STATIC AND DYNAMICS

384 CHAPTER 7. Circular motion A = 0.8 m SAMPLE 7.13 A dumbbell AB, made of two equal masses and a rigid rod AB of m negligible mass, is welded to a rigid arm OC such that OC is perpendicular to AB. y Arm OC rotates about O at a constant angular velocity ω = 10 rad/skˆ. At the instant when θ = 60o, find the relative velocity of B with respect to A. C B l=1m m θ = 60o O x Solution Since A and B are two points on the same rigid body (AB) and the body is spinning about point O at a constant rate, we may use the relative velocity formula Figure 7.38: (Filename:sfig4.3.1) v B/A ≡ v B − v A = ω × r B/A (7.44) y nˆ to find the relative velocity of B with respect to A. We are given ω = ωkˆ = 10 rad/skˆ. A λˆ Let λˆ and nˆ be unit vectors parallel to AB and OC respectively. Since OC⊥AB, we have nˆ ⊥ λˆ . Now we may write vector r B/A as 1 C B r B/A = λˆ Oθ ˆ nˆ = cos θıˆ + sin θˆ θ Substituting ω and r B/A in Eqn (7.44) we get ıˆ v B/A = ωkˆ × λˆ x = ω (kˆ × λˆ ) Figure 7.39: (Filename:sfig4.3.1a) nˆ 1 The vector r B/A may also be expressed = ω nˆ directly in terms of unit vectors ıˆ and ˆ, = ω (cos θ ıˆ + sin θ ˆ) √ = 10 rad/s(0.8 m)·( 1 ıˆ + 3 ˆ) but it involves a little bit more geometry. √2 2 Note how assuming λˆ and nˆ in the direc- = 4 m/s(ıˆ + 3ˆ) tions shown makes calculations easier and √ v B/A = 4 m/s(ıˆ + 3ˆ) cleaner. Comments: v B/A can also be obtained by adding vectors v B and −v A geometrically. vA A Since A and B execute circular motion with the same radius R = O A = O B, the magnitudes of v B and v A are the same (= ω R) and since the velocity in circular v B/A motion is tangential to the circular path, v A ⊥ O A and v B ⊥ O B. Then moving vB v A to point B, we can easily find v B − v A = v B/A. Its direction is found to be B perpendicular to AB, i.e., along OC. v B/A = v B − v A O Figure 7.40: (Filename:sfig4.3.1b)

7.3. Kinematics of a rigid body in planar circular motion 385 SAMPLE 7.14 For the same problem and geometry as in Sample 7.13, find the acceleration of point B relative to point A. Solution Since points A and B are on the same rigid body AB which is rotating at a constant rate ω = 10 rad/s, the relative acceleration of B is: a B/A = a B − a A = ω × (ω × r B/A) = ωkˆ × (ωkˆ × λˆ ) = ωkˆ × ω nˆ (since kˆ × λˆ = nˆ) = ω2 (kˆ × nˆ) = ω2 (−λˆ ) 1 Now we need to express λˆ in terms of known basis vectors ıˆ and ˆ. If you are 1 If a body rotates in a plane, i.e., ω = good with geometry, then by knowing that λˆ ⊥ nˆ and nˆ = cos θ ıˆ + sin θ ˆ you can ωkˆ , then ω × (ω × r ) = −ω2 r . Using immediately write this fact we can immediately write a B/A = −ω2 r B/A = −ω2 λˆ . λˆ = sin θıˆ − cos θˆ (so that λˆ ·nˆ = 0). nˆ Or you may draw a big and clear picture of λˆ , nˆ, ıˆ and ˆ and label the angles as θ shown in Fig 7.41. Then, it is easy to see that θ λˆ λˆ = sin θıˆ − cos θ ˆ. ˆ Substituting for λˆ in the expression for a B/A, we get θ ıˆ a B/A = −ω2 (sin θ ıˆ − cos θ ˆ) √ Figure 7.41: The geometry of vectors ıˆ rad2 31 and nˆ . = −100 s2 · 0.8 m( ıˆ − ˆ) λˆ = − cos θˆ + sin θıˆ 22 (Filename:sfig4.3.2) = −40 m/s2 √ − ˆ) ( 3ıˆ A √ a B/A = −40 m/s2( 3ıˆ − ˆ) Comments: We could also find a B/A using geometry and geometric addition aA C of vectors. Since A and B are going in circles about O at constant speed, their y a B/A accelerations are centripetal accelerations. Thus, a A points along AO and a B points -aA B along BO. Also |a A| = |a B | = ω2(OA). Now adding −a A to a B we get a B/A aB which is seen to be along BA. O x Figure 7.42: (Filename:sfig4.3.2a)

386 CHAPTER 7. Circular motion y SAMPLE 7.15 Test the velocity formula on something you know. The motor at O in B Fig. 7.43 rotates the ‘L’ shaped bar OAB in counterclockwise direction at an angular ω motor speed which increases at ω˙ = 2.5 rad/s2. At the instant shown, the angular speed ω = 4.5 rad/s. Each arm of the bar is of length L = 2 ft. O L (a) Find the velocity of point A. (b) Find the relative velocity v B/A (= ω × r B/A) and use the result to find the A absolute velocity of point B (v B = v A + v B/A). x (c) Find the velocity of point B directly. Check the answer obtained in part (b) against the new answer. L Figure 7.43: An ‘L’ shaped bar rotates at Solution (a) As the bar rotates, every point on the bar goes in circles centered at point O. speed ω about point O. Therefore, we can easily find the velocity of any point on the bar using circular motion formula v = ω × r . Thus, (Filename:sfig5.3.1a) v A = ω × r A = ωkˆ × Lıˆ = ωLˆ O rA vA = 4.5 rad/s · 2 ftˆ = 9 ft/sˆ. ω A The velocity vector v A is shown in Fig. 7.44. Figure 7.44: v A = ω × r A is tangential v A = 9 ft/sˆ to the circular path of point A. (b) Point B and A are on the same rigid body. Therefore, with respect to point A, (Filename:sfig5.3.1b) point B goes in circles about A. Hence the relative velocity of B with respect to A is v B/A B v B/A = ω × r B/A r B/A = ωkˆ × Lˆ = −ωLıˆ = −4.5 rad/s · 2 ftıˆ = −9 ft/sıˆ. A v B/A and v B = v A + v B/A vB vA = 9 ft/s(−ıˆ + ˆ). B These velocities are shown in Fig. 7.45. Figure 7.45: v B/A = ω × r B/A and v B = v A + v B/A. v B/A = −9 ft/sıˆ, v B = 9 ft/s(−ıˆ + ˆ) (Filename:sfig5.3.1c) (c) Since point B goes in circles of radius OB about point O, we can find its velocity directly using circular motion formula: vB B vB = ω× rB = ωkˆ × (Lıˆ + Lˆ) = ωL(ˆ − ıˆ) = 9 ft/s(−ıˆ + ˆ). rB The velocity vector is shown in Fig. 7.46. Of course this velocity is the same velocity as obtained in part (b) above. OA Figure 7.46: v B = ω × r B. v B = 9 ft/s(−ıˆ + ˆ) (Filename:sfig5.3.1d) Note: Nothing in this sample uses ω˙ !

7.3. Kinematics of a rigid body in planar circular motion 387 SAMPLE 7.16 Test the acceleration formula on something you know. Consider yB the ‘L’ shaped bar of Sample 7.15 again. At the instant shown, the bar is rotating at 4 rad/s and is slowing down at the rate of 2 rad/s2. L = 2 ft (i) Find the acceleration of point A. ω A (ii) Find the relative acceleration a B/A of point B with respect to point A and use motor x the result to find the absolute acceleration of point B (a B = a A + a B/A). O (iii) Find the acceleration of point B directly and verify the result obtained in (ii). L = 2 ft Solution We are given: Figure 7.47: The ‘L’ shaped bar is rotat- ω = ωkˆ = 4 rad/skˆ, and ω˙ = −ω˙ kˆ = −2 rad/s2kˆ. ing counterclockwise and is slowing down. (i) Point A is going in circles of radius L. Hence, (Filename:sfig5.3.2a) a A = ω˙ × r A + ω × (ω × r A) = ω˙ × r A − ω2 r A = −ω˙ kˆ × Lıˆ − ω2 Lıˆ = −ω˙ Lˆ − ω2 Lıˆ O rA A = −2 rad/s · 2 ftˆ − (4 rad/s)2 · 2 ftıˆ ω aA ω˙ × r A = −(4ˆ + 32ıˆ) ft/s2. -ω2 r A a A = −(4ˆ + 32ıˆ) ft/s2 Figure 7.48: (Filename:sfig5.3.2b) B (ii) The relative acceleration of point B with respect to point A is found by consid- ering the motion of B with respect to A. Since both the points are on the same r B/A a B/A rigid body, point B executes circular motion with respect to point A. Therefore, a B/A = ω˙ × r B/A + ω × (ω × r B/A) = ω˙ × r B/A − ω2 = −ω˙ kˆ × Lˆ − ω2 Lˆ = ω˙ Lıˆ − ω2 Lˆ = 2 rad/s2 · 2 ftıˆ − (4 rad/s)2 · 2 ftˆ = (4ıˆ − 32ˆ) ft/s2, and A aA B a B = a A + a B/A = (−28ıˆ − 36ˆ) ft/s2. a B/A a B = −(28ıˆ + 36ˆ) ft/s2 aB (iii) Since point B is going in circles of radius OB about point O, we can find the Figure 7.49: (Filename:sfig5.3.2c) acceleration of B as follows. -ω2 r B B a B = ω˙ × r B + ω × (ω × r B ) O ω˙ × r B = ω˙ × r B − ω2 r B = −ω˙ kˆ × (Lıˆ + Lˆ) − ω2(Lıˆ + Lˆ) rB = (−ω˙ L − ω2 L)ˆ + (ω˙ L − ω2 L)ıˆ = (−4 − 32) ft/s2ˆ + (4 − 32) ft/s2ıˆ aB = (−36ˆ − 28ıˆ) ft/s2. A This acceleration is, naturally again, the same acceleration as found in (ii) Figure 7.50: (Filename:sfig5.3.2d) above. a B = −(28ıˆ + 36ˆ) ft/s2

388 CHAPTER 7. Circular motion A SAMPLE 7.17 The dumbbell AB shown in the figure rotates counterclockwise y about point O with angular acceleration 3 rad/s2. Bar AB is perpendicular to bar OC. At the instant of interest, θ = 45o and the angular speed is 2 rad/s. L = 0.5 m (a) Find the velocity of point B relative to point A. Will this relative velocity be C different if the dumbbell were rotating at a constant rate of 2 rad/s? B (b) Without calculations, draw a vector approximately representing the acceleration of B relative to A. θ x O (c) Find the acceleration of point B relative to A. What can you say about the direction of this vector as the motion progresses in time? Figure 7.51: Relative velocity and accel- Solution eration: (a) Velocity of B relative to A: (Filename:sfig5.3.3) v B/A = ω × r B/A = θ˙kˆ × L(sin θ ıˆ − cos θ ˆ) = θ˙L(sin θ ˆ + cos θ ıˆ) = 2 rad/s · 0.5 m(sin 45oˆ + cos 45oıˆ) = 0.707 m/s(ıˆ + ˆ). y Thus the relative velocity is perpendicular to AB, that is, parallel to OC. A a B/A No, the relative velocity will not be any different at the instant of interest if the (a B/A)R dumbbell were rotating at constant rate. As is evident from the formula, the relative velocity only depends on ω and r B/A, and not on ω˙ . Therefore, v B/A r B/A (a B/A)θ will be the same if at the instant of interest, ω and r B/A are the same. (b) Relative acceleration vector: The velocity and acceleration of some point B B on a rigid body relative to some other point A on the same body is the same as the velocity and acceleration of B if the body is considered to rotate about point Ox A with the same angular velocity and acceleration as given. Therefore, to find the relative velocity and acceleration of B, we take A to be the center of rotation Figure 7.52: To draw the relative accel- and draw the circular path of B, and then draw the velocity and acceleration eration of B, a B/A, consider point B going vectors of B. in circles about point A. Since we know that the acceleration of a point under circular motion has tangen- tial (ω˙ × r or θ¨ Reˆθ in 2-D) and radial or centripetal (ω × (ω × r ) or −θ˙2 ReˆR (Filename:sfig5.3.3a) in 2-D ) components, the total acceleration being the vector sum of these com- ponents, we draw an approximate acceleration vector of point B as shown in y Fig. 7.52. A (c) Acceleration of B relative to A: r B/A eˆθ θ a B/A = ω˙ × r B/A + ω × (ω × r B/A) B eˆR = θ¨kˆ × LeˆR + θ˙kˆ × (θ˙kˆ × LeˆR) = L θ¨eˆθ − L θ˙2eˆR = 0.5 m · 3 rad/s2(cos 45oıˆ + sin 45oˆ) −0.5 m · (2 rad/s)2(sin 45oıˆ − cos 45oˆ) = 1.061 m/s2(ıˆ + ˆ) − 1.414 m/s2(ıˆ − ˆ) = (−0.353ıˆ + 2.474ˆ) m/s2. Ox a B/A = (−0.353ıˆ + 2.474ˆ) m/s2 Figure 7.53: The geometry of eˆ R and eˆθ for the imagined motion of B about A. eˆ R = sin θıˆ − cos θˆ and eˆθ = cos θıˆ + sin θˆ. (Filename:sfig5.3.3b)

7.4. Dynamics of a rigid body in planar circular motion 389 7.4 Dynamics of a rigid body in planar circular motion Our goal here is to set up and evaluate in simple terms, the momentum, angular momentum, and energy balance equations for a planar body that is rotating about one point. The most common application is the mechanical analysis of a part held in place by a hinge or bearing. Mechanics y If we can calculate the velocity and acceleration of every point in a system, we can RO dm eˆR evaluate all the momentum and energy terms in the equations of motion (inside cover), O R x namely: L, L˙ , HC, H˙ C , EK and E˙K for any reference point C of our choosing. For rotational motion these calculations are a little more complex than the special case of ω straight-line motion in chapter 6, where all points in a system had the same acceleration as each other. For circular motion of a rigid body, we just well-learned in the previous section that the velocities and accelerations are v = ω× r, a = ω˙ × r + ω × (ω × r ), where ω is the angular velocity of the body relative to a fixed frame and r is the position of a point relative to the axis of rotation. These relations apply to every point on a rotating rigid body. Example: Spinning disk FBD ˆ F ıˆ The round flat uniform disk in figure 8.18 is in the x y plane spinning at the constant rate ω = ωkˆ about its center. It has mass mtot and radius M R0. What force is required to cause this motion? What torque? What power? Figure 7.54: A uniform disk turned by a From linear momentum balance we have: motor at a constant rate. F i = L˙ = mtot acm = 0, (Filename:tfigure4.3.motordisk) Which we could also have calculated by evaluating the integral L˙ ≡ a dm instead of using the general result that L˙ = mtotacm. From angular momentum balance we have: Mi/O = H˙ /O ⇒ M = r /O × a dm = R0 2π m tot dA 0 π R2O (ReˆR) × (−Rω2eˆR) R dθ dR 0 dm = 0 dθ dR = 0. So the net force and moment needed are F = 0 and M = 0. Like that a particle moves at constant velocity with no force, it rotates at constant rate with no torque (at least in 2D). 2

390 CHAPTER 7. Circular motion We’d now like to consider the most general case that the subject of the section allows, an arbitrarily shaped 2D rigid body with arbitrary ω and ω˙ . Linear momentum For any system in any motion we know, as we have often used, that L = mtot vcm and L˙ = mtot acm. For a rigid body, the center of mass is a particular point G that is fixed relative to the body. So the velocity and acceleration of that point can be expressed the same way as for any other point. So, for a body in planar rotational motion about 0 L = mtotω × rG/0 and L˙ = mtot ω˙ × rG/0 + −ω2 rG/0) . If the center of mass is at 0 the momentum and its rate of change are zero. But if the center of mass is off the axis of rotation, there must be a net force on the object with a component parallel to r0/G (if ω = 0) and a component orthogonal to r0/G (if ω˙ = 0). This net force need not be applied at 0 or G or any other special place on the object. Angular momentum: H O and H˙ O The angular momentum itself is easy enough to calculate, HO = r × v dm (a) = all mass r × (ω × r ) dm (b) = ωkˆ r 2 dm (7.45) (c) ⇒ H0 = ω r 2 dm. (d) Here eqn. (7.45)c is the vector equation. But since both sides are in the kˆ direction we can dot both sides with kˆ to get the scalar moment equation eqn. (7.45)d, taking both Mnet and ω as positive when counterclockwise. To get the all important angular momentum balance equation for this system we could easily differentiate eqn. (7.45), taking note that the derivative is being taken relative to a fixed frame. More reliably, we use the general expression for H˙ O to write the angular momentum balance equation as follows. Net moment/0 = rate of change of angular momentum/0 (a) Mnet = H˙ O (b) = r × a dm (c) (d) all mass = r × −ω2 r + ω˙ kˆ × r dm = r × ω˙ kˆ × r dm (e) (7.46) = r × ω˙ kˆ × r dm (f) Mnet = ω˙ kˆ r 2 dm (g) ⇒ Mnet = ω˙ r 2 dm (h) We get from from eqn. (7.46)f to eqn. (7.46)g by noting that r is perpendicular to kˆ. Thus, using the right hand rule twice we get r × (kˆ × r ) = r 2kˆ. Eqn. 7.46g and eqn. (7.46)h are the vector and scalar versions of the angular momentum balance equation for for rotation of a planar body about 0.

7.4. Dynamics of a rigid body in planar circular motion 391 Power and Energy Although we could treat distributed forces similarly, lets assume that there are a set of point forces applied. And, to be contrary, lets assume the mass is continuously distributed (the derivation for rigidly connected point masses would be similar). The power balance equation for one rotating rigid body is (discussed below): Net power in = rate of change of kinetic energy (a) P = E˙K (b) Fi · vi = d 1v2 dm (c) dt all mass 2 all applied forces d1 F i · (ω × r i ) = dt (ω × r ) · (ω × r ) dm (d) 2 d 1 ω2r2 dm ω · (ri × Fi) = dt 2 (e) (7.47) ω· (ri × Fi) = d 1 ω2 r2 dm (f) dt 2 ω · Mi = ω˙ ω r 2 dm (g) ω · Mtot = ω˙ · ω r 2 dm (h) H/0 When not notated clearly, positions and moments are relative to the hinge at 0. Deriva- tion 7.47 is two derivations in one. The left side about power and the right side about kinetic energy. Lets discuss one at a time. On the left side of eqn. (7.47) we note in (c) that the power of each force is the dot product of the force with the velocity of the point it touches. In (d) we use what we know about the velocities of points on rotating rigid bodies. In (e) we use the vector identity A · B × C = B · C × A from chapter 2. In (f) we note that ω is common to all points so factors out of the sum. In (g) we note that r × F i is the moment of the force about pt O. And in (g) we sum the moments of the forces. So we find that The power of a set of forces acting on a rigid body is the product of thier net moment(about 0) and the body angular velocity, P = ω · Mtot. (7.48) On the right side of eqn. (7.47) we note in (c) that the kinetic energy is the sum of the kinetic energy of the mass increments. In (d) we use what we know about the velocities of these bits of mass, given that they are on a common rotating body. In (e) we use that the magnitude of the cross product of orthogonal vectors is the product of the magnitudes (|A × B| = AB) and that the dot product of a vector with itself is its magnitude squared (A · A = A2). In (f) we factor out ω2 because it is common to all the mass increments and note that the remaining integral is constant in time for a rigid body. In (g) we carry out the derivative. In (h) we de-simplify the result from (g) in order to show a more general form that we will find later in 3D mechanics. Eqn. (h) follows from (g) because ω is parallel to ω˙ for 2D rotations. Note that we started here with the basic power balance equation from the front inside cover. Instead, we could have derived power balance from our angular mo- mentum balance expression (see box 7.4 on 392).

392 CHAPTER 7. Circular motion 7.3 THEORY The relation between angular momentum balance and power balance For this system, angular momentum balance can be derived from The opposite derivation starts with the power balance power balance and visa versa. Thus neither is essentially more Fig. 7.47(g) fundamental than the other and both are reliable. First we can derive power balance from angular momentum balance as follows: ω · Mi = ω˙ ω r 2 dm (g) Mnet = ω˙ kˆ r 2 dm (7.49) ⇒ ω kˆ · Mi = ω˙ ω r 2 dm (7.50) ω · Mnet = ω · ω˙ kˆ r 2 dm . ⇒ kˆ · Mi = ω˙ r 2 dm P = E˙K That is, when we dot both sides of the angular momentum equation and, assuming ω = 0, divide by ω to get the angular momentum equation for planar rotational motion. with ω we get on the left side a term which we recognize as the power of the forces and on the right side a term which is the rate of change of kinetic energy.

7.4. Dynamics of a rigid body in planar circular motion 393 SAMPLE 7.18 A rod going in circles at constant rate. A uniform rod of mass m ˆ Bω and length is connected to a motor at end O. A ball of mass m is attached to the rod ıˆ m at end B. The motor turns the rod in counterclockwise direction at a constant angular speed ω. There is gravity pointing in the −ˆ direction. Find the torque applied by m the motor (i) at the instant shown and (ii) when θ = 0o, 90o, 180o. How does the torque change if the angular speed is doubled? O θ motor Solution The FBD of the rod and ball system is shown in Fig. 7.56(a). Since the system is undergoing circular motion at a constant speed, the acceleration of the ball Figure 7.55: A rod goes in circles at a as well as every point on the rod is just radial (pointing towards the center of rotation O) and is given by a = −ω2r λˆ where r is the radial distance from the center O to the constant rate. point of interest and λˆ is a unit vector along OB pointing away from O (Fig. 7.56(b)). (Filename:sfig4.5.5) Angular Momentum Balance about point O gives B MO = H˙ O mg MO = r G/O × (−mgˆ) + r B/O × (−mgˆ) + Mkˆ G = − cos θmgkˆ − cos θ mgkˆ + Mkˆ M mg 2 θ = (M − 3 mg cos θ)kˆ (7.51) Rx O 2 (7.52) Ry H˙ ball/O H˙ rod/O (a) FBD of rod+ball system H˙ O = r B/O × m a B + r dm/O × adm dm dl m l dm r dm/O adm O λˆ (b) Calculation of H˙ O of the rod = λˆ × (−mω2 λˆ ) + lλˆ × (−ω2lλˆ ) dm Figure 7.56: A rod goes in circles at a constant rate. m (Filename:sfig4.5.5a) =0 (since λˆ × λˆ = 0) (i) Equating (7.51) and (7.52) we get M = 3 cos θ. mg 2 M = 3 mg cos θ 2 (ii) Substituting the given values of θ in the above expression we get M (θ = 0o) = 3 , M(θ = 90o) = 0 M(θ = 180o) = − 3 mg mg 22 M(0o) = 3 m g , M(90o) = 0 M (180o ) = − 3 mg 2 2 The values obtained above make sense (at least qualitatively). To make the rod and the ball go up from the 0o position, the motor has to apply some torque in the counterclockwise direction. In the 90o position no torque is required for the dynamic balance. In 180o position the system is accelerating downwards under gravity; therefore, the motor has to apply a clockwise torque to make the system maintain a uniform speed. It is clear from the expression of the torque that it does not depend on the value of the angular speed ω! Therefore, the torque will not change if the speed is doubled. In fact, as long as the speed remains constant at any value, the only torque required to maintain the motion is the torque to counteract the moments at O due to gravity.

394 CHAPTER 7. Circular motion Idler ωout ωout SAMPLE 7.19 A compound gear train. When the gear of an input shaft, often D called the driver or the pinion, is directly meshed in with the gear of an output shaft, RD the motion of the output shaft is opposite to that of the input shaft. To get the output BC RC motion in the same direction as that of the input motion, an idler gear is used. If the idler shaft has more than one gear in mesh, then the gear train is called a compound RB gear train. ωin A Driver RA In the gear train shown in Fig. 7.57, the input shaft is rotating at 2000 rpm ωin and the input torque is 200 N-m. The efficiency (defined as the ratio of output power to input power) of the train is 0.96 and the various radii of the gears are: Figure 7.57: A compound gear train. RA = 5 cm, RB = 8 cm, RC = 4 cm, and RD = 10 cm. Find (Filename:sfig4.5.7) (a) the input power Pin and the output power Pout , (b) the output speed ωout , and (c) the output torque. Solution (a) The power: Pin = Minωin = 200 N·m · 2000 rpm = 400000 N·m · rev · 2π · 1 min min 1 rev 60 s = 41887.9 N m/s ≈ 42 kW. ⇒ pout = efficiency · Pin = 0.96 · 42 kW ≈ 40 kW Pin = 42 kW, pout = 40 kW (b) The angular speed of meshing gears can be easily calculated by realizing that the linear speed of the point of contact has to be the same irrespective of which ωout gear’s speed and geometry is used to calculate it. Thus, R vP = ωin RA = ωB RB vR ⇒ ωB = ωi n · RA P vP and RB ⇒ ωin But vR = ωC RC = ωout RD Figure 7.58: (Filename:sfig4.5.7a) ⇒ ωout = ωC · RC RD ωC = ωB ωout = ωi n · RA · RC RB RD = 2000 rpm · 5 · 4 = 500 rpm. 8 10 ωout = 500 rpm (c) The output torque, Mout = Pout = 40 kW ωout 500 rpm = 40 · 1000 N·m · min · 1 rev · 60 s 500 s rev 2π 1 min = 764 N·m. Mout = 764 N·m

7.4. Dynamics of a rigid body in planar circular motion 395 SAMPLE 7.20 At the onset of motion: A 2 × 4 rectangular plate of mass 20 lbm y is pivoted at one of its corners as shown in the figure. The plate is released from rest O in the position shown. Find the force on the support immediately after release. x Solution The free body diagram of the plate is shown in Fig. 7.60. The force F b = 2 ft G applied on the plate by the support is unknown. a = 4 ft The linear momentum balance for the plate gives Figure 7.59: A rectangular plate is re- F = maG (7.53) leased from rest from the position shown. F − mgˆ = m(θ¨ rG/O eˆθ − θ˙2 ReˆR) = m θ¨ rG/O eˆθ (since θ˙ = 0 at t = 0) (Filename:sfig5.4.3) Thus to find F we need to find θ¨. F The angular momentum balance for the plate about the fixed support point O gives O M O = H˙ O r G/O × mg(−ˆ) = Izz/O θ¨kˆ G ( a ıˆ − b ˆ) × mg(−ˆ) = m(a2 + b2) + m( a2 + b2 ) θ¨kˆ 22 12 4 4 r G/O parallel axis theorem mg (a) Free body diagram −mg a kˆ = m(a2 + b2) θ¨kˆ 23 ˆ ⇒ θ¨ = − 3g a ) ıˆ 2(a2 + b2 O = − 3 · 32.2 ft/s2 · 4 ft r G/O 2(16 + 4) ft2 θ eˆθ = −9.66 rad/s2. G eˆR Substituting this value of θ¨ in eqn. (7.53), we get F = mgˆ + m θ¨ rG/O eˆθ (b) Geometry of motion = 20 lbm · 32.2 ft/s2ˆ + 20 lbm · (−9.66 rad/s2) · ( 22 + 12 ft) (2ıˆ√+ 4ˆ) Figure 7.60: (a) The free body diagram 20 of the plate. (b) The geometry of mo- rG/O eˆθ tion. From the given dimensions, eˆ R = aıˆ−bˆ bıˆ+aˆ 20 · 9.66 and eˆθ = . 32.2 1 1 (a2+b2) 2 (a2+b2) 2 = 20 lbfˆ − (1ıˆ + 2ˆ) lbf (since 1 lbm · ft/s2 = 1 lbf) 32.2 (Filename:sfig5.4.3a) = (−3ıˆ + 14ˆ) lbf. F = (−3ıˆ + 14ˆ) lbf

396 CHAPTER 7. Circular motion ˆ SAMPLE 7.21 The swinging stick. A uniform bar of mass m and length is pinned O at one of its ends O. The bar is displaced from its vertical position by an angle θ and released (Fig. 7.61). kˆ ıˆ θ (a) Find the equation of motion using momentum balance. m (b) Find the reaction at O as a function of (θ, θ˙, g, m, ). A Solution First we draw a simple sketch of the given problem showing relevant geometry (Fig. 7.61(a)), and then a free body diagram of the bar (Fig. 7.61(b)). Figure 7.61: A uniform rod swings in the Ry plane about its pinned end O. Rx (Filename:sfig5.4.1a) O ˆ O ıˆ θ G mg G m AA (a) (b) Figure 7.62: (a) A line sketch of the swinging rod and (b) free body diagram of the rod. (Filename:sfig5.4.1b) We should note for future reference that ω = ωkˆ ≡ θ˙kˆ ω˙ = ω˙ kˆ ≡ θ¨kˆ (a) Equation of motion using momentum balance: We can write angular mo- mentum balance about point O as MO = H˙ O. Let us now calculate both sides of this equation: O MO = r G/O × mg(−ˆ) ω2 /2 ω˙ /2 = (sin θ ıˆ − cos θˆ) × mg(−ˆ) θ 2 θ rG = − mg sin θ kˆ. (7.54) 2 (7.55) G H˙ O = Izz/G ω˙ + r G × m a G A aG Figure 7.63: Radial and tangential com- = m2 ω˙ kˆ + r G × m(ω˙ kˆ × r G − ω2 r G ) 12 ponents of a G . Since the radial component 2 is parallel to r G , r G × a G = 4 ω˙ kˆ . = m 2 m 2 m 2 ω˙ kˆ + ω˙ kˆ, = ω˙ kˆ (Filename:sfig5.4.1c) 12 4 3

7.4. Dynamics of a rigid body in planar circular motion 397 where the last step, rG × maG = m2 ω˙ kˆ, should be clear from Fig. 7.63. 4 Equating (7.54) and (7.55) we get 2 (7.56) − m g sin θ = m ω˙ 23 or ω˙ + 3g sin θ = 0 2 or θ¨ + 3g sin θ = 0. 2 θ¨ + 3g sin θ = 0 2 (b) Reaction at O: Using linear momentum balance where F = m aG, F = Rx ıˆ + (Ry − mg)ˆ, and aG = ω˙ (cos θ ıˆ + sin θ ˆ) + ω2(− sin θ ıˆ + cos θ ˆ) 22 = [(ω˙ cos θ − ω2 sin θ )ıˆ + (ω˙ sin θ + ω2 cos θ )ˆ]. 2 Dotting both sides of F = m a G with ıˆ and ˆ and rearranging, we get Rx = m (ω˙ cos θ − ω2 sin θ ) 2 ≡ m (θ¨ cos θ − θ˙2 sin θ ), 2 Ry = mg + m (ω˙ sin θ + ω2 cos θ ) 2 ≡ mg + m (θ¨ sin θ + θ˙2 cos θ ). 2 Now substituting the expression for θ¨ from (7.56) in Rx and Ry, we get Rx = −m sin θ 3 g cos θ + θ˙2 , (7.57) 42 (7.58) Ry = mg 1 − 3 sin2 θ + m θ˙2 cos θ. 4 2 R = −m ( 3 g cos θ + 2 θ˙2) sin θ ıˆ + [mg(1 − 3 sin2 θ) + m 2 θ˙2 cos θ ]ˆ 4 4 Check: We can check the reaction force in the special case when the rod does not swing but just hangs from point O. The forces on the bar in this case have to satisfy static equilibrium. Therefore, the reaction at O must be equal to mg and directed vertically upwards. Plugging θ = 0 and θ˙ = 0 (no motion) in Eqn. (7.57) and (7.58) we get Rx = 0 and Ry = mg, the values we expect.

398 CHAPTER 7. Circular motion SAMPLE 7.22 The swinging stick: energy balance. Consider the same swinging stick as in Sample 7.21. The stick is, again, displaced from its vertical position by an angle θ and released (See Fig. 7.61). (a) Find the equation of motion using energy balance. (b) What is θ˙ at θ = 0 if θ (t = 0) = π/2? (c) Find the period of small oscillations about θ = 0. Solution (a) Equation of motion using energy balance: We use the power equation, E˙K = P, to derive the equation of motion of the bar. EK = 1 Izz/G ω2 + 1 m vG2 2 2 where 1 Izz/G ω2 = kinetic energy of the bar due to rotation 2 about the z-axis passing through the mass center G, and 1 mvG2 = kinetic energy of the bar due to translation 2 of the mass center. But vG = ωrG = ω 2 . Therefore, O EK = 1m 2 + 1 mω2 2 = 1 2ω2, m /2 2 12 ω2 2 4 G 6 θ and d (1m 1 1 dt 6 m m G'' E˙K = 2ω2) = 2ω ω˙ = 2θ˙θ¨. h 3 3 G' Calculation of power (P): There are only two forces acting on the bar, the reaction force, R(= Rx ıˆ + Ryˆ) and the force due to gravity, −mgˆ. Since the support point O does not move, no work is done by R. Therefore, A W = Work done by gravity force in moving from G to G. = −mgh Figure 7.64: Work done by the force of Note that the negative sign stands for the work done against gravity. Now, gravity in moving from G to G F ·d r = −mgˆ · hˆ = −mgh. h = OG − OG = − cos θ = (1 − cos θ ). 22 2 (Filename:sfig5.4.2a) Therefore, W = −mg (1 − cos θ ) 2 and P = W˙ = d W = −mg sin θθ˙. dt 2 Equating E˙K and P we get −m g sin θθ˙ = 1 2θ˙θ¨ m 23 or θ¨ + 3g sin θ = 0. 2 θ¨ + 3g sin θ = 0 2

7.4. Dynamics of a rigid body in planar circular motion 399 This equation is, of course, the same as we obtained using balance of angular O G' h= /2 1 θ = π/2 momentum in Sample 7.21. (b) Find ω at θ = 0: We are given that at t = 0, θ = π/2 and θ˙ ≡ ω = 0 G V=0 (released from rest). This position is (1) shown in Fig. 7.65. In position (2) θ = 0, i.e., the rod is vertical. Since there are no dissipative forces, the total energy of the system remains constant. Therefore, taking datum for potential energy as shown in Fig. 7.65, we may write EK1 +V1 = EK2 + V2 00 or mg = 1 Izz/G ω2 + 1 m vG2 2 2 2 vG = 1m 2 + 1 ω m 2 12 ω2 22 = 1 2ω2 2 θ=0 m 6 ⇒ ω = ± 3g Figure 7.65: The total energy between positions (1) and (2) is constant. (Filename:sfig5.4.2b) ω = ± 3g (c) Period of small oscillations: The equation of motion is θ¨ + 3g sin θ = 0. 2 For small θ, sin θ ≈ θ ⇒ θ¨ + 3g θ = 0 (7.59) or 2 where θ¨ + λ2θ = 0 λ2 = 3g . 2 Therefore, the circular frequency = λ = 3g , 2 and the time period T = 2π = 2π 2 . λ 3g T = 2π 2 3g [Say for g = 9.81 m/s2, = 1 m we get T = π 2 1 s = 0.4097 s] 4 2 3 9.81

400 CHAPTER 7. Circular motion SAMPLE 7.23 The swinging stick: numerical solution of the equation of motion. For the swinging stick considered in Samples 7.21 or 7.22, find the time that the rod takes to fall from θ = π/2 to θ = 0 if it is released from rest at θ = π/2? 1 You may solve equation (7.56) by nu- Solution π/2 is a big value of θ – big in that we cannot assume sin θ ≈ θ (obviously 1 = 1.5708). Therefore we may not use the linearized equation (7.59) to solve merical integration using any computer for t explicitly. We have to solve the full nonlinear equation (7.56) to find the software (e.g., MathCad, MATHEMAT- required time. Unfortunately, we cannot get a closed form solution of this equation ICA, Maple, MATLAB, etc..) or your using mathematical skills you have at this level. Therefore, we resort to numerical own program written in any language of integration of this equation. 1 your choice. To write your own pro- Here, we show how to do this integration and find the required time using the numerical solution. We assume that we have some numerical ODE solver, say gram, however, you will have to know odesolver, available to us that will give us the numerical solution given appropriate input. some method for numerical integration, e.g., Euler’s method, Runge-Kutta method, The first step in numerical integration is to set up the given differential equation et c.. of second or higher order as a set of first order ordinary differential equations. To do so for Eqn. (7.56), we introduce ω as a new variable and write θ˙ = ω (7.60) ω˙ = − 3g sin θ (7.61) 2 Thus, the second order ODE (7.56) has been rewritten as a set of two first order ODE’s (7.60) and (7.61). We may write these first order equations in vector form by assuming z = [θ ω]T . That is, z= z1 = θ ⇒ z˙ = θ˙ = z2 z2 ω ω˙ − 3g sin z1 2 To use any numerical integrator, we usually need to write a small program which will compute and return the value of z as output if t and z are supplied as input. Here is such a program called barpend, written as a pseudocode, for our equations. Anything in a line following a % sign is a comment. function zdot = barpend(t,z); % -------------------------------------------------- % function to compute zdot for a bar pendulum % where z = [theta; omega]; % -------------------------------------------------- g = 9.81; L = 1; % these values can be changed zdot = [z(2); -3*g/(2*L)*sin(z(1))]; Other than this function barpend, a numerical ODE solver will require the initial time t0, the final time t f , and initial conditions z0 as inputs. The following is a pseudocode, runpend, which contains all these input parameters and the commands to solve the ODEs and plot the results. t0 = 0; tf = 4; % specify initial and final times z0 = [pi/2; 0]; % initial conditions: % [theta=pi/2; omega=0] [t,z] = odesolver(’barpend’,t0,tf,z0); % run odesolver. The output is [t,z]

7.4. Dynamics of a rigid body in planar circular motion 401 plot(t,z) % plots t vs. theta % and t vs. omega together xlabel(’t’),ylabel(’theta and omega’) % label axes The results obtained from the numerical solution are shown in Fig. 7.66. 6theta and omega The problem of finding the time taken by the bar to fall from θ = π/2 to θ = 0 4 numerically is nontrivial. It is called a boundary value problem. We have only illustrated how to solve initial value problems. However, we can get fairly good 2 estimate of the time just from the solution obtained. We first plot θ against time t to get the graph shown in Fig. 7.67. We find the values of t and the corresponding 0 values of θ that bracket θ = 0. Now, we can use linear interpolation to find the value of t at θ = 0. Proceeding this way, we get t = 0.4833 (seconds), a little more than -2 we get from the linear ODE in sample 7.22 of 0.40975. Additionally, we can get by interpolation that at θ = 0 -4 ω = −5.4246 rad/s. -6 0 0.5 1 1.5 2 2.5 3 3.5 4 How does this result compare with the analytical value of ω from sample 7.22 (which t did not depend on the small angle approximates)? Well, we found that Figure 7.66: Numerical solution is shown by plotting θ and ω against time. (Filename:sfig5.4.4a) ω = − 3g = − 3 · 9.81 m/s2 = −5.4249 s−1. 2 1m 1.5 Thus, we get a fairly accurate value from numerical integration! theta 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t Figure 7.67: Graphic output of plot(t, z(:,1)), xlabel(’t’), ylabel(’theta’). (Filename:sfig5.4.4b)

402 CHAPTER 7. Circular motion SAMPLE 7.24 The swinging stick with a destabilizing torque. Consider the swinging stick of Sample 7.21 once again. (a) Find the equation of motion of the stick,if a torque M = Mkˆ is applied at end O and a force F = Fıˆ is applied at the other end A. (b) Take F = 0 and M = Cθ . For C = 0 you get the equation of free oscillations obtained in Sample 7.21 or 7.22 For small C, does the period of the pendulum increase or decrease? (c) What happens if C is big? Solution (a) A free body diagram of the bar is shown in Fig. 7.68. Once again, we can use MO = H˙ O to derive the equation of motion as in Sample 7.21. We calculated MO and H˙ O in Sample 7.21. Calculation of H˙ O remains the same in the present problem. We only need to recalculate MO. MO = Mkˆ + r G/O × mg(−ˆ) + r A/O × F = Mkˆ − mg sin θ kˆ + F cos θ kˆ 2 = (M + F cos θ − mg sin θ )kˆ 2 and H˙ O = 2 (see Sample 7.21) mθ¨ kˆ 3 Ry Therefore, from MO = H˙ O O Rx M 2 G M + F cos θ − mg sin θ = mθ¨ 23 mg F ⇒ θ¨ + 3g sin θ − 3F cos θ − 3M = 0. 2 m m2 A Figure 7.68: Free body diagram of the θ¨ + 3g sin θ − 3F cos θ − 3M =0 bar with applied torque M and force F 2 m m2 (Filename:sfig5.4.5a) (b) Now, setting F = 0 and M = C θ we get θ¨ + 3g sin θ − 3C θ = 0 (7.62) 2 m2 Numerical Solution: We can numerically integrate (7.62) just as in the previ- ous Sample to find θ (t). We will, however, need to modify function barpend to accommodate the new term due to M in the equation of motion. The new function, barpendf is as follows. function zdot = barpendf(t,z); % ------------------------------------------- % function to compute zdot for a bar pendulum % with applied torque M = C * theta. % where z = [theta; omega]; % ------------------------------------------- g = 9.81; L = 1; m = 1; C = 4; % parameters zdot = [ z(2) ; -(3*g/(2*L)) * sin(z(1)) + 3*C/(m*L^2) * z(1) ];

7.4. Dynamics of a rigid body in planar circular motion 403 Using this function we run odesolver just as in the previous case to find the response of the pendulum. Fig. 7.69 shows different responses for various values of C. Note that for C = 0, it is the same case as unforced bar pendulum considered above. From Fig. 7.69 it is clear that the bar has periodic motion 0.8 0.7 0.6 C=5 C = 4.905 C=2 C=4 0.5 C=0 C= 1 0.4 ω 0.3 0.2 0.1 0 -0.1 -0.2 0 1 2 3 4 5 6 7 8 9 10 t Figure 7.69: θ(t) with applied torque M = C θ for C = 0, 1, 2, 4, 4.905, 5. Note that for small C the motion is periodic but for large C (C ≥ 4.4) the motion becomes aperiodic. (Filename:sfig5.4.5b) for small C, with the period of motion increasing with increasing values of C. It makes sense if you look at Eqn. (7.62) carefully. Gravity acts as a restoring force while the applied torque acts as a destabilizing force. Thus, with the resistance of the applied torque, the stick swings more sluggishly making its period of oscillation bigger. (c) From Fig. 7.69, we see that at about C ≈ 4.9 the stability of the system changes completely. θ(t) is not periodic anymore. It keeps on increasing at faster and faster rate, that is, the bar makes complete loops about point O with ever increasing speed. Does it make physical sense? Yes, it does. As the value of C is increased beyond a certain value (can you guess the value?), the applied torque overcomes any restoring torque due to gravity. Consequently, the bar is forced to rotate continuously in the direction of the applied force.

404 CHAPTER 7. Circular motion 7.5 Polar moment of inertia: Izczm and IzOz 1 In fact the moment of inertia for a given We know how to find the velocity and acceleration of every bit of mass on a 2-D rigid object depends on what reference point is body as it spins about a fixed axis. So, as explained in the previous section, it is just a used. Most commonly when people say matter of doing integrals or sums to calculate the various motion quantities (momenta, ‘the’ moment of inertia they mean to use energy) of interest. As the body moves and rotates the region of integration and the the center of mass as the reference point. values of the integrands change. So, in principle, in order to analyze a rigid body one For clarity this moment of inertia matrix is has to evaluate a different integral or sum at every different configuration. But there is a shortcut. A big sum (over all atoms, say), or a difficult integral is reduced to a often notated as [I cm] in this book. If a dif- simple multiplication a simple multiplication. ferent reference point, say point O is used, The moment of inertia matrix [I ] 1 is defined to simplify the expressions for the angular momentum, the rate of change of angular momentum, and the energy of a rigid the matrix is notated as [I O]. body. For study of the analysis of flat objects in planar motion only one component of the matrix [I] is relevant, it is Izz, called just I or J in elementary physics courses. Here are the results. A flat object spinning with ω = ωkˆ in the x y plane has a mass distribution which gives, by means of a calculation which we will discuss shortly, a moment of inertia Izczm or just ‘I ’ so that: H cm = I ωkˆ (7.63) (7.64) H˙ cm = I ω˙ kˆ (7.65) EK/cm = 1 ω2 I. 2 The moments of inertia in 2-D : [Icm] and [IO]. We start by looking at the scalar I which is just the zz or 33 component of the matrix [I ] that we will study later. The definition of I cm is I cm ≡ = x2 + y2 dm = r2 for a uniform planar object dm r 2 mtot d A A fwf The mass per unit area. where x and y are the distances of the mass in the x and y direction measured from an origin, and r is the direct distance from that origin. If that origin is at the center of mass then we are calculating I cm, if the origin is at a point labeled C or O then we are calculating I C or I O . The term Izz is sometimes called the polar moment of inertia, or polar mass moment of inertia to distinguish it from the Ixx and Iyy terms which have little utility in planar dynamics (but are all important when calculating the stiffness of beams!).

7.5. Polar moment of inertia: Izczm and IzOz 405 What, physically, is the moment of inertia? It is a measure of the extent to which mass is far from the given reference point. Every bit of mass contributes to I in proportion to the square of its distance from the reference point. Note from, say, eqn. (7.46) on page 390 that I is just the quantity we need to do mechanics problems. Radius of gyration y Another measure of the extent to which mass is spread from the reference point, besides the moment of inertia, is the radius of gyration, rgyr . The radius of gyration is sometimes called k but we save k for stiffness. The radius of gyration is defined as: rgyr ≡ I /m ⇒ rg2yr m = I. That is, the radius of gyration of an object is the radius of an equivalent ring of mass that has the same I and the same mass as the given object. Other reference points y dm x For the most part it is I cm which is of primary interest. Other reference points are r useful x (a) if the rigid body is hinged at a fixed point O then a slight short cut in calculation O, C, or of angular momentum and energy terms can be had; and Figure 7.70: A general planar body. (b) if one wants to calculate the moment of inertia of a composite body about its center of mass it is useful to first find the moment of inertia of each of its parts (Filename:tfigure4.4.DefofI) about that point. But the center of mass of the composite is usually not the center of mass of any of the separate parts. The box 8.2 on page 458 shows the calculation of I for a number of simple 2 dimen- sional objects. The parallel axis theorem for planar objects The planar parallel axis theorem is the equation IzCz = Izczm + mtot rc2m/C . d2 In this equation d = rcm/C is the distance from the center of mass to a line parallel to the z-axis which passes through point C. See box 7.4 on page 407 for a derivation of the parallel axis theorem for planar objects. Note that IzCz ≥ Izczm , always. One can calculate the moment of inertia of a composite body about its center of mass, in terms of the masses and moments of inertia of the separate parts. Say the position of the center of mass of mi is (xi , yi ) relative to a fixed origin, and the moment of inertia of that part about its center of mass is Ii . We can then find the moment of inertia of the composite Itot about its center of mass (xcm, ycm) by the following sequence of calculations: (1) mtot = mi (2) xcm = xi mi /mtot ycm = yi mi /mtot (3) di2 = (xi − xcm )2 + (yi − ycm )2 (4) Itot = Ii + mi di2 .

406 CHAPTER 7. Circular motion Of course if you are mathematically inclined you can reduce this recipe to one grand formula with lots of summation signs. But you would end up doing the calculation in about this order in any case. As presented here this sequence of steps lends itself naturally to computer calculation with a spread sheet or any program that deals easily with arrays of numbers. The tidy recipe just presented is actually more commonly used, with slight modi- fication, in strength of materials than in dynamics. The need for finding area moments of inertia of strange beam cross sections arises more frequently than the need to find polar mass moment of inertia of a strange cutout shape. The perpendicular axis theorem for planar rigid bodies The perpendicular axis theorem for planar objects is the equation Izz = Ixx + Iyy which is derived in box 7.4 on page 407. It gives the ‘polar’ inertia Izz in terms of the inertias Ixx and Iyy. Unlike the parallel axis theorem, the perpendicular axis theorem does not have a three-dimensional counterpart. The theorem is of greatest utility when one wants to study the three-dimensional mechanics of a flat object and thus are in need of its full moment of inertia matrix.

7.5. Polar moment of inertia: Izczm and IzOz 407 7.4 THEORY The 2-D parallel axis theorem and the perpendicular axis theorem Sometimes, one wants to know the moment of inertia relative to Sometimes, people write the parallel axis theorem more simply the center of mass and, sometimes, relative to some other point O, as if the object is held at a hinge joint at O. There is a simple relation between these two moments of inertia known as the parallel axis I 0 = I cm + md2 or JO = Jcm + md2 theorem. using the symbol J to mean Izz. One thing to note about the parallel 2-D parallel axis theorem axis theorem is that the moment of inertia about any point O is always greater than the moment of inertia about the center of mass. For the two-dimensional mechanics of two-dimensional objects, For a given object, the minimum moment of inertia is about the our only concern is Izoz and Izczm and not the full moment of inertia center of mass. matrix. In this case, Izoz = r/2o dm and Izczm = r/2cm dm. Now, let’s prove the theorem in two dimensions referring to the figure. Why the name parallel axis theorem? We use the name because the two I ’s calculated are the moments of inertia about two parallel dm axes (both in the z direction) through the two points cm and O. y One way to think about the theorem is the following. The r /cm moment of inertia of an object about a point O not at the center of mass is the same as that of the object about the cm plus that of a r /O point mass located at the center of mass. If the distance from O to the cm is larger than the outer radius of the object, then the d2m r cm/O term is larger than Izczm . The distance of equality of the two terms is the radius of gyration, rgyr . Perpendicular axis theorem (applies to planar objects only) For planar objects, Ox IzOz = |r |2 dm IzOz = r/2O dm = (x/2O + y/2O )dm = (x/2O + y/2O )dm = x/2O dm + y/2O dm = [(xcm/O + x/cm )2 + (ycm/O + y/cm )2]dm = I O + IxOx yy x/O y/O Similarly, = [(xc2m/O + 2xcm/O x/cm + x/2cm ) + (yc2m/O + 2ycm/O y/cm + y/2cm )]dm Izczm = Ixcxm + I cm . yy = (xc2m/O + yc2m/O ) dm +2xcm/O x/cm dm + That is, the moment of inertia about the z-axis is the sum of the inertias about the two perpendicular axes x and y. Note that the objects must be planar (z = 0 everywhere) or the theorem would not be true. For example, Ixox = (y/2O + z/2O )dm = y/2O dm for a three-dimensional object. m 0 y 2 ycm / O y/cm dm + (x/2cm + y/2cm )dm = rc2m/O m + 0 y/O dm x (x/2cm + y/2cm )dm r /O O x/O Izczm = Izczm + rc2m/O m d2 The cancellation y/cm dm = x/cm dm = 0 comes from the definition of center of mass.

408 CHAPTER 7. Circular motion 7.5 Some examples of 2-D Moment of Inertia Here, we illustrate some simple moment of inertia calculations for A thin uniform rod two-dimensional objects. The needed formulas are summarized, in part, by the lower right corner components (that is, the elements in y 2 ds the third column and third row (3,3)) of the matrices in the table on the inside back cover. s 1O d m=ρ One point mass x dm = ρds ρ = mass per x2 + y2 = r2 y unit length r 1+ 2= Consider a thin rod with uniform mass density, ρ, per unit length, and length . We calculate Izoz as ρds Izoz = r2 dm 2 = s2ρds (s = r ) −1 = 1 ρs3 2 ρ ≡ const.) (since Ox 3 −1 If we assume that all mass is concentrated at one or more points, = 1ρ( 3 + 3 ). then the integral 3 1 2 Izoz = r/2o dm If either 1 = 0 or 2 = 0, then this expression reduces to Izoz = 1 2. 3 m If 1= 2, then O is at the center of mass and reduces to the sum Izoz = ri2/o m i Izoz = Izczm = 1ρ 3 3 = ml2 . 3 + 22 12 which reduces to one term if there is only one mass, Izoz = r 2m = (x2 + y2)m. We can illustrate one last point. With a little bit of algebraic histri- onics of the type that only hindsight can inspire, you can verify that So, if x = 3 in, y = 4 in, and m = 0.1 lbm, then Izoz = 2.5 lbm in2. the expression for Iz0z can be arranged as follows: Note that, in this case, Izczm = 0 since the radius from the center of Iz0z = 1ρ( 3 + 32) mass to the center of mass is zero. 3 1  2 = ρ( 1 + 2)  2− 1  + ρ ( 1 + 2)3 2 12 m m 2/12 d Two point masses 2 m2 y = md2 + m 12 r2 r1 = md2 + Izczm m1 That is, the moment of inertia about point O is greater than that about the center of mass by an amount equal to the mass times the distance from the center of mass to point O squared. This derivation of the parallel axis theorem is for one special case, that of a uniform thin rod. Ox In this case, the sum that defines Izoz reduces to two terms, so Izoz = ri2/omi = m1r12 + m2r22. Note that, if r1 = r2 = r , then Izoz = mtot r 2.

7.5. Polar moment of inertia: Izczm and IzOz 409 A uniform hoop Uniform rectangular plate y dm = ρRdθ y dθ dm = ρdxdy Ox b x R O m = ρab m = 2ρπ R a For a hoop of uniform mass density, ρ, per unit length, we might consider all of the points to have the same radius R. So, For the special case that the center of the plate is at point O, the center of mass of mass is also at O and Izoz = Izczm . Izoz = r 2dm = R2dm = R2 dm = R2m. Izoz = Izczm = r2dm Or, a little more tediously, ba dm Izoz = r2dm = 2 2 (x2 + y2) ρdxdy 2π − b − a 2 2 = R2ρRdθ b x3 + xy2 x = a 2 2 0 = ρ dy 2π − b 3 a 2 2 = ρR3 dθ x =− 0 x = a y= b 2 2 = 2πρ R3 = (2πρ R) R2 = m R2. ρ x3y + xy3 = 3 3 m x =− a b 2 y=− 2 This Izoz is the same as for a single point mass m at a distance R from the origin O. It is also the same as for two point masses if they both = ρ a3b + ab3 12 12 are a distance R from the origin. For the hoop, however, O is at the center of mass so Izoz = Izczm which is not the case for a single point = m (a2 + b2). mass. 12 A uniform disk Note that r 2dm = x2dm + y2dm for all planar objects (the y dm = ρ dA = ρr dr dθ perpendicular axis theorem). For a uniform rectangle, y2dm = ρ y2d A. But the integral y2d A is just the term often used for I , dθ the area moment of inertia, in strength of materials calculations for R rdθ the stresses and stiffnesses of beams in bending. You may recall that y2d A = ab3 = Ab2 for a rectangle. Similarly, x2d A = Aa2 . 12 12 12 1 Or x dA dr So, the polar moment of inertia J = Izoz = m 12 (a2 + b2) can be recalled by remembering the area moment of inertia of a rectangle m = ρπ R2 combined with the perpendicular axis theorem. Assume the disk has uniform mass density, ρ, per unit area. For a uniform disk centered at the origin, the center of mass is at the origin so Izoz = Izczm = r2dm R 2π = r2ρrdθdr 00 R = 2πρr 3dr 0 = 2πρ r 4 R = πρ R4 = (πρ R2) R2 40 2 2 = m R2 . 2 For example, a 1 kg plate of 1 m radius has the same moment of inertia as a 1 kg hoop with a 70.7 cm radius.

410 CHAPTER 7. Circular motion y x SAMPLE 7.25 A pendulum is made up of two unequal point masses m and 2m m connected by a massless rigid rod of length 4r . The pendulum is pivoted at distance rO r along the rod from the small mass. θ 3r (a) Find the moment of inertia IzOz of the pendulum. (b) If you had to put the total mass 3m at one end of the bar and still have the same IzOz as in (a), at what distance from point O should you put the mass? (This distance is known as the radius of gyration). Solution Here we have two point masses. Therefore, the integral formula for IzOz (IzOz = m r/2O dm) gets replaced by a summation over the two masses: 2m 2 Figure 7.71: (Filename:sfig4.5.1) IzOz = mi ri2/O i =1 = = m1r12/O + m2r22/O (a) For the pendulum, m1 = m, m2 = 2m, r1/O = r, r2/O = 3r . IzOz = mr 2 + 2m(3r )2 = 19mr 2 y IzOz = 19mr 2 O (b) For the equivalent simple pendulum of mass 3m, let the length of the massless θ rod (i.e., the distance of the mass from O) be rgyr . k (IzOz )simple = (3m)·rg2yr x Now we need (IzOz )simple = IzOz (from part (a)) ⇒ 3mrg2yr = 19mr 2 ⇒ rgyr = 19 r 3 = 2.52r 3m Thus the radius of gyration rgyr of the given pendulum is rgyr = 2.52 r . Figure 7.72: (Filename:sfig4.5.1a) rgyr = 2.52 r

7.5. Polar moment of inertia: Izczm and IzOz 411 SAMPLE 7.26 A uniform rigid rod AB of mass M = 2 kg and length 3 = 1.5 m A swings about the z-axis passing through the pivot point O. O (a) Find the moment of inertia IzOz of the bar using the fundamental definition 2 IzOz = m r/2O dm. y (b) Find IzOz using the parallel axis theorem given that Izczm = 1 m 2 where m = M total mass, and = total length of the rod. (You can 12 Izczm for many x find B commonly encountered objects in the table on the inside backcover of the text). Figure 7.73: (Filename:sfig4.5.2) A Solution (a) Since we need to carry out the integral, IzOz = m r/2O dm, to find IzOz , let us O consider an infinitesimal length segment d of the bar at distance from the ' pivot point O. (see Figure 7.74). Let the mass of the infinitesimal segment be dm. d ' dm Now the mass of the segment may be written as B dm = (mass per unit length of the bar) · (length of the segment) Figure 7.74: (Filename:sfig4.5.2a) = M Note: mass = total mass . A d unit length total length 3 3 2 We also note that the distance of the segment from point O, r/O = . Substi- O rO/cm tuting the values found above for r/O and dm in the formula we get cm IzOz = 2 ( )2 M d B −3 Figure 7.75: (Filename:sfig4.5.2b) r/2O dm =M 2 =M 32 3− 3 ( )2d 3− M 83 3 =M 2 = −− 3 33 = 2 kg·(0.5 m)2 = 0.5 kg· m2 IzOz = 0.5 kg· m2 (b) The parallel axis theorem states that IzOz = Izczm + M r 2 /cm . O Since the rod is uniform, its center of mass is at its geometric center, i.e., at distance 3 from either end. From the Fig 7.75 we can see that 2 rO/cm = AG − AO = 3 − = 2 2 Therefore, IzOz = 1 M(3 )2 +M( )2 12 2 Izczm = 9 2+M 2 2 M =M 12 4 = 0.5 kg· m2 (same as in (a), of course) IzOz = 0.5 kg· m2

412 CHAPTER 7. Circular motion SAMPLE 7.27 A uniform rigid wheel of radius r = 1 ft is made eccentric by cutting out a portion of the wheel. The center of mass of the eccentric wheel is at C, a distance e = r from the geometric center O. The mass of the wheel (after deducting 3 the cut-out) is 3.2 lbm. The moment of inertia of the wheel about point O, IzOz, is 1.8 e lbm· ft2. We are interested in the moment of inertia Izz of the wheel about points A CO A B and B on the perimeter. yr (a) Without any calculations, guess which point, A or B, gives a higher moment of inertia. Why? (b) Calculate IzCz, IzAz and IzBz and compare with the guess in (a). x Solution Figure 7.76: (Filename:sfig4.5.3) (a) The moment of inertia IzBz should be higher. Moment of inertia Izz measures the geometric distribution of mass about the z-axis. But the distance of the mass from the axis counts more than the mass itself (IzOz = m r/2O dm). The distance r/O of the mass appears as a quadratic term in IzOz. The total mass is the same whether we take the moment of inertia about point A or about point B. However, the distribution of mass is not the same about the two points. Due to the cut-out being closer to point B there are more “dm’s” at greater distances from point B than from point A. So, we guess that IzBz > IzAz (b) If we know the moment of inertia IzCz (about the center of mass) of the wheel, we can use the parallel axis theorem to find IzAz and IzBz. In the problem, we are given IzOz. But, r/3 B IzOz = IzCz + MrO2 /C (parallel axis theorem) A CO ⇒ IzCz = IzOz − MrO2 /C = 1.8 lbm ft2 − 3.2 lbm 1 ft 2 3 r O /C =e= r 3 = 1.44 lbm· ft2 rA/C rB/C Now, IzAz = IzCz + Mr 2 = IzCz + M 2r 2 A/C 3 2r/3 r/3 + r Figure 7.77: (Filename:sfig4.5.3a) = 1.44 lbm· ft2 + 3.2 lbm 2 ft 2 3 = 2.86 lbm· ft2 IzBz = IzCz + 2 = IzCz + r+r 2 B/C 3 and Mr M = 1.44 lbm· ft2 + 3.2 lbm 1 ft + 1 ft 2 3 = 7.13 lbm· ft2 IzCz = 1.44 lbm· ft2, IzAz = 2.86 lbm· ft2, IzBz = 7.13 lbm· ft2 Clearly, IzBz > IzAz, as guessed in (a).

7.5. Polar moment of inertia: Izczm and IzOz y 413 SAMPLE 7.28 A sphere or a point? A uniform solid sphere of mass m and radius radius r r is attached to a massless rigid rod of length . The sphere swings in the x y plane. mass m Find the error in calculating IzOz as a function of r/l if the sphere is treated as a point mass concentrated at the center of mass of the sphere. Solution The exact moment of inertia of the sphere about point O can be calculated using parallel axis theorem: IzOz = Izczm + ml2 +m l 2 . x = 2mr2 Figure 7.78: (Filename:sfig4.5.4) 5 See table 4.10 of the text. If we treat the sphere as a point mass, he moment of inertia IzOz is I˜zOz = ml2. Therefore, the relative error in IzOz is error = IzOz − I˜zOz = = IzOz 2 mr 2 + ml2 − ml2 5 2 mr 2 + ml2 5 2 r2 5 l2 2 r2 +1 5 l2 From the above expression we see that for r l the error is very small. From the graph of error in Fig. 7.79 we see that even for r = l/5, the error in IzOz due to approximating the sphere as a point mass is less than 2%. 30 % Error in IzOz 25 4 % Error 2 20 0 0 0.05 0.1 0.15 0.2 r/l 15 10 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 r/l Figure 7.79: Relative error in IzOz of the sphere as a function of r/ l. (Filename:sfig4.5.4a)

414 CHAPTER 7. Circular motion 7.6 Using Izczm and IzOz in 2-D circu- lar motion dynamics Once one knows the velocity and acceleration of all points in a system one can find all of the motion quantities in the equations of motion by adding or integrating using the defining sums from chapter 1.1. This addition or integration is an impractical task for many motions of many objects where the required sums may involve billions and billions of atoms or a difficult integral. As you recall from chapter 3.6, the linear momentum and the rate of change of linear momentum can be calculated by just keeping track of the center of mass of the system of interest. One wishes for something so simple for the calculation of angular momentum. It turns out that we are in luck if we are only interested in the two-dimensional motion of two-dimensional rigid bodies. The luck is not so great for 3-D rigid bodies but still there is some simplification. For general motion of non-rigid bodies there is no simplification to be had. The simplification is to use the moment of inertia for the bodies rather than evaluating the momenta and energy quantities as integrals and sums. Of course one may have to do a sum or integral to evaluate I ≡ Izczm or [I cm] but once this calculation is done, one need not work with the integrals while worrying about the dynamics. At this point we will assume that you are comfortable calculating and looking-up moments of inertia. We proceed to use it for the purposes of studying mechanics. For constant rate rotation, we can calculate the velocity and acceleration of various points on a rigid body using v = ω × r and a = ω × (ω × r ). So we can calculate the various motion quantities of interest: linear momentum L, rate of change of linear momentum L˙ , angular momentum H , rate of change of angular momentum H˙ , and kinetic energy EK. y Consider a two-dimensional rigid body like that shown in figure 7.80. Now let us consider the various motion quantities in turn. First the linear momentum L. The linear momentum of any system in any motion is L = v cmmtot . So, for a rigid body dm spinning at constant rate ω about point O (using ω = ωkˆ): r O x L = v cm mtot = ω × r cm/omtot . ω Similarly, for any system, we can calculate the rate of change of linear momentum L˙ as L˙ = acmmtot . So, for a rigid body spinning at constant rate, center of mass L˙ = acm mtot = ω × (ω × r cm/o)mtot . Figure 7.80: A two-dimensional body is rotating around the point O at constant rate ω. A differential bit of mass dm is shown. The center of mass is also shown. (Filename:tfigure4.2Dinertia) That is, the linear momentum is correctly calculated for this special motion, as it is for all motions, by thinking of the body as a point mass at the center of mass. Unlike the calculation of linear momentum, the angular momentum turns out to be something different than would be calculated by using a point mass at the center of mass. You can remember this important fact by looking at the case when the rotation is about the center of mass (point O coincides with the center of mass). In this case one can intuitively see that the angular momentum of a rigid body is not zero even

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 415 though the center of mass is not moving. Here’s the calculation just to be sure: HO = r /O × v dm (by definition of H O) = r /O × (ω × r /O ) dm (using v = ω × r ) = (x/O ıˆ + y/O ˆ) × (ωkˆ) × (x/O ıˆ + y/O ˆ) dm (substituting r /O and ω) = { (x/2O + y/2O ) dm}ωkˆ (doing cross products) = { r/2O dm}ωkˆ = IzOz ωkˆ fwf IzOz is the ‘polar’ moment of inertia. We have defined the ‘polar’ moment of inertia as Izoz = r/2o dm. In order to calculate 1 Note that the angular momentum about Izoz for a specific body, assuming uniform mass distribution for example, one must some other point than O will not be constant convert the differential quantity of mass dm into a differential of geometric quantities. unless the center of mass does not accelerate For a line or curve, dm = ρd ; for a plate or surface, dm = ρd A, and for a 3-D (i.e., is at point O). region, dm = ρd V . d , d A, and d V are differential line, area, and volume elements, respectively. In each case, ρ is the mass density per unit length, per unit area, or per unit volume, respectively. To avoid clutter, we do not define a different symbol for the density in each geometric case. The differential elements must be further defined depending on the coordinate systems chosen for the calculation; e.g., for rectangular coordinates, d A = d xd y or, for polar coordinates, d A = r dr dθ . Since H and ω always point in the kˆ direction for two dimensional problems people often just think of angular momentum as a scalar and write the equation above simply as ‘H = I ω,’ the form usually seen in elementary physics courses. The derivation above has a feature that one might not notice at first sight. The quantity called IzOz does not depend on the rotation of the body. That is, the value of the integral does not change with time, so IzOz is a constant. So, perhaps unsurprisingly, a two-dimensional body spinning about the z-axis through O has constant angular momentum about O if it spins at a constant rate. 1 H˙ O = 0. Now, of course we could find this result about constant rate motion of 2-D bodies somewhat more cumbersomely by plugging in the general formula for rate of change of angular momentum as follows: H˙ O = r /O × a dm (7.66) = r /O × (ω × (ω × r /O )) dm = (x/O ıˆ + y/O ˆ) × ωkˆ × (ωkˆ × (x/O ıˆ + y/O ˆ)) dm = 0. Finally, we can calculate the kinetic energy by adding up 1 m i vi2 for all the bits O 2 θ of mass on a 2-D body spinning about the z-axis: ˆ R EK = 1v2 dm = 1 (ωr )2 dm = 1 ω2 r2 dm = 1 Izoz ω2 . (7.67) ıˆ G 2 22 2 m If we accept the formulae presented for rigid bodies in the box at the end of chapter 7, we can find all of the motion quantities by setting ω = ωkˆ and α = 0. Example: Pendulum disk Figure 7.81: (Filename:tfigure5.3.pend.disk)

416 CHAPTER 7. Circular motion For the disk shown in figure 7.81, we can calculate the rate of change of angular momentum about point O as H˙ O = r G/O × m acm + Izczm αkˆ = R2mθ¨kˆ + Izczm θ¨kˆ = (Izczm + R2m)θ¨kˆ. Alternatively, we could calculate directly H˙ O = IzOz αkˆ = (Izczm + R2m) θ¨kˆ. wff by the parallel axis theroem 2 But you are cautioned against falling into the common misconception that the formula M = I α applies in three dimensions by just thinking of the scalars as vectors and matrices. That is, the formula H˙ O = [I O] · ω˙ (7.68) α is only correct when ω is zero or when ω is an eigenvector of [I/O ]. To repeat, the equation Moments about O = [I O] · α (7.69) is generally wrong, it only applies if there is some known reason to neglect ω × H 0. For example, ω × H 0 can be neglected when rotation is about a principal axis as for planar bodies rotating in the plane. The term ω × H 0 can also be neglected at the start or stop of motion, that is when ω = 0. The equation for linear momentum balance is the same as always, we just need to calculate the acceleration of the center of mass of the spinning body. L˙ = mtot acm = mtot ω × (ω × r cm/O ) + ω˙ × r cm/O (7.70) Finally, the kinetic energy for a planar rigid body rotating in the plane is: EK = 1 · ([I cm] · ω) + 1 vc2m . ω m 2 2 v cm = ω × r cm/O ¢¢

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 417

418 CHAPTER 7. Circular motion Idler ωout ωαoouut t SAMPLE 7.29 An accelerating gear train. In the gear train shown in Fig. 7.82, D αout RD RC the torque at the input shaft is Min = 200 N·m and the angular acceleration is αin = 50 rad/s2. The radii of the various gears are: RA = 5 cm, RB = 8 cm, RC = 4 cm, and RD = 10 cm and the moments of inertia about the shaft axis passing through their respective centers are: IA = 0.1 kg m2, IBC = 5IA, ID = 4IA. Find the output torque Mout of the gear train. αBin C RB ωin A ωαii n RA n Driver Figure 7.82: An accelerating compound Solution Since the difference between the input power and the output power is used in accelerating the gears, we may write gear train. Pin − Pout = E˙K (Filename:sfig5.6.1) Let Mout be the output torque of the gear train. Then, Pin − Pout = Min ωin − Mout ωout . (7.71) Now, E˙K = d ( EK ) (7.72) dt (7.73) = d (1 IA ωi2n + 1 IBC ω2BC + 1 ID ωo2ut ) dt 2 2 2 = I A ωin ω˙ in + IBC ωBC ω˙ BC + ID ωout ω˙ out = IA ωin αin + 5IA ωBC αBC + 4IA ωout αout . ωouαt out The different ω’s and the α’s can be related by realizing that the linear speed or the tangential acceleration of the point of contact between any two meshing gears has to be the same irrespective of which gear’s speed and geometry is used to calculate it. Thus, using the linear speed and tangential acceleration calculations for points P and R in Fig. 7.83, we find (aR)θ R vP = ωin RA = ωB RB vR ωBαB RA P (aP)θ vP ⇒ ωB = ωi n · RB (aP )θ = αin RA = αB RB ωαiinn ⇒ αB = αi n · RA . RB Similarly, Figure 7.83: The velocity or acceleration vR = ωC RC = ωout RD of the point of contact between two mesh- ⇒ ωout = ωC · RC ing gears has to be the same irrespective of RD which meshing gear’s geometry and motion is used to compute them. (aR)θ = αC RC = αout RD (Filename:sfig5.6.1a) RC . RD ⇒ αout = αC · But ωC = ωB = ωBC ⇒ ωout = ωi n · RA · RC RB RD

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 419 and αC = αB = αBC ⇒ αout = αi n · RA · RC . RB RD Substituting these expressions for ωout , αout , ωBC and αBC in equations (7.71) and (7.73), we get Pin − Pout = Min ωin − Mout ωin · RA · RC RB RD = ωi n Mi n − Mout · RA · RC . RB RD E˙K = IA ωin αin + 5ωi n αi n ( RA )2 + 4ωi n αi n ( RA · RC )2 RB RB RD = I Aωin αi n + 5αi n ( R A )2 + 4αi n ( R A · RC )2 . R B R B RD Now equating the two quantities, Pin − Pout and E˙K, and canceling ωin from both sides, we obtain Mout RA · RC = Min − IAαin 1 + 5( RA )2 + 4( RA · RC )2 RB RD RB RB RD Mout 5 · 4 = 200 N·m − 5 kg m2 · rad/s2 1 + 5( 5 )2 + 4( 5 · 4 )2 8 10 8 8 10 Mout = 735.94 N·m ≈ 736 N·m. Mout = 736 N·m

420 CHAPTER 7. Circular motion F SAMPLE 7.30 Drums used as pulleys. Two drums, A and B of radii Ro = 200 mm and Ri = 100 mm are welded together. The combined mass of the drums is M = A Ro 20 kg and the combined moment of inertia about the z-axis passing through their B O Ri common center O is Izz/O = 1.6 kg m2. A string attached to and wrapped around drum B supports a mass m = 2 kg. The string wrapped around drum A is pulled with a force F = 20 N as shown in Fig. 7.84. Assume there is no slip between the strings and the drums. Find (a) the angular acceleration of the drums, m (b) the tension in the string supporting mass m, and (c) the acceleration of mass m. Figure 7.84: Two drums with strings wrapped around are used to pull up a mass m. (Filename:sfig5.6.2) A F Solution The free body diagram of the drums and the mass are shown in Fig. 7.85 separately where T is the tension in the string supporting mass m and Ox and Oy are Oy the support reactions at O. Since the drums can only rotate about the z-axis, let ω = ωkˆ and ω˙ = ω˙ kˆ. Now, let us do angular momentum balance about the center of rotation O: MO = H˙ O DO Ox MO = T Ri kˆ − F Rokˆ = (T Ri − F Ro)kˆ. Mg Since the motion is restricted to the x y-plane (i.e., 2-D motion), the rate of change of angular momentum H˙ O may be computed as y T x H˙ O = Izz/cm ω˙ kˆ + r cm/O × a cm Mtotal = Izz/O ω˙ kˆ + r O/O × a cm Mtotal T 0 0 = Izz/O ω˙ kˆ. Setting MO = H˙ O we get mg T Ri − F Ro = Izz/O ω˙ . (7.74) Figure 7.85: Free body diagram of the Now, let us write linear momentum balance, F = m a, for mass m: drums and the mass m. T is the tension in (T − mg)ˆ = m a. the string supporting mass m and Ox and Oy are the reactions of the support at O. F (Filename:sfig5.6.2a) Do we know anything about acceleration a of the mass? Yes, we know its direction (±ˆ) and we also know that it has to be the same as the tangential acceleration (a D)θ of point D on drum B (why?). Thus, a = (a D)θ (7.75) = ω˙ kˆ × (−Ri ıˆ) = −ω˙ Ri ˆ.

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 421 Therefore, T − mg = −mω˙ Ri . (7.76) (a) Calculation of ω˙: We now have two equations, (7.74) and (7.76), and two unknowns, ω˙ and T . Subtracting Ri times Eqn.(7.76) from Eqn. (7.74) we get −F Ro + mg Ri = (Izz/O + m Ri2)ω˙ ⇒ ω˙ = −F Ro + mg Ri (Izz/O + m Ri2) = −20 N · 0.2 m + 2 kg · 9.81 m/s2 · 0.1m 1.6 kg m2 + 2 kg · (0.1 m)2 = −2.038 kg m2/ s2 1.62 kg m2 = −1.258 1 s2 ω˙ = −1.26 rad/s2kˆ (b) Calculation of tension T: From equation (7.76): T = mg − mω˙ Ri = 2 kg · 9.81 m/s2 − 2 kg · (−1.26 s−2) · 0.1 m = 19.87 N T = 19.87 N (c) Calculation of acceleration of the mass: Since the acceleration of the mass is the same as the tangential acceleration of point D on the drum, we get (from eqn. (7.75)) a = (a D)θ = −ω˙ Ri ˆ = −(−1.26 s−2) · 0.1 m = 0.126 m/s2ˆ a = 0.13 m/s2ˆ Comments: It is important to understand why the acceleration of the mass is the same as the tangential acceleration of point D on the drum. We have assumed (as is common practice) that the string is massless and inextensible. Therefore each point of the string supporting the mass must have the same linear displacement, velocity, and acceleration as the mass. Now think about the point on the string which is momentarily in contact with point D of the drum. Since there is no relative slip between the drum and the string, the two points must have the same vertical acceleration. This vertical acceleration for point D on the drum is the tangential acceleration (a D)θ .

422 CHAPTER 7. Circular motion SAMPLE 7.31 Energy Accounting: Consider the pulley problem of Sample 7.30 again. (a) What percentage of the input energy (work done by the applied force F) is used in raising the mass by 1 m? (b) Where does the rest of the energy go? Provide an energy-balance sheet. Solution (a) Let Wi and Wh be the input energy and the energy used in raising the mass by 1 m, respectively. Then the percentage of energy used in raising the mass is % of input energy used = Wh × 100. Wi Thus we need to calculate Wi and Wh to find the answer. Wi is the work done by the force F on the system during the interval in which the mass moves up by 1 m. Let s be the displacement of the force F during this interval. Since the displacement is in the same direction as the force (we know it is from Sample 7.30), the input-energy is Wi = F s. So to find Wi we need to find s. For the mass to move up by 1 m the inner drum B must rotate by an angle θ where 1m 1 m = θ Ri ⇒ θ = 0.1 m = 10 rad. Since the two drums, A and B, are welded together, drum A must rotate by θ as well. Therefore the displacement of force F is s = θ Ro = 10 rad · 0.2 m = 2 m, and the energy input is Wi = F s = 20 N · 2 m = 40J. Now, the work done in raising the mass by 1 m is Wh = mgh = 2 kg · 9.81 m/s2 · 1 m = 19.62J. Therefore, the percentage of input-energy used in raising the mass = 19.62 N·m × 100 = 49.05% ≈ 49%. 40 (b) The rest of the energy (= 51%) goes in accelerating the mass and the pulley. Let us find out how much energy goes into each of these activities. Since the initial state of the system from which we begin energy accounting is not prescribed (that is, we are not given the height of the mass from which it is to be raised 1 m, nor do we know the velocities of the mass or the pulley at that initial height), let us assume that at the initial state, the angular speed of the pulley is ωo and the linear speed of the mass is vo. At the end of raising the mass by 1 m from this state, let the angular speed of the pulley be ωf and the linear speed of the mass be vf . Then, the energy used in accelerating the pulley

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 423 is ( EK)pulley = final kinetic energy − initial kinetic energy = 1 I ω2f − 1 I ωo2 2 2 = 1 I (ω2f − ωo2) 2 fwf assuming constant accelera- tion, ω2f = ωo2 + 2αθ , or ω2f − ωo2 = 2αθ . = I α θ (from Sample 7.34, α = 1.258 rad/s2. ) = 1.6 kg m2 · 1.258 rad/s2 · 10 rad = 20.13 N·m = 20.13 J. Similarly, the energy used in accelerating the mass is ( EK)mass = final kinetic energy − initial kinetic energy = 1 mv2f − 1 mvo2 2 2 = 1 m(v2f − vo2) 2 2ah = mah = 2 kg · 0.126 m/s2 · 1 m = 0.25 J. We can calculate the percentage of input energy used in these activities to get a better idea of energy allocation. Here is the summary table: Activities Energy Spent in Joule as % of input energy In raising the mass by 1 m 19.62 49.05% In accelerating the mass 0.25 0.62 % In accelerating the pulley 20.13 50.33 % Total 40.00 100 % So, what would you change in the set-up so that more of the input energy is used in raising the mass? Think about what aspects of the motion would change due to your proposed design.

424 CHAPTER 7. Circular motion kk SAMPLE 7.32 A uniform rigid bar of mass m = 2 kg and length L = 1 m is pinned lm at one end and connected to two springs, each with spring constant k, at the other end. The bar is tweaked slightly from its vertical position. It then oscillates about its original position. The bar is timed for 20 full oscillations which take 12.5 seconds. Ignore gravity. (a) Find the equation of motion of the rod. (b) Find the spring constant k. (c) What should be the spring constant of a torsional spring if the bar is attached to one at the bottom and has the same oscillating motion characteristics? Figure 7.86: (Filename:sfig10.1.2) x = lsinθ ≈ lθ Solution (a) Refer to the free body diagram in figure 7.87. Angular momentum balance for kk the rod about point O gives lcosθ ≈ l θ θ is small MO = H˙ O m ˆ O l sin θ ıˆ (a) Geometry kx kx where MO = −2k x ·l cos θ kˆ θ = −2kl2 sin θ cos θ kˆ, and H˙ O = IzOz θ¨kˆ = 1ml2 . 3 IzOz Thus O 1 ml2 θ¨ = −2kl2 sin θ cos θ. 3 RO (b) Free body diagram However, for small θ, cos θ ≈ 1 and sin θ ≈ θ , Figure 7.87: (Filename:sfig10.1.2a) ⇒ θ¨ + 6k l 2 θ = 0. (7.77) ml2 θ¨ + 6k θ = 0 m (b) Comparing Eqn. (7.77) with the standard harmonic oscillator equation x¨ + λ2x = 0, we get angular frequency λ = 6k , m and the time period T = 2π λ = 2π m . 6k From the measured time for 20 oscillations, the time period (time for one oscillation) is T = 12.5 s = 0.625 s 20

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 425 Now equating the measured T with the derived expression for T we get 2π m = 0.625 s 6k ⇒ k = 4π 2· m s)2 6(0.625 4π 2·2 kg = 6(0.625 s)2 = 33.7 N/ m. k = 33.7 N/ m (c) If the two linear springs are to be replaced by a torsional spring at the bottom, we can find the spring constant of the torsional spring by comparison. Let ktor be the spring constant of the torsional spring. Then, as shown in the free body diagram (see figure 7.88), the restoring torque applied by the spring at an angular displacement θ is ktorθ . Now, writing the angular momentum balance about point O, we get MO = H˙ O θθ m ˆ −ktorθ kˆ = IzOz (θ¨kˆ) ıˆ ⇒ θ¨ + ktor θ = 0. IzOz Comparing with the standard harmonic equation, we find the angular frequency ktor O ktor ktor MO O λ= IzOz = 1 ml2 RO 3 FBD If this system has to have the same period of oscillation as the first system, the Figure 7.88: (Filename:sfig10.1.2b) two angular frequencies must be equal, i.e., ktor = 6k m 1 ml 2 3 ⇒ ktor = 6k· 1 l2 = 2kl2 3 = 2·(33.7 N/ m)·(1 m)2 = 67.4 N·m ktor = 67.4 N·m

426 CHAPTER 7. Circular motion d1 = O d2 = 2 B SAMPLE 7.33 Hey Mom, look, I can seesaw by myself. A kid, modelled as a point A mass with m = 10 kg, is sitting at end B of a rigid rod AB of negligible mass. The rod is supported by a spring at end A and a pin at point O. The system is in static k m = 10 kg equilibrium when the rod is horizontal. Someone pushes the kid vertically downwards C by a small distance y and lets go. Given that AB = 3 m, AC = 0.5 m, k = 1 kN/m; find Figure 7.89: (Filename:sfig3.4.3) (a) the unstretched (relaxed) length of the spring, (b) the equation of motion (a differential equation relating the position of the mass to its acceleration) of the system, and (c) the natural frequency of the system. If the rod is pinned at the midpoint instead of at O, what is the natural frequency of the system? How does the new natural frequency compare with that of a mass m simply suspended by a spring with the same spring constant? Solution ıˆ B (a) Static Equilibrium: The FBD of the (rod + mass) system is shown in Fig. 7.90. ˆ Let the stretch in the spring in this position be yst and the relaxed length of the AO spring be 0. The balance of angular momentum about point O gives: R mg M/o = H˙ /o = 0 (no motion) kyst ⇒ (kyst )d1 − (mg)d2 = 0 (a) Static equilibrium mg · d2 ⇒ yst = k d1 A' O θ = 10 kg · 9.8 m/s2 · 2 = 0.196 m ya R y 1000 N/ m · ky B' Therefore, 0 = AC − yst mg (b) Mass m is displaced downwards = 0.5 m − 0.196 m = 0.304 m. Figure 7.90: Free body diagrams 0 = 30.4 cm (Filename:sfig3.4.3a) (b) Equation of motion: As point B gets displaced downwards by a distance y, point A moves up by a proportionate distance ya. From geometry, 1 1 Here, we are considering a very small y so that we can ignore the arc the point mass y ≈ d2θ ⇒ y B moves on and take its motion to be just θ= vertical (i.e., sin θ ≈ θ for small θ ). d2 ya ≈ d1θ = d1 y d2 Therefore, the total stretch in the spring, in this position, y = ya + yst = d1 y + d2 mg d2 d1 k Now, Angular Momentum Balance about point O gives: M/o = H˙ /o (7.78) (7.79) M/o = r B × mgˆ + r A × k yˆ (7.80) = (d2mg − d1k y)kˆ H˙ /o = r B × m a = r B × m y¨ˆ = d2m y¨kˆ

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 427 Equating (7.78) and (7.80) we get d2mg − d1k y = d2m y¨ or d2mg − d1k d1 y + d2mg = d2m y¨ d2 d1k or d2mg − k d12 y − d2mg = d2m y¨ d2 or y¨ + k d12 y = 0 m d22 y¨ + k d12 y = 0 m d22 (c) The natural frequency of the system: We may also write the previous equation as d12 d22 y¨ + λy = 0 where λ = k . (7.81) m Substituting d1 = and d2 = 2 in the expression for λ we get the natural frequency of the system √ = 1 k =1 1000 N/ m = 5 s−1 λ 2 m 2 10 kg √ = 5 s−1 λ (d) Comparision with a simple spring mass system: When d1 = d2, the equation L O L ≡ k A B m m k C Figure 7.91: (Filename:sfig3.4.3b) of motion (7.81) becomes y¨ + k y = 0 m and the natural frequency of the system is simply √ k λ= m which corresponds to the natural frequency of a simple spring mass system shown in Fig. 7.91. In our system (with d1 = d2 ) any vertical displacement of the mass at B induces an equal amount of stretch or compression in the spring which is exactly the case in the simple spring-mass system. Therefore, the two systems are mechanically equivalent. Such equivalences are widely used in modeling complex physical systems with simpler mechanical models.

428 CHAPTER 7. Circular motion SAMPLE 7.34 Energy method: Consider the pulley problem of Sample 7.30 again. Use energy method to (a) find the angular acceleration of the pulley, and (b) the acceleration of the mass. Solution In energy method we use speeds, not velocities. Therefore, we have to be careful in our thinking about the direction of motion. In the present problem, let us assume that the pulley rotates and accelerates clockwise. Consequently, the mass moves up against gravity. (a) The energy equation we want to use is P = E˙ K . 1 There are other external forces on the The power P is given by P = F i · v i where the sum is carried out over system: the reaction force of the support all external forces. For the mass and pulley system the external forces that do point O and the weight of the pulley—both work are 1 F and mg. Therefore, forces acting at point O. But, since point O is stationary, these forces do no work. P = F · vA + mg · vm = Fıˆ · vAıˆ + (−mgˆ) · vDˆ vm = FvA − mgvD. The rate of change of kinetic energy is E˙ K = d( 1 m v2D + 1 Izz/O ω2 ) dt 2 2 K.E. of the mass K.E. of the pulley = m vD v˙D + Izz/O ω ω˙ . Now equating the power and the rate of change of kinetic energy, we get F vA − mg vD = m vD v˙D + Izz/O ω ω˙ From kinematics, vA = ω Ro, vD = ω Ri and v˙D ≡ (aD)θ = ω˙ Ri . Substituting these values in the above equation, we get ω(F Ro − mg Ri ) = ω ω˙ (m Ri2 + Izz/O ) ⇒ ω˙ = F Ro − mg Ri (Izz/O + m Ri2) = 20 N · 0.2 m − 2 kg · 9.81 m/s2 · 0.1m 1.6 kg m2 + 2 kg · (0.1 m)2 = 1.258 1 (same as the answer before.) s2 Since the sign of ω˙ is positive, our initial assumption of clockwise acceleration of the pulley is correct. ω˙ = 1.26 rad/s2 (b) From kinematics, am = (aD)θ = ω˙ Ri = 0.126 m/s2. am = 0.13 m/s2

7.6. Using Izczm and IzOz in 2-D circular motion dynamics 429 z SAMPLE 7.35 A flywheel of diameter 2 ft spins about the axis passing through its center and perpendicular to the plane of the wheel at 1000 rpm. The wheel weighs r = 1 ft 20 lbf. Assuming the wheel to be a thin, uniform disk, find its kinetic energy. Figure 7.92: (Filename:sfig7.4.1) Solution The kinetic energy of a 2-D rigid body spinning at speed ω about the z-axis passing through its mass center is EK = 1 Izczmω2 2 where Izczm is the mass moment of inertia about the z-axis. For the flywheel, Izczm = 1 m R2 (from table IV at the back of the book) 2 = 1 W R2 (where W is the weight of the wheel) 2g = 1 · ( 20 lbf ) · (1 ft)2 = 10 lbm· ft2 2g 20 lbm The angular speed of the wheel is ω = 1000 rpm = 1000· 2π rad/s 60 = 104.72 rad/s. Therefore the kinetic energy of the wheel is EK = 1 ·(10 lbm· ft2)·(104.72 rad/s)2 2 = 5.483 × 104 lbm· ft2/ s2 5.483 × 104 = lbf· ft 32.2 = 1.702 × 103 ft· lbf. 1.702 × 103 ft· lbf.

430 CHAPTER 7. Circular motion

8 Advanced topics in circular motion In the last chapter we discussed the motions of particles and rigid bodies that rotate Figure 8.1: A car crankshaft is a complex about a fixed axis from a planar point of view. In this chapter we again are going to think about fixed axis rotation, but now in three dimensions. The axis of rotation three-dimensional object which is well ap- might be in any skew direction and the rotating bodies might be arbitrarily complicated proximated for many purposes as rotating three-dimensional shapes. As a cartoon, imagine a rigid body skewered with a rigid about a fixed axis. The relative timing of rod and then turned by a motor that speeds up and slows down. More practically think the reciprocating pistons is controlled by of the crankshaft in a car engine (Fig. 8.1). Other applications include accelerating this complex shape. “Connecting rods” are or decelerating shafts of all kinds, gears, turbines, flywheels, pendula, and swinging pinned to the cylinders at one end and to doors. the short offset cylinders on the crankshaft at the other. To understand this motion we need to take a little more care with the kinematics because it now involves three dimensions, although in some sense the basic ideas (Filename:tfigure.crankshaft) are unchanged from the previous two-dimensional chapter. The three dimensional mechanics naturally gets more involved. As for all motions of all systems, the momentum balance equations apply to any system or any part of a system that has fixed axis rotation. So our mechanics results will be based on these familiar equations: Linear momentum balance: F i = L˙ , Angular momentum balance: Mi/O = H˙ O. and Power balance: P = E˙K. As always, we will evaluate the left hand sides of the momentum equations using the forces and moments in the free body diagram. We evaluate the right hand sides of 431

432 CHAPTER 8. Advanced topics in circular motion these equations using our knowledge of the velocities and accelerations of the various mass points. The chapter starts with a discussion of kinematics. Then we consider the me- chanics of systems with fixed axis rotation. The moment of inertia matrix is then introduced followed by a section where the moment of inertia is used as a shortcut in the evaluation of H˙ O. Finally we discuss dynamic balance, an important topic in machine design that is genuinely three-dimensional. 8.1 3-D description of circular mo- tion Let’s first assume each particle is going in circles around the z axis, as in the previous chapter. The figure below shows this two dimensional situation first two dimension- ally (left) and then as a two-dimensional motion in a three dimensional world (right). x 2D x 3D x axis v θ eˆR y rP ω θ O φ eˆθ R z v y y ω =ωkˆ Figure 8.2: When a point P is going in z circles about the z axis, we define the unit The velocity and acceleration are described in the same way for three dimensional motion about a fixed axis as for two dimensional motion. vector eˆ R to be pointed from the axis to the point P. We define the unit vector eˆθ to • the velocity is tangent to the circle it is going around and is proportional in magnitude to the radius of the circle and also to its angular speed. That is, the be tangent to the circle at P. Both of these direction of the velocity is in the direction eˆθ and has magnitude ω R, where ω vectors change in time as the point moves is the angular rate of rotation and R is the radius of the circle that the particle along its circular path. is going around. (Filename:tfigure4.2) • the acceleration can be constructed as the sum of two vectors. One is pointed to the center of the circle and proportional in magnitude to both the square of y the angular speed and to the radius. The other vector is tangent to the circle x and equal in magnitude to the rate of increase of speed. P These two ideas are summarized by the following formulas: O axis a z ω = ωkˆ v = ω Reˆθ and a = − ω2 R eˆR + Rθ¨ eˆθ (8.1) v2/R v˙ Figure 8.3: The acceleration is the sum with of two components. One directed towards v = ωR. (8.2) the center of the circle in the −eˆ R direc- The axis of rotation might not be the z-axis of a convenient x yz coordinate system. tion, and one tangent to the circle in the eˆθ So the x y plane of circles might not be the x y plane of the coordinate system you might want to use for some other reasons. Fortunately, we can write the formulas 8.1 direction. in a way that rids us of these problems. (Filename:tfigure5.3)

8.1. 3-D description of circular motion 433 Here are some formulas which are equivalent to the formulas 8.1 but which do not make use of the polar coordinate base vectors. v = ω× r, (8.3) a = ω × (ω × r ). (8.4) To check that equations 8.3 and 8.4 are really equivalent to 8.1? we need to verify that the vector ω × r is equal to ω Reˆθ and that the vector ω × (ω × r ) is equal to the vector −ω2 ReˆR. First, define r as the position of the point of interest relative to any point on the axis of rotation. If this point happens to be the center of the circle for some particle then r = R. But, in general, r = R. Now look at v = ω × r . Looking at figure 8.2 and using the right hand rule, it is clear that the direction of the cross product ω × r is in fact the eˆθ direction. What about the magnitude? The magnitude |ω × r | = |ω||r | sin φ. But |r | sin φ = R. So the magnitude of ω × r is ω R. That is, ω × r = (|ω × r |) · (unit vector in the direction of ω × r ) (8.5) = (ω R) · eˆθ (8.6) =v (8.7) Triple cross product The formula for acceleration of a point on a rigid body includes the centripetal term ω × (ω × r ). This expression is a special case of the general vector expression A × (B × C) which is sometimes called the ‘vector triple product’ because its value is a vector (as opposed to the scalar value of the ‘scalar triple product’). The primary useful identity with vector triple products is: A × (B × C) = (A · C)B − (A · B)C. (8.8) This formula may be remembered by the semi-mnemonic device ‘cab minus bac’ x θ since A · C = C · A and A · B = B · A. This formula is discussed in box 8.1 on axis eˆR page 437. rO P So now we can write O eˆθ R ω × (ω × r ) = ω × (ω R)·eˆθ = −ω2 ReˆR = −ω2R = a. (8.9) θ˙ R z y ω=ωkˆ Note that equations 8.3 and 8.4 are vector equations. They do not make use of any coordinate system. So, for example, we can use them even if ω is not in the z direction. Angular velocity of a rigid body in 3D Figure 8.4: A rigid body spinning about the z axis. Every point on the body, like point P at r , is going in circles. All of these circles have centers on the axis of rotation. All the points are going around at the same angular rate, θ˙ = ω. (Filename:tfigure4.3D)