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CLASSICAL MECHANICS
Chapter 1. Centres of Mass
1.1 Introduction and Some Definitions 1.2 Plane Triangular Lamina 1.3 Plane Areas 1.4 Plane Curves 1.5 Summary of the Formulas for Plane Laminas and Curves 1.6 The Theorems of Pappus 1.7 Uniform Solid Tetrahedron, Pyramid and Cone 1.8 Hollow Cone 1.9 Hemispheres 1.10 Summary
Chapter 2. Moment of Inertia
2.1 Definition of Moment of Inertia 2.2 Meaning of Rotational Inertia 2.3 Moments of Inertia of Some Simple Shapes 2.4 Radius of Gyration 2.5 Plane Laminas and Mass Points distributed in a Plane 2.6 Three-dimensional Solid Figures. Spheres, Cylinders, Cones 2.7 Three-dimensional Hollow Figures. Spheres, Cylinders, Cones 2.8 Toroids 2.9 Linear Triatomic Molecule 2.10 Pendulums 2.11 Plane Laminas. Product Moment. Translation of Axes (Parallel Axes Theorem). 2.12 Rotation of Axes 2.13 Momental Ellipse 2.14 Eigenvectors and Eigenvalues 2.15 Solid Body 2.16 Rotation of Axes - Three Dimensions 2.17 Solid Body Rotation. The Inertia Tensor 2.18 Determination of the Principal Axes 2.19 Moment of Inertia with Respect to a Point 2.20 Ellipses and Ellipsoids
Chapter 3 Systems of Particles
3.1 Introduction 3.2 Moment of a Force 3.3 Moment of Momentum 3.4 Notation
3.5 Linear Momentum 3.6 Force and Rate of Change of Momentum 3.7 Angular Momentum 3.8 Torque 3.9 Comparison 3.10 Kinetic energy 3.11 Torque and Rate of Change of Angular Momentum 3.12 Torque, Angular Momentum and a Moving Point 3.13 The Virial Theorem
Chapter 4. Rigid Body Rotation 4.1 Introduction 4.2 Angular Velocity and Eulerian Angles 4.3 Kinetic Energy 4.4 Lagrange’s Equations of Motion 4.5 Euler’s Equations of Motion 4.6 Force-free Motion of a Rigid Asymmetric Top 4.7 Nonrigid Rotator 4.8 Force-free Motion of a Rigid Symmetric Top 4.9 Centrifugal and Coriolis Forces 4.10 The Top
Chapter 5. Collisions 5.1 Introduction 5.2 Bouncing Balls 5.3 Head-on Collision of a Moving Sphere with an Initially Stationary Sphere 5.4 Oblique Collisions 5.5 Oblique (Glancing) Elastic Collisions, Alternative Treatment
Chapter 6. Motion in a Resisting Medium
6.1 Introduction 6.2 Uniformly Accelerated Motion 6.3 Motion in which the Resistance is Proportional to the Speed 6.4 Motion in which the Resistance is Proportional to the Square of the Speed.
Chapter 7. Projectiles
7.1 No Air Resistance 7.2 Air Resistance Proportional to the Speed 7.3 Air Resistance Proportional to the Square of the Speed
13.5 Acceleration Components 13.6 Slithering Soap in Conical Basin 13.7 Slithering Soap in Hemispherical Basin 13.8 More Examples 13.9 Hamilton’s Variational Principle
Chapter 14 Hamiltonian Mechanics
14.1 Introduction 14.2 A Thermodynamics Analogy 14.3 Hamilton’s Equations of Motion 14.4 Examples
Chapter 15. Special Relativity
15.1 Introduction 15.2 The Speed of Light 15.3 Preparation 15.4 Speed is Relative. The Fundamental Postulate of Special Relativity. 15.5 The Lorentz Transformations 15.6 But This Defies Common Sense 15.7 The Lorentz Transformation as a Rotation 15.8 Timelike and Spacelike 4-Vectors 15.9 The FitzGerald-Lorentz Contraction 15.10 Time Dilation 15.11 The Twins Paradox 15.12 A, B and C 15.13 Simultaneity 15.14 Order of Events, Causality and the Transmission of Information 15.15 Derivatives 15.16 Addition of Velocities 15.17 Aberration of Light 15.18 Doppler Effect 15.19 The Transverse and Oblique Doppler Effects 15.20 Acceleration 15.21 Mass 15.22 Momentum 15.23 Some Mathematical Results 15.24 Kinetic Energy 15.25 Addition of Kinetic Energies 15.26 Energy and Mass 15.27 Energy and Momentum 15.28 Units 15.29 Force
15.30 Electromagnetism
Chapter 16. Hydrostatics
16.1 Introduction 16.2 Density 16.3 Pressure 16.4 Pressure on a Horizontal Surface. Pressure at Depth z 16.5 Pressure on a Vertical Surface 16.6 Centre of Pressure 16.7 Archimedes’ Principle 16.8 Some Simple Examples 16.9 Floating Bodies
Chapter 17. Vibrating Systems
17.1 Introduction 17.2 The Diatomic Molecule 17.3 Two Masses, Two Springs and a Brick Wall 17.4 Double Torsion Pendulum 17.5 Double Pendulum 17.6 Linear Triatomic Molecule 17.7 Two Masses, Three Springs, Two brick Walls 17.8 Transverse Oscillations of Masses on a Taut String 17.9 Vibrating String 17.10 Water 17.11 A General Vibrating System 17.12 A Driven System 17.13 A Damped Driven System
Appendix A. Miscellaneous Problems
Appendix B. Solutions to Miscellaneous Problems
If the masses are distributed in three dimensional space, with m 1 at ( x 1 , y 1, z 1 ), etc,. the centre of mass is a point ( x , y z , )such that
M
m x x = ∑ i i M
m y y ∑ i i = M
m z z = ∑ i i 1.1.
In this case, (^) ∑ m xi i , (^) ∑ m yi i , (^) ∑ m zi i are the first moments of mass with respect to the y-z, z-x and
x-y planes respectively.
In either case we can use vector notation and suppose that r 1 , r 2 , r 3 are the position vectors of m 1 , m 2 , m 3 with respect to the origin, and the centre of mass is a point whose position vector r is defined by
M
∑^ m^ ii
r r 1.1.
In this case the sum is a vector sum and (^) ∑ mi r i , a vector quantity, is the first moment of mass
with respect to the origin. Its scalar components in the two dimensional case are the moments with respect to the axes; in the three dimensional case they are the moments with respect to the planes.
Many early books, and some contemporary ones, use the term "centre of gravity". Strictly the centre of gravity is a point whose position is defined by the ratio of the first moment of weight to the total weight. This will be identical to the centre of mass provided that the strength of the gravitational field g (or gravitational acceleration) is the same throughout the space in which the masses are situated. This is usually the case, though it need not necessarily be so in some contexts.
For a plane geometrical figure, the centroid or centre of area, is a point whose position is defined as the ratio of the first moment of area to the total area. This will be the same as the position of the centre of mass of a plane lamina of the same size and shape provided that the lamina is of uniform surface density.
Calculating the position of the centre of mass of various figures could be considered as merely a make-work mathematical exercise. However, the centres of gravity, mass and area have important applications in the study of mechanics.
For example, most students at one time or another have done problems in static equilibrium, such as a ladder leaning against a wall. They will have dutifully drawn vectors indicating the forces on the ladder at the ground and at the wall, and a vector indicating the weight of the ladder. They will have drawn this as a single arrow at the centre of gravity of the ladder as if the entire weight of the ladder could be "considered to act" at the centre of gravity. In what sense can we take this liberty and "consider all the weight as if it were concentrated at the centre of gravity"? In fact
the ladder consists of many point masses (atoms) all along its length. One of the equilibrium conditions is that there is no net torque on the ladder. The definition of the centre of gravity is such that the sum of the moments of the weights of all the atoms about the base of the ladder is equal to the total weight times the horizontal distance to the centre of gravity, and it is in that sense that all the weight "can be considered to act" there. Incidentally, in this example, "centre of gravity" is the correct term to use. The distinction would be important if the ladder were in a nonuniform gravitational field.
In dynamics, the total linear momentum of a system of particles is equal to the total mass times the velocity of the centre of mass. This may be "obvious", but it requires formal proof, albeit one that follows very quickly from the definition of the centre of mass.
Likewise the kinetic energy of a rigid body in two dimensions equals 21 MV^2 + 21 I ω^2 ,where M
is the total mass, V the speed of the centre of mass, I the rotational inertia and ω the angular speed, both around the centre of mass. Again it requires formal proof, but in any case it furnishes us with another example to show that the calculation of the positions of centres of mass is more than merely a make-work mathematical exercise and that it has some physical significance.
If a vertical surface is immersed under water (e.g. a dam wall) it can be shown that the total hydrostatic force on the vertical surface is equal to the area times the pressure at the centroid. This requires proof (readily deduced from the definition of the centroid and elementary hydrostatic principles), but it is another example of a physical application of knowing the position of the centroid.
1.2 Plane triangular lamina
Definition: A median of a triangle is a line from a vertex to the mid point of the opposite side.
Theorem I. The three medians of a triangle are concurrent (meet at a single, unique point) at a point that is two-thirds of the distance from a vertex to the mid point of the opposite side.
Theorem II. The centre of mass of a uniform triangular lamina (or the centroid of a triangle) is at the meet of the medians.
The proof of I can be done with a nice vector argument (figure I.1):
Let A , B be the vectors OA, OB. Then A + B is the diagonal of the parallelogram of which OA and OB are two sides, and the position vector of the point C 1 is 13 ( A + B ).
To get C 2 , we see that
C 2 = A + 23 (AM 2 ) = A + 23 ( M 2 − A ) = A + 23 ( 12 B − A ) = 13 ( A + B )
1.3 Plane areas.
Plane areas in which the equation is given in x-y coordinates
We have a curve y = y ( x ) (figure I.3) and we wish to find the position of the centroid of the area under the curve between x = a and x = b. We consider an elemental slice of width δ x at a distance x from the y axis. Its area is y δ x, and so the total area is
= (^) ∫
b a A ydx 1.3.
The first moment of area of the slice with respect to the y axis is xy δ x , and so the first moment
of the entire area is (^) ∫
b a xydx.
Therefore A
xydx
ydx
xydx x
b a b a
b a ∫
∫
∫ = = 1.3.
FIGURE I.
For y we notice that the distance of the centroid of the slice from the x axis is 12 y , and
therefore the first moment of the area about the x axis is 12 y.y δ x.
Therefore A
ydx y
b a 2
2 ∫ = 1.3.
Example. Consider a semicircular lamina, x^2 + y^2 = a^2 , x > 0 , see figure I.4:
We are dealing with the parts both above and below the x axis, so the area of the semicircle is
= (^) ∫
a A 2 0 ydx and the first moment of area is (^2) ∫
a 0 xydx.^ You^ should^ find x = 4 a /( 3 π) = 0. 4244 a.
Now consider the lamina x^2 + y^2 = a^2 , y > 0 (figure I.5):
FIGURE I.
FIGURE I.
∫
β α 31 r^3 cosθ^ d θ.
Therefore. 3
2 cos 2
3
∫
∫ β α
β α θ
θ θ
r d
r d x 1.3.
Similarly. 3
2 sin 2
3
∫
∫ β α
β α θ
θ θ
r d
r d y 1.3.
Example : Consider the semicircle r = a , θ = −π/2 to +π/2.
cos 3
cos 2 3
2 /^2
/ (^2) / 2 / 2
/ 2 / 2 ∫ ∫
∫ +π +π −π −π
+π −π π
θ θ = π
θ
θ θ
a d
a d
a d x 1.3.
The reader should now try to find the position of the centroid of a circular sector (slice of pizza!) of angle 2α. The integration limits will be −α to +α. When you arrive at a formula (which you should keep in a notebook for future reference), check that it goes to 4 a /(3π ) if α = π/2, and to 2 a /3 if α = 0.
1.4 Plane curves Plane curves in which the equation is given in x-y coordinates
FIGURE
I.
Figure I.7 shows how an elemental length δ s is related to the corresponding increments in x and y :
( ) [ 1 ( / )] [( / ) 1 ]. 2 21 /^221 /^221 /^2 δ s = δ x +δ y = + dy dx δ x = dx dy + δ y 1.4.
Consider a wire of mass per unit length (linear density) λ bent into the shape y = y x ( ) between x = a and x = b. The mass of an element ds is λ δ s , so the total mass is
[ 1 ( / )]. 2 1 /^2 ds dy dx dx
b ∫ λ^ =∫ a λ + 1.4.
The first moments of mass about the y - and x -axes are respectively
∫ λ [^ +^ (^ )]
b a x^ dy dx dx
21 /^2 1 / and [ 1 ( / ) )]. 2 1 /^2 y dy dx dx
b ∫ a λ^ + 1.4.
If the wire is uniform and λ is therefore not a function of x or y , λ can come outside the integral signs in equations 1.12 and 1.13, and we hence obtain
[ ( )]
[ ( )]
[ ( )]
[ ( )]
(^21) / 2
21 /^2
(^21) / 2
(^21) / 2
∫
∫
∫
∫
= (^) b a
b a b a
b a dy dx
y dy dx y dy dx dx
x dy dx dx x 1.4.
the denominator in each of these expressions merely being the total length of the wire.
Example : Consider a uniform wire bent into the shape of the semicircle x^2 + y^2 = a^2 , x > 0.
First, it might be noted that one would expect x > 0.4244 a (the value for a plane semicircular lamina).
The length (i.e. the denominator in equation 1.4.4) is just π a. Since there are, between x and x + δ x , two elemental lengths to account for, one above and one below the x axis, the numerator of the first of equation 1.4.4 must be
2 [ 1 ( / )].
1 / 2 0
ax dy dx (^2) dx ∫ +
[ ( ) ( )] [( ) ] [ 1 ( )]. 2 21 /^2221 /^221 /^2 δ s = δ r + r δθ = ddr θ^ + r δθ = + rdrd θ δ r 1.4.
The mass of the curve (between θ = α and θ = β) is
∫ λ[^ (^ ) + ]^ θ
β α θ^ ddr r d
2 21 /^2 .
The first moments about the y - and x -axes are (recalling that x = r cosθ and y = r sinθ )
∫ [^ (^ ) ]
β α θ^ λ θ + 2 21 /^2 r cos (^) ddr^ r and sin [( ) ]. 2 21 /^2 ∫ λ θ + θ
β α θ^ r (^) ddr r d
If λ is not a function of r or θ, we obtain
= (^) ∫ θ[ ( ) + ] θ = ∫ θ[( ) + ] θ
β α θ
β α θ^ x (^) L r ddr r d y L r ddr r d 2 21 /^2 (^1) cos 2 21 /^2 , (^1) sin 1.4.
where L is the length of the wire.
Example : Again consider the uniform wire of figure 8 bent into the shape of a semicircle. The equation in polar coordinates is simply r = a , and the integration limits are θ = −π / 2 to θ =+π/ 2 .The length is π a.
Thus [ ].
cos 0
1 /^21 /^2
/ 2
2 π
θ + θ = π
= (^) ∫
+π −π
a a a d a
x
The reader should now find the position of the centre of mass of a wire bent into the arc of a circle of angle 2α. The expression obtained should go to 2 a /π as α goes to π/2, and to a as α goes to zero.
1.5 Summary of the formulas for plane laminas and curves
y = y ( ) x r = r ( θ)
b x (^) A (^) axydx 1 θ
θ θ
β α
β α r d
r d x 2
3
2 cos
b A (^) a y (^) 21 y^2 dx
β α
β α θ
θ θ
r d
r d y 2
3
2 sin
y = y ( ) x r = r ( θ)
x x [ ( )] dx
b a dx
dy L
21 /^2 (^1 )
= ∫ + = ∫ θ[( ) + ] θ
β x (^) α r d θ r d dr L
2 21 /^2 (^1) cos
y y [ ( )] dx
b a dx
dy L
21 /^2 (^1 )
= ∫ + = ∫ θ[( ) + ] θ
β α θ^ y (^) L r ddr r d 2 21 /^2 (^1) sin
Uniform Plane Lamina
SUMMARY
Uniform Plane Curve
Consider an area A in the zx plane (figure I.9), and an element δ A within the area at a distance x from the z axis. Rotate the area through an angle φ about the z axis. The length of the arc traced by the element δ A in moving through an angle φ is x φ , so the volume swept out by δ A is x φδ A.
The volume swept out by the entire area is φ∫ xdA. But the definition of the centroid of A is such
that its distance from the z axis is given by xA = (^) ∫ xdA. Therefore the volume swept out by the
area is φ x A. But φ x is the distance moved by the centroid, so the first theorem of Pappus is proved.
Consider a curve of length L in the zx plane (figure I.10), and an element δ s of the curve at a distance x from the z axis. Rotate the curve through an angle φ about the z axis. The length of the arc traced by the element ds in moving through an angle φ is x φ , so the area swept out by δ s
is x φδ s. The area swept out by the entire curve is φ∫ xds. But the definition of the centroid is
such that its distance from the z axis is given by x L = (^) ∫ xds. Therefore the area swept out by the
curve is φ x L. But φ x is the distance moved by the centroid, so the second theorem of Pappus is
proved.
z
y
x
δ s
x
φ
FIGURE I.
Applications of the Theorems of Pappus.
Rotate a plane semicircular figure of area 12 π a^2 through 360 o^ about its diameter. The volume
swept out is 43 π a^3 , and the distance moved by the centroid is 2 π x. Therefore by the theorem of
Pappus, x = 4 a /( 3 π).
Rotate a plane semicircular arc of length 21 π a through 360o^ about its diameter. Use a similar
argument to show that x = 2 a / π.
Consider a right-angled triangle, height h , base a (figure I.11). Its centroid is at a distance a / from the height h. The area of the triangle is ah /2. Rotate the triangle through 360o^ about h. The distance moved by the centroid is 2π a /3. The volume of the cone swept out is ah /2 times 2 π a /3, equals π a^2 h /3.
Now consider a line of length l inclined at an angle α to the y axis (figure I.12). Its centroid is at a distance 21 l sinαfrom the y axis. Rotate the line through 360 o^ about the y axis. The distance
moved by the centroid is 2 π ×^12 l sinα=π l sinα. The surface area of the cone swept out is
l × π l sin α=π l^2 sinα.
FIGURE I.
