Chp 10.11 (Rotation)

Chp 10.11 (Rotation) - Physics 7A Discussion Notes Chapters...

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Unformatted text preview: Physics 7A Discussion Notes Chapters 10 and 11: Rotation Aaron Alpert April 2011 1 Introduction Axis of Rotation. Rotation can be a very tricky business. Extending translational motion to three dimensions is easy. All of the rules that hold in x also hold in y and z. However, rotation is another matter. How can we even describe a rotation? In 1D, it’s easy! Draw a line on the object, rotate it, and measure the angle between the original line and the rotated line. That’s the amount it’s rotated. But what about more complicated rotations? You must always begin with a “base configuration,” or the orientation that you will consider to be zero. Once you have that, you have to describe how it has been rotated. One approach to this is to talk about an axis of rotation. We represent this with the angular rotation vector θ. The magnitude of θ is the amount of rotation (i.e., θ = 2π means 1 full revolution). The direction of θ is along the axis of rotation.1 Along the axis of rotation (assuming no translation), the displacement of every particle is zero. At any other point, the displacement is ∆x = θ × r (1) where ∆x is the linear displacement caused by the rotation and r is the vector from the axis of rotation to the point. A very good way of representing this is with something called quaternions. Quaternions are mathematical objects defined by four parameters. Three of the parameters define a vector and the fourth defines a scalar. Together, they describe an axis of rotation and a rotation angle. Euler Angles. There is another way to describe rotations, discovered by Euler. Instead of thinking of it as one big rotation, he broke it down into 3 small rotations (each one an easy to visualize 1D rotation). He “glued” an orthonormal coordinate axis to the object to be rotated, usually of the familiar [i, j, k] basis. He would pick one of the three axes and rotate through an angle γ 1 about the chosen axis. He would call the newly rotated (though still orthonormal) basis [i￿ , j￿ , k￿ ]. He would pick a second axis, rotate i￿ , j￿ , and k￿ through γ 2 We tend to use a right-handed coordinate system. If you point your right thumb along the direction of θ and curl your finger, they will naturally curl in the direction of the rotation. 1 1 to [i￿￿ , j￿￿ , k￿￿ ], and finally, he would rotate through γ 3 about a third axis to [g1 , g2 , g3 ]. This final basis is orthonormal (just like [i, j, k]), but it is tilted. For an example, let us show the rotation of [i, j, k] to [i￿ , j￿ , k￿ ] as a rotation of ψ about k. In figure 1, it is clear that the new Figure 1: Rotation Through ψ About k to [i￿ , j￿ , k￿ ]. vectors [i￿ , j￿ , k￿ ] can be expressed in terms of [i, j, k] as follows: i￿ = cos ψ i + sin ψ j j￿ = − sin ψ i + cos ψ j k￿ = k This can be turned into a matrix equation. ￿ i cos ψ sin ψ 0 i j￿ = − sin ψ cos ψ 0 j k￿ 0 0 1 k The 3 × 3 matrix is called a rotation matrix. Any general rotation can be seen as the product of three rotation matrices. Alternatively, one can represent the Euler angles using a rotation tensor instead of a matrix. While they are the easiest conceptually, matrices have some very messy algebra and are prone to round-off errors (in computer calculations). Euler angles and tensors are subject to something called gimbal lock where they break down at certain angles. Quaternions are compact, numerically robust, and not subject to gimbal lock, but they are the most mathematically abstract. Example: Yaw Pitch Roll. When describing the motion of an aircraft, aeronautical engineers use yaw, pitch, and roll, as seen in the figure.2 The airplane is first turned right or left (yawed) about k, then lifted up or down (pitched) about j￿ , and finally rolled about i￿￿ . The net result is that the airplane is moved into a different configuration than it had on the ground. 2 Image courtesy Orientation.htm 2 Figure 2: Rotation of an Aircraft. Angular Velocity Vector. In three dimensions (and after a lot of algebra), the angular velocity vector3 is expressed as ω = γ 1 g1 + γ 2 g2 + γ 3 g3 ˙ ˙ ˙ (2) where each of the γ ’s represents one of the Euler angles and the g’s are the rotated orthonormal basis. Most of the time, we consider only one rotation. Thus, γ 2 = γ 3 = 0. For us, the rotation is very simply through θ about a fixed point O. In other words, we restrict the axis to not move. 2 2.1 Plane Rotation Kinematics Now that we have restrained ourselves to looking at a fixed axis of rotation, the rotational kinematics reduce to something very similar to what we had in 1D kinematics. Instead of “x,” we use θ, the angle of rotation. We futher define the angular velocity ω and the angular acceleration α. ω≡ dθ ˙ =θ dt dω d2 θ ¨ α≡ =ω= 2 =θ ˙ dt dt 3 Very technically, it’s the axial vector of R, the rotation tensor. 3 From this, we arrive at the “same” kinematics equations for constant angular acceleration. α = constant ω = ω0 + αt 1 θ = θ0 + ω0 t + αt2 2 2 ω 2 = ω0 + 2α(θ − θ0 ) Example: Washing Machine. Suppose the washing machine is on spin cycle at 3 rev/s. How many revolutions will it take for the machine to stop spinning if the motor is capable of affecting an angular acceleration of 4 rad/s2 ? Answer: Washing Machine. First, we convert 4rad/s2 to 4 21 = .637 rev/s2 . Then, we π simply use our kinematics relations. ω2 02 02 θ − θ0 = = = = 2 ω0 + 2α(θ − θ0 ) (3 rev/s)2 + 2(−.637 rev/s)(θ − θ0 ) (3 rev/s)2 + 2(−.637 rev/s)(θ − θ0 ) ∆θ = 14.1 rev 2.2 Torque and Moment of Inertia Torque. To begin, we ask a conceptual question. In figure 3, will it be easier to lift the Figure 3: Lifting a Boulder Using a Fulcrum (Lever). boulder if push at point A or point B? Instinctively, we know that it is point B. Here’s another way of thinking about it: By lifting the boulder, you increase its potential energy (we’ll assume the KE is negligible), and that increased PE came from the work you did via the lever. It doesn’t matter if you push at A or B, the change in boulder’s PE is the same, so the work done is the same. Although both point A and B rotate through the same angle, point B has a larger linear displacement. Thus, if you push at B, you are pushing over a greater distance to achieve the same amount of work, so you need less force. That is why it is “easier” to lift the boulder by pushing at B. 4 We define torque4 about a fixed point O as follows: Torque = τ = rF⊥ = rF sin θ = r × F (3) To determine the direction of torque (into or out of the page), we would normally need the right hand rule. This is done more generally later in these notes, but for now, we shall just consider torque to be positive counter-clockwise (this is the normal convention, though arbitrary). As in figure 4, the force F acts at an angle θ relative to the moment arm r. The Figure 4: Torque About a Fixed Axis at O. force perpendicular to the moment arm is F sin θ, and the resulting torque is rF sin θ. In the figure, the torque shown causes a counter-clockwise rotation. Moment of Inertia. What will be the acceleration of that particle? It rotates about O through some angle θ at an angular speed of ω and an angular acceleration of α. There is a centripetal acceleration inward, but there is also a tangential acceleration equal to atan = rα. We note that the force causing this is F⊥ , which can be expressed as τ /r. If we take the usual F = ma equation and substitute in our angular quantities, we find Ftan = matan τ = mrα r τ = mr2 α In the interest of making this equation as much like F = ma as possible, we give the quantity mr2 a name, moment of inertia.5 If there are a bunch of different masses, the total moment Somtimes you will hear the torque referred to as the moment. There is a subtle distinction (which mechanical engineers tend to make, but physicists don’t). The tendency of a force, in general, to cause a rotation is called a moment. That force may also cause a linear displacement. If the force does not cause a translation, but just pure rotation, it is called a torque. Sometimes it will be convenient to break a moment down into a force-couple. In a force-couple, you separte the purely rotational part (the torque) and the purely tranlational part (the force on the center of mass). However, I don’t make that distinction here; I use torque and moment interchangably. In fact, most dynamics classes use the word moment instead of torque, but it’s the same thing. In order to prevent confusion, they will refer to the moment of inertia as the rotational inertia instead. 5 Again, this is true only for 1D rotations. We are rotating about the z axis, so we are technically calculating Izz . If we want to expand this to a 3D rotation, suddenly the moment of inertia is a symmetric 4 5 of inertia is the sum of the moments of inertia for each little mass. Likewise, we can integrate over the volume of a rigid body. ￿ ￿ ￿ 2 2 2 I ≡ mr = m i ri = r dm = r2 ρ dV (4) i m V For our purposes, we can use the tables in the back of the book to find common moments of inertia. To find I for a more complicated rigid body (that is comprised of other simpler bodies), we simply add the moments of inertia. For example, a yo-yo can be thought of as three discs glued together coaxially, and to find I about the common axis, we simply add the I’s for the three discs. There are two other common occurences. When there is a “hole” in a rigid body, we can add “negative I” to account for the gap. Similarly, if the rigid body actually rotates at a different, but parallel axis, we can use the parallel axis theorem. Books tend to give the moment of inertia about the center of mass, so if the new axis is located at a distance d from the center of mass, the mass of the rigid body is m, and the original moment of inertia is ICM , then the new moment of inertia about the parallel axis is I = ICM + md2 (5) The same caveats that applied to F = ma apply to τ = I α. The τ must be the sum of all the torques acting on a system. The equation holds only in an inertial reference frame. Additionally, we must specify the axis of rotation. I have only shown that the sum of the torques about a fixed point O is proportional to the angular acceleration about O. We technically should write ￿ τO = I α O (7) We now have a very nice kinetics equation, similar to F = ma, but for rotation and torque. ￿ τ = Iα (6) Center of Mass. It is useful to talk about the center of mass, which is a mass-weighted average position of all the particles that make up a rigid body. It is (given for both discrete and continuous bodies): ￿N ￿N mi ri 1￿ 1￿ i=1 mi ri rCM = ￿N = i=1 = r dm = rρ dV (8) M Mm M i=1 mi V Furthermore, we can talk about the x and y components of the center of mass separately, as shown for just the x component of the CM in the equation below. ￿N ￿N 1￿ 1￿ i=1 mi xi i=1 mi xi xCM = ￿N = = x dm = xρ dV (9) M Mm M i=1 mi V 3 × 3 matrix with 6 independant components. This is just one more reason we restrict the motion to the simple case. 6 The center of mass is useful for several reasons. The first has to do with the center of gravity. The center of gravity is a point such that the sum of all the torques is zero. As it turns out, the center of gravity is equal to the center of mass. Thus, you can consider gravity to be a force −mg j that acts on a single point: the center of mass.6 The center of mass is also useful because when talking about Newton’s Second Law, the sum of all the forces acting on the rigid body is equal to the total mass of the rigid body times the acceleration of the center ￿ of mass : F = maCM = ma. There is one more very useful property. If, instead of a fixed axis O you use a moving axis on the center of mass, then you find that ￿ ￿ τCM = I αCM τ = Iα (10) This is a very special result. Remember, we showed that τ = I α for a fixed point O, but the center of mass moves! In general, if you try to do the torques about any moving point ￿ P , you find that τP = I αP + ρ × ma where ρ is the position of CM relative to P. As you can tell, this gets really tricky quite quickly, so we limit ourselves to two special cases: rotation about a fixed point and rotation about the center of mass. In general, if the rigid body is “pivoted” about a certain point, then we use the that pivot as our fixed point. If the body has three degrees of freedom (x, y, and θ), then we consider torques about the center of mass. 2.3 Energy and Angular Momentum ￿θ2 Work-Energy. The work done by a torque is W= τ dθ (11) θ1 When calculating the work done, you have to either consider it as force or a torque. You can’t “double dip” and count it twice! However, sometimes it will be easier one way or the other. Kinetic Energy. The kinetic energy of a rigid body must take into account a translational kinetic energy and a rotational kinetic energy. As before, kinetic energy of a particle is defined as 1 1 2 KEi ≡ mi vi · vi = mi vi (12) 2 2 Now, in the consideration of v, we must include both the rotation and the translation. We have v = v + ω × ρi , where v is the center of mass velocity, ω is the angular velocity, and ρi ¯ ¯ is the position of particle i relative to the center of mass. Plugging this in to the definition of kinetic energy, and expanding, yields: 1 1 1 KEi = mi (v + ω × ρi ) · (v + ω × ρi ) = mi v 2 + mi v · (ω × ρi ) + mi (ω × ρi ) · (ω × ρi ) (13) ¯ ¯ ¯ ¯ 2 2 2 6 I have written out the proof of this, which I will post on the website. 7 To find the total KE, we sum the KE for each particle. Let us see what what happens when we sum the first term. ￿1 1￿ 1 mi v 2 = v 2 ¯ ¯ mi = M v 2 ¯ (14) 2 2 2 i i where M is the total mass. Notice that we were able to “pull out” the v 2 term because it ¯ is the same for every particle in the system. This is the translational component of energy. Now, we look at the second term. ￿ ￿ ￿ ￿ m i v · ( ω × ρi ) = v · ω × ¯ ¯ m i ρi = v · ( ω × v ) = 0 ¯ ¯ (15) i i We are able to say that this equals zero because, by properties of the cross product, (ω × v) ¯ is perpendicular to v, and the dot product of v with something perpendicular to it is zero. ¯ ¯ More conceptually, circling about the center of mass, some particles are going up, some are going down, some are going right, and some are going left. In this term, they all cancel each other out. Finally, we look at the last term. ￿1 i 2 mi (ω × ρi ) · (ω × ρi ) = 1￿ 1 2￿ 1¯ 2 mi ρ2 ωi = ωi mi ρ2 = I ω 2 i i 2i 2 2 i (16) With this, we have used the definition of the moment of inertia. This is the rotational component of energy. Thus, we have 1 1¯ KEtot = KEtrans + KErot = M v 2 + I ω 2 ¯ 2 2 (17) Keep in mind that we are using the inertia about the center of mass! Now, what about fixed axis rotation? In that case, the velocity of a particle i is just vO + ω × ri , where vO is the velocity of point O and ri is the distance from the fixed axis O to the particle. Notice, however, that vO = 0, since the axis is fixed. Then, we can calulate energy easily. 1 1 KEi = mi vi · vi = mi (ω × ri ) · (ω × ri ) 2 2 Summing over all the particles gives us, KE = ￿ i (18) KEi = ￿1 i 1￿ 1 2 mi ( ω × ri ) · ( ω × ri ) = ω 2 m i ri = I 0 ω 2 2 2 2 i (19) Notice that equations 17 and 19 are very similar in form. In fact, they give the same answer numerically. What, then, is the difference (aside from the obvious extra term of 1 M v 2 )? ¯ 2 The moments of inertia are different! One is calculated about the center of mass; the other is calculated about a fixed axis point O. If you are doing a problem with fixed axis rotation, ¯ it is better to use the 1 IO ω 2 representation. If there is no fixed axis, use 1 M v 2 + 1 I ω 2 . ¯ 2 2 2 8 Work-Energy Theorem. With this in mind, the work-kinetic energy theorem holds.7 1 1¯ 1 ∆KE + ∆P E = W where KE = M v 2 + I ω 2 or IO ω 2 ¯ (20) 2 2 2 Angular Momentum. This is a very brief and incomplete introduction to angular momentum. To really get into it, we need to talk about it as a vector. This will be done later. In the very simplest of cases, when a rigid body rotates about a fixed axis, the angular momentum L is equal to I ω . It is positive counter-clockwise (by the usual convention of right-handedness). Be careful! This only applies when ω is the angular velocity about an axis of rotation through CM. Also, L = I ω disregards the directional nature of L. However, it can still be useful. Suppose a pivoted rigid body is struck by a mass travelling in straight line. Just as τ = rF sin θ, the mass’s linear momentum p, becomes absorbed into the rotation of the rigid body, and change in the angular momentum is ∆L = rp sin θ. We never talk about an angular impulse, but the time derivative of the angular momentum is the torque. 2.4 Examples Figure 5: The Bottle Rocket Lab Set-Up. The Bottle Rocket. In Hesse Hall, there is a lab for Mechanical Engineering 107 called “The Bottle Rocket.” Basically, a large, upright disc is pivoted about its center. A 2 liter soda bottle is attached to the edge of the disc as shown in figure 5. The bottle is filled with water and pressurized. As, the highly pressurized water exits the bottle, the whole apparatus begins to spin. Given the following parameters, find the initial angular acceleration of the system, if the bottle is lined up vertically with the bottleneck pointing toward the ground (as shown in the figure). Mass and radius of the disc = M, R, mass of the bottle + the initial amount of water = mb , density of water = ρ, area of the bottleneck = A, initial speed at which the water exits the bottle = vw . Note that mb and vw change over time as the water exits the bottle; these values are given for the initial moment. PE is still potential energy, due to gravity and so on. We have yet to meet a force that depends only on rotation angle. There are torsional springs that do this, but there are few other examples. Universal gravitation is not a “conservative moment,” and neiter is a constant moment (unless you restrict the motion to planar rotation). The end of the story is that for moments, we will calculate using work rather than trying to think of “rotational PE.” 7 9 Figure 6: The Bottle Rocket Lab, Free Body Diagram and Thrust Due to Water. Answer: The Bottle Rocket. Begin by drawing a free body diagram, as shown in figure 6. For now, we will denote the force exerted by the pressurized water as Fw , but later, we will calculate it in terms of the given parameters. This is basic fixed axis rotation about ⊗. Calculate the torque about the axis. Notice that I have arbitrarily set clockwise as the positive direction of rotation. ￿ ￿ τ⊗ = r × F = −Rmb g + RFw + (0)M g = R(Fw − mb g ) = I⊗ α (21) R, the radius, happens to be the moment arm. If the bottle were not located at the outer edge of the disc, this moment arm would be less than R. We must now calculate the moment of inertia. To do that, we simply add the moment of inertia of the disc and of the bottle (which is a “point particle” pivoted at a radius R). ￿ ￿ 1 1 2 2 2 I⊗ = Ibottle + Idisc = mb R + M R = R mb + M (22) 2 2 Before we can go further, we must calculate the force of the water Fw . Imagine that in a very short time ∆t, a small volume of water ∆V with mass ∆m leaves the bottle. Since the water is leaving the bottle at vw and it travels for time ∆t, the length of the water that has exited is vw ∆t. The volume of water is ∆V = Avw ∆t, and the mass is ∆m = ρ∆V = ρAvw ∆t. Since everything was initially stationary, the change in velocity of the water was vw − 0 = vw , 2 and the change in momentum is ∆p = vw ∆m = ρAvw ∆t. The force of the water, then, is Fw = p = lim ˙ 2 ∆p (∆m)(∆v ) ρAvw ∆t 2 = lim = lim = ρAvw t →0 ∆ t t→0 t →0 ∆t ∆t (23) We substitute everything into the τ = I α equation and solve for α. ￿ ￿ 1 F w − mb g 2 ￿ ￿ R ( F w − mb g ) = I ⊗ α = R mb + M α → α = 2 R mb + 1 M 2 (24) Rolling with Bowling. A bowling ball with mass m and radius r is initially sliding without rolling at a velocity v0 . The floor has coefficient of dynamic friction µ. (A) Find the time t = t∗ when the ball stops slipping. (B) For t ≤ t∗ , plot the following: v vs. t, ω vs. t, KE vs. t. ¯ 10 Figure 7: Bowling Ball Rolling and Sliding on a Rough Surface. Answer: Rolling with Bowling. (A) Draw a force diagram, as seen in figure 7. Because the ball exhibits no vertical motion, ay = 0, which implies N = mg . Now, we know that the frictional force f = µmg in the −x direction. This is the only force along x. Let us calculate the acceleration in x. ￿ Fx = −µmg = max → ax = −µg ¯ ¯ (25) Notice that the over bar indicates this is the center of mass’s acceleration. Because ax is ¯ constant, we can find the velocity as a function of time using our basic kinematics formula. vx (t) = v0,x + ax t = v0 − µgt ¯ ¯ ¯ (26) Now, we wish to calculate the angular speed. Because it’s rolling and slipping, the angular and linear speed are not coupled. Thus, we must use torques to calculate the angular acceleration and kinematics to find the angular velocity. We take torques about the center of mass, noting that clockwise rotation has been arbitaritly designated as “positive.” ￿ τ= ¯ ￿ 2 ✿ ✘ ✘0 ✿ ✘1 ¯ r × F = 0 × mg + ✘✘sin(0) + r(µmg )✘✘✘✘ = rµmg = I α = mr2 α rN ✘✘ sin(90◦ ) 5 5µg . 2r (27) Thus, we can see that α = time using kinematics. From here, we can find the angular velocity as a function of ω (t) = ω0 + αt = 5µg t (28) 2r The condition for rolling without slipping is v = rω . At time t∗ , this condition will be true. v (t∗ ) = r ω (t∗ ) → v0 − µgt∗ = r 5µg ∗ t 2R → t∗ = 2 v0 7µg (29) (B) For the vvs.t plot, the velocity decreases linearly. It begins at v0 , of course, but what is the v (t∗ )? We can just plug in t∗ into the v (t) equation. ￿ ￿ 2 v0 5 ∗ vx (t ) = v0 − µg ¯ = v0 (30) 7µg 7 11 Thus, the ball’s linear velocity decreases to 5 of its original value. We can do the same for 7 angular velocity. ￿ ￿ 5µg ∗ 5µg 2v0 5 v0 ∗ ω (t ) = t= = (31) 2r 2r 7µg 7r We can calculate the KE in two ways. First, we can simply use expression for KE in terms of velocity. ￿ ￿2 1 2 1¯ 2 1 1 2 2 5µg 12 9 2 KE = mv + I ω = m (v0 − µgt) + mr t = mv0 − mv0 µgt + m (µgt)2 2 2 2 25 2r 2 4 (32) 2 2 Of course, KE (t = 0) = 1 mv0 , but the KE at t∗ is about .398mv0 . Also, note that KE 2 is parabolic. You can pick one more point to fully define the parabola, or you can simply realize that the initial slope is zero. The second way would be to calculate the work done by the frictional force, which I leave as an exercise. Figure 8: Plots of Bowling Ball’s Velocity v , Angular Velocity ω , and Kinetic Energy Over ¯ Time. The Swashbuckler. In an attempt to escape death / find treasure / get girls, Captain Jack Sparrow grabs the end of a rope connected to a pulley, and he cuts it with his saber. Also attached to the pulley is a sandbag with mass equal to the Captain’s mass m. However, the pulley is comprised of two pieces. The first is a disc of radius R and mass equal to the Captain, and the second has radius 2R with mass equal to one quarter of the Captain’s 12 mass. Sparrow’s rope is connected to the outer radius, and the mass is connected to the inner radius. Find the Captain’s acceleration using forces and torques. Repeat the problem using energy. Figure 9: Captain Jack Sparrow on a Pulley. Answer: The Swashbuckler. We begin by finding the moment of inertia of the pulley. We have to add the two moments of inertia for each of the two discs. We note that, in general, for a disc of mass M and radius R, I = 1 M R2 . 2 1m 1 (2R)2 + m(R)2 = mR2 24 2 If we call counter-clockwise rotation positive, and acceleration up positive, we have the following relationships between the accelerations of the pulley, the sand bag, and Jack. I = Iouter + Iinner = 2Rα = −aJack Rα = abag aJack = −2abag We now do the equation for the sum of the torques. ￿ τ = Iα 2RF2 − RF1 = mR2 α 2F2 − F1 = mRα We now do the sum of the forces on Jack and the sand bag. ￿ F = ma F2 − mg = maJack F1 − mg = mabag 13 Now, we combine the three equations by solving for F1 and F2 and plugging that into the torque equation. 2(mg + maJack ) − (mg + mabag ) = mRα Since we want Jack’s acceleration, we substitue abag and α with the appropriate geometric relations. ￿ ￿ 1 −aJack 2(mg + maJack ) − (mg − maJack ) = mR 2 2R Now solve for aJack , which we can just call a. 1 1 2mg − mg + 2ma + ma = − ma 2 2 mg + 2.5ma = −.5ma g a=− 3 We now do the same problem, but with energy. First, we have similar relationships between the velocities and between the positions. 2Rθ = −yJack 2Rω = −vJack Rθ = ybag Rω = vbag yJack = −2ybag vJack = −2vbag Now, we express the energy of motion as the sum of KE + RE . 12121 KE + RE = mvJ + mvb + I ω 2 2 2 2 The potential energy is due to gravity. P E = mgyJack + mgybag To find the total energy, we combine kinetic, rotational, and potential energy. 12121 E = KE + RE + P E = mvJ + mvb + I ω 2 + mgyJack + mgybag 2 2 2 Now, we put everything in terms of Jack’s velocity and position, which we abbreviated y ˙ and y , respectively. ￿ ￿2 ￿ ￿2 ￿￿ 121 1 1 y ˙ −y 2 E= my + m − y + mR − ˙ ˙ + mgy + mg 2 2 2 2 2R 2 321 = my + mgy ˙ 4 2 This expression contains velocity and position terms. We want acceleration. Thus, we take ˙ the time derivative. Since no work is done, energy is constant and E = 0. (Make sure to remember the chain rule when differentiating!) 3 1 ˙ E= 2my y + mg y = 0 ˙¨ ˙ 4 2 0 = 3my + mg ¨ g y=− ¨ 3 14 The Toppling Stick. A stick of mass m and length L is placed upright with one end, which we label point P , resting on a table. The stick begins to topple with negligible initial velocity. We want to analyze what happens as the stick falls. (A) In this case, the point P does not slip on the table. What is the speed of the tip vtip as the stick crashes onto the table? (B) What is the mininum coefficient of static friction µs such that P will never slip? (C) Now, assume that the table is frictionless. Express the velocity vectors of the tip (vtip ), the center of mass (v), and the point P (vP ) as functions of the angle of rotation. ¯ Assume that when the stick is upright, θ = 0, and that when the stick hits the table, θ = π . For this problem, it is more useful to define θ as the angle from the x axis and 2 counter-clockwise as positive rotation, as is the standard convetion. The Toppling Stick. (A) For this question, it is simple enough to use the conservation of energy. We use the fixed axis expression for energy (where P is our fixed axis), KEP = 1 IP ω 2 . 2 For a rod rotated about its end, Irod,end = 1 mL2 . For potential energy, we must note that 3 the change in height is the change in height of the center of mass. The CM is initially located at L/2 directly above P (which is its initial height) and then falls to 0. We write out the conservation of energy. ∆KE + ∆P E = 0 (33) ￿ ￿ ￿ 11 L mL2 ω 2 + mg − =0 (34) 23 2 ￿ 3g ω= (35) L To find the velocity of the tip, we just use our handy relationship between angular and linear velocity. Remember, vP = 0 because the stick is not slipping. ￿ ￿ 3g ✯0 vtip = ￿✟P + ω × rP →tip ￿ = ω rP →tip = v✟ ·L= 3gL (36) L (B) We must draw a force diagram, as shown in figure 10. Since P is a fixed point (assuming that stick does not slip), we can take the torque about that point. We will call clockwise rotation positive for this problem. ￿ ￿ ￿ L τP = 0 × f + 0 × N + ¯ × −mg j = r⊥ mg = r sin θ mg (37) 2 We now calculate the angular acceleration. ￿ mgL 1 3g τP = I P α → sin θ = mL2 α → α = sin θ (38) 2 3 2L From here, we can calculate the acceleration of the center of mass. We must be careful to note the x and y components of the acceleration. ￿ L 3g 3g ￿ ¯ = α × ¯ = rα (cos θi − sin θj) = · a r¯ sin θ (cos θi − sin θj) = sin θ cos θi − sin2 θj 2 2L 4 (39) ￿ 15 Figure 10: Free Body Diagram of a Toppling Stick, No Slipping at P . At this point, we can use Newton’s second law to relate the forces in the FBD to the acceleration of the center of mass, which we just found. ￿ ￿ ￿ ￿ ￿ 3g 32 2 Fy = N − mg = may = −m sin θ → N = mg 1 − sin θ (40) 4 4 ￿ 3 3 Fx = f = max = mg sin θ cos θ = mg sin(2θ) (41) 4 8 Although the frictional force obtains a maximum at θ = π , this might not tell us where we 4 need our maximum µs . If the normal force is particularly strong at that point, we can have a weaker µs and still have no-slip. Instead, we use the fact that f ≤ µs N , and correspondingly, f µs ≥ N . This must be true for every possible rotation angle, so we find what the maximum value of f /N is, and that is the minimum value for µs . Unfortunately, this requires some fairly ugly arithmetic. ￿￿ 3 sin(2θ) f df 3 cos(2θ) 9 sin(θ) cos(θ) cos(2θ) 8 ￿; ￿+ =￿ =￿ ￿ ￿2 = 0 (42) 2 2 3 3 N dθ N 1 − 4 sin θ 4 1 − 4 sin θ 16 1 − 3 sin2 θ 4 This is, of course, horrendous to try to solve by hand. At this point, I have used￿ √CAS a ￿ (specifically,, to find that the maximum occurs at θ = 2 arctan 52−1 and that this maximum is ￿f ￿ N max 3 . 4 (C) There is a trick to this problem. If there is no friction, then the only two forces on the stick are the normal force and gravity. These forces are both in the vertical direction, meaning that there are no forces in x. Thus, we conclude that linear momentum in the x direction is conserved. This means that ax = 0, and because the motion starts from rest, ¯ vx = 0. Thus, the CM falls straight down. We also know that P is constrained to move on ¯ the table, so it cannot move up and down. Thus, vP,y = 0. We express the positions and = 3 . Thus, in order to prevent slipping, µs ≥ 4 16 velocities of P and CM as follows. L L ˙ ¯ = sin(θ)j r v = cos(θ)θj = −v j ¯ ¯ (43) 2 2 L L ˙ rP = − cos(θ)i vP = sin(θ)θi = −vP i (44) 2 2 where v and vP are the speeds of the CM and P, respectively, and θ is the angle as measured ¯ counter-clockwise from the +x axis. Note that because the angle decreases from π /2 to 0 as ˙ the stick falls, θ is negative, and the signs are correct. We also have the relative position of P with respect to the CM. L rP,rel = rP − ¯ = − [cos(θ)i + sin(θ)j] r (45) 2 ˙ We now realize that the angular velocity is ω = thetak = ω k. It is important to realize that ω is negative, since the rod actually rotates clockwise. We set up the usual relationship, L L L [cos(θ)i + sin(θ)j] = −v j − ω cos(θ)j + ω sin(θ)i ¯ (47) 2 2 2 By matching terms, we see that 1 1 v = − ω L cos θ ¯ vP = − ω L sin θ (48) 2 2 Note that despite the negative signs above, both v and vP are positive because ω is negative. ¯ Finally, we establish energy conservation as follows. − vP i = − v j + ω k × − ¯ P Ei + KEi = P Ef + KEf (49) L L 1 1¯ mg + 0 = mg sin θ + mv 2 + I ω 2 ¯ (50) 2 2 2 2 ￿ ￿￿ ￿2 L 1211 2¯ v 2 mg (1 − sin θ) = mv + ¯ mL − (51) 2 2 2 12 L cos θ ￿ ￿ ￿ ￿ 1 1 + 3 cos2 θ 2 2 gL(1 − sin θ) = v 1 + ¯ =v ¯ (52) 3 cos2 θ 3 cos2 θ ￿ ￿ 2 ) This means that v = 3gL(1−sin θ2 cos θ , or, when θ = 0, v = 3 gL. Thus, from equation 48, ¯ ¯ 1+3 cos θ 4 we know that ω = −2¯/ cos θ = −2¯, and thus vP = − 1 ω L sin θ = 0. Finally, to find vtip , v v 2 ￿ ￿ 3 3 L vtip = v + ω × rtip = − ¯ gLj + (−2) gLk × i 4 4 2 ￿ ￿ ￿ 3 3 3 =− gLj − L gLj = −(1 + L) gLj (53) 4 4 4 Thus, the final velocities, in vector form, are: ￿ ￿ 3 3 vP = 0 v=− ¯ gLj vtip = −(1 + L) gLj (54) 4 4 17 vP = v + ω × rP,rel ¯ (46) Conservation of Momenta. A bullet of mass m is fired at velocity v1 . It strikes a uniform thin rod of mass m and length d halfway between the center of mass and the end point. Assuming the bullet sticks to the rod, what are the linear and angular velocities after the collision? Answer: Conservation of Momenta. First, let us conserve linear momentum. p1 = p2 mv1 = (2m)v 1 v= v1 2 This is the same result as if two carts of equal mass had collided and stuck. Notice that we found the velocity of the center of mass. This is the CM of the new bullet-rod system. We can calculate the location by using the center of mass equation: ￿ mi yi m(d/2) + m(d/4) 5d y= ¯ = = M 2m 8 In other words, the new center of mass is half-way between the bullet and the rod’s CM. Now we￿must calculate the new moment of inertia. For the bullet, this is easy. Ibullet = mr2 = ￿2 2 m d = md . For the rod, we consider it to be two rods, one of length 3d/8 and mass 3m/8 8 64 and the other of length 5d/8 and mass 5m/8, both pivoted about their edge (which is on the CM). Thus ￿ ￿ ￿ ￿2 ￿ ￿ ￿ ￿2 15 5d 13 3d 19 Irod = m + m = md2 38 8 38 8 192 11 2 The total moment of inertia is 96 md . Now we must also conserve the angular momentum. Initially, stick has no angular momentum. The final angular momentum comes from the bullet. We “convert” the linear momentum of the ball by multiplying the moment arm (CM to the point of impact) times the component of linear momentum perpendicular to the rod. d d L1 = × m(v1 )⊥ = mv1 8 8 ¯ The angular momentum after the collision L2 is equal to I ω . We just set that equal to L1 and solve for ω . L 1 = L2 d 11 ¯ mv1 = I ω = md2 ω 8 96 12v1 ω= 11d Just a few things to think about... What if the bullet had struck the rod at an angle? How does the amount of energy lost in the collision relate to how far from the CM the bullet struck? (Think: If it hit the CM, what would the final velocities have been, and would that have accounted for more or less final energy?) 18 3 3.1 General Rotation Math Review. Figure 11: Cross Product of Vectors a and b. The cross product of two vectors produces a third vector, T : R3 × R3 → R3 . The cross product is defined to have the following properties. a×b ⊥ a a×b ⊥ b ￿a × b￿ = ￿a￿ · ￿b￿ sin θ (55) (56) (57) With these three properties, the three components of a × b can be determined. The easiest way to remember them is as a “determinant”. (Cross products have nothing to do with the determinant of a matrix; this is just a little trick for remembering!) ￿ ￿ ￿i j k￿ ￿ ￿ a × b = ￿ ax ay az ￿ (58) ￿ ￿ ￿ bx by bz ￿ Note that as in figure 11, the cross product is shown perpendicular to the original two vectors with magnitude equal to the product of the magnitude of the other two vectors times the sine of the angle between them. However, there are two vectors that satisfy this equation. One points “up” in figure 11, and the other (shown with a dotted line) points “down.” We artibrarily select one of them to be the actual cross product a × b. The other one is −(a × b), or, equivalently, b × a. We use the right hand rule to determine this. There are several versions of the right hand rule. The first says, “Using your right hand, point your index finger in the direction of a, point your middle finger in the direction of b, and your thumb will point toward a × b.” I prefer a slight variation on this: “Using your right hand, point all of your fingers toward a, curl the fingers toward b, and your thumb will point toward a × b.” 19 3.2 Kinematics: 3D Angular Velocity and Accelertion If a point on the rigid body is located at rO from a fixed point O, and the angular velocity of the rigid body about that point is ω , then the linear velocity is v = ω × rO ω = rO × v . (59) If the center of mass is moving at v, then the velocity at any point located at ρ, the position relative to the CM, is v = ω × ρ + v. (60) Thus, the acceleration of a point is ˙ a = v = α × rO + ω × v = α × rO + ω × ( ω × rO ) ˙ a = v = α × ρ + ω × ρ + a = α × ρ + ω × ( ω × ρ) + a ˙ (61) (62) 3.3 Kinetics: Angular Momentum Definition of Angular Momentum. The angular momentum of an element is the cross product of the moment arm and the linear moment of the particle. The net angular momentum is the sum of all of these little angular momenta. ￿ ￿ Angular Momentum = LO = ri × pi = ri × mv i (63) i i When we consider a rigid body, this equation reduces to ￿ ￿ LO = r × p dm = r × ρv dV R R (64) The angular momentum about the center of mass is ￿ ￿ L = π × p dm = π × ρv dV R R (65) A little bit of algebra shows that L O = L + r × p = L + r × mv If we take the time derivative of the angular momentum about O, we get ￿ ￿ ￿ ￿ ˙O = d L r × mv dm = v × mv + r × ma dm = r × F dm = τO dt ￿ R R R (66) (67) Similarly, dL = τCM . The moment of inertia for this actually appears as a 3 × 3 matrix, dt and L = Iω . Unfortunately, this means that ω is not necessarily parallel to L.8 However, in the special case of rotation in a plane, L = I ω . 8 In fact, this is true only if ω is an eigenvector of L. 20 ˙ Conservation of Angular Momentum. Since L = τ , if there are no external torques on a system, the angular momentum is constant. If the rotation is limited to the planar case, then the quantity I ω is conserved for the system. Let’s Shift Gears... As a biker switches gears, the chain disengages from one gear and switches to another. The pedal-bottom bracket-front gear assembly for her bike has a moment of inertia of .18 kg · m2 , and the front gear has a radius of .16 m. The back gear has a moment of inertia of .04 kg · m2 and a radius of .04 m. If the pedal assembly initially has a angular velocity of 12 rad/s and the back gear is initially at rest, what is the angular velocity of each wheel after the chain engages? Answer: Let’s Shift Gears... First of all, since the chain is inextensible, the velocities of the two points that make contact with the chain are the same. v1 = v2 → r1 ω1 = r2 ω2 → ω1 r2 .04 1 = = = ω2 r1 .16 4 ω2 + .04ω2 4 (68) Also, we can use the fact that angular momentum is conserved. I1 ωi = I1 ω1 + I2 ω2 → (.18)(12) = .18ω1 + .04ω2 = .18 → (69) This leads to ω2 = 25.4 rad/s and ω1 = 6.35 rad/s . Example: Aaron Dancing. One of Aaron’s hobbies is Israeli Dancing, and one of his favorite dances is Hora Mechudeshet. During the end of this dance, you spin around with your leg sticking out, and then you bring it in. Assuming that Aaron’s leg makes up 10% of his total mass and that the length of his leg is approximately twice as long as his “diameter” (assuming Aaron is roughly cyclindrical), by what factor does his angular velocity change? Did his kinetic energy change, and if so, where did the energy come from / go to? Answer: Aaron Dancing. We need to find the change in Aaron’s moment of inertia. If he has mass m, then when he has his leg sticking straight out, his body’s moment of inertia is roughly the same as a disc: Ibody = 1 (.9m)R2 . His leg is like a thin rod rotating 2 about an axis R from the end of the rod. We know that a rod pivoted about it’s CM is 1 ICM = 12 mL2 , but we must use the parallel axis theorem to pivot about Aaron’s central 1 axis. Irod = 12 (.1m)(4R)2 + (.1m)(3R)2 = 1.033mR2 and the net moment of inertia is (1.033 + .45)mR2 = 1.483mR2 . The initial angular momentum, therefore is L1 = I1 ω1 = 1.483mR2 ω1 . After he brings his leg in, Aaron is a cylinder of mass m and radius R, so I2 = .5mR2 , and his final angular momentum is L2 = I2 ω2 = .5mR2 ω2 . Since momentum is conserved, L1 = L 2 1.483mR ω1 = .5mR2 ω2 ω2 = 2.966ω1 2 21 His angular velocity increased 2.966 times. The extra energy came from the work he did to pull his leg in. Gyroscopic Motion. We shall explore this by way of an example. A top is (roughly) a cone. Suppose we have a top of radius R and height h = 3R. The mass is m. The cone is ˙ inclined at some angle φ and spinning about it’s primary axis at θ. We wish to find the rate at which it will precess. In other words, what is the rate that the primary axis will rotate about a vertical axis? Figure 12: Precession of a Spinning Top. 3 We first note that the cone’s moment of inertia is I = 10 mR2 and that it’s center of mass is located 3 h = 9 R along the primary axis as measured from the tip of the cone. We now 4 4 notice that the cone experiences only two forces: gravity and normal. Since the top does not fall N = mg . Calculating the torque about the center of mass, 9 ￿τCM ￿ = ￿r × −mg j￿ = Rmg sin φ 4 We recall the relationship between momentum and torque: dL = τ dt = 9 Rmg sin φ dt. This 4 change in momentum vector is perpendicular to the original momentum vector. Thus, the magnitude of the angular momentum remains constant, but the angle changes. As can be seen in figure 12, there is a small change in ψ . We can relate L and ψ by ￿ ￿ 9 9 dψ ˙ = 4 Rmg dψ (L sin φ) = dL = τ dt = Rmg sin φ dt → =ψ 4 dt L ˙ Finally, we do say that L = I θ = 3 ˙ mR2 θ, 10 which is an approximation. = 45g ˙ 6R θ ˙ ψ= 9 Rmg 4 3 ˙ mR2 θ 10 In this calculation, we have made several approximations. When are they valid? 22 1. We assumed that angular momentum is perpendicular to the primary axis of the top. In fact, this is not true. If it were, the axis would not move, and the top would stay at the same ψ (clearly there would have to be another torque to counteract the gravitynormal couple). However, as it turns out, the change in angular momentum is still perpendicular to L when the precession is taken into account. ˙ 2. We assumed that L = I θ. First, it is not necessarily true that L and omega point in the same direction. However, because θ is the rotation about one of the (presumably) principle axes, this is a special case in which it is true. Second, we have ignored the ˙ ˙ ˙ contribution of ψ . We justify this in cases when θ >> ψ . 3.4 Center of Percussion: The Sweet Spot If a rigid body is struck, the center of mass will move and the whole rigid body will start to rotate. However, there will be one point on the object that, as a result of the combined effect of translation and rotation, will have no acceleration or change in velocity (at least initially). In many cases, this is desirable. For people in a car, you would like them to be seated such that they do not experience a “jolt” if the car hits a bump. A tennis player would like to hold the racket at a point where he or she cannot feel a “jolt” when the ball is struck. If you strike a rigid body at the center of percussion, the velocity experienced is zero. Example: Conservation of Momenta, Part II. Earlier in these notes, we shot a bullet at a rod. The bullet had mass m and muzzle speed v1 . The rod had mass m and length d. The bullet hid the rod halfway between the center of mass and the edge of the rod. We found that the resulting center of mass velocity was v = 1 v1 i and the angular velocity was 2 12v1 (−k). If the bullet struck the rod at the center of percussion, find the zero velocity point. 11d Figure 13: After the Rod was Struck by the Bullet. Answer: Conservation of Momenta, Part II. At a position r below the CM, as indicated in figure 13, the velocity relative to the CM is ω × r. In this case, it is − 12vd1 ri. 11 The velocity of the CM is .5v1 i. Thus, the net velocity at that point is 1 12v1 v = v + ω × ρ = v1 i − ri 2 11d 23 11 d . At this point, .4583 of the total length 24 below the CM, the velocity instantaneoulsly experienced by the rod is zero. Setting this equal to zero, we find that r = 3.5 Extra Practice Figure 14: Hammerhead Shark Ride at Six Flags Discovery Kingdom. Hammerhead Shark. As shown in the image9 of figure 14, the Hammerhead Shark ride at Six Flags Discovery Kingdom consists of two arms that are suspended from a central support. A big motor forces the arms to turn. We will use a simplified approach where the arms have length λ and the distance of the arms from the central support is also λ. We will assume all of the mass is concentrated in the rider seating areas, and that each of these has mass m. In reality, the two arms rotate in opposite directions. We shall call θ the spin angle and ψ the angle about the vertical. (A) Why do the arms swing about in opposite directions? Keep in mind that in reality, the support does not allow the arm to precess and that the spin velocity changes. (B) Now, the arms were swung in the same direction. The arms start from rest and reach their maximum speed of 13rpm (1.361 rad/s) in 12.5 seconds. If, m = 7500kg, δ = 8 m, what is the bending moment (i.e., the torque that tends to cause a rod to bend) applied to the rod? (C) Consider if there is only one arm, but the arm is now allowed to pivot about the ˙ support (i.e., ψ ￿= 0). For this portion of the problem, we will assume that the motor acts to ˙ keep spin angle rate θ constant. At the moment when the arm is completely horizontal and rising, calculate the angular momentum vector and the angular velocity vector. Are they parallel, and if so, why? If this ride were in outer space, would the arm precess? On planet ˙ earth, what is the precession rate ψ in terms of the specified variables? Answer: Hammerhead Shark. (A) If the arms both swing counter-clockwise, the direction of their angular momenta will be in the same direction. If the spin velocity increases, 9 Courtesy 24 the angular momenta will increase by the same amount in the same direction, causing a torque (the “equal and opposite” torque) on the support. When they swing in opposite directions, the angular momenta are equal and opposite, causing the net angular momentum to be always zero, and thus, there is no net torque on the support.10 (B) This is a fixed axis rotation, as can be seen in figure 15. From the given information, Figure 15: Hammerhead Shark Ride Model: Arms Swinging in the Same Direction. we can calculate the angular acceleration. We use the kinematics equation involving speed, acceleration, and time to solve. ω = ω0 + αt 1.362 rad/s = α(12.5 s) α = .1090 rad/s2 In this specific case, LO = IO ω . This is true because it is a fixed axis rotation. Thus, ˙ ˙ LO = I θ = mδ 2 θ. To find the torque, we simply take the time derivative of the angular momentum. dLO d ￿ ˙￿ ¨ τO = = I θ = mδ 2 θ = (7500)(8)2 (.1090) = 52.3 kN · m dt dt This torque is called a bending moment. That is, it causes the support beam to bend in the direction shown in figure 15. When the arms swing in opposite directions, there is no bending moment. (C) First, we want to find the linear velocity of the mass. We shall say that the mass is at point S, and we shall define another point Q for convenience. Q is located at the “bend” in the arm, as shown in figure 16. We do this because it is easy to calculate the velocity of Q and the velocity of S relative to Q. Even when the ride spins, point Q does not move up and down. Point Q can only move around the central axis (dotted line). It does so at a rate ˙ equal to the precession ψ . Thus, the position of Q is rQ = δ (i cos ψ + j sin ψ ) = δ j, and the ˙ ˙ velocity of Q is simply the derivative of this, vQ = δ ψ (−i sin ψ + j cos ψ ) = −δ ψ i. What is the velocity of S relative to Q? Relative to Q, S just travels in a circle, so rS rel Q = δ (er and Of course, there are two equal and opposite torques, one from each arm, acting on the support. This causes the metal in the suport to be compressed. However, metal can withstand compression much better than it can stand bending, as in the case when the arms swing in the same direction. 10 25 Figure 16: Hammerhead Shark Ride Model: One Arm Allowed to Precess. ˙ vS rel Q = δ θ(eθ . Note that er and eθ are vectors that point from Q to S and tangent to the circle about which S “orbits” Q. These vectors change with ψ and θ, but at this particular ˙ moment, rS rel Q = −δ i and vS rel Q = δ θk. We can now compose the vectors rS and vS , the position and velocity of S. rS = rQ + rS rel Q = δ j + −δ i = δ (−i + j) ˙ ˙ ˙ ˙ vS = vQ + vS rel Q = −δ ψ i + δ θk = δ (−ψ i + θk) To find the angular momentum of a particle, we cross the moment arm rS with the linear momentum pS = mvS . ￿ ￿ ￿ ￿ ˙ ˙ ˙ ˙ L O = rO × mv = m δ ( − i + j ) × δ ( − ψ i + θ k ) = mδ 2 θ ( i + j ) + ψ k ￿ ˙ ˙ The magnitude of this is mδ 2θ2 + ψ 2 . To find the angular velocity, we can just compose ￿ ˙ ˙ the two rotations, as earlier was stated that ω = 3=1 γi gi . Therefore, ω = θj + ψ k. Clearly ˙ i the momentum and velocity are not parallel in this case. To find the precession more easily, ˙ ˙ we’re going to make the approximation that θ >> ψ , and our angular momentum reduces 2˙ to mδ θ(i + j). What is the torque acting on the system? The force is −mg k, so τ = r × F = δ (−i + j) × −mg k = mg δ i + j We know that dL = τ dt, and, as shown in figure 16, this is perpendicular to L. Thus, dL = mg δ dt. We can relate the little change in precession angle dψ to dL by L dψ = dL. ￿ ￿ ￿ 2 ˙ ˙ L dψ = mδ 2θ2 + ψ 2 dψ = dL = mg δ dt √ ˙ ≈ mδ 2 θ 2 dψ = mg δ dt dψ mδ g g ˙ √= √ =ψ = ˙ ˙ dt mδ 2 θ 2 mδ θ 2 In space, however, there would be no gravity and thus no precession. These notes ©2010 by Aaron Alpert. 26 ...
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