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CHAPTER 1

Introductory Concepts • Elements of Vector Analysis • Newton’s Laws • Units • The basis of Newtonian Mechanics • D’Alembert’s Principle

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Science of Mechanics: It is concerned with the motion of material bodies.

• Bodies have different scales: Microscropic, macroscopic and astronomic scales. In mechanics - mostly macroscopic bodies are considered. • Speed of motion - serves as another important variable - small and high (approaching speed of light).

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• In Newtonian mechanics - study motion of bodies much bigger than particles at atomic scale, and moving at relative motions (speeds) much smaller than the speed of light. • Two general approaches: – Vectorial dynamics: uses Newton’s laws to write the equations of motion of a system, motion is described in physical coordinates and their derivatives; – Analytical dynamics: uses energy like quantities to define the equations of motion, uses the generalized coordinates to describe motion. 3

1.1 Vector Analysis: • Scalars, vectors, tensors: – Scalar: It is a quantity expressible by a single real number. Examples include: mass, time, temperature, energy, etc. – Vector: It is a quantity which needs both direction and magnitude for complete specification. – Actually (mathematically), it must also have certain transformation properties. 4

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These properties are: vector magnitude remains unchanged under rotation of axes. ex: force, moment of a force, velocity, acceleration, etc.

– geometrically, vectors are shown or depicted as directed line segments of proper magnitude and direction.

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e (unit vector) A A = A e – if we use a coordinate system, we define a basis set (iˆ , ˆ j , k ˆ ): we can write

A = Axi + Ay j + Azk Z or, we can also use the A three components and Y define X T {A}={Ax,Ay,Az}

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– The three components Ax , Ay , Az can be used as 3-dimensional vector elements to specify the vector.

– Then, laws of vector-matrix algebra apply.

– Tensors: scalar - an array of zero dimension vector - an array of one dimension

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– quantities which need arrays of two or higher dimension to specify them completely - called tensors of appropriate rank. Again - to be a tensor, the object must also satisfy certain transformation properties of rotation and translation. Exs: Second-order tensors: stress at a point in deformable body - stress tensor has nine components (a 3x3 matrix in a representation when the basis is defined), inertia tensor (again, a 3x3 matrix in usual notation)

expressing mass distribution in a rigid body.8

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• TYPES OF VECTORS: Consider a force F acting on a body at point B P. The force has a line F of action AB. This force P can lead to translation of the rigid body, rotation A of the rigid body about some point, as well as deformation of the body.

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F F or P2 P1

The same force F is now acting at two different points P1 , P2 of the body, i.e., the lines of action are distinct. – same translational effect – the translational effect depends only on magnitude and direction of the force, not on its point of application or the line of action- free vectors 10

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B B F

P1 F or P A A 2 The force F has the same line of action AB in the two cases. The points of application (P1 and P2) are different but moment about every point is the same → same rotational effect (as well as translational effect): effect of vector F depends on magnitude, direction as well as line of action - sliding vectors 11

If the body is deformable, the effect of force is different depending its point of application; whether the force acts at point P1 or P2. Thus, in such a case, the point of application is also crucial - bound vectors.

B B F P or 1 F P A A 2 deformation

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• Equality of vectors: For free vectors A a n d B , A = B if and only if A a n d B have the same magnitude and the same direction.

• Unit vectors: If A is a vector with magnitude A, A / A is a vector along A with unit length → e A = A /A or A =AeA.

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e unit vector in A |1|

eA the direction A = A e of A.

• Addition of vectors: Consider two vectors A and B. Their addition is a vector C given by C = A + B . Also C = B + A (addition is commutative). The result is also a vector.

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C = A+B Graphically, one can use the parallelogram rule of vector addition. B C B For more than two vectors, one can add A sequentially - polygon of vectors. Consider The addition of vectors A, B, and D . C = A + B , E = (A + B) + D = C + D D or B E E = (A+ D)+ B C B

= A+(B+ D). O A 15

• COMPONENTS OF A VECTOR: Consider the vector addition for A and B: graphically: C = A + B C B

O A We can interpret A and B to be components of the vector C . Clearly, the components of C are non-unique. As an F B example, E and F are E C also components. O A

We can make it more systematic. 16

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• Let e 1 , e 2 , e 3 - three linearly independent unit vectors (not necessarily orthogonal) and let A be a vector.

We can write A = A 1 + A 2 + A 3 where Ai , i = 1, 2, 3 are components of A along the directions specified by unit vectors

e i ,i = 1, 2, 3. e3 A Then : e 2 A3 A2 A= A1e1 +A2e2 +A3e3 A1

e1 A , i = 1, 2, 3 are unique scalar components i 17

Let B = B 1 e 1 + B 2 e 2 + B 3 e 3 be another vector, with components expressed in same unit vectors. We can then write the sum as

C= A+B=(A1 +B1)e1 +(A2 +B2)e2 + (A3 + B3)e3 The components of the vector C are then

→ C1 = A1 + B1 , C2 = A2 + B2 and

C3 = A3 + B3.

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• The more familiar case of unit vectors is the Cartesian coordinate system - (x, y, z) Let i , j , k - unit vectors along x, y and z directions. Then

A=Ax i+Ay j+A2 k where Ax , Ay , Az are components of A Z A k along axes. z

A Ay j Y Ax i

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•SCALAR PRODUCT: Definition: (DOT) For two vectors A and B, the dot product is defines as A ⋅ B = A B c o s θ Dot product is Commutative, i.e., A ⋅B = B⋅ A If 3 3 B A = ƒ Ai ei , B = ƒ Bi ei , i=1 i=1 θ A then A⋅B = A1B1 + A2B2 + A3B3 |B|cosθ provided the unit vectors are an orthogonal set, i.e., e1 ⋅ e 2 = e 2 ⋅ e3 = e3 ⋅ e1 = 0. 20

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•VECTOR PRODUCT: Let A , B be two vectors that make an angle θ with respect to each other. Then, the vector or cross product is defined as a vector C with magnitude C = A× B = A B sinθ .

B Let k be the unit vector normal to the k θ plane formed by vectors A and B. It is A fixed by the right hand screw rule. Then A× B = A B sinθk 21

Some properties of cross product are : A×B=-B×A Consider unit vectors for the Cartesian coordinate system (x,y,z), (i, j, k): Then i i × j = j × k = k × i = 1 i × i = j × j = k × k = 0 k j i × j = k ; j× k = i right-hand rule k × i = j

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Now, consider cross-product again. When vectors A and B are expressed in component form: A=Ax i +Ay j+Az k B= Bxi+By j+Bzk, The cross product is evaluated by the operation i j k

A × B = A x A y A z B x B y B z

≡ (Ay Bz − Az By )i + (Az Bx − Ax Bz ) j

+ (Ax By − Ay Bx )k

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• SCALAR TRIPLE PRODUCT: Consider three vectors A, B, and C. The scalar triple product is given by R=A •(B×C) = (A×B)• C

Note that the result is the same scalar quantity. It can be interpreted as the A volume of the C parallelepiped having the vectors A , B a n d C as B the edges. The sign can be +ve or -ve. 24

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• VECTOR TRIPLE PRODUCT: Consider vectors A, B, and C. Then, vector triple product is defined as a vector D, given by D = A × ( B × C ) Note that A×(B×C) ≠ (A×B)×C One can show that D=A×(B×C)=(A•C)B-(A•B)C • DERIVATIVE OF A VECTOR: Suppose that a vector A is a function of a scalar u, i.e., A = A ( u ) . We can then consider change in vector A associated with change in the scalar u.

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Let A = A ( u ) and A (u + ∆u) ≡ A(u) + ∆A dA A(u) + ∆A -A(u) Then = lim du ∆u→0 ∆u dA ∆A or = lim du ∆u→0 ∆u This is the derivative of A with respect to u. Ex: The position vector r(t) for a particle moving depends on time. We define the dr ∆r velocity to be v(t) ≡ = lim dt ∆t→0 ∆t 26

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s ex: O’ z ∆r O P P’ r (s) x y OP ∆s Consider a particle P moving along a curved path. Its position depends on distance from some landmark, O’, i.e. r O P ≡ r O P ( s ) where ‘s’ is the distance along the curve. dr (s) We shall consider O P later in the next ds chapter. 27

Some useful properties and rules of d dA dB differentiation are: (A + B) = + du du du d dA dg(u) (g(u)A(u)) = g(u) + A(u) du du du d dB(u) dA(u) (A(u) • B(u)) = A(u) • + • B(u) du du du d dB(u) dA(u) (A(u)× B(u)) = A(u)× + × B(u) du du du

Finally, if A = A1e1 + A2 e2 + A3 e3 = ƒ Ai ei , dA dA de = ƒ ( i )e + ƒ A ( i ) then du du i i du 28

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Some more useful properties: Concept of a Dyad and Dyadic: Consider two vectors a and b Dyad: It consists of a pair of vectors a b for two vectors a a n d b . a - called an antecedent, b - called a consequent. Dyadic: It is a sum of dyads. Suppose that the vectors a a n d b are expressed in terms of a set of unit basis vectors e 1 , e 2 , e 3 , so that a = a1 e1 + a2 e2 + a3 e3 and b = b1 e1 + b2 e2 + b3 e3 Then, A ≡ ab = (a e + a e + a e )(b e + b e + b e ) 1 1 2 2 3 3 1 1 2 2 3 3 29

3 3 3 3 and, A ≡ a b = ƒƒ aibj ei e j = ƒƒ Ai j ei e j i=1 j=1 i=1 j=1

In the Text of Greenwood, the unit vectors e1 , e2 ,e3 are mostly limited to the Cartesian basis (i, j, k)

A Conjugate Dyadic for the dyadic A is obtained by interchanging the order of vectors a a n d b and is denoted by AT. Thus, 3 3 3 3 T A ≡ b a = ƒƒbia j ei e j = ƒƒ Aji ei e j i=1 j=1 i=1 j=1 T A dyadic is symmetric if A = A , that is Aij = Aji

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An example of a symmetric dyadic is the inertia dyadic: 3 3

I 𠟟 Ii j ei e j = I xx i i + I xy i j + I xz i k i=1 j=1

+ I yx j i + I yy j j + I yz j k

+ I zx k i + I zy k j + I zz k k T A dyadic is skew-symmetric if A = − A , that is, it is negative of its conjugate. Note that symmetry property of a dyadic is independent of the unit vector basis or its orthogonality.

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Some operational properties: 1. The sum of two dyadics is a dyadic obtained by adding the corresponding elements in the same basis: C = A+ B if and onlyif Ci j = Ai j + Bi j 2. The dot product of a dyadic with a vector is a vector. Consider vectors a a n d b , and the derived dyadic A = a : b . The dot product with the vector c is the vector d given by d ≡ A⋅c = (ab)⋅c = a(b ⋅c) which is a vector in the direction of vector a . Note that e ≡ c ⋅ A = c ⋅(ab) = (c ⋅b)c ≠ d 32

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Ω In general, pre-multiplying a vector by a dyadic ≠ post-multiplying the vector by the same dyadic. For a symmetric dyadic, the order does not matter. Consider the product of Inertia dyadic with the angular velocity vector:

I ⋅ω = (I xx i i + I xy i j + I xz i k + I yx j i + I yy j j + I yz j k

+ I zx k i + I zy k j + I zz k k)⋅ω = I ⋅(ωx i + ωy j + ωk k)

Now, note that I xx i i ⋅(ωx i +ωy j +ωk k)

= I xx i(ωx i ⋅i +ωy i ⋅ j +ωk i ⋅k) = I xxωx i etc.

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Thus, I ⋅ω = (I xxωx + I xyωy + I xzωz )i

+ (I yxωx + I yyωy + I yzωz ) j

+ (I zxωx + I zyωy + I zzωz )k = ω ⋅ I Since dot product of a dyadic and a vector is a new vector, dyadic is really an operator acting on a vector. A interesting symmetric dyadic is the unit dyadic: U ≡ i i + j j + k k

For any vector a, U ⋅a ≡ i (i ⋅a) + j ( j ⋅a) + k( k ⋅a) = a ⋅U = a So, it leaves every vector unchanged. 34

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Finally, consider the cross product of a dyadic with a vector: 3 3 c× A ≠ A×c. If A = ƒƒ Ai j ei e j , then i=1 j=1 3 3 3 3 c× ƒƒ Ai j ei e j = ƒƒ Ai j (c×ei )e j i=1 j=1 i=1 j=1 − another dyadic

See Section 7.5 for Greenwood.

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1.2 Newton’s Laws: There are three laws 1. Every body continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by forces acting upon it.

2. The time rate of change of linear momentum of a body is proportional to the force acting upon it and occurs in the direction in which the force acts.

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3. To every action there is an equal and opposite reaction. • Laws of motion for a particle: Let F = applied force, m = mass of the particle, v = velocity at an instant, p ≡ mv = linear momentum of the particle. d d Then, F=k (p)=k (mv) dt dt =kma (for a body with constant mass)

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Here, k > 0 constant; it is chosen such that k = 1 depending on the choice of units. d d F= (p)= (mv) → dt dt =ma (for constant mass system) Since, F a n d a are vectors, we can express them in the appropriate coordinate system. ex: in a Cartesian coordinate system (x,y,z):

F=Fx i + Fy j+Fz k=m(a x i + a y j+a zk) or F =ma , F =ma , F =ma x x y y z z 38

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Imp: Newton’s laws are applicable in a special reference frame - called the inertial reference frame. Note that in practice, any fixed reference frame or rigid body will suffice. READING ASSIGNMENT #1: The discussion in text. 1.3 UNITS: One first introduces dimensions associated with each quantity.

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Note that the quantities related by Newton’s Laws are: F - force, M - mass, L - length, T - time. Since there is one relation among the four quantities (second law), three of the units are independent, and the fourth fixed by the requirement of principle of dimensional homogeneity. • Absolute system - mass, length, and time are fundamental quantities, where as force is considered a derived quantity.

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In the absolute system, the units of various quantities are: Mass - kilogram (kg); time - second (sec.); length - meter (m) Now, consider the dimensional relation: F = ML/T2 = kg.m/s2 This unit is called a Newton: it is the force needed to give an object of mass 1 kg an acceleration of 1 m/s2. • Gravitational system - In this case, length, time, and force are assumed fundamental, 2 where as mass is derived. 1 slug = lb.s /ft 41

READING ASSIGNMENT #2:

1.4. The Basics of Newtonian Mechanics.

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1.5 D’ALEMBERT’S PRINCIPLE: Consider Newton’s second law: F=ma If we write ( F-ma)=0, one can imagine (-ma) to represent another force, the so called inertia force. Then, we just have summation of forces = 0, that is, an equivalent statics problem. We will see that this principle has profound significance when considering derivation of Lagrange’s

equations. 43

Ex 1.1: (Text) O → a Consider a massless rigid rod l ↓g suspended from point O in a θ box which is accelerating to the right at a constant m acceleration ‘a’. Find: The tension in the Free body diagram: cable and the angle θ, θ when rod has reached a ma T steady position relative mg to the box.

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Ex 1.1 O → a Free body diagram: l ↓g θ θ ma T y m mg x

D’Alembert’s principle:
ƒ Fx -ma x =0: Tsin -ma=0;
ƒ Fy -ma y =0: Tcos -mg=0; Ω T=m (a 2 +g2 ), tan
=a/g

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Ex 1.2: (to clearly point out the difference in components and orthogonal projections) • Consider the vector A with A = 5, at an angle of 60° with the horizontal. j A θ=60° +ve horizontal i

• Suppose that we want to express it in terms of unit vectors e 1 , e 2 where E e1 = →( i ), e2 = 1 135° (=(- i +j)/ 2). 46

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The vector sum can be represented as A=A1+A2

A2 A or A=A1e1 +A2 e2 135° e2 60° Now e1 A1 A=5(cos60 i +sin60 j)=A1 e1 +A2 e2

=A1 i1 +A2 (- i +j)/ 2

→ (5 i +5 3 j)/2=A1 i1 +A2 (-i +j)/ 2 or i :5/ 2 = A1 − A2 / 2 ; j : A2 = 5 6 / 2

→ A = [5(1+ 3) e1 + 5 6 e2 ]/ 2 47

In the above,

component of Aalong e1 is 5(1+ 3) e1 / 2;

component of Aalong e 2 is5 6 e 2 / 2.

• What about orthogonal projections of A?

A2 A 135° |A1| e2 60° |A2|

e1 A1 48

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