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# 10 rotation of a rigid object about a fixed axis

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• 1. P U Z Z L E RDid you know that the CD inside thisplayer spins at different speeds, depend-ing on which song is playing? Why wouldsuch a strange characteristic be incor-porated into the design of every CDplayer? (George Semple)c h a p t e r Rotation of a Rigid Object About a Fixed Axis Chapter Outline 10.1 Angular Displacement, Velocity, 10.5 Calculation of Moments of and Acceleration Inertia 10.2 Rotational Kinematics: Rotational 10.6 Torque Motion with Constant Angular 10.7 Relationship Between Torque Acceleration and Angular Acceleration 10.3 Angular and Linear Quantities 10.8 Work, Power, and Energy in 10.4 Rotational Energy Rotational Motion292
• 3. 294 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis y Q ,tf r P, ti θf In a short track event, such as a 200-m or θi 400-m sprint, the runners begin from stag- x gered positions on the track. Why don’t O they all begin from the same line?Figure 10.2 A particle on a rotat-ing rigid object moves from P to Qalong the arc of a circle. In thetime interval ⌬t ϭ t f Ϫ t i , the ra- As the particle in question on our rigid object travels from position P to positiondius vector sweeps out an angle Q in a time ⌬t as shown in Figure 10.2, the radius vector sweeps out an angle ⌬␪ ϭ ␪f⌬␪ ϭ ␪f Ϫ ␪i . Ϫ ␪i . This quantity ⌬␪ is deﬁned as the angular displacement of the particle: ⌬␪ ϭ ␪f Ϫ ␪i (10.2) We deﬁne the average angular speed ␻ (omega) as the ratio of this angular dis- placement to the time interval ⌬t: ␪f Ϫ ␪i ⌬␪ Average angular speed ␻ϵ ϭ (10.3) tf Ϫ ti ⌬t In analogy to linear speed, the instantaneous angular speed ␻ is deﬁned as the limit of the ratio ⌬␪/⌬t as ⌬t approaches zero: ⌬␪ d␪ Instantaneous angular speed ␻ ϵ lim ϭ (10.4) ⌬t:0 ⌬t dt Angular speed has units of radians per second (rad/s), or rather secondϪ1 (sϪ1) because radians are not dimensional. We take ␻ to be positive when ␪ is in- creasing (counterclockwise motion) and negative when ␪ is decreasing (clockwise motion). If the instantaneous angular speed of an object changes from ␻i to ␻f in the time interval ⌬t, the object has an angular acceleration. The average angular ac- celeration ␣ (alpha) of a rotating object is deﬁned as the ratio of the change in the angular speed to the time interval ⌬t : ␻f Ϫ ␻i ⌬␻ Average angular acceleration ␣ϵ ϭ (10.5) tf Ϫ ti ⌬t
• 4. 10.1 Angular Displacement, Velocity, and Acceleration 295 ω Figure 10.3 The right-hand rule for deter- ω mining the direction of the angular velocity vector. In analogy to linear acceleration, the instantaneous angular acceleration isdeﬁned as the limit of the ratio ⌬␻/⌬t as ⌬t approaches zero: ⌬␻ d␻ ␣ ϵ lim ϭ (10.6) Instantaneous angular ⌬t:0 ⌬t dt accelerationAngular acceleration has units of radians per second squared (rad/s2 ), or just sec-ondϪ2 (sϪ2 ). Note that ␣ is positive when the rate of counterclockwise rotation isincreasing or when the rate of clockwise rotation is decreasing. When rotating about a ﬁxed axis, every particle on a rigid object rotatesthrough the same angle and has the same angular speed and the same an-gular acceleration. That is, the quantities ␪, ␻, and ␣ characterize the rotationalmotion of the entire rigid object. Using these quantities, we can greatly simplifythe analysis of rigid-body rotation. Angular position (␪), angular speed (␻), and angular acceleration (␣) areanalogous to linear position (x), linear speed (v), and linear acceleration (a). Thevariables ␪, ␻, and ␣ differ dimensionally from the variables x, v, and a only by afactor having the unit of length. We have not speciﬁed any direction for ␻ and ␣. Strictly speaking, thesevariables are the magnitudes of the angular velocity and the angular accelera-tion vectors ␻ and ␣, respectively, and they should always be positive. Becausewe are considering rotation about a ﬁxed axis, however, we can indicate the di-rections of the vectors by assigning a positive or negative sign to ␻ and ␣, as dis-cussed earlier with regard to Equations 10.4 and 10.6. For rotation about a ﬁxedaxis, the only direction that uniquely speciﬁes the rotational motion is the di-rection along the axis of rotation. Therefore, the directions of ␻ and ␣ arealong this axis. If an object rotates in the xy plane as in Figure 10.1, the direc-tion of ␻ is out of the plane of the diagram when the rotation is counterclock-wise and into the plane of the diagram when the rotation is clockwise. To illus-trate this convention, it is convenient to use the right-hand rule demonstrated inFigure 10.3. When the four ﬁngers of the right hand are wrapped in the direc-tion of rotation, the extended right thumb points in the direction of ␻. The di-rection of ␣ follows from its deﬁnition d␻/dt. It is the same as the direction of␻ if the angular speed is increasing in time, and it is antiparallel to ␻ if the an-gular speed is decreasing in time.
• 6. 10.3 Angular and Linear Quantities 297 TABLE 10.1 Kinematic Equations for Rotational and Linear Motion Under Constant Acceleration Rotational Motion About a Fixed Axis Linear Motion ␻f ϭ ␻ i ϩ ␣t vf ϭ vi ϩ at ␪f ϭ ␪i ϩ ␻ i t ϩ 1 2 ␣t 2 xf ϭ xi ϩ v i t ϩ 1 2 at 2 ␻f ϭ ␻ i ϩ 2␣(␪f Ϫ ␪i ) 2 2 vf ϭ v i ϩ 2a(xf Ϫ xi ) 2 2 y10.3 ANGULAR AND LINEAR QUANTITIESIn this section we derive some useful relationships between the angular speed and vacceleration of a rotating rigid object and the linear speed and acceleration of an Parbitrary point in the object. To do so, we must keep in mind that when a rigid ob-ject rotates about a ﬁxed axis, as in Figure 10.4, every particle of the object moves rin a circle whose center is the axis of rotation. θ We can relate the angular speed of the rotating object to the tangential speed x Oof a point P on the object. Because point P moves in a circle, the linear velocityvector v is always tangent to the circular path and hence is called tangential velocity.The magnitude of the tangential velocity of the point P is by deﬁnition the tangen-tial speed v ϭ ds/dt, where s is the distance traveled by this point measured along Figure 10.4 As a rigid object ro-the circular path. Recalling that s ϭ r ␪ (Eq. 10.1a) and noting that r is constant, tates about the ﬁxed axis throughwe obtain O, the point P has a linear velocity v that is always tangent to the circu- ds d␪ vϭ ϭr lar path of radius r. dt dtBecause d␪/dt ϭ ␻ (see Eq. 10.4), we can say Relationship between linear and v ϭ r␻ (10.10) angular speedThat is, the tangential speed of a point on a rotating rigid object equals the per-pendicular distance of that point from the axis of rotation multiplied by the angu-lar speed. Therefore, although every point on the rigid object has the same angu-lar speed, not every point has the same linear speed because r is not the same forall points on the object. Equation 10.10 shows that the linear speed of a point onthe rotating object increases as one moves outward from the center of rotation, aswe would intuitively expect. The outer end of a swinging baseball bat moves muchfaster than the handle. QuickLab We can relate the angular acceleration of the rotating rigid object to the tan- Spin a tennis ball or basketball andgential acceleration of the point P by taking the time derivative of v: watch it gradually slow down and stop. Estimate ␣ and at as accurately dv d␻ as you can. at ϭ ϭr dt dt Relationship between linear and at ϭ r ␣ (10.11) angular accelerationThat is, the tangential component of the linear acceleration of a point on a rotat-ing rigid object equals the point’s distance from the axis of rotation multiplied bythe angular acceleration.
• 9. 300 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis We simplify this expression by deﬁning the quantity in parentheses as the moment of inertia I:Moment of inertia I ϵ ⌺ mi ri 2 (10.15) i From the deﬁnition of moment of inertia, we see that it has dimensions of ML2 (kgи m2 in SI units).1 With this notation, Equation 10.14 becomesRotational kinetic energy KR ϭ 1I␻2 2 (10.16) Although we commonly refer to the quantity 2I␻ 2 as rotational kinetic energy, 1 it is not a new form of energy. It is ordinary kinetic energy because it is derived from a sum over individual kinetic energies of the particles contained in the rigid object. However, the mathematical form of the kinetic energy given by Equation 10.16 is a convenient one when we are dealing with rotational motion, provided we know how to calculate I. It is important that you recognize the analogy between kinetic energy associ- ated with linear motion 1mv 2 and rotational kinetic energy 1 I␻ 2. The quantities I 2 2 and ␻ in rotational motion are analogous to m and v in linear motion, respectively. (In fact, I takes the place of m every time we compare a linear-motion equation with its rotational counterpart.) The moment of inertia is a measure of the resis- tance of an object to changes in its rotational motion, just as mass is a measure of the tendency of an object to resist changes in its linear motion. Note, however, that mass is an intrinsic property of an object, whereas I depends on the physical arrangement of that mass. Can you think of a situation in which an object’s mo- ment of inertia changes even though its mass does not? EXAMPLE 10.3 The Oxygen Molecule Consider an oxygen molecule (O2 ) rotating in the xy plane about the z axis. The axis passes through the center of the ϭ 1.95 ϫ 10 Ϫ46 kgиm2 molecule, perpendicular to its length. The mass of each oxy- gen atom is 2.66 ϫ 10Ϫ26 kg, and at room temperature the This is a very small number, consistent with the minuscule average separation between the two atoms is d ϭ 1.21 ϫ masses and distances involved. 10Ϫ10 m (the atoms are treated as point masses). (a) Calcu- (b) If the angular speed of the molecule about the z axis is late the moment of inertia of the molecule about the z axis. 4.60 ϫ 1012 rad/s, what is its rotational kinetic energy? Solution This is a straightforward application of the def- Solution We apply the result we just calculated for the mo- inition of I. Because each atom is a distance d/2 from the z ment of inertia in the formula for K R : axis, the moment of inertia about the axis is KR ϭ 1 I␻2 2 ΂2΃ ΂2΃ d 2 d 2 I ϭ ⌺ mi ri 2 ϭ m ϩm ϭ 1md 2 2 ϭ 1(1.95 ϫ 10Ϫ46 kgиm2)(4.60 ϫ 1012 rad/s)2 2 i ϭ 1(2.66 ϫ 10 Ϫ26 kg)(1.21 ϫ 10 Ϫ10 m)2 ϭ 2.06 ϫ 10 Ϫ21 J 2 1 Civil engineers use moment of inertia to characterize the elastic properties (rigidity) of such struc- tures as loaded beams. Hence, it is often useful even in a nonrotational context.
• 11. 302 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis into Equation 10.17 gives Iϭ ͵␳r 2 dV If the object is homogeneous, then ␳ is constant and the integral can be evaluated for a known geometry. If ␳ is not constant, then its variation with position must be known to complete the integration. The density given by ␳ ϭ m/V sometimes is referred to as volume density for the obvious reason that it relates to volume. Often we use other ways of expressing density. For instance, when dealing with a sheet of uniform thickness t, we can de- ﬁne a surface density ␴ ϭ ␳t, which signiﬁes mass per unit area. Finally, when mass is distributed along a uniform rod of cross-sectional area A, we sometimes use linear density ␭ ϭ M/L ϭ ␳A, which is the mass per unit length. EXAMPLE 10.5 Uniform Hoop Find the moment of inertia of a uniform hoop of mass M and y radius R about an axis perpendicular to the plane of the hoop and passing through its center (Fig. 10.9). dm Solution All mass elements dm are the same distance r ϭ R from the axis, and so, applying Equation 10.17, we obtain x for the moment of inertia about the z axis through O: O ͵ ͵ R Iz ϭ r 2 dm ϭ R 2 dm ϭ MR 2 Note that this moment of inertia is the same as that of a sin- gle particle of mass M located a distance R from the axis of rotation. Figure 10.9 The mass elements dm of a uniform hoop are all the same distance from O. Quick Quiz 10.3 (a) Based on what you have learned from Example 10.5, what do you expect to ﬁnd for the moment of inertia of two particles, each of mass M/2, located anywhere on a circle of ra- dius R around the axis of rotation? (b) How about the moment of inertia of four particles, each of mass M/4, again located a distance R from the rotation axis? EXAMPLE 10.6 Uniform Rigid Rod Calculate the moment of inertia of a uniform rigid rod of Substituting this expression for dm into Equation 10.17, with length L and mass M (Fig. 10.10) about an axis perpendicu- r ϭ x, we obtain ͵ ͵ ͵ lar to the rod (the y axis) and passing through its center of L/2 L/2 M M mass. Iy ϭ r 2 dm ϭ x2 dx ϭ x 2 dx ϪL/2 L L ϪL/2 Solution The shaded length element dx has a mass dm ΄ x3 ΅ M 3 L/2 equal to the mass per unit length ␭ multiplied by dx : ϭ ϭ 1 12 ML 2 L ϪL/2 M dm ϭ ␭ dx ϭ dx L
• 12. 10.5 Calculation of Moments of Inertia 303 y′ y dx x O x Figure 10.10 A uniform rigid rod of length L. The moment of in- L ertia about the y axis is less than that about the yЈ axis. The latter axis is examined in Example 10.8. EXAMPLE 10.7 Uniform Solid Cylinder A uniform solid cylinder has a radius R, mass M, and length cylindrical shells, each of which has radius r, thickness dr, and L. Calculate its moment of inertia about its central axis (the z length L, as shown in Figure 10.11. The volume dV of each axis in Fig. 10.11). shell is its cross-sectional area multiplied by its length: dV ϭ dAи L ϭ (2␲ r dr)L. If the mass per unit volume is ␳, then the Solution It is convenient to divide the cylinder into many mass of this differential volume element is dm ϭ ␳dV ϭ ␳2␲ rL dr. Substituting this expression for dm into Equation z 10.17, we obtain dr r Iz ϭ ͵ r 2 dm ϭ 2␲␳L ͵0 R r 3 dr ϭ 1␲␳LR 4 2 Because the total volume of the cylinder is ␲R 2L, we see that R ␳ ϭ M/V ϭ M/␲R 2L. Substituting this value for ␳ into the above result gives L (1) Iz ϭ 1 2 MR 2 Note that this result does not depend on L, the length of the cylinder. In other words, it applies equally well to a long cylin- der and a ﬂat disc. Also note that this is exactly half the value Figure 10.11 Calculating I about the z axis for a uniform solid we would expect were all the mass concentrated at the outer cylinder. edge of the cylinder or disc. (See Example 10.5.) Table 10.2 gives the moments of inertia for a number of bodies about speciﬁcaxes. The moments of inertia of rigid bodies with simple geometry (high symme-try) are relatively easy to calculate provided the rotation axis coincides with an axisof symmetry. The calculation of moments of inertia about an arbitrary axis can becumbersome, however, even for a highly symmetric object. Fortunately, use of animportant theorem, called the parallel-axis theorem, often simpliﬁes the calcula-tion. Suppose the moment of inertia about an axis through the center of mass ofan object is ICM . The parallel-axis theorem states that the moment of inertia aboutany axis parallel to and a distance D away from this axis is I ϭ ICM ϩ MD 2 (10.18) Parallel-axis theorem
• 13. 304 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis TABLE 10.2 Moments of Inertia of Homogeneous Rigid Bodies with Different Geometries Hoop or Hollow cylinder cylindrical shell I CM = 1 M(R 12 + R 22) R1 I CM = MR 2 R 2 R2 Solid cylinder Rectangular plate or disk R I CM = 1 M(a 2 + b 2) I CM = 1 MR 2 12 2 b a Long thin rod Long thin with rotation axis rod with through center rotation axis through end I CM = 1 ML 2 L L 12 I = 1 ML 2 3 Solid sphere Thin spherical shell I CM = 2 MR 2 5 I CM = 2 MR 2 3 R R Proof of the Parallel-Axis Theorem (Optional). Suppose that an object rotates in the xy plane about the z axis, as shown in Figure 10.12, and that the coordinates of the center of mass are x CM , y CM . Let the mass element dm have coordinates x, y. Because this element is a distance r ϭ √x 2 ϩ y 2 from the z axis, the moment of in- ertia about the z axis is Iϭ ͵ r 2 dm ϭ ͵ (x 2 ϩ y 2) dm However, we can relate the coordinates x, y of the mass element dm to the coordi- nates of this same element located in a coordinate system having the object’s cen- ter of mass as its origin. If the coordinates of the center of mass are x CM , y CM in the original coordinate system centered on O, then from Figure 10.12a we see that the relationships between the unprimed and primed coordinates are x ϭ xЈ ϩ x CM
• 14. 10.5 Calculation of Moments of Inertia 305 y dm x, y z Axis through y′ r Rotation CM axis y y CM xCM, yCM O CM yCM D x O x xCM x′ x (a) (b)Figure 10.12 (a) The parallel-axis theorem: If the moment of inertia about an axis perpendic-ular to the ﬁgure through the center of mass is ICM , then the moment of inertia about the z axisis Iz ϭ I CM ϩ MD 2. (b) Perspective drawing showing the z axis (the axis of rotation) and the par-allel axis through the CM.and y ϭ yЈ ϩ y CM . Therefore, Iϭ ͵ [(xЈ ϩ x CM)2 ϩ (yЈ ϩ y CM)2] dm ϭ ͵ [(xЈ)2 ϩ (yЈ)2] dm ϩ 2x CM ͵ xЈ dm ϩ 2y CM ͵ yЈ dm ϩ (x CM2 ϩ y CM 2) ͵ dmThe ﬁrst integral is, by deﬁnition, the moment of inertia about an axis that is par-allel to the z axis and passes through the center of mass. The second two integralsare zero because, by deﬁnition of the center of mass, ͵ xЈ dm ϭ ͵ yЈ dm ϭ 0. Thelast integral is simply MD 2 because ͵ dm ϭ M and D 2 ϭ x CM2 ϩ y CM2. Therefore,we conclude that I ϭ ICM ϩ MD 2 EXAMPLE 10.8 Applying the Parallel-Axis Theorem ΂L΃ Consider once again the uniform rigid rod of mass M and 2 length L shown in Figure 10.10. Find the moment of inertia I ϭ ICM ϩ MD 2 ϭ 12 ML 2 ϩ M 1 ϭ 1 3 ML 2 2 of the rod about an axis perpendicular to the rod through one end (the yЈaxis in Fig. 10.10). So, it is four times more difﬁcult to change the rotation of a rod spinning about its end than it is to change the motion of Solution Intuitively, we expect the moment of inertia to one spinning about its center. be greater than ICM ϭ 12ML 2 because it should be more difﬁ- 1 cult to change the rotational motion of a rod spinning about Exercise Calculate the moment of inertia of the rod about an axis at one end than one that is spinning about its center. a perpendicular axis through the point x ϭ L/4. Because the distance between the center-of-mass axis and the yЈ axis is D ϭ L/2, the parallel-axis theorem gives Answer I ϭ 48 ML 2. 7
• 15. 306 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis 10.6 TORQUE Why are a door’s doorknob and hinges placed near opposite edges of the door? 7.6 This question actually has an answer based on common sense ideas. The harder we push against the door and the farther we are from the hinges, the more likely we are to open or close the door. When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque ␶ (tau). Consider the wrench pivoted on the axis through O in Figure 10.13. The ap- plied force F acts at an angle ␾ to the horizontal. We deﬁne the magnitude of the torque associated with the force F by the expressionDeﬁnition of torque ␶ ϵ r F sin ␾ ϭ Fd (10.19) where r is the distance between the pivot point and the point of application of F and d is the perpendicular distance from the pivot point to the line of action of F. (The line of action of a force is an imaginary line extending out both ends of the vector representing the force. The dashed line extending from the tail of F in Fig- ure 10.13 is part of the line of action of F.) From the right triangle in Figure 10.13 that has the wrench as its hypotenuse, we see that d ϭ r sin ␾. This quantity d isMoment arm called the moment arm (or lever arm) of F. It is very important that you recognize that torque is deﬁned only when a reference axis is speciﬁed. Torque is the product of a force and the moment arm of that force, and moment arm is deﬁned only in terms of an axis of rotation. In Figure 10.13, the only component of F that tends to cause rotation is F sin ␾, the component perpendicular to r. The horizontal component F cos ␾, because it passes through O, has no tendency to produce rotation. From the deﬁnition of torque, we see that the rotating tendency increases as F increases and as d in- creases. This explains the observation that it is easier to close a door if we push at the doorknob rather than at a point close to the hinge. We also want to apply our push as close to perpendicular to the door as we can. Pushing sideways on the doorknob will not cause the door to rotate. If two or more forces are acting on a rigid object, as shown in Figure 10.14, each tends to produce rotation about the pivot at O. In this example, F2 tends to F1 F sin φ F r d1 φ F cos φ O O φ Line of d2 d action Figure 10.13 The force F has a F2 greater rotating tendency about O as F increases and as the moment Figure 10.14 The force F1 tends arm d increases. It is the compo- to rotate the object counterclock- nent F sin ␾ that tends to rotate the wise about O, and F2 tends to ro- wrench about O. tate it clockwise.
• 16. 10.7 Relationship Between Torque and Angular Acceleration 307 rotate the object clockwise, and F1 tends to rotate it counterclockwise. We use the convention that the sign of the torque resulting from a force is positive if the turn- ing tendency of the force is counterclockwise and is negative if the turning ten- dency is clockwise. For example, in Figure 10.14, the torque resulting from F1 , which has a moment arm d 1 , is positive and equal to ϩ F1 d 1 ; the torque from F2 is negative and equal to Ϫ F2 d 2 . Hence, the net torque about O is ⌺ ␶ ϭ ␶1 ϩ ␶2 ϭ F 1d 1 Ϫ F 2d 2 Torque should not be confused with force. Forces can cause a change in lin- ear motion, as described by Newton’s second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change de- pends on both the forces and the moment arms of the forces, in the combination that we call torque. Torque has units of force times length — newton и meters in SI units — and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. EXAMPLE 10.9 The Net Torque on a Cylinder A one-piece cylinder is shaped as shown in Figure 10.15, with Solution The torque due to F1 is Ϫ R 1 F1 (the sign is nega- a core section protruding from the larger drum. The cylinder tive because the torque tends to produce clockwise rotation). is free to rotate around the central axis shown in the drawing. The torque due to F2 is ϩ R 2 F 2 (the sign is positive because A rope wrapped around the drum, which has radius R 1 , ex- the torque tends to produce counterclockwise rotation). erts a force F1 to the right on the cylinder. A rope wrapped Therefore, the net torque about the rotation axis is around the core, which has radius R 2 , exerts a force F2 down- ward on the cylinder. (a) What is the net torque acting on the ⌺ ␶ ϭ ␶1 ϩ ␶2 ϭ ϪR 1F 1 ϩ R 2F 2 cylinder about the rotation axis (which is the z axis in Figure 10.15)? We can make a quick check by noting that if the two forces are of equal magnitude, the net torque is negative because y R 1 Ͼ R 2 . Starting from rest with both forces acting on it, the cylinder would rotate clockwise because F 1 would be more ef- fective at turning it than would F 2 . F1 R1 (b) Suppose F 1 ϭ 5.0 N, R 1 ϭ 1.0 m, F 2 ϭ 15.0 N, and R 2 ϭ 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate from rest? R2 x O ⌺ ␶ ϭ Ϫ(5.0 N)(1.0 m) ϩ (15.0 N)(0.50 m) ϭ 2.5 Nиm z Because the net torque is positive, if the cylinder starts from rest, it will commence rotating counterclockwise with increas- F2 ing angular velocity. (If the cylinder’s initial rotation is clock- Figure 10.15 A solid cylinder pivoted about the z axis through O. wise, it will slow to a stop and then rotate counterclockwise The moment arm of F1 is R 1 , and the moment arm of F2 is R 2 . with increasing angular speed.) 10.7 RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION In this section we show that the angular acceleration of a rigid object rotating7.6 about a ﬁxed axis is proportional to the net torque acting about that axis. Before discussing the more complex case of rigid-body rotation, however, it is instructive
• 18. 10.7 Relationship Between Torque and Angular Acceleration 309portional to the angular acceleration of the object, with the proportionality factorbeing I, a quantity that depends upon the axis of rotation and upon the size andshape of the object. In view of the complex nature of the system, it is interesting tonote that the relationship ⌺␶ ϭ I␣ is strikingly simple and in complete agreementwith experimental observations. The simplicity of the result lies in the manner inwhich the motion is described. Although each point on a rigid object rotating about a ﬁxed axis may not expe- Every point has the same ␻ and ␣ rience the same force, linear acceleration, or linear speed, each point experi- ences the same angular acceleration and angular speed at any instant. There- fore, at any instant the rotating rigid object as a whole is characterized by QuickLab speciﬁc values for angular acceleration, net torque, and angular speed. Tip over a child’s tall tower of blocks. Try this several times. Does the tower “break” at the same place each time? Finally, note that the result ⌺␶ ϭ I␣ also applies when the forces acting on the What affects where the tower comes apart as it tips? If the tower weremass elements have radial components as well as tangential components. This is made of toy bricks that snap together,because the line of action of all radial components must pass through the axis of what would happen? (Refer to Con-rotation, and hence all radial components produce zero torque about that axis. ceptual Example 10.11.) EXAMPLE 10.10 Rotating Rod A uniform rod of length L and mass M is attached at one end compute the torque on the rod, we can assume that the gravi- to a frictionless pivot and is free to rotate about the pivot in tational force acts at the center of mass of the rod, as shown the vertical plane, as shown in Figure 10.18. The rod is re- in Figure 10.18. The torque due to this force about an axis leased from rest in the horizontal position. What is the initial through the pivot is angular acceleration of the rod and the initial linear accelera- tion of its right end? ␶ ϭ ⌴g ΂L΃ 2 Solution We cannot use our kinematic equations to ﬁnd ␣ With ⌺␶ ϭ I␣, and I ϭ 1 ML 2 for this axis of rotation (see 3 or a because the torque exerted on the rod varies with its po- Table 10.2), we obtain sition, and so neither acceleration is constant. We have enough information to ﬁnd the torque, however, which we ␶ ⌴g (L/2) 3g can then use in the torque – angular acceleration relationship ␣ϭ ϭ ϭ (Eq. 10.21) to ﬁnd ␣ and then a. I 1͞3 ⌴L 2 2L The only force contributing to torque about an axis All points on the rod have this angular acceleration. through the pivot is the gravitational force Mg exerted on To ﬁnd the linear acceleration of the right end of the rod, the rod. (The force exerted by the pivot on the rod has zero we use the relationship a t ϭ r␣ (Eq. 10.11), with r ϭ L: torque about the pivot because its moment arm is zero.) To a t ϭ L␣ ϭ 3 2g L/2 This result — that at Ͼ g for the free end of the rod — is rather interesting. It means that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does! Pivot Mg Other points on the rod have a linear acceleration that is less than 3 g. For example, the middle of the rod has 2 Figure 10.18 The uniform rod is pivoted at the left end. an acceleration of 3 g. 4
• 19. 310 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis CONCEPTUAL EXAMPLE 10.11 Falling Smokestacks and Tumbling Blocks When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground, as shown in Figure 10.19. The same thing happens with a tall tower of children’s toy blocks. Why does this happen? Solution As the smokestack rotates around its base, each higher portion of the smokestack falls with an increasing tangential acceleration. (The tangential acceleration of a given point on the smokestack is proportional to the dis- tance of that portion from the base.) As the acceleration in- creases, higher portions of the smokestack experience an acceleration greater than that which could result from gravity alone; this is similar to the situation described in Example 10.10. This can happen only if these portions are being pulled downward by a force in addition to the gravi- tational force. The force that causes this to occur is the shear force from lower portions of the smokestack. Eventu- ally the shear force that provides this acceleration is greater than the smokestack can withstand, and the smokestack breaks. Figure 10.19 A falling smokestack. EXAMPLE 10.12 Angular Acceleration of a Wheel A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle, as shown in Figure M 10.20. A light cord wrapped around the wheel supports an object of mass m. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord. O Solution The torque acting on the wheel about its axis of rotation is ␶ ϭ TR, where T is the force exerted by the R cord on the rim of the wheel. (The gravitational force ex- erted by the Earth on the wheel and the normal force ex- T erted by the axle on the wheel both pass through the axis of rotation and thus produce no torque.) Because ⌺␶ ϭ I␣, we obtain ⌺ ␶ ϭ I␣ ϭ TR T TR (1) ␣ϭ I Now let us apply Newton’s second law to the motion of the m object, taking the downward direction to be positive: ⌺ F y ϭ mg Ϫ T ϭ ma mg Ϫ T (2) aϭ m Equations (1) and (2) have three unknowns, ␣, a, and T. Be- mg cause the object and wheel are connected by a string that does not slip, the linear acceleration of the suspended object Figure 10.20 The tension in the cord produces a torque about is equal to the linear acceleration of a point on the rim of the the axle passing through O.
• 20. 10.7 Relationship Between Torque and Angular Acceleration 311wheel. Therefore, the angular acceleration of the wheel and gthis linear acceleration are related by a ϭ R␣. Using this fact aϭ 1 ϩ I/mR 2together with Equations (1) and (2), we obtain TR 2 mg Ϫ ⌻ (3) a ϭ R␣ ϭ ϭ ␣ϭ a ϭ g I m R R ϩ I/mR mg (4) Tϭ Exercise The wheel in Figure 10.20 is a solid disk of M ϭ mR 2 2.00 kg, R ϭ 30.0 cm, and I ϭ 0.090 0 kgи m2. The suspended 1ϩ I object has a mass of m ϭ 0.500 kg. Find the tension in the cord and the angular acceleration of the wheel.Substituting Equation (4) into Equation (2), and solving fora and ␣, we ﬁnd that Answer 3.27 N; 10.9 rad/s2.EXAMPLE 10.13 Atwood’s Machine RevisitedTwo blocks having masses m1 and m 2 are connected to each Substituting Equation (6) into Equation (5), we haveother by a light cord that passes over two identical, friction- [(m 1 Ϫ m 2)g Ϫ (m 1 ϩ m 2)a]R ϭ 2I␣less pulleys, each having a moment of inertia I and radius R,as shown in Figure 10.21a. Find the acceleration of each Because ␣ ϭ a/R, this expression can be simpliﬁed toblock and the tensions T1 , T2 , and T3 in the cord. (Assume ano slipping between cord and pulleys.) (m 1 Ϫ m 2)g Ϫ (m 1 ϩ m 2)a ϭ 2I R2Solution We shall deﬁne the downward direction as posi-tive for m 1 and upward as the positive direction for m 2 . This (m 1 Ϫ m 2)gallows us to represent the acceleration of both masses by a (7) aϭ Isingle variable a and also enables us to relate a positive a to a m1 ϩ m2 ϩ 2 R2positive (counterclockwise) angular acceleration ␣. Let uswrite Newton’s second law of motion for each block, using This value of a can then be substituted into Equations (1)the free-body diagrams for the two blocks as shown in Figure10.21b: (1) m 1g Ϫ T1 ϭ m 1a (2) T3 Ϫ m 2g ϭ m 2a T2 T1 T3 Next, we must include the effect of the pulleys on the mo- T1 T3 +tion. Free-body diagrams for the pulleys are shown in Figure m1 m210.21c. The net torque about the axle for the pulley on the m1 m1gleft is (T1 Ϫ T2 )R, while the net torque for the pulley on the m2 m2gright is (T2 Ϫ T3 )R. Using the relation ⌺␶ ϭ I␣ for each pul- +ley and noting that each pulley has the same angular acceler- (a)ation ␣, we obtain (b) (3) (T1 Ϫ T2)R ϭ I␣ n1 T2 T2 n2 (4) (T2 Ϫ T3)R ϭ I␣ We now have four equations with four unknowns: a, T1 ,T2 , and T3 . These can be solved simultaneously. Adding T1 mpg mpgEquations (3) and (4) gives T3 (5) (T1 Ϫ T3)R ϭ 2I␣ (c)Adding Equations (1) and (2) gives Figure 10.21 (a) Another look at Atwood’s machine. (b) Free-body diagrams for the blocks. (c) Free-body diagrams for T3 Ϫ T1 ϩ m 1g Ϫ m 2g ϭ (m 1 ϩ m 2)a the pulleys, where m p g represents the force of gravity acting on each (6) T1 Ϫ T3 ϭ (m 1 Ϫ m 2)g Ϫ (m 1 ϩ m 2)a pulley.
• 21. 312 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis and (2) to give T1 and T3 . Finally, T2 can be found from pulleys accelerate counterclockwise. If m 1 Ͻ m 2 , then all the Equation (3) or Equation (4). Note that if m 1 Ͼ m 2 , the ac- values are negative and the motions are reversed. If m 1 ϭ m 2 , celeration is positive; this means that the left block acceler- then no acceleration occurs at all. You should compare these ates downward, the right block accelerates upward, and both results with those found in Example 5.9 on page 129. 10.8 WORK, POWER, AND ENERGY IN ROTATIONAL MOTION F In this section, we consider the relationship between the torque acting on a rigid object and its resulting rotational motion in order to generate expressions for the power and a rotational analog to the work – kinetic energy theorem. Consider the rigid object pivoted at O in Figure 10.22. Suppose a single external force F is ap- φ plied at P, where F lies in the plane of the page. The work done by F as the object ds rotates through an inﬁnitesimal distance ds ϭ r d␪ in a time dt is P dθ dW ϭ Fؒds ϭ (F sin ␾)r d ␪ r O where F sin ␾ is the tangential component of F, or, in other words, the component of the force along the displacement. Note that the radial component of F does no work because it is perpendicular to the displacement.Figure 10.22 A rigid object ro-tates about an axis through O un- Because the magnitude of the torque due to F about O is deﬁned as rF sin ␾der the action of an external force by Equation 10.19, we can write the work done for the inﬁnitesimal rotation asF applied at P. dW ϭ ␶ d ␪ (10.22) The rate at which work is being done by F as the object rotates about the ﬁxed axis is dW d␪ ϭ␶ dt dt Because dW/dt is the instantaneous power ᏼ (see Section 7.5) delivered by the force, and because d␪/dt ϭ ␻, this expression reduces to dW Power delivered to a rigid object ᏼϭ ϭ ␶␻ (10.23) dt This expression is analogous to ᏼ ϭ Fv in the case of linear motion, and the ex- pression dW ϭ ␶ d␪ is analogous to dW ϭ Fx dx. Work and Energy in Rotational Motion In studying linear motion, we found the energy concept — and, in particular, the work – kinetic energy theorem — extremely useful in describing the motion of a system. The energy concept can be equally useful in describing rotational motion. From what we learned of linear motion, we expect that when a symmetric object rotates about a ﬁxed axis, the work done by external forces equals the change in the rotational energy. To show that this is in fact the case, let us begin with ⌺␶ ϭ I␣. Using the chain rule from the calculus, we can express the resultant torque as d␻ d␻ d ␪ d␻ ⌺ ␶ ϭ I␣ ϭ I dt ϭI d ␪ dt ϭI d␪ ␻
• 22. 10.8 Work, Power, and Energy in Rotational Motion 313 TABLE 10.3 Useful Equations in Rotational and Linear Motion Rotational Motion About a Fixed Axis Linear Motion Angular speed ␻ ϭ d␪/dt Linear speed v ϭ dx/dt Angular acceleration ␣ ϭ d␻/dt Linear acceleration a ϭ dv/dt Resultant torque ⌺␶ ϭ I␣ Resultant force ⌺ F ϭ ma Ά Ά If ␻f ϭ ␻ i ϩ ␣t If vf ϭ vi ϩ at ␣ ϭ constant ␪f Ϫ ␪i ϭ ␻ i t ϩ 1 ␣t 2 2 a ϭ constant xf Ϫ xi ϭ v i t ϩ 1 at 2 2 ␻f 2 ϭ ␻ i 2 ϩ 2␣(␪f Ϫ ␪i ) vf 2 ϭ v i 2 ϩ 2a(xf Ϫ xi ) Work W ϭ ͵ ␪f ␪i ␶ d␪ Work W ϭ ͵xi xf Fx dx Rotational kinetic energy K R ϭ 1I␻ 2 2 Kinetic energy K ϭ 1mv 2 2 Power ᏼ ϭ ␶␻ Power ᏼ ϭ Fv Angular momentum L ϭ I␻ Linear momentum p ϭ mv Resultant torque ⌺␶ ϭ dL/dt Resultant force ⌺ F ϭ dp/dtRearranging this expression and noting that ⌺␶ d␪ ϭ dW, we obtain ⌺ ␶ d ␪ ϭ dW ϭ I␻ d␻Integrating this expression, we get for the total work done by the net externalforce acting on a rotating system ⌺W ϭ ͵⌺␪f ␪i ␶ d␪ ϭ ͵ ␻f ␻i I␻ d␻ ϭ 1I␻f 2 Ϫ 1I␻i 2 2 2 (10.24) Work – kinetic energy theorem for rotational motionwhere the angular speed changes from ␻i to ␻f as the angular position changesfrom ␪i to ␪f . That is, the net work done by external forces in rotating a symmetric rigid object about a ﬁxed axis equals the change in the object’s rotational energy. Table 10.3 lists the various equations we have discussed pertaining to rota-tional motion, together with the analogous expressions for linear motion. The lasttwo equations in Table 10.3, involving angular momentum L, are discussed inChapter 11 and are included here only for the sake of completeness.Quick Quiz 10.4For a hoop lying in the xy plane, which of the following requires that more work be done byan external agent to accelerate the hoop from rest to an angular speed ␻ : (a) rotationabout the z axis through the center of the hoop, or (b) rotation about an axis parallel to zpassing through a point P on the hoop rim?
• 23. 314 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis EXAMPLE 10.14 Rotating Rod Revisited A uniform rod of length L and mass M is free to rotate on a energy is entirely rotational energy, 1 I␻ 2, where I is the mo- 2 frictionless pin passing through one end (Fig 10.23). The rod ment of inertia about the pivot. Because I ϭ 1 ML 2 (see Table 3 is released from rest in the horizontal position. (a) What is its 10.2) and because mechanical energy is constant, we have angular speed when it reaches its lowest position? Ei ϭ Ef or 1 2 MgL ϭ 1 I␻ 2 ϭ 1 (1 ML 2)␻ 2 Solution The question can be answered by considering 2 2 3 the mechanical energy of the system. When the rod is hori- √ 3g zontal, it has no rotational energy. The potential energy rela- ␻ϭ tive to the lowest position of the center of mass of the rod L (OЈ) is MgL/2. When the rod reaches its lowest position, the (b) Determine the linear speed of the center of mass and the linear speed of the lowest point on the rod when it is in the vertical position. O E i = U = MgL/2 Solution These two values can be determined from the re- lationship between linear and angular speeds. We know ␻ L/2 from part (a), and so the linear speed of the center of mass is L O′ v CM ϭ r␻ ϭ ␻ϭ 1 2 √3gL 2 Because r for the lowest point on the rod is twice what it is for the center of mass, the lowest point has a linear speed equal 1 ω E f = K R = – Iω 2 to 2 Figure 10.23 A uniform rigid rod pivoted at O rotates in a vertical 2v CM ϭ √3gL plane under the action of gravity. EXAMPLE 10.15 Connected Cylinders Consider two cylinders having masses m1 and m 2 , where m1 ϶ inertia I about its axis of rotation. The string does not slip m2 , connected by a string passing over a pulley, as shown in on the pulley, and the system is released from rest. Find the Figure 10.24. The pulley has a radius R and moment of linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this time. Solution We are now able to account for the effect of a massive pulley. Because the string does not slip, the pulley ro- tates. We neglect friction in the axle about which the pulley rotates for the following reason: Because the axle’s radius is small relative to that of the pulley, the frictional torque is R much smaller than the torque applied by the two cylinders, provided that their masses are quite different. Mechanical en- ergy is constant; hence, the increase in the system’s kinetic en- ergy (the system being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy. Because m2 K i ϭ 0 (the system is initially at rest), we have h h ⌬K ϭ K f Ϫ K i ϭ (1m 1v f 2 ϩ 1m 2v f 2 ϩ 1I␻f 2) Ϫ 0 2 2 2 m1 where vf is the same for both blocks. Because vf ϭ R␻f , this expression becomes Figure 10.24 2 ΂ ⌬K ϭ 1 m 1 ϩ m 2 ϩ I v 2 R2 f ΃
• 24. Summary 315 From Figure 10.24, we see that the system loses potential en- Because v f ϭ R␻f , the angular speed of the pulley at this in- ergy as cylinder 2 descends and gains potential energy as stant is cylinder 1 rises. That is, ⌬U 2 ϭ Ϫm 2 gh and ⌬U 1 ϭ m 1gh. Ap- 2(m 2 Ϫ m 1)gh ΄΂ ΅ plying the principle of conservation of energy in the form vf 1 1/2 ⌬K ϩ ⌬U 1 ϩ ⌬U 2 ϭ 0 gives ␻f ϭ ϭ ΃ R R I m1 ϩ m2 ϩ 1 2 ΂ I ΃ m 1 ϩ m 2 ϩ 2 v f 2 ϩ m 1gh Ϫ m 2gh ϭ 0 R R2 Exercise Repeat the calculation of vf , using ⌺␶ ϭ I␣ ap- 2(m 2 Ϫ m 1)gh ΄΂ ΅ 1/2 vf ϭ plied to the pulley and Newton’s second law applied to the ΃ I two cylinders. Use the procedures presented in Examples m1 ϩ m2 ϩ 10.12 and 10.13. R2SUMMARYIf a particle rotates in a circle of radius r through an angle ␪ (measured in radi-ans), the arc length it moves through is s ϭ r ␪. The angular displacement of a particle rotating in a circle or of a rigid ob-ject rotating about a ﬁxed axis is ⌬␪ ϭ ␪ f Ϫ ␪ i (10.2) The instantaneous angular speed of a particle rotating in a circle or of arigid object rotating about a ﬁxed axis is d␪ ␻ϭ (10.4) dt The instantaneous angular acceleration of a rotating object is d␻ ␣ϭ (10.6) dtWhen a rigid object rotates about a ﬁxed axis, every part of the object has thesame angular speed and the same angular acceleration. If a particle or object rotates about a ﬁxed axis under constant angular accel-eration, one can apply equations of kinematics that are analogous to those for lin-ear motion under constant linear acceleration: ␻f ϭ ␻i ϩ ␣t (10.7) ␪f ϭ ␪i ϩ ␻ i t ϩ 2 ␣t 1 2 (10.8) ␻f 2 ϭ ␻i 2 ϩ 2␣(␪f Ϫ ␪i) (10.9)A useful technique in solving problems dealing with rotation is to visualize a linearversion of the same problem. When a rigid object rotates about a ﬁxed axis, the angular position, angularspeed, and angular acceleration are related to the linear position, linear speed,and linear acceleration through the relationships s ϭ ru (10.1a) v ϭ r␻ (10.10) at ϭ r␣ (10.11)