Chapter 3: Modeling of Rigid-Body Mechanical Systems PDF

118 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems that we treat the case of general motion in a plane, involving simultaneous translation and rotation. In Chapter 4 we will consider systems that have nonrigid, or elastic, behavior. ■ 3.1 TRANSLATIONAL MOTION A particle is a mass of negligible dimensions. We can consider a body to be a particle if its dimensions are irrelevant for specifying its position and the forces acting on it. For example, we normally need not know the size of an earth satellite to describe its orbital path. Newton’s ﬁrst law states that a particle originally at rest, or moving in a straight line with a constant speed, will remain that way as long as it is not acted upon by an unbalanced external force. Newton’s second law states that the acceleration of a mass particle is proportional to the vector resultant force acting on it and is in the direction of this force. Newton’s third law states that the forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. The third law is summarized by the commonly used statement that every action is opposed by an equal reaction. For an object treated as a particle of mass m, the second law can be expressed as dv =f (3.1.1) ma = m dt where a and v are the acceleration and velocity vectors of the mass and f is the force vector acting on the mass (Figure 3.1.1). Note that the acceleration vector and the force vector lie on the same line. If the mass is constrained to move in only one direction, say along the direction of the coordinate x, then the equation of motion is the scalar equation dv = f (3.1.2) ma = m dt or f dv = =a (3.1.3) dt m It will be convenient to use the following abbreviated “dot” notation for time derivatives: d2x dx ẍ(t) = 2 ẋ(t) = dt dt Thus, we can express the scalar form of Newton’s law as m v̇ = f . y Figure 3.1.1 Particle motion showing the coordinate system, the applied force f, and the resulting acceleration a, velocity v, and path. a v m f z x 3. 1 Translational Motion If we assume that the object is a rigid body and we neglect the force distribution within the object, we can treat the object as if its mass were concentrated at its mass center. This is the point mass assumption, which makes it easier to obtain the translational equations of motion, because the object’s dimensions can be ignored and all external forces can be treated as if they acted through the mass center. If the object can rotate, then the translational equations must be supplemented with the rotational equations of motion, which are treated in Sections 3.2 and 3.4. 3.1.1 MECHANICAL ENERGY Conservation of mechanical energy is a direct consequence of Newton’s second law. Consider the scalar case (3.1.2), where the force f can be a function of displacement x. m v̇ = f (x) Multiply both sides by v dt and use the fact that v = d x/dt. dx mv dv = v f (x) dt = f (x) dt = f (x) d x dt Integrate both sides. mv 2 = mv dv = f (x) d x + C (3.1.4) 2 where C is a constant of integration. Work is force times displacement, so the integral on the right represents the total work done on the mass by the force f (x). The term on the left-hand side of the equal sign is called the kinetic energy (KE). If the work done by f (x) is independent of the path and depends only on the end points, then the force f (x) is derivable from a function V (x) as follows: dV (3.1.5) f (x) = − dx Then, in this case, f (x) is called a conservative force. If we integrate both sides of the last equation, we obtain V (x) = dV = − f (x) d x or from (3.1.4), mv 2 + V (x) = C (3.1.6) 2 This equation shows that V (x) has the same units as kinetic energy. V (x) is called the potential energy (PE) function. Equation (3.1.6) states that the sum of the kinetic and potential energies must be constant, if no force other than the conservative force is applied. If v and x have the values v0 and x0 at the time t0 , then mv02 + V (x0 ) = C 2 Comparing this with (3.1.6) gives mv 2 mv02 − + V (x) − V (x0 ) = 0 (3.1.7) 2 2 which can be expressed as KE + PE = 0 (3.1.8) 119 120 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems where the change in kinetic energy is KE = m(v 2 − v02 )/2 and the change in potential energy is PE = V (x)− V (x0 ). For some problems, the following form of the principle is more convenient to use: mv 2 mv02 + V (x0 ) = + V (x) 2 2 (3.1.9) In the form (3.1.8), we see that conservation of mechanical energy states that the change in kinetic energy plus the change in potential energy is zero. Note that the potential energy has a relative value only. The choice of reference point for measuring x determines only the value of C, which (3.1.7) shows to be irrelevant. Gravity is an example of a conservative force, for which f = −mg. The gravity force is conservative because the work done lifting an object depends only on the change in height and not on the path taken. Thus, if x represents vertical displacement, V (x) = mgx and mv 2 + mgx = C 2 mv 2 mv02 − + mg(x − x0 ) = 0 2 2 (3.1.10) (3.1.11) Speed of a Falling Object E X A M P L E 3.1.1 ■ Problem Figure 3.1.2 A falling object. m52 An object with a mass of m = 2 slugs drops from a height of 30 ft above the ground (see Figure 3.1.2). Determine its speed after it drops 20 ft to a platform that is 10 ft above the ground. ■ Solution Measuring x from the ground gives x0 = 30 ft and x = 10 ft at the platform. If the object is dropped from rest, then v0 = 0. From (3.1.11) 209 Platform m 2 (v − 0) + mg(10 − 30) = 0 2 109 Ground √ or v 2 = 40g. Using g = 32.2 ft /sec2 , we obtain v = 392.4 = 19.81 ft /sec. This is the speed of the object when it reaches the platform. Note that if we had chosen to measure x from the platform instead of the ground, then v0 = 0, x0 = 20, and x = 0 at the platform. Equation (3.1.11) gives m 2 (v − 0) + mg(0 − 20) = 0 2 or v 2 = 40g, which gives the same answer as before. When x is measured from the platform, (3.1.10) gives C = 20mg. When x is measured from the ground, C = 30mg, but the two values of C are irrelevant for solving the problem because it is the change in kinetic and potential energies that governs the object’s dynamics. 3. 1 Translational Motion 121 3.1.2 CONSTANT FORCE CASE For the point mass model, ma = f , (3.1.9) can be used to ﬁnd the speed v as a function of displacement x. If f is a constant, then mv02 mv 2 = f (x − x0 ) + 2 2 (3.1.12) Noting that work equals force times displacement, the work done on the mass by the force f is f (x − x 0 ). Thus (3.1.12) says that the ﬁnal energy of the mass, mv 2 /2, equals the initial energy, mv02 /2, plus the work done by the force f . This is a statement of conservation of mechanical energy. 3.1.3 DRY FRICTION FORCE Not every constant force is conservative. A common example of a non-conservative force is the dry friction force. This force is non-conservative because the work done by the force depends on the path taken. The dry friction force F is directly proportional to the force N normal to the frictional surface. Thus F = μN . The proportionality constant is μ, the coefﬁcient of friction. The dry friction force that exists before motion begins is called static friction (sometimes shortened to stiction). The static friction coefﬁcient has the value μs to distinguish it from the dynamic friction coefﬁcient μd , which describes the friction after motion begins. Dynamic friction is also called sliding friction, kinetic friction, or Coulomb friction. In general, μs > μd , which explains why it is more difﬁcult to start an object sliding than to keep it moving. We will use the symbol μ rather than μd because most of our applications involve motion. Because Coulomb friction cannot be derived from a potential energy function, the conservation of mechanical energy principle does not apply. This makes sense physically because the friction force dissipates the energy as heat, and thus mechanical energy, which consists of kinetic plus potential energy, is not conserved. Total energy, of course, is conserved. Equation of Motion with Friction ■ Problem Derive the equation of motion (a) for the block of mass m shown in Figure 3.1.3a and (b) for the mass m on an incline, shown in Figure 3.1.3b. In both cases, a force f 1 , which is not the friction force, is applied to move the mass. ■ Solution a. The free body diagrams are shown in Figure 3.1.3a for the two cases: v > 0 and v < 0. The normal force N is the weight mg. Thus the friction force F is μN , or F = μmg. If v > 0, the equation of motion is m v̇ = f 1 − μmg v>0 (1) Dry friction always opposes the motion. So, for v < 0, m v̇ = f 1 + μmg v<0 Equations (1) and (2) are the desired equations of motion. (2) E X A M P L E 3.1.2 122 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Motion Motion Figure 3.1.3 Motion with friction a) on a horizontal surface and b) on an inclined plane. mg v m f1 mg m F f1 F N N g f1 m (a) Motion F v mg cos ␾ mg sin ␾ m ␾ f1 Motion mg sin ␾ m N mg cos ␾ m f1 F N f1 g (b) b. The friction force depends on the force normal to the surface. For the mass m shown in Figure 3.1.3b, the normal force N must equal mg cos φ as long as the mass is in contact with the surface. The free body diagrams are shown in the ﬁgure for the two cases: v > 0 and v < 0. Newton’s second law applied in the direction parallel to the surface gives, for v > 0, m v̇ = f 1 − mg sin φ − μmg cos φ v>0 (3) m v̇ = f 1 − mg sin φ + μmg cos φ v<0 (4) For v < 0, Equations (3) and (4) are the desired equations of motion. E X A M P L E 3.1.3 Motion with Friction on an Inclined Plane ■ Problem For the mass shown in Figure 3.1.3b, m = 2 kg, φ = 30◦ , v(0) = 3 m/s, and μ = 0.5. Determine whether the mass comes to rest if (a) f 1 = 50 N and (b) f 1 = 5 N. ■ Solution Because the velocity is initially positive [v(0) = 3], we use equation (3) of Example 3.1.2: 2v̇ = f 1 − (sin 30◦ + 0.5 cos 30◦ )(2)(9.81) = f 1 − 18.3 For part (a), f 1 = 50 and thus v̇ = (50 − 18.3)/2 = 15.85 and the acceleration is positive. Thus, because v(0) > 0, the speed is always positive for t ≥ 0 and the mass never comes to rest. For part (b), f 1 = 5, v̇ = (5 − 18.3)/2 = −6.65, and thus the mass is decelerating. Because v(t) = −6.65t + 3, the speed becomes zero at t = 3/6.65 = 0.45 s. 3. 2 Rotation About a Fixed Axis 3.2 ROTATION ABOUT A FIXED AXIS In this section, we consider the dynamics of rigid bodies whose motion is constrained to allow only rotation about an axis through a nonaccelerating point. In Section 3.4 we will treat the case of rotation about an axis through an accelerating point. For planar motion, which means that the object can translate in two dimensions and can rotate only about an axis that is perpendicular to the plane, Newton’s second law can be used to show that I O ω̇ = M O (3.2.1) where ω is the angular velocity of the mass about an axis through a point O ﬁxed in an inertial reference frame and attached to the body (or the body extended), I O is the mass moment of inertia of the body about the point O, and M O is the sum of the moments applied to the body about the point O. This situation is illustrated in the Figure 3.2.1. The angular displacement is θ , and θ˙ = ω. The term torque and the symbol T are often used instead of moment and M. Also, when the context is unambiguous, we use the term “inertia” as an abbreviation for “mass moment of inertia.” 3.2.1 CALCULATING INERTIA The mass moment of inertia I about a speciﬁed reference axis is deﬁned as I = r 2 dm (3.2.2) where r is the distance from the reference axis to the mass element dm. The expressions for I for some common shapes are given in Table 3.2.1. If the rotation axis of a homogeneous rigid body does not coincide with the body’s axis of symmetry, but is parallel to it at a distance d, then the mass moment of inertia about the rotation axis is given by the parallel-axis theorem, I = Is + md 2 (3.2.3) where Is is the inertia about the symmetry axis (see Figure 3.2.2). Figure 3.2.1 An object rotating about a ﬁxed axis. Figure 3.2.2 Illustration of the parallel-axis theorem. Axis ␪ d I Symmetry ␻ M Rotation 123 124 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Table 3.2.1 Mass moments of inertia of common elements. IG = Sphere 2 m R2 5 R G IO = m R2 Mass rotating about point O m R O 1 m(R 2 + r 2 ) 2 1 m(3R 2 + 3r 2 + L 2 ) I y = Iz = 12 Ix = Hollow cylinder y G r z R L x Ix = Rectangular prism y 1 m(b2 + c2 ) 12 c b G z x a E X A M P L E 3.2.1 A Single-Drum Hoist ■ Problem In Figure 3.2.3a, a motor supplies a torque T to turn a drum of radius R and inertia I about its axis of rotation. The rotating drum lifts a mass m by means of a cable that wraps around the drum. The drum’s speed is ω. Discounting the mass of the cable, use the values m = 40 kg, R = 0.2 m, and I = 0.8 kg·m2 . Find the acceleration v̇ if the torque T = 300 N·m. 3. 2 Rotation About a Fixed Axis 125 Figure 3.2.3 (a) A single-drum hoist. (b) Free body diagram. DRUM ␻ T F R ␻ T I v m v m g mg (a) (b) ■ Solution The free body diagrams are shown in Figure 3.2.3b. Let F be the tension in the cable. Because the drum rotation axis is assumed to be ﬁxed, we can use (3.2.1). Summing moments about the drum center gives M O = 300 − 0.2F. Because I O = I = 0.8, we have 0.8ω̇ = 300 − 0.2F (1) Summing vertical forces on the 40-kg mass m gives 40v̇ = F − 40(9.81) (2) Solve (2) for F and substitute for F in (1) to obtain 0.8ω̇ = 300 − 8v̇ − 8(9.81) (3) Note that v = Rω = 0.2ω to obtain v̇ = 0.2ω̇. Substitute this into (3) and collect terms to obtain 12v̇ = 300 − 8(9.81) or v̇ = 18.46 m/s2 . Pendulum with a Concentrated Mass ■ Problem The pendulum shown in Figure 3.2.4a consists of a concentrated mass m a distance L from point O, attached to a rod of length L. (a) Obtain its equation of motion. (b) Solve the equation assuming that θ is small. ■ Solution a. Because the support at point O is assumed to be ﬁxed, we can use (3.2.1). The free body diagram is shown in Figure 3.2.4b, where we have resolved the weight into a component parallel to the rod and one perpendicular to the rod. This makes it easier to compute the moment about point O caused by the weight. The parallel component has a line of action through point O and thus contributes nothing to M O . The moment arm of the perpendicular component is L, and thus M O = −mgL sin θ . The negative sign is required because the moment acts in the negative θ direction. From Table 3.2.1, I O = m L 2 , and thus the equation of motion is m L 2 θ̈ = −mgL sin θ E X A M P L E 3.2.2 126 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Figure 3.2.4 (a) Pendulum with a concentrated mass. (b) Free body diagram. O T ␪ m m L g mg sin ␪ mg cos ␪ (a) (b) The product mL can be factored out of both sides of the equation to give L θ̈ = −g sin θ (1) which is nonlinear and not solvable in terms of elementary functions. b. If θ is small enough and measured in radians, then sin θ ≈ θ (for example, if θ = 32◦ = 0.56 rad, sin 0.56 = 0.5312, an error of 5%). Then (1) can be replaced by the linear equation √ L θ̈ = −gθ (2) The characteristic roots are s = ± j g/L = ± jωn , where ωn = θ (t) = θ (0) cos ωn t + √ g/L. The solution is θ̇ (0) sin ωn t ωn The pendulum will swing back and forth with a radian frequency of ωn . Using the results of Examples 2.8.1 and 2.8.2 in Chapter 2, we can obtain the following expression for the oscillation amplitude: θmax = θ 2 (0) + θ̇ 2 (0) ωn2 So, before accepting any predictions based on this linear model, we should ﬁrst check to see whether sin θmax ≈ θmax . 3.2.2 ENERGY AND ROTATIONAL MOTION The work done by a moment M causing a rotation through an angle θ is θ W = M dθ (3.2.4) 0 Multiply both sides of (3.2.1) by ω dt, and note that ω = dθ/dt. I ω dω = M dθ Integrating both sides gives θ 1 2 I ω dω = I ω = M dθ (3.2.5) 2 0 0 We thus see that the work done by the moment M produces the kinetic energy of rotation: KE = I ω2 /2. ω 3. 2 Rotation About a Fixed Axis 127 Figure 3.2.5 Pulley forces. 3.2.3 PULLEY DYNAMICS Pulleys can be used to change the direction of an applied force or to amplify forces. In our examples, we will assume that the cords, ropes, chains, and cables drive the pulleys without slipping and are inextensible; if not, then they must be modeled as springs. Figure 3.2.5 shows a pulley of inertia I whose center is ﬁxed to a support. Assume that the tension forces in the cable are different on each side of the pulley. Then application of (3.2.1) gives ␪ R I θ̈ = RT1 − RT2 = R(T1 − T2 ) An immediate result of practical signiﬁcance is that the tension forces are approximately equal if I θ̈ is negligible. This condition is satisﬁed if either the pulley rotates at a constant speed or if the pulley inertia is negligible compared to the other inertias in the system. The pulley inertia will be negligible if either its mass or its radius is small. Thus, when we neglect the mass, radius, or inertia of a pulley, the tension forces in the cable may be taken to be the same on both sides of the pulley. The force on the support at the pulley center is T1 + T2 + mg. If the mass, radius, or inertia of the pulley are negligible, then the support force is 2T1 . Energy Analysis of a Pulley System T1 T2 E X A M P L E 3.2.3 ■ Problem Figure 3.2.6a shows a pulley used to raise the mass m 2 by hanging a mass m 1 on the other side of the pulley. If pulley inertia is negligible then it is obvious that m 1 will lift m 2 if m 1 > m 2 . How does a nonnegligible pulley inertia I change this result? Also, investigate the effect of the pulley inertia on the speed of the masses. ■ Solution Deﬁne the coordinates x and y such that x = y = 0 at the start of the motion. If the pulley cable is inextensible, then x = y and thus ẋ = ẏ. If the cable does not slip, then θ˙ = ẋ/R. Because we were asked about the speed and because the only applied force is a conservative force (gravity), this suggests that we use an energy-based analysis. If the system starts at rest at x = y = 0, then the kinetic energy is initially zero. We take the potential energy to be zero at x = y = 0. Thus, ␪ ␪ R R I Figure 3.2.6 A pulley system for lifting a mass. I T2 y m2 T1 m2 m2 g y x m1 m1 m1g (a) (b) x 128 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems the total mechanical energy is initially zero, and from conservation of energy we obtain 1 1 1 m 1 ẋ 2 + m 2 ẏ 2 + I θ˙2 + m 2 gy − m 1 gx = 0 2 2 2 Note that the potential energy of m 1 has a negative sign because m 1 loses potential energy when x is positive. Substituting y = x, ẏ = ẋ, and θ˙ = ẋ/R into this equation and collecting like terms gives KE + PE = 1 2 m1 + m2 + and thus I R2 ẋ = ẋ 2 + (m 2 − m 1 )gx = 0 2(m 1 − m 2 )gx m 1 + m 2 + I /R 2 (1) The mass m 2 will be lifted if ẋ > 0; that is, if m 2 < m 1 . So the pulley inertia does not affect this result. However, because I appears in the denominator of the expression for ẋ, the pulley inertia does decrease the speed with which m 1 lifts m 2 . In Example 3.2.3, it is inconvenient to use an energy-based analysis to compute x(t) or the tensions in the cable. To do this it is easier to use Newton’s law directly. E X A M P L E 3.2.4 Equation of Motion of a Pulley System ■ Problem Consider the pulley system shown in Figure 3.2.6a. Obtain the equation of motion in terms of x and obtain an expression for the tension forces in the cable. ■ Solution The free body diagrams of the three bodies are shown in part (b) of the ﬁgure. Newton’s law for mass m 1 gives m 1 ẍ = m 1 g − T1 (1) m 2 ÿ = T2 − m 2 g (2) Newton’s law for mass m 2 gives Equation (3.2.1) applied to the inertia I gives I θ̈ = RT1 − RT2 = R(T1 − T2 ) (3) Because the cable is assumed inextensible, x = y and thus ẍ = ÿ. We can then solve (1) and (2) for the tension forces. T1 = m 1 g − m 1 ẍ = m 1 (g − ẍ) (4) T2 = m 2 ÿ + m 2 g = m 2 ( ÿ + g) = m 2 (ẍ + g) (5) Substitute these expressions into (3). I θ̈ = (m 1 − m 2 )g R − (m 1 + m 2 )R ẍ Because x = Rθ , ẍ = R θ̈, and (6) becomes ẍ I = (m 1 − m 2 )g R − (m 1 + m 2 )R ẍ R (6) 3. 2 which can be rearranged as I m1 + m2 + 2 R Rotation About a Fixed Axis 129 ẍ = (m 1 − m 2 )g (7) This is the desired equation of motion. We can solve it for ẍ and substitute the result into equations (4) and (5) to obtain T1 and T2 as functions of the parameters m 1 , m 2 , I , R, and g. Equation (7) can be solved for ẋ(t) and x(t) by direct integration. Let A= (m 1 − m 2 )g R 2 (m 1 + m 2 )R 2 + I Then (7) becomes ẍ = A, whose solutions are ẋ = At + ẋ(0) and x = At 2 /2 + ẋ(0)t + x(0). Note that if we use the solutions to express ẋ as a function of x, we will obtain the same expression as equation (1) in Example 3.2.3. 3.2.4 PULLEY-CABLE KINEMATICS Consider Figure 3.2.7. Suppose we need to determine the relation between the velocities of mass m A and mass m B . Deﬁne x and y as shown from a common reference line attached to a ﬁxed part of the system. Noting that the cable lengths wrapped around the pulleys are constant, we can write x + 3y = constant. Thus ẋ + 3 ẏ = 0. So the speed of point A is three times the speed of point B, and in the opposite direction. Figure 3.2.7 A multiple-pulley system. y x B A mA mB In many problems, we neglect the inertia of the pulleys so as to keep the resulting model as simple as possible. As we will see, the models of many practical engineering applications are challenging enough without introducing pulley dynamics. Example 3.2.4 illustrates such an application. A Pulley System ■ Problem The two masses shown in Figure 3.2.8a are released from rest. The mass of block A is 60 kg; the mass of block B is 40 kg. Disregard the masses of the pulleys and rope. Block A is heavier than block B, but will block B rise or fall? Find out by determining the acceleration of block B by (a) using free body diagrams and (b) using conservation of energy. E X A M P L E 3.2.5 130 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Figure 3.2.8 (a) A pulley system. (b) Free body diagrams. F D T T T T D DATUM xA B C 40g xB C 2T A A B 60g (a) (b) ■ Solution a. The free body diagrams are given in part (b) of the ﬁgure. We take the acceleration of each block to be positive downward. For block A, Newton’s law gives 60ẍ A = 60g − 2T (1) where T is the tension in the rope. For block B, 40ẍ B = 40g − T (2) For the massless pulley C, 2T = T + T , which gives no information. For the massless pulley D, F = 2T , which is useful only if we need to calculate the support force F. If the unwrapped rope length below the datum line is L, then we have 2x A + x B = L. Differentiating this twice gives ẍ B = −2ẍ A . Substituting this into (1) gives T = 30g + 15ẍ B , and from (2) we have 11ẍ B = 2g or ẍ B = 2g/11 = 1.784 m/s . The acceleration is positive, which means that block B is accelerating downward. b. Choosing gravitational potential energy to be zero at the datum line, the total energy in the system is 1 1 m A ẋ 2A + m B ẋ B2 − m A gx A − m B gx B = constant 2 2 Differentiate both sides with respect to time to obtain 2 m A ẋ A ẍ A + m B ẋ B ẍ B − m A g ẋ A − m B g ẋ B = 0 However, because the unwrapped rope length below the datum line is L, then we have 2x A + x B = L. This gives ẋ A = −ẋ B /2 and ẋ A = −ẋ B /2. Substitute these and cancel to 3. 3 obtain m A + m B ẍ B − m A Equivalent Mass and Inertia 131 − mg g = 0 4 2 Substituting the given values for m A and m B gives the same answer as obtained in part (a): 11ẍ B = 2g 3.3 EQUIVALENT MASS AND INERTIA Some systems composed of translating and rotating parts whose motions are directly coupled can be modeled as a purely translational system or as a purely rotational system, by using the concepts of equivalent mass and equivalent inertia. These models can be derived using kinetic energy equivalence. Equivalent mass and equivalent inertia are complementary concepts. A system should be viewed as an equivalent mass if an external force is applied, and as an equivalent inertia if an external torque is applied. Examples 3.3.1–3.3.7 will illustrate this approach. 3.3.1 MECHANICAL DRIVES Gears, belts, levers, and pulleys transform an input motion, force, or torque into another motion, force, or torque at the output. For example, a gear pair can be used to reduce speed and increase torque, and a lever can increase force. Several types of gears are used in mechanical drives. These include helical, spur, rack-and-pinion, worm, bevel, and planetary gears. Other mechanical drives use belts or chains. We now use a spur gear pair, a rack-and-pinion gear pair, and a belt drive to demonstrate the use of kinetic energy equivalence to obtain a model. This approach can be used to analyze other gear and drive types. A pair of spur gears is shown in Figure 3.3.1. The input shaft (shaft 1) is connected to a motor that produces a torque T1 at a speed ω1 , and drives the output shaft (shaft 2). One use of such a system is to increase the effective motor torque. The gear ratio N is deﬁned as the ratio of the input rotation θ1 to the output rotation θ2 . Thus, N = θ1 /θ2 . From geometry we can see that N is also the speed ratio N = ω1 /ω2 . Thus, the pair is a speed reducer if N > 1. The gear ratio is also the diameter ratio N = D2 /D1 , and the gear tooth ratio N = n 2 /n 1 , where n is the number of gear teeth. If the gear inertias are negligible or if there is zero acceleration, and if we neglect energy loss due to friction, such as that between the gear teeth, then the input work T1 θ1 Figure 3.3.1 A spur-gear pair. I1 T1 ␻1 ␻ N 5 ␻12 ␻2 T2 I2 132 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems must equal the output work T2 θ2 . Thus, under these conditions, T2 = T1 (θ1 /θ2 ) = N T1 , and the output torque is greater than the input torque for a speed reducer. For cases that involve acceleration and appreciable gear inertia, the output torque is less than N T1 . E X A M P L E 3.3.1 Equivalent Inertia of Spur Gears ■ Problem Consider the spur gears shown in Figure 3.3.1. Derive the expression for the equivalent inertia Ie felt on the input shaft, and obtain the equation of motion in terms of the speed ω1 . ■ Solution Let I1 and I2 be the total moments of inertia on the shafts. Note that ω2 = ω1 /N . The kinetic energy of the system is then 1 1 1 1 ω1 KE = I1 ω12 + I2 ω22 = I1 ω12 + I2 2 2 2 2 N or 2 1 1 I1 + 2 I2 ω12 2 N KE = Therefore the equivalent inertia felt on the input shaft is Ie = I1 + I2 N2 (1) This means that the dynamics of the system can be described by the model Ie ω̇1 = T1 . The torque T2 is not the torque on the load shaft due to the torque T1 . Rather, T2 is due to external causes. For example, if I1 represents a motor, and I2 represents a vehicle wheel, then T2 would be due to road forces, or gravity. If the vehicle were going downhill, gravity would act to accelerate the vehicle (ω2 > 0), and the resulting torque T2 would be positive. If the vehicle were going uphill, gravity would act to decelerate the vehicle (ω2 < 0), and the resulting torque T2 would be negative. E X A M P L E 3.3.2 A Speed Reducer ■ Problem For the geared system shown in Figure 3.3.1, the inertias in kg·m2 are I1 = 0.1, for the motor shaft and I2 = 0.4 for the load shaft. The motor speed ω1 is ﬁve times faster than the load speed ω2 , so this device is called a speed reducer. Obtain the equation of motion (a) in terms of ω1 and (b) in terms of ω2 , assuming that the motor torque T1 and load torque T2 are given. ■ Solution For the given speed information, we have that N = ω1 /ω2 = 5. a. Referencing both inertias to shaft 1 gives the equivalent inertia I2 0.4 = 0.116 = 0.1 + 2 N 25 Note that the moment felt on shaft 1 due to T2 is T2 /N . Summing moments on shaft 1 gives Ie = I1 + Ie ω̇1 = T1 + T2 N 3. 3 Equivalent Mass and Inertia 133 or T2 (1) 5 b. Starting from basic principles, we express the kinetic energy of the system as 1 1 1 1 1 1 KE = I1 ω12 + I2 ω22 = I1 (5ω2 )2 + I2 ω22 = (2.5 + 0.4) ω22 = (2.9) ω22 2 2 2 2 2 2 Thus the equivalent inertia referenced to shaft 2 is 2.9, and the equation of motion is 0.116ω̇1 = T1 + 2.9ω̇2 = N T1 + T2 = 5T1 + T2 Of course, we could have obtained this result directly from (1) with the substitution ω1 = 5ω2 , if (1) were already available. Note that with a speed reducer, the load speed is slower than the motor speed, but the effect of the motor torque on the load shaft is increased by a factor equal to the gear ratio N . A Three-Gear System E X A M P L E 3.3.3 ■ Problem For the system shown in Figure 3.3.2, assume that the shaft inertias are small. The remaining inertias in kg·m 2 are I1 = 0.005, I2 = 0.01, I3 = 0.02, I4 = 0.04, and I5 = 0.2. The speed ratios are ω2 3 ω1 = =2 ω2 2 ω3 Obtain the equation of motion in terms of ω3 . The torque T is given. ■ Solution Note that 3 3 ω1 = ω2 = 2ω3 = 3ω3 2 2 So this system is a speed reducer. Also note that ω2 = 2ω3 . Either we may reference all inertias to shaft 1 and then use the relation ω1 = 3ω3 , or we may reference everything to shaft 3 and note Figure 3.3.2 A three-gear system. I4 I1 ␻1 I2 T ␻2 I3 ␻3 I5 134 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems that the torque on shaft 3 due to T is T /3. Choosing the latter approach, we express the kinetic energy as KE = 1 1 1 1 1 1 1 1 I4 ω12 + I1 ω12 + I2 ω22 + I3 ω32 + I5 ω32 = (I4 + I1 ) ω12 + I2 ω22 + (I3 + I5 ) ω32 2 2 2 2 2 2 2 2 or KE = 1 1 1 1 1 1 (I4 + I1 ) (3ω3 )2 + I2 (2ω3 )2 + (I3 + I5 ) ω32 = (0.405) ω32 + (0.04) ω22 + (0.22) ω32 2 2 2 2 2 2 which reduces to KE = 1 (0.665) ω32 2 Thus the equivalent inertia referenced to shaft 3 is 0.665. Because the speed is reduced by a factor of 3 going from shaft 1 to shaft 3, the torque is increased by a factor of 3. Thus the equation of motion is 0.665ω̇3 = 3T A spur gear pair consists of only rotating elements. However, a rack-and-pinion consists of a rotating component (the pinion gear) and a translating component (the rack). The input to such a device is usually the torque applied to the shaft of the pinion. If so, then we should model the device as an equivalent inertia. The following example shows how to do this. E X A M P L E 3.3.4 Equivalent Inertia of a Rack-and-Pinion ■ Problem A rack-and-pinion, shown in Figure 3.3.3, is used to convert rotation into translation. The input shaft rotates through the angle θ as a result of the torque T produced by a motor. The pinion rotates and causes the rack to translate. Derive the expression for the equivalent inertia Ie felt on the input shaft. The mass of the rack is m, the inertia of the pinion is I , and its mean radius is R. ■ Solution The kinetic energy of the system is (neglecting the inertia of the shaft) KE = Figure 3.3.3 A rack-andpinion gear. R T ␪ I x m 1 2 1 2 m ẋ + I θ˙ 2 2 3. 3 Equivalent Mass and Inertia 135 where ẋ is the velocity of the rack and θ˙ is the angular velocity of the pinion and shaft. From ˙ Substituting for ẋ in the expression for KE, we obtain geometry, x = Rθ, and thus ẋ = R θ. 1 1 1 2 m R θ˙ + I θ˙2 = m R 2 + I θ˙2 2 2 2 Thus the equivalent inertia felt on the shaft is KE = Ie = m R 2 + I (1) and the model of the system’s dynamics is Ie θ̈ = T , which can be expressed in terms of x as Ie ẍ = RT . Belt and chain drives are good examples of devices that can be difﬁcult to analyze by direct application of Newton’s laws but can be easily modeled using kinetic energy equivalence. Equivalent Inertia of a Belt Drive E X A M P L E 3.3.5 ■ Problem Belt drives and chain drives, like those used on bicycles, have similar characteristics and can be analyzed in a similar way. A belt drive is shown in Figure 3.3.4. The input shaft (shaft 1) is connected to a device (such as a bicycle crank) that produces a torque T1 at a speed ω1 , and drives the output shaft (shaft 2). The mean sprocket radii are r1 and r2 , and their inertias are I1 and I2 . The belt mass is m. Derive the expression for the equivalent inertia Ie felt on the input shaft. ■ Solution The kinetic energy of the system is 1 1 1 I1 ω12 + I2 ω22 + mv 2 2 2 2 If the belt does not stretch, the translational speed of the belt is v = r1 ω1 = r2 ω2 . Thus we can express KE as KE = r 1 ω1 1 1 KE = I1 ω12 + I2 2 2 r2 2 1 + m r 1 ω1 2 2 2 r1 1 = I1 + I2 2 r2 + mr12 ω12 Therefore, the equivalent inertia felt on the input shaft is Ie = I1 + I2 2 r1 + mr12 r2 (1) This means that the dynamics of the system can be described by the model Ie ω̇1 = T1 . Figure 3.3.4 A belt drive. v r2 I2 T1 ␻ 1 ␻2 r1 I1 m 136 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems E X A M P L E 3.3.6 A Conveyer System ■ Problem Conveyor systems are used to produce translation of the load, as shown in Figure 3.3.5. The reducer is a geared system that reduces the motor speed by a factor of 3:1. The motor inertia is I1 = 0.002 kg·m2 . Disregard the inertias of the reducer and the tachometer, which are used to measure the speed for control purposes. Also discount the inertias of the two sprockets, the chain, and all shafts. The only signiﬁcant masses and inertias are the inertias of the four drive wheels (0.02 kg·m2 each), the two drive chains (8 kg each), and the load mass (10 kg). The radius of sprocket 1 is 0.05 m, and that of sprocket 2 is 0.1 m. The drive wheel has a radius of 0.1 m. The load friction torque measured at the drive shaft is 1.2 N·m. a. Derive the equation of motion of the conveyor in terms of the motor velocity, with the motor torque T1 as the input. b. Suppose the motor torque is constant at 1.6 N·m. Determine the resulting motor angular acceleration and load acceleration. ■ Solution a. The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocket diameters, which 0.1/0.05 = 2. So the sprocket drive is a speed reducer. Thus, in going from the motor to the drive shaft, there is a total speed reduction of 3(2) = 6. Therefore, when referencing the signiﬁcant inertias to the motor shaft, we must divide the inertias by 62 = 36. Thus, the equivalent inertia felt at the motor shaft due to the four drive wheels is 4(0.02)/36 = 0.2. To compute the equivalent inertias of the drive chains and load mass, we ﬁrst express their kinetic energies in terms of the motor speed ω1 . Let v be the translational velocity of the drive chains and the load mass. Because a drive wheel has a radius of 0.1 m, the angular velocity ωd of the drive shaft is related to v as 0.1ωd = v. We are now ready to express the kinetic energy drive chains and load mass as follows. Each 8 kg drive chain has Figure 3.3.5 A conveyor drive system. Load Drive chains Tachometer Drive wheels Sprocket 2 Drive shaft Chain Sprocket 1 Reducer Motor 3. 3 Equivalent Mass and Inertia 137 a kinetic energy of KEchain = 1 2 1 1 8v = 8(0.1ωd )2 = (0.08)ωd2 2 2 2 But ωd is related to the motor speed ω1 as ωd = ω1 /6. Thus KEchain ω 2 1 1 1 = (0.08) = 2 6 2 0.08 36 ω12 1 = 2 0.02 9 ω12 Because the kinetic energy of an inertia I rotating at a speed ω is I ω2 /2, we see that the equivalent inertia of the four drive wheels is 4(0.02)/9 = 0.08/9. We use a similar method to compute the equivalent inertia of the 10 kg load mass. KEload 1 1 1 1 = 10v 2 = 10(0.1ωd )2 = (0.1)ωd2 = 2 2 2 2 0.1 36 ω12 Thus the equivalent inertia of the load mass 0.1/36 = 0.025/9. The total equivalent inertia Ie felt at the motor shaft is the sum of the equivalent inertias of the motor, the drive wheels, the drive chain, and the load mass: Ie = 0.002 + 0.125 0.02 0.08 0.025 + + = 0.002 + = 0.0159 kg · m2 9 9 9 9 The 1.2 N·m friction torque acts against the motor torque and is reduced 6 because of the net gear ratio of 3(2) = 6. Thus the equation of motion in terms of the motor speed is 0.0159ω̇1 = T1 − b. 1.2 = T1 − 0.2 6 If T1 = 1.6, then the motor angular acceleration is ω̇1 = 1.6 − 0.2 = 88.05 rad/s2 0.0159 The acceleration v̇ of the load is related to ω̇1 by the drive wheel radius (0.1) and the net gear ratio (6). Thus v̇ = 0.1ω̇1 /6 = 1.4675 m/s2 . 3.3.2 SLIDING VERSUS ROLLING MOTION Wheels are common examples of systems undergoing general plane motion with both translation and rotation. The wheel shown in Figure 3.3.6 can have one of three possible motion types: 1. Pure rolling motion. This occurs when there is no slipping between the wheel and the surface. In this case, v = Rω. 2. Pure sliding motion. This occurs when the wheel is prevented from rotating (such as when a brake is applied). In this case, ω = 0 and v = Rω. 3. Sliding and rolling motion. In this case ω = 0. Because slipping occurs in this case, v = Rω. The wheel will roll without slipping (pure rolling) if the tangential force f t is smaller than the static friction force μs N , where N is the force of the wheel normal to the surface. In this case, the tangential force does no work because it does not act through a distance. If the static friction force is smaller than f t , the wheel will slip. Figure 3.3.6 Motion of a wheel. v R ft ␻ N f 138 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems E X A M P L E 3.3.7 A Wheeled Vehicle ■ Problem An inextensible cable with a tension force f = 400 N is used to pull a two-wheeled cart on a horizontal surface (Figure 3.3.7a). The wheels roll without slipping. The cart has a mass m c = 100 kg. Each wheel has a radius Rw = 0.3 m. Disregard the mass of the axle. The center of mass of the system is at point G. We wish to solve for the translational acceleration v̇ of the cart. a. b. Solve the problem assuming that the wheel masses are negligible. Assuming that each wheel has a mass m w = 10 kg and an inertia I = 0.45 kg·m2 about its center, use the free body diagram method to solve the problem. c. Solve the problem assuming that each wheel has a mass m w = 10 kg and an inertia Iw = 0.45 kg·m2 about its center. Use the energy equivalence method to solve the problem. ■ Solution a. Discounting the wheel mass means that the wheel moment of inertia is also zero. The free body diagrams are shown in part (b) of the ﬁgure (treating the axle and the two wheels as a single rigid object). The forces Rx and R y are the reaction forces between the axle and the cart body. For the cart mass, in the x direction m c v̇ = f + Rx (1) In the y direction, assuming that the wheels do not bounce so their vertical acceleration is zero, we have 0 = Ry − mc g (2) which gives R y = m c g. For the massless wheel unit, the effect of the ground on each wheel can be resolved into a tangential component f t and a normal component N . The Figure 3.3.7 A two-wheeled cart. v Ry mc mc f y Rw mw G␻ f G Rx Ry 2ft x mcg 2N (b) (a) 2mwg Ry mc f G Rx Ry 2ft mcg 2N (c) Rx Rx 3. 3 Equivalent Mass and Inertia equations of Statics give f x = −Rx − 2 f t = 0 (sum of forces in the x direction is zero) (3) f y = 2N − R y = 0 (sum of forces in the y direction is zero) (4) M G = 2 f t Rw = 0 (sum of moments about point G is zero) This implies that f t = 0, because Rw = 0, and thus (3) implies that Rx = 0. From (2) and (4), we see that the normal force on each wheel is N = m c g/2. Finally, from (1), the desired expression for the translational acceleration v̇ of the cart is m c v̇ = f or f 400 = 4 m/s2 . = mc 100 Note that this equation is identical to the equation of motion of a wheel-less mass m c being pulled by a force f on a frictionless surface. b. The free body diagrams are shown in part (b) of the ﬁgure (again treating the axle and the two wheels as a single rigid object). For the cart mass, (1) and (2) still apply. However, for the axle-wheel system, we now have v̇ = f x = −Rx − 2 f t = 2m w v̇ (5) f y = 2N − R y − 2m w g = 0 (this assumes the wheels do not bounce) (6) MG = 2 f t Rw = 2Iw ω̇ (7) From (1), Rx = m c v̇ − f (8) Because v = Rw ω, (7) gives ft = Iw ω̇ Iw v̇ = 2 Rw Rw Using this result and (8) in (5), we obtain 2Iw m c + 2m w + 2 Rw (9) v̇ = f Substituting the given values gives the ﬁnal answer: 130v̇ = 400, or v̇ = 3.077 m/s2 . The effect of the wheel mass is to increase the vehicle mass by 20% from 100 to 120, but the wheel inertias increase the equivalent mass by 30% from 100 to 130. If needed, the expressions for the forces Rx , f t , and N can be obtained by substituting this value of v̇ into (6), (8), and (9), using R y = m c g. c. Using the energy equivalence method, we write the kinetic energy expression for the entire vehicle. 1 1 KE = m c v 2 + 2m w v 2 + 2Iw ω2 2 2 Because Rw ω = v, this expression becomes KE = 1 1 m c v2 + 2 2 2m w v 2 + 2Iw v2 Rw2 = 1 2 m c + 2m w + 2Iw Rw2 v2 139 140 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Thus the equivalent mass is m e = m c + 2m w + 2Iw = 130 Rw2 and the equation of motion is m e v̇ = f , which gives v̇ = 3.077 m/s2 , the same answer obtained in part (b). Example 3.3.7 shows that by assuming that the mass of the wheels is small compared with the total vehicle mass, we can ignore the rolling and translational motions of the wheels and thus treat the vehicle as a single rigid body translating in only one direction, with no rotating parts. In addition, it is important to realize that the assumption of negligible wheel mass implies that there is no tangential force on the wheels. We can see this from equation (9), which shows that f t is zero if Iw is zero. In Example 3.3.7, we modeled the wheeled vehicle as an equivalent translating mass. We can “lump” the rotating elements with the translating elements only when the translational and rotational motions are directly related to one another. If the vehicle’s wheels slip, we cannot express the vehicle’s translational speed in terms of the wheels’ rotational speed. We must then treat the vehicle body and the wheels as separate masses, and apply Newton’s laws to each separately. We must also take this approach if we need to compute any forces internal to the system. Modeling Insight 3.4 GENERAL PLANAR MOTION In Section 3.1 we limited our attention to systems undergoing pure translation, and in Section 3.2 we analyzed systems rotating about a single nonaccelerating axis. As demonstrated by the examples in Section 3.3, we can use energy equivalence to model a system as if it were in pure translation or pure rotation, but only if the motions of the rotating and translating components are directly coupled. We now consider the general case of an object undergoing translation and rotation about an accelerating axis. We will restrict our attention to motion in a plane. This means that the object can translate in two dimensions and can rotate only about an axis that is perpendicular to the plane. Many practical engineering problems can be handled with the plane motion methods covered here. The completely general motion case involves translation in three dimensions, and rotation about three axes. This type of motion is considerably more complex to analyze, and is beyond the scope of our coverage. 3.4.1 FORCE EQUATIONS Newton’s laws for plane motion are derived in basic dynamics references, and we will review them here. We assume that the object in question is a rigid body that moves in a plane passing through its mass center, and that it is symmetrical with respect to that plane. Thus it can be thought of as a slab with its motion conﬁned to the plane of the slab. We assume that the mass center and all forces acting on the mass are in the plane of the slab. We can describe the motion of such an object by its translational motion in the plane and by its rotational motion about an axis perpendicular to the plane. Two force equations describe the translational motion, and a moment equation is needed to describe the rotational motion. Consider the slab shown in Figure 3.4.1, where we arbitrarily assume that three external forces f 1 , f 2 , and f 3 are acting on the slab. Deﬁne an x-y coordinate system as shown with its origin located at a nonaccelerating point. Then the 3. 4 aG General Planar Motion Figure 3.4.1 Planar motion of a slab. f3 d O ␣ G r f1 141 aP P y f2 x two force equations can be written as f x = maG x (3.4.1) f y = maG y (3.4.2) where f x and f y are the net forces acting on the mass m in the x and y directions, respectively. The mass center is located at point G. The quantities aG x and aG y are the accelerations of the mass center in the x and y directions relative to the ﬁxed x-y coordinate system. 3.4.2 MOMENT EQUATIONS Recall that in Section 3.2 we treated the case where an object is constrained to rotate about an axis that passes through a ﬁxed point O. For this case, we can apply the following moment equation: IO α = MO (3.4.3) where α is the angular acceleration of the mass about the axis through point O, I O is the mass moment of inertia of the body about the point O, and M O is the sum of the moments applied to the body about the point O. The following moment equation applies regardless of whether or not the axis of rotation is constrained: MG = IG α (3.4.4) where MG is the net moment acting on the body about an axis that passes through the mass center G and is perpendicular to the plane of the slab. IG and α are the mass moment of inertia and angular acceleration of the body about this axis. The net moment MG is caused by the action of the external forces f 1 , f 2 , f 3 , . . . and any couples applied to the body. The positive direction of MG is determined by the right-hand rule (counterclockwise if the x-y axes are chosen as shown). Note that point G in the preceding equation must be the mass center of the object; no other point may be used. However, in many problems the acceleration of some point P is known, and sometimes it is more convenient to use this point rather than the mass center or a ﬁxed point. The following moment equation applies for an accelerating 142 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems point P, which need not be ﬁxed to the body: M P = IG α + maG d (3.4.5) where the moment M P is the net moment acting on the body about an axis that passes through P and is perpendicular to the plane of the slab, aG is the magnitude of the acceleration vector aG , and d is the distance between aG and a parallel line through point P (see Figure 3.4.1). An alternative form of this equation is M P = I P α + mr x a Py − mr y a Px (3.4.6) where a Px and a Py are the x and y components of the acceleration of point P relative to the x-y coordinate system. The terms r x and r y are the x and y components of the location of G relative to P, and I P is the mass moment of inertia of the body about an axis through P. Note that in general, M P does not equal MG , and I P does not equal IG . If point P is ﬁxed at some point O, then a Px = a Py = 0, and the moment equation (3.4.6) simpliﬁes to (3.4.3), because M O = M P and I O = I P . Note that the angular acceleration α is the same regardless of whether point O, G, or P is used. E X A M P L E 3.4.1 A Rolling Cylinder ■ Problem A solid cylinder of mass m and radius r starts from rest and rolls down the incline at an angle θ (Figure 3.4.2). The static friction coefﬁcient is μs . Determine the acceleration of the center of mass aG x and the angular acceleration α. Assume that the cylinder rolls without bouncing, so that aG y = 0. Assume also that the cylinder rolls without slipping. Use two approaches to solve the problem: (a) Use the moment equation about G, and (b) use the moment equation about P. (c) Obtain the frictional condition required for the cylinder to roll without slipping. ■ Solution The free body diagram is shown in the ﬁgure. The friction force is F and the normal force is N . The force equation in the x direction is maG x = f x = mg sin θ − F (1) maG y = f y = N − mg cos θ (2) In the y direction, If the cylinder does not bounce, then aG y = 0 and thus from (2), N = mg cos θ Figure 3.4.2 A cylinder rolling down an inclined plane. y r (3) mg ␣ G mg cos ␪ G mg sin ␪ P x ␪ F (a) (b) N 3. 4 a. General Planar Motion The moment equation about the center of mass gives IG α = MG = Fr (4) Solve for F and substitute it into (1): maG x = mg sin θ − IG α r (5) If the cylinder does not slip, then aG x = r α (6) Solve this for α and substitute into (5): maG x = mg sin θ − I G aG x r2 Solve this for aG x : mgr 2 sin θ mr 2 + IG 2 For a solid cylinder, IG = mr /2, and the last expression reduces to 2 aG x = g sin θ 3 b. We could have used instead the moment equation (2.4.5) about the point P, which is accelerating. This equation is aG x = (7) (8) M P = IG α + maG x d where d = r and M P = (mg sin θ )r . Thus aG x + maG x r r we obtain the same expression as (7). The angular acceleration is mgr sin θ = IG Solving this for aG x found from (6). c. The maximum possible friction force is Fmax = μs N = μs mg cos θ . From (4), (6), and (7), IG α I G aG x IG mg sin θ F= = = 2 r r IG + mr 2 If Fmax > F, the cylinder will not slip. The condition of no slip is therefore given by IG sin θ μs cos θ > (9) IG + mr 2 For a solid cylinder, this reduces to μs cos θ > 1 sin θ 3 (10) Simplifying Vehicle Models Example 3.3.7 treated a wheeled vehicle being pulled by an external force, such as from a cable tow. The example showed that by neglecting the mass of the wheels relative to the rest of the vehicle mass, we can ignore the rolling and translational motions of the wheels and thus treat the vehicle as a single rigid body translating in only one direction, with no rotating parts. The example also showed that the assumption of negligible wheel mass implies that there is no tangential force on the wheels. We will use this insight in the next example to simplify the analysis of the undriven wheels of a vehicle. However, the situation is a little different for the driven wheels. In this case, a tangential traction force between the tire and the road is necessary to propel the vehicle, but if we neglect the driven wheel mass, we can ignore the rolling and translational motions of those wheels. Thus the simpliﬁed vehicle model consists of a single translating mass. 143 144 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Because it enables us to ignore the wheel motions, the assumption of negligible wheel mass also enables us to ignore the effects of the vehicle suspension. These concepts are explored in the following example. E X A M P L E 3.4.2 Maximum Vehicle Acceleration ■ Problem It is required to determine, as a function of the friction coefﬁcient μs , the maximum acceleration of the rear-wheel drive vehicle shown in Figure 3.4.3. The vehicle mass is 1800 kg, and its dimensions are L A = 1.3 m, L B = 1 m, and H = 0.5 m. Assume that the mass of the wheels is small compared with the total vehicle mass and neglect the effects of the vehicle suspension. These assumptions enable us to ignore the rolling and vertical motions of the wheels and thus to treat the vehicle as a single rigid body translating in only one direction, with no rotating parts. The assumption of negligible wheel mass implies that there is no tangential force on the front wheels. Assume that each front wheel experiences the same reaction force N A /2. Similarly, each rear wheel experiences the same reaction force N B /2. Thus the traction force μs N B is the total traction force due to both driving wheels. The traction force is due to the torque T applied from the engine through the axles to the rear wheels. Thus, when the rear wheels are on the verge of slipping, μs N B = T /R, where R is the radius of the rear wheels. ■ Solution The key to solving this problem is to recognize that the maximum traction force, and therefore the maximum acceleration, is obtained when the driving tires are just on the verge of slipping relative to the road surface. In this condition, the friction force, which is the traction force, is given by μs N B . From the free body diagram shown in Figure 3.4.3b, Newton’s law applied in the x direction gives f x = maG x or μs N B = maG x (1) In the y direction, if the vehicle does not leave the road, Newton’s law gives f y = maG y Figure 3.4.3 Vehicle acceleration. or N A + N B − mg = 0 (2) LB LA H A B (a) y P G x B A mg NA (b) ␮s NB NB 3. 5 Additional Examples 145 The moment equation about the mass center G gives MG = IG α or N B L B − μs N B H − N A L A = 0 (3) because α = 0 if the vehicle body does not rotate. Equation (1) shows that we need to ﬁnd N B to determine the acceleration aG x . Equations (2) and (3) can be solved for N B . The solution is NB = 9.8(1.3)m mgL A 25.5 = = m L A + L B − μs H 1.3 + 1 − 0.5μs 4.6 − μs (4) and thus the maximum acceleration is aG x = 25.5μs μs N B = m 4.6 − μs An alternative approach to the problem is to use the moment equation (3.4.5) about the accelerating point P shown in the ﬁgure. This approach avoids the need to solve two equations to obtain N B . This equation gives M P = IG α + maG d or μs N B H − N B (L A + L B ) + mgL A = 0 (5) because α = 0 and d = 0. This gives the same solution as equation (4). The analysis in Example 3.4.2 ignored the effects of the vehicle suspension. This simpliﬁcation results in the assumption that the vehicle body does not rotate. You may have noticed, however, that a vehicle undergoing acceleration will have a pitching motion that is made possible because of the suspension springs. So a complete analysis of this problem would include this effect. However, it is always advisable to begin with a simpliﬁed version of a problem, to make sure you understand the problem’s basic features. If you cannot solve the problem with the suspension effects ignored, then you certainly will not be able to solve the more complex problem that includes the suspension dynamics! Vehicle Suspension Effects 3.5 ADDITIONAL EXAMPLES This section provides additional practice in modeling mechanical systems by using six examples based on more realistic and more complex versions of some examples presented in earlier sections. Example 3.5.1 treats a more realistic, distributed-mass pendulum model. Example 3.5.2 analyzes a system for raising a mast and illustrates how a potentially complex model can be simpliﬁed by neglecting pulley inertia. Examples 3.5.2 and 3.5.3 return to the wheeled vehicle problem for a case where the vehicle must climb an incline. They illustrate two approaches to the problem: the energy analysis approach and the force analysis approach. The latter shows how much more complex the model becomes if we must determine the internal reaction forces. Example 3.5.4 shows how to use the equivalent inertia approach to obtain a simpliﬁed model of a realistic example of a robot arm link containing a gear drive. The ﬁnal example analyzes a personal transporter whose model must describe the dynamics of a translational element connected to a rotational one. A Rod-and-Bob Pendulum ■ Problem The pendulum shown in Figure 3.5.1a consists of a concentrated mass m C (the bob) a distance L C from point O, attached to a rod of length L R and inertia I RG about its mass center. (a) Obtain its E X A M P L E 3.5.1 146 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems Figure 3.5.1 A rod-and-bob pendulum. LC O O L L LR C ␪ mg sin ␪ ␪ O G ␪ mC mg cos ␪ mC O L mC mC g mg mg ␪ m g (b) (a) G mR g m 5 m R 1 mC G g LR 2 L (c) (d) equation of motion. (b) Discuss the case where the rod’s mass m R is small compared to the concentrated mass. ■ Solution a. For the pendulum shown in Figure 3.5.1a, the inertia of the concentrated mass m C about point O is m C L C2 (see Table 3.2.1). From the parallel-axis theorem, the rod’s inertia about point O is I R O = I RG + m R 2 LR 2 and thus the entire pendulum’s inertia about point O is I O = I R O + m C L C2 = I RG + m R LR 2 2 + m C L C2 With the moment equation (3.2.1) about point O, the moment M O is caused by the perpendicular component of the weight mg acting through the mass center at G (see Figure 3.5.1b). Thus the desired equation of motion is I O θ̈ = −mgL sin θ (1) The distance L between point O and the mass center G of the entire pendulum is not given, but can be calculated as follows (Figure 3.5.1c). If the entire pendulum mass were concentrated at G, the weight force would produce the same moment about point O as the real pendulum. Thus, taking moments about point O, we have mgL = m C gL C + m R g LR 2 where m = m C + m R . Solve for L to obtain L= b. m C L C + m R (L R /2) mC + m R (2) If we neglect the rod’s mass m R compared to the concentrated mass m C , then we can take m R = I RG = 0, m = m C , L = L C , and I O = m L 2 . In this case, the equation of motion reduces to L θ̈ + g sin θ = 0 (3) This is a model for a pendulum whose mass is concentrated at a distance L from the pivot point, like that shown in Figure 3.5.1d. Note that this equation of motion is independent of the value of m. This is the same result obtained in Example 3.2.2. 3. 5 Additional Examples 147 For small angles, sin θ ≈ θ if θ is in radians. Substituting this approximation into (3) gives L θ̈ + gθ = 0 (4) This equation was solved in Example 3.2.2. Lifting a Mast E X A M P L E 3.5.2 ■ Problem A mast weighing 500 lb is hinged at its bottom to a ﬁxed support at point O (see Figure 3.5.2). The mast is 70 ft long and has a uniform mass distribution, so its center of mass is 35 ft from O. The winch applies a force f = 380 lb to the cable. The mast is supported initially at the 30◦ angle, and the cable at A is initially horizontal. Derive the equation of motion of the mast. You may assume that the pulley inertias are negligible and that the pulley diameter d is very small compared to the other dimensions. ■ Solution Part (b) of the ﬁgure shows the geometry of the mast at some angle θ > 30◦ , with the diameter d neglected. From the law of sines, sin φ = P sin(180◦ − μ − θ ) sin(μ + θ ) =P Q Q From the law of cosines, Q= P 2 + L 2 − 2P L cos(180◦ − μ − θ ) = where H = 20 ft, W = 5 ft, and μ = tan−1 H W P= = tan−1 √ 20 5 P 2 + L 2 + 2P L cos(μ + θ ) (1) = 76◦ = 1.33 rad H 2 + W 2 = 20.6 m The moment equation about the ﬁxed point O is I O θ̈ = −mg R cos θ + f LP sin(μ + θ ) Q (2) Figure 3.5.2 A system for lifting a mast. f sin ␾ A Q d D D f ␾ A 380 lb mg P H H 5 209 O 308 ␮ L 5 409 ␪ R 5 359 L 5 409 WO W 5 59 (a) mg cos ␪ (b) 148 CHAPTER 3 Modeling of Rigid-Body Mechanical Systems The moment of inertia is 1 1 500 2 m(70)2 = 70 = 25,400 slug-ft2 3 3 32.2 The for

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