FZX: Personal Lecture Notes from Daniel W. Koon St. Lawrence University Physics Department CHAPTER 8

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1 FZX: Personal Lecture Notes from Daniel W. Koon St. Lawrence University Physics Department Copyright 1995, 011, D. W. Koon. All Rights reserved CHAPTER 8 Please report any glitches, bugs or errors to the author: dkoon at stlawu.edu. 8. Rotational Dynamics Digging Deeper: rotational energy and power Seeing the Connections Rotational Equilibrium page 1

2 FZX, Chapter 8: ROTATIONAL DYNAMICS Copyright 1995, 011, D. W. Koon. All Rights reserved There is more than one way to slice up two-dimensional space. When we first considered two-dimensional motion, we said that we could break it up into x-motion and y-motion. This was fine for free-fall problems, where the only force acting was the weight, which points straight down. For such problems there was no acceleration in the x-direction, and so the x-motion was the simplest kind of motion to describe -- constantvelocity motion. But not all objects travel in nice straight lines and not all objects have constant acceleration. The Moon circles the Earth and the Earth circles the Sun, and in the course of 4 hours, our own motion is more circular than linear: most of us end up at the same place we started, most nights. For this reason, we should shop around for other ways of slicing up the two-dimensional space that we have been working in. Slicing space into two directions is a lot like dividing humanity into two types of people: the particular choice of how to divvy things up depends on what you are going to do with the results. For studying things that travel in circles, we can choose to divide our two-dimensional space into radial and angular axes. Divide a sheet of paper into fourths by drawing a horizontal and a vertical line. Now draw diagonal lines in an X that cross these lines in the center where they cross. These are the radial axes. Now draw evenly spaced concentric circles on the page, each having its center where those radial lines meet. You have just drawn a polar coordinate system, not unlike our usual rectangular coordinate system. Both systems have their own merits, and for different problems one or the other may prove to be more convenient. Why bother? Consider your motion if you sit on a merry-go-round. Your motion is simple: you travel with a constant speed in a simple path. In terms of x- and y-coordinates, however, your motion is quite complex. Your acceleration is a centripetal acceleration, so it points toward the center of the circle you follow. This means that your acceleration changes direction as you change direction, and so the components of acceleration, a x and a y are constantly changing. We have not studied how to analyze non-uniform acceleration: in general it is NOT pretty, although we will consider this special case of it in a few chapters, when we study simple harmonic motion. If we consider an angular position which changes at a constant rate (no acceleration) and a radial position that doesn t change at all, we can describe the motion of the merry-go-round much more simply. To save effort, and to try to see the relation between angular and linear motion, let s make a special effort to relate all of the following rotational terms we are about to introduce to the linear terms we already know. To help remember the connection, I will show the related linear term in curly brackets after each new term that I introduce. The first angular concept we introduce is that of angular position. This is nothing more than angle, θ. We shall measure angle in units of radians. These are the units we get if we divide the distance that an object travels as it travels in a circle by the distance between the object and the center of rotation: θ = s / r. [Angle] { Linear analogue: x } Once around a circle is thus equivalent to π radians. Notice that θ is dimensionless if we use the definition above. If your calculator displays the letters DEG (for degrees of angle), or something similar, in its display, then you can probably get it to do its angle calculations in radians too. If you don t know how to switch between DEG mode and RAD mode, make sure to find out. The rate at which angle changes is the angular velocity, ω, ω = Δθ / Δt, [Angular velocity] {Linear analogue: v = x/ t} } (ω is the lower case form of the Greek character omega.) and the rate at which angular velocity changes is the angular acceleration, α: α = Δω/Δt. [Angular acceleration] {Linear analogue: a = Δv / Δt } (α is a lower case Greek letter alpha.) These last two quantities are related to the components of linear velocity or acceleration in the direction of motion by page

3 Copyright 1995, 011, D. W. Koon. All Rights reserved and v t = rω, [Tangential velocity] a t = rα, [Tangential acceleration] where vt is called the tangential velocity, and at is called the tangential acceleration. The tangential velocity tells how fast you are travelling along the circumference of the circle, and the tangential acceleration tells the rate at which the tangential speed is changing. If you are on a merry-go-round, your tangential speed and acceleration, vt and at, will be larger at the edge than for someone closer to the axis around which it spins. The two of you will share the same angular speed and acceleration, ω and α. If the M-G-R spins with a constant speed, then a = 0 = a t. Let me add two things to further connect these quantities to quantities we ve studied before. ω is related to the frequency at which something turns. If a circle contains π radians of angle, and if something is travelling at f rotations per second, then it is covering πf radians of angle per second: ω = πf Secondly, the tangential acceleration is quite a different thing than the centripetal acceleration. Think of the tangential acceleration as telling whether something is speeding up or slowing down. Think of the centripetal acceleration as telling whether you are changing direction -- which you are ALWAYS doing if you are travelling in a circle. page 3

4 Copyright 1995, 011, D. W. Koon. All Rights reserved If we wish to consider some object which is brought into angular motion, we need to be able to define its angular inertia, the angular analogue of mass or linear inertia. The term usually used for this quantity is moment of inertia, and we call it I: I [Moment of Inertia] {Linear analogue: m } Since mass is a measure of how hard it is to change the linear motion of an object, the moment of inertia must tell us how hard it is to get something to turn. What determines how hard it is to get something to turn? Take a meter stick or ruler, and see how easy it is to spin it around its center, like twirling a baton in a marching band. Now compare this to spinning it about its end. It is about ¼ as hard to spin it about its center. Why? An object has a larger moment of inertia when its mass is distributed far away from the axis of rotation because you need to move mass further distances in rotating it. By the way, if we had asked you to twirl a piece of lead pipe instead, it would have been harder, because of the large mass. We can show that the moment of inertia can be expressed ( ) avg I = m r, [General expression, moment of inertia] where ( r ) avg is the average value of the squared distance to the axis of rotation. We get this average by adding together r for every single molecule of the material making up the object being considered. Notice that if we change the axis, the moment of inertia may change as well. This is why it is easier to twirl a baton about its center than about its end. Convince yourself that ( r is larger for one of these cases than for the other. ) avg What is the combined moment of inertia when two spinning objects become one? If we drop a vinyl phonograph record onto the turntable of a record player, the two will spin together. How do we calculate the total moment of inertia of the system? Well, if two masses collide into each other, the combined mass of the new system is just the sum of the two page 4

5 Copyright 1995, 011, D. W. Koon. All Rights reserved masses. If moment of inertia is just an angular mass, we should be able to just add the moments of inertia of the two objects. Now, how do we find the moment of inertia for an object? How do we calculate ( r ) avg? In general this is a difficult calculation, but it has been done for the most commonly encountered shapes, and the moments of inertia for these shapes are easily available in tables. So rather than calculating ( r ) avg for a given shape rotated about a particular axis, we usually just look up the appropriate formula. If we spin a disc of mass M and radius R as an LP or a compact disc or a wheel spins, for example, we find that I = 1 MR. [disc spun about center] If we spin a rod of length L and mass M, first about its center but then about its end, we find that and I = 1 1 MR [rod spun about center] I = 1 3 MR. [rod spun about end] Well, if I tells us how difficult it is to change the angular motion, we also need to consider what it is that can change that motion. Force can change the linear motion of an object: we defined it as that which can change the velocity of an object. We need to define some quantity which will be that which can change the angular velocity of an object. Let s call it torque, τ. To determine how torque is related to force, let s consider how hard it can be to get a door to open or close. If we grab the doorknob from both sides of a door and push with a force directed toward the hinges of the door, the door will not budge, regardless of how hard we push. If, however, we push in the direction in which we want the door to move, we can get it to move. So, direction is important in defining a torque. If we take a bathroom scale and measure the force we need to move the door, we find that we need a larger force to budge the door if we apply our force closer to the hinges. (It helps to put a doorstop on the other side of the door to make the required force large enough to be easily measured.) From all of this, we find that the torque is equal to τ = rf sinθ, [Torque] {Linear analogue: F } where r is the distance between where the force is applied and the axis of rotation, is the force exerted, and θ is the angle between the direction of the force and a line from the axis to the point where the force is applied. (τ is the lower case form of the Greek letter tau.) While we re talking torque, we can address a question that surely puzzled you once or twice, long ago: WHY DO THINGS FALL OVER? Floors are extremely intelligent beings that sense how much force they need to exert to keep you from falling into the basement, and so they provide just that amount of normal force, and no more, because any more would cause you to spontaneously accelerate into the sky. The floor also knows where to apply that normal force. The force gravity exerts on you -- your weight -- can be considered to be applied at your center of mass. If you lean too far to the side, your center of mass is no longer above your feet, and, since a normal force can only be exerted where two objects meet, the normal is applied at your feet. If the weight and the normal are not directly above each other, there will be a net torque on you, and you will start to rotate. This is exactly what is happening to you when you fall over. Notice that this torque will increase as you fall, as the angle θ we defined above gets bigger. F page 5

[Effect of torque on angular acceleration] r r {Linear analogue: F = ma } What if there is no net external torque?

6 Copyright 1995, 011, D. W. Koon. All Rights reserved What effect does a torque have if an object starts to rotate? Just working from analogy, we can rewrite Newton s Second Law in angular form : τ = Iα. [Effect of torque on angular acceleration] r r {Linear analogue: F = ma } What if there is no net external torque? We should expect similar results to what happens when a system has no net external force acting on it. In the linear case, there is conservation of momentum, so in the torqueless case, we should expect conservation of angular momentum. (First of course, we have to define angular momentum.) If we define angular momentum as L = Iω, [Angular momentum] { Linear analogue: p = mv } then it will be conserved in the case in which the net external torque is zero. This is the angular equivalent of a collision. If an LP drops down onto a turntable, the total momentum is conserved. This can only happen if the turntable slows down. If a skater who is spinning on ice brings her arms inward, so that her moment of inertia decreases, her angular momentum must be conserved. This can only happen if her angular speed increases. If you haven t observed a skater doing this, try it yourself sometime on an ice rink or on a rotating platform or chair. page 6

Copyright 1995, 011, D. W. Koon. All Rights reserved If an object is spinning, it must have some kinetic energy, even if the average linear velocity of the object is zero.

[Rotational kinetic energy] { Linear analogue: KE 1 linear = mv } KE linear Notice that we use, or linear (or translational) kinetic energy to stand for what we have simply called kinetic energy

8 Copyright 1995, 011, D. W. Koon. All Rights reserved If an object is spinning, it must have some kinetic energy, even if the average linear velocity of the object is zero. If we add up the kinetic energy of every piece of the object, we find that the kinetic energy due to its rotational motion is KE 1 rot = Iω. [Rotational kinetic energy] { Linear analogue: KE 1 linear = mv } KE linear Notice that we use, or linear (or translational) kinetic energy to stand for what we have simply called kinetic energy before. In order to accurately measure the kinetic energy of a car, for example, we not only have to know its mass and speed to get the linear kinetic energy,, but we also have to know the moment of inertia and angular KE 1 linear = mv KE rot = 1 ω speed of the wheels in order to calculate the rotational kinetic energy,. The total kinetic energy is just the sum of the two. (In the past we fudged things by assuming that the rotational kinetic energy was much smaller than the linear, as you should be able to convince yourself must be the case for a car, for example. In the future this may not be true for objects we consider.) I page 8

9 Copyright 1995, 011, D. W. Koon. All Rights reserved DIGGING DEEPER (Rotational Energy and Power): If there is rotational kinetic energy, then it must be possible to do work in rotation problems, so we can introduce, by way of analogy, the work done by a torque: W = τδθ [Work due to torque] {Linear analogue: W = FxΔx } Likewise a force which exerts a torque can supply power, P = ωτ. [Power from torque] { Linear analogue: P = vf cosθ } Just about any equation that you can write for linear quantities can be extended to the angular analogue simply by making the right substitutions of variables. The basic kinematics equations become θ = θ 0 + ω avg t { Linear analogue: x x + vavgt ω = ω 0 + αt { Linear analogue: v v + at 1 1 θ = θ0 + ω0t + αt { Linear analogue: x x0 + v0t + at ω = ω + αδθ v = v + aδx 0 = 0 } = 0 } = } { Linear analogue: 0 } [Rotational kinematics] Notice what we ve just done here: we have developed the kinematics ( the how ) and the dynamics ( the why ) of rotational motion in a much quicker fashion than we did for the linear case. See, it does get easier! SEEING THE CONNECTIONS: Why bother? The results we have derived have application well beyond circular motion. If we apply a force in the direction something is moving, we increase its speed, if we apply a force opposite the motion, it slows down, if we apply a force perpendicular to the motion, it maintains the same speed, but changes direction. It is possible to have a force that changes the speed and changes the direction of motion, if it is at applied at some angle other than 0 o, 90 o, or 180 o, relative to the direction of motion. What we are doing in this case is dividing the force into a tangential component (parallel to the motion) and a centripetal component (perpendicular to the motion). We also can gain some insight into the equation for work. Only the component of the force which is parallel or antiparallel to the motion changes the kinetic energy (speeds things up or slows them down). is the component of the force parallel to the motion. If it is in the direction of the motion, then it does positive work ( F < 0 ), and the object speeds up. If it is opposite the motion, then it does negative work ( x ), and the object slows down. If it points perpendicularly to the direction of motion, then it does no work ( F x = 0 ), and it just changes the direction of motion, making the object curve. F x F > 0 x ROTATIONAL EQUILIBRIUM When we said that the net force on an object must be zero in order for it to remain in equilibrium, that was only part of the story. If the net force on an object is zero, its center of mass will not move, but if there is a nonzero net torque, it will rotate. Therefore, for an object in static equilibrium, page 9

10 F r = 0 τ = 0 Copyright 1995, 011, D. W. Koon. All Rights reserved [Conditions for equilibrium] We can choose any arbitrary pivot point for measuring the torque about: it will be true for any pivot. We have said that the magnitude, the direction, and the position of a force are all important in evaluating what torque it will provide, and thus, how it changes the rotational motion of an object. This means that we must now be careful, in drawing our free-body diagrams, to draw each force precisely at the point on the body where it is acting. That s easy enough to do if the force is a contact force acting between two objects that only touch in a single point, but what about your weight, acting on all parts of your body? What do we choose for the point at which this force is applied? We can average the position at which the various small components of force that make up your weight act. We can further idealize your body, treat it like a point object, and assume that gravity only pulls on you at one point. Fortunately, the algebra allows us to get away with this simplification. But how do we find this magic spot, which we will call your center of mass? If you support yourself on a stool (so that the normal force supporting you is confined to a relatively small region), you can keep from falling over only if your center of gravity is above the stool. Stretch out horizontally with your body touching the stool only in one place, without touching the floor, the walls, or anything else. The choice of what part of your body you can place on the stool is constrained by the requirement that the net torque on you must be zero. If your center of mass is not above the stool, then the normal and the weight will be applied at different positions, and you will rotate (falling off the stool). In doing rotational equilibrium problems, the most important step will be drawing a good free-body diagram. The rules for this will be a little different, however, from the rules we had just learned for free-body diagrams. Now it IS important to draw the forces acting where they act on the body being considered. This means we will always draw the weight acting on the center of mass of the body. It also means that contact forces, like friction, tension, and normal forces, must be drawn right where the other object is touching the body being considered. The second most important part of doing a rotational equilibrium problem is to write down the correct expressions for the sum of the force components in the x- and y-directions, as well as the equation for the sum of the torques. The least important part of doing a rotational equilibrium problem is solving the algebra that results from writing down these equations: setting it all up is the hardest and most important part. A very useful exercise at this point in the course is to go to some standard college physics text, turn to the problems at the end of this chapter, and practice drawing free-body diagrams for as many problems as you can stand. (If you can t stand too many, do some more the next day. Build up your tolerance!) This is a good time to compare notes with a classmate, or your instructor. If you re confident of your diagrams, then write down your force and torque equations for these problems. You should go through the algebra only if you have plenty of time left, or feel you really need a lot of practice with algebra. Remember, the algebra is the LEAST important part of the problem. A few words of advice: 1) Diagrams get overcrowded with information very quickly. (If they don t, you are probably not using your diagrams effectively.) If they get too cluttered, they become more of a hindrance than a help. You may find it useful to redraw your diagram once or twice in order to keep it clear. You may wish to have one diagram include only the dimensions of the problem -- lengths and angles -- and have another diagram include only vectors. Whatever helps you see what s going on. ) The choice of a good pivot point is important, but not essential. Theory says that whatever point you choose as a pivot point, the torque about it, for a rotational equilibrium problem, is zero. This does make sense, since the object ain t rotating about ANY point. However, if you choose a pivot point which is a point where two or more forces are acting, none of these forces will exert a torque about that pivot, and you will have two fewer torques to have to calculate than if you have chosen a less auspicious pivot point. In other words, using a bad pivot point, you will still arrive at the right answer: it just may take longer. 3) In hunting down torques for your τ equation, remember that for each torque, τ = rf sinθ. For each term, ask yourself what is r, the radius from the pivot point at which the force is applied, what is F, the force exerted at that point, and what is θ, the angle between the radius and the force. page 10

11 Copyright 1995, 011, D. W. Koon. All Rights reserved 4) This shortcut is almost too much of a time-saver to be legal, but it is valid. Draw a dotted line along the direction of the force whose torque you want to calculate. When you reach the point where this line comes as close to the pivot point that the line will go, redraw the force, with the tail end of the force at this point, and the vector pointing the same way it did when you first drew it. The torque exerted by this force acting at this point is the same as for its original position. The angle, however, is now 90 o, which makes your calculation easier. The reason this works is that r sinθ for the original placement of the vector is equal to the distance between the pivot point and the new spot you have located. Check out the geometry and convince yourself that this is so. We call the new radius, l, the moment arm or lever arm of the torque, and so τ = lf [ Other definition of torque] page 11

CIRCULAR MOTION AND ROTATION

CIRCULAR MOTION AND ROTATION 1. UNIFORM CIRCULAR MOTION So far we have learned a great deal about linear motion. This section addresses rotational motion. The simplest kind of rotational motion is an object moving in a perfect circle