Notes Meriam Kraige PDF

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These are the notes for the dynamics course , it has all the farmulas and the teqniques to solve the problems of this course ...

Description

1

Introduction

1.1 Books

2

1.1 Books

1P3 Dynamics 8 lectures Hilary 2012 Tutorial sheets 1P3H/J Course page

Mark Cannon [email protected] www.eng.ox.ac.uk/∼conmrc/dyn

Favourites are • Meriam and Kraige Engineering Mechanics Volume 2 Dynamics 5th edition, SI version, Wiley, 2003. • Meriam Dynamics 2nd edition, SI version, Wiley, 1975. Other possibilities are • Hibbeler Engineering Mechanics - Dynamics SI edition, Prentice Hall, 1997.

1

Introduction

A vital aspect of engineering design is the calculation of forces, and hence material stresses. In many cases, for example bridges and buildings, the laws of statics may be sufficient to calculate these forces. In more mechanically oriented design, the forces due to the motion of components of machines or structures may become far more significant. To calculate these forces, we need to understand the laws of dynamics. In this course we consider kinematics, which is about space and movement, and ways of describing the motion of a particle or a body. We also consider dynamics, which is about the forces and moments that must be

• Bedford and Fowler Engineering Mechanics - Dynamics SI edition, AddisonWesley, 1996. • Soutas-Little and Inman Engineering Mechanics - Dynamics SI edition, Prentice Hall, 1999. • Shames Engineering Mechanics. • Goodman and Warner Dynamics. • Housner and Hudson Dynamics. • Greenwood Principles of Dynamics. • Pytel and Kiusalaas Engineering Mechanics - Dynamics second edition, Brooks-Cole, 1999 (imperial units).

applied to the particle or body in order to cause motion. Fundamental to this course is Newton’s second law . In fact, it is not too much of an exaggeration to say that Newton’s second law is about the only thing you need to know to succeed in dynamics – you just have to know it well!

1.2 Syllabus The Course Handbook gives the following expanded syllabus: Plane kinematics of particles: rectilinear and curvilinear motion in rectangular, normal-tangential, and polar coordinates; relative motion (translating, not rotating, axes). Plane kinematics of rigid bodies: translation, rotation, and general plane motion; relative motion; rotation about a fixed axis.

3

Introduction

Dynamics of particles: Newton’s second law; work, energy, power; impulse and momentum (linear and angular); conservation of energy and momentum (linear and angular); impact; central-force motion. Dynamics of rigid bodies: equations of motion for translation and fixedaxis rotation; moment of inertia; work and energy; impulse and momentum (linear and angular). Simple variable mass problems (i.e. rockets).

1.3 Learning Outcomes

4

• analyse elastic and inelastic collisions between particles • understand the concepts of work, energy, and power • use the principle of conservation of energy to analyse the motion of particles • describe planar motion of a particle in rectangular, normal-tangential, and polar coordinates • understand the concepts of moment and moment of momentum

Lectures

• analyse the motion of a particle orbiting under the action of a central force

There are eight lectures. We begin by looking at force and momentum as applied to particles, then go on to consider work, power and energy. Next we consider circular motion of a particle, and also gravity, both of which are applied to the analysis of satellite orbits. Finally we extend the work on particles to the case of rigid bodies under the action of steady or impulsive forces and torques.

• find the instantaneous centre of rotation of a rigid body

Example Sheets

• determine the motion of a planar rigid body using the principles of conservation of angular momentum and conservation of energy, where appropriate

There are two example sheets: 1P3H concentrates on the dynamics of particles, and 1P3J covers the dynamics of rigid bodies .

• analyse the translation and rotation of a planar rigid body under the action of a steady or impulsive force or moment.

1.3

Learning Outcomes

After attending the lectures and doing the accompanying example sheets in tutorials, you should be able to: • understand and use the definitions of velocity and acceleration

• calculate the moment of inertia of a planar rigid body from first principles or from standard cases

• express general planar rigid body motion as the combination of rotation and translation

Lecture Notes These lecture notes (and also the lecture slides) are provided as handouts in lectures. Both the notes and slides are available on weblearn and on the course web page: http://www.eng.ox.ac.uk/∼conmrc/dyn.

• analyse straight-line motion of a particle with variable acceleration

The lecture notes draw on material prepared in the past for this and related courses by previous lecturers, particularly Dr Colin Wood, Dr Peter

• use Newton’s second law to analyse planar motion of a particle under the action of a steady or impulsive force

McFadden and Dr Yiannis Ventikos. Their contributions are gratefully acknowledged.

• understand the principle of conservation of momentum and its application to particles

Please send any comments or corrections to [email protected]

5

CONTENTS

Contents

CONTENTS

6

5.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 50 5.3 Moment of Momentum . . . . . . . . . . . . . . . . . . . . . . . 54

1 Introduction 1.1 Books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 2

1.2 Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

1.3 Learning Outcomes . . . . . . . . . . . . . . . . . . . . . . . .

3

6 Gravity and Satellite Orbits

58

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 6.2 Newton’s Law of Gravitation . . . . . . . . . . . . . . . . . . . . 60 6.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 61

2 Force and Momentum

7

6.4 Moment of Momentum and Energy . . . . . . . . . . . . . . . . 61

7

6.5 Calculations Without Full Analysis . . . . . . . . . . . . . . . . 63

2.2 Impulse and Momentum . . . . . . . . . . . . . . . . . . . . . . 12

6.6 Full Analysis of Satellite Orbits . . . . . . . . . . . . . . . . . . 67

2.1 Force and Motion . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.4 Problems involving Continuous Flow . . . . . . . . . . . . . . . 19

7 Rigid Bodies

72

7.1 Moment of Momentum of a Particle . . . . . . . . . . . . . . . . 72 3 Work and Energy

22

7.2 Moments of Momentum and Inertia of a Rigid Body . . . . . . . 74

3.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7.3 Kinetic Energy of Body Rotating about Fixed Axis . . . . . . . . 77

3.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7.4 Inertia Properties for Thin Flat Bodies . . . . . . . . . . . . . . 78

3.3 Reconciliation with Thermodynamics . . . . . . . . . . . . . . . 31

7.5 Further Examples . . . . . . . . . . . . . . . . . . . . . . . . . 82

3.4 Examples of Work and Energy Conservation . . . . . . . . . . 31

7.6 Combined rotation and translation . . . . . . . . . . . . . . . . 83

3.5 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.6 Forces and Moments . . . . . . . . . . . . . . . . . . . . . . . . 36 4 Examples I

39

4.1 Problems Requiring Full Equations of Motion . . . . . . . . . . 39 4.2 Problems Using Energy and/or Momentum . . . . . . . . . . . 42 5 Circular and General Curvilinear Motion

49

5.1 Normal-tangential Coordinates . . . . . . . . . . . . . . . . . . 49

8 Examples II

87

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 8.2 Examples Using Moment of Momentum . . . . . . . . . . . . . 87 8.3 Examples Using Conservation of Energy . . . . . . . . . . . . . 90

7

Force and Momentum

2 Force and Momentum 2.1

2.1 Force and Motion

8

time. If the acceleration a is constant, this gives the familiar formulae: V (t ) = V0 + at

Force and Motion

1 x (t ) = x0 + V0 t + at 2 . 2

The expression for acceleration in (2.1) could also be written another way: 2.1.1

Particles and Rigid Bodies

d x˙ dt dV d x˙ dx d x˙ =V = x˙ = dx dx dt dx which can be integrated as follows: d �v2 � dV =a = V dx dx � 2 x 1 2 1 2 V − V0 = a(x) dx , 2 2 x0 a = x¨ =

In dynamics, the term particle is used to describe any body which we decide to treat as a discrete mass concentrated at a point. We do this when we wish to consider the linear movement or translation of the body, but not its rotation. Typical applications would be the analysis of the motion of a spacecraft orbiting the earth, or the trajectory of a golf ball after it has been struck. Later in this course, for problems where rotation matters as well as translation, we shall consider rigid bodies with a defined mass distribution.

(2.2)

and hence allows V to be determined if a is known as a function of x . If the acceleration a is constant, then clearly this gives 2.1.2

Motion in a Straight Line with Variable Acceleration

0

P

P

1 2 1 2 V − V0 = a(x − x0 ) 2 2 which is the same as the familiar equation V 2 = V0 2+ 2as.

x

!x

2.1.3

Newton’s Second Law

Figure 1: Particle moving in a straight line For a particle, Newton’s second law can be written as The average velocity of a particle moving a distance ∆x in time ∆t is defined Vav = ∆x/∆t . As ∆t gets smaller, this quantity tends to the instantaneous velocity: V = lim∆t→0 ∆x/∆t . The instantaneous velocity V (and similarly acceleration a) are therefore V =

dx = x˙ dt

a=

where momentum = mass × velocity Expressing this in algebraic form, we write the vector equation

dV = x¨ dt

(2.1) F=

The above equations can be integrated to give � t � t V (t ) = V0 + a(t ) dt x (t ) = x0 + V (t ) dt 0

force = rate of change of momentum

dV dm d mV = m +V dt dt dt

We will see many examples where the mass is invariant in which case d m/dt = 0 and the familiar, simple form of Newton’s second law emerges

0

which allows V and x to be determined when a is known as a function of

(2.3)

F=m

dV dt

or

F = ma

(2.4)

9

Force and Momentum

2.1 Force and Motion

2.1.5

or in words force = mass × acceleration We will also see some examples, such as rockets, where the mass changes, and so the full form of Newton’s second law must be used. Note that these are vector statements. Force, velocity and momentum are all vector quantities that have both magnitude and direction. When we are only concerned with forces and motion in one direction, we can use the corresponding scalar forms given by

d dV dm mV = m +V dt dt dt

(2.5)

and scalar acceleration so we will have to deduce this from the context.

Example – Motion in a Straight Line

Problem: A boat of mass 1500 kg is launched from a trolley on a sloping ramp. The trolley is allowed to run down the ramp at 1 m s−1 until the boat is

and F =m

2.1.4

Velocity and Speed

It is a useful convention to use the term velocity when referring to a vector which has both speed and direction. The term speed then refers to the magnitude only and is a scalar . In these notes, vectors will be represented in bold upright font, and scalars in italic font e.g. V is velocity and V is speed. Unfortunately, we do not have words in English to distinguish between vector

2.1.6 F =

10

dV dt

or

F = ma

(2.6)

Force Mass and Weight

just afloat. The trolley then stops and the boat continues to move at 1 m s−1 . Once afloat, a crew member of mass 70 kg stops the boat by pulling steadily on a rope with a force equal to 30 % of his own weight, as shown in figure 2. How long will the boat take to stop, and what length of rope must be allowed to slip? 1 ms

!1

force

If we drop an object, it accelerates downwards at g = 9.81 m s−2 , assuming that we are at or near the earth’s surface. If we do not drop the object, we must exert an upward force on it to oppose the downward force of gravity. This force is the weight of the object and it is the weight that causes the acceleration if it is dropped. weight = mass × gravitational acceleration In the SI system the unit of force is the Newton. It is defined so that unit force causes unit acceleration when it acts upon unit mass i.e. 1 Newton acting on 1 kg produces 1 m s−2 acceleration. Using these units, the weight of a 1 kg mass at or near the earth’s surface must be 9.81 Newtons. Thus weight (Newtons) = mass (kg) × 9.81 (m s−2 )

Figure 2: Launching a boat Solution: We are interested here in motion and forces in one direction only for a mass-invariant system, so we can use the scalar form of Newton’s second law, F = ma, as given in equation 2.6. Substituting the values above gives −70 × 9.81 × 0.3 = 1500 × a from which we can find the acceleration a=

dV = −0.137 m s−2 dt

11

Force and Momentum

By separating the variables dV a and integrating we get the time to stop the boat dt =

t=

V2 − V1 0−1 = 7.28 s = −0.137 a

To find the distance we write a=

dV dV dx dV = =V dx dt dx dt

We then separate the variables

12

Solution: Consider the free body diagram in figure 3. The only forces acting are the weight force mg downwards and the aerodynamic drag D upwards. � Using Newton’s second law F = ma we get mg −

1 dV CD ρV 2 A = m dt 2

(2.7)

Terminal velocity Vt occurs when dV /dt = 0 in equation 2.7 so Vt 2 =

2mg 2 × 75 × 9.81 = ρACD 1.007 × 0.8 × 1.2

which gives Vt = 39.01 m s−1

V dV = a dx

To calculate the variation of velocity V with distance, we re-write equation 2.7 recalling that a = V dV /dx :

and integrate

1 2 (V − V12 ) = a (x2 − x1 ) 2 2 Hence the distance to stop the boat is 1 02 − 12 s = x2 − x1 = = 3.64 m 2 −0.137

2.1.7

2.2 Impulse and Momentum

Example – Terminal Velocity in Free Fall

Problem: A free-fall parachutist has a mass m = 75 kg, and a frontal area A = 0.8 m2 . If the air density at 2000 m is ρ = 1.007 kg m−3 (HLT page 68) and the aerodynamic drag is given by D = 21CD ρV 2 A where the drag coefficient is CD = 1.2, find the terminal speed Vt . How far will the parachutist fall before reaching 90 % of this terminal speed?

V

dV ρACD = dx 2m

�

2mg − V2 ρACD

�

Recognising the terms which correspond to Vt 2, we can write � � dV g V = 2 V 2t − V 2 dx Vt Separating the variables and integrating between limits of V = 0 at x = 0 and V = 0.9Vt at x = −s (i.e. after falling a distance s) gives � 2 � � V V 2 0.9Vt V2 V − (0.9Vt )2 −s − 0 = t dV = t ln t 2 2 2 2g − V g 0 − 0 V Vt � � t 1 = 128.8 m =⇒ s = 77.58 ln 1 − 0.81 2.2

Impulse and Momentum

D

In the above example, the force acting on the body was constant and we knew its value. In some instances, the force will vary with time during the interaction with the body, and we may not always know in what way it varies, mg

Figure 3: Parachutist in free fall

nor its magnitude at any time. This is particularly true during impacts and collisions. To deal with these problems, we use the concept of impulse, the integral of the force on the body over the duration of the interaction, and relate it to the change in the momentum of the body.

13

Force and Momentum

Given that F=m

2.2.2

dV dt

for a body of mass m, then �

2.2 Impulse and Momentum

2

F dt = m (V2 − V1 ) 1

The integral of force is called the impulse, and the product mV is called the momentum. The verbal equivalent of this equation is impulse = change of momentum The convenience of this form is that it requires no details of the time-variation of the force. An impulse can describe an impact involving a very large force with a very short duration, as illustrated in figure 4. The shaded area under the graph gives the magnitude of the impulse.

14

Example – Impulse with Constant Force

In the example in section 2.1.6, the time required to stop the boat could have been found by considering impulse and momentum instead of the equation of motion. In this case the force F is constant, so the form of the calculation would then have been � 2 � 2 F dt = F dt = F (t2 − t1 ) = m (V2 − V1 ) 1

1

giving t = t2 − t1 = m

2.2.3

V2 − V1 0−1 = 7.28 s = 1500 × −70 × 9.81 × 0.3 F

Example – Impulse and Momentum as Vectors

force impulse = area under force-time graph

time 0

Figure 4: Impulse = area under force-time graph

Problem: A cannon of mass M is free to roll without friction on horizontal ground. An explosive charge projects a ball of mass m at speed v relative to the barrel, which is inclined upward at angle θ as shown in figure 5(a). At the instant after the ball leaves the muzzle find (a) the backward recoil speed u of the gun (b) the absolute velocity components of the ball (c) the magnitude and direction of any external impulse acting on the system. v

2.2.1

u

(b)

(a)

Example – Impulse with Unknown Force Variation

v

!

Problem: During a game of cricket, the batsman is accidentally struck by a cricket ball of mass 0.15 kg travelling at 40 m s−1 . The ball is stopped by the impact. Can you estimate the force exerted by the ball? Solution: The duration of the impact is not known, nor do we know the variation of force during that brief instant, but the impulse is m∆V = 0.15 × 40

v sin !

u

! v cos ! " u

= 6.0 kg m s−1 . The shorter the time, the greater the force and the greater the

Figure 5: Components of velocity of cannon ball

pain. The wicket keeper, catching a similar ball, will flex his arms to increase the duration of the impact. The impulse on his hands is still the same, but by increasing its duration he reduces the magnitude of the force.

Solution: Instead of using vector algebra, here we shall try using scalar equations in the x (horizontal) and y (vertical) directions. The relative velocity and

15

Force and Momentum

2.3 Collisions

16

(a)

absolute velocity components ...