- •Preface
- •Contents
- •1 Introduction
- •1.1 Physics
- •1.2 Mechanics
- •1.3 Integrating Numerical Methods
- •1.4 Problems and Exercises
- •1.5 How to Learn Physics
- •1.5.1 Advice for How to Succeed
- •1.6 How to Use This Book
- •2 Getting Started with Programming
- •2.1 A Python Calculator
- •2.2 Scripts and Functions
- •2.3 Plotting Data-Sets
- •2.4 Plotting a Function
- •2.5 Random Numbers
- •2.6 Conditions
- •2.7 Reading Real Data
- •2.7.1 Example: Plot of Function and Derivative
- •3 Units and Measurement
- •3.1 Standardized Units
- •3.2 Changing Units
- •3.4 Numerical Representation
- •4 Motion in One Dimension
- •4.1 Description of Motion
- •4.1.1 Example: Motion of a Falling Tennis Ball
- •4.2 Calculation of Motion
- •4.2.1 Example: Modeling the Motion of a Falling Tennis Ball
- •5 Forces in One Dimension
- •5.1 What Is a Force?
- •5.2 Identifying Forces
- •5.3.1 Example: Acceleration and Forces on a Lunar Lander
- •5.4 Force Models
- •5.5 Force Model: Gravitational Force
- •5.6 Force Model: Viscous Force
- •5.6.1 Example: Falling Raindrops
- •5.7 Force Model: Spring Force
- •5.7.1 Example: Motion of a Hanging Block
- •5.9.1 Example: Weight in an Elevator
- •6 Motion in Two and Three Dimensions
- •6.1 Vectors
- •6.2 Description of Motion
- •6.2.1 Example: Mars Express
- •6.3 Calculation of Motion
- •6.3.1 Example: Feather in the Wind
- •6.4 Frames of Reference
- •6.4.1 Example: Motion of a Boat on a Flowing River
- •7 Forces in Two and Three Dimensions
- •7.1 Identifying Forces
- •7.3.1 Example: Motion of a Ball with Gravity
- •7.4.1 Example: Path Through a Tornado
- •7.5.1 Example: Motion of a Bouncing Ball with Air Resistance
- •7.6.1 Example: Comet Trajectory
- •8 Constrained Motion
- •8.1 Linear Motion
- •8.2 Curved Motion
- •8.2.1 Example: Acceleration of a Matchbox Car
- •8.2.2 Example: Acceleration of a Rotating Rod
- •8.2.3 Example: Normal Acceleration in Circular Motion
- •9 Forces and Constrained Motion
- •9.1 Linear Constraints
- •9.1.1 Example: A Bead in the Wind
- •9.2.1 Example: Static Friction Forces
- •9.2.2 Example: Dynamic Friction of a Block Sliding up a Hill
- •9.2.3 Example: Oscillations During an Earthquake
- •9.3 Circular Motion
- •9.3.1 Example: A Car Driving Through a Curve
- •9.3.2 Example: Pendulum with Air Resistance
- •10 Work
- •10.1 Integration Methods
- •10.2 Work-Energy Theorem
- •10.3 Work Done by One-Dimensional Force Models
- •10.3.1 Example: Jumping from the Roof
- •10.3.2 Example: Stopping in a Cushion
- •10.4.1 Example: Work of Gravity
- •10.4.2 Example: Roller-Coaster Motion
- •10.4.3 Example: Work on a Block Sliding Down a Plane
- •10.5 Power
- •10.5.1 Example: Power Exerted When Climbing the Stairs
- •10.5.2 Example: Power of Small Bacterium
- •11 Energy
- •11.1 Motivating Examples
- •11.2 Potential Energy in One Dimension
- •11.2.1 Example: Falling Faster
- •11.2.2 Example: Roller-Coaster Motion
- •11.2.3 Example: Pendulum
- •11.2.4 Example: Spring Cannon
- •11.3 Energy Diagrams
- •11.3.1 Example: Energy Diagram for the Vertical Bow-Shot
- •11.3.2 Example: Atomic Motion Along a Surface
- •11.4 The Energy Principle
- •11.4.1 Example: Lift and Release
- •11.4.2 Example: Sliding Block
- •11.5 Potential Energy in Three Dimensions
- •11.5.1 Example: Constant Gravity in Three Dimensions
- •11.5.2 Example: Gravity in Three Dimensions
- •11.5.3 Example: Non-conservative Force Field
- •11.6 Energy Conservation as a Test of Numerical Solutions
- •12 Momentum, Impulse, and Collisions
- •12.2 Translational Momentum
- •12.3 Impulse and Change in Momentum
- •12.3.1 Example: Ball Colliding with Wall
- •12.3.2 Example: Hitting a Tennis Ball
- •12.4 Isolated Systems and Conservation of Momentum
- •12.5 Collisions
- •12.5.1 Example: Ballistic Pendulum
- •12.5.2 Example: Super-Ball
- •12.6 Modeling and Visualization of Collisions
- •12.7 Rocket Equation
- •12.7.1 Example: Adding Mass to a Railway Car
- •12.7.2 Example: Rocket with Diminishing Mass
- •13 Multiparticle Systems
- •13.1 Motion of a Multiparticle System
- •13.2 The Center of Mass
- •13.2.1 Example: Points on a Line
- •13.2.2 Example: Center of Mass of Object with Hole
- •13.2.3 Example: Center of Mass by Integration
- •13.2.4 Example: Center of Mass from Image Analysis
- •13.3.1 Example: Ballistic Motion with an Explosion
- •13.4 Motion in the Center of Mass System
- •13.5 Energy Partitioning
- •13.5.1 Example: Bouncing Dumbbell
- •13.6 Energy Principle for Multi-particle Systems
- •14 Rotational Motion
- •14.2 Angular Velocity
- •14.3 Angular Acceleration
- •14.3.1 Example: Oscillating Antenna
- •14.4 Comparing Linear and Rotational Motion
- •14.5 Solving for the Rotational Motion
- •14.5.1 Example: Revolutions of an Accelerating Disc
- •14.5.2 Example: Angular Velocities of Two Objects in Contact
- •14.6 Rotational Motion in Three Dimensions
- •14.6.1 Example: Velocity and Acceleration of a Conical Pendulum
- •15 Rotation of Rigid Bodies
- •15.1 Rigid Bodies
- •15.2 Kinetic Energy of a Rotating Rigid Body
- •15.3 Calculating the Moment of Inertia
- •15.3.1 Example: Moment of Inertia of Two-Particle System
- •15.3.2 Example: Moment of Inertia of a Plate
- •15.4 Conservation of Energy for Rigid Bodies
- •15.4.1 Example: Rotating Rod
- •15.5 Relating Rotational and Translational Motion
- •15.5.1 Example: Weight and Spinning Wheel
- •15.5.2 Example: Rolling Down a Hill
- •16 Dynamics of Rigid Bodies
- •16.2.1 Example: Torque and Vector Decomposition
- •16.2.2 Example: Pulling at a Wheel
- •16.2.3 Example: Blowing at a Pendulum
- •16.3 Rotational Motion Around a Moving Center of Mass
- •16.3.1 Example: Kicking a Ball
- •16.3.2 Example: Rolling down an Inclined Plane
- •16.3.3 Example: Bouncing Rod
- •16.4 Collisions and Conservation Laws
- •16.4.1 Example: Block on a Frictionless Table
- •16.4.2 Example: Changing Your Angular Velocity
- •16.4.3 Example: Conservation of Rotational Momentum
- •16.4.4 Example: Ballistic Pendulum
- •16.4.5 Example: Rotating Rod
- •16.5 General Rotational Motion
- •Index
332 |
11 Energy |
xlabel(’U/Uˆ{\ast}’), ylabel(’x/b’) subplot(2,1,2); plot(x,t,’-b’,x[i],t[i],’o’); ylabel(’t/tˆ{\ast}’), xlabel(’x/b’)
11.4 The Energy Principle
So far we have addressed processes where the work is done by conservative forces only. Let us now use what we have learned about potential energy to also discuss processes with non-conservative forces. Non-conservative forces are typically forces that not only depend on position, but for example also on velocity, such as air resistance or friction. Let us examine the motion of an object subject to both conservative and non-conservative forces.
We decompose the net force acting on an object into conservative forces, F j , and a non-conservative force, f :
Fnet = F j + f , |
(11.91) |
j
If the object moves from r(t0) = r0 at t0 to r(t1) = r1 at t1, the net work is:
W0net,1 |
= |
t0 |
Fnet · v d t = |
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F j + f |
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j |
t0 |
F j · v d t + |
t0 |
f · v d t |
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(11.92) |
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[U (r0) − U (r1)] + W0,1 |
= U0 − U1 + W0,1 . |
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The work-energy theorem relates the net work to the change in kinetic energy for the object:
W0net,1 = U0 − U1 + W0f,1 = K1 − K0 . |
(11.93) |
We group terms related to position 1 on the left hand side:
U1 + K1 = U0 + K0 + W0f,1 . |
(11.94) |
If we introduce the term E1 = U1 + K1 for the total (mechanical) energy of the object, we get:
E1 = E0 + W0f,1 , |
(11.95) |
or |
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E1 − E0 = E = W0f,1 . |
(11.96) |
The change in the (mechanical) energy of the object is equal to the work done by the non-conservative force. This law is a completely general law of nature usually called the second law of thermodynamics or the energy principle:
11.4 The Energy Principle |
333 |
Energy principle: |
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E = W . |
(11.97) |
The change in (mechanical) energy is equal to the work done by nonconservative forces (W ).
We will later see how we can relate this energy principle for a single object to a more general principle for a system of several objects interacting with its environment through work and heat (thermal energy transfer).
11.4.1 Example: Lift and Release
Problem: If you lift a book from the floor to a height h using a constant force F , and release it, you have increased the total energy of the book, but the book was affected by a constant force. How can we explain this using the energy principle?
Solution: When you are not holding the book, and the book is falling, it is affected by the gravitational force alone. The total mechanical energy of the book (when falling) is therefore:
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1 |
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E = U + K = mgy + |
2 mv2 . |
(11.98) |
Initally, the book was lying on the floor with zero velocity. In this case, the total mechanical energy is:
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E0 = mgy0 + |
2 mv02 = 0 J . |
(11.99) |
If we choose the potential energy to be zero when the book lies on the floor, that is, when y = 0 m and v0 = 0 m/s. When the book is at rest (v1 = 0 m/s) at y1 = h, the total mechanical energy is:
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E1 = mgh + |
2 mv12 = mgh . |
(11.100) |
The change in mechanical energy is therefore:
E = E1 − E0 = mgh . |
(11.101) |
where did this energy come from? It came from the work done by the constant force F lifting the book to the height h.
W0F,1 = F (y1 − y0) = F h . |
(11.102) |
334 |
11 Energy |
Because F is a non-conservative force, the energy principle gives that:
E = W0F,1 = F h = mgh . |
(11.103) |
The work done by F corresponds to the work done by a constant force mg. But, you protest, the force F is a conservative force, since it is a constant. Something must be wrong with the energy principle!
Not so fast. The force F was constant only when you lifted the book. Afterwards, you released the book, and this applied force therefore became zero. The force F is therefore not that constant after all. This is indeed how we should regard such a force: The force F lifting the book is a time-dependent force. It only acted when the book was lifted. It does not act when the book is released afterwards. Then it is only affected by gravity. This is why we call it a non-conservative force. We could look at it in a different way: The force F does not only depend on the position y of the book, because when we lift the book, it is affected by the lifting force, but afterwards, when the book is released, it may be falling through the same position without being affected by the lifting force. Thus the lifting force is not a conservative force, and we must use the energy principle to understand the change in mechanical energy as caused by the work done by non-conservative forces.
11.4.2 Example: Sliding Block
Problem: A block is sliding down an inclined plane forming an angle α with the horizontal. The kinetic coefficient of friction for the contact between the block and the plane is µ. What is the velocity of the block as a function of the length L it has slid?
Identify: We use a coordinate system oriented along the plane, as illustrated in Fig. 11.15. The block slides from position x0, with initial velocity v0 = 0, to x1. Our task is to find the velocity v1 as a function of x1.
Model: Figure 11.15 shows that the block is affected by the normal force N , friction, f , and by gravity, G. Because the block is affected by a non-conservative force, f , we cannot use energy conservation to find the velocity as a function of position, but we can still use the energy principle: The change in energy is equal to the work done by the non-conservative forces.
E = W f , |
(11.104) |
where the change in energy is |
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E = (K1 + U1) − (K0 + U0) = W0f,1 . |
(11.105) |
The work done by the constant friction force, f = −µN , is |
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11.4 The Energy Principle |
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Fig. 11.15 Left Illustration of a block sliding from x0 to x1 along an inclined plane. Right Free-body diagram for the block
W0f,1 |
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f d x = −µN (x1 − x0) , |
(11.106) |
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where L = x1 − x0 is the distance moved along the plane. Because the block is not moving in the y-direction we find the normal force from Newton’s second law:
Fy = N − G y = N − mg cos α = may = 0 , |
(11.107) |
where Fig. 11.15 shows that G y = mg cos α. The work done by friction is therefore:
W0f,1 = −µmg L cos α . |
(11.108) |
To find the change in energy, E , we need the change in potential energy of the block, U1 − U0. Since the normal force does no work, the potential energy is the same as for motion under gravity alone: U = mgh, where h is the vertical position
of the block. Figure 11.15 shows that |
h = L sin α, and |
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U1 − U0 = mg |
h = −mg L sin α . |
(11.109) |
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We insert these results in E = W f : |
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E = E1 − E0 = U1 + K1 − (U0 + K0) = K1 − K0 + U1 − U0 |
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mv12 − 0 − mg L sin α = W0f,1 = −mg L cos α . |
(11.110) |
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We rearrange the equation to find: |
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v12 = mg L (sin α − µ cos α) v1 = |
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mg L (sin α − µ cos α) |
(11.111) |