inertia?
2
For a rigid
body
ML 
2
I   r 2 dm
kg m
2
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
15
Rotational Kinetic Energy
y
vi
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
K i  mi vi2  mi ri 2 
Kinetic energy of a masslet, mi,
2
2
moving at a tangential speed, vi, is
mi
ri

O
x
Since a rigid body is a collection of masslets, the total kinetic energy of the
rigid object is
1
1
2 
2 
K R   Ki   mi ri     mi ri 
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Wednesday, May 5, 2004
1 
K R  I
2
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
16
Example for Moment of Inertia
In a system consists of four small spheres as shown in the figure, assuming the radii are
negligible and the rods connecting the particles are massless, compute the moment of
inertia and the rotational kinetic energy when the system rotates about the y-axis at .
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
M
x
b
m
I   mi ri2  Ml2 Ml 2 m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
1 2 1
K R  I  2 Ml 2  2  Ml 2 2
2
2
Why are some 0s?

Thus, the rotational kinetic energy is

Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.

2
2
2
I   mi ri 2  Ml 
Ml 2 mb2 mb 2  2 Ml  mb
i
Wednesday, May 5, 2004

1
1
K R  I 2  2 Ml 2  2mb2  2  Ml 2  mb2  2
2
2
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
17
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down a hill without slipping.
R

h

vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
1
1
2
2 2
K  I CM   MR 
2
2
2
1
 vCM   1 Mv 2
 I CM 
CM

2
2
 R 
1I
 2
  CM2  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

vCM 
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2 gh
1  I CM / MR 2
18
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?
dL
t

ext
0

dt
L  const
Angular momentum of the system before and
after a certain change is the same.
ur
ur
L i  L f  constant
Three important conservation laws
for isolated system that does not get
affected by external forces
Wednesday, May 5, 2004
Ki  U i  K f  U f
ur
ur
pi  p f
ur
ur
Li  L f
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Mechanical Energy
Linear Momentum
Angular Momentum
19
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Wednesday, May 5, 2004
Linear
Mass
M
Distance
r
t
v
a
t
L
v
Rotational
Moment of Inertia
I  mr 2
Angle  (Radian)

t


t

ur
r
Force F  ma
Work W  Fd cos
r ur
Torque t  I 
Work W  t
P  Fv cos
P  t
p  mv
L  I
Kinetic
K
1
mv 2
2
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Rotational
KR 
1
I 2
2
20
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
t 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, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
CM
21
More on Conditions for Equilibrium
To simplify the problem, 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
t  0 t
z
0
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
r’
O
r5
O’
any arbitrary axis.
Why is this true?
Because the object is not moving, no matter what
the rotational axis is, there should not be a motion.
Wednesday, May 5, 2004
PHYS 1441-004,
Spring 2004
It is simply
a matter
of mathematical calculation.22
Dr. Jaehoon Yu
Example for Mechanical Equilibrium
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, May 5, 2004
t  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 1441-004, Spring 2004
Dr. Jaehoon Yu
23
Example for Mech. Equilibrium Cont’d
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg
t
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
t  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, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
24
Example for Mechanical Equilibrium
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 t  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, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
25
Example for Mechanical Equilibrium
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
00N
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

T  313N
And the magnitude of R is
R sin   T sin 53.0  600 N  200 N
 800  313  sin 53.0 
  71.7
  tan 

 313 cos 53.0

Wednesday, May 5, 2004
 R sin   T sin 53.0  600N  200N  0
t  T sin 53.0  8.00  600N  2.00  200N  4.00m  0
R cos  T cos 53.0
1
 R cos  T cos 53.0  0
T cos 53.0 313  cos 53.0
R

 582 N

cos 71.1
cos
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
26
Example 9 – 9
A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is
uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not),
determine the forces exerted on the ladder by the ground and the wall.
FW
FBD
mg
FGy
O
FGx
First the translational equilibrium,
using components
 Fx FGx  FW  0
 F  mg  F
y
0
Gy
Thus, the y component of the force by the ground is
FGy  mg  12.0  9.8N  118N
The length x0 is, from Pythagorian theorem
x0  5.02  4.02  3.0m
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
27
Example 9 – 9 cont’d
From the rotational equilibrium
t
O
 mg x0 2  FW 4.0  0
Thus the force exerted on the ladder by the wall is
mg x0 2 118 1.5
FW 

 44 N
4.0
4.0
Tx component of the force by the ground is
F
x
 FGx  FW  0
Solve for FGx
FGx  FW  44 N
Thus the force exerted on the ladder by the ground is
FG  FGx2  FGy2  442  1182  130N
The angle between the  tan 1  FGy 
1  118 
o

tan

70





ladder and the wall is
 44 
 FGx 
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
28
How do 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 from as many of the unknown
forces become 0.
Write down torque equation with proper signs
Solve the equations for unknown quantities
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
29
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. (Ultimate
strength of a material)
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, May 5, 2004
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 1441-004, Spring 2004
Dr. Jaehoon Yu
30
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+L
Tensile strain
Tensile Strain 
F
Y
ex
Tensile Stress
A


Tensile Strain L L
i
L
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, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
31
Bulk Modulus
F
Bulk Modulus characterizes the response of a substance to uniform
squeezing or reduction of pressure.
V
After the pressure change
F
F
V’
F
Normal Force
F
Volume stress
Pressure 

Surface Area the force applies
A
=pressure
If the pressure on an object changes by P=F/A, the object will
undergo a volume change V.
Bulk Modulus is
defined as
Because the change of volume is
reverse to change of pressure.
Wednesday, May 5, 2004
F
P
Volume Stress  
A 
B
V
V
Volume Strain
Vi
V
i
Compressibility is the reciprocal of Bulk Modulus
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
32
Example for Solid’s Elastic Property
A solid brass sphere is initially under normal atmospheric pressure of 1.0x105N/m2. The
sphere is lowered into the ocean to a depth at which the pressures is 2.0x107N/m2. The
volume of the sphere in air is 0.5m3. By how much its volume change once the sphere is
submerged?
Since bulk modulus is
P
B
V
Vi
The amount of volume change is
V  
PVi
B
From table 12.1, bulk modulus of brass is 6.1x1010 N/m2
The pressure change P is
P  Pf  Pi  2.0 107 1.0 105  2.0 107
Therefore the resulting
2.0 107  0.5
4
3

V

V

V




1
.
6

10
m
f
i
volume change V is
6.11010
The volume has decreased.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
33
Density and Specific Gravity
Density, r (rho) , of an object is defined as mass per unit volume
3
kg / m
Unit?
3
[
ML
]
Dimension?
M
r 
V
Specific Gravity of a substance is defined as the ratio of the density
of the substance to that of water at 4.0 oC (rH2O=1.00g/cm3).
r substance
SG 
rH O
2
What do you think would happen of a
substance in the water dependent on SG?
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Unit?
None
Dimension? None
SG  1 Sink in the water
SG  1 Float on the surface
34
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.
This force by the fluid on an object usually is expressed in the form of P  F
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, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
35
Example for Pressure
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  rW 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, May 5, 2004
F mg 1.18 10 4
3




2
.
95

10
P A A
4.00
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
36
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
Let’s imagine a liquid contained in a cylinder with height h and
cross sectional area A immersed in a fluid of density r at rest, as
shown in the figure, and the system is in its equilibrium.
h
Mg
PA
If the liquid in the cylinder is the same substance as the fluid,
the mass of the liquid in the cylinder is
M  rV  rAh
Since the system is in its equilibrium
Therefore, we obtain P  P0  rgh
Atmospheric pressure P0 is
1.00atm  1.013 105 Pa
Wednesday, May 5, 2004
PA  P0 A  Mg  PA  P0 A  rAhg  0
The pressure at the depth h below the surface of a fluid
open to the atmosphere is greater than atmospheric
pressure by rgh.
What else can you learn from this?
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
37
Pascal’s Principle 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  rgh
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?
Since the pressure change caused by the
F1 F2
d1


d2 the force F1 applied on to the area A1 is P
A1
A1 A2
F2
transmitted to the F2 on an area A2.
In other words, the force gets multiplied by
A2
Therefore, the resultant force F2 is F2  A F1 the ratio of the areas A2/A1 and is
1
transmitted to the force F2 on the surface.
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.
F1
A2
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
38
Example for Pascal’s Principle
In a car lift used in a service station, compressed air exerts a force on a small piston
that has a circular cross section and a radius of 5.00cm. This pressure is transmitted by
a liquid to a piston that has a radius of 15.0cm. What force must the compressed air
exert to lift a car weighing 13,300N? What air pressure produces this force?
Using the Pascal’s principle, one can deduce the relationship between the
forces, the force exerted by the compressed air is
2
A2
p 0.15
4
3

F


1
.
33

10

1
.
48

10
N
F1
2
2
A1
p 0.05
Therefore the necessary pressure of the compressed air is
P
F1 1.48 103
5



1
.
88

10
Pa
2
A1 p 00
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
39
Example for Pascal’s Principle
Estimate the force exerted on your eardrum due to the water above
when you are swimming at the bottom of the pool with a depth 5.0 m.
We first need to find out the pressure difference that is being exerted on
the eardrum. Then estimate the area of the eardrum to find out the
force exerted on the eardrum.
Since the outward pressure in the middle of the eardrum is the same
as normal air pressure
P  P0  rW gh  1000  9.8  5.0  4.9 10 4 Pa
Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we obtain
F  P  P0 A  4.9 10 4 1.0 10 4  4.9 N
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
40
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  rgh
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
atmospheric pressure due to changes in P0 depending on the PG  P  P0 
environment. This is called gauge or relative pressure.
rgh
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
P0  rgh  (13.595 103 kg / m3 )(9.80665m / s 2 )(0.7600m)
 1.013 105 Pa  1atm
If one measures the tire pressure with a gauge at 220kPa the actual pressure is 101kPa+220kPa=303kPa.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
41
Finger Holds Water in Straw
pinA
You insert a straw of length L into a tall glass of your favorite
beverage. You place your finger over the top of the straw so that
no air can get in or out, and then lift the straw from the liquid. You
find that the straw strains the liquid such that the distance from the
bottom of your finger to the top of the liquid is h. Does the air in the
space between your finder and the top of the liquid have a pressure
P that is (a) greater than, (b) equal to, or (c) less than, the
atmospheric pressure PA outside the straw?
Less
What are the forces in this problem?
Gravitational force on the mass of the liquid
Fg  mg  r A  L  h  g
Force exerted on the top surface of the liquid by inside air pressure Fin  pin A
mg
Force exerted on the bottom surface of the liquid by outside air
Since it is at equilibrium Fout  Fg  Fin  0
p AA
Cancel A and
solve for pin
Wednesday, May 5, 2004
pin  pA  r g  L  h 
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Fout  p A A
 pA A  r g  L  h  A  pin A  0
So pin is less than PA by rgh.
42
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 rgh.
h
Mg B
Therefore, P  B / A  rgh
Where Mg is the
B  PA  rghA  rVg
weight of the fluid.
B  Fg  rVg  Mg
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
43
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 r0, is
immersed in the fluid with density rf .
The magnitude of the buoyant force is
B  r f Vg
The weight of the object is Fg  Mg  r 0Vg
h
Mg B


Therefore total force of the system is F  B  Fg  r f  r 0 Vg
What does this tell you?
Wednesday, May 5, 2004
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 1441-004, Spring 2004
Dr. Jaehoon Yu
44
More Archimedes’ Principle
Case 2: Floating object
h
Mg B
Let’s consider an object of mass M, with density r0, is in
static equilibrium floating on the surface of the fluid with
density rf , 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
What does this tell you?
Wednesday, May 5, 2004
B  r fVf g
Fg  Mg  r 0V0 g
F  B  Fg  r f V f g  r 0V0 g
0
r f V f g  r0V0 g
r0
Vf

rf
V0
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 1441-004, Spring 2004
Dr. Jaehoon Yu
45
Example for Archimedes’ Principle
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  r wVw g  r 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
r w g 1000  9.8
Therefore the density of
mc mc g 7.84
7.84
3
3





8
.
3

10
kg
/
m
r

4
the crown is
c V
Vc g Vc g 1.0 10  9.8
c
3kg/m
3, this Spring
Wednesday,
May 5,of
2004
PHYS
1441-004,
46
Since
the density
pure gold is 19.3x10
crown2004
is either not made of pure gold or hollow.
Dr. Jaehoon Yu
Example for Buoyant Force
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  riVi 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  r 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
riVi g  r wVw g
Vw r i 917kg / m3


 0.890
3
Vi r w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
47
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water dynamics?? Hydro-dynamics
•Streamline or Laminar flow: Each particle of the fluid
Two main
types of flow follows a smooth path, a streamline w/o friction
•Turbulent flow: Erratic, small, whirlpool-like circles called
eddy current or eddies which absorbs a lot of energy
Flow rate: the mass of fluid that passes a given point per unit time m / t
m1
r1V1 r1 A1l1


 r1 A1v1
t
t
t
since the total flow must be conserved
m1 m2
r1 A1v1  r 2 A2 v2

t
t
Equation of Continuity
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
48
Example for Equation of Continuity
How large must a heating duct be if air moving at 3.0m/s along it can
replenish the air every 15 minutes, in a room of 300m3 volume?
Assume the air’s density remains constant.
Using equation of continuity
r1 A1v1 r 2 A2 v2
Since the air density is constant
A1v1  A2 v2
Now let’s imagine the room as
the large section of the duct
A2l2 / t
V2
A2 v2
300



A1 
 0.11m 2
v1
v1  t
v1
3.0  900
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
49
Bernoulli’s Equation
Bernoulli’s Principle: Where the velocity of fluid is high, the
pressure is low, and where the velocity is low, the pressure is high.
Amount of work done by the force, F1,
that exerts pressure, P1, at point 1
W1  F1l1  P1 A1l1
Amount of work done on the other
section of the fluid is
W2   P2 A2 l2
Work done by the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
50
Bernoulli’s Equation cont’d
Since
1
1
r A2 l2 v22  r A1l1v12  P1 A1l1  P2 A2 l2  r A2 l2 gy2  r A1l1 gy1
2
2
We
obtain
Reorganize P1 
1 2 1 2
r v2  r v1  P1  P2  r gy2  r gy1
2
2
1 2
1 2
Bernoulli’s
r v1  r gy1  P2  r v2  r gy2 Equation
2
2
Thus, for any two
points in the flow
1 2
P1  r v1  r gy1  const.
2
For static fluid P2  P1  r g  y1  y2   P1  r gh
1
2
2
P

P

r
v

v

2
1
1
2
For the same heights
2
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
51
Example for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under
a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter
pipe on the second 5.0m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1v1 p r12 v1
 0.020 

v2 

0.5

 1.2m / s
2


A2
p r2
 0.013 
Using Bernoulli’s equation, the pressure in the pipe on the second floor is


1
P2  P1  r v12  v22  r g  y1  y2 
2
1
5
 3.0 10  1103 0.52  1.22  1103  9.8   5 
2
 2.5 105 N / m 2

Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu

52
Vibration or Oscillation
What are the things
that vibrate/oscillate?
•
•
•
•
•
Tuning fork
A pendulum
A car going over a bump
Building and bridges
The spider web with a prey
So what is a vibration or oscillation?
A periodic motion that repeats over the same path.
A simplest case is a block attached at the end of a coil spring.
When a spring is stretched from its equilibrium
position by a length x, the force acting on the mass is
F  kx
The sign is negative, because the force resists against the
change of length, directed toward the equilibrium position.
Acceleration is proportional to displacement from the equilibrium
Acceleration is opposite direction to displacement
Wednesday, May 5, 2004
This system is doing a simple harmonic motion (SHM).
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
53
Vibration or Oscillation Properties
The maximum displacement from
the equilibrium is
Amplitude
One cycle of the oscillation
The complete to-and-fro motion from an initial point
Period of the motion, T
The time it takes to complete one full cycle
Unit?
s
Frequency of the motion, f
The number of complete cycles per second
Unit?
s-1
Relationship between
period and frequency?
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
f

1
T
or
T

1
f
54
Vibration or Oscillation Properties
Amplitude?
•
•
•
•
•
Wednesday, May 5, 2004
A
When is the force greatest?
When is the velocity greatest?
When is the acceleration greatest?
When is the potential energy greatest?
When is the kinetic energy greatest?
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
55
Example 11-1
Car springs. When a family of four people with a total mass of 200kg step into their
1200kg car, the car’s springs compress 3.0cm. (a) What is the spring constant of the car’s
spring, assuming they act as a single spring? (b) How far will the car lower if loaded with
300kg?
(a) What is the force on the spring?
From Hooke’s law
F  mg  200  9.8  1960 N
F  kx  k  0.03  mg  1960 N
Fg mg 1960
k 


 6.5  104 N / m
x
x
0.03
(b) Now that we know the spring constant, we can solve for x in the force equation
F  kx  mg  300  9.8
mg 300  9.8
2
x


4.5

10
m
4
k
6.5 10
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
56
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

mv 2
KE
harmonic oscillator is
2
1 2

kx
The elastic potential energy stored in the spring
2
1 2 1 2
Therefore the total mechanical energy

mv  kx
E  KE  PE
2
2
of the harmonic oscillator is
PE
Total mechanical energy of a simple harmonic oscillator is
proportional to the square of the amplitude.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
57
Energy of the Simple Harmonic Oscillator cont’d
1
1
2
KEmax  mvmax  k 2
2
2
k
vmax 
A
m
Maximum KE is
when PE=0
Maximum speed
The speed at any given
point of the oscillation
E
v
1
1 2 1
2
2

mv

kx

k

 KE  PE
2
2
2

 k m A x
2
2

  vmax
KE/PE
-A
Wednesday, May 5, 2004
x
1  
 A
2
E=KE+PE=kA2/2
A
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
x
58
Example 11-3
Spring calculations. A spring stretches 0.150m when a 0.300-kg mass is hung from it.
The spring is then stretched an additional 0.100m from this equilibrium position and
released.
(a) Determine the spring constant.
From Hooke’s law
F  kx  mg  0.300  9.8 N
mg 0.300  9.8
k

 19.6 N / m
x
0.150
(b) Determine the amplitude of the oscillation.
Since the spring was stretched 0.100m from
equilibrium, and is given no initial speed, the
amplitude is the same as the additional stretch.
A  0.100m
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
59
Example cont’d
(c) Determine the maximum velocity vmax.
k
A
m
vmax 
19.6
0.100  0.808m / s
0.300
(d) Determine the magnitude of velocity, v, when the mass is 0.050m from equilibrium.
vv
max
 x
1  
 A
2
 0.050 
 0.808 1  

 0.100 
2
 0.700m / s
(d) Determine the magnitude of the maximum acceleration of the mass.
Maximum acceleration is at the point where the mass is stopped to return.
F  ma  kA
Solve for a
kA 19.6  0.100

 6.53m / s 2
a 
m
0.300
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
60
Example for Energy of Simple Harmonic Oscillator
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
From the problem statement, A and k are
The total energy of
the cube is
E
 KE  PE 
1 2 1
2
kA  20.0 0.03  9.00 10 3 J
2
2
Maximum speed occurs when kinetic energy is the same as the total energy
KEmax
vmax
Wednesday, May 5, 2004
1 2
 mvmax
2
E
1 2
 kA
2
k
20.0
A
 0.03
 0.190m / s
m
0.500
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
61
Example for Energy of Simple Harmonic Oscillator
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.
Kinetic
energy, KE
Potential
energy, PE
Wednesday, May 5, 2004
KE 
1 2
mv
2
PE 
1
2
 0.500  0.141  4.97 10 3 J
2
1 2 1
kx  20.0  0.022  4.00 10 3 J
2
2
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
62
Sinusoidal Behavior of SHM
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
63
The Period and Sinusoidal Nature of SHM
Consider an object moving on a circle with a constant angular speed 
v0
A2  x 2
x
 1  
A
 A
v
sin   
v0
Since it takes T to
complete one full
circular motion
From an energy
relationship in a
spring SHM
Thus, T is
v0 
2
x
v  v0 1   
 A
2p A
 2p Af
T
1
1
mv02  kA2
2
2
T  2p
m
k
f 
T 
2p A
v0
v0 
k
A
m
1
1

T 2p
k
m
Natural Frequency
If you look at it from the side, it looks as though it is doing a SHM
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
64
Example 11-5
Car springs. When a family of four people with a total mass of 200kg step into their
1200kg car, the car’s springs compress 3.0cm. The spring constant of the spring is
6.5x104N/m. What is the frequency of the car after hitting the bump? Assume that the
shock absorber is poor, so the car really oscillates up and down.
T  2p
1400
m
 2p
 0.92s
4
6.5 10
k
1
1
f  
2p
T
Wednesday, May 5, 2004
k
1

m 2p
6.5 10
 1.09 Hz
1400
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
65
Example 11-6
Spider Web. A small insect of mass 0.30 g is caught in a spider web of negligible mass.
The web vibrates predominantly with a frequency of 15Hz. (a) Estimate the value of the
spring constant k for the web.
f  1
2p
k
 15Hz
m
Solve for k
k  4p mf  4p  3  10  15   2.7 N / m
2
2
4
2
2
(b) At what frequency would you expect the web to vibrate if an insect of mass 0.10g were
trapped?
1
f 
2p
Wednesday, May 5, 2004
k
1

m 2p
2.7
 26 Hz
4
110
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
66
The SHM Equation of Motion
The object is moving on a circle with a constant angular speed 
How is x, its position at any given time expressed with the known quantities?
v0
x  A cos since    t and   2p f
x  A cos t  A cos 2p ft
k
How about its velocity v at any given time?
v0 
m
A
v  v0 sin   v0 sin  t   v0 sin  2p ft 
How about its acceleration a at any given time?
From Newton’s
Wednesday, May 5, 2004
2nd
law a 
F  kx
 kA 




 cos  2p ft   a0 cos  2p ft 
m
m
 m
kA
a0 
m
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
67
Sinusoidal Behavior of SHM
x  A cos  2p ft 
v  v0 sin  2p ft 
a  a0 cos  2p ft 
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
68
The Simple Pendulum
A simple pendulum also performs periodic motion.
The net force exerted on the bob is
 Fr  T  mg cos A  0
F
t
 mg sin  A  mg
Since the arc length, x, is
x  L
mg
F   Ft  
x
L
Satisfies conditions for simple harmonic motion!
It’s almost like Hooke’s law with. k  mg
L
The period for this motion is
T  2p
m
 2p
k
mL
 2p
mg
L
g
The period only depends on the length of the string and the gravitational acceleration
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
69
Example 11-8
Grandfather clock. (a) Estimate the length of the pendulum in a grandfather clock that
ticks once per second.
Since the period of a simple
pendulum motion is
T  2p
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
L
g
T 2 g 1 9.8
L

 0.25m
2
2
4p
4p
(b) What would be the period of the clock with a 1m long pendulum?
T  2p
Wednesday, May 5, 2004
L
1.0
 2p
 2.0s
g
9.8
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
70
Damped Oscillation
More realistic oscillation where an oscillating object loses its mechanical
energy in time by a retarding force such as friction or air resistance.
How do you think the
motion would look?
Amplitude gets smaller as time goes on since its energy is spent.
Types of damping
A: Overdamped
B: Critically damped
C: Underdamped
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
71
Forced Oscillation; Resonance
When a vibrating system is set into motion, it oscillates with its natural
frequency f0.
1
k
f0 
2p
m
However a system may have an external force applied to it that has
its own particular frequency (f), causing forced vibration.
For a forced vibration, the amplitude of vibration is found to be dependent
on the different between f and f0. and is maximum when f=f0.
A: light damping
B: Heavy damping
The amplitude can be large when
f=f0, as long as damping is small.
This is called resonance. The natural
frequency f0 is also called resonant frequency.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
72
Wave Motions
Waves do not move medium rather carry energy
from one place to another
Two forms of waves
– Pulse
– Continuous or
periodic wave
Mechanical Waves
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
73
Characterization of Waves
• Waves can be characterized by
– Amplitude: Maximum height of a crest or the depth of a trough
– Wave length: Distance between two successive crests or any
two identical points on the wave
– Period: The time elapsed by two successive crests passing by
the same point in space.
– Frequency: Number of crests that pass the same point in space
in a unit time
• Wave velocity: The velocity at which any part of the
wave moves
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
v

T
f
74
Waves vs Particle Velocity
Is the velocity of a wave moving along a cord the
same as the velocity of a particle of the cord?
No. The two velocities are
different both in magnitude
and direction. The wave
on the rope moves to the
right but each piece of the
rope only vibrates up and
down.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
75
Speed of Transverse Waves on Strings
How do we determine the speed of a transverse pulse traveling on a string?
If a string under tension is pulled sideways and released, the tension is responsible for
accelerating a particular segment of the string back to the equilibrium position.
The acceleration of the
So what happens when the tension increases?
particular segment increases
Which means?
The speed of the wave increases.
Now what happens when the mass per unit length of the string increases?
For the given tension, acceleration decreases, so the wave speed decreases.
Which law does this hypothesis based on?
Based on the hypothesis we have laid out
v
above, we can construct a hypothetical
formula for the speed of wave
Is the above expression dimensionally sound?
Wednesday, May 5, 2004
Newton’s second law of motion
T


T
m L
T: Tension on the string
: Unit mass per length
T=kg m/s2. =kg/m
(T/)1/2=[m2/s2]1/2=m/s
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
76
Example for Traveling Wave
A uniform cord has a mass of 0.300kg and a length of 6.00m. The cord passes over a
pulley and supports a 2.00kg object. Find the speed of a pulse traveling along this cord.
5.00m
1.00m
M=2.00kg
Since the speed of wave on a string with line v  T

density  and under the tension T is
0.300kg
The line density  is  
 5.00 10  2 kg / m
6.00m
The tension on the string is
provided by the weight of the
object. Therefore
T  Mg  2.00  9.80  19.6kg  m / s 2
Thus the speed of the wave is
v
Wednesday, May 5, 2004
T


19.6
 19.8m / s
2
5.00 10
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
77
Type of Waves
• Two types of waves
– Transverse Wave : A wave whose media particles move
perpendicular to the direction of the wave
– Longitudinal wave: A wave whose media particles move
along the direction of the wave
• Speed of a longitudinal wave
For solid v 
E
r
Wednesday, May 5, 2004
E:Young’s modulus
r: density of solid
v
Elastic Force Factor
inertia factor
liquid/gas v 
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
B
r
E: Bulk Modulus
r: density
78
Example 11 – 11
Sound velocity in a steel rail. You can often hear a distant train approaching by putting
your ear to the track. How long does it take for the wave to travel down the steel track if
the train is 1.0km away?
The speed of the wave is
v
E
r

2.0 10 N / m
3
 5.110 m / s
3
3
7.8 10 kg / m
11
2
The time it takes for the wave to travel is
1.0 10 m
l
t  
 0.20s
3
v
5.110 m / s
3
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
79
Energy Transported by Waves
Waves transport energy from one place to another.
As waves travel through a medium, the energy is transferred as vibrational energy
from particle to particle of the medium.
1 2
For a sinusoidal wave of frequency f, the particles move in
E  kA
SHM as a wave passes. Thus each particle has an energy
2
Energy transported by a wave is proportional to the square of the amplitude.
energy / time power
Intensity of wave is defined as the power transported across

I 
unit area perpendicular to the direction of energy flow.
area
area
Since E is
2
2
proportional to A .
I1
I2
power
P
For isotropic medium, the I 

wave propagates radially
area
4p r 2
Ratio of intensities at I 2 P 4p r22  r  2
A2 r1
1


Amplitude

2 
two different radii is
A1 r2
I1 P 4p r1  r2 
IA
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
80
Example 11 – 12
Earthquake intensity. If the intensity of an earthquake P wave 100km from the source is
1.0x107W/m2, what is the intensity 400km from the source?
Since the intensity decreases as the square of the distance from the source,
I 2  r1 
 
I1  r2 
2
The intensity at 400km can be written in terms of the intensity at 100km
2
2
 r1 
I 2    I1   100km  1.0 107 W / m 2  6.2 105W / m 2


r
400
km
 2


Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
81
Reflection and Transmission
A pulse or a wave undergoes various changes when the medium
it travels changes.
Depending on how rigid the support is, two radically different reflection
patterns can be observed.
1.
2.
The support is rigidly fixed (a): The reflected pulse will be inverted to the original
due to the force exerted on to the string by the support in reaction to the force on
the support due to the pulse on the string.
The support is freely moving (b): The reflected pulse will maintain the original
shape but moving in the reverse direction.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
82
2 and 3 dimensional waves and the Law of
Reflection
•
•
•
Wave fronts: The whole width of
wave crests
Ray: A line drawn in the direction
of motion, perpendicular to the
wave fronts.
Plane wave: The waves whose
fronts are nearly straight
The Law of Reflection:
The angle of reflection
is the same as the angle
of incidence.
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
i=r
83
Transmission Through Different Media
If the boundary is intermediate between the previous two extremes,
part of the pulse reflects, and the other undergoes transmission,
passing through the boundary and propagating in the new medium.
When a wave pulse travels from medium A to B:
1. vA> vB (or A<B), the pulse is inverted upon reflection
2. vA< vB(or B), the pulse is not inverted upon reflection
Wednesday, May 5, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
84
Superposition Principle of Waves
If two or more traveling waves are moving through a
medium, the resultant wave function at any point is the
algebraic sum of the wave functions of the individual waves.
Superposition
Principle
The waves that follow this principle are called linear waves which in general have
small amplitudes. The ones that don’t are nonlinear waves with larger amplitudes.
Thus, one can write the
resultant wave function as
Wednesday, May 5, 2004
n
y  y1  y2      yn   yi
i 1
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
85
Wave Interferences
Two traveling linear waves can pass through each other without being destroyed or altered.
What do you think will happen to the water
waves when you throw two stones in the pond?
They will pass right through each other.
The shape of wave will
change Interference
Constructive interference: The amplitude increases when the waves meet
What happens to the waves at the point where they meet?
Destructive interference: The amplitude decreases when the waves meet
Wednesday, May 5, 2004
In phase  constructive
PHYS 1441-004, Spring 2004
Out of phase
by p/2 Yu
 destructive
Dr. Jaehoon
Out of phase not86by p/2
 Partially destructive