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
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 T The angular acceleration, α, is a c a vector that points out of the page. α 32
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
8 At which point is the magnitude of the centripetal acceleration the greatest? A B B A C E D D C E 34
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: a T a total a c 35
9 A child pushes, with a constant force, a merrygoround 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 merrygoround's tangential acceleration? A 0.25 m/s 2 B 0.35 m/s 2 Answer C 0.62 m/s 2 D 0.75 m/s 2 E 0.94 m/s 2 36
10 A child pushes, with a constant force, a merrygoround 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 merrygoround's centripetal acceleration? A 7.5 m/s 2 B 19 m/s 2 Answer C 23 m/s 2 D 38 m/s 2 E 46 m/s 2 37
11 A child pushes, with a constant force, a merrygoround 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 merrygoround's total linear acceleration? A 5.6 m/s 2 B 12 m/s 2 Answer C 23 m/s 2 D 38 m/s 2 E 46 m/s 2 38
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
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 0.40 s Answer B 1.3 rev/s 0.77 s C 2.6 rev/s 0.39 s D 2.6 rev/s 0.78 s E 3.9 rev/s 0.78 s 40
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
Translational Kinematics Here are the equations. If you would like, please review their derivations in the Kinematics section of this course. 42
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
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
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
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
Rolling Without Slipping d A 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. 47
Rolling Without Slipping d A 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. Use only for rolling motion without slipping. 48
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
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 Answer C 40 rad/s D 50 rad/s E 60 rad/s 50
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 Answer C 15 m/s D 10 m/s E 5 m/s 51
15 A bicycle wheel with a radius of 0.300 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 Answer C 230 rad D 251 rad E 273 rad 52
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 Answer C 43 rev D 41 rev E 39 rev 53
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 Answer C 160 m D 180 m E 210 m 54
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 Answer C π/3 rad/s 2 D π/4 rad/s 2 E π/5 rad/s 2 55
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 Answer C 96π rad D 112π rad E 164π rad 56
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 Answer C 36π rad D 42π rad E 48π m 57
Rotational Dynamics Return to Table of Contents 58
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
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. http://commons.wikimedia.org/wiki/File What if a force is applied at %3AAction_meca_equivalence_force_couple_plan.svg point B? 60
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? http://commons.wikimedia.org/wiki/File %3AAction_meca_equivalence_force_couple_plan.svg What is different? 61
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. Which force will cause F 1 the door to open F 2 quicker (have a greater angular acceleration)? 62
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. It will also allow a smaller F 1 force to have the same impact F 2 as a greater force applied at an angle different than 90 0 . 63
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 F 2 θ r 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
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 F 2 θ r When you get to vector calculus, the equation becomes more elegant involving the cross product: 65
Torque The units of torque are Newtonmeters. Note how this is similar to work and energy and in those cases, we replace Nm 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
21 Assume a force is applied perpendicular to the axis of rotation of the wrenchnut system. At which point will the nut experience the greatest torque? B D C A A Answer B C D E All result in the same torque. http://commons.wikimedia.org/wiki/File %3AAction_meca_equivalence_force_couple_plan.svg 67
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 Nm B 25 Nm Answer C 20 Nm D 15 Nm E 10 Nm 68
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 counterclockwise, and a negative value if it accelerates it in the clockwise direction. 69
Balance Here is a balance with two masses attached to it. Block A has a mass of r A r B 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 A B the balance beam is perfectly horizontal and is not moving? Is this a translational motion or a rotational motion problem? 70
SeeSaw (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 SeeSaw" 1873 A side note there are many connections between physics and art this would be a good project to research. 71
Balance This is a rotational motion problem, so in order to have the beam not rotate, the r A r B sum of the torques applied to it must equal zero (Newton's Second Law). A 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
Balance and SeeSaw Again, many elementary school r A r B students already know this! If you're on a seesaw and you have to balance several students on the other side, you A B position yourself further from the fulcrum than they are. Winslow Homer, "The SeeSaw" 1873 73
23 John and Sally are sitting the on the opposite sides of a SeeSaw, 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 SeeSaw is not moving? Use g = 10 m/s 2 . A 450 N Answer B 400 N C 350 N D 300 N E 250 N 74
24 A student wants to balance two masses of 2.00 kg and 0.500 kg on a meter stick. The 0.500 kg mass is 0.500 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 0.500 m B 0.333 m Answer C 0.250 m D 0.125 m E 0.075 m 75
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 Answer B 0.85 m C 1.7 m D 2.0 m E 2.7 m 76
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
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
The ladder against the wall f wl y F Nw f wl ≡ static friction force of wall on ladder F Nw ≡ normal force of wall on ladder mg ≡ gravitational force on ladder mg F Ng θ x f gl f gl ≡ static friction force of ground on ladder F Ng ≡ normal force of ground on ladder 79
The ladder against the wall Choose a point about which the f wl torques will be calculated. y F Nw 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 reference point choose the center of mass of the ladder. This will remove the mass from the torque equation. mg F Ng And, as it will turn out, we won't need the sum of the forces in the θ y direction to solve the problem. x f gl 80
The ladder against the wall f wl y The sum of the forces in the x F Nw and y direction equal zero. θ mg F Ng The sum of the torques about the center of mass equal zero. mg does not θ contribute a torque as it acts at the center x f gl of mass. 81
The ladder against the wall f wl To find the minimum angle at which the ladder will not slip, the static friction forces will be at y F Nw their maximum value: Use the forces in the x direction: mg F Ng θ x f gl 82
The ladder against the wall f wl y F Nw replacing the friction forces with the normal forces. mg F Ng θ replacing F Nw with F Ng And we get: x f gl 83
The ladder against the wall f wl y F Nw mg The angle is independent of F Ng the mass of the ladder much like a mass sliding down an incline. θ x f gl 84
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
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
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
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
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
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? Answer L 90
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
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
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 L We'll need to use the R cylindrical symmetry to substitute for dm. 93
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 2 πrLdr. z dr r L R 94
Moment of Inertia Calculation z dr r Express ρ as M/V. The volume of the cylinder is L πR 2 L. Substitute for ρ. R Moment of Inertia of a Cylinder about its z (long) axis. 95
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
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
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
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
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 Nm B 0.5 Nm Answer C 0.4 Nm D 0.3 Nm E 0.2 Nm 100
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