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UNESCO-NIGERIA TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
SCIENCE LABORATORY TECHNOLOGY
PHYSICAL PENDULUM
MECHANICS
COURSE CODE: STP 111
YEAR I- SE MESTER I
THEORY
Version 1: December 2008
1
TABLE OF CONTENT (THEORY)
1.
Rotational motion …………………………………..……..3
2.
Rotational motion (continued)……………………………..7
3.
Rotational motion (continued) …………………………… 8
4.
Rotational motion (continued) ……………………………10
5.
Rotational motion (continued) ……………………………12
6.
Surface tension ……………………………………….…..20
7.
Surface tension (continued) ………………………………22
8.
Periodic motion …………………………………………...25
9.
Periodic motion (continued …………………………….…29
10.
Periodic motion (continued) ……………………………....33
11.
Periodic motion (continued) ……………………………....41
12.
Viscosity …………………………………………….……..43
13.
Viscosity (continued) …………………………….………..44
14.
Viscosity (continued) ………………………………….…..46
15.
Viscosity (continued) ……………………………………...49
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WEEK 1 ROTATIONAL MOTION
1.1 Explain moment of inertia about an axis
The Moment of Inertia (I) of a body is a measure of the Rotational Inertia of the body. If
an object that is free to rotate about an axis is difficult to set into rotation, its Moment of Inertia
about that axis is large. An object with small I has little Rotational Inertia.
If a body is consider to be made up of tiny masses, M1, M2, M3, …, at respective distance,
r1, r2, r3 …, from an axis, its Moment of Inertia about the axis is
I = M1r12 + M2r2 + M3r33…
= ΣM1r12
The unit of I are Kg.m2
Parallel axes theorem:
The Moment of Inertia I of a body about any axis is equal to the Moment of Inertia
ICM of the body about a parallel axis through its center of mass plus the mass M of
the body multiplied by the square of the perpendicular distance l between the axes
.
I = I C.M. + Ml2
Let the axis about which the moment of inertia is to be taken be the Z – axis which is
perpendicular to the XY plane with origin at O.
Let the co-ordinates of a typical particle of mass m be (x, y, z) and the co-ordinate of the center
of gravity be (x, y, z ) and the co-ordinate of the typical particle referred to the parallel axes
through G be (x1, y1, z1);
then x = x + x1;
y = y + y1 ;
z = z + z1
3
The moment of inertia about OZ is
Σ m (x2 + y2) =
Σ m [ ( x + x1)2 + ( y + y1)2 ]
= Σ m (x12 +y12) + (x2 + y2) Σ m
= IG + Md2
where IG is the moment of inertia about a parallel axis through G and d is the distance between
the axes.
Perpendicular axes
Z
0
X
P
Y
Fig. 1.1
If OX and OY are two perpendicular axes in plane of lamina and the moments of inertia about
these axes are A and B, then the moment of inertia of the lamina about the axis OZ perpendicular
to OX and OY is A and B.
4
Let m be the mass of a particle P of lamina and (x, y) be the co-ordinates of P referred to the axes
OX and OY, where OP = r, then r2 = x2 + y2
mr2 = mx2 + my2 and this is true for all particles of the lamina.
Σ mr2 = Σ mx2 + Σ my2
1.2 Expression for moment of inertia for various bodies (rotational motion continued)
The Moment of Inertia (about an axis through the centre of mass of several uniform object each
of mass M are shown).
Analogous linear and angular quantity
mass (Inertia) m
force F
Moment of Inertia I
torque τ
5
Fig. 1.2
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WEEK 2 ROTATIONAL MOTION (CONTINUED)
1.3
Radius of gyration
Moment of inertia of an object about an axis Σmr2 is sometimes written as mk2, where m is the
mass of the object and k is a quantity called the radius of gyration about the axis. It is also
convenient to define a radius of gyration (k) for an object about an axis by relation I = Mk2 ,
where M is the total mass of the object. Hence k is the distance a point mass m must be from the
axis if the point mass is to have the same I as the object.
1. 4 Calculate Radius of gyration
The moment of inertia of a rod about an axis through one end = ML2/3 = m(L/53)2 thus the
radius of gyration k = 1/√3 = 0.58L, the moment of inertia of a sphere about it centre = 2/5ma2 0
m x L (√2/5a)2.
Thus the radius of gyration k =√2/5a = 0.63a in this case.
1.5
Torque
The quantitative measure of the tendency of a surface of course to change a body’s rotational
motion is called Torque. In general, for a force of magnitude F whose line of action is a
perpendicular distance, L, from O,
Torque is τ = FL
A Torque τ acting on a body of Inertia I produces in it an angular acceleration a given by τ = Iα
1.6
Angular momentum
Angular momentum in a vector quantity that has magnitude, Iω, and is directed along the
axis of rotation, if the net torque on a body is zero, its angular momentum will remain unchanged
in both magnitude and direction. This is the law of conservation of linear momentum.
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WEEK 3 ROTATIONAL MOTION (CONTINUED)
1.7 Relationship between torque and angular momentum.
The analog of linear momentum of a particle is angular momentum, a vector quantity denoted as
L. I ts relationship to momentum P is exactly the same as the relationship of torque to force.
T
=
r
x
F
For a particle with constant mass m, velocity v, linear momentum, p = mv, and position vector r
respective to the origin O of an Inertia frame, we define angular momentum L as
L=
r
x
p =
r
x
mv
The value of L depend on the choice of origin O, since it involves the particles position vector
relative to O, the unit of angular momentum are Kg m2s-1
Conservation of angular momentum like conservation of energy and of linear momentum,
principle of conservation of angular momentum is a universal conservation law valid at all scales
from atomic and nuclear system to the motion of galaxies. This principle follow directly from
Στ = dL
dt
if Στ = 0, then dL/dt =0
and L is constant.
When the net external torque acting as a system is zero, the total angular momentum of the
system is constant (conserved)
1.8 Law of conservation of angular momentum.
The angular momentum of a system is constant in both magnitude and direction. If the resultant
external torque acting on the system is zero, i. e., if the system is isolated.
1.9 Reduction in speed of a rotating body when stroke by small mass
The earth is an object which rotates about an axis passing through its geographic north
and south poles with a period of 1 day. If it is struck by meteorites, then since action and reaction
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are equal, no external couple acts on the earth and meteorites. Their total angular momentum is
thus conserved, neglecting the angular momentum of the meteorites about the north axis before
collision compared with that of the earth.
Angular momentum of earth poles meteorites after collision = angular momentum of
earth before collision.
Since the effective mass of the earth has increased after collision the moment of inertia
has increase. Hence the earth will show up slightly.
Similarly if a mass is dropped gently on to a turn table rotating freely at a steady speed,
the conservation of angular momentum leads to a reduction in speed of the table
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WEEK FOUR: ROTATIONAL MOTION (CONTINUED)
1.10 Write and explain the expression for the kinetic energy of rotation of a rigid body
A rotating rigid body consist of mass in motion, so it has K.E. We can express this K.E in
term of the body’s angular speed and a new quantity, called Moment of Inertia, that depends on
the body’s mass and how the mass is distributed.
We think of a body as being made up of a large number of particles, with masses m1,
m2… at distances r1, r2… from the axis of rotation, we label the particles with the index i; the
mass of the ith particles is Mi and its distance from the axis of rotation is ri which is the
perpendicular distance from the axis to the ith particle.
When a rigid body rotates about a fixed axis, the speed V; of the ith particle is given by
Vi = ri ω, where ω is he body’s angular speed. Different particle have different value of r, but ω
is the same for all (otherwise, the body wouldn’t be rigid). The kinetic energy of ith particle can
be expressed as
The total kinetic energy of the body is the sum of the K.E of all its particles.
=
Taking the common factor
out of this equation, we get
The quantity in parenthesis, obtained by multiplying the mass of each particle by the square of its
distance from the axis of rotation and adding these product is donated by I and is called the
Moment of Inertia of the body for this rotation axis.
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The word ’moment’ mean that I depend on how the mass is distributed in space. For a body with
a given rotation axis and a given total mass, the greater the distance from the axis to the particles
that make up the body, the greater the moment of inertia in the rigid body. The distance ri are all
constant and I is independent of how the body rotates around the given axis. The S.I unit of
moment of inertia is kilogram – metre2 (kg m2). In term of moment of inertia in the rotational
kinetic energy K of a rigid body is
the formula above is not a new form of energy. It’s simply the sum of kinetic energies of the
individual particles that makes up the rotating rigid body. To use the equation above, ω must be
measured in radians per second, not revolution or degrees per second, to give K in Joules. Since
we used
in our derivation, the above equation gives a single physical interpretation
of Moment of Inertia; the greater the Moment of Inertia, the greater the kinetic energy of a rigid
body rotating with a given angular speed, ω. The kinetic energy of a body equals the amount of
work done to accelerate that body from rest. So the greater a body’s Moment of Inertia, the
harder it is to start the body rotating if it is at rest and the harder it is to stop its rotation if it is
already rotating.
For this reason, I is called Rotational Inertia.
WEEK 5: ROTATIONAL MOTION (CONTINUED)
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1.11 Calculate moment of inertia about some axis
i) Uniform thin rod, axis perpendicular to length
Fig below show a slender uniform rod with mass M and length L - it might be a baton held by a
twirler in a matching band (less the rubber end caps). Compute its moment of inertia about an
axis through O at an arbitrary distance h from one end.
Fig. 1.3
Solution
The rod is a continuous distribution of mass so we must use integration to find the moment of
Inertia, we choose as an element of mass a short section of rod with length dx at a distance x
from point O.
The ratio of the mass dm of an element to the total mass M is equal to the ratio of it
length dx to the total length L
so
We’ll use I=
with r replaced by x.
Fig show that the integration limit on x are from - h to (L – h) hence we obtain
12
I = ∫ x 2dm =
Evaluation- from this general expression we can find the moment of Inertia about an axis
through any point on the rod e.g. if the axis is at the left end h = 0 and
If the axis is at right end we should get the same result. Putting h = L
We again get
If the axis passes through the centre, the usual place for a twirled baton, then h = L/2 and
ii) Moment of inertia of a thin uniform rod about an axis through its centre and perpendicular to
its length
y
G
2l
-
x
+
O
Fig. 1.4
Let the mass of the rod be M and its length be 2l.. then the mass per unit length of the rod σ is
σ = M/2l
13
Consider a small element of the rod of length δx whose distance from G is x. The moment of
inertia of this small element about Gy is σδx . x2 . The total moment of inertia is the limit of
Σσ.x2 .δx as δx → 0
=
]l
=[
= M
-l
- (M = 2σl )
iii) Moment of inertia of a rectangle
y
2a
2b
x
O
Fig 1.5
Sides 2a and 2b; mass M about one end.
Let σ be mass per unit area = M/4ab
Consider a typical strip of width δx as shown.
= .M
(M = 4. σ. a. b)
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iv) Hollow or solid cylinder, rotating about axis of symmetry
Fig shows a hollow, uniform cylinder with length L inner radius R1 and outer radius R2. It might
be a steel cylinder in a printing press or a sheet-steel rolling mill. Find the moment of Inertia
about the axis of symmetry of the cylinder.
Fig. 1.6
Solution:
We must use integral to find the moment of Inertia, but now we choose as a volume
element a thin cylindrical shell of radius r, thickness dr, and length L. all part of this element are
at very nearly the same distance from the axis.
The volume of the element is very nearly equal to that of a flat sheet with thickness dr,
length L, and width 2πr ( the circumference of the shell). Then;
We will use this expression
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I = ∫ r2dm
And integrate from r = R1 to r = R2
The moment of Inertia is given by
I = ∫ r2 dm =
=
ρ(2πrLdr)
−R
=
is usually more convenient to express the moment of Inertia in term of the total mass M of the
body, which is its density P multiplied by the total volume V. the volume is
V=
So the total mass M is
Hence the moment of Inertia is
If the cylinder is solid such as a lawn roller R1 = θ
Calling the outer radius R2 simply R1 we find that the moment of Inertia of a solid cylinder of
radius R is
16
If the cylinder has a very thin wall (like a pipe) R1 and R2 are very nearly equal, if R represents
this common radius.
I = MR2
We could have predicted this result, in a thin walled cylinder all the mass is the same distance r =
R from the axis so
I = ∫rdm = R2∫dm
= MR2
v) Uniform sphere with radius r, axis through centre
Find the moment of inertia of a solid uniform sphere (like a billiard ball or ball bearing)
about an axis though it centre.
Fig. 1.7
Solution
To calculate the moment of Inertia we divide the sphere into thin disks of thickness dx
whose moment of inertia we know from previous example we’ll integrate over these to find the
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total moment of inertia. The only tricky point is that the radius and mass of disk depend on it
distance x from the centre of the sphere.
Setup: the radius r of the disk is
r = √R2 – x2
it volume is
dv = πr2 dx = π(R2 – x2) dx
dm = ρdv = πρ(R2 – x2)dx
the moment of inertia of a disk of radius r and mass dm is
=
Integrating this expression from x = 0 to x = R given the moment of inertia of the right
hemisphere, the total I for the entire sphere include both hemisphere, is just twice this
o
I=
Carrying out the integration, we obtain
I = 8πρ R5
15
The mass M of the sphere of volume V
M
By comparing the expressions for
I and M we find
18
Note that the moment of Inertia of a solid sphere of mass M and radius R is less than the moment
of Inertia of a solid cylinder of the same mass and radius
I = MR2,
The reason is that more of the sphere’s mass is located close to the axis.
a) (Uniform rod) Thus if the mass of the rod is 60g and its length is 20cm m = 6x10-2 kg, L= 0.2m, and I
= 6x10-2 x 0.22/12 = 2x10-4kgm2.
b)(Circular disc)
Thus if the disc weigh 60g and has a radius of 10cm, m=60g = 60 x 10-2kg, a = 0.1m, so that I
= 6x10-2 x 0.12/2 = 3 x 10-4kgm2
(Sphere)
Thus if the sphere has mass kg and a radius of 0.2m, the moment of inertia = 2/5 x 4 x 0.22 =
0.064kgm2.
WEEK 6: SURFACE TENSION
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2.1 Explain surface tension
An object less dense than water, such as an air filled beach ball, floats with part of it
volume below the surface. Conversely, a paper clip can rest atop a water surface even though it
density is several time that water. This is an example of surface tension.
2.2 Explain surface tension using molecular theory
The surface of the liquid behave like a membrane under tension, surface tension arises because
the molecules of the liquid exert attractive forces on each other. There is zero net force in a
molecules inside the volume of the liquid, but a surface molecule is drawn into the volume.
Thus, the liquid tend to minimize it surface area, just as a stretched membrane does.
Fig. 2.1
2.3 Coefficient of surface tension
20
The coefficient of surface tension is define as the force per unit length acting in the surface at
right angle to one sides of a line drawn in the surface.
The unit of γ is newton-metre-1 (N m-1)
2.4 Adhesion and cohesion
Cohesion is the force of attraction between molecules of the same kind e.g the molecules of
water Adhesion is the force of attraction between molecules of water and glass
Cohesion and adhesion explain the different action of water and mercury when spilled on
a clean glass surface.
Because of the adhesion of water molecules to glass is stronger than that cohesion
between water molecules, water spread out on a clean glass surface when sprinkled on it and wet
the glass this not so in the case of mercury which does not spread on glass or wet it. The
cohesion of the mercury molecules to those of glass, as a result the mercury forms into goblet or
spherical beads on the glass surface.
2.5 Angle of contact
Liquid has no definite shape so it assumes the shape of the container in which it is kept
though the material of the container also effect the shape. The angle of contact θ is the angle
between the tangent to the liquid surface and the tangent to the solid surface, both drawn in the
place at right angle to both surfaces and meeting at the point of contact measured within the
liquid.
For a liquid in which the cohesive force is greater than the adhesive force, the angle of
contact is obtuse e.g. mercury. If the adhesive force exceeds the cohesive force of the liquid
molecules, the angle of contact is very small or almost zero, like water.
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WEEK 7 SURFACE TENSION (CONTINUED)
2.6 Capillarity
Capillarity or capillary action is the tendency of a liquid to rise or fall in a narrow tube. In both
the water and soap solution the surface of the liquid or its meniscus curves upward. But in
mercury the meniscus is curved downwards away from the glass tube. Cohesion and adhesion as
well as surface tension forces are responsible for the capillarity of liquids. Water wets glass tube,
spreading a thin film of water on the inner surface of the tube. In capillary tube, water is held up
as it creeps by surface tension force acting on the circumference of the meniscus. The water, thus
keeps rising in the tube until the weight of water drawn up in the tube balances the surface
tension acting at the top of the column of water.
The adhesion of mercury to glass is much less than the cohesion of mercury molecules,
hence, mercury is depressed in capillary tubes.
Capillarity also explain
1) Why a drop of water remains for a time on the surface of a dry cloth before being
absorbed, but a similar drop on a damp cloth is rapidly absorbed.
2) Liquid candle wax rises up the wick of a candle or kerosene rises up the wick of a lamp.
3) Water rises in the stem of plants
4)
And a blotting paper absorbs ink - when blotting paper is used to dry ink, the liquid rises
up the pores of the paper when it is pressed on the ink.
2.7 Variation of surface tension with temperature
It was determined by Jaegers method. Experiment shows that the surface tension of liquid
and water in particular decreases with increase in temperature along a fair smooth curve. The
decrease of surface tension with temperature may be attributed to the greater average
22
separation of the molecules at higher temperature. The force of attraction between molecules
is then reduced.
2.8 Surface tension in terms of surface energy
Molecules in the surface of a liquid have potential energy. A molecules in the back of
the liquid form bonds with more neighbours than one in the surface. Thus bond must be
broken i.e. work must be done, to bring a molecules into the surface. Molecules in the surface
of the liquid hence have more potential energy than those in the back.
2.9 Surface tension and specific latent heat
Inside a liquid molecules moves about in all direction continuously breaking and reforming
bonds with neighbours.
If a molecules in the surface passes into the vapour outside a definite amount of energy is
needed to permanently break the bonds with molecules in the liquid.
This amount of energy is the work done in over coming the inward force on a molecules
in the surface. The energy needed to evaporate a liquid is related to its surface energy or
surface tension.
The latent heat of vaporization, which is the energy needed to change liquid to vapour at
the boiling point, is therefore related to surface energy.
2.10 Calculation
A soap bubble in a vacuum has a radius of 3cm and another soap bubble in the vacuum has a
radius of 6cm. if the two bubbles coalesce under isothermal conditions, calculate the radius of
the bubble formed.
23
Since the bubbles coalesce under isothermal conditions, the surface tension y is constant.
Suppose R is the radius in cm, R x 10-2m, of the bubble formed.
Then
work done = γ x surface area = γ x 8πR2 x 10-4
But
original work done = (γ x 8π.32 + γ x 8π.62) x 10-4.
γ x 8πR2 = γ x 8π.32 + γ.8π.62.
R2 = 32 + 62.
R = 32 + 62 = 6.7cm.
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WEEK 8 PERIODIC MOTION
3.1 (i) Periodic Motion
Fig. 3.1
Many kind of motion repeat them selves over and over - the vibration of a quartz crystal in a
water, the swing sound vibrations produced by a clarinet or an organ pipe and the back and forth
motion of the positions of a car engine. This kind of motion are called periodic motion or
oscillation
A body that undergoes periodic motion always has a stable equilibrium position. When it
is moved away from this position and released, a force or torque come into a play to pull it back
toward equilibrium. But by the time it gets there, it has pulled up some kinetic energy, so it
overshoots, stopping somewhere on the other side and it again pulled back toward equilibrium,
but by the time it gets there, it has packed up some kinetic energy, soft overshoots, stopping
somewhere on the other side, and is again pulled back toward equilibrium, picture a ball rolling
back and forth in a round bows or a pendulum that swing back and forth past its straight dawn
position.
(ii) Simple harmonic motion is the vibratory motion which a system that obeys Hooke’s law
undergoes. Because of the resemblance of it graph to a sine, or cosine curve simple harmonic
25
motion is frequently called sinusoidal motion. A central feature of simple harmonic motion is
that the system oscillates at a single constant frequency, that is, what makes it ‘simple’ harmonic.
A special kind of periodic motion occur in mechanical system when the force acting on an
object is proportional to the position of the object relative to some equilibrium position. If this
force is always directed toward the equilibrium position, the motion is called simple harmonic
motion.
3.2
List example of system performing simple harmonic motion
When the bob of a pendulum moves to and fro through a small angle, the bob is said to be
moving with simple harmonic motion. The prongs of a sounding turning fork, and the layers of
air near it, are moving with simple harmonic motion, and light waves can be considered due to
simple harmonic motion.
3.3
Amplitude
Here are some term that we will use in discussing periodic motion of all kind the
Amplitude of the motion, denoted by A, is from maximum magnitude of displacement from
equilibrium that is, the maximum value of /X/
DIAGRAM
26
Fig. 3.2
27
The Period T is the time for one cycle. It is always positive. The SI unit is the second, but
is some times expressed as” seconds per cycle.”
The frequency f is the number of cycle in a unit of time. It is always positive. The SI unit
of frequency is the hertz.
I hertz = 1Hz = 1cycle/s = 1 s-1
The Angular frequency, ω, is 2π times the frequency:
ω=2πf
ω represents the rate of change of angular quantity and is always measure in radians, so its unit
are rad/s. From the definition of period T and frequency f we see that each is the reciprocal of the
other:
The relationship between f and T also from the definition of ω,
ω = 2πf =
28
WEEK 9: PERIODIC MOTION (CONTINUED)
3.4
State and explain the expression for the period of oscillation of the following:
i) simple pendulum
ii) compound pendulum
iii) loaded elastic spring
It is an idealized model consisting of a point mass suspended by a light string. When the point
mass is pulled to one side of its straight-down equilibrium position and released, it oscillation
about the equilibrium position.
Familiar situation such as a wrecking back on a cranes cable or a person on a swing can
be modeled as simple pendulum.
29
Fig. 3.3 30
The path of the point mass (pendulum bob) is not a straight line but the arc of a circle with radius
L equal to the length of the string (fig above). We use as our coordinate the distance X measured
along the arc, if the motion is S.H.M, the restoring force must be directly proportional to X.
(Fig below) we represent the forces on the mass in term of tangential and radial component. The
restoring force Fθ is a tangential component of the net force.
Fθ = - mgSinθ
The restoring force is provided by gravity; the Tension T merely act
to make the point mass move in an arc, the restoring for is proportional not to θ, but to
the
motion
is
not
S.
H.
M
.
How
ever,
if
the
angle
is
, so
θ
in
Fig. 3.4
radians, for example, when θ=0.1 rad., sin θ=0.0998, a different of only 0.2% with this
approximation
31
Fθ = - mg sin θ becomes
Fθ = -mg θ = - mgx/l
or Fθ= - mgx/l
The restoring force is then proportional to the coordinate for small displacement, and the
force constant is K=mg/l
(S H M)
=
(simple pendulum, small amplitude)
The corresponding frequency and period relationship are
L
Note that the expression do not involve the mass of the particle, this is because, the restoring
forces, a component of the particles weight is proportional to mass. Thus the mass appears on
both side of F=ma and cancels out (this is the same physics that explain why bodies of different
masses fall with the same acceleration in vacuum.)
32
WEEK 10 PERIODIC MOTION (CONTINUED)
3.4(i) Simple pendulum
For small oscillations, the period of pendulum for a value of g is determined entirely by its
length. ct. A long pendulum has a longer period than a shorter one. Increasing g increase the
restoring force, causing the frequency to increase and the period to decrease.
ii. Physical pendulum (compound pendulum)
A physical pendulum is any real pendulum that uses an extended body, as contrasted to
the idealized model of the simple pendulum with all the mass concentrated at a single point. For
Fig. 3.5
33
PHYSICAL PENDULUM
small oscillation, analyzing the motion of a real physical pendulum is almost as easy as for a
single pendulum.
show a body of irregular shape pivoted so that it can turn without friction about an axis through
point O. in the equilibrium position the centre of gravity is directly below pivot. In the position
shown in the fig, the body is displaced from equilibrium by an angle θ, which we use as a
coordinate for the system. The distance from θ to the centre of gravity is d. the moment of inertia
of the body about the axis of rotation through θ is I, and the total mass is m, when the body is
displaced as shown, the weight mg causes a restoring torque.
T2 = - (mg) (dsinθ)
The negative sign shows that the restoring torque is clockwise when the displacement is
counted clockwise, and vise versa.
When the body is released, it oscillates about it equilibrium position. The motion is not
∫.H because the torque Tz is proportional to Sinθ rather than to θ itself. However, if θ is small,
34
we can approximate sinθ in radian just as we did in analyzing the simple pendulum. Then the
motion is approximately simple harmonic with this approximation.
Tz= - (mgd) θ
The equation of motion is
ΣTz = Iαz= I d so
- (mgd) θ = Iαz = I d2θ
dt2
Comparing this equation above with this equation below
ax = d2x = - k x (S.H.M)
dt2
m
we see that the role of (K/m) for spring mass system is played here by the quantity (mg d/I). thus
the angular frequency is
ω = √mgd
(physical pendulum, small amplitude)
I
The frequency F is 1/2π time this, and the period T is
T = 2π √ I
mgd
The above equation is the basics of a common method for experimentally determining the
moment of Inertia of a body with a complicated shape < first located the centre of gravity of the
body by balancing, then suspend the body so that it is free to oscillate about an axis, and measure
the period T of small amplitude oscillations. Finally use the above equation to calculate the
moment of inertia I of the body about this axis from T, the body’s mass m, and the distance d
from the axis to the centre of gravity.
35
Bio mechanics researchers use this method to find the moment of inertia of an animal limbs.
(iii)
Torsional pendulum. (loaded elastic spring)
Fig. 3.6
Show a rigid object suspended by a wire attached at the top to a fixed support. When the object is
twisted through some angle θ. The twisted wire exert on the object a restoring torque that is
proportionally to the angular position that is;
T = -Kθ
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where K is called the torsion constant of the support wire. The value of K can be obtained by
applying a known torque to twist the wire through a measurable angle, θ. Applying Newton
second law of rotational motion, we find
T = - Kθ
= I d2θ
dt2
d2θ = -K θ
dt2
I
again, this is the equation of motion for a simple harmonic oscillator with
ω = √K/I
T = 2π √ I
K
This system is called a torsional pendulum. There is no small angle restriction in this situation as
long as the elastic limit of the wire is not exceeded.
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Fig. 3.7
3.5 Draw and explain the graph of potential energy, kinetic energy and total energy
against distance
A block attached to a spring moving on a frictionless surface (a) when the block is displaced
to the right of equilibrium (x > O) the force exerted by the spring act to the left, (b) when the
block is at its equilibrium position (x = O), the force exerted by the spring is zero, (c) when the
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block is displaced to the left of equilibrium ( x < O), the force exerted by the spring act to the
right.
Let us examine the mechanical energy of the block-spring illustrated in figure above,
because the surface is frictionless, we expect the total mechanical energy of the system to be
constant, we assume a light spring, so the kinetic energy of the system corresponds only to that
of the block. From the relation v =
=
K=
= - ω A sin (ωt + φ)
mw2 A2 Sin2 (wt +φ)
(kinetic energy SHM)
The elastic potential energy stored in the spring for any elongated X is given by
X(t) = A cos (ωt + φ)
U=
=
k A2 Cos2(ωt + φ)
(potential energy of S H M )
We see that K and U are always positive quantities. Because
E = K+U =
= K/M ,we can express
KA2 Sin2 (ωt + φ) +Cos2(ωt+φ)]
From the identity sin2 θ + cos2 θ =1
We see that the quantities in square bracket, is unity, therefore, this equation reduce to
E=
Total energy of Simple Harmonic Motion, that is, the total mechanical energy of, S.H.M
is a constant of the motion and is proportional to the square of the amplitude. U is small when K
is large and vice versa because the sum must be constant.
In fact the total mechanical energy is equal to the maximum potential energy stored in the
spring when X =
A because v = 0 at these point and thus there is no kinetic energy. At the
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equilibrium position, where U = O because x = 0, the total energy, all in the form of kinetic
energy is again
That is
(at
Fig. 3.8
40
)
WEEK 11 : PERIODIC MOTION (CONTINUED)
3.6. Calculate velocities of bodies in periodic motion
i) A steel strip clamped at one end, vibrates with frequency of 20Hz and an amplitude of 5mm at
the free end, where a small mass of 2g is positioned. Find
(a). The velocity of the end when passing through the zero position.
(b) the acceleration of maximum displacement.
(c) the maximum kinetic and potential energy of the mass.
Suppose
y = r sin ωt represent the vibration of the strip where r is the amplitude.
Solution:
(a.) the velocity V = ω √(r2 – y2), when the end of the strip passes through the zero position y =
0 and the maximum speed m1 is given by
Vm = ωr
Now
ω = 2πf
= 2π x 20 and r = 0.005m.
Vm = 2π x 20 x 0.005 = 628 ms-1
(b). The acceleration = - ω2y = -ω2r at maximum displacement
am = (2π x 20)2 x 0.005
= 79 ms-1
(c. )
m = 2g = 2x10-3 kg, Vm = 0.628ms-1
maximum K. E. = m
2
= ½ x (2x10-3) x 0.06282 = 3.9 x 10-4 J
maximum P. E. (V – O) = maximum K. E. = 3.9 x 10-4 J
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ii)
A small bob of mass 20g oscillates as a simple pendulum, with amplitude 5cm and period
2 second.
Find the velocity of the bob and tension in the supporting thread, when the velocity of the
bob is a maximum.
Solution:
The velocity V of the bob is a maximum when it passes through its original position. The
maximum velocity Vm is given by
Vm = ωr
where r is the amplitude of 0.05. Since
T = 2π/ω and
Vm = ωr = π x 0.05 = 0.16 ms-1
If P is the tension in the thread the net force toward the centre along which the bob move is
then given by (P – mg) the acceleration toward the centre of the circle, which is the point of
suspension is
2
/L where l is the length of the pendulum.
P – mg = mVm2
L
P = mg + mVm2
L
L=
now
g
gT2 = g x 4
4π2
4π2
Since m = 0.02kg, g = 9.8 m
T = 2π √ L
= g/π2
, it follow from above that
P = 0.02 x 9.8 + 0.02 x (0.05π)2 x π2
9.8
= 19.65 x 10 -2 N
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WEEK 12: VISCOSITY
4.1 Explain viscosity applying molecular theory
A viscous fluid always tend to clinch to a solid surface in contact with it. There is always
a thin boundary layer of fluid near the surface in which the fluid is nearly at rest with respect to
the surface. That is why dust particle can cling to a fan blade even when it is rotating rapidly, and
why you can’t get all the dirt off your car by just squirting a hose at it.
Viscosity is internal friction, viscosity opposes the motion of one portion of a fluid
relative to another. Viscosity is the reason it take effort to paddle a canoe through calm water,
but it is also the reason the paddle work. Viscous effect are important in the flow of fluids in
pipe, the flow of blood, the lubrication of engine part, any many other situations.
Fluid that flow steadily, such as water, have smaller viscosities than do thick liquid, such
as honey or motor oil. Viscosities of all fluid are strongly temperature dependent, increasing for
gases and decreasing for liquid, as the temperature increase. Lava is an example of viscosity
fluid, the viscosity decrease with increasing temperature. The hotter the lava, the more easily it
can flow.
An important goal in the design of oils for engine lubrication is to reduce the temperature
variation of viscosity as much as possible.
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WEEK 13 VISCOSITY (CONTINUED)
4.2 Define velocity gradient in a fluid
4.4 State and explain Newton’ s formula for viscosity
When water flow slowly and steady through a pipe, the layer A of the liquid in contact with the
pipe is practically stationary, but the central part C of the water is moving relatively fast. At
other layer between A and C such as B, the water has a velocity less than at C, the magnitude of
the velocity is represented by the length of the arrow line in the figure below.
A
B
C
B
A
Fig. 4.1
Now as in the case of two solid surfaces moving over each other, a frictional force is exerted
between two liquid layers when they move against each other. Thus because of the velocities of
neighbouring layers are different, frictional force occur between the various layer of the liquid
when flowing through a pipe. F is the frictional force in a liquid, Newton saw that the larger the
area of the surface of liquid considered, the greater was the frictional force F. He also stated that
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F was directly proportional to velocity gradient at the part of the liquid considered. If V1, V2 are
the velocity of C, B respectively, and h their distance apart, the velocity gradient between the
liquid is define as (V1 – V2)/h. The velocity gradient can thus be expressed in
(m/s)/m
or
as
‘s- 1’
Thus if A is the area of the liquid surface considered, the frictional force F on the surface is given
by
F ∞ A x velocity gradient
F = K x A x velocity gradient
F = ηA x velocity gradient
Where η is a constant of the liquid known as coefficient of viscosity.
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WEEK 14 VISCOSITY (CONTINUED)
4.3 Distinguish between streamline and turbulent flow
laminar- when fluid is in motion, its flow can be characterized as being one of two main type the flow is said to be steady or laminar, if each particle of the fluid follows a smooth path, such
that the path of different particle never cross each other in steady flow, the velocity of fluid
particle passing any point remain constant in time.
Fig. 4.2
turbulent - above a certain critical speed fluid flow becomes turbulent. Turbulent flow is
irregular flow characterized by small whirlpool-like region. Hot gases from a cigarette made
visible by smoke particle, the smoke first moves in laminar flow at the bottom and then in
turbulent flow as shown in the diagram above.
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4.5 Define the coefficient of viscosity stating the unit.
The coefficient of viscosity, η is defined as force acting normally in liquid per unit area per unit
velocity gradient;
η=
force
Area x Velocity gradient
4.6 Poiseulle’s law
The fluid through a cylindrical pipe of length L and cross-sectional radius R is given by
Volume per second = πR4 (Pi - Po)
8ηL
Where P1-Po is the pressure difference between the two end of the pipe (input minus output)
Poiseulle investigated steady flow of liquid through a pipe and derived an expression for the
volume of liquid issuing per second from the pipe.
Deriving the formula by dimension
The volume of liquid issuing per second from the pipe depend on
i.
The coefficient of viscosity η
ii.
The radius R, of the pipe
iii.
The pressure gradient of set up along the pipe. The pressure gradient = P/L
P = Pressure difference between the end of the pipe and L is its length.
xyz are indices require to be found
Suppose
The dimension of volume per second are L3T – 3 for η are Ml-1T-1, a is L and g are
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or MLT-2
L xL
which is
[pressure] or [force]
[Length]
[area][Length]
2
ML-2T-2
From (1) equating dimension on both sides
L3T-1
(ML-1T-1) x Ly (ML-2T-2)z
Equating the respective indices of M, LT on side we have
-x + y = 2z = 3
x+2=o
x + 2z = 1
solving we obtain x = -1 z=1 y=4
from (1) volume per second
=
η
we cannot obtain the numerical factor K from the method of dimensions. The factor π/8
enter into the formula which is
volume per second
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WEEK 15 VISCOSITY (CONTINUED)
4.7 Describe and explain the motion of a small spherical body falling
through a viscous fluid
4.8 Explain terminal velocity
When a stone falls through a viscous liquid it is subjected to three forces;
It weight (W)
acting downward, the up thrush (U) of the liquid (V) opposing its motion, the viscous force
acting opposite to the motion of the stone is upward
Fig. 4.3
49
We can therefore write the equation of the stone as
W – U –V = ma
Where a is the acceleration of the stone through the liquid and m is the mass of the stone. The
viscous force V increases with the speed of the stone. As the stone falls faster and faster through
the liquid, the viscous force opposing speed, the viscous drag balances the down ward force of
the weight of stone. At this point the stone moves with constant velocity (termed TERMINAL
VELOCITY) because it acceleration a is zero. Hence, our equation now becomes
W-V-U= Ma =0
Or
V=W-U
Rain drop whose motion is opposed by the Viscosity of air attains a terminal velocity
before
reaching
the
ground.
Similarly,
any
object
released
It was Stokes who first discovered that the terminal velocity of small particle is proportional to
their weight. This relation is known as Stokes laws.
4.9 State and explain Stokes law.
4.10 Write the expression for the terminal velocity of a small spherical ball
Stoke formula can be used to compare the coefficient of viscosity of very viscous liquid as
glycerin. A tall glass vessel G is filled with the liquid, and a small ball bearing P is dropped
gently in to the liquid so that it falls along the axis of G toward the middle of the liquid, P
reaches its terminal velocity Vo, which is measured by timing its fall through a distance AB or
BC.
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The Up thrust, U, on the bearing, Z = 4 πa3δ g/3, where a is the radius of ρ and δ is the density of
the liquid. The weight, W, of ρ is 4πa3ρg/3 where ρ is density of the bearing’s material. The net
downward force is thus 4πa3g (ρ – δ)/3. When the opposing frictional force grows to this
magnitude, the resultant force on the bearing is zero and it attains its terminal velocity, vo.
6πηavo = 4 πa3g (ρ– δ)
3
η = 2ga2(ρ-δ)
9vo
Repeating the experiment with liquid of coefficient of viscosity η, and density δ, then
η1 = 2ga2(ρ-δ)
9v1
v1 is the terminal velocity divide both equation
η = v1(ρ-δ)
η1
vo(ρ-δ1)
4 .12 Explain the importance of viscosity in lubricant.
Viscosity has an important effect on the flow of liquid through pipes, including the flow of
blood in the circular system.
Oils, grease, and air are used as lubricants because of their viscosity. Engine Oils are used in
lubricating engines and other machines so as to keep metal surfaces from rubbing against each
other to prevent wear and tear in machine part.
4.13 Explain the effect of temperature on the viscous of liquid
The viscosity of Oils and grease decrease with temperature hence their lubricating effect are
lowered at high temperature. Water is not used as a lubricant because it has low viscosity.
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4.14
Derive Bernoulli’s equation.
Bernoulli’s equation for the steady flow of a continuous stream of fluid.
Consider two different point along the stream path, let point 1 be at height, hi, and let Vi,
ρi and Pi be the fluid speed, density, and pressure at that point. Similarly define h2, V2, ρ2 and P2,
for point 2, then provided the fluid is incompressible and has negligible viscosity
where
and g is the accelaration due to gravity.
4.15 List some application of Bernoulli’s principle.
A filter pump has a narrow section in the middle, so that a jet of water from the tap flows
faster here. This causes a drop in pressure near it and air therefore flows in from the side tube to
which a vessel is connected. The air and water together are expelled through the bottom of the
filter pump.
2. A suction effect is experienced by a person standing close to the platform at a station when a
fast train passes. The fast moving air between the person and train produces a decrease in
pressure and the excess air pressure on the other side pushes the person toward the train.
3. . The curved shape of an aerofoil creates flow of air over its top surface than the lower one.
Fig. 4.4
52
This is shown by the closeness of streamlines above the aerofoil compared with those below
from Bernoulli’s principle, the pressure of air below is greater than that above, and this produce
the lift on aerofoil
4.16
State the dimension of coefficient viscosity
The dimension of force ( = mass x acceleration = mass x velocity change/time) are MLT-2
The dimension of velocity gradient
= velocity change
=
Distance
η
=
F
A x velocity gradient
dimension of
η = MLT-2
L2 x 1/T
= ML -1 T-1
Thus η may be expressed in unit ‘kgm-1 S-1.
53