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Bilingual Mechanics Chapter 6 Rotation and Angular Momentum 制作 张昆实 谢 丽 Yangtze University Chapter 6 Rotation and Angular Momentum 6-1 6-2 6-3 6-4 6-5 What Is Physics? Equilibrium The Rotational Variables Are Angular Quantities Vectors Relating the Linear and Angular Variables 6-6 Kinetic Energy of Rotation Chapter 6 Rotation and Angular Momentum 6-7 Calculating the Rotational Inertia 6-8 Newton’s Second Law for Rotation 6-9 Work and Rotational Kinetic Energy 6-10 Rolling as Translation and Rotation Combined 6-11 The Kinetic Energy of Rolling 6-12 The Forces of Rolling Chapter 6 Rotation and Angular Momentum 6-13 Torque Revisited 6-14 Angular Momentum 6-15 Newton’s Second Law in Angular Form 6-16 The Angular Momentum of a System of Particles 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 6-18 Precession of a Gyroscope 6-1 What Is Physics ★ So far we have examined only the motion of translation. In which an object moves along a straight or curved line. ★ Now we deal with rotational motion of a rigid body (a body with definite shape that does’t change) about a fixed axis. ★ In this chapter we learn what is physics through studying rotation of a regid body and the related angular momentum. 6-2 Equilibrium Consider these objects: (1) a book resting on a table, (2) a hockey puck sliding across a frictionless surface with constant velocity, (3) the rotating blades of a ceiling fan, (4) the wheel of a bicycle that is traveling along a straight path at constant speed. Question 1 What is the linear momentum P and the angular momentum L of each of these four objects? 6-2 Equilibrium • For each of these four objects: 1. The linear momentum P of its center of mass is constant. 2. Its angular momentum L about its center of mass, or about any other point, is also constant. P a constant , L a constant (6-1) ★ two requirements for equilibrium objects satisfying Eq.(13-1) are said to be in equilibrium. 6-2 Equilibrium ★ two requirements for equilibrium P a constant , L a constant (6-1) ● Static equilibrium P0 L0 (constant =0) objects in static equilibrium are not moving in any way: either in translation or in rotation ( object (1)) ● stable static equilibrium: If a body returns to a state of static equilibrium after having been displaced from it. ● unstable static equilibrium: If a small force can displace the body and end the equilibrium. 6-2 Equilibrium The analysis of static equilibrium is very important in engineering practice. Landing gear 6-2 Equilibrium Static equilibrium in Building designing is very important especially in the earthquake area ! 6-2 Equilibrium The Yangtze River bridge at JingZhou(荆州) The analysis of static equilibrium is also very important in designing bridgies 6-2 Equilibrium ★ The translational motion of a body is governed by Newton's second law in its linear momentum form : dP Fnet dt (6-2) If the body is in translational equilibrium Fnet 0 (6-3) P is a constant dP dt 0 ★ The (balance of forces) rotational motion of a body is governed by Newton's second law in its angular momentum form : net If the body is in rotational equilibrium net 0 L is a constant dL dt 0 dL dt (6-4) (6-5) (balance of torques) 6-2 Equilibrium ★ The two Requirements for a body to be in Equilibrium 1. The vector sum of all the external forces that act on the body must be zero. Vector equation Fnet 0 (balance of forces) Fnet , x 0 Fnet , y 0 Fnet , z 0 Component equations (6-6) 2. The vector sum of all the external torques that act on the body, measured about an possible point, must also be zero. Vector equation net 0 (balance of torques) net , x 0 net , y 0 net , z 0 Component equations (6-6) 6-2 Equilibrium Consider the only simplifying situations: the forces only act on the body lie in the xy plane. then the only torques that can act on the body must tend to cause rotation around an axis parallel to the z axis. With this assumption, we can eliminate one force equation and two torque equations from Eqs. 6-6 Fnet , x 0 √ Fnet , y 0 √ Fnet , z 0 × net , x 0 × net , y 0 × net , z 0 √ (6-6) ★ Equilibrium equations for a body lie in the xy plane: (balance of forces) (6-7) Fnet , x 0 Fnet , y 0 net , z 0 (balance of forces) (6-8) (balance of torques) (6-9) 6-2 Equilibrium The gravitational force on an extended body is the vector sum of the gravitational forces acting on the individual elements of the body. cog com Fg ★ The gravitational force Fg on a body acts at a specific point, called the center of gravity (cog) of the body. ★ If g is the same for all elements of a body, then the body’s center of gravity (cog) is coincident with the body’s center of mass (com). xcog xcom 6-3 The Rotational Variables rigid body : a body that can rotate with all its parts locked together and without any change in its shape. A rigid body is rotating about a fixed axis In pure rotation, every point of the body moves in a circle whose center lies on the axis of rotation, and every point moves through the same angle during a particular time interval. Rotation axis Rigid body 6-3 The Rotational Variables ★ Angular Position A reference line (fixed in the body, perpendicular to the rotation axis) is rotating with a rigid body. The angular position of this line is the angle of the line relative to a fixed direction (positive x axis, the zero angular position ) s (radian measure) r 1rev 360 2r r 2 rad reference line r zero angular position (6-10) (6-11) 1rad 57.3 0.159rev (6-12) s x 6-3 The Rotational Variables ★ Angular Displacement 2 1 ω (6-13) An angular displacement in the counterclockwise direction is positive “+”, and one in the clockwise direction is negative “ ” ★ Angular Velocity average angular velocity avg z 2 1 t 2 t1 t r t2 . 1 . t1 O x z z (6-14) 0 0 6-3 The Rotational Variables The instantaneous angular velocity d lim t 0 t dt (6-15) unit (rad/s) or (rev/s) ★ Angular Acceleration The average angular acceleration avg 2 1 t2 t1 t (6-16) The instantaneous angular acceleration d lim t 0 t dt unit (rad/s2) or (rev/s2) (6-17) 6-4 Are Angular Quantities Vectors Can we treat the angular displacement, velocity, And acceleration of a rotating body as vectors? For angular velocity, there are only two directions. ★ Right hand rule Curl you right hand about rotating disk, your fingers pointing in the direction of rotation. Your extended thumb points in the direction of the angular velocity vector. z In pure rotation, a vector defines an axis of rotation, not a direction in which something moves ! 6-4 Are Angular Quantities Vectors caution Angular displacements (unless they are very small) cannot be treated as vectors! A vector must obey the rules of vector addtion, one of which says that if you add two vectors, the order in which you add them does not matter. Angular displacements fail this test ! 6-4 Are Angular Quantities Vectors In pure rotation, the case of constant angular acceleration is very important ! Equations of Motion for Constant Linear Acceleration and for Constant Angular Acceleration Equation Number (2-43) (2-47) (2-48) v v0 at Linear Equation v v0 at x x0 v0t 12 at 2 Angular Equation 0 t 0 0t 12 t 2 2 2 2 ( 0 ) 0 v v 2a ( x x0 ) 2 2 0 (2-49) x x0 12 (v0 v)t (2-50) x x0 vt 12 at 2 Three replacements: Equation Number (6-18) (6-19) (6-20) 0 12 (0 )t (6-21) 0 t 12 t 2 (5-22) x , v , a 6-5 Relating the Linear and Angular Variables Relate the linear variables s, v and to the angular variables , and ★ The Position s r (radian measure) (6-23) v a M at s ar a r ★ The Speed M0 Differentiating Eq6-23 with respect to t d ds d ds is the linear speed, r dt dt dt dt So we can obtain : v r is the angular speed (radian measure) (6-24) Caution: The angle and the angular speed measured in radians. must be 6-5 Relating the Linear and Angular Variables The period of revolution T for the motion of each point and for the rigid body itself is 2 r T v (6-25) Substituting for v from Eq.6-24 and canceling r, we can find 2 T (radian measure) (6-26) et ar r v at ★ The Acceleration Tangential acceleration differentiating v r dv d (6-27) at r r (6-28) dt dt radial acceleration v 2 ( r )2 ar 2 r (6-29) r r (radian measure) 6-6 Kinetic Energy of Rotation Treating a rotating rigid body as a collection of particles with different speeds. ★ The kinetic energy of a rotating body 1 2 1 1 2 2 K m1v1 m2 v2 .... mi vi (6-37) 2 2 2 O z ri m i mi : the mass of the ith particle, vi: Its speed Substituting for v from v r, we can get 1 1 2 K mi ( ri ) ( mi ri 2 ) 2 2 2 (6-38) ★ The rotational inertia ( moment of inertia) I mi ri 2 (6-39) 1 K I 2 2 (radian measure) (6-40) 6-6 Kinetic Energy of Rotation ★ The rotational inertia ( moment of inertia) I mi ri 2 (6-39) The SI unit for I is kilogram-square meter ( kg m 2 ) ★ The rotational inertia of a rigid body depends on: (1) the mass of the rigid body ; (2) the mass distribution; (3) the rotation axis (a) Parallel-Axis Theorem (b) Perpendicular-Axis Theorem 6-7 Calculating the Rotational Inertia If a rigid body consists of a great many adjacent par2 ticles, replace the sum in I mi ri with an integral: I r dm 2 ( rotational inertia, continuous body ) (6-41) Example: There is a uniform circular disk with mass m and radius R , what is the rotational inertia of the disk about the axis through the center o , perpendicular to the disk? ( 漆安慎力学:P223 例题1 ) m h o R 6-7 Calculating the Rotational Inertia Solution: The key idea here is that we divide the disk into infinite number of thin rings, the mass of the ring at radius r with width dr is dm 2 rhdr ( h: thickness , : density) then the rotational inertia of the ring can be found dI r 2 dm 2 hr 3dr Through integration we get R I r dm 0 2 R 0 1 4 2 hr dr hR 2 3 h m oo r dr Since m h R 2, the rotational inertia of the disk is 1 I mR 2 2 R mass element 6-7 Calculating the Rotational Inertia TABLE 6-2 Some Rotational Inertias 6-7 Calculating the Rotational Inertia ★ Parallel-Axis Theorem As show in the figure, an axis zc goes through the center of mass of a rigid body, the rotational inertia of the body about this axis is Icom . If another axis z paralle to zc, it can be proved that the rotational inertia of the rigid body about the z axis is I I com Mh 2 z Zc (6-42) (parallel-axis theorem) Where M is the mass of the rigid body, h is the distance between the two parallel axes. M h C 6-7 Calculating the Rotational Inertia ★ Proof of the Parallel-Axis Theorem Let O be the center of mass of the arbitrary shaped body. Consider an axis through O perpendicular to the plane, and another axis through point P parallel to the first axis. Let dm be a mass element with the general coordinate x and y. The rotational inertia of the body about the axis through P is I r 2 dm [( x a )2 ( y b )2 ]dm axis y Rotation through P r I ( x y )2 dm 2a xdm 2b ydm ( a b )2 dm 0 R2 0 I R dm h dm I com Mh 2 2 h2 2 dm y- b p • x-a h b o a Rotation axis through center of mass x 6-7 Calculating the Rotational Inertia Sample Problem 6-3 A thin, uniform rod of mass M and length L lies on an x axis with the origin at the rod’s center. (a) What is the rotational inertia of the rod about the perpendicular rotation axis through the center? (b) What is the rod’s rotational inertia I about a new rotation axis that is perpendicular to the rod and through the left end? M com L 2 L 2 (a) x M L (b ) x 6-7 Calculating the Rotational Inertia Solution: (a) for a continuous object using I r 2 dm dm mass element dm ML dx I com L2 L 2 M rx x ( )dx 2 M L 1 12 L 0 1 3 L 2 M x L 2 dm 2 1 3 I x ( )dx ML 2 M L x ML I ML M ( ) ML L 2 2 dx (a) 2 I I com Mh 2 com L 2 Solution: (b) using parallel-axis theorem 2 1 12 x 2 x M dx x L (b ) 6-7 Calculating the Rotational Inertia ★ Perpendicular-Axis Theorem ( 漆安慎力学:P225 ) The rigid body is a very thin plate, a coordinate system is chosen as the figure shows with the x and y axes lie in the plate and the z axis perpendicular to the plate. The rotational inertia of the rigid body about the z axis is I z mi ri 2 z I z mi x mi y 2 i Iz Ix 2 i (perpendicularI y axis theorem) Where I x , I y is the rotational inertia about the x, y axis, respectively. xi x o ri yi mi y 6-7 Calculating the Rotational Inertia z ★ Perpendicular-Axis Theorem Iz Ix I y m (perpendicular-axis theorem) x Example: What is the rotational inertia of a disk about any diameter? x , y axes are lie in the plane of the disk, y z m z axis is perpendicular to them. C R I z I x I y 2I x I z mR (from symmetry) I x I y I z mR 1 2 2 1 2 x 1 4 2 y 6-7 Calculating the Rotational Inertia ★ Torque A force F is applied on a body (only shown a cross section) F Ft radial component FF rr tangential component torque F Ft (r )( F sin ) F Rotation axis o F Line of r r r action of F Moment arm of F Two equivalent ways of computing the torque ( r )( F sin ) rFt ( r sin )( F ) r F r : the moment arm 6-7 Calculating the Rotational Inertia torque (r )( F sin ) ( r )( F sin ) rFt ( r sin )( F ) Frr F Ft F Rotation axis o F F Line of r r r action of F The ability of F to rotate the body Moment arm of F depends not only on the magnitude of its tangential component Ft , but SI unit : N• m also on just how far from o the force is applied. Torques obey the superposition principle : when several torques act on a body, the net torque (or resultant torque) is the sum of individual torques. net torque: counterclockwise positive; clockwise negative 6-8 Newton’s Second Law for Rotation ★ Newton’s Second low for rotation Fnet ma net I Fnet net , m I , y F (6-45) Ft a Fr m r A force act on the particle ,only the Rod tangential component Ft of the force can accelerate the particle along the o Rotation axis circular path. So with the Newton’s second law we can write From x at r Ft mat The torque acting on the particle is Ft r mat r m( r )r ( mr ) 2 I net I (6-47) ( radian measure ) (6-48) 6-9 Work and Rotational Kinetic Energy with the work- kinetic energy theorem (Eq 4-10) ,we relate the change K in kinetic energy to the work W, writing ★ 1 2 1 2 K K f Ki mv f mvi W 2 2 W xf xi Fdx (work, one-dimensional) (6-52) (6-53) The rate at which the work is done is the power, which we can find with Eqs 4-41 and 4-46 dW P Fv dt (power, one-dimensional) (6-54) 6-9 Work and Rotational Kinetic Energy ★A rotational situation is similar, so we can write a rotational kinetic energy as 1 2 1 2 K K f Ki I f I i W (work-kinetic energy theorem) (6-55) 2 2 We can calculate the work with a rotational equivalent of Eq6-53 W f d (work, rotation about fixed axis) (6-56) i is constant, Eq 6-56 reduces to W ( f i ) (work, constant torque) When (6-57) The rate at which the work is done is the power dW P dt (power, rotation about fixed axis) (6-58) 6-9 Work and Rotational Kinetic Energy ★ Proof of Equation 6-55 through 6-58 During the rotation, force F does work on the body. Let’s assume that the only energy of the body that changed by F is the kinetic energy. Then we can apply the work-kinetic energy theorem K K f K i W Using k 12 mv 2 and K Substituting v r (6-59) we can rewrite the Eq6-59 1 2 2 1 2 2 mr f mr i W 2 2 I mr 2 (6-60) into Eq 6-60 yields 1 1 2 K I f I 2i W 2 2 6-9 Work and Rotational Kinetic Energy We write that work dW as Ft ds. However, we can replace ds with Thus we have dW Ft rd From d . (6-61) Ft r ,we can rewrite Eq 6-61 as dW d (6-62) The work done during a finite angular displacement from i to f is then We can find power P for rotational motion from Eq 6-62 W f d i dW d P dt dt 6-9 Work and Rotational Kinetic Energy 6-3 From Left to Right : Five replacements x , v , a , m I , F 6-10 Rolling as Translation and Rotation Combined Consider only objects that roll smoothly along a surface without slipping. Rolling can be treated as a combination of the translation of the center of mass and rotation of the rest of the object around that center. 6-10 Rolling as Translation and Rotation Combined A bicycle wheel rolls along a street. During a time interval t , both O and the contact point P move a distance s . The wheel rotates through an angle about the center of the wheel, The initial contact point P moves through arc length s : s R (6-66) R Differentiating Eq. 6-66 with respect to time (with R held constant) gives us vcom R (6-67) (smooth rolling motion) P 6-12 P 6-10 Rolling as Translation and Rotation Combined + = T P The combination of Figs (a) and (b) yields the actual rolling motion of the wheel, Fig. (c). The portion of the wheel at the bottom (at point P) is stationary and the portion of the wheel at the top (at point T ) is moving at speed 2vcom , faster than any other portion of the wheel. The motion of any round body rolling smoothly over a surface can be separated into purely rotational and purely translational motions. 6-10 Rolling as Translation and Rotation Combined ★ Rolling as Pure Rotation: Another way to look at the rolling motion of the wheel We consider the rolling motion to be pure rotation about an axis passing through point P in Fig 6-22c and perpendicular to the plane of the figure. The vectors in Fig.6-24 then represent the instantaneous velocities of points on the rolling wheel. P 6-24 What is the angular speed of the rolling wheel about the new axis ? The same angular speed as that the wheel rotates about its “ COM ” axis ! vtop ( )( 2 R ) 2( R ) 2Vcom vcom ( )( R ) R v p ( )( 0 ) 0 6-22 Check this answer by calculating the lineal speed at the top point T, and the point O and P We get the same results ! 6-11 The Kinetic Energy of Rolling If we view the rolling as pure rotation about an axis through P in Fig. 6-24, then from Eq. 6-40 1 K I 2 2 1 2 K I P 2 (6-68) where is the angular speed of the wheel and I P is the rotational inertia of the wheel about the axis through P. From the parallel-axis theorem, we have I P I com MR 2 (6-69) where M is the mass of the wheel, I com is its rotational inertia about an axis through its center of mass. 6-11 The Kinetic Energy of Rolling 1 K I P 2 2 I P I com MR 2 1 1 2 2 2 K I com MR 2 2 vcom R , yields 1 1 2 2 K I com Mvcom Konig’s theorem 2 2 Using the relation (6-70) A rolling object has two type of kinetic energy : 2 1 a rotational kinetic energy ( 2 I com ) due to its rotation about its center of mass and a translational kinetic energy ( 1 Mv 2 ) due to translation com 2 of its center of mass. 6-12 The Forces of Rolling ★ Friction and Rolling If a net force acts on the rolling wheel to speed it up cause acom and . tend to make the wheel slide at P. Tendency of slide fs A frictional force must acts on the wheel at P to oppose that sliding tendency. If the wheel does not slide, that frictional force is a static frictional force f , the s motion is smooth rolling. Differentiating vcom R with respect to time we have acom R (smooth rolling motion) (6-71) 6-12 The Forces of Rolling ★ Rolling down a Ramp A round uniform body of radius R rolls smoothly down a ramp at angle , along an axis. x N Fg sin R ★ The gravitational force Fg , acting at the COM of the body. Its component along the ramp is Mg sin ★ A normal force Fg fs P Fg cos , acting at point P but is shifted to the COM of the body. N ★ A static frictional force f s , acting at point P and is directed up the ramp. We can write Newton’s second law for components along the x axis as Fnet , x max f s Mg sin Macom, x (6-72) x 6-12 The Forces of Rolling ★ Rolling down a Ramp N Next, check all the external torques acting on the rolling body : Fg sin R ★ the torque of the frictional force : Rf s ( counterclockwise ) ★ the torque of the gravitational force : F (0) ( zero moment arm ) Fg fs x P Fg cos g ★ the torque of the normal force : N (0) ( zero moment arm ) Apply Newton’s second law in angular form to the body’s rotation about its center of mass. net I Rf s I com (6-73) 6-12 The Forces of Rolling Because the body is rolling smoothly, we can use Eq.6-71 (acom R ) but we must be cautious because here is negative ( in the com negative direction of x axis ) and is positive (counterclockwise). Thus substitute a / R for com Then, solving for f s I com fs , we obtain acom , x R2 a (6-74) f s Mg sin Macom, x Substituting the right side of Eq. 6-74 for acom, x fs in Eq. 6-72, we then find g sin 1 I com / MR 2 (6-75) We can use this equation to find the linear acceleration any body rolling along an incline of angle (6-72) acom , x of with horizontal. 6-13 Torque Revisited Torque Torque fixed axis; counterclockwise : + ; clockwise: fixed point; may have any direction! (expanded defination) A , r is its position A single force F acts on the A particle is at point vector. particle. z ★ The torque acts on the particle relative to the fixed point O is a vector quantity defined as rF z F F o r O r * A P (6-76) We use the right-hand rule for vector products, sweeping the fingers of the right hand from r (the first vector in the product) into F (the second vector). The outstretched right thumb then gives the direction of . 6-13 Torque Revisited Determine the magnitude of a torque rF ( Torque defined ) rF sin F F sin (6-77) r r sin rF r F (6-78) : z o F r r F F A (6-79) r Moment arm of F 6-14 Angular Momentum A particle of mass m passes through point A with linear momentum p( mv ), r is the position vector of the particle. right-hand-rule The angular momentum of the particle with respect to the origin O is a vector quantity r P m(r v) z (6-80) p o The SI unit of is (kg.m2/s) or (J. s) x rmv sin r A Using the right-hand rule we can determine the direction of the angular momentum . x (6-81) p z o The magnitude of p y y r r A p 6-14 Angular Momentum rmv sin v v sin (6-81) rp rmv (6-82) magnitude r r sin right-hand-rule r p r mv z (6-83) Note: 1. to have angular momentum about O, the particle does not Itself have to rotate around O ! p o x p r A 2. Angular momentum has meaning only with respect to a specified origin ! 3. The direction of the angular momentum vector is always perpendicular to the plane formed by the position and linear momentum vectors ! Sample problem 6-7 y p z o x y r r A p 6-15 Newton’s Second Law in Angular Form Newton’s second law written in the form F net dP dt (single particle) (6-84) We have seen enough of the parallelism between linear and angular quantities. Guided by Eq.6-84, we guess the relation between torque and angular momentum must be net d dt (single particle) (6-85) ★ The (vector) sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle. Note: defined with the same origin ! The torques and the angular momentum must be 6-15 Newton’s Second Law in Angular Form Prove of Eq.6-85 net We know the definition of the angular momentum of a particle: d dt m(r v) Differentiating each side with respect to time t yields d dv dr m( r v) dt dt dt d m( r a v v ) dt v v 0 (6-86) d m( r a ) dt ma Fnet We now use Newton’s Second law, obtaining (6-87) d r F net (r F ) dt net d dt (6-85) 6-16 The Angular Momentum of a System of Particles The total angular momentum L of a system of particles is the (vector) sum of the angular momenta of the individual particles: n L 1 2 3 .... n i (6-88) i 1 By taking the time derivative of Eq.6-88. The change in be found (6-89) dL dt n d i 1 dt dL dt L can n net ,i (6-90) i 1 The only torques that can change the total angular momentum L of the system are the external torques acting on the system. The sum of the internal torques due to the interaction between the particles is zero ! 6-16 The Angular Momentum of a System of Particles Let net represent the net external torque. Then Eq.6-90 becomes dL (system of particles) (6-91) net dt This is the Newton’s second law in angular form. The net extenal torque net acting on a system of particles is equal to the time rate of change of the system’s total angular momentum L . caution 1. Torques and system’s angular momentum must be measured relative to the same origin. 2. If the center of mass of the system is not accelerating relative to an inertia frame, that origin can be any point ; if the center of mass of the system is accelerating, the origin can be only at that center of mass . 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis A rigid body rotates about the z axis with angular speed . a mass element of mass mi within the body is located by the position vector ri . It moves about the z axis in a circle with radius r i and has the linear momentum pi . ri The magnitude of the angular momentum of this mass element, with respect to O is pi i ri iz i i ( ri )( pi )(sin 900 ) ( ri )( mi vi ) The angular momentum vector for the i mass element is Shown in Fig.b; its direction must be perpendicular to those of ri and pi . 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis The iz z cmponent of i is i sin (ri sin )(mi vi ) ri mi vi Because v r , we may write n LZ i 1 n iz n n mi vi ri mi (ri )ri ( mi r2i ) i 1 i 1 i 1 n The quantity (6-92) 2 is the rotational inertia m r I of the i i i 1 body about the fixed Axis .Thus Eq. 6-92 reduces to Lz Iz z ( rigid body, fixed axis ) (6-93) 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 6-4 6-3 From Left to Right : six replacements x , v , a , m I , F , p PL 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis six repla- 6-3 cements x , v , a , m I, 6-4 F , p 6-3 PL 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis The law of conservation of angular momentum If no net external torque acts on the system or net dL dt L = a constant net angular momentum at some initial time i t Li L f dL dt 0 ( isolated system ) (6-94) = net angular momentum at some later time f ( isolated system ) t (6-95) ★ The law of conservation of angular momentum If the net external torque acting on a system is zero, the angular momentum L of the system remains constant, no matter what changes take place within the system. 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis The law of conservation of angular momentum L = a constant or Li L f ( isolated system ) (6-94) (6-95) ★ If the component of the net external torque on a system along a certain axis ( z ) is zero, then the component of the angular momentum of the system along that axis ( Lz ) cannot change, no matter what changes take place within the system. Substituting L I into Li L f ,we get I ii I f f (6-96) 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis The law of conservation of angular momentum Li L f I ii I f f Like the other two conservation laws , The law of conservation of angular momentum holds beyond the limitations of Newtonian mechanics. It holds for particles whose speeds approach that of light ( where the theory of relativity reigns) and it remains true in the world of subatomic particles ( where quantum mechanics reigns). No exceptions to it have ever been found ! 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis Examples 1. The spinning volunteer A student + a stool rotate around z axis. No external torque acts on the system, the angular momentum of the system must remain constant, At the beginning, the student stretches out his both arms with each hand holding a dumbbell, Ii i I ii I f f I f f when the person drops both arms to reduce the moment of inertia the rotational angular velocity of the person and the stool is increased. 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 2. The springboard diver The diver leaves the high-diving board with outstretched arms and legs and some initial angular velocity about his centre of gravity. His angular momentum ( I ) remains constant since no external torques act on him (gravity exerts no torque about his centre of gravity). To make a somersault he must increase his angular velocity. He does this by pulling in his legs and arms so that decreases I and therefore increases . By extending his arms and legs again, the diver can enter the water with little splash. I i i I ii I f f I f f 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 3. Spacecraft orientation The spacecraft + fiywheel form an isolated system. Therefore, if the system's total angular momentum L is zero, it must remain zero. To change the orientation of the spacecraft, the flywheel is made to rotate. The spacecraft will start to rotate in the opposite sense to maintain the system's angular momentum at zero. When the flywheel is then brought to rest, the spacecraft will also stop rotating but will have changed its orientation. 6-17 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis 4. The incredible shrinking star When the nuclear fire in the core of a star bums low, the star may eventually begin to collapse. I1 1 beforeshrinking shrinking before The collapse may go so far as to reduce the radius of the star from something like that of the Sun to the incredibly small value of a few kilometers. The star then becomes a neutron star. During this shrinking process, the star is an isolated system and its angular momentum L cannot change. Because its rotationaI inertia is greatly reduced, its angular speed is correspondingly greatly increased, to as much as 600 to 800 revolutions per second. I11 I 22 2 I2 after shrinking The neutron star 6-18 Precession of a Gyroscope A nonspinning gyroscope fslls downword about the tip of the support O because of the torque: Mgr sin 90 Mgr net L A rapidly spinning gyroscope rotates horizontally about a vertical axis through support point O in a motion called precession. Spinning wheel : L I , Mg produce a torque: dL dt dL dt Only change the direction of L ! 6-18 Precession of a Gyroscope Mgr sin 90 Mgr dL dt L Only change the direction of L ! dL // L rotates about z axis, Precession! dL dt Mgrdt d Mgr dt I dL Mgrdt d L I (precession rate) (6-107)