484 CHAPTER 8. Advanced topics in circular motion SAMPLE 8.15 The calculation of H˙ using the moment of inertia matrix. For the crooked plate considered in Sample 8.7, find the rate of change of angular momentum H˙ O, using the moment of inertia [I O], calculated in the x yz coordinate system Solution We calculate H˙ O using the following formula: H˙ O = [I O] · ω˙ HO 0 +ω × ([I O] · ω . Note that pointO is the center of mass of the plate. Let us write the expression for the angular momentum H O in matrix form. Hx/O Ixx Ixy Ixz 0 = Hy/O Ixy Iyy Iyz 0 . Hz/O Ixz Iyz Izz ω It is clear from the components of ω that we only need the last column of [I ] matrix since other elements of [I ] will multiply with zeros of ω vector. Carrying out the multiplication we get H0x Ixzω H0y = Iyzω H0z Izzω or, in vector form H O = Ixzωıˆ + Iyzωˆ + Izzωkˆ. Therefore, H˙ O = ωkˆ × ω(Ixzıˆ + Iyzˆ + Izzkˆ) = −ω2 Iyzıˆ + ω2 Ixzˆ. But, Ixz = − x z dm and Iyz = − yz dm mm and x = w sin φ, y = y, z = w cos φ (see Fig. 8.29), hence H˙ O = −ω2 (−wy cos φıˆ + w2 sin φ cos φˆ) dm m = b/2 /2 ω2(−w2 sin φ cos φˆ + wy cos φıˆ) m dw d y b −b/2 − /2 which is the same inetgral as obtained in Sample 8.7 for H˙ O. Therefore, the result is also the same: − mω2 2 12 H˙ O = sin φ cos φˆ. H˙ O = − mω2 2 sin φ cos φˆ 12
8.4. Mechanics using [I cm] and [I O] 485 SAMPLE 8.16 Direct application of formula: A rectangular plate is mounted on a massless shaft with the center of mass of the plate on the shaft axis. The shaft rotates about its axis with angular acceleration ω˙ = 0.5 rpm/ s(ıˆ + ˆ). At the instant of interest, the angular velocity is ω = 100 rpm(ıˆ+ˆ) and the components of the moment of inertia matrix of the plate are Ixx = 2 kg·m2, Iyy = 4 kg·m2, Izz = 6 kg·m2 and Ixy = Iyz = Ixz = 0. (a) Find the angular momentum of the plate about its mass-center and show that it is not in the same direction as the angular velocity. (b) Find the net moment acting on the plate. Solution We are given the angular velocity, the angular acceleration, and the moment of inertia matrix of the plate: ω = 100 rpm(ıˆ + ˆ) = 10.47 rad/s(ıˆ + ˆ) ω˙ = 5 rpm/ s(ıˆ + ˆ) = 0.52 rad/s2(ıˆ + ˆ) 200 [I cm] = 0 4 0 kg·m2. 006 (a) The angular momentum: The angular momentum of the plate is H cm = [I cm] · ω 200 10.47 = 0 4 0 kg·m2 · 10.47 rad/s 0 0 6 0 = (20.94ıˆ + 41.88ˆ) kg·m2 · s−1 N·m· s = (20.94ıˆ + 41.88ˆ) N·m · s. H cm = (20.94ıˆ + 41.88ˆ) N·m · s There are many ways of showing that H cm is not parallel to ω. We can simply draw the two vectors and show that they are not parallel We can, alternatively, take the cross product of the two vectors: H cm × ω = (20.94ıˆ + 41.88ˆ) N·m · s × 10.47 rad/s(ıˆ + ˆ) ıˆ ˆ kˆ = 20.94 41.88 0 N·m 10.47 10.47 0 = kˆ(219.24 − 438.48) N·m = −20.94 N·mkˆ which is not zero, implying that the two vectors are not parallel. (b) The net moment: The net moment on the plate can be found by applying angular momentum balance: M/cm = H˙ cm = ω × [I cm] · ω +[I cm] · ω˙ H cm = 20.94 N·mkˆ + 1.04 N·mıˆ + 2.08 N·mˆ ω×H cm [I cm]·ω˙ M/cm = (1.04ıˆ + 2.08ˆ + 20.94kˆ) N·m
486 CHAPTER 8. Advanced topics in circular motion g B y' SAMPLE 8.17 A rod swings around a tipped axis in 3-D. A rigid shaft of negligible z C' φ C y mass is connected to frictionless hinges at A and B. The shaft is tipped from the z' horizontal plane (x y-plane) such that the shaft axis makes an angle γ with the vertical Looking axis. A uniform rod OC of mass m and length is welded to the shaft. At time t = 0, γ O down the the rod is tipped by a small angle φ from its position OC in the yz (or y z ) plane. γ y'-axis Find the equation of motion of the rod. z' A φ x,x' x' Solution Let ıˆ , ˆ , kˆ be the basis vectors associated with the x y z coordinate Figure 8.54: A rod, welded to a tipped system. Since y z axes can be obtained by rotating yz axes counterclockwise about shaft AB, swings around the shaft axis if it the x-axis by an angle 90o − γ , we can relate the basis vectors of the two coordinate is tipped slightly from the vertical plane yz. systems with the help of Fig. 8.55: (Filename:sfig5.5.2) ıˆ = ıˆ, ˆ = sin γ ˆ + cos γ kˆ, and kˆ = − cos γ ˆ + sin γ kˆ. (8.51) z' z γ γ y' The free body diagram of the shaft with rod OC is shown in Fig. 8.56. We γ y γ can write angular momentum balance for this system about any point on the axis of x,x' rotation AB. However, rather than writing angular momentum balance about a point, Figure 8.55: y z axes are obtained by let us write angular momentum balance about axis AB. This ‘trick’ will eliminate rotating yz axes counterclockwise about the reactions R A and RB from our equations. Angular Momentum Balance about axis x-axis by an angle 90o − γ . AB is: λˆ AB · [ MO = H˙ O] (Filename:sfig5.5.2a) or ˆ · MO = ˆ · H˙ O. Calculation of (ˆ · MO) : Since R A and RB pass through axis AB, they do not produce any moment about this axis. Therefore, z' y' φC B ˆ · MO = ˆ · [r G/O × mg(−kˆ)] G RB = ˆ · L (sin φıˆ + cos φkˆ ) × mg(−kˆ) 2 O x' mg ˆ using Eqn.(8.51) A = (sin γ ˆ + cos γ kˆ) · L mg(sin φˆ + cos φ cos γ ıˆ)] RA [ 2 Looking z' down the = mg sin φ sin γ φ y'-axis 2 G x' Calculation of ˆ · H˙ O : Since the rod rotates about axis AB, we may write mg ω = φ˙ˆ , Figure 8.56: Free body diagram of the ω˙ = φ¨ˆ . bar and shaft system. In the inset, only the Now, H˙ O = [I O]{ω˙ } + ω × [I O]{ω} weight of the bar, mg, is shown for clarity where of geometry. (Filename:sfig5.5.2b) [I O]{ω˙ } = [I O]x y z {ω˙ }x y z
8.4. Mechanics using [I cm] and [I O] 487 Ix x Ix y Ix z 0 Iy y φ¨ = Ix y Iy z 0 Iz z Ix z Iy z Ix y φ¨ Iy y φ¨ = Iy z φ¨ , similarly, Ix y φ˙ ⇒ Iy y φ˙ Iy z φ˙ [I O]{ω} = , ω × [I O]{ω} = φ˙ˆ × [Ix y φ˙ıˆ + Iy y φ˙ˆ + Iy z φ˙kˆ ] = −Ix y φ˙2kˆ + Iy z φ˙2ıˆ . Therefore, ˆ · H˙ O = ˆ · [(Ix y φ¨ıˆ + Iy y φ¨ˆ + Iy z φ˙kˆ ) + (−Ix y φ˙2kˆ + Iy z φ˙2ıˆ )] = Iy y φ¨. Now, setting ˆ · MO = ˆ · H˙ O, we get L mg sin φ sin γ = Iy y φ¨ 2 or φ¨ = mg (L/2) sin γ sin φ Iy y or φ¨ = C sin φ where C = mg (L/2) sin γ . For rod OC, Iy y Iy y = (x 2 + z 2) dm = L l2 m dl = 1mL2. 0L 3 m Therefore, the equation of motion of the rod is φ¨ − mg (L/2) sin γ sin φ = 0 (1/3)m L2 φ¨ − 3g sin γ sin φ = 0. 2L φ¨ − 3g sin γ sin φ = 0 2L
488 CHAPTER 8. Advanced topics in circular motion y SAMPLE 8.18 Kinetic energy in 3-D rotation. A thin rod of mass 2 kg and length L = 1 m is welded to a massless shaft at an angle θ = 45o. The shaft rotates about 2 its longitudinal axis (y-axis) at 100 rpm. The moment of inertia matrix of the rod ω about the weld point O is 45o rod of O m = 2 kg 1 −1 0 & L = 1/2m [I O] = 0.08 kg· m2 −1 1 0 . x 0 02 z shaft Find the kinetic energy of the rod. Figure 8.57: (Filename:sfig7.4.2) Solution The rod rotates about the fixed point O with angular velocity ω = ωˆ where ω = 100 rpm = 10.47 rad/s. The kinetic energy of a rigid body rotating about a fixed point O is given by EK = 1 ω· [I O]·ω 2 1 This step is simply to facilitate compu- For the given problem, let us write the moment of inertia matrix [I O] of the rod as 1 tation. We carry out the multiplication and at the end substitute the value of K0. 1 −1 0 where K0 = 0.08 kg· m2. [I O] = K0 −1 1 0 0 02 Now ω = ωˆ = {0 ω 0}T . 1 −1 0 0 −ω Therefore, [I O]·ω = K0 −1 0 1 0 ω = K0 ω 0 2 0 0 = −K0ωıˆ + K0ωˆ. Substituting the expression in the formula for EK we get EK = 1 ωˆ·(− K 0ωıˆ + K0ωˆ) 2 = 1 K 0ω2 2 = 1 ·0.08 kg· m2·(10.47 rad/s)2 2 = 4.38 N·m = 4.38J EK = 4.38J
8.5. Dynamic balance 489 8.5 Dynamic balance Sometimes when something spins it can shake the structure that holds it. A familiar example is an ‘unbalanced’ car tire. But also important are unbalanced rotors in all kinds of machines. Most often one seeks to eliminate or minimize the imbalance. For example the strange design of car crank-shafts is due in large part to an attempt to minimize its imbalance. But what does it mean, in the language of mechanics, to say a rotating body is ‘unbalanced’? It means that forces and/or torques are required to hold the axis still while the object spins at constant rate. Going back to the basic momentum balance equations: Linear momentum balance: F = L˙ and MC = H˙ C, Angular momentum balance: we see that forces and torques are required if L˙ = 0 or if H˙ C = 0. So imbalance means L˙ and/or H˙ C is not zero. We break the concept of balance into the following two concepts: (1) An object is said to be statically balanced with respect to a given axis of rotation if L˙ = 0 when it spins at constant rate about that axis. (2) An object is said to be dynamically balanced with respect to a given axis of rotation if both L˙ = 0 and H˙ = 0 for constant rate rotation about that axis. The origin of the words ‘static’ balance and ‘dynamic’ balance is in how the im- balance can be measured. Static imbalance can be measured with a statics experiment, dynamic imbalance requires a dynamics experiment. Static balance If the center of mass of a rigid body is on the fixed axis of rotation then it will not accelerate. Thus, L˙ = acmmtot = 0 and the object is statically balanced. An equivalent definition of static balance is that the net force on the spinning body is zero. Whether or not this net force is so can be tested with a statics experiment. Put the axis of rotation on good bearings and see if the mass hangs down in any preferred direction. In a tire shop, this kind of balancing is something called bubble balancing because of the bubble in the level measuring device. Dynamic balance The first condition for dynamic balance of an object with respect to spinning about an axis is that the object be statically balanced. The center of mass must lie on the axis of rotation. The second condition for dynamic balance, that H˙ = 0, is a little more subtle. To make things more specific, let’s calculate the rate of change of angular momentum H˙ O with respect to a point O that is on the axis of rotation. So, for constant rate rotation about a fixed axis we have: H˙ O = ω × H O (8.52) because H O spins with the body. H˙ O is evidently zero if the angular momentum, H O , is parallel to the angular velocity, ω. However, H O = [I O] · ω which means H O is parallel to ω only when ω is an eigenvector of [I O].
490 CHAPTER 8. Advanced topics in circular motion 1 Caution: A common misperception So an object which is spinning about a fixed axis is dynamically balanced about angular momentum is that it is always if its center of mass is on the axis (static balance) and the angular velocity parallel to angular velocity. Sometimes an- vector ω is an eigenvector of the moment of inertia matrix. gular momentum is parallel to angular ve- If we restrict 1 our attention to cases where the axis of rotation is the z-axis then locity. When? When ω is an eigenvector of this condition is easy to recognize. It is when Ixz = Iyz = 0, that is, when the only the moment of inertia matrix [I O], which is non-zero element in the third column and in the third row of the [I O] matrix is IzOz in the lower right corner. always the case in 2-D . But in general, an- gular momentum is not parallel to angular Since these terms cause wobbling if one of the coordinate axis is an axis of velocity. rotation, the terms that are off the main diagonal in the moment of inertia matrix are sometimes called the ‘imbalance’ terms. These terms lead to dynamic imbalance 2 When you go to ACME garage and get when the object is spun about the x, y, or z-axes. The off diagonal terms are also your car tires balanced you can politely ask sometimes called the ‘centrifugal’ terms. This naming has an intuitive basis. One the mechanic: “Ma’am, could you make way to understand dynamic imbalance is to think of it being due to the unbalanced sure that one of the eigenvectors of my centrifugal pull of bits of mass that are spinning in circles. 2 wheel’s moment of inertia matrix is paral- lel to the car axle? Thanks.” The mechanic, Often, but not always, you can tell by inspection if something is dynamically if she is a decent person, will make sure balanced for rotation about a certain axis. If, for every bit of mass that is not on the that it is the appropriate eigenvector that is axis there is another equal bit of mass that is exactly opposite, with respect to the axis, parallel to the axle. Otherwise the wheel then the object is balanced. Some objects that do not meet this symmetry condition would point straight sideways with the axle are also dynamically balanced, however. piercing the rubber. Example: Some common objects All of the figures in figure 2.58 on page 65 are dynamically balanced about all the axes shown and about any axis perpendicular to any pair of these axes. 2 Example: A cube = a sphere Both a cube and a sphere have moment of inertia matrices proportional to the identity matrix 1 0 0 [I cm]cube m 2 0 1 0 0 = 60 1 1 0 0 [I cm]sphere = m D2 0 1 0 . 0 10 0 1 So, for any ω, we get that H cm is parallel to ω and, thus, both of these objects are dynamically balanced for rotation about any axis through their centers of mass. Here is an example where the mathematics contradicts intuition; it does not seem like a cube should be balanced for rotation about a random skewed axis through its center of mass! 2
8.5. Dynamic balance 491 SAMPLE 8.19 Static and dynamic balance in 2-D motion. A system with two equal yB masses at A and B rotates about point O at a constant rate ω = 10 rpm. The system is shown in the figure. The rod connecting the two masses has negligible mass. r2 = 0.5 m m = 1 kg x (a) Is the system statically balanced? If not, suggest a way to balance it. (b) Is the system dynamically balanced? If not, suggest a way to balance it. r1 = 0.3 m O ω = 10 rpm Solution A m = 1 kg (a) Since the two masses are equal, the center of mass of the system is at the geometric center of rod AB, i.e., at a point O , halfway between A and B. Clearly, Figure 8.58: (Filename:sfig4.7.DH1) O is not on the axis of rotation which passes through O and is perpendicular to the plane of motion. Therefore, the system is not statically balanced. B To balance the system, we must move the center of mass to O or pivot the r2 system at O . Say we cannot move the pivot point. So, to move the center of mass O, we add mass m to m at A. From the definition of center of mass: O' r1 O (m + m )r1 = mr2 ⇒ m = m (r2 − r1) = 1 kg(0.5 m − 0.3 m) A m' r1 0.3 m = 0.67 kg. Figure 8.59: (Filename:sfig4.7.DH2) Add 0.67 kg to mass at A. y m2 eˆR (b) The system, as given, is not dynamically balanced since it is not statically balanced. Let us check if it is dynamically balanced after adding m to A as suggested above. MO = H˙ O. Let us calculate H˙ O to see if it is zero: H˙ O = r 1 × m1 a1 + r 2 × m2 a2 = (−r1eˆR) × m1(ω2r1eˆR) + (r2eˆR) × (−ω2r2eˆR)m2 = −m1ω2r12(eˆR × eˆR) − m2ω2r22(eˆR × eˆR) 0 0 ω r2 O x = 0. The net torque on the system is zero, therefore it is dynamically balanced. Note r1 that H˙ O = 0 irrespective of the values of m1, m2, r1, and r2. Thus H˙ O = 0 m1 even for the system as given. But the system, as given, is not dynamically Figure 8.60: (Filename:sfig4.7.DH3) balanced because static balance is a necessary condition for dynamic balance.
492 CHAPTER 8. Advanced topics in circular motion m1 x SAMPLE 8.20 Imbalance of a rotor due to rotating masses. Two masses m1 and m2 are attached to a massless shaft AB by massless rigid rods of length h each. The h two masses are separated by distance b along the shaft and are in the same plane but A on the opposite sides of the shaft. The shaft is rotating with a constant angular speed ω. The shaft is free to move along the z-axis at point B. Ignore gravity. yb ω B (a) Is the system statically balanced? (b) What is the torque required to keep the motion going about the z-axis (i.e., z m2 Mz) ? (c) What are the reactions at the support points of the shaft? Figure 8.61: Imbalance of a rotor due to Solution point masses spinning at a constant rate. (a) A simple line sketch and the free body diagram of the system are shown in (Filename:sfig4.7.1) Fig. 8.62. The linear momentum balance ( F = m a ) for the system gives: m1 C m1 C Ah Az Ay ω ıˆ Mz ˆ kˆ Ax b B By D m2 D m2 Bx (a) (b) Figure 8.62: (a) A simple line diagram of the system. (b) Free Body Diagram of the system. (Filename:sfig4.7.1a) x ( Ax + Bx )ıˆ + ( Ay + By)ˆ + Azkˆ = m1a1 + m2a2 m1 v1 = m1ω2h(−ıˆ) + m2ω2hıˆ a1 ⇒ Ax + Bx = ω2h(m2 − m1) (8.53) y a2 Ay + By = 0 (8.54) v2 m2 Az = 0 Figure 8.63: Accelerations of m1 and m2. Clearly, F = 0 which means the system is not statically balanced. (Filename:sfig4.7.1b) (b) Mz can be easily calculated by writing angular momentum balance about axis AB. In fact, we do not even need to write the equation in this case; since all the forces pass through this axis and the accelerations of m1 and m2 also pass through it, M/AB(due to reaction forces) = 0 and H˙/AB = 0. But M/AB = H˙/AB or Mz + M/AB(due to reaction forces) = H˙/AB ⇒ Mz = 0. Mz = 0
8.5. Dynamic balance 493 (c) For the four unknown reactions at A and B (we have already found Az and Mz) we have two scalar equations so far (from Linear Momentum Balance ). Angular Momentum Balance about point A gives: M/A = H˙ /A Now, M/A = r B/A × F B = (2 + b)kˆ × (Bx ıˆ + Byˆ) = Bx (2 + b)ˆ − By(2 + b)ıˆ H˙ /A = r C/A × m1 a 1 + r D/A × m2 a 2 = ( kˆ + hıˆ) × m1(−ω2hıˆ) + (( + b)kˆ − hıˆ) × m2(ω2hıˆ) = ω2h(m2( + b) − m1 )ˆ Equating M/A and H˙ /A we get Bx (2 + b)ˆ − By(2 + b)ıˆ = ω2h(m2( + b) − m1 )ˆ Dotting both sides of the equation with ıˆ and ˆ, we get By = 0 ω2h and Bx = (2 + b) [m2( + b) − m1 ]. Substituting Bx and By in (8.53) and (8.54) we find Ax = ω2h − m1( + b)] (2 + b) [m2 Ay = 0 Ax = ω2h [m 2 − m1( + b)] (2 +b) Bx = ω2h [m 2 ( + b) − m1 ] (2 +b) Ay = By = Az = Mz = 0
494 CHAPTER 8. Advanced topics in circular motion m x SAMPLE 8.21 Balancing a rotor with spinning masses. Consider the same system as in the previous sample problem (Sample 8.20). Assume that m1 = m2 = m. h A (a) Set m1 = m2 in the reactions calculated. Is the system statically balanced now? Explain. yb ω B (b) Is the system dynamically balanced? Explain. (c) Balance the system dynamically by (a) adjusting the geometry, (b) adding two z m masses to the system. Figure 8.64: Imbalance of a rotor due to Solution (a) Recall from the solution of Sample 8.20 that point masses spinning at a constant rate. ω2h (Filename:sfig4.7.2) Ax = (2 + b) [m2 − m1( + b)] m ω2h Bx = (2 + b) [m2( + b) − m1 ] h Ay = By = Az = Mz = 0 cm B Setting m1 = m2 = m in the above expressions for the reactions, we get A ω2h ω2h h Ax = (2 + b) (−mb) and Bx = (2 + b) (mb). m m Thus, Ax + Bx = 0. Therefore, F = 0 which means the system is statically Ax balanced. Bx Static Balance: When the two masses are equal, the center of mass of the m system is on the axis of rotation. Therefore, the static reactions are zero irrespective of the orientations of the two masses. Figure 8.65: Reaction forces at A and B The requirement for static balance is F = 0 in any static orientation of the system. As long as the center of mass of the system lies on the axis are required to keep the motion going. of rotation, the system will be in static balance. (Filename:sfig4.7.2a) Dynamic Balance: When the masses are rotating at constant speed, reaction forces at A and B are required to keep the motion going. Clearly, M = 0 now. Therefore, the system is not in dynamic balance. At the instant shown the reaction forces are Ax and Bx as shown in Fig. 8.65, but they change directions as the position of the masses changes. For example, if the masses are in the y–z plane the reaction forces will be Ay and −By. The magnitudes of these forces will remain the same. Thus, we have rotating reaction forces at A and B. These rotary forces cause wear in the bearings and induce vibrations in the frame of the machine and the supporting structure. Therefore, these forces are undesirable. If we can somehow make these dynamic reactions zero, the bearings, the machine frame, the supporting structure, the machine operator and the company will all be very happy. So, how do we do it? Read on. (b) There are many ways in which the rotor can be dynamically balanced: 1 Balancing a rotor by adding two masses • Trivial solution: Remove both the masses. Of course, there is nothing left in two selected transverse (to the shaft axis) to produce any nonzero H˙ . Thus, M/any point = 0. Also, F = 0. planes is quite a general method. Even if the rotor has more than two spinning masses the • Adjust geometry: Set b = 0. The center of mass is still on net angular momentum due to all masses F = 0. In addition, H˙ /A = can be reduced to the angular momentum the axis of rotation ⇒ produced by an equivalent system with just two masses such as the one under consider- 0⇒ M/A = 0. ation. • Add two masses: If b is required to be nonzero, we can balance the rotor by adding two masses in any two selected transverse (to the shaft-axis) planes to make H˙ /A = 0. 1 For example:
8.5. Dynamic balance 495 – We can take two equal masses m and m, and place them in the m m m m' opposite directions of the masses already on the shaft as shown in A BA b B Fig. 8.66(a). It should be clear that the net angular momentum about any point on the shaft will be zero now. mm m' m (a) c – We can add two masses m and m different from the masses attached (b) to the shaft and place them a distance c apart on the shaft. Let the length of the connecting rods of the new masses be h . For the net angular momentum to be zero we need H˙ due to m s H˙ due to ms m ω2h c = mω2hb Figure 8.66: Dynamic balancing of the ⇒ m h c = mhb rotor by adding two masses. (a) The two added masses (shown by dotted circles) are Thus, we have the freedom 1 of selecting any combination of m , h the same as the original masses on the shaft. (b) The two added masses are different from and c to give the required product. Here are two examples: the original masses (m = m). (Filename:sfig4.7.2b) (i) Let m = m/2, c = b/2, then h = mhb = 4h. mc mhb (ii) Let m = m/2, c = 2b, then h = mc = h. 1 In practice, this freedom is usually re- stricted by geometric and space constraints. m/2 m 4h m m/2 h b/2 hb h Ah B Ah hB m m/2 m 4h 2b m/2 (a) (b) Figure 8.67: Dynamic balancing of the rotor by adding two masses. (a) m = m/2, c = b/2 and h = 4h. (b) m = m/2, c = 2b and h = h. (Filename:sfig4.7.2c)
496 CHAPTER 8. Advanced topics in circular motion
9 General planar motion of a rigid body Many parts of practical machines and structures move in ways that can be idealized Figure 9.1: Planar motion of a 3D car. as straight-line or parallel motion (Chapter 6) or circular motion (Chapters 7 and 8). But often an engineer must analyze parts with more general motions a plane in (Filename:tfigure.2D3Dcar) unsteady flight or a connecting rod in a car engine. Of course, the same basic laws of mechanics still apply. The chapter starts with the kinematics of a planar rigid body in two dimensions and then progresses to the mechanics and analysis of motions. Almost throughout, we will use two modeling approximations: • The objects are planar, or symmetric with respect to a plane; and • They have planar motions in that plane. A planar object is one where the whole body is flat and all its matter is confined to one plane, say the x y plane. This is surely a palatable approximation for a flat cut piece of sheet metal. For more substantial real objects, like a full car, the approximation seems at a glance to be terrible. But, it turns out, so long as the motion is planar the squishing of the car into the x y plane has no effect on the dynamics. A planar motion is one where the velocities of all points are in the same constant plane, say a fixed x y plane, at all times. Note that the positions of the points do not have to be in same plane for a planar motion. Example: A car going over a hill A car driving on a road is going over a bump. Assume the road is straight in map view, say in the x direction. So long as the bump crosses the road (so the road surface is made of line segments in the z direction). Although 497
498 CHAPTER 9. General planar motion of a rigid body the car is clearly not planar, the car motion is likely well modeled as planar, with the velocities of all points in the car in the x y plane. 2 Example: Skewered sphere A sphere skewered and rotating about a fixed axes in the kˆ direction has a planar motion. The points on the object do not live in a common plane. But all of the velocities are orthogonal to kˆ and thus in the x y plane. This problem does fit in with the methods of this chapter. The symmetry of the sphere with respect to the x y plane makes it so that the three-dimensional mass distribution does not invalidate the two-dimensional analysis. 2 Example: Skewered plate Imagine a flat rectangular plate with normal nˆ. Imagine it has a fixed axis of rotation in the direction λˆ that makes a 45o to nˆ. This is a planar object (a plane normal to nˆ) in planar motion (all velocities are in the plane normal to λˆ . But the plane of motion is not the plane of the mass distribution. So this example does not fit into the discussion of this chapter. [A two-dimensional analysis of the plate in this example we would project all the plates mass into the plane normal to the λˆ direction. Effects of static imbalance would be correctly predicted, but effects of dynamic imbalance are essentially three-dimensional, and would be lost in this projection.] 2 Of course no real object is exactly planar and no real motion is exactly a planar motion. But some objects are relatively flat or thin and, more importantly, many motions are well approximated as planar motions. Thus many, if not most, engineering analysis are based on the planar motion approximation and thus use, as a convenience of thought but not actually a further approximation, a planar mass distribution as well. For bodies that are approximately symmetric about the x y plane of motion (such as a car, if the asymmetrically placed driver’s mass is neglected), there is no loss in doing a two-dimensional rather than full three dimensional analysis. But even if one desires in the end to do a three-dimensional analysis, the easiest way to gain the needed skills is to master the analysis of two-dimensional models using flat bodies in planar motions. 9.1 Kinematics of planar rigid- body motion The approximation that an object moves as if was were rigid is common in mechanics. And useful too, that’s why we have a whole book on the topic. We now continue our study of rigid body mechanics by considering the kinematics of planar motion. Namely, the motions of points. We study how points on a rigid body move for two, somewhat independent, reasons. First, one needs to know about the motion of mass points in order to apply the momentum-balance equations. Second, formulas involving the motion of points on a rigid body are useful to understand mechanisms, machines where rigid bodies are attached in various ways to each other. The velocities of points on a rigid body are not arbitrary. That the various points maintain constant distance with respect to each other restricts the relative velocities
9.1. Kinematics of planar rigid-body motion 499 and accelerations of the points. In this section you will learn three important concepts about this restriction: the concept of angular velocity, the vector formula for relative velocity of points on a rigid body, and the vector formula for relative acceleration of points on a rigid body. All of these are familiar topics, just now in a broader context. Displacement and rotation 1 That’s Oo longitude and Oo latitude. When a planar body (read, say, machine part) B moves from one point in the plane Motion P θ it has a displacement and a rotation. For definiteness, we start in some reference rP/0 position. We mark a reference point on the body at a point that, in the reference P configuration, coincides with a fixed reference point, say 0. We mark a (directed) line rP/0ref 0' on the body that, in the reference configuration, coincides with a fixed reference line, 0 r0 /0 say the positive x axis. When a plane flies from New York to Bombay its position is measured relative to a point in the Gulf of Guinea 1000 miles west of Gabon, 1 even Figure 9.2: The displacement and rota- though the plane never goes there (nor does anyone want it to). Similarly, a body never needs to be at its reference position or orientation as it moves. tion of a planar body relative to a reference configuration. We could measure rotation by measuring the rotations of any line that connected two points fixed to the body. For each line we keep track of the angle the line makes (Filename:tfigure.dispandrot) with a line fixed in space, say the positive x or y axis. And, as for the case of pure rotation, its simplest just to stick to the convention that counter-clockwise rotations are positive (Fig. 9.3). The angles θ1, θ2 . . ., all change with time and are all different from each other. But all the angles change at the the same amount, just like in section 7.3. We lose nothing by picking any line we like for reference and measuring the body rotation by the rotation of that line. So the net motion of a rigid planar body is described by the vector displacement of a reference point from a reference position ro /o = roo and a rotation of the body from the reference position, θ. The position of a point on a moving rigid body. Given that a point P on the body is at a known position rP/0ref in the reference configuration, where is it after the body has been displaced by r0 /0 and rotated an angle θ ? An easy way treat this is to write the position rP of P as rP/0 = r0 /0 + rP/0 . Figure 9.3: Rotation of body B is mea- One way to think of this is as the body first displacing without rotation and then sured by the rotation of real or imagined lines marked on the body. The lines make rotating without displacing. Don’t mind that this is not the route by which the body different angles: θ1 = θ2, θ2 = θ3 etc, but θ˙1 = θ˙2 = θ˙3 = . . .. Angular velocity is moved to its new configuration, the end state is the same. The displacement of the defined as ω = ωkˆ with ω ≡ θ˙1 = θ˙2 = body is given by r0 /0 so we already have half our answer. The other term rP/0 we find by rotating rP/0ref as we did in Sect. 7.3. Thus, we can describe the coordinates θ˙3 = . . .. of a point as, (Filename:tfigure.2Drotation) rP/0 x y = r0 /0 + R(θ ) rP/0 . (9.1) xy xy displacement rotation where R(θ ) = cos θ sin θ . As the motion progresses the displacement − sin θ cos θ changes with time as does the rotation angle θ.
500 CHAPTER 9. General planar motion of a rigid body Angular velocity Because all lines body B rotate at the same rate (at a given instant) B’s rotation rate is the single number we call ωB (‘omega b’). In order to make various formulas work out we define a vector angular velocity with magnitude ωB which is perpendicular to the x y plane: ωB = ωB kˆ θ˙ where θ˙ is the rate of change of the angle of any line marked on body B. So long as you are careful to define angular velocity by the rotation of line segments and not by the motion of individual particles, the concept of angular velocity in general motion is defined exactly as for a body rotating about a fixed axis. A legitimate way to think about planar motion of a rigid body is that any given point is moving in circles about any other given point (relative to that point). When a rigid body moves it always has an angular velocity (possible zero). If we call the body B (script B), we then call the body’s angular velocity ωB. The angular velocity ωB is a vector that describes the rate of rotation of the body. Relative velocity of two points on a rigid body For any two points A and B glued to a rigid body, B the relative velocity of the points (‘the velocity of B relative to A’) is given by the cross product of the angular velocity of the body with the relative position of the two points: v B/A = v B − v A = ωB × r B/A. (9.2) This formula says that the relative velocity of two points on a rigid body is the same as would be predicted for one of the points if the other were stationary. Absolute velocity of a point on a rigid body If one knows the velocity of one point on a rigid body and one also knows the angular velocity of the body, then one can find the velocity of any other point. How? v B = v A + (v B − v A) = v A + v B/A = v A + ωB × r B/A r B−r A That is, the absolute velocity of the point B is the absolute velocity of the point A plus the velocity of the point B relative to the point A. Because B and A are on the same rigid body, their relative velocity is given by formula 9.2 above. The equation is valid no matter what values are known and which are not known. It can be used in various ways depending on what is known about the motion of the body being studied and what is known about the motion of the chosen points A and B.
9.1. Kinematics of planar rigid-body motion 501 An alternative approach is to differentiate the coordinate expression eqn. (9.1) as follows. d (9.3) vP/0 x y = dt rP/0 x y (9.4) (9.5) =d x0 /0 +d cos θ sin θ x (9.6) dt y0 /0 dt − sin θ cos θ y (9.7) = x˙0 /0 + −θ˙ sin θ θ˙ cos θ x y˙0 /0 −θ˙ cos θ −θ˙ sin θ y = x˙0 /0 + 0ω cos θ sin θ x y˙0 /0 −ω 0 − sin θ cos θ y = x˙0 /0 + 0ω rP/0 . y˙0 /0 −ω 0 xy Thus, matrix product 0ω rP/0 is equivalent to the vector product ω × −ω 0 xy rP/0 and the matrix 0ω is sometimes called the angular velocity matrix. It −ω 0 is an example of a so-called skew symmetric matrix because it is the negative of its own transpose. Angular acceleration We define the angular acceleration α of a rigid body as the rate of change of angular velocity, α = ω˙ . The angular acceleration of a body B is called αB. For two- dimensional bodies moving in the plane both the angular velocity and the angular acceleration are always perpendicular to the plane. That is ω = ωkˆ and α = αkˆ = ω˙ kˆ. In this case the angular acceleration is only due to the speeding up or slowing down of the rotation rate; i.e., α = ω˙ = θ¨. Relative acceleration of two points on a rigid body For any two points A and B glued to a rigid body B, the acceleration of B relative to A is d a B/A = dt vB/A = d ωB × rB/A dt = ω˙ B × r B/A + ωB × (vB/A), = ω˙ B × r B/A + ωB × (ωB × rB/A), = αB kˆ × r B/A + −ωB2 rB/A), (9.8) If point A has no acceleration, this formula is the same as that for the acceleration of a point going in circles from chapter 7. On a rigid body in 2D all two points on rigid body can do relative to each other is to go in circles.
502 CHAPTER 9. General planar motion of a rigid body Absolute acceleration of a point on a rigid body If one knows the acceleration of one point on a rigid body and the angular velocity and acceleration of the body, then one can find the acceleration of any other point. How? a B = a A + (a B − a A) = a A + a B/A (9.9) = a A + ωB × (ωB × r B/A) + ω˙ B × r B/A = a A − ωB2 r B/A + αB kˆ × r B/A Equation 9.9 is often called the three term acceleration formula. The acceleration of a point B on a rigid body is the sum of three terms. The first, a A, is the acceleration of some point A on the body. The second term, ωB × (ωB × r B/A), is the centripetal acceleration of B going in circles relative to A. It is directed from B towards A. The third term, ω˙ B × r B/A, is due to the change in the magnitude of the angular velocity and is in the direction normal to the line from A to B. Summary of the kinematics of one rigid body in general 2D motion The key idea in this section are that if you know the position, velocity, and acceleration of one point on a rigid body, and you know the rotation angle, angular rate and angular acceleration then you can find the position, velocity and acceleration of any point on the body. Thus, when we study the mechanics of a rigid body, we can use the position of one reference point and the rotation of the body as simple kinematic measures of the entire motion of the body.
9.1. Kinematics of planar rigid-body motion 503 SAMPLE 9.1 Velocity of a point on a rigid body in planar motion. A plate ABC, of an equilateral triangle geometry, is in motion in the x-y plane. At the instant shown in the figure, point B has velocity vB = 0.3 m/sıˆ + 0.6 m/sˆ and the plate has angular velocity ω = 2 rad/skˆ. Find the velocity of point A. Solution We are given vB and ω, and we need to find vA, the velocity of point A on the same rigid body. We know that, vA = vB + ω × rA/B Thus, to find vA, we need to find rA/B. Now, rA/B = rA − rB = 0 − (0.2 mıˆ) = −0.2 mıˆ Thus, vA = vB + ω × rA/B = (0.3ıˆ + 0.6ˆ) m/s + 2 rad/skˆ × (−0.2ıˆ) m = (0.3ıˆ + 0.6ˆ) m/s − 0.4ˆ m/s = (0.3ıˆ + 0.2ˆ) m/s. vA = (0.3ıˆ + 0.2ˆ) m/s SAMPLE 9.2 The instantaneous center of rotation. A point C on a rigid body, 1 The point with zero velocity is called the instantaneous center of rotation. Some- rotating with angular velocity ω = 5 rad/skˆ has velocity vC = 2 m/sıˆ − 5 m/sˆ. times this point may lie outside the body. Find a point on the body, assuming it exists 1 , that has zero velocity. Solution Let the point of zero velocity be O, with position vector rO/C with respect to point C. Since vO = vC + ω × rO/C, for vO to be zero, ω × rO/C must be parallel to and in the opposite direction of vC. Since ω is out of plane, rO/C must be normal to vC for the cross product to be parallel to vC. Now, let vC = vC λˆ . Then, rO/C = r nˆ where nˆ ⊥ λˆ and r = |rO/C|. Thus, vC λˆ + ωkˆ × r nˆ = vO = 0 (9.10) Dotting eqn. (9.10) with λˆ , we get vC √ ω 29 m/s vC = ωr ⇒ r = = 5 rad/s = 1.08 m. Since λˆ = vC/|vC| = 0.37ıˆ − 0.93ˆ, nˆ = 0.93ıˆ + 0.37ˆ. Thus rO/C = r nˆ = 1.08 m(0.93ıˆ + 0.37ˆ) = 1 mıˆ + 0.4 mˆ. rO/C = 1 mıˆ + 0.4 mˆ Note: It is also possible to find rO/C purely by vector algebra. Assuming rO/C = (xıˆ + yˆ) m and plugging into vO = vC + ω × rO/C along with the given values, we get 0 = (2 − 5y) m/sıˆ + (−5 + 5x) m/sˆ. Dotting this equation with ıˆ and ˆ, we get 2 − 5y = 0 and −5 + 5x = 0, which give x = 1 and y = 0.4. Thus, rO/C = 1 mıˆ + 0.4 mˆ as obtained above.
504 CHAPTER 9. General planar motion of a rigid body vA SAMPLE 9.3 A cheerleader throws her baton up in the air in the vertical x y-plane. yA At an instant when the baton axis is at √θ = 60o from the horizontal, the velocity of µ = 60o end A of the baton is v A = 2 m/sıˆ + 3 m/sˆ. At the same instant, end B of the G baton has velocity in the negative x-direction (but |v B| is not known). If the length 1 of the baton is L = 2 m and the center of mass is in the middle of the baton, find the velocity of the center of mass. Solution vB B √ Figure 9.4: (Filename:sfig7.1.1) We are given: v A = (2ıˆ + 3ˆ) m/s and v B = −vB ıˆ x where vB = |v B| is unknown. We need to find v G . Using the relative velocity formula for two points on a rigid body, we can write: vG = v A + ω × r G/A (9.11) Aµ Here, v A and r G/A are known. Thus, to find v G , we need to find ω, the angular velocity of the baton. Since the motion is in the vertical x y-plane, let ω = ωkˆ. Then, v B = v A + ω × r A/B = v A + ωkˆ × L(− cos θ ıˆ + sin θ ˆ) √ r A/B or − vB ıˆ = (2ıˆ + 3ˆ) m/s − ωL(cos θ ˆ + s√in θ ıˆ) = (2ıˆ + √ m/s − ω· 1 m·( 1 ˆ + 3 ıˆ) L 22 2 ˆ 3ˆ) ıˆ B Dotting both sides of this equation with ˆ we get: r A/B = L(-cos θ ıˆ + sin θ ˆ) 0 = √ − ω m· 1 Figure 9.5: (Filename:sfig7.1.1a) 3 m/s 22 A √ √ µ ⇒ ω = 3 m · 4 = 4 3 rad/s. L/2 s 1m ˆ Now substituting the appropriate values in Eqn 9.11 we get: ıˆ B vG = vA + ωkˆ × L (cos θ ıˆ − sin θ ˆ) r G/A = L/2(cos θ ıˆ - sin θ ˆ) 2 r G/A = vA + ωL (cos θ ˆ + sin θ ıˆ) √ = (2ıˆ + 2 3 ıˆ) √ m/s + √ m/s·( 1 ˆ + 3ˆ) 3 √2 2 3 3 √ = (2 + ) m/sıˆ + ( 3 + ) m/sˆ 22 = 3.5 m/sıˆ + 2.6 m/sˆ v G = (3.5ıˆ + 2.6ˆ) m/s Figure 9.6: (Filename:sfig7.1.1b)
9.1. Kinematics of planar rigid-body motion 505 SAMPLE 9.4 Falling ladder: The ends of a ladder of length L = 3 m slip along the A frictionless wall and floor shown in Figure 9.7. At the instant shown, when θ = 60o, the angular speed θ˙ = 1.15 rad/s and the angular acceleration θ¨ = 2.5 rad/s2. Find the absolute velocity and acceleration of end B of the ladder. Solution Since the ladder is falling, it is rotating clockwise. From the given infor- L mation: frictionless ıˆ ω = θ˙kˆ = −1.15 rad/skˆ ˆ ω˙ = θ¨kˆ = −2.5 rad/s2kˆ. µ = 60o We need to find v B, the absolute velocity of end B, and a B, the absolute acceleration xB of end B. Since the end A slides along the wall and end the B slides along the floor, we know y the directions of v A, v B , a A and a B . Let v A = vAˆ, a A = aAˆ, v B = vB ıˆ and a B = aB ıˆ where the scalar quan- Figure 9.7: (Filename:sfig7.2.2) tities vA, aA, vB and aB are unknown. Now, v A = v B + v A/B = v B + ω × r A/B or vAˆ = vBıˆ + θ˙kˆ × L(− cos θ ıˆ − sin θ ˆ) r A/B = (vB + θ˙ L sin θ )ıˆ − θ˙L cos θ ˆ. Dotting both sides of the equation with ıˆ, we get: A vA ˆ·ıˆ = (vB + θ˙ L sin θ ) ıˆ·ıˆ +θ˙ L cos θ ˆ·ıˆ 0 10 ⇒ 0 = vB + θ˙ L sin θ √ ⇒ vB = −θ˙L sin θ = −(−1.15 rad/s)·3 m· 3 r A/B -L sin θ ˆ 2 = 2.99 m/s. v B = 2.9 m/sıˆ µ B Similarly, ıˆ -L cos θıˆ ˆ −ω2 r A/B r A/B = -L cos θ ıˆ - L sin θ ˆ a A = a B + ω˙ × r A/B + ω × (ω × r A/B ) Figure 9.8: (Filename:sfig7.2.2b) aAˆ = aBıˆ + θ¨kˆ × L(− cos θ ıˆ − sin θ ˆ) − θ˙2 L(− cos θ ıˆ − sin θ ˆ) = (aB + θ¨ L sin θ + θ˙2 L cos θ )ıˆ + (−θ¨ L cos θ + θ˙2 L sin θ )ˆ. Dotting both sides of this equation with ıˆ (as we did for velocity) we get: 0 = aB + θ¨ L sin θ + θ˙2 L cos θ ⇒ aB = −θ¨ L sin θ − θ˙2 L cos√θ = −(−2.5 rad/s2·3 m· 3 ) − (−1.15 rad/s)2·3 m· 1 22 = 4.51 m/s2. a B = 4.51 m/s2
506 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.5 A board in the back of an accelerating truck. A 10 ft long board B AB rests in the back of a flat-bed truck as shown in Fig. 9.9. End A of the board L ˆ is hinged to the bed of the truck. The truck is going on a level road at 55 mph. In GC preparation for overtaking a vehicle in the front the trucker accelerates at a constant Aθ ıˆ rate 3 mph/s. At the instant when the speed of the truck is 60 mph, the magnitude of the relative velocity and relative acceleration of end B with respect to the bed of the truck are 10 ft/s and 12 ft/s2, respectively. There is wind and at this instant, the Figure 9.9: (Filename:sfig7.2.2a) board has lost contact with point C. If the angle θ between the board and the bed is 45o at the instant mentioned, find (a) the angular velocity and angular acceleration of the board, (b) the acceleration of the center of mass of the board. Solution At the instant of interest v A = velocity of the truck = 60 mph ıˆ = 88 ft/s ıˆ a A = acceleration of the truck = 3 mph/s = 4.4 ft/s2 ıˆ |v B/A| = vB/A = magnitude of relative velocity of B = 10 ft/s |a B/A| = aB/A = magnitude of relative acceleration of B = 12 ft/s2. Let ω = ωkˆ be the angular velocity and ω˙ = ω˙ kˆ be the angular acceleration of the board. (a) The relative velocity of end B of the board with respect to end A is v B/A = ω × r B/A = ωkˆ × L(cos θ ıˆ + sin θ ˆ) = ωL(cos θ ˆ − sin θ ıˆ) ⇒ |v B/A| = ωL ⇒ ω = |v B/A| = vB/A LL = 10 ft/s = 1 rad/s. 10 ft Note that we have taken the positive value for ω because the board is rotating counterclockwise at the instant of interest (it is given that the board has lost contact with point C). Similarly, we can compute the angular acceleration: a B/A = ω˙ × r B/A − ω2 r B/A = ω˙ kˆ × L(cos θ ıˆ + sin θ ˆ) − ω2 L(cos θ ıˆ + sin θ ˆ) = ω˙ L(cos θ ˆ − sin θ ıˆ) − ω2 L(cos θ ıˆ + sin θ ˆ) ⇒ |a B/A| = (ω˙ L)2 + (ω2 L)2 = aB/A (given) ⇒ a2B/A = (ω˙ L)2 + (ω2 L)2 ⇒ ω˙ = a 2B / A − ω4 L2
9.1. Kinematics of planar rigid-body motion 507 = 12 ft/s2 2 10 ft − (1 rad/s)4 = ±0.663 rad/s2. Once again, we select the positive value for ω˙ since we assume that the board accelerates counterclockwise. ω = 1 rad/skˆ, ω˙ = 0.663 rad/s2kˆ (b) Now, we can easily calculate the acceleration of the center of mass as follows. aG = a A + aG/A = aAıˆ + ω˙ × r G/A − ω2 r G/A = a A ıˆ + ω˙ kˆ × L (cos θıˆ + sin θ ˆ) − ω2 L (cos θ ıˆ + sin θ ˆ) 2 2 = a A ıˆ + ω˙ L (cos θ ˆ − sin θ ıˆ) − ω2 L (cos θ ıˆ + sin θ ˆ) 2 2 = 4.4 ft/s2ıˆ + 0.663 rad/s2 · 10 ft · ( 1 ˆ − 1 ıˆ) √ √ 222 −(1 rad/s)2 · 10 ft · ( √1 ıˆ + √1 ˆ) 2 22 = −1.48 ft/s2ıˆ − 1.19 ft/s2ˆ. a G = −(1.48ıˆ + 1.19ˆ) ft/s2 Comments: This problem is admittedly artificial. We, however, solve this problem to show kinematic calculations.
508 CHAPTER 9. General planar motion of a rigid body 9.2 Unconstrained dynamics of 2- D rigid-body planar motion 1 Advanced aside: What we call “simple We will now use the kinematics of a rigid body (Sect. 9.1) to allow us to “evaluate” kinematic measures” are examples of “gen- the motion quantities, namely linear momentum, angular momentum, kinetic energy, eralized coordinates” in more advanced and their rates of change. By “evaluate” we mean that we would like to express the treatments. The idea sounds intimidating, motion quantities in terms of simple kinematic measures 1 rather than as sums over but is simply this. If something can only Avogadro’s number of particles (There are on the order of 1023 atoms in a typical move so many ways, you should only keep engineering part.). Even neglecting atoms and viewing matter as continuous we could track of the motion with so many variables. be stuck with integrals over complex domains. The simple measures of motion will be the displacement, velocity and acceleration of one reference point 0 on the body (r0 , v0 and a0 ) and the rotation, angular velocity, and angular acceleration of the body (θ, ω and α). We will treat all bodies as if they are squished into the plane and moving in the plane. But the analysis is perfectly sensible for bodies that are symmetric with respect to the plane containing the velocities (see Box 9.2 on page 509). Momenta balance for a rigid body As always, once you have defined the system and the forces acting on it by drawing a free body diagram, the basic momentum balance equations are applicable (and exact for engineering purposes). Namely, Linear momentum balance: F i = L˙ and Angular momentum balance: Mi/0 = H˙ O. Where the same point 0, any point, is used on both sides of the angular momentum balance equation. We also have the power balance equations, which, for a system with no internal energy or dissipation is Power balance: P = E˙K. The left hand sides of the momentum balance equations are evaluated the same way, whether the system is composed of one body or many, whether the bodies are deformable or not, and whether the points move in straight lines, circles, hither and thither, or not at all. It is the right hand sides of the momentum equations that involve the motion. Similarly, in the energy balance equations the applied power P only depends on the position of the forces and the motions of the material points they touch. But the kinetic energy EK and its rate of change depend on the motion of the whole system. You already know how to evaluate the momenta and energy, and their rates of change, for a variety of special cases, namely • Systems composed of particles where all the accelerations are known (Chap- ter 5); • Systems where all points have the same acceleration. That is, the system moves like a rigid body that does not rotate (Chapter 6); and • Systems where all points move in circles about the same fixed axis. That is, the system moves like a rigid body that is rotating about a fixed skewer (Chapters 7 and 8). Now we go on to consider the general 2-D motions of a planar rigid body. Its now a little harder to evaluate L, L˙ , H O, H˙ O, EK and E˙K. But not much.
9.2. Unconstrained dynamics of 2-D rigid-body planar motion 509 Summary of the motion quantities As for most of the special cases described in the chapters, table I in the back of the book describes the motion quantities for the planar motions we consider in this chapter. Most relevant is row (7). The basic idea is that the momenta for general motion, which involves translation and rotation, is the sum of the momenta (both linear and angular, and their rates of change too) from those two effects. Namely, the linear momentum is described, as for all systems, by the motion of the center of mass L = mtot vcm and L˙ = mtot acm, and the angular momentum has two contributions, one from the motion of the center of mass and one from rotation about the center of mass, H O = rcm/O × (mtot vcm) + Izczmω and H˙ O = rcm/O × (mtot acm) + Izczmω˙ . ¢¢ wff Angular momentum due to Angular momentum due to motion of the center of mass motion relative to the center of mass The kinetic energy and its rate of change are given by EK = 1 m totvc2m + 1 Izczmω2 and E˙K = mtotvv˙ + Izczmωω˙ 2 2 ¢¢ fwf kinetic energy from kinetic energy relative to the center of mass motion center of mass 9.1 THEORY The 2D mechanics of 3D objects Precisely speaking, the situation is this. For a three dimensional analysis of this chapter is the exact projection of a proper three di- object with planar motion in the x y plane then the two dimensional mensional analysis. All forces and displacements are projected onto the x y plane and all moment terms are projected in the z direction.
510 CHAPTER 9. General planar motion of a rigid body 9.2 THEORY The work of a moving force and of a couple The work of a force acting on a body from state one to state two is the train and 9.13 does not. Simultaneously, the train does no work on your hand since your hand is not moving as both 9.13 and 9.14 t2 correctly give. W12 = Pdt. t1 But sometimes we like to think not of the time integral of the power, Work of an applied torque but of the path integral of the moving force. So we rearrange this integral as follows. By thinking of an applied torque as really a distribution of forces, the work of an applied torque is the sum of the contributions of the W12 = t2 (9.12) applied forces. If a collection of forces equivalent to a torque is = (9.13) applied to one rigid body the power of these forces turns out to be = Pdt M · ω. At a given angular velocity a bigger torque applies more. t1 And a given torque applies more power to a faster spinning object. t2 Here’s a quick derivation for a collection of forces Fi that add F · vdt to zero acting at points with positions ri relative to a reference point t1 d r on the body o’. vr2 F ·dr r1 P= Fi · vi (9.15) The validity of equation 9.13 depends on the force acting on the same = Fi · vo + ω × ri/o (9.16) material point of the moving body as it moves from position 1 to = vo · Fi + Fi · ω × ri/o (9.17) position 2; i.e., the force moves with the body. If the material point of force application changes with time, equation 9.13 is senseless 0 and should be replaced with the following more generally applicable equation: = ω · ri/o × Fi = ω · ri/o × Fi W12 ≡ t2 t2 Pdt (9.14) = ω · Mo (9.18) P dt = F · vdt = (9.19) (9.20) t1 t1 (9.21) where v is the velocity of the material point at the instantaneous Work of a general force distribution location of the applied force. A general force distribution has, by reasoning close to that above, a The distinction between 9.13 and 9.14 is subtle. As an example power of: think of standing still and dragging your hand on a passing train. P = Ftot · vo + ω · Mo . (9.22) Your hand slows down the train. It does (negative) work on the (9.23) train. Naively you might think that the work of your hand on the train is zero because your hand is not moving. But if you read the For a given force system applied to a given body in a given motion paragraph above carefully you will see that the power of the hand any point o’ can be used. The terms in the formula above will depend force on the train is the force on the train dotted with the velocity on o’, but the sum does not. of the train (not with the velocity of your hand). Thus, your hand does negative work on the train. In this case equation 9.14 applies to
9.2. Unconstrained dynamics of 2-D rigid-body planar motion 511 SAMPLE 9.6 A plate tumbling in space. A rectangular plate of mass m = d a 0.5 kg, Izczm = 2.08 × 10−3 kg · m2, and dimensions a = 200 mm and b = 100 mm P is pushed by a force F = 0.5 Nıˆ, acting at d = 30 mm away from the mass-center, as shown in the figure. Assume that the force remains constant in magnitude and F direction but remains attached to the material point P of the plate. There is no gravity. ˆγ (a) Find the initial acceleration of the mass-center. ıˆγ (b) Find the initial angular acceleration of the plate. (c) Write the equations of motion of the plate (for both linear and angular motion). Solution The only force acting on the plate is the applied force F . Thus, Fig. 9.10 is also the free body diagram of the plate at the start of motion. (a) From the linear momentum balance we get, F = m acm b ⇒ acm = F Figure 9.10: (Filename:sfig9.tumblingplate1) m = 0.5 Nıˆ = 1 m/s2ıˆ 0.5 kg which is the initial acceleration of the mass-center. acm = 1 m/s2ıˆ (b) From the angular momentum balance about the mass-center, we get Mcm = H˙ cm Fdkˆ = Izczmω˙ ⇒ ω˙ = Fd kˆ Izczm 0.5 N · 0.03 m = 2.08 kg · m2 = 7.2 rad/s2kˆ which is the initial angular acceleration of the plate. y ω˙ = 7.2 rad/s2kˆ θ (c) To find the equations of motion, we can use the linear momentum balance a and the angular momentum balance as we have done above. So, why aren’t the equations obtained above for the linear acceleration, acm = F/mıˆ, and α x the angular acceleration, ω˙ = Fd/Izczmkˆ, qualified to be called equations of motion? Because, they are not valid for a general configuration of the plate Fd rP/cm during its motion. The expressions for the accelerations are valid only in the initial configuration (and hence those are initial accelerations). P Let us first draw a free body diagram of the plate in a general configuration during its motion (see Fig. 9.11). Assume the center of mass to be displaced ˆγ b by xıˆ and yˆ, and the longitudinal axis of the plate to be rotated by θ kˆ with ıˆγ respect to the vertical (inertial y-axis). The applied force remains horizontal and attached to the material point P, as stated in the problem. The linear momentum balance gives F = m acm F Figure 9.11: (Filename:sfig9.tumblingplate1.a) m ⇒ acm =
512 CHAPTER 9. General planar motion of a rigid body or x¨ıˆ + y¨ˆ = F ıˆ m ⇒ x¨ = F m y¨ = 0 Since F/m is constant, the equations of motion of the center of mass indi- cate that the acceleration is constant and that the mass-center moves in the x -direction. Similarly, we now use angular momentum balance to determine the rotation (angular motion) of the plate. The angular momentum balance about the mass- center give Mcm = H˙ cm rP/cm × F = Izczmθ¨kˆ Now, rP/cm = −r [cos(θ + α)ıˆ + sin(θ + α)ˆ] Thus, F = Fıˆ ⇒ rP/cm × F = Fr sin(θ + α)kˆ = F(r sin α cos θ + r cos α sin θ )kˆ db = F(d cos θ + b sin θ )kˆ θ¨ = Fd cos θ + b sin θ . Izczm d Thus, we have got the equations of motion for both linear and the angular motion. x¨ = F , y¨ = 0, θ¨ = Fd cos θ + b sin θ m Izczm d
9.3. Special topics in planar kinematics 513 9.3 Special topics in planar kine- matics Pure rolling in 2-D In this section, we would like to add to the vocabulary of special motions by consid- ering pure rolling. Most commonly, one discusses pure rolling of round objects on flat ground, like wheels and balls, but we will also mention more advanced topics. 2-D rolling of a round wheel on level ground The simplest case, the no-slip rolling of a round wheel, is an instructive starting point. ˆ First, we define the geometric and kinematic variables as shown in Fig. 9.12. For y convenience, we pick a point D which was at xD = 0 at the start of rolling, when xC = 0. The key to the kinematics is that: D R ıˆ The arc length traversed on the wheel is the distance traveled by the wheel center. sD φ C sD = xC That is, O rolling xC A contact xC = sD Bx = Rφ Figure 9.12: Pure rolling of a round ⇒ vC = x˙C = Rφ˙ ⇒ aC = v˙C = x¨C = Rφ¨ wheel on a level support. (Filename:tfigure7.2D.pure.rolling) So the rolling condition amounts to the following set of restrictions on the position of C, r C , and the rotations of the wheel φ: r C = Rφıˆ + Rˆ, v C = Rφ˙ıˆ, aC = Rφ¨ıˆ, ω = −φ˙kˆ, and α = ω˙ = −φ¨kˆ. If we want to track the motion of a particular point, say D, we could do so by using the following parametric formula: rD = r C + r D/C (9.24) ⇒ vD = R(φıˆ + ˆ) + R(− sin φıˆ − cos φˆ) ⇒ aD = R (φ − sin φ)ıˆ + (1 − cos φ)ˆ) = R (φ˙(1 − cos φ)ıˆ + φ˙ sin φˆ) = Rφ˙2(sin φıˆ + cos φˆ). fwf assuming φ˙= constant Note that if φ = 0 or 2π or 4π, etc., then the point D is on the ground and eqn. (9.24) correctly gives that 1 v D = R φ˙(1 − cos(2nπ ) ıˆ + φ˙ sin(2nπ ) ˆ = 0. 00
514 CHAPTER 9. General planar motion of a rigid body Instantaneous Kinematics Instead of tracking the wheel from its start, we could analyze the kinematics at the instant of interest. Here, we make the observation that the wheel rolls without slip. Therefore, the point on the wheel touching the ground has no velocity relative to the ground. Velocity of point on the wheel Velocity of ground = 0 touching the ground ffx ¢¢ vA = vB (9.25) Now, we know how to calculate the velocity of points on a rigid body. So, v A = vC + v A/C , where, since A and C are on the same rigid body (Fig. 9.12), we have from eqn. (8.14) that v A/C = ω × r A/C . Putting this equation together with eqn. (9.25), we get vA = vB ⇒ vC + ω × r A/C = 0 vC ıˆ ωkˆ −Rˆ ⇒ vC ıˆ + ω Rıˆ = 0 ⇒ vC = −ω R . We use v C = vıˆ since the center of the wheel goes neither up nor down. Note that if you measure the angle by φ, like we did before, then ω = −φ˙kˆ so that positive rotation rate is in the counter-clockwise direction. Thus, vC = −ω R = −(−φ˙)R = φ˙ R. Since there is always some point of the wheel touching the ground, we know that vC = −ω R for all time. Therefore, aC = v˙C ıˆ = −ω˙ Rıˆ. O Rolling of round objects on round surfaces For round objects rolling on or in another round object, the analysis is similar to that for rolling on a flat surface. A common application is the so-called epicyclic, hypo- cyclic, or planetary gears (See Box 9.3 on planetary gears on page 516). Referring to Fig. 9.13, we can calculate the velocity of C with respect to a fixed frame two ways and compare: R1 θ˙ A B R2 vC = v B + vC/B vB C eˆθ B vC = v A + v B/A +vC/B. 00 reference line θ˙(R1 + R2)eˆθ = ωB R2eˆθ Figure 9.13: (Filename:tfigure7.rolling.on.another) ⇒ ωB = θ˙(R1 + R2) = θ˙(1 + R1 ). R2 R2
9.3. Special topics in planar kinematics 515 Example: Two quarters. The formula above can be tested in the case of R1 = R2 by using two quarters or two dimes on a table. Roll one quarter, call it B, around around another quarter pressed fast to the table. You will see that as the rolling quarter B travels around the stationary quarter one time, it makes two full revolutions. That is, the orientation of B changes twice as fast as the angle of the line from the center of the stationary quarter to its center. Or, in the language of the calculation above, ωB = 2θ˙. 2
516 CHAPTER 9. General planar motion of a rigid body 9.3 The Sturmey-Archer hub In 1903, the year the Wright Brothers first flew powered Look at the velocity of point C in two ways. First, airplanes, the Sturmey-Archer company patented the internal-hub transmission. This marvel of engineering was sold on the best A point on the spider = A point on the planetary gear bikes until finicky but fast racing bicycles using derailleurs started = to push them out of the market in the 1960’s. The internal- ffx ¢¢ hub Sturmey-Archer gear utilizes a system called planetary gears, vC vC gears which roll around other gears. See the figure below. ⇒ ωS × r C v B +ωP × r C/B In order to understand 0 ⇒ ωSrC = ωP RP ⇒ ωP = rC ωS (9.26) RP Next, let’s look at point D and E: vD = vE v A + v D/A = ωR × r R 0 + ωP × r D/A = ωRkˆ × r R this gear system, we need to understand its kinematics—the motion ωP (2RP )eˆθ = ωRrR eˆθ of its parts. Referring to figure above, the central ‘sun’ gear F is 2RP rR stationary, at least we treat it as stationary in this discussion since it ⇒ ωR = ωP is fixed to the bike frame, so it is fixed in body F . The ‘planet’ gears roll around the sun gear. Let’s call one of these planets P . The spider fwf S connects the centers of the rolling planets. Finally, the ring gear R ωP = rC ωS RP rotates around the sun. Ring R Sun F (fixed frame) (rolls around rC = rS + RP planets) ¢¢ ωR = 2RP rC ωS rR RP = 2(rS + RP ) ωS rR Spider S Planet P (connects (rolls on sun) planets) fwf The gear transmission steps up the angular velocity when the spider rR = rS + 2RP S is driven and ring R, which moves faster, is connected to the wheel. The transmission steps down the angular velocity when the ωR 1+ RP ωS 1+ rS ring gear is driven and the slower spider is connected to the wheel. ⇒ = 2 2RP = angular velocity step-up. rS The third ‘speed’ in the three-speed gear transmission is direct drive (the wheel is driven directly). Typically, the gears What are the ‘gear ratios’ in the planetary gear system? The ‘trick’ ˆ is to recognize that for rolling contact that the contacting points have F R ıˆ O rR the same velocity, v A = v B and v D = v E . Let’s define some terms. ωS = ωS kˆ angular velocity of the spider rS A BC Rolling ωP = ωP kˆ angular velocity of the planet DE contact: ωR = ωRkˆ angular velocity of the ring rC S eˆθ vA = vB vD = vE P RP vC = vC have radius ratio of RP = 3 which gives a gear ratio of 5 . Thus, rS 2 4 the ratio of the highest gear to the lowest gear on a Sturmey-Archer hub is 5 / 4 = 25 = 1.5625. You might compare this ratio to that of 4 5 16 a modern mountain bike, with eighteen or twenty-one gears, where Now, we can find the relation of these angular velocities as follows. the ratio of the highest gear to the lowest is about 4:1.
9.3. Special topics in planar kinematics 517 SAMPLE 9.7 A cylinder of diameter 500 mm rolls down an inclined plane with C uniform acceleration (of the center of mass) a = 0.1 m/s2. At an instant t0, the ˆ mass-center has speed v0 = 0.5 m/s. ıˆ (a) Find the angular speed ω and the angular acceleration ω˙ at t0. Figure 9.14: (Filename:sfig9.rolling.may00) (b) How many revolutions does the cylineder make in the next 2 seconds? (c) What is the distance travelled by the center of mass in those 2 seconds? Solution This problem is about simple kinematic calculations. We are given the velocity, x˙, and the acceleration, xddot, of the center of mass. We are supposed to find angular velocity ω, angular acceleration ω˙ , angular displacement θ in 2 seconds, and the corresponding linear distance x along the incline. The radius of the cylinder R = diameter/2 = 0.25 m. (a) From the kinematics of pure rolling, ω = x˙ = 0.5 m/s = 2 rad/s, R 0.25 m ω˙ = x¨ = 0.1 m/s2 = 0.4 rad/s2. R 0.25 m ω = 2 rad/s, ω˙ = 0.4 rad/s2 (b) We can find the number of revolutions the cylinder makes in 2 seconds by solving for the angular dispcament θ in this time period. Since, θ¨ ≡ ω˙ = constant, we integrate this equation twice and substitute the initial conditions, θ˙(t = 0) = ω = 2 rad/s and θ(t = 0) = 0, to get θ (t) = ωt + 1 ω˙ t2 2 ⇒ θ (t = 2 s) = (2 rad/s) · (2 s) + 1 (0.4 rad/s) · (4 s2) 2 = 4.8 rad = 4.8 rev = 0.76 rev. 2π θ = 0.76 rev (c) Now that we know the angular displacement θ , the distance travelled by the mass-center is the arc-length corresponding to θ, i.e., x = Rθ = (0.25 m) · (4.8) = 1.2 m. x = 1.2 m Note that we could have found the distance travelled by the mass-center by integrating the equation x¨ = 0.1 m/s2 twice
518 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.8 Condition of pure rolling. A cylinder of radius R = 20 cm rolls on a flat surface with absolute angular speed ω = 12 rad/s under the conditions shown in the figure (In cases (ii) and (iii), you may think of the ‘flat surface’ as a conveyor belt). In each case, (a) Write the condition of pure rolling. (b) Find the velocity of the center C of the cylinder. C ˆ P C ıˆ C (i) Fixed base P v0 = 1m/s P v0 = 1m/s (iii) Base moves to the left (ii) Base moves to the right Figure 9.15: (Filename:sfig7.rolling1) C Solution At any instant during rolling, the cylinder makes a point-contact with the ˆ flat surface. Let the point of instantaneous contact on the cylinder be P, and let the corresponding point on the flat surface be Q. The condition of pure rolling, in each ıˆ P case, is v P = v Q, that is, there is no relative motion between the two contacting points (a relative motion will imply slip). Now, we analyze each case. Q Case(i) In this case, the bottom surface is fixed. Therefore, Figure 9.16: The cylinder rolls on the (a) The condition of pure rolling is: v P = v Q = 0. (b) Velocity of the center: flat surface. Instantaneously, point P on the cylinder is in contact with point Q on v C = v P + ω × r C/P = 0 + (−ωkˆ) × Rˆ the flat surface. For pure rolling, points P = ω Rıˆ = (12 rad/s) · (0.2 m)ıˆ = 2.4 m/sıˆ. and Q must have the same velocity. Case(ii) In this case, the bottom surface moves with velocity v = 1 m/sıˆ. Therefore, (Filename:sfig7.rolling1a) v Q = 1 m/sıˆ. Thus, (a) The condition of pure rolling is: v P = v Q = 1 m/sıˆ. (b) Velocity of the center: v C = v P + ω × r C/P = v0ıˆ + ω Rıˆ = 1 m/sıˆ + 2.4 m/sıˆ = 3.4 m/sıˆ. Case(iii) In this case, the bottom surface moves with velocity v = −1 m/sıˆ. There- fore, v Q = −1 m/sıˆ. Thus, (a) The condition of pure rolling is: v P = v Q = −1 m/sıˆ. (b) Velocity of the center: v C = v P + ω × r C/P = −v0ıˆ + ω Rıˆ = −1 m/sıˆ + 2.4 m/sıˆ = 1.4 m/sıˆ. (a): (i) v P = 0, (ii) v P = 1 m/sıˆ, (iii) v P = −1 m/sıˆ, (b): (i) v C = 2.4 m/sıˆ, (ii) v C = 3.4 m/sıˆ, (iii) v C = 1.4 m/sıˆ
9.3. Special topics in planar kinematics 519
520 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.9 Motion of a point on a disk rolling inside a cylinder. A uniform disk of radius r rolls without slipping with constant angular speed ω inside a fixed cylinder of radius R. A point P is marked on the disk at a distance ( < r ) from the center of the disk. at a general time t during rolling, find R (a) the position of point P, (b) the velocity of point P, and (c) the acceleration of point P r P Figure 9.17: A uniform disk of radius r Solution Let the disk be vertically below the center of the cylinder at t = 0 s such that point P is vertically above the center of the disk (Fig. 9.18). At this instant, Q rolls without slipping inside a fixed cylin- der. is the point of contact between the disk and the cylinder. Let the disk roll for time t such that at instant t the line joining the two centers (line OC) makes an angle φ (Filename:sfig6.5.3) with its vertical position at t = 0 s. Since the disk has rolled for time t at a constant angular speed ω, point P has rotated counter-clockwise by an angle θ = ωt from its original vertical position P . yy P' O φ O x φ x rC θ θ C φ ω r P/C C DP P l P C Q (a) (b) Figure 9.18: Geometry of motion: keeping track of point P while the disk rolls for time t, rotating by angle θ = ωt inside the cylinder. (Filename:sfig6.5.3a) (a) Position of point P: From Fig. 9.18(b) we can write r P = r C + r P/C = (R − r )λˆ OC + λˆ C P where λˆ OC = a unit vector along OC = − sin φıˆ − cos φˆ, λˆ C P = a unit vector along C P = − sin θ ıˆ + cos θ ˆ. Thus, r P = [−(R − r ) sin φ − sin θ ]ıˆ + [−(R − r ) cos φ + cos θ ]ˆ. We have thus obtained an expression for the position vector of point P as a function of φ and θ. Since we also want to find velocity and acceleration of point P, it will be nice to express r P as a function of t. As noted above, θ = ωt; but how do we find φ as a function of t? Note that the center of the disk C is going around point O in circles with angular velocity −φ˙kˆ. The disk,
9.3. Special topics in planar kinematics 521 however, is rotating with angular velocity ω = ωkˆ about the instantaneous center of rotation, point D. Therefore, we can calculate the velocity of point C in two ways: vC = vC or ω × r C/D = −φ˙kˆ × r C/O or ωkˆ × r (−λˆ OC ) = −φ˙kˆ × (R − r )λˆ OC or − ωr (kˆ × λˆ OC ) = −φ˙(R − r )(kˆ × λˆ OC ) ⇒ R r r ω = φ˙ . − Integrating the last expression with respect to time, we obtain φ = R r r ωt . − Let r − q = R r , then, the position vector of point P may now be written as r P = [−(R − r ) sin(qωt) − sin(ωt)]ıˆ + [−(R − r ) cos(qωt) + cos(ωt)]ˆ. (9.27) (b) Velocity of point P: Differentiating Eqn. (9.27) once with respect to time we get v P = −ω[(R−r )q cos(qωt)+ cos(ωt)]ıˆ+ω[(R−r )q sin(qωt)− sin(ωt)]ˆ. Substituting (R − r )q = r in v P we get v P = −ωr [{cos(qωt) + r cos(ωt)}ıˆ − {sin(qωt) − r sin(ωt)}ˆ]. (9.28) (c) Acceleration of point P: Differentiating Eqn. (9.28) once with respect to time we get a P = −ω2r [−{q sin(qωt) + r sin(ωt)}ıˆ − {q cos(qωt) − r cos(ωt)}ˆ]. (9.29) r P = [−(R − r ) sin(qωt) − sin(ωt)]ıˆ + [−(R − r ) cos(qωt) + cos(ωt)]ˆ v P = −ωr [{cos(qωt) + r cos(ωt)}ıˆ − {sin(qωt) − r sin(ωt)}ˆ] a P = −ω2r [−{q sin(qωt) + r sin(ωt)}ıˆ − {q cos(qωt) − r cos(ωt)}ˆ]
522 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.10 The rolling disk: instantaneous kinematics. For the rolling disk in Sample 9.9, let R = 4 ft, r = 1 ft and point P be on the rim of the disk. Assume that at t = 0, the center of the disk is vertically below the center of the cylinder and point P is on the vertical line joining the two centers. If the disk is rolling at a constant speed ω = π rad/s, find (a) the position of point P and center C at t = 1 s, 3 s, and 5.25 s, (b) the velocity of point P and center C at those instants, and (c) the acceleration of point P and center C at the same instants as above. Draw the position of the disk at the three instants and show the velocities and accel- erations found above. Solution The general expressions for position, velocity, and acceleration of point P obtained in Sample 9.9 can be used to find the position, velocity, and acceleration of any point on the disk by substituting an appropriate value of in equations (9.27), (9.28), and (9.29). Since R = 4r , q = r = 1. R−r 3 Now, point P is on the rim of the disk and point C is the center of the disk. Therefore, for point P: = r, for point C: = 0. Substituting these values for , and q = 1/3 in equations (9.27), (9.28), and (9.29) we get the following. (a) Position: ωt ωt r C = −3r sin 3 ıˆ + cos 3 ˆ , r P = r C + r −sin (ωt) ıˆ + cos (ωt) ˆ . (b) Velocity: vC = −ωr cos ωt ıˆ − sin ωt ˆ , 3 3 v P = −ωr cos ωt + cos (ωt) ıˆ − sin ωt − sin (ωt) ˆ . 33 (c) Acceleration: aC = ω2r sin ωt ıˆ + cos ωt ˆ , 3 3 3 a P = ω2r 1 ωt + sin (ωt) ıˆ + 1 ωt − cos (ωt) ˆ . sin cos 33 33 We can now use these expressions to find the position, velocity, and acceleration of the two points at the instants of interest by substituting r = 1 ft, ω = π rad/s, and appropriate values of t. These values are shown in Table 9.1. The velocity and acceleration of the two points are shown in Figures 9.19(a) and (b) respectively. It is worthwhile to check the directions of velocities and the accelerations by thinking about the velocity and acceleration of point P as a vector sum of the velocity (same for acceleration) of the center of the disk and the velocity (same for acceleration) of point P with respect to the center of the disk. Since the motions involved are circular motions at constant rate, a visual inspection of the velocities and the accelerations is not very difficult. Try it.
9.3. Special topics in planar kinematics 523 t 1s 3s 5.25 s r C ( ft) √ r P ( ft) 3 1 3( √1 ıˆ − √1 ˆ) v C ( ft/s) 3(− 2 ıˆ − 2 ˆ) 3ˆ v P ( ft/s) 22 aC ( ft/s2) r C − ˆ r C − ˆ v P ( ft/s2) 4( √1 ıˆ − √1 ˆ) 1 √ 2 3 22 π (− ıˆ + 2 ˆ) π ıˆ π(− √1 ıˆ − √1 ˆ) √ 1 3 22 π ( 2 ıˆ + 2 ˆ) 2π ıˆ 0 π2 √ π2 π2 √1 ˆ) 3 3 1 3 3 √1 ( 2 ıˆ + 2 ˆ) − ˆ (− 2 ıˆ + 2 11.86(.24ıˆ + .97ˆ) 2π 2 ˆ 13.16(− √1 ıˆ + √1 ˆ) 3 22 Table 9.1: Position, velocity, and acceleration of point P and point C t=3s C t=3s C P P vC P t = 5.25 s aC P t = 5.25 s C vP C C aP C C C P P P P t=1s t=1s t=0s t=0s (a) (b) Figure 9.19: (a) Velocity and (b) Acceleration of points P and C at t = 1 s, 3 s, and 5.25 s. (Filename:sfig6.5.4a)
524 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.11 The rolling disk: path of a point on the disk. For the rolling disk in Sample 9.9, take ω = π rad/s. Draw the path of a point on the rim of the disk for one complete revolution of the center of the disk around the cylinder for the following conditions: (a) R = 8r , (b) R = 4r , and (c) R = 2r . Solution In Sample 9.9, we obtained a general expression for the position of a point on the disk as a function of time. By computing the position of the point for various values of time t up to the time required to go around the cylinder for one complete cycle, we can draw the path of the point. For the various given conditions, the variable that changes in Eqn. (9.27) is q. We can write a computer program to generate the path of any point on the disk for a given set of R and r . Here is a function written in MATLAB to generate the required path according to Eqn. (9.27). MATLAB function to plot the path of a point on the disk: program rollingdisk %------------------------------------------------------------- % This program will plots the path of any point on a disk of radius % ’r’ rolling with speed ’w’ inside a cylinder of radius ’R’. % The point of interest is distance ’l’ away from the center of % the disk. The coordinates x and y of the specified point P are % calculated according to the relation mentioned above. %-------------------------------------------------------------- phi = pi/50*[1,2,3,...,100] % make a vector phi from 0 to 2*pi x = R*cos(phi) % create points on the outer cylinder y = R*sin(phi) plot y vs x % plot the outer cylinder hold this plot % hold to overlay plots of paths q = r/(R-r) % calculate q. T = 2*pi/(q*w) % calculate time T for going aound- % the cylinder once at speed ’w’. t = T/100*[1,2,3, ..., 100] % make a time vector t from 0 to T- % taking 101 points. rcx = -(R-r)*(sin(q*w*t)) % find the x coordinates of pt. C. rcy = -(R-r)*(cos(q*w*t)) % find the y coordinates of pt. C. rpx = rcx-l*sin(w*t) % find the x coordinates of pt. P. rpy = rcy + l*cos(q*t) % find the y coordinates of pt. P. plot rpy vs rpx % plot the path of P and the path plot rcy vs rcx % of C. For path of C Once coded, we can use this program to plot the paths of both the center and the point P on the rim of the disk for the three given situations. Note that for any point on the rim of the disk l = r (see Fig 9.18).
9.3. Special topics in planar kinematics 525 (a) Let R = 4 units. Then r = 0.5 for R = 8r . To plot the required path, we run 4 our program rollingdisk with desired input, R=4 3 r = 0.5 w = pi 2 l = 0.5 execute rollingdisk 1 Path of point P Path of The plot generated is shown in Fig.9.20 with a few graphic elements added for 0 center C illustrative purposes. yy y R = 8r (b) Similarly, for R = 4r we type: -1 R=4 r=1 -2 P 4 w = pi -3 l=1 -4-4 -2 C execute rollingdisk 02 x to plot the desired paths. The plot generated in this case is shown in Fig.9.21 Figure 9.20: Path of point P and the cen- (c) The last one is the most interesting case. The plot obtained in this case by typing: ter C of the disk for R = 8r . R=4 r=2 (Filename:sfig6.5.5a) w = pi l=2 4 execute rollingdisk is shown in Fig.9.22. Point P just travels on a straight line! In fact, every point 3 on the rim of the disk goes back and forth on a straight line. Most people find this motion odd at first sight. You can roughly verify the result by cutting a 2 whole twice the diameter of a coin (say a US quarter or dime) in a piece of Path of Path of cardboard and rolling the coin around inside while watching a marked point on the perimeter. 1 point P center C A curiosity. We just discovered something simple about the path of a point on the 0 edge of a circle rolling in another circle that is twice as big. The edge point moves R = 4r in a straight line. In contrast one might think about the motion of the center G of a straight line segment that slides against two straight walls as in sample 9.20. A -1 problem couldn’t be more different. Naturally the path of point G is a circle (as you -2 P can check physically by looking at the middle of a ruler as you hold it as you sliding against a wall-floor corner). -3 C 4 -4 02 -4 -2 x Figure 9.21: Path of point P and the cen- ter C of the disk for R = 4r . (Filename:sfig6.5.5b) 4 3 Path of 2 center C Path of R = 4r 1 point P 0 P -1 -2 C -3 -4-4 -2 0 2 4 x Figure 9.22: Path of point P and the cen- ter C of the disk for R = 2r . (Filename:sfig6.5.5c)
526 CHAPTER 9. General planar motion of a rigid body 9.4 Mechanics of contacting bodies: rolling and sliding A typical machine part has forces that come from contact with other parts. In fact, with the major exception of gravity, most of the forces that act on bodies of engineering interest come from contact. Many of the forces you have drawn in free body diagrams have been contact forces: The force of the ground on an ideal wheel, of an axle on a bearing, etc. We’d now like to consider some mechanics problems that involve sliding or rolling contact. Once you understand the kinematics from the previous section, there is nothing new in the mechanics. As always, the mechanics is linear momentum balance, angular momentum balance and energy balance. Because we are considering single rigid bodies in 2D the expressions for the motion quantities are especially simple (as you can look up in Table I at the back of the book): L˙ = mtotacm, H˙ C = rcm/C × (mtot acm) + I ω˙ kˆ (where I = Izczm), and EK = mtotvc2m /2 + I ω2/2. The key to success, as usual, is the drawing of appropriate free body diagrams (see Chapter 3 pages 88-91 and Chapter 6 pages 328-9). The two cases one needs to consider as possible are rolling, where the contact point has no relative velocity and the tangential reaction force is unknown but less than µN , and sliding where the relative velocity could be anything and the tangential reaction force is usually assumed to have a magnitude of µN but oppose the relative motion. For friction forces in rolling refer to chapter 2 on free body diagrams. Note that in pure rolling contact, the contact force does no work because the material point of contact has no velocity. However, when there is sliding mechanical energy is dissipated unless there is no friction. Work-energy relations and impulse-momentum relations are useful to solve some problems. As for various problems occurring earlier in the text, it is often a savings of calculation to use angular momentum balance relative to a contact point.
9.4. Mechanics of contacting bodies: rolling and sliding 527 SAMPLE 9.12 A rolling wheel with mass. Consider the wheel with mass m shown λˆ φ in figure ??. The free body diagram of the wheel is shown here again. Write the equation of motion of the wheel. Solution We can write the equation of motion of the wheel in terms of either the RF ˆ center of mass position x or the angular displacement of the wheel θ. Since in pure rolling, these two variables share a simple relationship (x = Rθ ), we can easily get cm ıˆ the equation of motion in terms of x if we have the equation in terms of θ and vice versa. Since all the forces are shown in the free body diagram , we can easily write mg C the angular momentum balance for the wheel. We choose the point of contact C as Ffriction our reference point for the angular momentum balance (because the gravity force, N −mgˆ, the friction force −Ff rictionıˆ, and the normal reaction of the ground N ˆ, all pass through the contact point C and therefore, produce no moment about this point). Figure 9.23: FBD of a wheel with mass We have m. Force F is applied by the axle. (Filename:tfigure2.wheel.mass.lhs) where MC = H˙ C Rˆ MC = r cm/C ×(Fλˆ ) = Rˆ × F(− cos φıˆ − sin φˆ) = F R cos φkˆ and H˙ C = rcm/C × m a cm + Izczmω˙ = Rˆ × m x¨ ıˆ − Izczmω˙ kˆ ω˙ R = −mω˙ R2kˆ − Izczmω˙ kˆ = −(Izczm + m R2)ω˙ kˆ. Thus, F R cos φkˆ = −(Izczm + m R2)ω˙ kˆ ⇒ ω˙ ≡ θ¨ = F R cos φ Izczm + m R2 which is the equation of motion we are looking for. Note that we can easily substitute θ¨ = x¨/R in the equation of motion above to get the equation of motion in terms of the center of mass displacement x as x¨ = FR 2 cos φ . Izczm + mR2 θ¨ = F R cos φ Izczm+m R2 Comments: We could have, of course, used linear momentum balance also to derive the equation of motion. Note, however, that the linear momentum balance will essentially give two scalar equations in the x and y directions involving all forces shown in the free body diagram . The angular momentum balance , on the other hand, gets rid of some of them. Depending on which forces are known, we may or may not need to use all the three scalar equations. In the final equation of motion, we must have only one unknown.
528 CHAPTER 9. General planar motion of a rigid body cm SAMPLE 9.13 Energy and power of a rolling wheel. A wheel of diameter 2 ft and r mass 20 lbm rolls without slipping on a horizontal surface. The kinetic energy of the Figure 9.24: (Filename:sfig7.4.1a) wheel is 1700 ft· lbf. Assume the wheel to be a thin, uniform disk. (a) Find the rate of rotation of the wheel. (b) Find the average power required to bring the wheel to a complete stop in 5 s. Solution (a) Let ω be the rate of rotation of the wheel. Since the wheel rotates without slip, its center of mass moves with speed vcm = ωr . The wheel has both translational and rotational kinetic energy. The total kinetic energy is EK = 1 m vc2m + 1 I cmω2 2 2 = 1 mω2r 2 + 1 I cmω2 22 = 1 (mr 2 + I cm )ω2 2 1 2 mr 2 = 3 mr2ω2 4 ⇒ ω2 = 4EK 3mr 2 = 4 × 1700 ft· lbf 3 × 20 lbm·1 ft2 4 × 1700 × 32.2 lbm· ft/ s2 = 3 × 20 lbm· ft = 3649.33 1 s2 ⇒ ω = 60.4 rad/s. ω = 60.4 rad/s Note: This rotational speed, by the way, is extremely high. At this speed the center of mass moves at 60.4 ft/s! (b) Power is the rate of work done on a body or the rate of change of kinetic energy. Here we are given the initial kinetic energy, the final kinetic energy (zero) and the time to achieve the final state. Therefore, the avarage power is, P = EK1 − EK2 t = 1700 ft· lbf − 0 = 340 ft· lbf/ s 5s = 340 ft· lbf/ s · 1 hp 550 ft· lbf/ s = 0.62 hp P = 0.62 hp
9.4. Mechanics of contacting bodies: rolling and sliding 529 SAMPLE 9.14 Equation of motion of a rolling wheel from energy balance. Consider λˆ φ the wheel with mass m from figure ??. The free body diagram of the wheel is shown F here again. Derive the equation of motion of the wheel using energy balance. R cm ˆ mg ıˆ C Ffriction N Solution From energy balance, we have Figure 9.25: FBD of a rolling wheel. P = E˙K (Filename:tfigure2.wheel.mass.lhs.en) where P= Fi · vi 00 vıˆ vıˆ = Ff rictionıˆ · v C +N ˆ · v C −mgˆ · v cm +Fλˆ · v cm = −mgv(ıˆ · ˆ) + Fv( λˆ · ıˆ ) 0 − cos φ = −Fv cos φ and (x˙ / R )2 E˙ K = d ( 1 mx˙2 + 1 Izczm ω˙ 2 ) dt 2 2 = 1 d [(m + Izczm / R 2 )x˙ 2 ] 2 dt = (m + Izczm/R2)x˙ x¨. Thus, −Fv cos φ = (m + Izczm/R2)x˙ x¨ or − F x˙ cos φ = (m + Izczm/R2)x˙ x¨ ⇒ x¨ = − m F cos φ . + Izczm / R 2 We can also write the equation of motion in terms of θ by replacing x¨ with θ¨ R giving, θ¨ = F R cos φ . m + Izczm/R2 x¨ = − F cos φ 2 m+ Izczm / R Comments: In the equations above (for calculating P), we have set v C = 0 because in pure rolling, the instantaneous velocity of the contact point is zero. Note that the force due to gravity is normal to the direction of the velocity of the center of mass. So, the only power supplied to the wheel is due to the force Fλˆ acting at the center of mass.
530 CHAPTER 9. General planar motion of a rigid body C SAMPLE 9.15 Equation of motion of a rolling disk on an incline. A unifrom circular ˆ disk of mass m = 1 kg and radius R = 0.4 m rolls down an inclined shown in the ıˆ figure. Write the equation of motion of the disk assuming pure rolling, and find the distance travelled by the center of mass in 2 s. Figure 9.26: (Filename:sfig9.rolling.incline1) Solution The free body diagram of the disk is shown in Fig. 9.27. In addition to the base unit vectors ıˆ and ˆ, let us use unit vectors λˆ and nˆ along the plane and perpendicular to the plane, respectively, to express various vectors. We can write the equation of motion using linear momentum balance or angular momentum balance. However, note that if we use linear momentum balance we have two unknown forces in the equation. On the otherhand, if we use angular momentum balance about the contact point C, these forces do not show up in the equation. So, let us use angular momentum balance about point C: R where MC = H˙ C O (cm) and MC = rO/C × mg = Rnˆ × (−mgˆ) C mg ˆ Thus, = −Rmg sin αkˆ N F nˆ ıˆ Rnˆ Rω˙ λˆ α λˆ H˙ C = −Izczmω˙ kˆ + rO/C ×m acm Figure 9.27: (Filename:sfig9.rolling.incline1a) = −Izczmω˙ kˆ + m R2ω˙ (nˆ × λˆ ) = −(Izczm + m R2)ω˙ kˆ. −Rmg sin αkˆ = −(Izczm + m R2)ω˙ kˆ g sin α ⇒ ω˙ = R[1 + Izczm/(m R2)] ω˙ = g sin α R[1+Izczm/(m R2)] Note that in the above equation of motion, the right hand side is constant. So, we can solve the equation for ω and θ by simply integrating this equation and substituting the initial conditions ω(t = 0) = 0 and θ (t = 0) = 0. Let us write the equation of motion as ω˙ = β where β = g sin α/R(1 + Izczm/m R2). Then, ω ≡ θ˙ = βt + C1 θ = 1βt2 + C1t + C2. 2 Substituting the given initial condtions θ˙(0) = 0 and θ (0) = 0, we get C1 = 0 and 1 C2 = 0, which implies that θ = 2 β t 2. Now, in pure rolling, x = Rθ. Therefore, x (t ) = Rθ (t) = 1βt2 = R · 1 g sin α t 2 2 2 + Izczm/m R(1 R 2) = 1 g sin α t2 = 1 (g sin α)t2 2 3 1+ 1 m R2 2 m R2 x(2 s) = 1 · 9.8 m/s2 · sin(30o) · (2 s)2 = 6.53 m 3 x(2 s) = 6.53 m
9.4. Mechanics of contacting bodies: rolling and sliding 531 SAMPLE 9.16 Using Work and energy in pure rolling. Consider the disk of Sam- ple 9.15 rolling down the incline again. Suppose the disk starts rolling from rest. Find the speed of the center of mass when the disk is 2 m down the inclined plane. Solution We are given that the disk rolls down, starting with zero initial velocity. We C are to find the speed of the center of mass after it has travelled 2 malong the incline. ˆ We can, of course, solve this problem using equation of motion, by first solving for the ıˆ time t the disk takes to travel the given distance and then evaluating the expression for speed ω(t) or x(t) at that t. However, it is usually easier to use work energy Figure 9.28: (Filename:sfig9.rolling.incline2) principle whenever positions are specified at two instants, speed is specified at one of those instants, and speed is to be found at the other instant. This is because we can, presumably, compute the work done on the system in travelling the specified distance and relate it to the change in kinetic energy of the system between the two instants. In the problem given here, let ω1 and ω2 be the initial and final (after rolling down by d = 2 m) angular speeds of the disk, respectively. We know that in rolling, the kinetic energy is given by (ω R)2 R O (cm) EK = 1 vc2m + 1 Izczmω2 = 1 (m R2 + Izczm)ω2. C mg ˆ m 2 2 2 Therefore, F nˆ ıˆ N α λˆ EK = EK2 − EK1 = 1 (m R2 + Izczm)(ω22 − ω12) (9.30) 2 Now, let us calculate the work done by all the forces acting on the disk during the displacement of he mass-center by d along the plane. Note that in ideal rolling, the Figure 9.29: (Filename:sfig9.rolling.incline2a) contact forces do no work. Therefore, the work done on the disk is only due to the gravitational force: − sin α (9.31) W = (−mgˆ) · (dλˆ ) = −mgd( ˆ · λˆ ) = mgd sin α From work-energy principle (integral form of power balance, P = E˙K ), we know that W = EK. Therefore, from eqn. (9.30) and eqn. (9.31), we get mgd sin α = 1 (m R2 + Izczm)(ω22 − ω12) 2 ⇒ ω22 = ω12 + 2mgd sin α = ω12 + 2gd sin α m R2 + Izczm R2 1+ Izczm m R2 = ω12 + 4gd sin α 3R2 Substituting the values of g, d, α, R, etc., and setting ω1 = 0, we get ω22 = 4 · (9.8 m/s2) · (2 m) · (sin(30o) = 81.67/s2 3 · (0.4 m)2 ⇒ ω2 = 9.04 rad/s. The corresponding speed of the center of mass is vcm = ω2 R = 9.04 rad/s · 0.4 m = 3.61 m/s. vcm = 3.61 m/s
532 CHAPTER 9. General planar motion of a rigid body SAMPLE 9.17 Impulse and momentum calculations in pure rolling. Consider the disk of Sample 9.15 rolling down the incline again. Find an expression for the rolling speed (ω) of the disk after a finite time t, given the initial rolling speed ω1. C ˆ ıˆ Figure 9.30: (Filename:sfig9.rolling.incline3) Solution Once again, this problem can be solved by integrating the equation of motion (as done in Sample 9.15). However, we will solve this problem here using impulse-momentum relationship. Note that we need the speed of the disk ω2, after a finite time t, given the initial speed ω1. Since the forces acting on the disk do not change during this time (assuming pure rolling), it is easy to calculate impulse and then relate it to the change in the momenta of the disk between the two instants. Now, from the linear impulse momentum relatioship, F · t = L2 − L1, we have R (−Fλˆ + N ıˆ − mgˆ) t = m(v2 − v1)λˆ (9.32) O (cm) C mg ˆ Dotting eqn. (9.32) with λˆ gives N F nˆ ıˆ (−F − mg( ˆ · λˆ )) t = m(v2 − v1) α λˆ − sin α (9.33) (−F + mg sin α) t = m R(ω2 − ω1) Figure 9.31: (Filename:sfig9.rolling.incline3a) Similarly, the angular impulse-momentum relationship about the mass-center, MO t = (H O)2 − (H O)1, gives (−F Rkˆ) t = −Izczm(ω2 − ω1)kˆ (9.34) ⇒ F R t = Izczm(ω2 − ω1) Note that the other forces (N and mg) do not produce any moment about the mass- center as they pass through this point. We can now eliminate the unknown force F from eqn. (9.33) and eqn. (9.34) by multiplying eqn. (9.33) with R and adding to eqn. (9.34): mg R sin α t = (Izczm + m R2)(ω2 − ω1) or g sin α t = R 1 + Izczm (ω2 − ω1) m R2 ⇒ ω2 = ω1 + g sin α t R 1 + Izczm m R2 ω2 = ω1 + g sin α t R 1+ Izczm m R2
9.4. Mechanics of contacting bodies: rolling and sliding 533 SAMPLE 9.18 Work and energy calculations in sliding. A block of mass m = 2.5 kg 1 slides down a frictionless incline from a 5 m height. The block encounters a frictional bed AB of length 1 m on the ground. If the speed of the block is 9 m/s at point B, 5m find the coefficient of friction between the block and the frictional surface AB. m 2 AB 1m Figure 9.32: (Filename:sfig2.9.1a) Solution We can divide the problem in two parts: We first find the speed of the block as it reaches point A using conservation of energy for its motion on the inclined surface, and then use work-energy principle to find the speed at B. Let the ground level be the datum for P.E. and let v be the speed at A. For the motion on the incline; EK1 + EP1 = EK2 + EP2 0 + mgh = 1mv2 + 0 2 ⇒ v = 2gh = 2 · 9.81 m/s2 · 5 m = 9.90 m/s. Now, as the block slides on the surface AB, a force of friction = µ N = µmg (since ˆ mg N = mg, from linear momentum balance in the vertical direction) acts in the opposite direction of motion (see Fig. 9.33. Work done by this force on the block is, W = F·r ıˆ m = −µmgıˆ · Lıˆ µkN = −µmgL From work energy relationship we have, N W = EK = EK2 − EK1 Figure 9.33: Free body diagram of the ⇒ EK2 = EK1 + W block when the block is on the rough sur- face. 1 1 mv2 − µmgL 2 2 (Filename:sfig2.9.1b) m v2B = −µm g L = 1 m (v2B − v2) 2 ⇒ µ = 1 (v2B − v2) 2g L = (9.90 m/s)2 − (9 m/s)2 2 · 9.81 m/s2 · 1 m = 0.87 µ = 0.87
Search
Read the Text Version
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
- 167
- 168
- 169
- 170
- 171
- 172
- 173
- 174
- 175
- 176
- 177
- 178
- 179
- 180
- 181
- 182
- 183
- 184
- 185
- 186
- 187
- 188
- 189
- 190
- 191
- 192
- 193
- 194
- 195
- 196
- 197
- 198
- 199
- 200
- 201
- 202
- 203
- 204
- 205
- 206
- 207
- 208
- 209
- 210
- 211
- 212
- 213
- 214
- 215
- 216
- 217
- 218
- 219
- 220
- 221
- 222
- 223
- 224
- 225
- 226
- 227
- 228
- 229
- 230
- 231
- 232
- 233
- 234
- 235
- 236
- 237
- 238
- 239
- 240
- 241
- 242
- 243
- 244
- 245
- 246
- 247
- 248
- 249
- 250
- 251
- 252
- 253
- 254
- 255
- 256
- 257
- 258
- 259
- 260
- 261
- 262
- 263
- 264
- 265
- 266
- 267
- 268
- 269
- 270
- 271
- 272
- 273
- 274
- 275
- 276
- 277
- 278
- 279
- 280
- 281
- 282
- 283
- 284
- 285
- 286
- 287
- 288
- 289
- 290
- 291
- 292
- 293
- 294
- 295
- 296
- 297
- 298
- 299
- 300
- 301
- 302
- 303
- 304
- 305
- 306
- 307
- 308
- 309
- 310
- 311
- 312
- 313
- 314
- 315
- 316
- 317
- 318
- 319
- 320
- 321
- 322
- 323
- 324
- 325
- 326
- 327
- 328
- 329
- 330
- 331
- 332
- 333
- 334
- 335
- 336
- 337
- 338
- 339
- 340
- 341
- 342
- 343
- 344
- 345
- 346
- 347
- 348
- 349
- 350
- 351
- 352
- 353
- 354
- 355
- 356
- 357
- 358
- 359
- 360
- 361
- 362
- 363
- 364
- 365
- 366
- 367
- 368
- 369
- 370
- 371
- 372
- 373
- 374
- 375
- 376
- 377
- 378
- 379
- 380
- 381
- 382
- 383
- 384
- 385
- 386
- 387
- 388
- 389
- 390
- 391
- 392
- 393
- 394
- 395
- 396
- 397
- 398
- 399
- 400
- 401
- 402
- 403
- 404
- 405
- 406
- 407
- 408
- 409
- 410
- 411
- 412
- 413
- 414
- 415
- 416
- 417
- 418
- 419
- 420
- 421
- 422
- 423
- 424
- 425
- 426
- 427
- 428
- 429
- 430
- 431
- 432
- 433
- 434
- 435
- 436
- 437
- 438
- 439
- 440
- 441
- 442
- 443
- 444
- 445
- 446
- 447
- 448
- 449
- 450
- 451
- 452
- 453
- 454
- 455
- 456
- 457
- 458
- 459
- 460
- 461
- 462
- 463
- 464
- 465
- 466
- 467
- 468
- 469
- 470
- 471
- 472
- 473
- 474
- 475
- 476
- 477
- 478
- 479
- 480
- 481
- 482
- 483
- 484
- 485
- 486
- 487
- 488
- 489
- 490
- 491
- 492
- 493
- 494
- 495
- 496
- 497
- 498
- 499
- 500
- 501
- 502
- 503
- 504
- 505
- 506
- 507
- 508
- 509
- 510
- 511
- 512
- 513
- 514
- 515
- 516
- 517
- 518
- 519
- 520
- 521
- 522
- 523
- 524
- 525
- 526
- 527
- 528
- 529
- 530
- 531
- 532
- 533
- 534
- 535
- 536
- 537
- 538
- 539
- 540
- 541
- 542
- 543
- 544
- 545
- 546
- 547
- 548
- 549
- 550
- 551
- 552
- 553
- 554
- 555
- 556
- 557
- 558
- 559
- 560
- 561
- 562
- 563
- 564
- 565
- 566
- 567
- 568
- 569
- 570
- 571
- 572
- 573
- 574
- 575
- 576
- 577
- 578
- 579
- 580
- 581
- 582
- 583
- 584
- 585
- 586
- 587
- 588
- 589
- 590
- 591
- 592
- 593
- 594
- 595
- 596
- 597
- 598
- 599
- 600
- 601
- 602
- 603
- 604
- 605
- 1 - 50
- 51 - 100
- 101 - 150
- 151 - 200
- 201 - 250
- 251 - 300
- 301 - 350
- 351 - 400
- 401 - 450
- 451 - 500
- 501 - 550
- 551 - 600
- 601 - 605
Pages: