- •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
190 |
7 Forces in Two and Three Dimensions |
Fig. 7.4 The motion of a ball thrown across the lecture room
7.3.1 Example: Motion of a Ball with Gravity
Throughout this chapter we will follow a particular problem as we gradually increase the complexity of the physical model: The motion of a ball thrown across the classroom. The experiment is illustrated in Fig. 7.4, which illustrates the observed path of a ball in an experiment. How can we develop a realistic model for the motion of this ball? We start from the simplest description: The motion when affected by gravity alone.
Problem: A ball is thrown from a height h above the ground at an angle α with the horizontal with an initial speed v0. Find the velocity and position of the ball at a time t .
Identify and Sketch: In this exercise we address the motion of the ball, described by the position r(t ) as a function of time. At the time t0, the ball was thrown. We place the coordinate system so that gravity is acting in the y-direction, and we place the x -axis so that the ball is thrown in the positive x -direction. The origin is placed at the ground, directly below the initial position of the ball at t = t0. The initial position vector is therefore r(t0) = r0 = h j. The initial velocity is directed at an angle α with the horizontal, this means that the initial velocity is v(t0) = v0 cos(α) i + v0 sin(α) j. The situation is illustrated in Fig. 7.5.
Fig. 7.5 a Illustration of the motion of the ball.
b Free-body diagram of the ball
(A)
y v0
α
h
x
(B)
FD
y |
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v |
x G
7.3 Force Model—Constant Gravity |
191 |
Model: The motion of the ball is determined by the forces acting on it. The only contact force acting on the ball is air resistance, FD , but we will here assume that this force is negligible. The only long-distance force acting on the ball is gravity, G, as illustrated in the free-body diagram in Fig. 7.5.
Newton’s second law is applied to both the x - and the y-component of the forces independently. In the x -direction Newton’s second law gives:
Fx = max = 0 . |
(7.9) |
There are no horizontal forces. The sum of the forces in the horizontal, x -direction is therefore zero. Consequently, the acceleration in the x -direction, ax is also zero.
Newton’s law of motion in the y-direction gives:
Fy = G = −mg = may , (7.10)
where we have used that the gravitational force from the Earth is mg, and that it acts in the negative y-direction.
The acceleration of the ball is therefore: |
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d2r |
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a = dt 2 |
= −g j , |
(7.11) |
and the initial conditions are r(t0) = h j and v(t0) = v0.
Solve: We find the motion of the ball by solving the differential equation in (7.11). Since the acceleration is constant, we can solve it by direct integration for each of the components.
In the x -direction, the acceleration is zero, and the velocity in this direction is therefore constant.
vx (t ) = vx (t0) = v0 cos(α) . |
(7.12) |
The x -position is given by direct integration:
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vx (t ) = |
d x |
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(7.13) |
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dt |
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t |
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t |
d x |
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vx (t )dt = |
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dt |
(7.14) |
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dt |
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t |
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t0 |
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t0 |
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v0 cos(α)dt = |
x |
(t )d x |
(7.15) |
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t0 |
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x (t0) |
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v0 cos(α)(t − t0) = x (t ) − x (t0) |
(7.16) |
that is, we have recovered motion with constant velocity:
x (t ) = x (t0) = v0 cos(α)(t − t0) , |
(7.17) |
192 |
7 Forces in Two and Three Dimensions |
In the y-direction, the acceleration is constant, ay = −g. We can find the velocity by direct integration:
t
t0
vy (t ) −
dv |
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t |
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dt = |
−gdt |
(7.18) |
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dt |
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t0 |
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vy (t0) = −g(t − t0) |
(7.19) |
which gives:
vy (t ) = (v0 sin(α) − g(t − t0)) |
(7.20) |
We find the position by integrating once more, using that vy (t ) = d y/dt , and that v0,y = v0 sin(α):
d y = dt
t d y
dt =
t0 dt
v0,y − g(t − t0) |
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(7.21) |
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t |
v0,y − g(t − t0) |
dt |
(7.22) |
t0 |
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y(t ) − y(t0) = t |
v0,y dt − g t (t − t0)dt |
(7.23) |
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t0 |
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t0 |
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y(t ) − y(t0) = v0,y (t − t0) − |
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g(t − t0)2 |
(7.24) |
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which gives |
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y(t ) = h + v0 sin(α)(t − t0) − |
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g(t − t0)2 , |
(7.25) |
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where y(t0) = h is the launch height of the projectile.
Analyze: We notice that the motion in the x - and y-directions are independent of each other. The motion in the x -direction is simply a motion with constant velocity. The motion in the y-direction is the same as for the one-dimensional problem. If the ground is flat, it is the motion in the y-direction that determines how long time it takes to reach the ground. We can therefore answer questions about flight time, and maximum height by just studying the one-dimensional motion along the y-direction.
We have now found the complete solution for the motion of a ball subject only to gravity. From this solution, we can answer any complicated question, such as how far the projectile travels or what choise of initial direction gives the maximum length.
7.4 Force Model—Viscous Force
For an object moving through a fluid, such as a projectile flying through the air, a meteor entering the Earth’s atmosphere, or a tiny microrobot navigating through your bloodstream, there is a contact force on the object due to the motion of the
7.4 Force Model—Viscous Force |
193 |
object relative to the fluid. The fluid has to flow around the object when the object moves, as a result the fluid exerts a force on the object. This force is distributed: It acts everywhere on the surface of the object, and it may also vary in magnitude and direction along the surface of the object. Usually, we will simplify the effect of the this distribution of forces into a single force acting in a single point on the object, and we will call this force the drag force or the “fluid resistance” acting on the object. For most purposes this is a sufficiently precise description of the interaction with the fluid.
The form of the drag force depends on the velocity of the object relative to the fluid. We discern between a behavior at low velocities, where the drag force is proportional to the velocity, and high velocities, where the drag force depends on the square of the velocity:
The drag force on an object moving at a velocity v relative to a fluid moving with a velocity w is:
FD |
−kv (v − w) at small velocities |
(7.26) |
−D |v − w| (v − w) at high velocities |
The constant kv depends on the object’s size, shape and surface, as well as on the (dynamic) viscosity of the fluid. For a sphere Stokes found that
kv = 6π R η |
(7.27) |
where R is the radius of the sphere, and η is the viscosity of the fluid. The viscosity of air is η = 1.82 × 10−5 Nsm−2 and for water it is η = 1.00 × 10−3 Nsm−2, both at room temperature.
Experimental data indicates an approximative value for D for a spherical object:
D 12.0 ρ R2 . |
(7.28) |
where v = |v|, ρ is the density of the fluid, and R is the radius of the sphere.
Versatility of the viscous force model: The viscous force model
FD = −kvv , |
(7.29) |
is much more versatile than suggested by its application to fluid drag forces. It is often used as a general damping term—a term reducing relative motion that also introduces dissipation and heat generation. For example, you will find that the viscous force model used to model damping of vibrations in solid object, to model the damping of vibrations in macroscopic objects and macroscopic springs, and to model surface forces in nano-scale surface contact. The viscous force model is a first order model to study any velocity-dependent force that tends to reduce velocity differences.