PHYSICS. Chapter 12 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.

PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 12 Lecture RANDALL D. KNIGHT

Chapter 12 Rotation of a Rigid Body IN THIS CHAPTER, you will learn to understand and apply the physics of rotation. Slide 12-2

Chapter 12 Preview Slide 12-3

Chapter 12 Preview Slide 12-4

Chapter 12 Preview Slide 12-5

Chapter 12 Preview Slide 12-6

Chapter 12 Preview Slide 12-7

Chapter 12 Preview Slide 12-8

Chapter 12 Reading Questions Slide 12-9

Reading Question 12.1 When floating in space with no external forces acting on it, an object tends to rotate about its A. Axle. B. Center of mass. C. Edge. D. Geometrical center. E. Pivot point. Slide 12-10

Reading Question 12.2 A rotating object has some rotational kinetic energy. If its angular speed is doubled, but nothing else changes, the rotational kinetic energy A. Increases by a factor of 4. B. Doubles. C. Does not change. D. Halves. E. Decreases by a factor of 4. Slide 12-12

Reading Question 12.3 A single particle has a mass, m, and it is at a distance, r, away from the origin. The moment of inertia of this particle about the origin is A. mr B. m 2 r C. m 2 r 2 D. m 2 r 4 E. mr 2 Slide 12-14

Reading Question 12.4 When discussing torque, the line of action is A. The line connecting the pivot with the point where the force acts. B. The line along which the force acts. C. The line that passes through the center of mass. D. The line along which motion occurs. E. The axis around which the object rotates. Slide 12-16

Reading Question 12.5 A new way of multiplying two vectors is introduced in this chapter. What is it called? A. The dot product. B. The scalar product. C. The tensor product. D. The cross product. E. The angular product. Slide 12-18

Reading Question 12.6 Moment of inertia is A. The rotational equivalent of mass. B. The point at which all forces appear to act. C. The time at which inertia occurs. D. An alternative term for moment arm. Slide 12-20

Reading Question 12.7 A rigid body is in equilibrium if A. B. τ net = 0 C. Neither A nor B D. Either A or B E. Both A and B Slide 12-22

Reading Question 12.7 A rigid body is in equilibrium if A. B. τ net = 0 C. Neither A nor B D. Either A or B E. Both A and B Slide 12-23

Chapter 12 Content, Examples, and QuickCheck Questions Slide 12-24

The Rigid-Body Model Slide 12-25

Three Basic Types of Motion of a Rigid Body Slide 12-26

Rotational Motion Review Recall that angular velocity is If the rotation is speeding up or slowing down, its angular acceleration is All points on a rotating rigid body have the same ω and the same α. Slide 12-27

Rotational Motion Review Rotational kinematics for constant angular acceleration The signs of angular velocity and angular acceleration. Slide 12-28

QuickCheck 12.1 Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The angular velocity of A is twice that of B. B. The angular velocity of A equals that of B. C. The angular velocity of A is half that of B. Slide 12-29

QuickCheck 12.2 The fan blade is speeding up. What are the signs of ω and α? A. ω is positive and α is positive. B. ω is positive and α is negative. C. ω is negative and α is positive. D. ω is negative and α is negative. Slide 12-31

Rotation About the Center of Mass An unconstrained object (i.e., one not on an axle) on which there is no net force rotates about a point called the center of mass. The center of mass remains motionless while every other point in the object undergoes circular motion around it. Slide 12-33

Finding the Center of Mass Consider an object made of particles. Particle i has mass m i. The center-of-mass is at Calculating center of mass is much like calculating your grade-point average. Particles of higher mass count more than particles of lower mass. Slide 12-34

Example 12.1 The Center of Mass of a Barbell Slide 12-35

Example 12.1 The Center of Mass of a Barbell Slide 12-36

Example 12.1 The Center of Mass of a Barbell Slide 12-37

Example 12.1 The Center of Mass of a Barbell Slide 12-38

Finding the Center of Mass by Integration Divide a solid object into many small cells of mass Δm. As Δm 0 and is replaced by dm, the sums become Before these can be integrated: dm must be replaced by expressions using dx and dy. Integration limits must be established. Slide 12-39

Example 12.2 The Center of Mass of a Rod Slide 12-40

Example 12.2 The Center of Mass of a Rod Slide 12-41

Example 12.2 The Center of Mass of a Rod Slide 12-42

QuickCheck 12.3 A baseball bat is cut in half at its center of mass. Which end is heavier? A. The handle end (left end) B. The hitting end (right end) C. The two ends weigh the same. Slide 12-43

QuickCheck 12.3 A baseball bat is cut in half at its center of mass. Which end is heavier? A. The handle end (left end) B. The hitting end (right end) C. The two ends weigh the same. Slide 12-44

Rotational Energy A rotating object has kinetic energy because all particles in the object are in motion. The kinetic energy due to rotation is called rotational kinetic energy. Adding up the individual kinetic energies, and using v i = r i ω: Slide 12-45

Rotational Energy Define the object s moment of inertia: Then the rotational kinetic energy is simply The units of moment of inertia are kg m 2. Moment of inertia depends on the axis of rotation. Mass farther from the rotation axis contributes more to the moment of inertia than mass nearer the axis. This is not a new form of energy, merely the familiar kinetic energy of motion written in a new way. Slide 12-46

Calculating the Moment of Inertia As we did for center of mass, divide a solid object into many small cells of mass Δm and let Δ m 0. The moment of inertia sum becomes where r is the distance from the rotation axis. The procedure is much like calculating the center of mass. One rarely needs to do this integral because moments of inertia of common shapes are tabulated. Slide 12-47

Moments of Inertia Slide 12-48

Consequences of Moment of Inertia Easier to spin up Harder to spin up Slide 12-49

QuickCheck 12.4 Which dumbbell has the larger moment of inertia about the midpoint of the rod? The connecting rod is massless. A. Dumbbell A. B. Dumbbell B. C. Their moments of inertia are the same. Slide 12-50

The Parallel-Axis Theorem You do sometimes need to know the moment of inertia about an axis in an unusual position. You can find it if you know the moment of inertia about a parallel axis through the center of mass. Slide 12-52

Example 12.4 The Speed of a Rotating Rod Slide 12-53

Example 12.4 The Speed of a Rotating Rod Slide 12-54

Example 12.4 The Speed of a Rotating Rod Slide 12-55

Example 12.4 The Speed of a Rotating Rod Slide 12-56

QuickCheck 12.5 The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3 Slide 12-57

QuickCheck 12.5 The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3 Your intuition likely led you to choose F 1. The reason is that F 1 exerts the largest torque about the hinge. Slide 12-58

Torque Torque measures the effectiveness of the force at causing an object to rotate about a pivot. Torque is the rotational equivalent of force. On a bicycle, your foot exerts a torque that rotates the crank. Slide 12-59

Torque The effectiveness of a force at causing a rotation is called torque. Torque is the rotational equivalent of force. We say that a torque is exerted about the pivot point. Slide 12-60

Torque Mathematically, we define torque τ (Greek tau) as SI units of torque are N m. English units are footpounds. The ability of a force to cause a rotation depends on 1. the magnitude F of the force. 2. the distance r from the point of application to the pivot. 3. the angle at which the force is applied. Slide 12-61

Torque Torque has a sign. Slide 12-62

Moment Arm In practice, torque is often calculated using the moment arm or lever arm. The torque is τ = df. But this is only the absolute value. You have to provide the sign of τ, based on which direction the object would rotate. What is the sign of τ here? Slide 12-63

QuickCheck 12.6 Which third force on the wheel, applied at point P, will make the net torque zero? Slide 12-64

QuickCheck 12.6 Which third force on the wheel, applied at point P, will make the net torque zero? Slide 12-65

Example 12.8 Applying a Torque Slide 12-66

Example 12.8 Applying a Torque Slide 12-67

Example 12.8 Applying a Torque Slide 12-68

Example 12.8 Applying a Torque Slide 12-69

Net Torque The figure shows the forces acting on the crankset of a bicycle. axle exerts a force on the axle to balance the other forces and keep. The net torque about the axle is the sum of the torques due to the applied forces: Slide 12-70

Gravitational Torque The torque due to gravity is found by treating the object as if all its mass is concentrated at the center of mass. Slide 12-71

Rotational Dynamics What does a torque do? For linear motion, a net force causes an object to accelerate. For rotation, a net torque causes an object to have angular acceleration. In the absence of a net torque (τ net = 0), the object either does not rotate (ω = 0) or rotates with constant angular velocity (ω = constant). Slide 12-72

Analogies Between Linear and Rotational Dynamics Slide 12-73

QuickCheck 12.7 A student gives a quick push to a puck that can rotate in a horizontal circle on a frictionless table. After the push has ended, the puck s angular speed A. Steadily increases. B. Increases for awhile, then holds steady. C. Holds steady. D. Decreases for awhile, then holds steady. E. Steadily decreases. Slide 12-74

Problem-Solving Strategy: Rotational Dynamics Problems Slide 12-76

Example 12.11 Starting an Airplane Engine Slide 12-77

Example 12.11 Starting an Airplane Engine Slide 12-78

Example 12.11 Starting an Airplane Engine Slide 12-79

Constraints Due to Ropes and Pulleys A rope passes over a pulley and is connected to an object in linear motion. The rope does not slip as the pulley rotates. Tangential velocity and acceleration of the rim of the pulley must match the motion of the object: Slide 12-80

The Constant-Torque Model Slide 12-81

Static Equilibrium A rigid body is in static equilibrium if there is no net force and no net torque. An important branch of engineering called statics analyzes buildings, dams, bridges, and other structures in total static equilibrium. For a rigid body in total equilibrium, there is no net torque about any point. Slide 12-82

The Static Equilbrium Model Slide 12-83

Example 12.15 Will the Ladder Slip? Slide 12-84

Example 12.15 Will the Ladder Slip? Slide 12-85

Example 12.15 Will the Ladder Slip? Slide 12-86

Example 12.15 Will the Ladder Slip? Slide 12-87

Balance and Stability Stability depends on the position of the center of mass. Slide 12-88

Balance and Stability This dancer balances en pointe by having her center of mass directly over her toes, her base of support. Slide 12-89

QuickCheck 12.8 Which object is in static equilibrium? Slide 12-90

QuickCheck 12.8 Which object is in static equilibrium? Slide 12-91

QuickCheck 12.9 What does the scale read? A. 500 N B. 1000 N C. 2000 N D. 4000 N Answering this requires reasoning, not calculating. Slide 12-92

QuickCheck 12.9 What does the scale read? A. 500 N B. 1000 N C. 2000 N D. 4000 N Slide 12-93

Rolling Without Slipping Rolling is a combination of rotation and translation. For an object that rolls without slipping, the translation of the center of mass is related to the angular velocity by Slide 12-94

Rolling Without Slipping Taking the time derivative of position of a particle i in a rolling object: The velocity of the particle is the velocity of the center of mass of the whole object plus the velocity of particle i relative to the center of mass. Slide 12-95

Rolling Without Slipping The figure below shows how the velocity vectors at the top, center, and bottom of a rolling wheel are found. v top = 2v cm v bottom = 0 The point on the bottom of a rolling object is instantaneously at rest. Slide 12-96

QuickCheck 12.10 A wheel rolls without slipping. Which is the correct velocity vector for point P on the wheel? Slide 12-97

QuickCheck 12.10 A wheel rolls without slipping. Which is the correct velocity vector for point P on the wheel? Slide 12-98

Kinetic Energy of Rolling The kinetic energy of a rolling object is In other words, the rolling motion of a rigid body can be described as a translation of the center of mass (with kinetic energy K cm ) plus a rotation about the center of mass (with kinetic energy K rot ). Slide 12-99

The Vector Description of Rotational Motion One-dimensional motion uses a scalar velocity v and force F. A more general understanding of motion requires vectors and. Similarly, a more general description of rotational motion requires us to replace the scalars ω and τ with the vector quantities and. Doing so will lead us to the concept of angular momentum. Slide 12-100

The Angular Velocity Vector The magnitude of the angular velocity vector is ω. The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated. Slide 12-101

The Cross Product of Two Vectors The dot product is one way to multiply two vectors, giving a scalar. A different way to multiple two vectors, giving a vector, is called the cross product. If vectors and have angle α between them, their cross product is the vector: Slide 12-102

The Right-Hand Rule The cross product is perpendicular to the plane of and. The right-hand rule for the direction comes in several forms. Try them all to see which works best for you. Note that. Instead,. Slide 12-103

Example 12.16 Calculating a Cross Product Slide 12-104

The Torque Vector We earlier defined torque τ = rfsinϕ. r and F are the magnitudes of vectors, so this is a really a cross product: A tire wrench exerts a torque on the lug nuts. Slide 12-105

Angular Momentum of a Particle A particle of mass m is moving. The particle s momentum vector makes an angle β with the position vector. We define the particle s angular momentum vector relative to the origin to be Slide 12-106

Angular Momentum of a Particle Why this definition? If you take the time derivative of and use the definition of the torque vector (see book for details), you find Torque causes a particle s angular momentum to change. This is the rotational equivalent of and is a general statement of Newton s second law for rotation. Slide 12-107

Angular Momentum of a Rigid Body For a rigid body, we can add the angular momenta of all the particles forming the object. If the object rotates on a fixed axle, or about an axis of symmetry then it can be shown that And it s still the case that Slide 12-108

Analogies Between Linear and Angular Momentum and Energy Slide 12-109

Conservation of Angular Momentum An isolated system that experiences no net torque has and thus the angular momentum vector constant: is a Slide 12-110

Conservation of Angular Momentum As an ice skater spins, external torque is small, so her angular momentum is almost constant. By drawing in her arms, the skater reduces her moment of inertia I. To conserve angular momentum, her angular speed ω must increase. Slide 12-111

Example 12.19 Two Interacting Disks Slide 12-112

Example 12.19 Two Interacting Disks Slide 12-113

Example 12.19 Two Interacting Disks Slide 12-114

Example 12.19 Two Interacting Disks Slide 12-115

Example 12.19 Two Interacting Disks Slide 12-116

QuickCheck 12.11 Two buckets spin around in a horizontal circle on frictionless bearings. Suddenly, it starts to rain. As a result, A. The buckets speed up because the potential energy of the rain is transformed into kinetic energy. B. The buckets continue to rotate at constant angular velocity because the rain is falling vertically while the buckets move in a horizontal plane. C. The buckets slow down because the angular momentum of the bucket + rain system is conserved. D. The buckets continue to rotate at constant angular velocity because the total mechanical energy of the bucket + rain system is conserved. E. None of the above. Slide 12-117

Advanced Topic: Precession of a Gyroscope Consider a horizontal gyroscope, with the disk spinning in a vertical plane, that is supported at only one end of its axle, as shown. You would expect it to simply fall over but it doesn t. Instead, the axle remains horizontal, parallel to the ground, while the entire gyroscope slowly rotates in a horizontal plane. This steady change in the orientation of the rotation axis is called precession, and we say that the gyroscope precesses about its point of support. The precession frequency Ω is much less that the disk s rotation frequency ω. Slide 12-119

Gravity on a Nonspinning Gyroscope Shown is a nonspinning gyroscope. When it is released, the net torque is entirely gravitational torque. Initially, the angular momentum is zero. Gravity acts to increase the angular momentum gradually in the direction of the torque, which is the -direction. This causes the gyroscope to rotate around x and fall. Slide 12-120

Gravity on a Spinning Gyroscope Shown is a gyroscope initially spinning around the z-axis. Initially, gravity acts to increase the angular momentum slightly in the direction of the torque, which is the -direction. This causes the gyroscopes angular momentum to shift slightly in the horizontal plane. The gravitational torque vector is always perpendicular to the axle, so dl is always perpendicular to L. Slide 12-121

Advanced Topic: Precession of a Gyroscope The precession frequency of a gyroscope, in rad/s, is Here M is the mass of the gyroscope, I is its moment of inertia, and d is the horizontal distance of the center of mass from the support point. The angular velocity of the spinning gyroscope is assumed to be much larger than the precession frequency: ω >> Ω Slide 12-122