- •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
450 |
14 Rotational Motion |
We call such a condition a kinematic condition. We will frequently use such conditions when we solve problems in mechanics.
For each of the wheels, the velocity at the point of contact is related to the angular velocity by:
v A = R A ωA , vB = R B ωB . |
(14.26) |
Since the velocities at the point of contact are equal and oppositely directed, we find that the angular velocities also are related:
v A = −vB |
ωB = − |
R A |
ωA . |
(14.27) |
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R B |
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This relation is general and does not requires the velocities to be constant. We can therefore find the angular accelerations by taking the time derivatives on each side:
αB = |
d |
ωB = − |
d |
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R A |
ωA = − |
R A |
αA . |
(14.28) |
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d t R B |
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d t |
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R B |
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14.6 Rotational Motion in Three Dimensions
You now know how to describe rotations in a plane: Rotations around the origin in the x y-plane are described by the rotation angle θ , the angular velocity ω, and the angular acceleration α. We found that a point, P , on the rotating object follows a circular path with a constant distance R from the rotation axis as illustrated in Fig. 14.8, and that the velocity of P is:
v = Rω , |
(14.29) |
directed along the tangent to the circle. Can we find a simple expression for both the direction and magnitude of the velocity? Yes! Since we know that the tangent
Fig. 14.8 An illustration of rotation around the z-axis. The direction of the velocity vector is given by v = ω × r
ω
ρ V
ω
γ R
R R
V
14.6 Rotational Motion in Three Dimensions |
451 |
is normal to the radius vector, r, and since it is in the x y-plane, the tangent is also normal to the unit vector in the z-direction. This allows us to write the velocity vector of the point P as:
v = ω k × r . |
(14.30) |
This expression provides the correct magnitude and direction of the velocity vector: It points in the positive rotational direction if ω is positive, and in the negative rotational direction if ω is negative. We can therefore introduce the angular velocity as a vector:
ω = ω k (angular velocity vector) |
(14.31) |
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The angular velocity vector, ω, points along the axis of the rotation. For rotation in the x y-plane, the rotation axis is the z-axis.
The expression in (14.30) is valid not only for rotation in the x y-plane, but for any rotation around the axis given by the ω vector. This is illustrated in the right part of Fig. 14.8, where the velocity of a point at r going in a circular orbit around the rotation axis is given by the radius of the circle, ρ, and the angular velocity, ω:
vθ = ω ρ . |
(14.32) |
We see from the geometry that the radius ρ is r sin(γ ), where γ is the angle between the position vector, r and the angular velocity vector, ω. Therefore
vθ = ω r sin(γ ) . |
(14.33) |
The direction of the velocity is tangential to the circular orbit: orthogonal to the position vector, r, and to the angular velocity vector, ω. This means that we can write the velocity as
v = ω × r , |
(14.34) |
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which gives both the correct magnitude: |v| = ωr sin(γ ), and the correct direction. However, you can only use this expression when the origin of the position vector is on the rotation axis.
Let us use (14.34) to find the acceleration of the point P , which moves in a circular orbit with constant radius ρ = R sin(γ ) around the rotation axis given by the angular velocity vector ω. The acceleration vector is the time derivative of the velocity vector:
a = |
d v |
= |
d |
ω × r = |
d ω |
× r + ω × |
d r |
, |
(14.35) |
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d t d t |
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d t |
d t |
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but we know that: |
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d r |
= v = ω × r , |
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(14.36) |
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d t |
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452 |
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14 Rotational Motion |
and we introduce the angular acceleration vector, α as: |
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α = |
d ω |
, |
(14.37) |
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d t |
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which inserted back into (14.35) gives: |
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a = α × r + ω × (ω × r) . |
(14.38) |
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For a motion with constant angular velocity, that is with constant speed, the angular acceleration vector is zero: α = 0, and the acceleration vector is:
a = ω × (ω × r) . |
(14.39) |
This is the sentripetal acceleration on vector form.
•The direction of the vector is correct. If you use the right-hand rule to find the vector product, you realize the vector points orthogonally inwards towards the rotation axis.
•The magnitude of the acceleration is |a| = a = ω2ρ = (v/ρ)2 ρ = v2/ρ, which is the magnitude of the sentripetal acceleration we found in (14.35).
14.6.1Example: Velocity and Acceleration of a Conical Pendulum
Problem: A conical pendulum consists of a mass in a string of length L . The mass rotates with angular velocity ω in a circular orbit with radius R, as illustrated in Fig. 14.9. Find the velocity and acceleration of the mass.
Solution: The vector velocity of the conical pendulum is given as v = ω × r. We find the velocity when the pendulum crosses the x -axis. In this case, the position vector is
Fig. 14.9 Illustration of a |
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conical pendulum |
ω |
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L
R R V
14.6 Rotational Motion in Three Dimensions |
453 |
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r = R i + |
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k . |
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L 2 − R2 |
(14.40) |
The rotation is about the z-axis. We therefore introduce the angular velocity vector as ω = ω k.
We find the vector velocity: |
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v = ω × r = ω k × R i + |
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k |
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L 2 − R2 |
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= ω R (k × i) = ω R j . |
(14.41) |
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Similarly, we find the acceleration vector: |
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a = α × r + ω × (ω × r) = 0 + ω k × (ω R j) = −ω2 R i . |
(14.42) |
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The acceleration vector points in towards the axis of rotation, as expected. |
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Summary
Description of rotation:
•The rotation of an object is described by the angle θ (t )
•The angular velocity of the object is: ω(t ) = d θ /d t
•The angular acceleration of the object is: α(t ) = d ω/d t = d 2/d t 2.
Rotation and translation:
•A point on the rotating body at a distance R from the rotational axis has a tangential velocity: v = R ω.
Solving rotational motion:
•We solve problems in rotations using the same structured approach as for linear motion.
•In the “Solver” we solve the equation: d 2θ /d t 2 = α(t , θ , d θ /d t ) with the initial conditions θ (t0) = θ0 and ω (t0) = ω0.
Numerical solution:
•Numerically, we solve the equation using an iterative approach starting from the initial conditions. For example, we can use Euler-Cromer’s method:
ω (ti + |
t ) = ω(ti ) + t α(θ (ti ), ω (ti ), ti ) |
θ (ti + |
t ) = θ (ti ) + t ω(ti + t ) |
454 |
14 Rotational Motion |
Analytical solution:
•When the angular acceleration, α = α(t ), is only a function of time, t , we can solve the equations by direct integration:
ω(t ) = ω(t0) + t |
α(t ) d t , θ (t ) = θ (t0) + t |
ω (t ) d t , |
t0 |
t0 |
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A typical example is motion with constant angular acceleration.
•For motion with constant angular velocity the solution is:
θ(t ) = θ (t0) + ω (t − t0) .
•For motion with constant angular acceleration the solution is:
ω(t ) = ω(t0) + α (t − t0)
θ (t ) = θ (t0) + ω(t0) (t − t0) + 1 α (t − t0)2 .
2
Rotational motion in three dimensions:
•Generally, rotations occur around an axis, given by the angular velocity vector, ω.
•The velocity of a point on the rotating object at position r is: v = ω × r
•The acceleration of a point on the rotating object at position r is: a = α × r + ω ×
(ω × r)
Exercises
Discussion Questions
14.1Flywheel. What is the acceleration of a point at a rim of a rotating flywheel when the flywheel is rotating at a constant rate and when the flywheel is speeding up?
14.2Spin cycle. Explain the working of the spin cycle of a washing machine in terms of acceleration components.
14.3Degrees and radians. Why do we use radians and not degrees to describe angles in rotational motion?
14.6 Rotational Motion in Three Dimensions |
455 |
Problems
14.4Flywheel position. The angular position of a flywheel on an engine is given as θ = c1 (t /t1) + c2 (t /t2)2, where c1, and c2 are dimensionless constants, and t1 and t2 are characteristic times.
(a) Find an expression for the flywheel’s angular velocity.
(b) Find an expression for the flywheel’s angular acceleration.
14.5Unbalanced wheel. The angular position of an unbalanced wheel is given as
θ= 5.0 rad sin (t /(2 s)).
(a) Find an expression for the wheel’s angular velocity.
(b) Find an expression for the wheel’s angular acceleration.
14.6Earth and Sun. (a) What is the angular velocity of the Earth in its orbit around the Sun?
(b) What is the angular velocity of the Earth in its rotation about its own axis?
14.7Engine. A car engine accelerates from 1000 to 2000 rpm at a constant rate during 15 s.
(a) Find the angular acceleration of the engine.
(b) Find the number of rotations the engine revolves from it starts at 1000 rpm until it has accelerated to 2000 rpm.
14.8Spinning down. You start a spinning wheel by rotating it with an angular acceleration of 10 rad/s2 for 3 s.
(a) What is the angular velocity of the wheel as a function of time for the first 3 s.
(b) Find the angle of the wheel as a function of time for the first 3 s.
After releasing the spinning wheel, it slows down at a constant rate of 0.1 rad/s2. (c) What is the angular velocity of the wheel as a function of time after the first 3 s. (d) Find the angle of the wheel as a function of time after the first 3s.
(e) How long time does the wheel take to stop?
(f ) How much longer would the wheel take to stop if you spun it for 6 seconds instead?
14.9A slippery wheel. You are testing the behavior of a car wheel in a river bed.
The angular acceleration of the wheel spinning semi-saturated in water is α = −kω ω, where kω = 0.1 s−1 for the wheel you are testing.
(a) The wheel starts with the angular velocity ω0 = 10 rad/s when you put the wheel into the water. Find the angular velocity of the wheel as a function of time.
(b) How long time does it take until the wheel has 1/10th of its initial angular velocity?
14.10Running the curve. A sprinter is running through a circular curve with radius 50 m with a constant speed of 10 m/s.
(a) What is the angular velocity of the sprinter?
(b) What is the angular acceleration of the sprinter? (c) What is the linear acceleration of the sprinter?
456 14 Rotational Motion
14.11 Rotating Earth. The radius of the Earth is approximately 6378 km. (a) What is the angular velocity of the Earth as it rotates around its own axis? (b) What is the angular velocity of a person on the equator?
(c) What is the linear velocity of a person on the equator? (d) What is the angular velocity of a person at 60◦ North? (e) What is the linear velocity of a person at 60◦ North?
(f ) What is the angular acceleration of a person on the equator and at 60◦ North? (g) What is the linear acceleration of a person on the equator? Compare with g = 9.8 m/s2 and comment.
(h) What is the linear acceleration of a person at 60◦ North? (Find both the magnitude and direction of the acceleration.) Comment on the results.
14.12 Rolling wheel. A wheel of radius R is rolling without slipping along a flat surface. The center of the wheel is moving with the constant horizontal velocity v. (a) Show that for the wheel not to slip, the angular velocity of the wheel must be
ω = v/ R.
(b) What is the velocity of the point on the wheel that is in contact with the ground? (Measured in a coordinate system on the ground.)
(c) What is the velocity of a point on the top of the wheel?
(d) What is the acceleration of the point on the wheel that is in contact with the ground? (Relative to the ground).
(e) What is the acceleration of the point on the top of the wheel?