Pendulum Rotation Analysis PDF

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 ac...

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