- •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
336 |
11 Energy |
Comment: This demonstrates how to use the energy principle for practical calculations. We use the same procedure as for energy conservation, but include the work of non-conservative forces. In most cases, we need to know the path, r(t ) taken by the object in order to calculate the work done by the non-conservative forces. This method is therefore often not that practical.
11.5 Potential Energy in Three Dimensions
So far we have restricted ourselves to one-dimensional motions. Let us now introduce the potential energy for a three-dimensional motion and a three-dimensional force.
We call a force F(r) conservative if the work done by the force from a point 0 to a point 1 is independent of the path taken. That is, we call a force F(r) conservative, if (and only if) there is a function U (r) so that:
1
W0 to 1 = F(r) · d r = U (r0) − U (r1) , (11.112)
0
for all (possible) paths between r0 and r1.
It is trivial to show that the work is independent of the path if a function U (r) with the property of (11.112) exists. What about the other way—if the work along all paths are the same, can we prove that there must exist a function U (r)? This is also not difficult, since the function is simply defined as the work integral between the two end-points.
How is the force F(r) and the function U (r) related? Let us look at the work done
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W = 0 1 F · d r = 0 |
1 Fx d x + F |
y d y + Fz d |
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Because U (r) = U (x , y, z), we can write:
dU (x , y, z) = |
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∂U |
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d z = −Fx d x − Fy d x − Fz d z , (11.114) |
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that is:
11.5 Potential Energy in Three Dimensions |
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If the force F is conservative, we can find a function U so that |
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F = − U , |
(11.116) |
We call the function U (r) the potential energy for the force (field) F(r).
Criterion for a conservative force: This means that the criterion for a force to be conservative is a bit stronger in twoand three-dimensions than in one dimension. In one dimension, a force F (x ) is conservative if F (x ) is a function of position alone.
In twoand three-dimensions, it is a necessary condition that the force F(r) is only a function of the position, for the force to be conservative. But it is not a sufficient condition. There are forces F(r) that are only functions of position, but that still are not conservative. In order for the force to be conservative, it must be the gradient of a potential:
F = − U . |
(11.117) |
From calculus, we know that a force F(r) can be written as a gradient of a function, if and only if, the curl of F(r) is zero everywhere (for all r):
× F = 0 ( for all r) , |
(11.118) |
because in this case, the integral of all closed curves is zero.
11.5.1 Example: Constant Gravity in Three Dimensions
Problem: Find the potential energy U (r) for the gravitational force G = −mg j. Solution: We want to find a function U (r) that satisfies:
G = −mg j = − U , |
(11.119) |
Let us try the solution we already know from one dimension:
U (x , y, z) = mgy , |
(11.120) |
We find the gradient of U :
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∂∂x i + ∂ y j + ∂ z k |
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11 Energy |
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U = |
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This means that the graviational force G is the gradient of the potential U = mgy, that U = mgy is a potential energy (function) also in three dimensions, and that the constant gravitational force is conservative.
11.5.2 Example: Gravity in Three Dimensions
Problem: The gravitational force on an object of mass m at position r from an object of mass M in the origin is:
F = − |
Gm M r |
(11.123) |
r 2 r . |
Is this force conservative, and can you find the potential energy for this force? Approach: We know that the force is conservative if the work on the object (of mass m) does not depend on the path. Let us find the work done along a path, and demonstrate that it is only dependent on the displacement and not on the path. Solution: The work done on an object when it is moved along a path r(t ) from r(t0) = r0 to r(t1) = r1 is:
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F · d r = 0 |
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− G r 3 r · d r |
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We introduce a common trick for such integrals: We use that d (r · r) = 2r · d r, and therefore that r · d r = d ((1/2)r · r), and write the integral as:
W0,1 = 0 |
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(11.125) This integral does not depend on the path, only on the end-points. The gravitational force is therefore conservative.
We solve the integral to find the potential energy function: |
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The potential energy function is therefore: |
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11.5 Potential Energy in Three Dimensions |
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Analyze: We can check this result by calculating the gradient of the potential energy function:
U (r) = −Gm M r |
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∂ x i + |
∂ y j + |
∂ z k r |
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Let us calculate the x -component, using that r = x 2 + y2 + z2 1/2:
U · i = −Gm M ∂ x r |
= −Gm M − r 2 |
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And similarly for the y and the z components. This shows that:
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(11.129)
(11.130)
which proves that U (r) is the potential energy function for the gravitational energy.
11.5.3 Example: Non-conservative Force Field
Problem: Show that the force:
F = −y i + x j , |
(11.131) |
is not conservative, even though it depends on the position r alone.
Approach: We can prove that the force is not conservative in several ways. First, we recall the definition of a conservative force: A force is conservative if the work done by the force from point 0 to a point 1 is independent of the path. Our plan is to calculate the work along two different paths. If they are not the same we can conclude that the force is not conservative.
Solution: We study two different paths, A and B, from r0 = x0 i + y0 j = 0 to r1 = x1 i + y1 j as illustrated in Fig. 11.16. Let us calculate the work done along each path:
Path A: Path A goes as a straight line first from (0, 0) to (x1, 0) (path A1), and then from (x1, 0) to (x1, y1) (path A2). The work alongthese two subpaths are:
W A1 |
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F · d r = |
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(−y i + x j) · i d x = 0 |
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− y0 d x = 0 . (11.132) |
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Fig. 11.16 Illustration of the two paths A and B between the points 0 and 1 are shown. The arrows indicate the force field, F(r)
11 Energy
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(−y i + x j) · j d y = 0 |
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x1 d y = x1 y1 . (11.133) |
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W A = W A1 + W A2 = 0 + x1 y1 = x1 y1 . |
(11.134) |
Path B: Path B goes as a straight line first from (0, 0) to (0, y1) (path B1), and then from (0, y1) to (x1, y1) (path B2). The work along these two subpaths are:
WB1 = B1 |
F · d r = y0 |
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(−y i + x j) · j d y = |
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x0 d y = 0 . |
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(−y i + x j) · i d x = 0 |
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(11.136) |
The work done by F along path B is therefore:
WB = WB1 + WB2 = 0 − x1 y1 = −x1 y1 . |
(11.137) |
We see that the work done by F along the two paths A and B between the two points 0 and 1 are not the same. The work therefore depends on the path, and the force is not conservative!
Analyze: The force is conservative if and only if it can be written as the gradient of a potential. A necessary condition for that, is that the curl of F is zero. We can therefore alos determine is the force is conservative by calculating the curl of F: