Mechanics And Mechanisms

Introduction to Mechanics and Mechanisms Maithripala D. H. S., Dept. of Mechanical and Manufacturing Engineering, Faculty of Engineering, University of Ruhuna, Sri Lanka. August 20, 2007

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Contents 1 Mechanics 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Particle Motion in Euclidian Space . . . . . . . . . . . . 1.2.1 Kinematics of a Particle Moving in Space . . . . . 1.2.2 Kinetics of a Particle Moving in Space . . . . . . 1.3 Relative Motion . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Parallely Translating Frames . . . . . . . . . . . . 1.3.2 Rotating Frames . . . . . . . . . . . . . . . . . . 1.3.3 General Moving Frames . . . . . . . . . . . . . . 1.4 Rigid Body Motion . . . . . . . . . . . . . . . . . . . . . 1.4.1 Two-dimensional Rigid Body Rotational Motion . 1.4.2 Three-dimensional Rigid Body Rotational Motion 1.4.3 Three-dimensional General Rigid Body Motion . 1.5 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . 1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Practice exercises . . . . . . . . . . . . . . . . . . . . . .

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5 5 6 6 10 11 12 14 17 18 19 21 22 23 25 33

2 Mechanisms 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 1 Mechanics

Figure 1.1: Mechanics gone wrong: Tacoma Narrows bridge minutes before collapse.

1.1

Introduction

Historically it is the the study of mechanics that evolved into what we call Engineering today. The following quote by Wikipedia exemplifies this belief: “Important aspects of the fields of mechanical engineering, aerospace engineering, civil engineering, structural engineering, materials engineering, 5

6 biomedical engineering and biomechanics were spawned from the study of mechanics”. Mechanics deals with the scientific description of the world as we perceive it. There are two major branches of mechanics: classical mechanics and quantum mechanics. The latter has only a history of about hundred years while the former goes beyond the period of written history. Classical Mechanics is concerned with describing macroscopic phenomena while quantum mechanics is mainly concerned with describing microscopic phenomena. It is widely believed that quantum mechanics at “large” scales approximates classical mechanics. Since at this stage of undergraduate Engineering we are mostly interested in macroscopic systems we will only consider classical mechanics. The subject of classical mechanics is generally divided into three branches; rigid-body mechanics, deformable-body mechanics and fluid mechanics. In this class we will concentrate on learning the basics of rigid-body mechanics while elsewhere in the mechanical Engineering program you will learn the basics of the other two branches. The study of rigid-body mechanics begins with describing the motion of a mathematically abstracted point particle that does not occupy any volume in space. A general rigid body is considered to consist of a large number of point particles where the distance between each of the particles remain fixed. The geometric description of the motion of these objects is what is generally known as Kinematics while the study of the cause of motion is referred to as Kinetics.

1.2

Particle Motion in Euclidian Space

In classical mechanics, to describe the motion of a point particle we need to define and accept three fundamental concepts; mass, time and space. We associate with each point a quantity called mass that in a certain sense describes the resistance to change of motion. Time is the quantity that measures the distance between two separate events and we assume that it has an independent meaning. We assume that space is a three-dimensional Euclidean space, that is a space where Euclidean geometry holds.

1.2.1

Kinematics of a Particle Moving in Space

Co-ordinates and Degrees of Freedom First we will demonstrate how to describe the position of a given point P in a three-dimensional Euclidean space. Euclidean space is a space where the Euclidean geometric notions such as parallel lines and normal lines have the usual meaning. Thus to describe a point P in space we begin by picking a point O in space and setting three mutually perpendicular unit length axis at O. Such a set of axis is called a ortho-normal frame of reference. Labelling the axis e1 , e2 , e3 to give a right hand orientation we can symbolically represent the frame as a matrix e = [e1 e2 e3 ]. Using such a frame any point P in space can be described using three measurements (numbers) x1 , x2 and

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

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x3 . These three numbers describe respectively the distance to the point along the e1 , e2 , and e3 axis. Symbolically we describe this as OP = x1 e1 + x2 e2 + x3 e3 = The matrix



h

e 1 e2 e 3

i





x1    x2  = e x. x3



x1  x=  x2  x3

is referred to as the representation of the position of the point P and the components are referred to as the position components of P with respect to the frame e. If Q is another point in space and if its representation y with respect to e is given by 



y1  y=  y2  , y3 then the distance between P and Q is assumed to be given by, d(P, Q) =

q

(x1 − y1 )2 + (x2 − y2 )2 + (x3 − y3 )2 := ||x − y||.

This notion of distance is essentially what it means to be a Euclidean space. It allows us to mathematically define what a right angle is and also when two straight lines are parallel. Observe that the introduction of the ortho-normal frame has allowed us to uniquely identify points in three dimensional Euclidean space with, R3 , the set of ordered triple of numbers. Namely by identifying each point P with the uniquely corresponding ordered triple (x1 , x2 , x3 ). It has to be kept in mind that this identification depends heavily on the choice of the ortho-normal frame e. For instance consider figure 1.2. The position of the particle P is described by the two different matrices x = [x1 x2 x3 ]T and X = [X1 X2 X3 ]T with reference to the two frames e and b respectively. Orhto-normal frames are also not the only means of identifying Euclidean space with R3 . Curvilinear frames can also be used for this identification. For example, as shown in figure 1.3, we may use the quantities (r, θ, φ) to describe the point P . All three sets of ordered triples (x1 , x2 , x3 ), (X1 , X2 , X3 ) and (r, θ, φ) describe the same point P but note that they are in general different. Thus the three descriptions provide three different identifications with R3 . If the point P is moving freely in space (moves without any geometric constraint) then at each time t, the specification of the three independent measurements is necessary to un-ambiguously describe the position of the point P at each time t. Since, at each time instant t, three independent

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Figure 1.2:

quantities (measurements) are needed to describe the position we say that a free particle in space has three degrees of freedom. In general, a collection of independent measurements (quantities) required to uniquely describe the position of a moving point particle are called the co-ordinates of that point and the number of such co-ordinates are called the degrees of freedom (DOF) of the particle. In the unconstrained situation the three components of the position representation matrix provide a set of co-ordinates with respect to the frame e. Co-ordinates that describe the position with respect to an ortho-normal frame are called Euclidean co-ordinates. From the discussion in the previous paragraph it is clear that Euclidean co-ordinates are not the only type of co-ordinates. The quantities (r, θ, φ) used to describe the point P are called spherical-polar coordinates. The relationship between various co-ordinates can be found and are called co-ordinate transformations. For instance the Euclidean co-ordinates of the free particle P are related to the spherical polar co-ordinates by 







x1 r sin φ cos θ    x =  x2  =  r sin φ sin θ  . r cos φ x3 If the motion of the particle is geometrically constrained so that the quantities (x1 (t), x2 (t), x3 (t)) have to satisfy some scalar expression h(x1 (t), x2 (t), x3 (t)) = 0 then the particle is said to be Holonomically constrained and it can be shown that only two independent measurements are required to uniquely describe the position of the particle. Thus the constraint has reduced the DOF of the particle from three to two. If the number of such constraints are two then the DOF of the particle reduces to one. For instance if the particle is constrained to move on a sphere of radius equal to one then the constraint equation in the spherical-polar co-ordinates is h(r, θ, φ) = r − 1 = 0

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Figure 1.3: Description of the point P using Spherical-Polar Co-ordinates.

and the position matrix x will become [sin φ cos θ sin φ sin θ cos φ]T . Therefore the two independent measurements (θ, φ) are sufficient to describe the position uniquely and constitutes the co-ordinates of the constrained P . Since only two measurements (co-ordinates) are required to uniquely describe the position of P we say that P has two DOF. On the other hand if the particle is constrained to move in a circle of radius one that lies in the e1 , e2 plane then there are two constraints, h1 (r, θ, φ) = r − 1 = 0, h2 (r, θ, φ) = φ − π/2 = 0 and the position matrix x will become [cos θ sin θ 0]T . Therefore the single measurement θ is sufficient to describe the position uniquely and constitutes the co-ordinate of the constrained P and thus the DOF of P is one. The general expression n = 3 − f, relates the DOF, n, of a particle moving in three dimensional space to the number of holonomic constraints, f , imposed on that particle. If N points are moving in space then the total degrees of freedom of the system of points is n = 3N − f, where now f is the total number of holonomic constraints of the system of points. An important fact to keep in mind is that what ever the co-ordinates we use to describe the point we will always use the matrix x to represent the point in space and that to do so we first need to specify an ortho-normal frame e. However, the components of this matrix x1 , x2 and x3 will be functions of the co-ordinates.

10 Velocity and Acceleration of a Particle The velocity, v(t), and acceleration, a, of a point P (t) moving in space is defined with respect to the ortho-normal frame e with respect to which the position of P (t) is described. That is by definition the velocity measured in the e frame is the quantity v(t) = x(t), ˙ and the acceleration measured in the e frame is the quantity a(t) = v(t) ˙ = x¨(t). Example 1 Consider a particle constrained to move on a sphere of radius one (refer to figure 1.3). The representation of P in an ortho-normal frame e fixed at the center of the sphere is 



sin φ cos θ  x =  sin φ sin θ  . cos φ Note that the components are a function of the co-ordinates (θ, φ). Then the velocity is 

 φ˙ cos φ cos θ − θ˙ sin φ sin θ   v = x˙ =  φ˙ cos φ sin θ + θ˙ sin φ cos θ  . −φ˙ sin φ

and the acceleration is 

 φ¨ cos φ cos θ − φ˙ 2 sin φ cos θ − 2φ˙ θ˙ cos φ sin θ − θ˙2 sin φ cos θ − θ¨ sin φ sin θ ¨ cos φ sin θ − φ˙ 2 sin φ sin θ + 2φ˙ θ˙ cos φ cos θ − θ˙2 sin φ sin θ + θ¨ sin φ cos θ  a = v˙ =   φ . 2 ¨ ˙ −φ sin φ − φ cos φ

1.2.2

Kinetics of a Particle Moving in Space

Kinetics deal with the apparent causes of motion. In this section we investigate the cause of motion in classical mechanics. Given the present “state” of a particle, the primary interest in mechanics is to be able to completely predict the future “state” of it. More precisely we need to know what measurements are needed at the present time in order to be able to uniquely describe the position of a point of mass m for all future times. For macroscopic motions, the answer to this deep and profound question was provided by Newton. He postulated that for macroscopic motions the acceleration a(t) of a given particle was not an independent (arbitrary) quantity and

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that it was completely specified. More precisely in his second law he stated that in a given frame e m¨ x = m a(t) = f e (t),

(1.1)

where f e (t) was completely known. For macroscopic motion this notion has been verified experimentally. In this expression the mass of the point is taken to be a fundamental property of the point and is further assumed to be a constant. From a mathematical perspective equation (1.1) describes a second order differential equation and solving it for x(t) will enable us to predict the position for all future time. To solve this equation we need to know the right hand side at every time instant t as well as know the initial conditions x(t0 ) = x0 and v(t0 ) = v0 . Thus knowing the right hand side turned out to be a crucial step in describing the position and as such was termed to be the cause of motion. Newton termed this constraint on mass times acceleration, the Force acting on the particle in the ortho-normal frame e. Thus the force “felt” in a frame e is nothing but mass times the acceleration in that frame. The force can not be directly measured and can only be inferred indirectly by resorting to other physical reasoning and justifications. Example 2 Consider the problem of a horizontal spring with one end fixed to a support and the other end fixed to point mass P , of mass m. If we give an initial horizontal displacement to the point mass we know empirically that the point mass will exhibit a simple harmonic motion if the air and other resistances on the particle is negligible and the motion is small. That is if x(t) is the displacement of the mass from the unstretched position and if the air resistance on the particle is negligible and the motion is small the position x(t) of the particle P at a given time t is described by the second order differential equation m¨ x(t) = −k x(t). Observing this expression it is evident that the mass times the acceleration of the particle is constrained and is equal to −k x(t). Thus we would call −k x(t) the force exerted by the spring on the particle. Observe that we can not directly measure this instead we can only infer it by measuring the deflection of the spring.

1.3

Relative Motion

In section 1.2.1 it was pointed out that choosing different frames of reference resulted in different representations for the position of a given particle. In this section we will see how these different descriptions with respect to different frames are related to each other.

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Figure 1.4:

1.3.1

Parallely Translating Frames

Let e be an ortho-normal frame of reference and b(t) be another ortho-normal frame of reference that is parallely translating with respect to e. We will also refer to b(t) as a moving frame. Let the origin O0 (t) of the translating frame have the representation o(t) with respect to the frame e. Consider a point P (t) moving in Euclidian space. This point can be expressed in two different ways using the two different frames. Let x(t) and X(t) be the representations of the point P (t) in the e and b(t) frames respectively (refer to figure 1.4). In Euclidian space the meaning of b(t) being parallel to e is that the two representations x(t) and X(t) of P are related by, x(t) = o(t) + X(t). Observing that symbolically this means OP (t) = e x(t) = e o(t) + b(t) X(t) you will notice that this is what you traditionally wrote down as OP (t) = OO0 (t) + O0 P (t), for addition of “vectors”.

(1.2)

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The velocity of the point P (t) measured in frame e is v(t) = x(t), ˙ and measured in the moving frame is

˙ V (t) = X(t).

Differentiating (1.2) we see that the two measured velocities are related by ˙ v(t) = x(t) ˙ = o(t) ˙ + X(t) = o(t) ˙ + V (t).

(1.3)

The acceleration of the point P (t) measured in the e frame is a(t) = v(t) ˙ = x¨(t), and measured in the moving frame is ¨ A(t) = V˙ (t) = X(t). Differentiating (1.3) we see that the two measured accelerations are related by a(t) = v(t) ˙ = o¨(t) + V˙ (t) = o¨(t) + A(t).

(1.4)

From (1.4) and (1.1) we have o(t) + A(t)), f e (t) = m a(t) = m(¨ and hence that the total force experienced in the moving frame is F b (t) = m A(t) = −m o¨(t) + f e (t).

(1.5)

Fapp = −m o¨(t).

(1.6)

That is an observer moving with the translating frame will, in addition to the force f e (t) felt in the e frame, also experience another apparent force

This particular type of apparent forces that arise as a consequence of the translational accelerating motion of the reference frame are called Einstein forces. Recall that we defined at the beginning that force felt in a frame is simply the constraint on mass times the acceleration of the particle in that frame. Do exercise 11 at this moment. In exercise 11 you are asked to explain why a person standing on a scale inside an elevator sees his or her weight doubled as the elevator accelerates up at a rate of g and sees the weight reduced to zero if the elevator decelerates at a rate of g where g is the gravitational acceleration. You are also asked to show that if, for some reason, the gravitational force field vanished and that the elevator was moving up at an acceleration of g then the scale would show the correct weight of the person. This last observation shows that a person inside the elevator can not distinguish between the following two cases:

14 a.) Gravity is present and the elevator is standing still (or moving at constant velocity). b.) Gravity is absent and the elevator is accelerating upwards at a rate of g. It is this observation that led Einstein to the conclusions of General Relativity and in particular that gravity is an apparent force !!!

1.3.2

Rotating Frames

Figure 1.5: Let e be an ortho-normal frame of reference and b(t) be an ortho-normal frame of reference that is rotating with respect to e. We will consider the case where both origins coincide at the point O in space (rotating frames). Then the two frames are related by b(t) = e R(t),

(1.7)

where R(t) is a special orthogonal matrix (ie RT R = I and det(R) = 1). If we are considering 2-dimensional Euclidian motions then e = [e1 e2 ], b(t) = [b1 (t) b2 (t)] and R(t) is a 2 × 2 matrix while if it is 3-dimensional Euclidian motions then e = [e1 e2 e3 ], b(t) = [b1 (t) b2 (t) b3 (t)] and R(t) is a 3 × 3 matrix (refer to figure 1.5). In exercise 8 you are asked to show that the 2 × 2 special orthogonal matrices can be identified with points on the unit circle and hence can be parameterized by a single angle co-ordinate θ and that the 3 × 3 special orthogonal matrices can be parameterized using three angle co-ordinates known as Euler angles.

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The point P (t) can be expressed in both fames as OP (t) = e x(t) = b(t) X(t) = e R(t)X(t),

(1.8)

where x(t) is the representation of the point P (t) with respect to the frame e and X(t) is the representation of the point P (t) with respect to the moving frame b(t). Thus the two representations are related by x(t) = R(t)X(t). (1.9) The velocity of the point P (t) measured in the e frame is v(t) = x(t), ˙ and measured in the moving ˙ frame b(t) is V (t) = X(t). Now differentiating (1.9) we have that 



˙ ˙ ˙ v(t) = x(t) ˙ = R(t)X(t) + R(t)X(t) = R(t) RT (t)R(t)X(t) + V (t) = R(t)Vb (t).

(1.10)

The quantity Vb (t) is given by ˙ Vb (t) = RT (t)R(t)X(t) + V (t).

(1.11)

Observe that Vb is the velocity v(t) expressed using the moving frame b(t). This is only a derived quantity as compared to the other measured quantities. Since RT (t)R(t) = I, it follows that R˙ T R + RT R˙ = 0, and hence that

ˆ ˙ T = Ω, RT R˙ = −(RT R)

ˆ is a skew symmetric matrix. where Ω ˆ is a 2 × 2 skew-symmetric matrix while it is a 3 × 3 For 2-dimensional Euclidian motion Ω skew-symmetric matrix for 3-dimensional Euclidian motion. In exercise 9 you are asked to show that 2 × 2 skew-symmetric matrices can be identified with the set of real numbers, R, using the ˆ where identification Ω 7→ Ω " # 0 −Ω ˆ Ω= , (1.12) Ω 0 and in exercise 10 you are asked to show that the space of 3 × 3 skew-symmetric matrices can ˆ be identified with, the set of ordered triple of real numbers, R3 , using the identification Ω 7→ Ω where   0 −Ω3 Ω2 ˆ = 0 −Ω1  (1.13) Ω  Ω3 , −Ω2 Ω1 0

16 and Ω = [Ω1 Ω2 Ω3 ]T . In 2-dimensional Euclidian motions the quantity Ω corresponds to an instantaneous rotation of the point P about the fixed point O by an amount equal to Ω. Similarly it can be shown that in the case of 3-dimensional Euclidian motion the quantity Ω corresponds to an instantaneous rotation of the point P about the axis Ω through O by an amount equal to the magnitude ||Ω||. Thus in both cases Ω is defined to be the angular velocity of the point P . ˙ T =Ω ˆ in (1.11) the velocity v expressed using the moving frame Substituting RT R˙ = −(RT R) b(t) is given by ˆ + V (t). (1.14) RT (t)v(t) = Vb (t) = Ω(t)X(t) Therefore the two velocities, v(t) and V (t) are explicitly related by 



ˆ v(t) = R(t) Ω(t)X(t) + V (t) .

(1.15)

The acceleration of the point P (t) measured in the e frame is a(t) = v(t) ˙ = x¨(t), and measured ¨ in the moving frame is A(t) = V˙ (t) = X(t). Differentiating (1.10) we have that 



T ˙ ˙ ˙ ˙ a(t) = R(t)V b (t) + R(t)Vb (t) = R(t) R (t)R(t)Vb (t) + Vb (t) ,





ˆ ˙ = R(t) Ω(t)V b (t) + Vb (t) = R(t)Ab (t),

(1.16)

where ˆ ˙ Ab (t) = Ω(t)V b (t) + Vb (t), ˆ 2 (t)X(t) + 2 Ω(t)V ˆ ˆ˙ = Ω (t) + Ω(t)X(t) + A(t).

(1.17)

Observe that Ab is the acceleration of the particle in the e frame expressed in the moving frame b(t). This too is only a derived quantity as compared to the other measured quantities. From (1.16) and (1.17) the two accelerations, a(t) and A(t) are explicitly related by 



ˆ 2 (t)X(t) + 2 Ω(t)V ˆ ˆ˙ a(t) = R(t) Ω (t) + Ω(t)X(t) + A(t) .

(1.18)

From (1.17) and (1.1) we have that e

f (t) = m a(t) = m(R(t)Ab (t)) = m R(t)





ˆ ˆ˙ ˆ 2 (t)X(t) + 2 Ω(t)V (t) + Ω(t)X(t) + A(t) , (1.19) Ω

and hence in the rotating moving frame we have that ˆ 2 (t)X(t) − 2m Ω(t)V ˆ ˆ˙ F b (t) = m A(t) = −m Ω (t) − m Ω(t)X(t) + RT f e (t).

(1.20)

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Thus an observer moving with the rotating frame will, in addition to the rotating frame version of the force felt in the e frame f e (t), feel the effect of the apparent force: ˆ 2 (t)X(t) − 2m Ω(t)V ˆ ˆ˙ Fapp = −m Ω (t) − m Ω(t)X(t).

(1.21)

ˆ ˆ 2 (t)X(t) is known as the Centrifugal force, the second term −2m Ω(t)V (t) The first term −m Ω ˙ˆ is known as the Coriolis force and the third term −m Ω(t)X(t) is known as the Euler force.

1.3.3

General Moving Frames

Let e be an ortho-normal frame of reference and b(t) be a moving ortho-normal frame of reference with origins O and O0 respectively. In this section we consider the case where the frame b(t) is rotating with respect to the frame e as well as translating with respect to e (ie. the origin O0 of the b(t) frame is also moving with respect to the frame e). Then a point P (t) can be expressed in both fames as OP (t) = e x(t) = e o(t) + b(t) X(t) = e (o(t) + R(t)X(t)) , (1.22) where x(t) and o(t) are the representations of the points P (t) and O0 (t) in the frame e while X(t) is the representation of P (t) in the moving frame b(t). Thus the two representations of the point P are related by x(t) = o(t) + R(t)X(t). (1.23) The velocity of the point P (t) measured in the frame e is v(t) = x(t), ˙ and measured in the moving ˙ frame is V (t) = X(t). Now differentiating (1.23) we have that 



ˆ ˙ + Ω(t)X(t) + V (t) = R(t)Vb (t), v(t) = R(t) RT (t)o(t)

(1.24)

where the quantity Vb (t) ˆ Vb (t) = RT (t)o(t) ˙ + Ω(t)X(t) + V (t),

(1.25)

is the velocity in e expressed with respect to the moving frame b(t). Let Vo (t) = RT (t)o(t) ˙ be the 0 velocity of the point O (t) as measured in b(t). The acceleration of the point P (t) measured in the frame e is a(t) = v(t) ˙ = x¨(t), and measured ˙ ¨ in the moving frame is A(t) = V (t) = X(t). Differentiating (1.24) we have that 



˙ ˙ ˙ ˆ a(t) = R(t)V b (t) + R(t)Vb (t) = R(t) Ω(t)Vb (t) + Vb (t) , = R(t)Ab (t),

(1.26)

18 where ˆ ˆ2 ˆ ˆ˙ Ab (t) = V˙ o (t) + Ω(t)V o (t) + Ω (t)X(t) + 2 Ω(t)V (t) + Ω(t)X(t) + A(t),

(1.27)

is the acceleration of the particle in e expressed in the moving frame b(t). From (1.27) and (1.1) we have that e

f (t) = m a(t) = m R(t)





ˆ ˆ2 ˆ ˆ˙ V˙ o (t) + Ω(t)V o (t) + Ω (t)X(t) + 2 Ω(t)V (t) + Ω(t)X(t) + A(t) , (1.28)

and hence in the general moving frame we have that T e ˆ ˆ2 ˆ ˆ˙ F b (t) = m A(t) = −m V˙ o (t) − m Ω(t)V o (t) − m Ω (t)X(t) − 2m Ω(t)V (t) − m Ω(t)X(t) + R f (t). (1.29)

1.4

Rigid Body Motion

Consider a rigid body that is free to rotate about a fixed point O. Let e be a “fixed” ortho-normal frame and let b(t) be a body-fixed ortho-normal frame (which we will call the body frame). Both frames have coinciding origins at O. The position of a point P in the body, at a time t, is given by the co-ordinates x(t) in the frame e and by the co-ordinates X in the body frame b(t). Observe that since the frame b(t) is fixed in the body, the co-ordinates X are independent of time. At t = 0, the two frames coincide (ie. b(0) = e). We have already seen that the two frames are related by b(t) = e R(t) and hence that x(t) = R(t)X, where R(t) is a special orthogonal matrix. It can be shown that in this fashion any given configuration of the body can be uniquely identified with a special orthogonal matrix R. Recall that, ˆ since RT R˙ = Ω, ˆ R˙ = R Ω, ˆ is given by (1.12) for 2-dimensional Euclidian motions and is given by (1.13) for 3where Ω dimensional Euclidian motions. In 2-dimensional Euclidian motions the quantity Ω corresponds to an instantaneous rotation of the body about the fixed point by an amount equal to Ω. Similarly it can be shown that in the case of 3-dimensional Euclidian motion the quantity Ω corresponds to an instantaneous rotation of the body about the axis Ω by an amount equal to the magnitude ||Ω||. Thus in both cases Ω is defined to be the body angular velocity of the rigid body.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

19

Assume that the body is made up of a very large number of point masses. Isolate a particular point P of mass m and let us analyze the mechanics of this point mass. The velocity and acceleration of the point P at time t as measured in the inertial frame e is v(t) = R(t)Vb (t) and a(t) = R(t)Ab (t) respectively where from (1.14) ˆ Vb (t) = Ω(t)X,

(1.30)

ˆ 2 (t)X + Ω(t)X, ˆ˙ Ab (t) = Ω

(1.31)

and from (1.17)

are their body representations. Let fpe (t) = fpr (t) + fpext (t) be the co-ordinates of the total force acting on the point mass P expressed in the inertial frame. Here fpr is the force that maintains rigidity and arises due to the interaction of the neighboring particles and fpext (t) is the resultant external force expressed in the inertial frame. Their respective representations in the body frame are RT (t)fpr (t) = Fpr (t) and RT (t)fpext (t) = Fpext (t). Due to the assumption of rigidity it follows that Fpr the interaction force as seen in the body is a constant. Applying Newton’s equations we have that 



ˆ 2 (t)X + Ω(t)X ˆ˙ , fpe (t) = m a(t) = m R(t)Ab = m R(t) Ω 



ˆ 2 (t)X + Ω(t)X ˆ˙ . RT (t)fpe (t) = m Ω

1.4.1

(1.32)

Two-dimensional Rigid Body Rotational Motion

In this section we only consider the case where the body is fixed at a point O and is free to rotate about that point in a plane. Chose a body frame b(t) and an inertial frame e such that their origins coincide at O. In the case of the two-dimensional Euclidian space, the rotation matrix R is a 2 × 2 special orthogonal matrix and " # 0 −Ω ˆ= Ω , Ω 0 where the real number Ω is defined to be the body angular velocity of the body about O. Thus using the results shown in exercise 12, Newton’s equation (1.32) in two dimension reduces to 



ˆ˙ = RT (t)fpI (t) = RT (t)(Fpr + Fpext (t)). m −Ω2 (t)X + Ω(t)X

(1.33)

20 ˜ where for X = [ X1 X2 ]T Now if we pre-multiply both sides by X ˜= X

h

−X2 X1

and sum over all the points in the body we obtain X

This reduces to



i

,



X ˜ r + F ext (t)). ˜ +X ˜ Ω(t)X ˆ˙ X(F m −Ω (t)XX = p p 2

˙ Ω(t)

X

m ||X||2 =

X

˜ pr + Fpext (t)), X(F

2 ˜ Ω(t)X ˆ˙ ˙ because it can be easily shown that the first term becomes zero and X . = Ω||X|| Let us individually consider each of these terms. The internal constraint forces that maintain rigidity are of the nature that they satisfy Newton’t third law, action equals reaction. Thus

Let

X X

˜ r = 0. XF p

˜ ext (t) = T ext (t), XF p

(1.34)

˜ ext (t) is defined to be the moment of the force F ext (t) about the axis of rotation and where XF p p therefore T ext (t) as the resultant moment of the external forces about the axis of rotation. If we define X I= m ||X||2 , (1.35) as the Moment of Inertia of the 2-dimensional body about the axis of rotation then the Kinetics of a 2-dimensional rigid body is given by I Ω˙ = T ext (t). Thus the complete motion of the rigid body in 2-dimensions is governed by ˆ R˙ = R Ω, I Ω˙ = T ext (t).

(1.36) (1.37)

These equations are referred to as Euler’s 2-dimensional rigid body equations. Recall that Ω was defined to be the body angular velocity. Consequently the quantity Π = IΩ is defined to be the body angular momentum. From equation (1.37) it follows that if the external ˙ = I Ω˙ = 0. This is known as the law of conservation of body angular moment T e = 0 then Π momentum.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

1.4.2

21

Three-dimensional Rigid Body Rotational Motion

In this section we consider the case where a point O in the body is fixed and the body is free to rotate about that point. Chose a body frame b(t) and an inertial frame e such that their origins coincide at O. ˆ and sum over Let us pre-multiply both sides of (1.32) by the 3 × 3 skew symmetric matrix X all the points in the body X





X ˆ 2 (t)X + Ω(t)X ˆ r + F ext (t)). ˆ Ω ˆ˙ X(F mX = p p

(1.38)

Consider individually each of the terms in the above equation. The internal constraint forces that maintain rigidity are of the nature that they satisfy Newton’t third law, action equals reaction. Thus X X ˆ pr = X × Fpr = 0. XF Let

X

ˆ pext (t) = XF

X

X × Fpext (t) = T ext (t),

(1.39)

ˆ pext (t) = X × Fpext (t) is defined to be the moment of the force Fpext (t) about the fixed where XF point of rotation O and therefore T ext (t) as the resultant moment of the external forces about O. In the exercises you are asked to show that

and where

X X

ˆ Ω(t)X ˆ˙ ˙ mX = I Ω(t),

ˆΩ ˆ 2 (t)X = − IΩ(t) × Ω(t), mX

I=

X





m ||X||2 I3×3 − XX T ,

(1.40)

is defined to be the inertia tensor of the body about the fixed point O. Then we have that (1.50) reduces to I Ω˙ = IΩ × Ω + T ext . Thus the complete motion of a rigid body in 3-dimensions is governed by ˆ R˙ = R Ω, I Ω˙ = IΩ × Ω + T ext .

(1.41) (1.42)

Recall that Ω(t) was defined to be the body angular velocity. The quantity Π(t) = I Ω(t) is defined to be the body angular momentum. Using properties of cross products and dot products it can be easily shown that IΩ ∙ (IΩ × Ω) = 0.

22 Consequently if the resultant external moment of the forces is zero, T ext = 0, we have that d d√ IΩ ∙ I Ω˙ = 0. ||Π(t)|| = IΩ ∙ IΩ = √ dt dt IΩ ∙ IΩ This implies that if the resultant external moment about the fixed point is zero then the magnitude of the body angular momentum is a constant. This is known as the law of conservation of momentum for a rigid body. The inertia tensor (27) is a symmetric matrix and it can be shown that it is positive definite. Consequently it is always possible to find a body frame such that the Inertia tensor is diagonalized. In particular if the the body is axi-symmetric and the body frame is aligned along the axis of symmetry then   I1 0 0  I= (1.43)  0 I2 0  , 0 0 I3 and (1.42) takes the form

I1 Ω˙ 1 = (I2 − I3 )Ω2 Ω3 + T1e , I2 Ω˙ 2 = (I3 − I1 )Ω3 Ω1 + T2e , I3 Ω˙ 3 = (I1 − I2 )Ω1 Ω2 + T3e ,

(1.44) (1.45) (1.46)

where now I1 , I2 , I3 are called the principle moments of inertia.

1.4.3

Three-dimensional General Rigid Body Motion

In this section we consider the case where the body describes a general Euclidean motion. Chose a body frame b(t) and an inertial frame e. From (1.29) a point P fixed in the moving frame satisfies the following relationship. r ext ˆ ˆ2 ˆ˙ m V˙ o (t) + m Ω(t)V o (t) + m Ω (t)X(t) + m Ω(t)X(t) = Fp + Fp (t).

(1.47)

First let us sum (1.47) overall the points in the body. X



m V˙ o (t)+

X



ˆ ˆ2 m Ω(t)V o (t)+ Ω (t)

X

X X X ˆ˙ m X(t)+ Ω(t) m X(t) = Fpr + Fpext (t). (1.48)

Now if we choose the b(t) frame so that its origin O coincides with the center of mass of the rigid body, then the last two terms on the left hand side of (1.48) become zero and we have ext ˆ (t), M V˙ o (t) + M Ω(t)V o (t) = F

(1.49)

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306 P

P

23

P

where M = m, F ext (t) = Fpext (t) and we have set Fpr = because the constraint forces occur in equal and opposite pairs. ˆ and sum Let us now pre-multiply both sides of (1.47) by the 3 × 3 skew symmetric matrix X over all the points in the body X

X X ˆ Ω(t)X(t) ˆ˙ ˆ r+ ˆ ext (t). mX = XF XF p p (1.50) If the body frame b(t) is fixed at the center of mass of the body then the fist two terms become zero and we end up with the rigid body equations (1.42) for a purely rotational rigid body derived in section 1.4.2, Thus the complete equations of motion for general rigid body motion where the body frame b(t) is fixed at the center of mass of the body are

ˆ V˙ o (t) + mX

X

ˆ Ω(t)V ˆ mX o (t) +

X

˙ R(t) o(t) ˙ ˙ M Vo (t) I Ω˙

1.5

ˆΩ ˆ 2 (t)X(t) + mX

= = = =

X

ˆ R(t)Ω(t), R(t)V0 (t), ext ˆ −M Ω(t)V (t), o (t) + F ext IΩ × Ω + T .

(1.51) (1.52) (1.53) (1.54)

Kinetic Energy

The kinetic energy KEp of a particle of mass m is defined with respect to an inertial frame and is defined to be 1 KEp = m||v(t)||2 , (1.55) 2 where || ∙ || is the Euclidian norm in R3 and is defined to be ||X||2 = X T X = X ∙ X (physically this gives the length of the vector X). Since ||RX|| = ||X|| with respect to a frame b(t) rotating purely with respect to the inertial frame we also have that 1 1 1 KEp = m||v(t)||2 = m||RT (t)v(t)||2 = m||Vb (t)||2 . 2 2 2 Hence from (1.14) we have that,  1 1  ˆ 2 X + 2V T ΩX ˆ + ||V ||2 . KEp = m||Vb (t)||2 = m −X T Ω 2 2

(1.56)

˙ If the particle P is fixed with respect to the purely rotating frame then V (t) = X(t) = 0 and hence  1 1  ˆ 2X . (1.57) KEp = m||Vb (t)||2 = m −X T Ω 2 2

24 To compute the total energy of a purely rotating rigid body we sum up the kinetic energy of all the particles to obtain KE =

X

Tp =

 1X 1X  ˆ 2X . m||Vb (t)||2 = m −X T Ω 2 2

(1.58)

In 2-dimensions it is easy to show that X

1 ˆ 2 X = 1 IΩ2 , −m X T Ω 2 2

and hence that

1 KE = IΩ2 . 2 In exercise 27 you are asked to show that in 3-dimensions X

(1.59)

1 ˆ 2 X = 1 ΩT IΩ, −m X T Ω 2 2

and hence that

1 KE = ΩT IΩ. (1.60) 2 On the other hand with respect to a general rotating and translating body where the body frame is fixed at the center of mass of the body we can show that the kinetic energy is given by KE =

M 1 ||V0 ||2 + ΩT IΩ. 2 2

(1.61)

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

1.6

25

Exercises

Exercise 1 A bead is constrained to move on a frictionless wire that lies on a horizontal plane. The wire is bent to a shape of a parabola (ie. y = x2 ). Find the constraint forces that keep the bead on the wire and the equation of motion of the bead. Exercise 2 Two point masses P1 and P2 are moving in three-dimensional Euclidean space such that P1 moves on a sphere and the distance between P1 and P2 remain fixed. What is the degrees of freedom of this system of particles and how many co-ordinates would you need to uniquely describe the motion of the system. Also specify the configuration space and a suitable set of coordinates for the system. Exercise 3 In a Euclidean frame e, with no external force fields present, a particle P is constrained to move such that ||x||2 = xT x = constant, where x is the representation of the particle in the e frame (observe that this corresponds to a particle motion on a sphere if the Euclidean space is three-dimensional while it corresponds to a particle motion on a circle if the Euclidean space is two-dimensional). Show that the velocity of the particle is always orthogonal to the position x, ie. xT x˙ = 0. Differentiating this expression show that the motion of the particle in the e frame is described by ||x|| ˙ 2 1 x, x¨ = fce (t) = − ||x||2 m where fce (t) is the representation of the constraint force in the e frame. Let b(t) be a frame such that its origin coincides with that of e and the particle P appears to be fixed with respect to b(t). Let b(t) = e R(t) and x(t) = R(t)X where X is the representation of the point P in b(t) and is a constant. T

ˆ2

T (RT Ω ˆ 2 R)x

x Ω X 1. Show that the constraint force is fce (t) = m X||X|| 2 RX = m angular velocity of the particle.

||x||2

x, where Ω is the

2. Exercise 4 In a 2D-Euclidean frame e, with no external force fields present, a particle is constrained to move on a circular wire with no friction. If x is the representation of the particle in the e frame show that the motion of the particle is described by x¨ = −Ω2 x, where Ω is the angular velocity of the particle. How would you relate this to the results of exercise 6.

26 Exercise 5 In a 2D-Euclidean frame e, with no external force fields present, a particle is constrained to move on a circular wire with no friction. If x is the representation of the particle in the e frame show that the motion of the particle is described by x¨ = −Ω2 x, where Ω is the angular velocity of the particle. How would you relate this to the results of exercise 6. Exercise 6 A bead is constrained to move on a frictionless wire that lies on a horizontal plane. The wire is bent to a shape of a circle. Picking suitable co-ordinates find the constraint forces that keep the bead on the circle and the equation of motion of the bead. Exercise 7 A cannon is fired with an initial velocity of 1 ms and a firing angle of 45o . Find the horizontal distance to the point of landing of the cannon. Exercise 8 Show that 2 × 2 special orthogonal matrices R can be identified with points on the unit circle S 1 and hence that they can be parameterized using the single angle co-ordinate θ. Also show how to parameterize 3 × 3 special orthogonal matrices R using Euler angles. Exercise 9 Show that the space of 2 × 2 skew-symmetric matrices can be identified with R using ˆ where the identification Ω 7→ Ω " # 0 −Ω ˆ= Ω . (1.62) Ω 0 ˆ 2 = −Ω2 I2×2 where I2×2 is the 2 × 2 identity matrix. Using this identification show that Ω Exercise 10 The space of 3 × 3 skew-symmetric matrices can be identified with R3 using the ˆ where identification Ω 7→ Ω   0 −Ω3 Ω2 ˆ = 0 −Ω1  Ω (1.63)  Ω3 , −Ω2 Ω1 0 and Ω = [Ω1 Ω2 Ω3 ]T . Using this identification show that ˆ = Ω × X. ΩX Exercise 11 — Einstein’s box experiment: Explain why a person standing on a scale inside an elevator sees his or her weight doubled as the elevator accelerates up at a rate of g and sees the weight reduced to zero if the elevator decelerates at a rate of g. Also show that if, for some reason, the gravitational force field vanished and the elevator was moving up at an acceleration of g then the scale would still show the correct weight of the person.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

27

Exercise 12 — Particle motion in 2-dimensions: Consider a particle P moving in 2-dimensional Euclidian space. Using the results of section 1.3.2 describe the inertial velocity and inertial acceleration of the particle P as expressed in the rotating ortho-normal frame b(t) = [b1 (t) b2 (t)] (refer to figure 1.6). The frame e = [e1 e2 ] is an ortho-normal inertial frame. Write down the apparent forces acting on a particle as observed in the moving frame.

Figure 1.6:

Exercise 13 — Motion of a particle fixed in a rotating frame: If a particle P appears to be fixed in the moving frame of exercise 12 describe the inertial velocity and inertial acceleration of the particle P as observed in the rotating frame. What are the constraint forces required to keep the particle fixed in the rotating frame. If the rate of rotation of b(t) is a constant in the counter ˙ clockwise direction, θ(t) = ψ, describe the motion of the particle P in the inertial frame e and show that the motion corresponds to a constant rate counter clockwise circular motion. Verify your conclusions using MATLAB simulations. Exercise 14 — Observing a fixed particle in a rotating frame: A particle P is fixed in an orthonormal inertial frame, e, at a distance l from the origin. Show that an observer moving with a rotating ortho-normal frame, with origin O coinciding with that of e, rotating at a constant rate of ψ in the counter clockwise direction will see it rotating clockwise in a circle at a constant angular rate of ψ. Verify your answer using MATLAB simulations. Show that the observer in the moving frame will think that a constant radial force equal to lψ 2 is acting on the particle. Exercise 15 — 2-dimensional particle motion in polar co-ordinates: Using the results of Exercise 12, for a particle moving in 2-dimensions, write down explicitly the inertial velocity and inertial acceleration as expressed in the moving frame b(t) = [er (t) eθ (t)] (refer to figure 1.7). Additionally ˙ = ψ, a constant, then write down the inertial velocity and inertial acceleration of the particle if θ(t)

28

Figure 1.7:

Figure 1.8:

as expressed in the moving frame b(t) = [er (t) eθ (t)]. In each case write down the apparent forces acting on a particle as observed by an observer in the moving frame. Exercise 16 A point P (t) appears to be fixed in the moving frame b(t) = [b1 (t) b2 (t) b3 (t)] write down the forces acting on the particle as observed by an observer in the moving frame. Exercise 17 Consider a ball of mass m constrained to move as shown in figure 1.8. The disk is rotating at a constant angular rate of Ω. Write down the equations of motion of the ball and also the constraint forces. Exercise 18 Consider a ball of mass m constrained to move as shown in figure 1.9. The Disk is rotating at a constant angular rate of Ω. Write down the equations of motion of the ball.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

29

Figure 1.9:

Exercise 19 Consider the bead on a rotating hoop shown in figure 1.10. The hoop is rotating about the vertical axis in the counter clockwise direction at a constant angular rate of Ω. Neglecting friction between the bead and the hoop write down the governing equations that describe the motion of the bead in a co-ordinate frame fixed on the hoop. Also write down the constraint forces acting on the bead. Exercise 20 Consider the systems in figure 1.11. The shaft with the arm rotates about the vertical axis in the counter clockwise direction. The spring is pivoted to the arm so that it can freely rotate in the plane of the shaft and the arm. Write down the equations that describe the motion of the point P in a co-ordinate frame that is fixed on the shaft. The mass of point P is m. Exercise 21 For the system shown in figure 1.12, (i.) Derive the equations of motion as observed by an observer in an inertial frame. (ii.) Derive the equations of motion as observed by an observer moving with the box (ie. in the moving frame co-ordinates).

30

Figure 1.10:

Exercise 22 For each of the systems shown in figures 1.13 to 1.18, derive the governing differential equations using Newton’s law. Exercise 23 Consider the ball inside the swinging hoop shown in figure 1.19. The hoop has a mass distribution of ρ per unit length and a radius of r. Neglecting friction between the ball and the hoop write down the constraint forces acting on the ball. Write down Euler’s rigid body equations for the hoop and using this expression eliminate the constraint forces appearing in the equations for the ball and find the equations that govern the motion of the entire system. Exercise 24 — Simulation and Experimentation: Consider the case of two observers E and B. Observer E is positioned in a 2-dimensional ortho-normal inertial frame, e = [e1 e2 ] and the observer B is positioned in a rotating ortho-normal frame b(t) = [b1 (t) b2 (t)]. Both frames have coinciding origins. Discuss the following for the case where the frame b(t) with observer B is ˙ = ψ in the counter clockwise direction. rotating at a constant angular rate of θ(t) a.) Observer B takes a particle of unit mass and slowly releases it on the rotating plane along the b1 (t) axis at a unit distance away from the origin. Neglecting friction between the plane and the particle, simulate using MATLAB, the motion of the particle as seen by B. b.) Observer B takes a particle of unit mass and constraints it so that it can move only along the b1 (t) axis. He then slowly releases it at time t = 0 at a unit distance away from the origin. Find the constraint force at a time t.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

31

Figure 1.11:

c.) However observer B is aware that no external force fields are present. Therefore having learnt Newtonian mechanics in his undergraduate years B realizes that the frame in which he is making the measurements is rotating. Device and carry out an experiment the observer B can use to determine if the frame is rotating at a constant speed or not and find the rate of rotation if the answer is yes. (Hint: You can use a setup similar to that of the Centrifugal force apparatus in the Applied Mechanics lab.) d.) Describe using MATLAB the motion of the particle in part a.) as seen by the observer E. e.) Assume that a central inward force field is acting in the inertial frame where the intensity is equal to r12 Newtons per unit mass. A particle of unit mass is released in the rotating frame at a distance of 0.5 units away from the origin and with zero velocity. Using results of exercise 12 predict the behavior of the particle as observed in the rotating frame and using MATLAB simulate the motion of the particle in the rotating frame. f.) Explain using the results of e.) why a Hurricane that has formed in the Northern hemisphere has a counter clockwise rotation and a Hurricane that has formed in the Southern hemisphere has a clockwise rotation (see figure 1.20). Exercise 25 — Self study on vibrating MEMS Gyroscopes: The operation of a vibrating MEMS Gyroscope relies on the presence Coriolis forces. Write a two page report on vibrating MEMS gyroscopes. Include in the report an explanation of its working principle, its use, and methods of fabrication.

32

Figure 1.12:

Figure 1.13:

Exercise 26 Show that x(t) × fpe (t) = R(t) (X × Fpe (t)). (Hint: First show using the properties of cross product that R(a × b) = Ra × Rb) Exercise 27 Using the following properties of cross products in R3 1. A × (B × C) = (A ∙ C)B − (B ∙ A)C 2. A ∙ (B × C) = C ∙ (A × B) = B ∙ (C × A) 3. (Ω × X) ∙ (ω × X) = (Ω ∙ ω)||X||2 − (Ω ∙ X)(ω ∙ X) show that where

X

I=

1 ˆ 2 X = 1 ΩT IΩ, −m X T Ω 2 2

X





m ||X||2 I3×3 − XX T ,

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

33

Figure 1.14:

Figure 1.15:

is defined to be the inertia tensor of the body about the fixed point O, and

and

X X

ˆ Ω(t)X ˆ˙ ˙ mX = I Ω,

ˆΩ ˆ 2 (t)X = − IΩ × Ω. mX

Exercise 28 — Gimbal Gyroscope: Consider the Gyroscope shown in figure 1.21. Show using the rigid body equations for an axi-symmetric body, (1.44) — (1.46), that a.) if no external torques are applied on the system the spin axis direction will remain a constant. b.) an external torque applied on the frame about the z axis will induce a motion about the x axis. Verify your answers using the Gyroscopic experimental setup in the applied mechanics lab.

1.7

Practice exercises

1. For the single bar pendulum shown in figure 1.22 derive the equations of motion using Euler’s 2-dimensional rigid body equations and also write down the kinetic energy of the system.

34

Figure 1.16:

Figure 1.17:

2. For the double bar pendulum shown in figure 1.23 write down the kinetic energy of the system.

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

35

Figure 1.18:

Figure 1.19:

(a) A Southern Hurricane (Catarina).

(b) A Northern Polar Hurricane.

Figure 1.20: Figures show the clockwise and anti-clockwise rotation of respectively a southern hemisphere and northern hemisphere formed hurricane. Figures are courtesy of Wikipedia.

36

Figure 1.21: The Gimbal Gyroscope

Figure 1.22:

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

Figure 1.23:

37

38

Chapter 2 Mechanisms 2.1

Introduction

A coupling of rigid bodies through various types of connections is called a kinematic chain. The various types of connections are called kinematic pairs while each constituent rigid body is called a link. A kinematic chain is said to be open if at least one link is coupled to the others only through one kinematic pair and is said to be closed otherwise. Kinematic pairs are classified by their degree of movability s (ie. the degrees of freedom of the relative motion of the two links joined by the pair). Revolute, prismatic and screw are examples of s = 1 while cylindric and roll-slide pairs are examples of s = 2. Higher pairs of movability are spheric and planar — s = 3, cylinder-plane — s = 4, sphere-plane — s = 5. The degrees of freedom (DOF), W , of a kinematic chain is given by f = 6N −

5 X

s=0

(6 − s)ns ,

(2.1)

where N is the total number of links and ns is the total number of kinematic pairs with movability s. When the chain is constrained to move in a two-dimensional Euclidian plane then (2.1) reduces to f = 3N −

2 X

s=0

(3 − s)ns .

(2.2)

A kinematic chain is said to be a structure if the couplings are such that no mobility remains in the chain. A kinematic chain is said to be a mechanism if at least one link is “fixed”. A mechanism is capable of transmitting force and/or motion from one place to another. The fixed link of a mechanism is usually referred to as the frame and the motion of the rest of the links are expressed with respect to this link. By fixing different links of a given chain we can obtain different mechanisms. This is called kinematic inversion. It can be seen that the relative motion 39

40 of all kinematic inversions are identical. The number of DOF of a mechanism is also given by (2.1) and (2.2) with N being replaced by N − 1 to account for the loss of DOF due to fixing one link. The first step in the design or analysis of a mechanism is to sketch an equivalent skeleton or kinematic diagram. Conventionally in this diagram the links are numbered and the joints are lettered with the fixed link always numbered as 1. The next step is to represent the mechanism using a graph whose vertices correspond to links and whose edges correspond to kinematic pairs where the number of edges joining two nodes correspond to the number of DOF, ns , of the kinematic pair that joins the two links represented by the nodes. These graphs enable one to easily calculate the number of DOF and furthermore allows the use of graph theory to design and analyze mechanisms. The next step in the design and analysis process is the problem of geometric analysis. If a given mechanism has n DOF then there exists n number of independent quantities q = (q1 , q2 , ∙ ∙ ∙ , qn ) called configuration co-ordinates that uniquely describe the configuration of the mechanism. The output of a mechanism are a m number of functions yk = φ(q) k = 1, 2, ∙ ∙ ∙ , m of the configuration co-ordinates q = (q1 , q2 , ∙ ∙ ∙ , qn ). It is only rarely that one can find these functions in explicit form. Typically one finds m number of constraint equations between the qk ’s and the yk ’s that need to be solved using numerical methods. Sometimes the inverse problem of finding the qk for a given yk is the primary concern. The next step in the design and analysis is to obtain the corresponding relationships between the instantaneous velocities. The last step is to find the relationship between the generalized input forces and the generalized output forces. This is obtained by neglecting the dynamics of the mechanism and considering energy conservation. For two bodies in relative motion it can be shown that there exists a common point with zero instantaneous velocity. Such a point is called an instantaneous center. An important result applicable for two-dimensional mechanisms is what is known as Kennedy’s theorem. This states that three rigid bodies in relative motion with each other will have their respective instantaneous centers lying in a straight line. x = Y + R(φ)X, ˆ˙ = 0, x˙ = Y˙ + R(φ)φX ˆ˙ φX = −R(−φ)Y˙ , 1 ˆ˙ X = φR(−φ)Y˙ φ˙ 2

(2.3) (2.4) (2.5) (2.6)

Proof of Kennedy’s theorem x = R(θ)X1 = Y + R(φ)X2 ,

(2.7)

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306 ˆ˙ , ˙ 1 = R(φ)φX x˙ = R(θ)ˆθX 2 ˆθR(θ)X ˆ˙ ˙ 1 = φR(φ)X2 , ˆ˙ ˆθ˙ (Y + R(φ)X ) = φR(φ)X 2 2, ˙θ R(φ)X2 = Y, ˙ (φ˙ − θ) x =

2.2

φ˙ Y, ˙ (φ˙ − θ)

41 (2.8) (2.9) (2.10) (2.11) (2.12)

Exercises

1. Obtain the corresponding graph of the mechanisms shown in figures figure 2.1 — figure 2.8 and find their number of DOF. 2. For the slider crank mechanism shown in figure 2.4

Figure 2.1: Digger Mechanism

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Figure 2.2: Aircraft Door Mechanism

Figure 2.3: Hood Mechanism

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

Figure 2.4: Windshield Wiper Mechanism

Figure 2.5: Windshield Wiper Mechanism

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44

Figure 2.6: Crank and Rocker Mechanism

Figure 2.7: Cam and Follower Mechanism

Mechanics of Machines: Class notes for ME2204, ME3305, ME4306

Figure 2.8: Plier Mechanism

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Bibliography [1] Haim Baruh, Analytical Dynamics McGraw Hill, 1999. [2] R. Abraham and J. E. Marsden, Foundations of Mechanics, Second Ed. Westview, 1978. [3] V. I. Arnold, Mathematical Methods of Classical Mechanics, Second Ed., Springer-Verlag, New York 1989. [4] J. E. Marsden and T. S. Ratiu, Introduction to Mechanics and Symmetry, Second Ed. SpringerVerlag, New York 1999. [5] F. Bullo and A. D. Lewis, Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems, Springer-Verlag, New York 2004.

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