AP Physics C Mechanics

Transcription

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2 AP Physics C Mechanics Rotational Motion

3 Table of Contents Click on the topic to go to that section Rotational Kinematics Review Rotational Dynamics Rotational Kinetic Energy Angular Momentum 3

4 Rotational Kinematics Review Return to Table of Contents 4

5 What is Rotational Motion? Linear Kinematics and Dynamics dealt with the motion of "point particles" moving in straight lines. Life is more complex than that. Objects have an underlying structure, so it cannot be assumed that a force applied anywhere on the object results in the same kind of motion. In football (American and the rest of the world), if a player is struck from the front, at the ankle, he will fall forward, pivoting about his foot. The same force applied in his midsection will push him backwards. The player is rotating about his ankle in the first case, and translating (linear) backwards in the second case. 5

6 Axis of Rotation In the first case, the player was rotated about his axis of rotation a line parallel to the ground and through his ankle, perpendicular to the force. Assume that all the objects that we have been dealing with are rigid bodies, that is, they keep their shape and are not deformed in any way by their motion. Here's a sphere rotating about its axis of rotation the vertical red line. Does the axis of rotation have to be part of the rigid body? "Rotating Sphere". Licensed under Creative Commons Attribution Share Alike 3.0 via Wikimedia Commons File:Rotating_Sphere.gif#mediaviewer/File:Rotating_Sphere.gif 6

7 Axis of Rotation No if you were to spin this donut around its center, the axis of rotation would be in the donut hole, pointing out of the page. "Chocolate dip, " by Pbj2199 Own work. Licensed under Creative Commons Attribution Share Alike 3.0 via Wikimedia Commons jpg#mediaviewer/File:Chocolate_dip, jpg 7

8 Rotational Equivalents to Linear (Translational) Motion Linear motion is described by displacement, velocity and acceleration. Rotational motion is described by angular displacement, angular velocity and angular acceleration. It is also important to define the axis of rotation objects will have different properties depending on this choice. Once the choice is made, it must remain the same to describe the motion. If the axis is changed, then a different rotational motion occurs. 8

9 Angular Displacement There are several theories why a circle has 360 degrees. Here's a few if you're interested, there's a lot of information that can be found on the web. Some people believe it stems from the ancient Babylonians who had a number system based on 60 instead of our base 10 system. Others track it back to the Persian or biblical Hebrew calendars of 12 months of 30 days each. But it has nothing to do with the actual geometry of the circle. There is a more natural unit the radian, which is the rotational equivalent of displacement (Δs). 9

10 Angular Displacement As a rigid disc (circle) rotates about its center, the angle of rotation, θ, is defined as the ratio of the subtended arc length, s, to the radius of the circle, and is named angular displacement. B A Rotational angles are now defined in geometric terms, not based on an old calendar or arbitrary numbering system. 10

11 Radian Angular displacement is unit less since it is the ratio of two distances. But, we will say that angular displacement is measured in radians (abbreviated "rad"). A point moving around the entire circumference will travel 2πr as an angle of is swept through. Using the angular displacement definition: B A 11

12 Angular Displacement Something interesting look at the three concentric circles drawn on the rigid disc the radii, arc lengths s r and the points A r and B r. B 1 B 2 B 3 As the disk rotates, each point A moves to point B, covering the SAME angle θ, but covering a different arc length s r. A 1 A 2 A 3 Thus, all points on a rotating rigid disc move through the same angle, θ, but different arc lengths s, and different radii, r. 12

13 1 What is the angular displacement for an arc length (s) that is equal to the radius of the circular rigid body? A 0.5 rad B 1.0 rad C 0.5π rad D 1.0π rad Answer E 1.5π rad 13

14 2 A record spins 4 times around its center. It makes 4 revolutions. How many radians did it pass? A π rad B 2π rad C 4π rad D 8π rad 14

15 3 A record spins 4 times around its center. It makes 4 revolutions. How many radians did it pass? A π rad B 2π rad C 4π rad Answer D 6π rad E 8π rad 15

16 4 A circular hoop of radius 0.86 m rotates π/3 rad about its center. A small bug is on the hoop what distance does it travel (arc length) during this rotation? A 0.30 m B 0.90 m C 1.4 m Answer D 2.7 m E 3.2 m 16

17 5 There are two people on a merry go round. Person A is 2.3 m from the axis of rotation. Person B is 3.4 m from the axis of rotation. The merry go round moves through an angular displacement of π/4. What linear displacement (arc length) is covered by both people? Compare and contrast these motions. Answer 17

18 Other Angular Quantities Angular displacement has now been be related to Linear displacement. Next, let's relate angular velocity and angular acceleration to their linear equivalents: 18

19 Other Angular Quantities Define average Angular Velocity (ω avg ) as the change in angular displacement over time, and take the limit as Δt approaches zero for the instantaneous angular velocity (ω). Similarly for average Angular Acceleration (α) and instantaneous angular acceleration (α). 19

20 Other Angular Quantities Since ω and α are both related to θ, and θ is the same for all points on a rotating rigid body, ω and α are also the same for all points on a rotating rigid body. But the same is not true for linear velocity and acceleration. Let's relate the linear velocity and acceleration to their angular equivalents. 20

21 Angular and Linear Velocity Start with angular velocity, and substitute in the linear displacement for the angular displacement. r is constant, so move it outside the derivative This confirms what you feel on a merry go round the further away from the center you move, the faster you feel you're going you're feeling the linear velocity! 21

22 Angular and Linear Acceleration Start with angular acceleration, and substitute in the linear velocity for the angular velocity. r is constant, so move it outside the derivative 22

23 Angular and Linear Quantities Summary Angular Linear Relationship Displacement Velocity Acceleration 23

24 Angular Velocity sign We're familiar with how to assign positive and negative values to displacement. Typically if you move to the right or up, we give that a positive displacement (of course, this is arbitrary, but it's a pretty standard convention). How do we assign a "sign" for angular displacement, velocity or acceleration? Does "right" or "up" work with rotational motion? Does it work with a circle? B A 24

25 Angular Velocity sign Not really. Not at all. From your math classes, you know that the horizontal axis through a circle is labeled as 0 0, and angles are measured in a counter clockwise direction. Once we agree on that, let's look at the definition of angular velocity: B A 25

26 Angular Velocity sign As the disc rotates in a counter clockwise fashion, θ increases, so Δθ is positive. Since dt is also positive, dθ/dt must be positive. Thus, Counter clockwise rotations result in a positive ω. B A 26

27 6 Explain why a disc that rotates clockwise leads to a negative value for angular velocity, ω. You can use an example to show your point. Answer 27

28 Angular Acceleration sign Start with the definition of angular acceleration: There are four cases here, all similar to the case of linear acceleration. Can you figure them out? Here's a hint for linear acceleration, if an object's velocity is increasing in the same direction of its displacement, the acceleration is positive. If the object's velocity is decreasing in the direction of its displacement, then its acceleration is negative. 28

29 Angular Acceleration sign Here they are: ω increases in counter clockwise direction α is positive ω decreases in the counter clockwise direction α is negative ω increases in clockwise direction α is negative ω decreases in the clockwise direction α is positive These conventions are arbitrary, and in this case are based on the axis of rotation, but once chosen, they need to stay consistent for solving problems. There will be problems where the choice is not so obvious, but just make a choice and stay consistent. 29

30 7 What is the angular acceleration of a ball that starts at rest and increases its angular velocity uniformly to 5 rad/s in 10 s? A 0.5 rad/s 2 B 1.0 rad/s 2 C 1.5 rad/s 2 Answer D 2.0 rad/s 2 E 2.5 rad/s 2 30

31 Three types of Acceleration So, we have two types of displacement, velocity and acceleration linear and rotational. There is a third type of acceleration centripetal. This is the acceleration that a point on a rotating disc experiences when it moves in a circular direction. The v in the above equation is linear velocity, and for an object moving in a circle, it is the velocity tangent to the circle. 31

32 Three types of Acceleration Look at the disc below which shows the linear acceleration that we've defined as well as the centripetal acceleration. But, centripetal acceleration is also linear! So we will rename the linear acceleration we've been dealing with in this chapter as tangential acceleration. a c a T α The angular acceleration, α, is a vector that points out of the page. 32

33 Three types of Acceleration We now have three accelerations associated with a rigid rotating body: tangential, angular and centripetal, or a T, α, and a C. They are related to the angular velocity, ω, as follows: 33

34 8 At which point is the magnitude of the centripetal acceleration the greatest? A B B C A D C E D E 34

35 Total Linear Acceleration There are two flavors of linear acceleration tangential and centripetal. They are perpendicular to each other; using vector addition and the Pythagorean Theorem, the total linear acceleration is found: atotal a T a c 35

36 9 A child pushes, with a constant force, a merry go round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry go round's tangential acceleration? A 0.25 m/s 2 B 0.35 m/s 2 C 0.62 m/s 2 D 0.75 m/s 2 Answer E 0.94 m/s 2 36

37 10 A child pushes, with a constant force, a merry go round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry go round's centripetal acceleration? A 7.5 m/s 2 B 19 m/s 2 C 23 m/s 2 D 38 m/s 2 Answer E 46 m/s 2 37

38 11 A child pushes, with a constant force, a merry go round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry go round's total linear acceleration? A 5.6 m/s 2 B 12 m/s 2 C 23 m/s 2 D 38 m/s 2 Answer E 46 m/s 2 38

39 Angular Velocity and Frequency Frequency, f, is defined as the number of revolutions an object makes per second and f = 1/T, where T is the period. A rotating object moves through an angular displacement, θ = 2πr rad, in one revolution. The linear velocity of any point on a rotating object is given by the following equation. THESE ARE VERY VALUABLE EQUATIONS for solving rotational motion problems. 39

40 12 A mining truck tire with a radius of 2.1 m rolls with an angular velocity of 8.0 rad/s. What is the frequency of the tire's revolutions? What is the period? A 1.3 rev/s B 1.3 rev/s C 2.6 rev/s D 2.6 rev/s E 3.9 rev/s 0.40 s 0.77 s 0.39 s 0.78 s 0.78 s Answer 40

41 Translational Kinematics Now that all the definitions for rotational quantities have been covered, we're ready to discuss the motion of rotating rigid objects Rotational Kinematics. First, let's review the equations for translational kinematics this was one of the first topics that was covered in this course. *Important* These equations are ONLY valid for cases involving a constant acceleration. This allows gravitational problems to be solved (a = g, a constant) and makes it possible to solve kinematics equations without advanced calculus. When acceleration changes, the problems are more complex. 41

42 Translational Kinematics Here are the equations. If you would like, please review their derivations in the Kinematics section of this course. 42

43 Rotational Kinematics For each of the translational kinematics equations, there is an equivalent rotational equation. We'll derive one of them and just present the others basically each translational variable is replaced with its rotational analog. 43

44 Rotational Kinematics Here they are. We've been using Δs for linear displacement, but Δx is just fine. Similar to the translational equations, these ONLY work for constant angular acceleration. 44

45 Rolling Without Slipping Rolling Motion is a practical application of rotational kinematics. The best example is a tire rotating as the car moves. Another example is a rotating pulley (Atwood's machine). The key in both examples is there is no slipping either the tire on the road or the string on the pulley. In these cases, the equations derived relating angular velocity and acceleration to linear velocity and acceleration hold. 45

46 Rolling Without Slipping Here's a rotating circle that represents a car tire moving down the road. d A A A The tire makes one complete revolution, so point A on the tire moves a total arc length of s = 2πr. If the tire does not slip on the road, then the linear distance moved on the road during the full revolution, d, is equal to s. 46

47 Rolling Without Slipping d A A The tangential velocity of point A is s/t while the linear velocity of the wheel is d/t. Recognize s/t as v T = rω. That gives us v linear = v T = rω. By similar reasoning, we have a linear = rα. These equations will come in very handy when solving rolling motion problems. A 47

48 Rolling Without Slipping d A A CAUTION. If the object is rolling with slipping then these two equations cannot be used. If a car tire is spinning on ice, and the car is not moving, then v linear = 0, and ω can be very large. A Use only for rolling motion without slipping. 48

49 Kinematics Equations & Rolling Motion The rotational motion equations also work as part of the equations for rolling motion. Rolling motion is simply rotational motion and linear motion combined. Think of a wheel on a car: It is rotating, showing rotational motion. It also moves forward as it rotates, therefore showing linear motion. Thus, we use the translational and rotational kinematics equations to solve rolling motion problems. 49

50 13 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s 2 for 11 s. What is its final angular velocity? A 20 rad/s B 30 rad/s C 40 rad/s D 50 rad/s Answer E 60 rad/s 50

51 14 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s 2 for 11 s. What is its final linear velocity? A 25 m/s B 20 m/s C 15 m/s D 10 m/s Answer E 5 m/s 51

52 15 A bicycle wheel with a radius of m starts from rest and accelerates at a rate of 4.50 rad/s 2 for 11.0 s. What is its angular displacement? A 195 rad B 200 rad C 230 rad D 251 rad Answer E 273 rad 52

53 16 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s 2 for 11 s. How many revolutions did it make? A 50 rev B 45 rev C 43 rev D 41 rev Answer E 39 rev 53

54 17 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s 2 for 11 s. What is its linear displacement? A 40 m B 82 m C 160 m D 180 m Answer E 210 m 54

55 18 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular acceleration? A π rad/s 2 B π/2 rad/s 2 C π/3 rad/s 2 Answer D π/4 rad/s 2 E π/5 rad/s 2 55

56 19 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular displacement? A 56π rad B 82π rad C 96π rad D 112π rad Answer E 164π rad 56

57 20 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What linear displacement, s, will a point on the outside edge of the wheel have traveled during that time? A 24π rad B 30π rad C 36π rad Answer D 42π rad E 48π m 57

58 Rotational Dynamics Return to Table of Contents 58

59 Rotational Dynamics Just like there are rotational analogs for Kinematics, there are rotational analogs for Dynamics. Kinematics allowed us to solve for the motion of objects, without caring why or how they moved. Dynamics showed how the application of forces causes motion and this is summed up in Newton's Three Laws. These laws also apply to rotational motion. The rotational analogs to Newton's Laws will be presented now. 59

60 Torque Forces underlie translational dynamics. A new term, related to force is the foundation of causing rotational motion. It is called Torque. The wrench is about to turn a nut by a force applied at point A. The nut is on a bolt that is attached to a wall.will the nut turn, and in which direction? Consider the axis of rotation to be in the middle of the nut, and pointed into the page. What if a force is applied at point B? %3AAction_meca_equivalence_force_couple_plan.svg 60

61 Torque A force at point A will turn the nut in a clockwise direction recall that "left" and "right" have no meaning in rotational motion. This is good a force causes motion from rest which implies an acceleration (but its an angular acceleration, so it's a little new). A force applied at point B will not rotate the nut. Assuming the forces were the same, why does the force at point B not result in motion? %3AAction_meca_equivalence_force_couple_plan.svg What is different? 61

62 Torque The distance between where the force was applied and the axis of rotation. When the force was applied at the axis of rotation, the nut did not move. There's just one more variable to consider for torque. Here's a picture of a door (the horizontal line) and its door hinge (the dot). Let's say you want to open the door by pushing it rotating it about the hinge. F 1 and F 2 have the same magnitude. F 2 F 1 Which force will cause the door to open quicker (have a greater angular acceleration)? 62

63 Torque You've probably known this since you first started walking F 1 gives the greater angular acceleration. Pushing perpendicular to a line connecting the force to the axis of rotation results in the greater angular acceleration. F 2 F 1 It will also allow a smaller force to have the same impact as a greater force applied at an angle different than

64 Torque This combination of Force applied at a distance and an angle from the axis of rotation that causes an angular accleration is called Torque, which is represented by the Greek letter tau: F 1 r F 2 θ To continue our sign convention, if the object rotates counter clockwise, the torque is positive. Negative values of torque cause clockwise rotations. Torque is a vector, so its direction must be specified along with its magnitude. 64

65 Torque The sin θ takes into account that the closer the force is applied perpendicular to the line connected to the axis of rotation results in a greater torque a greater angular acceleration. θ is the angle between the applied force and the line connecting it to the axis of rotation. F 1 r F 2 θ When you get to vector calculus, the equation becomes more elegant involving the cross product: 65

66 Torque The units of torque are Newton meters. Note how this is similar to work and energy and in those cases, we replace N m with Joules. Torque, however is a vector depending on its direction (positive or negative), it causes a different direction of motion. Work and energy are scalars. So, to emphasize this difference, torque is never expressed in terms of Joules. 66

67 21 Assume a force is applied perpendicular to the axis of rotation of the wrench nut system. At which point will the nut experience the greatest torque? A B B D C A Answer C D E All result in the same torque. %3AAction_meca_equivalence_force_couple_plan.svg 67

68 22 A force of 300 N is applied a crow bar 20 cm from the axis of rotation at an angle of 250 to a line connecting it to the axis of rotation. Calculate the torque. A 30 N m B 25 N m C 20 N m Answer D 15 N m E 10 N m 68

69 Torque Applications What if more than one more torque is applied to an object? Just as in linear dynamics, where multiple forces are added to find the linear acceleration of the object, multiple torques are added (in a vector fashion) to find the angular acceleration of the object as it rotates. Assign a positive value to the individual torque if it accelerates the object in the counter clockwise, and a negative value if it accelerates it in the clockwise direction. 69

70 Balance Here is a balance with two masses attached to it. Block A has a mass of 0.67 kg and is located 0.23 m from the pivot point. Block B has a mass of 1.2 kg. Where should it be located so that the balance beam is perfectly horizontal and is not moving? A r A r B B Is this a translational motion or a rotational motion problem? 70

71 See Saw (balance) Here's an artistic hint. The two boys on the left represent mass A and the two slightly larger boys on the right represent mass B. Winslow Homer, "The See Saw" 1873 A side note there are many connections between physics and art this would be a good project to research. 71

72 Balance This is a rotational motion problem, so in order to have the beam not rotate, the sum of the torques applied to it must equal zero (Newton's Second Law). A r A r B B Note the signs mass A is trying to rotate the beam counterclockwise so its torque is positive. Mass B is trying to rotate the beam clockwise negative torque. 72

73 Balance and See Saw Again, many elementary school students already know this! If you're on a see saw and you have to balance several students on the other side, you position yourself further from the fulcrum than they are. A r A r B B Winslow Homer, "The See Saw"

74 23 John and Sally are sitting the on the opposite sides of a See Saw, both of them 5.0 m distant from the fulcrum. Sally has a mass of 45 kg. What is the gravitational force on John if the See Saw is not moving? Use g = 10 m/s 2. A 450 N B 400 N C 350 N Answer D 300 N E 250 N 74

75 24 A student wants to balance two masses of 2.00 kg and kg on a meter stick. The kg mass is m from the fulcrum in the middle of the meter stick. How far away should she put the 2.00 kg mass from the fulcrum? A m B m C m Answer D m E m 75

76 25 Sally wants to lift a rock of 105 kg off of the ground. She has a 3.00 m long metal bar. She positions the bar on a fulcrum (actually, a thick tree log 1.00 m away from the rock) and wedges it under the rock. She can exert a force of 612 N with her arms. How far away from the fulcrum should she exert this force to move the rock? Use g = 10 m/s 2. A 0.17 m B 0.85 m C 1.7 m Answer D 2.0 m E 2.7 m 76

77 The ladder against the wall For an object to be in equilibrium (not moving), there are two requirements that need to be satisfied. The sum of the torques on the object needs to be zero and the sum of the forces needs to be zero. This is a branch of engineering called statics as nothing is moving, which is a nice thing to happen if you're planning a bridge or building. Let's apply this to a classic physics problem a ladder leaning against a wall. 77

78 The ladder against the wall A ladder of length L and mass m, is leant against a wall. The coefficient of static friction between the ladder and the ground is 0.6 (μ sg ). The coefficient of static friction between the ladder and the wall is 0.3 (μ sw ). At what minimum angle, θ, from the ground should the ladder be placed so that it doesn't slip?. First step draw a coordinate system at the ladder/ground interface and sketch the forces acting on the ladder. L θ 78

79 The ladder against the wall y F Nw f wl f wl static friction force of wall on ladder F Nw normal force of wall on ladder mg mg gravitational force on ladder F Ng θ f gl x f gl static friction force of ground on ladder F Ng normal force of ground on ladder 79

80 The ladder against the wall F Ng y F Nw f wl reference point mg θ f gl x Choose a point about which the torques will be calculated. Normally, a point is chosen to minimize the amount of torques created by the forces. But since this problem is asking for θ, we'll choose the center of mass of the ladder. This will remove the mass from the torque equation. And, as it will turn out, we won't need the sum of the forces in the y direction to solve the problem. 80

81 The ladder against the wall The sum of the forces in the x and y direction equal zero. y F Nw f wl θ mg The sum of the torques about the center of mass equal zero. mg does not contribute a torque as it acts at the center of mass. F Ng θ f gl x 81

82 The ladder against the wall To find the minimum angle at which the ladder will not slip, the static friction forces will be at their maximum value: y F Nw f wl Use the forces in the x direction: F Ng mg θ f gl x 82

83 The ladder against the wall y F Nw f wl replacing the friction forces with the normal forces. F Ng mg And we get: replacing F Nw with F Ng θ x f gl 83

84 The ladder against the wall y F Nw f wl The angle is independent of the mass of the ladder much like a mass sliding down an incline. F Ng θ mg x f gl 84

85 Torque Torque is the rotational analog to Force. Newton's Second Law states that F = ma. Time to derive the angular version of this law. Newton's Second Law has been applied assuming it was dealing with point particles that don't rotate. We now have an extended rigid body, but let's start with the external force on one small piece of the rigid body as it's rotating: since a i = r i α, where r i is the distance between the small piece of the object and the axis of rotation 85

86 Torque Assume that the external force is applied perpendicular to the line connected to the axis of rotation (sinθ = 1): from the previous page Now sum up the torques due to the external force acting on all the little pieces of the rigid body: α is factored out of the summation because it is constant for all points on a rigid object. Why are we not including the torques due to the internal forces? 86

87 Moment of Inertia From linear dynamics, we know that internal forces form action reaction pairs and the net force is equal to zero. Similar logic applies to the sum of the internal torques their forces are equal and opposite and act along the same displacement r, so they all cancel. Define the net external torque as: Where I is the moment of inertia of the rigid body: 87

88 Moment of Inertia This looks very similar to Newton's Second Law in translational motion. ΣF has been replaced by Στ, a has been replaced by α, and m has been replaced by I. The mass of a given rigid object is always constant. Can the same be said for the moment of inertia? 88

89 Moment of Inertia No it depends on the configuration of the rigid object, and where its axis of rotation is located. Newton's Second Law for rotational motion: Moment of Inertia: If more of the little pieces of the rotating object are located further away from the axis of rotation, then its moment of inertia will be greater than an object of the same mass where more of the mass is concentrated near the axis of rotation. 89

90 26 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. What is the moment of inertia of this configuration if the axis of rotation is located at one of the masses? What is the moment of inertia if the axis of rotation is located at the midpoint of the rod? Which is easier to rotate? Why? L Answer 90

91 Moment of Inertia The moment of inertia was easily calculated when there were a limited number of objects that could be represented as point particles. For rigid objects of given shapes, we will need to add an infinite amount of small masses times their distance from the axis of rotation. For objects with symmetries, we can use integral calculus to calculate their moments of inertia. Each small mass will rotate about the axis of rotation due to an external tangential force providing a tangential acceleration. 91

92 Moment of Inertia Label each small mass, dm, that it is acted on by an external tangential force, df T, producing a tangential acceleration a T. Working with the magnitudes and Newton's Second Law: The tangential force provides a torque about the axis of rotation: since α is constant at all points on the rigid body The moment of inertia can be calculated now: 92

93 Moment of Inertia Calculation Find the moment of inertia of a solid cylinder of radius R, length L, and mass M rotating about the z axis through its center of mass. Express dm as ρdv, where ρ is the volume mass density and dv is a small piece of volume, as it is easier to work with volumes than masses. z dr r We'll need to use the cylindrical symmetry to substitute for dm. R L 93

94 Moment of Inertia Calculation Draw a thin shell of thickness dr that is centered on, and a distance r from the axis of rotation. It extends from the top of the cylinder to the bottom, so its length is L. The cross sectional area of this shell is 2πrdr. The volume, dv, is 2πrLdr. dr z r R L 94

95 Moment of Inertia Calculation Express ρ as M/V. The volume of the cylinder is πr 2 L. Substitute for ρ. dr z r R L Moment of Inertia of a Cylinder about its z (long) axis. 95

96 Moment of Inertia Here are various moments of inertia for other symmetrical objects. All of the objects have the same mass, M, but different shapes and different axes of rotation. Take some time to see how and why the moments of inertia change. 96

97 Moment of Inertia Just as mass is defined as the resistance of an object to accelerate due to an applied force, the moment of inertia is a measure of an object's resistance to angular acceleration due to an applied torque. The greater the moment of inertia of an object, the less it will accelerate due to an applied torque. The same torque is applied to a solid cylinder and a rectangular plate. Which object will have a greater angular acceleration? You can go back a slide to find their moments of inertia. 97

98 Moment of Inertia Since the rectangular plate has a smaller moment of inertia, it will have a greater angular acceleration. The angular acceleration can be different for the same object, if a different axis of rotation is chosen. A long thin rod is easier to rotate if it is held in the middle. By rotating it about that axis, as compared to rotating it at the end, it has a smaller moment of inertia, so for a given torque, it will have a greater angular acceleration it will be easier to rotate. This was solved earlier in the chapter with a rod with a mass at either end. It's nice that physics is consistent. 98

99 Parallel Axis Theorem If the moment of inertia about the center of mass of a rigid body is known, the moment of inertia about any other axis of rotation of the body can be calculated by use of the Parallel Axis Theorem. Often, this simplifies the mathematics. This theorem will be presented without proof: where D is the distance between the new axis of rotation from the axis of rotation through the center of mass (I cm ), and M is the total mass of the object. 99

100 27 A child rolls a 0.02 kg hoop with an angular acceleration of 10 rad/s 2. If the hoop has a radius of 1 m, what is the net torque on the hoop? A 0.6 N m B 0.5 N m C 0.4 N m Answer D 0.3 N m E 0.2 N m 100

101 28 A construction worker spins a square sheet of metal of mass kg with an angular acceleration of 10.0 rad/s 2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is N m? A 42 m x 42 m B 42 m x 39 m C 39 m x 39 m Answer D 30 m x 39 m E 30 m x 42 m 101

102 29 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. The moment of inertia about the center of mass of the system is 1/2 ml 2. Use the parallel axis theorem to find the moment of inertia of the system if it is rotated about one of the masses. L Answer 102

103 Rotational Kinetic Energy Return to Table of Contents 103

104 Rotational Kinetic Energy After Kinematics and Dynamics were covered, your physics education continued with the concept of Energy. Energy is hard to define, but since it is a scalar quantity, it is great for solving problems. By now, you should be ready for the concept of a rotational version of the linear (also called translational) kinetic energy. Given that KE = 1/2 mv 2, what substitution should be made? 104

105 Rotational Kinetic Energy Use the rotational equivalent of velocity and substitute it into the Kinetic Energy equation. The second line shows where one piece of a rotating object is being considered, and it is now a rotational energy since v is replaced by ω. Add up all the KE R 's for each small mass 105

106 Work done by Torque The concept of Work was introduced in parallel with Energy, where. A shortcut will now be taken to determine how much work is done by an applied torque the rotational equivalents of force and displacement will just be substituted into the above equation: Since torque is always perpendicular to the angular displacement, the vector dot product reduces to multiplying the values for torque and angular displacement (in radians). 106

107 Total Kinetic Energy This course has so far discussed three types of energy GPE, KE and EPE (Elastic Potential Energy). All three of those were dealt with in terms of a point mass there was no rotation. Now that we have rigid rotating bodies, KE will be redefined as a total kinetic energy that sums up the linear and rotational aspects. 107

108 Total Kinetic Energy A common rotational motion problem is having two different objects, each with the same mass and determining which object will get to the bottom of an incline first. If they were just blocks, or even spheres, that slid down without rotating, they would both get to the bottom at the same time, since objects fall at the same rate. By using an extended mass (not a point particle), and rotation, it's quite different. Why? Think about Energy. At the top of the ramp, both objects have the same energy it's all potential. How about at the bottom? 108

109 Total Kinetic Energy For a non rotating object, all of the gravitational potential energy is transferred into translational kinetic energy. If the object rotates, then part of that energy goes into rotational kinetic energy. Rotational kinetic energy depends on the shape of the object which affects the moment of inertia. Here's an example. A sphere and a hoop of the same mass and radius both start at rest, at the top of a ramp. Assume that both objects roll without slipping, which one gets to the bottom first? 109

110 Here's the sketch. Total Kinetic Energy Hoop Sphere h What is the equation, using the Conservation of Energy, that will give the velocity of the objects at the bottom of the ramp? Solve the equation. Qualitatively, how can that be used to determine which object reaches the bottom first? 110

111 Total Kinetic Energy Hoop Sphere h This equation works for both the hoop and the sphere. 111

112 Total Kinetic Energy Hoop Sphere h Did you question why it was stated that the sphere and the hoop roll without slipping? When that occurs, then v linear = v = rω. We will use this fact to replace ω in the above equation with v/r. 112

113 Total Kinetic Energy Hoop Sphere h Hoop Sphere 113

114 Total Kinetic Energy Hoop Sphere h The velocity of the sphere is 1.2 times as much as the velocity of the hoop. The sphere gets to the bottom of the ramp first because at any point on the incline, the sphere has a greater linear velocity (h is just the height above ground). Solving the kinematics equations would also give the same answer. One more question why did this happen? 114

115 Total Kinetic Energy Hoop Sphere h Both the sphere and the hoop started with the same GPE. And, at the bottom of the ramp, both had zero GPE and a maximum KE again, both the same. But the hoop, with its greater moment of inertia, took a greater amount of the total kinetic energy to make it rotate so there was less kinetic energy available for linear motion. Hence, the sphere has a greater linear velocity than the hoop and reaches the bottom first. 115

116 30 A spherical ball with a radius of 0.50 m rolls down a 10.0 m high hill. What is the velocity of the ball at the bottom of the hill? Use g = 10 m/s 2. A 10 m/s B 11 m/s C 12 m/s Answer D 13 m/s E 14 m/s 116

117 31 A solid cylinder with a radius of 0.50 m rolls down a hill. What is the height of the hill if the final velocity of the block is 10.0 m/s? Use g = 10 m/s 2. A 7.0 m B 7.5 m C 8.0 m Answer D 8.5 m E 9.0 m 117

118 32 Consult a table that lists the moments of inertia of various symmetric objects. If each of the following have the same mass and radius, rotate as described, and are let go from the top of an incline, which object reaches the bottom first? A Thin walled cylinder about its central axis B Thick walled cylinder about its central axis C Solid sphere about its center Answer D Hollow sphere about its center E Solid cylinder about its central axis 118

119 Angular Momentum Return to Table of Contents 119

120 Angular Momentum As was done in the last chapter, we won't derive angular momentum from linear momentum we will substitute in the angular values for the linear (translational) ones. Linear momentum (p): Angular momentum (L): Angular momentum is also described in terms of linear momentum as follows: 120

121 Conservation of Angular Momentum The conservation of (linear) momentum states that in the absence of external forces, momentum is conserved. This came from the original statement of Newton's Second Law: If there are no external forces, then: Thus there is no change in momentum; it is conserved. 121

122 Conservation of Angular Momentum Starting with Newton's Second Law for rotational motion, the conservation of angular momentum will be derived. If there are no external torques, then: Thus there is no change in angular momentum; it is conserved. 122

123 Conservation of Angular Momentum The standard example used to illustrate the conservation of angular momentum is the ice skater. Start by assuming a frictionless surface ice is pretty close to that although if it were totally frictionless, the skater would not be able to stand up and skate! The skater starts with his arms stretched out and spins around in place. He then pulls in his arms. What happens to his rotational velocity after he pulls in his arms? Click on the below site and observe what happens

124 Conservation of Angular Momentum Assuming a nearly frictionless surface implies there is no net external torque on the skater, so the conservation of angular momentum can be used. The human body is not a rigid sphere or a hoop or anything simple. But, knowing what the definition of moment of inertia is, at what point in the spinning does the skater have a greater moment of inertia? 124

125 Conservation of Angular Momentum When the skater's arms are stretched out, more of his mass is located further from his axis of rotation (the skates on the ice), so he has a greater moment of inertia then when he tucks his arms in. In the video, is greater than. Thus, ω f is greater than ω 0 and the skater rotates faster with his arms tucked in. 125

126 33 A 2.0 kg solid spherical ball with a radius of 0.20 m rolls with an angular velocity of 5.0 rad/s down a street. Magneto shrinks the radius of the ball down to 0.10 m, but does not change the ball's mass and keeps it rolling. What is the new angular velocity of the ball? A 10 rad/s B 15 rad/s C 20 rad/s Answer D 25 rad/s E 30 rad/s 126

127 34 An ice skater extends her 0.5 m arms out and begins to spin. Then she brings her arms back to her chest. Which statement is true in this scenario? A Her angular velocity increases. B Her angular velocity decreases. C Her moment of inertia increases. D Her angular acceleration decreases. Answer E Her angular acceleration is zero. 127

128 Kepler's Second Law Kepler's Second Law states that a radius vector from the sun to one of its planets (like earth) sweeps out an equal area in an equal time during the planet's orbit. Since the orbits are ellipses with the sun at one of the foci (Kepler's First Law), the planets move faster when they are further away from the sun. In the Energy unit of this course, the Work Energy Equation was used to explain the speed change. Conservation of Angular Momentum can also be used. By Kepler2.gif: Illustration by RJHall using Paint Shop Pro derivative work: Talifero (Kepler2.gif) [CC BY SA 2.0 at ( via Wikimedia Commons 128

129 Kepler's Second Law In orbit, there are no external non conservative forces acting on the earth. The gravitational force is an internal force to the system and is accounted for with its potential energy. Thus, the Conservation of Angular Momentum applies. Select points A and B, recognizing that r A > r B means that I A > I B and using the moment of Inertia formulation: A B The earth has a smaller angular velocity at point A. By Kepler2.gif: Illustration by RJHall using Paint Shop Pro derivative work: Talifero (Kepler2.gif) [CC BY SA 2.0 at ( via Wikimedia Commons 129

130 Kepler's Second Law Using the formula that relates angular momentum to linear momentum and recognizing that r A > r B : The earth has a smaller linear (tangential) momentum, and hence, linear velocity at point A. Since v = rw, it is a good thing that we get the same results using both equations. A B By Kepler2.gif: Illustration by RJHall using Paint Shop Pro derivative work: Talifero (Kepler2.gif) [CC BY SA 2.0 at ( via Wikimedia Commons 130