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Chapter 9-10 rotation of rigid bodies Rotational motion dynamics kinematics Instantaneous effect 1. model 2.Angular quantity theorem of rotation Space effect 1.work 2.kinetic and potential energy 3.conservation of energy Time effect 1.Theorem of angular momentum 2.conservatio n of angular momentum Chapter 9-10 rotation of rigid bodies Key terms: rigid body rotation translation radianrad angular velocity angular acceleration angular speed moment of inertia rotational kinetic energy parallel-axis theorem Chapter 9-10 rotation of rigid bodies Key terms: torque line of action lever arm rotational analog of Newton’s second law combined translation and rotation angular momentum conservation of angular momentum Types of Motion Translation The linear position of a body changes with respect to a fixed frame, but its angular orientation remains unchanged. Rotation The angular orientation of a body changes about a fixed frame of reference, but its linear position remains fixed. Complex Simultaneous combination of translation and rotation motions 1. Ideal Model-----Rigid Body we neglect the deformations and assume that the body has a perfectly definite and unchanging shape and size. Definite shape and definite size. The distance between any two points on a rigid body remain unchanging. A rigid body can be considered as a special collection of particles with mass continuously distributed. A rigid body in general can have rotational as well as translational motion. 2. Angular Quantities let’s think first about a rigid body that rotates about a fixed axis. (The axis passes through point O and is perpendicular to the x-y plane) y P r o x We can use the linear quantities ( r, v, a) as well as the angular quantities ( , , ) to describe the motion of the rigid body. 2.1 Angular Coordinate s rad 2 1 r 2.2 Angular Velocity d rad / s v r dt Counterclockwise(+) 2.3 Angular Acceleration y d d 2 2 rad / s 2 dt dt P s r If is a constant (=c): 0 t o x 0 t 1 / 2t 2 2 2 2 0 2.4 Relating Linear and Angular Kinematics We can use linear quantities ( r, v, a) or angular quantities ( , , ) to describe the rotational motion. The relationships between them are: r s r an 2 r r 2 d at r dt y v r P s o x The linear quantities of any point on the rigid body are different, while angular quantities are the same. 3. Theorem of Rotation 3.1 torque (/moment) of a force r F force times lever-arm, tends to produce rotation. rF sin Fd z Vector: right-hand rule. o x d r y F mi discussion : F r , F // r , F // axis Net torque is the vector sum of individual torques: i r1 F1 r1 F2 ri Fi i i Example: a force on particle F ( xi yj ) find the torque on particle if the fixed axis is in original point. Solution: (1) r F r ( r ) 0 ( 2) r F ( xi yj ) ( xi yj ) 0 3.2 Rotation Analog of Newton’s Second Law I Proof: A rigid body can be looked as a special collection of particles. i ri Fi ri The net torque on mi to point O: z o f in ri Fex j( i j ) The net torque on the rigid body to point O: i ri Fi ri i mi f ji i ri i i f ji 0 j( i j ) f ji j( i j ) i ri Fi ri i i i f ji j( i j ) i ri Fi ri Fi ,tan i i i ri ( mi ai ,tan ) ri mi ( ri ) i z o ri i j(i j ) f ji 0 Fi ,tan mi ai ,tan ai ,tan ri i ( ri2 mi ) “”s are the same for every part ri i of the rigid body. Fex I ri2 mi i mi I Rotational analog of Newton’s law for a rigid body. I Caution: 1) 2) 3) Means external torque, the torque offered by a pair of internal force equals to zero The angular acceleration of rigid body with fixed axis 0 0 The rigid body keep rest or rotate in uniform angular velocity 4) I moment of inertial, measurement of rotational inertial 1 Given , I 3.3 Calculation of Moment of Inertial I mi ri2 i 1) A few point masses: I 2 m r ii i 2) Continuous distribution of mass: I r dm 2 Moment of inertial depends on the distribution of mass in the system. Moment of inertia depends on the location and orientation of the axis. Example: An engineer is designing a one-piece machine part consisting of three heavy connectors linked by light molded struts. The connectors can be considered as massive particles connected by massless rods. a) What is the moment of inertia of this body about an axis through point A, perpendicular to the plane of the diagram? b) What is the moment of inertia about an axis coinciding with rod BC? (see page 277) B mB=0.10kg Solution: 0.50m I mi ri2 i 0.30m A mA=0.30kg mC=0.20kg 0.40m C Solution: a) I a 2 2 2 m r ( 0 . 10 kg )( 0 . 50 m ) ( 0 . 20 kg )( 0 . 40 m ) ii i 0.057 kg .m 2 a) b) I b 2 2 2 ( 0 . 30 kg )( 0 . 40 m ) m r 0 . 048 kg . m ii i B mB=0.10kg 0.50m 0.30m A mA=0.30kg mC=0.20kg 0.40m C Example: A slender uniform rod with mass M and length L. It might be a baton held by a twirler in a marching band. Compute its moment of inertia about an axis through center of mass point O. Solution: I 2 r dm I r dm 2 x dm 2 M L L 2 L 2 2 x dx 1 2 ML 12 C o dm x dx x Uniform thin loop (M ,R): I m c R 2 dm MR 2 R Uniform thin disc(M, R) R M Ic r 2 2rdr 0 R 2 1 MR 2 2 r dr 2) Parallel-Axis Theorem: (p281) I p I cm Md 2 p Icm: about axis through CM M: total mass d : distance between CM and point p. Icm Ip d O CM I p r 2dm [( x a )2 ( y b)2 ]dm I p ( x 2 y 2 )dm 2a xdm 2b ydm (a 2 b2 )dm xc xdm 0 M yc ydm o I p I cm Md M 2 Example: If the rod rotates about the axis through one end, Use parallel-axis theorem: I I cm Md 2 1 1 2 ML ML2 12 4 1 ML2 3 o 3.3 Application of rotational analog of Newton’s second law I Caution: 1) The left is vector sum of external torque about a fixed axis, and the direction is parallel to axis. 2) “ I ” is the Moment of inertial 3) “” means the angular acceleration to fixed axis, same direction with torque Example: In judo, a weaker and smaller fighter who understands physics can defeat a stronger and larger fighter who doesn’t. The fact is demonstrated by the basic ‘ hip throw ’, in which a fighter rotates his opponent around his hip. Solution: Fd1 I F d1 mgd2 I FF Example: The massive shield door at a neutron test facility at Lawrence Livermore Laboratory is the world’s heaviest hinged door M=44000kg, I=8.7*104 kgm2, and face width 2.4m find the smallest force to move the door from rest through an angle of 900 in 30s. Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the acceleration of the object of mass m and the angular acceleration of the cylinder N R For m: O M T m Mg T y h Solution: mg mg T ma (1) For M: TR ( 1 MR 2 ) 2 a R (3) (2) Example: An Atwood’s machine. There is no slipping between the cord and the surface of the wheel. Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord a) If weight of cylinder is negligible ; b) If weight of cylinder is not negligible. T Solution: a) The weight of cylinder is negligible … T’2 R C O T’1 b)If weight of cylinder is not negligible: For m1: m1 g T1 m1a1 y For m2: m2 g T2 m2 a 2 y (1) (2) T1 T2 mB m2g mA m1g y For m1: For m2: m1 g T1 m1a1 y m2 g T2 m2 a2 y 1 For M: T R T R ( MR 2 ) 2 ' 2 ' 1 (1) (2) (3) a1 y a2 y ( 4 ) a1 y R ( 5 ) T O From (1),(2),(3),(4),(5) : m1 m 2 a1 y a 2 y g m1 m 2 M 2 m1 ( 2 m 2 M 2 ) m2 ( 2m 1 M 2 ) T1 g , T2 g m1 m 2 M 2 m1 m 2 M 2 T’2 R C T’1 T1 T2 m2 m2g m1 m1g y Solution: mg L cos I 2 L d d mg cos I I 2 dt d L mg 2 cosd 0 Id 2 L 1 mg (1 sin ) I 2 2 2 Example: A grindstone 1.0 m in diameter, of mass 50 kg, is rotating at 900 rev.min-1. A tool is pressed against the rim with a normal force of 200N, and the grindstone comes to rest in 10s. Find the coefficient of friction between the tool and the grindstone. Neglect friction in the bearings. Solution: I , 0 t f k R , f k N k 0 900 2 / 60 30 ( rad / s ) t 0 , t 10 s ( t 0 ) / t 3 ( rad / s 2 ) 1 I MR 2 2 k o .59 Example: A wheel rotates about a fixed axis from rest. A constant torque,=20N.m, acts on it for 10 seconds. =100rev/min at t=10s. Then it takes 100 seconds for the wheel to come to stop again. Find the moment of inertial of the wheel with respect to the axis. Solution: =I, = o+ t acted by external torque: 20=I r=I 20= 1/t (for: o=0) o=0) 1, 1, 1=1/t 1 (for: When release external torque: - r=I2 , 2=- /t2 (1) (2) t1=10s , t2=100s , =(100×2)/60=10.5rad/s, From (1)、(2): I=17.3kg.m2 。 4. Energy in Rotational Motion 4.1 Rotational Kinetic Energy The kinetic energy of mi : 1 1 2 K i m i v i m i ri2 2 2 2 The rigid body : vi O r P mi Ki i i 1 1 mi v i2 ( mi ri2 ) 2 2 2 i I mi ri2 i vi ri i 1 K I 2 2 Example: Each of the helicopter rotor blades is 5.2m long, m=240kg. The rotor is rotating at 350rev/min. Find 1) the rotational inertial of the rotor assemble about the axis of rotation. 2)what is the total kinetic energy of rotation. 4.2 Gravitational Potential Energy The gravitational potential energy is the same as though all the mass were concentrated at the center of mass of the rigid body. U mi yi g Mgy cm i m y i i i My cm 4.3 Work and Power in Rotational Motion Work of the torque Suppose a force F acts on the rigid body. The rigid body rotates through an infinitesimal angle d about a fixed axis. The work dW done by the force while point P moves distance ds is: dW F ds Fr sind d 2 d r P ds F W d 1 work done by a torque 2 W d 1 d d W d I 1 1 dt 2 1 1 2 Id I 2 I 12 1 2 2 2 2 I d CAUTION: a work done by the forces doesn’t equal to the work done by the torques. 1 1 For a particle: W F dr m 2 2 m1 2 2 2 dt Compare with the work-energy theorem for collection of particles: Wtot 1 1 2 2 I 2 I 1 2 2 Wtot 1 1 2 2 mv 2 mv 1 2 2 Power of a torque: dW d P dt dt p F v 4.4 Conservation of Mechanical Energy If the work done by all forces other than conservative force equals zero. The total mechanical energy of the rigid body is conservative. Wext Winnon 0 E C Example: We wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity. Find the speed of the falling object and the cylinder just as the object strikes the floor. See page 280, example 9-9. N R No work is done by friction, so O M T m Mg T y h Solution: mg 1 v 2 2 1 1 2 mgh mv ( MR )( ) R 2 2 2 Solution: L L 1 2 mg mg sin I 2 2 2 1 I ml 2 3 We get ( 2) (1) Example: A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley . Find the speed of the object when it falls a distance h from rest. E mgh (1) (2) v s R v p r From the conservation law of energy Example: the given is as follows,find the velocity of the object and the acceleration of M。 Solution: 1) 1 1 1 Mgh Mv 2 I 2 kh2 2 2 2 1 I mr 2,v r 2 2 Mgh kh2 v 1 M m 2 k M h Example: the given is as follows,find the velocity of the object and the acceleration of M。 Solution: 2) 1 1 2 1 2 2 Mgh Mv I kh 2 2 2 k Make a derivation mgv mva I khv a mg ma I 2 kh R mg kh We get: a I m 2 R M h Alternative solution: mg T1 ma (1) (T1 T2 ) R I (2) T2 kh ( 3) a R (4) We get: k M h mg kh a I m 2 R Example: The power output of an automobile engine is advertised to be 200 hp at 6000 rpm. What is the corresponding torque? Solution: 1 HP = 746 W 746W P 200hp 200hp( ) 1.49 105W 1hp 6000rev / min 628rad / s 1.49 10 N m / s 237 N m 628rad / s P 5 5. Angular Momentum Coordinate(x) --- Angular Coordinate() Distance(s) --- Angular Displacement() Velocity(v) --- Angular Velocity() Acceleration(a) --- Angular Acceleration() Momentum(P) --- ? M, L P mi v i 0 i ' v ? v m 5.1 Angular Momentum of a particle L r p r mv Unit: kg m2/s z L r o x d m y mv Magnitude: L rmv sin mvd Direction: determined by right-hand rule for vector products. z o x d r m Compare with torque y F r F Caution: The Value of L and both depends on the choice of origin O. z L r o x L r p r mv d m y mv mv Angular momentum of collection of particles: L Li ri pi ri mv i i i O R O i P d mv To point O: L=mvd L=mvR To point P: L=0 5.2 Angular Momentum of a rigid body The angular momentum of mi to point O: v i ri Li ri mi vi mi ri2 The angular momentum of the rigid body to the fixed axis (z-axis): z Lz Li ( mi ri2 ) i r O i x mi vi y I mi ri2 i i Lz I z CAUTION: Giving angular momentum you must always specify about axis 5.3 The Theorem of Angular Momentum for one single particle: L r p r mv dp dL d ( r p ) dp dr F , r p dt dt dt dt dt dr v, dL dt r F dt v mv 0 The rate of change of angular momentum of a particle equals the torque of the net Force acting on It. t2 t1 L2 dt dL L2 L1 L1 for collection of particles: i i d ( Li ) i i ri Fi ri i i ri dt dL dt i f ji o i j( i j ) i ri Fi i i for rigid body: dLz d ( I z ) I z z dt dt t2 t1 L2 dt d ( I ) I 2 2 I11 L1 Lz I z f ji j( i j ) L2 L1 t1 dt d L L2 L1 t2 P2 Fdt dp P2 P1 t2 t1 dL dt P1 If the torque of all the external forces equals 0: i 0 , L2 L1 i ------conservation of the angular momentum for collection of particles or one single particle. Compare with conservation of momentum: F Fi 0 , i i 0 , i P2 P1 L2 L1 Example: the disk rotate on the table, angular velocity and friction coefficient is , how much time it will take for the disk to stop its motion Solution : Theorem of angular momentum dt I I 2 1 d dmgr 2r gdr 2 R 2 d RMg 3 0 t I I 3 R t 4 g r dr 5.4 Conservation of Angular Momentum When the net external torque acting on a system is zero, the total angular momentum of the system is constant. ex 0 , L2 L1 , I11 I 2 2 Example: The position of a object with mass m is given by r=acost i+bsint j (a, b, and are all constant), find the angular moment of the object and the net torque acting on it. Solution: L r mv r a cos tî b sin tĵ dr v a sin tî b cos tĵ dt î ĵ k̂ L a cos t b sint 0 ma sint mb sint 0 mabk r F r ma 2 d r 2 a r 2 dt 2 r ( r ) 0 Example: A small block on a frictionless horizontal surface has a mass of m. it is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of r from the hole with an angular speed 0. Pull the cord from below slowly, shorten the radius of the circle to r/2. How much work was done in pulling the cord? See page 322 10-33, 10-76, 10-87 Solution: net to O is zero conservation of L : L1 L2 mr 2 0 m( r 2 )2 0 m r o 40 1 r 2 2 1 3 2 2 W m( ) mr 0 mr 2 0 2 2 2 2 2 F Example: see page 314 example 10-16 Solution: Consider the door and bullet as a system, net torque about the axis through O equals zero, so angular momentum is conserved. l initial L : L1 mv 2 final L : L2 ( I bul I rod ) 1 l 2 I bul m( ) , I rod Ml 2 , 3 2 0.50 conservation of momentum m 1.00 L1 L2 m v 400 m / s so : 0.40 rad / s Example: see page 327 10-75 Solution: Consider A and B as a system, when remove the accelerating torque from A, angular momentum is conserved. I A 0 ( I A I B ) The thermal energy equals the lose of the total rotational kinetic energy: 1 Q I A 02 1 ( I A I B ) 2 2 2 S S A C B 1 Q 2400 J , I A I B 3 so : 1 K A0 I A 02 3200 J 2 Example:a block with spring was in a frictionless surface, k=100N/m, a spring was fixed in o point,m=1kg, the natural length of the spring l0=0.2m, the velocity of the block is v0=5m/s, perpendicular to the spring。When the spring rotate 900, the length of l=0.5m find the velocity of the block. Solution: mv0l0 mvl sin 1 1 2 1 2 mv 0 mv k ( l l0 )2 2 2 2 We get:v =4m/s, =300 l O l0 d v0 v Example: a rocket’s mass is m, velocity is v0, oc=3R。 Find the angle between v and v0。(the earth mass is M、 radius=R) Solution: 1 Mm 1 2 Mm 2 mv 0 G mv G Conservation of the energy 2 R 2 3R mv 0 R mv 3 R sin Conservation of the angular momentum We get: 2 A m O M R Rv 0 sin 2 3( 3 Rv 0 4GM ) v0 3R O’ C v Example: A man of mass m runs around the edge of a horizontal turntable that is mounted on a frictionless vertical axis through its center. The velocity of the man, relative to the disk, is v. The turntable is rotating in the opposite direction with an angular velocity of , relative to the earth. The radius of the turntable is R, and its mass is M. Find the final angular velocity of the system if the man comes to rest, relative to the turntable. Solution: Consider the man and turntable as a system, angular momentum is conserved. ' Rm I disc ( I disc I man ) The equation is wrong, because conservation of L is only valid in inertial reference frame R ME MD D E ME v R ' ( I I ) I I man ME disc man disc I man 1 mR , I disc MR 2 , 2 2 mv / R m M / 2 mM / 2 ' R Summary: Instantaneous Effect Fx ma x F ma Fy ma y (One particle) Fz ma z I (Rigid body) I m i ri i I r 2 dm v2 Fn ma n , ( a n R ) Ft ma t , ( a t dv ) dt 2 I p I cm Md 2 Accumulation Effect of Space 1 1 2 2 (One particle) W F d r mv 2 mv 1 tot a 21 2 1 2 (particles) Wex Win K 2 K 1 mi v i 2 mi v i21 i 2 i 2 Wcon ( U 2 U 1 ) b Wex Win ,non ( K 2 U 2 ) ( K 1 U 1 ) E2 E1 if Wex Win ,non 0 , E2 E1 (Rigid body) 1 2 1 2 d I 2 I 1 a Wtot b 2 1 2 E mgy cm I 2 2 时间积累 t2 (One particle) J F dt P2 P1 mv2 mv1 t1 t2 (particles) J Fdt P2 P1 mi v i 2 mi v i 1 t1 i i if F 0 , P2 P1 dLz dL dL z dt dt dt (One particle) (particles) (Rigid body) t2 矩冲量= t dt L2 L1 Li 2 Li 1 1 i i if 0 , L2 L1 or I 1 1 I 2 2 6* Rigid-body Rotation about a Moving Axis 6.1 Combined translation and rotation: Energy Relations For mi: ' v i vcm v i 1 ' ' K i mi ( vcm v i ) ( vcm v i ) 2 1 ' ' 2 2 mi ( vcm 2vcm v i v i ) 2 cm v cm ri mi ' vi vi v cm 1 2 For mi: K i m i ( v cm 2v cm v i' v'i 2 ) 2 2 For body: K K i ( 1 m i vcm ) 2 1 ' '2 m v v ( m v i cm i 2 i i ) ' ' mi vcm vi vcm ( mi vi ) 0 1 1 2 K ( m i v cm ) ( m i v'i 2 ) 2 2 1 1 2 Mv cm I cm 2 2 2 2. Rolling without slipping vcm R 1 1 2 K Mv 1 I 12 2 2 1 0 ( I cm MR 2 )2 2 1 1 2 2 I cm Mv cm 2 2 F M a Center of mass: i cm Rotation about axis through CM: Icm 3 ' v3 v v 2 cm ' v2 2 v cm R v cm ' 1 v1 vcm v cm v1 0 v3 4 v cm ' v v4 4 Example13: See page 303 example 10-6. Use two ways to find the speed vcm of solid cylinder after it has dropped a distance h. 1 1 2 Mv cm I cm 2 2 2 Solution2: Mg T Ma (1) cm Solution1: Mgh TR I cm (2) vcm R I cm From (1),(2): a 2 g ,T 1 mg cm 3 3 v 2 cm 2acm h vcm 4 gh 3 1 MR 2 2 h T Mg 7* Gyroscopes and precession p dL L I r O mg r mg dL mg dt d mg dL L sind mg d p dt I sin