Rotational Kinematics and Dynamics. UCVTS AIT Physics

Transcription

1 Rotational Kinematics and Dynamics UCVTS AIT Physics

2 Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin

3 Angular Position, 2 Point P will rotate about the origin in a circle of radius r Every particle on the disc undergoes circular motion about the origin, O Polar coordinates are convenient to use to represent the position of P (or any other point) P is located at (r, q) where r is the distance from the origin to P and q is the measured counterclockwise from the reference line

4 Angular Position, 3 As the particle moves, the only coordinate that changes is q As the particle moves through q, it moves though an arc length s. The arc length and r are related: s = q r

5 Radian This can also be expressed as q is a pure number, but commonly is given the artificial unit, radian One radian is the angle subtended by an arc length equal to the radius of the arc

6 Conversions Comparing degrees and radians 1 rad = = 57.3 Converting from degrees to radians θ [rad] = [degrees]

7 Angular Displacement The angular displacement is defined as the angle the object rotates through during some time interval q q q This is the angle that the reference line of length r sweeps out f i

8 Average Angular Speed The average angular speed, ω, of a rotating rigid object is the ratio of the angular displacement to the time interval q q q f i t t t f i

9 Instantaneous Angular Speed The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero lim q dq t 0 t dt

10 Angular Speed, final Units of angular speed are radians/sec rad/s or s -1 since radians have no dimensions Angular speed will be positive if θ is increasing (counterclockwise) Angular speed will be negative if θ is decreasing (clockwise)

11 Average Angular Acceleration The average angular acceleration, a, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change: a f i t t t f i

12 Instantaneous Angular Acceleration The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0 a lim t 0 t d dt

13 Directions, details Strictly speaking, the speed and acceleration (, a) are the magnitudes of the velocity and acceleration vectors The directions are actually given by the right-hand rule

14 Rotational Kinematic Equations with constant angular acceleration at f i q q t 1 f i i 2 at 2 a ( q q ) 2 2 f i f i 1 q q 2 2 ) f i i f t

15 Comparison Between Rotational and Linear Equations

16 Relationship Between Angular and Linear Quantities Displacements s Speeds v q r Accelerations a r a r Every point on the rotating object has the same angular motion Every point on the rotating object does not have the same linear motion

17 Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration Therefore, each point on a rotating rigid object will experience a centripetal acceleration a C v r 2 2 r

18 Rotational Motion Example For a compact disc player to read a CD, the angular speed must vary to keep the tangential speed constant (v t = r) At the inner sections, the angular speed is faster than at the outer sections

19 Inertia of Rotation Consider Newton s second law for the inertia of rotation to be patterned after the law for translation. F = 20 N a = 4 m/s 2 F = 20 N R = 0.5 m a = 2 rad/s 2 Linear Inertia, m 24 N m = 4 m/s 2 = 5 kg Rotational Inertia, I t (20 N)(0.5 m) I = = = 2.5 kg m a 4 m/s 2 2 Force does for translation what torque does for rotation:

20 Common Rotational Inertias L L I 1 3 ml 2 I 1 12 ml 2 R R R I = mr 2 I = ½mR 2 I 2 5 mr Hoop Disk or cylinder Solid sphere 2

21 Moments of Inertia of Various Rigid Objects

22 Rotational Kinetic Energy Consider tiny mass m: v = R K = ½mv 2 K = ½m(R) 2 m 1 m m 4 m 3 K = ½(mR 2 ) 2 axis m 2 Sum to find K total: K = ½(SmR 2 ) 2 (½ 2 same for all m ) Object rotating at constant. Rotational Inertia Defined: I = SmR 2

23 Example 1: What is the rotational kinetic energy of the device shown if it rotates at a constant speed of 600 rpm? First: I = SmR 2 I = (3 kg)(1 m) 2 + (2 kg)(3 m) 2 + (1 kg)(2 m) 2 2 kg 3 m 2 m 3 kg 1 m 1 kg I = 25 kg m 2 = 600 rpm = 62.8 rad/s K = ½Iw 2 = ½(25 kg m 2 )(62.8 rad/s) 2 K = 49,300 J

24 Example 2: A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias. I mr 2 (3 kg)(0.2 m) 2 R I = kg m 2 I = mr 2 Hoop R I mr (3 kg)(0.2 m) I = ½mR 2 I = kg m 2 Disk

25 Parallel-Axis Theorem In the previous examples, the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis, the parallel-axis theorem often simplifies calculations The theorem states I = I CM + MD 2 I is about any axis parallel to the axis through the center of mass of the object I CM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis

26 Parallel-Axis Theorem Example The axis of rotation goes through O The axis through the center of mass is shown The moment of inertia about the axis through O would be I O = I CM + MD 2

27 Moment of Inertia for a Rod Rotating Around One End The moment of inertia of the rod about its center is ICM D is ½ L Therefore, I I MD CM ML 2 2 L I ML M ML

28 Definition of Torque Torque is defined as the tendency to produce a change in rotational motion. Examples:

29 Torque Torque, t, is the tendency of a force to rotate an object about some axis Torque is a vector t = r F sin f = F d F is the force f is the angle the force makes with the horizontal d is the moment arm (or lever arm)

30 Torque is Determined by Three Factors: The magnitude of the applied force. The direction of the applied force. The location of the applied force. Each The forces 40-N of the force nearer 20-N the produces forces end of has the twice wrench a different the torque have greater as due does to torques. the the 20-N direction force. of force. Direction Magnitude Location of of Force of force force 20 N q q N 20 N N 20 N 40 N 20 N 20 N

31 Torque, cont The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force d = r sin Φ

32 Torque, final The horizontal component of F (F cos f) has no tendency to produce a rotation Torque will have direction If the turning tendency of the force is counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will be negative

33 Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. Positive torque: Counter-clockwise, out of page ccw cw Negative torque: clockwise, into page

34 Net Torque The force F 1 will tend to cause a counterclockwise rotation about O The force F 2 will tend to cause a clockwise rotation about O St t 1 t 2 F 1 d 1 F 2 d 2

35 Torque vs. Force Forces can cause a change in linear motion Described by Newton s Second Law Forces can cause a change in rotational motion The effectiveness of this change depends on the force and the moment arm The change in rotational motion depends on the torque

36 Torque Units The SI units of torque are N. m Although torque is a force multiplied by a distance, it is very different from work and energy The units for torque are reported in N. m and not changed to Joules

37 The Moment Arm The moment arm of a force is the perpendicular distance from the line of action of a force to the axis of rotation. F 1 r F 2 r r F 3

38 Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r. r = 12 cm sin 60 0 = 10.4 cm t = (80 N)(0.104 m) = 8.31 N m

39 Alternate: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. positive 12 cm Resolve 80-N force into components as shown. Note from figure: r x = 0 and r y = 12 cm t = (69.3 N)(0.12 m) t = 8.31 N m as before

40 Example 2: Find resultant torque about axis A for the arrangement shown below: Find t due to each force. Consider 20-N force first: 30 N m 2 m 40 N negative r A m 20 N r = (4 m) sin 30 0 = 2.00 m t = Fr = (20 N)(2 m) = 40 N m, cw The torque about A is clockwise and negative. t 20 = -40 N m

41 Example 2 (Cont.): Next we find torque due to 30-N force about same axis A. Find t due to each force. Consider 30-N force next. 30 N m 2 m 40 N r negative A m 20 N r = (8 m) sin 30 0 = 4.00 m t = Fr = (30 N)(4 m) = 120 N m, cw The torque about A is clockwise and negative. t 30 = -120 N m

42 Example 2 (Cont.): Finally, we consider the torque due to the 40-N force. Find t due to each force. Consider 40-N force next: 30 N m 2 m 40 N positive r A m 20 N r = (2 m) sin 90 0 = 2.00 m t = Fr = (40 N)(2 m) = 80 N m, ccw The torque about A is CCW and positive. t 40 = +80 N m

43 Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: Resultant torque is the sum of individual torques. 30 N m 2 m 40 N A m 20 N t R = t 20 + t 20 + t 20 = -40 N m -120 N m + 80 N m t R = - 80 N m Clockwise

44 Torque and Angular Acceleration Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force F t The tangential force provides a tangential acceleration: F t = ma t

45 Torque and Angular Acceleration, Particle cont. The magnitude of the torque produced by F t around the center of the circle is t = F t r = (ma t ) r The tangential acceleration is related to the angular acceleration t = (ma t ) r = (mra) r = (mr 2 ) a Since mr 2 is the moment of inertia of the particle, t = Ia The torque is directly proportional to the angular acceleration and the constant of proportionality is the moment of inertia

46 Torque and Angular Acceleration, Extended cont. From Newton s Second Law df t = (dm) a t The torque associated with the force and using the angular acceleration gives dt = r df t = a t r dm = ar 2 dm Finding the net torque 2 2 t ar dm ar dm This becomes St Ia

47 Torque and Angular Acceleration, Extended final This is the same relationship that applied to a particle The result also applies when the forces have radial components The line of action of the radial component must pass through the axis of rotation These components will produce zero torque about the axis

48 Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. x f m A resultant force F produces negative acceleration a for a mass m. t I R 4 kg o 50 rad/s t = 40 N m A resultant torque t produces angular acceleration a of disk with rotational inertia I. F ma t I a

49 Newton s 2nd Law for Rotation How many revolutions required to stop? t = Ia F R 4 kg o 50 rad/s R = 0.20 m F = 40 N a FR = (½mR 2 )a 2F 2(40N) mr (4 kg)(0.2 m) q 0 2aq 2 f - 2 o (50 rad/s) 2 2a 2(100 rad/s ) a = 100 rad/s 2 q = 12.5 rad = 1.99 rev

50 Example 3: What is the linear acceleration of the falling 2-kg mass? Apply Newton s 2nd law to rotating disk: t Ia TR = (½MR 2 )a T = ½MRa a T = ½MR( ) ; R but a = ar; a = and T = ½Ma Apply Newton s 2nd law to falling mass: mg - T = ma mg - ½Ma T = ma (2 kg)(9.8 m/s 2 ) - ½(6 kg) a = (2 kg) a 19.6 N - (3 kg) a = (2 kg) a a = 3.92 m/s 2 a R R = 50 cm M 6 kg a =? R = 50 cm 6 kg +a 2 kg T T mg 2 kg

51 Torque and Angular Acceleration, Wheel Example The wheel is rotating and so we apply St Ia The tension supplies the tangential force The mass is moving in a straight line, so apply Newton s Second Law SF y = ma y = mg - T

52 Torque and Angular Acceleration, Multi-body Ex., 1 Both masses move in linear directions, so apply Newton s Second Law Both pulleys rotate, so apply the torque equation

53 Torque and Angular Acceleration, Multi-body Ex., 2 The mg and n forces on each pulley act at the axis of rotation and so supply no torque Apply the appropriate signs for clockwise and counterclockwise rotations in the torque equations

54 Work in Rotational Motion Find the work done by F on the object as it rotates through an infinitesimal distance ds = r dq dw = F. d s = (F sin f) r dq dw = t dq The radial component of F does no work because it is perpendicular to the displacement

55 Power in Rotational Motion The rate at which work is being done in a time interval dt is Power dw dq t t dt dt This is analogous to P = Fv in a linear system

56 Work-Kinetic Energy Theorem in Rotational Motion The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object s rotational kinetic energy f W I d I f Ii i 2 2

57 Work and Power for Rotation Work = Fs = FRq Work = tq t FR q s F Work Power = = t tq t = q t s = Rq F Power = t Power = Torque x average angular velocity

58 Example 4: The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s. Work = tq = FR q s 20 m q = = = 50 rad R 0.4 m F = mg = (2 kg)(9.8 m/s 2 ); F = 19.6 N Work = (19.6 N)(0.4 m)(50 rad) s q 2 kg 6 kg F F=W s = 20 m Work = 392 J Work Power = = t 392 J 4s Power = 98 W

59 The Work-Energy Theorem Recall for linear motion that the work done is equal to the change in linear kinetic energy: 2 ½ 2 f 0 Fx ½mv mv Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: 2 ½I 2 f 0 tq ½I

60 Two Kinds of Kinetic Energy Kinetic Energy of Translation: K = ½mv 2 Kinetic Energy of Rotation: K = ½I 2 R P v Total Kinetic Energy of a Rolling Object: T K mv I

61 Angular/Linear Conversions In many applications, you must solve an equation with both angular and linear parameters. It is necessary to remember the bridges: Displacement: sqr s q R Velocity: vr v R Acceleration: v ar a a R

62 Example 5: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v. Compare the kinetic energies. Two kinds of energy: K T = ½mv 2 K r = ½I 2 v v Total energy: E = ½mv 2 + ½I 2 = v R ½ ½ ½ ) v E mv mr 2 R ½ ½ ) v E mv mr 2 R Disk: E = ¾mv 2 Hoop: E = mv 2

63 Conservation of Energy The total energy is still conserved for systems in rotation and translation. However, rotation must now be considered. Begin: (U + K t + K R ) o = End: (U + K t + K R ) f Height? Rotation? velocity? mgh o ½I o 2 ½mv o 2 = mgh f ½I f 2 ½mv f 2 Height? Rotation? velocity?

64 Example 6: Find the velocity of the 2-kg mass just before it strikes the floor. R = 50 cm mgh o ½I o 2 ½mv o 2 = mgh f ½I f 2 ½mv f 2 6 kg 2 kg h = 10 m mgh mv I v mgh0 2 mv 2 ( 2 MR ) 2 R (2)(9.8)(10) (2) v (6) v I MR 2 2.5v 2 = 196 m 2 /s 2 v = 8.85 m/s

65 Applying the Work-Energy Theorem: What work is needed to stop wheel rotating: Work = K r F R 4 kg o 60 rad/s R = 0.30 m F = 40 N First find I for wheel: I = mr 2 = (4 kg)(0.3 m) 2 = 0.36 kg m 2 0 tq ½I ½I Work = -½I 2 o 2 2 f 0 Work = -½(0.36 kg m 2 )(60 rad/s) 2 Work = -648 J

66 Summary of Useless Equations

67 Pure Rolling Motion In pure rolling motion, an object rolls without slipping In such a case, there is a simple relationship between its rotational and translational motions

68 Rolling Object, Center of Mass The velocity of the center of mass is ds dq v R R CM dt dt The acceleration of the center of mass is dv dt d dt CM acm R R a

69 Rolling Object, Other Points A point on the rim, P, rotates to various positions such as Q and P At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs

70 Combined Rotation and Translation v cm v cm v cm Now consider a ball rolling without slipping. The angular velocity about the point P is same as for disk, so that we write: v R First consider a disk sliding without friction. The velocity of any part is equal to velocity v cm of the center of mass. R Or v R P v

71 Translation or Rotation? If you are to solve for a linear parameter, you must convert all angular terms to linear terms: q s R v R a a R I (?) mr 2 If you are to solve for an angular parameter, you must convert all linear terms to angular terms: s qr v R var

72 Example (a): Find velocity v of a disk if given its total kinetic energy E. Total energy: E = ½mv 2 + ½I 2 E mv I ; I mr ; v E mv mr E mv mv R ) ; 2 4 E 2 3mv 4 E or v 4 3m v R

73 Example (b) Find angular velocity of a disk given its total kinetic energy E. Total energy: E = ½mv 2 + ½I 2 E mv I ; I mr ; v R ) E m( R) mr ; E mr mr E 2 2 3mR 4 E or 4 3mR 2

74 Example 7: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m? mgh o = ½mv 2 + ½I 2 Hoop: I = mr v mgh0 ½mv ½( mr ) 2 R mgh o = ½mv 2 + ½mv 2 ; mgh o = mv 2 20 m 2 v gh 0 (9.8 m/s )(20 m) Hoop: v = 14 m/s Disk: I = ½mR 2 ; mgh o = ½mv 2 + ½I v mgh0 ½mv ½(½ mr ) 2 R v gh v = 16.2 m/s

75 Angular Momentum Defined Consider a particle m moving with velocity v in a circle of radius r. Define angular momentum L: L = mvr Substituting v= r, gives: L = m(r) r = mr 2 For extended rotating body: L = (Smr 2 ) v = r axis m 1 m m 4 m 3 m 2 Object rotating at constant. Since I = Smr 2, we have: L = I Angular Momentum

76 Example 8: Find the angular momentum of a thin 4-kg rod of length 2 m if it rotates about its midpoint at a speed of 300 rpm. L = 2 m m = 4 kg For rod: I = ml 2 = (4 kg)(2 m) 2 I = 1.33 kg m 2 rev 2 rad 1 min rad/s min 1 rev 60 s L = I (1.33 kg m 2 )(31.4 rad/s) 2 L = 1315 kg m 2 /s

77 Angular Momentum Consider a particle of mass m located at the vector position r and moving with linear momentum p r F t r Adding the term t d( r p) dt dr dt dp dt p

78 Angular Momentum, cont The instantaneous angular momentum L of a particle relative to the origin O is defined as the cross product of the particle s instantaneous position vector r and its instantaneous linear momentum p L = r x p

79 Torque and Angular Momentum The torque is related to the angular momentum Similar to the way force is related to linear momentum t d L dt This is the rotational analog of Newton s Second Law St and L must be measured about the same origin This is valid for any origin fixed in an inertial frame

80 More About Angular Momentum The SI units of angular momentum are (kg. m 2 )/ s Both the magnitude and direction of L depend on the choice of origin The magnitude of L = mvr sin f f is the angle between p and r The direction of L is perpendicular to the plane formed by r and p

81 Angular Momentum of a Particle, Example The vector L = r x p is pointed out of the diagram The magnitude is L = mvr sin 90 o = mvr sin 90 o is used since v is perpendicular to r A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path

82 Angular Momentum of a System of Particles The total angular momentum of a system of particles is defined as the vector sum of the angular momenta of the individual particles L tot = L 1 + L L n = SL i Differentiating with respect to time dl dt tot i dl dt i i t i

83 Angular Momentum of a Rotating Rigid Object Each particle of the object rotates in the xy plane about the z axis with an angular speed of The angular momentum of an individual particle is L i = m i r i 2 L and are directed along the z axis

84 Angular Momentum of a Rotating Rigid Object, cont To find the angular momentum of the entire object, add the angular momenta of all the individual particles 2 Lz Li miri I i i This also gives the rotational form of Newton s Second Law t ext dl dt d dt z I Ia

85 Angular Momentum of a Rotating Rigid Object, final The rotational form of Newton s Second Law is also valid for a rigid object rotating about a moving axis provided the moving axis: (1) passes through the center of mass (2) is a symmetry axis If a symmetrical object rotates about a fixed axis passing through its center of mass, the vector form holds: L = I where L is the total angular momentum measured with respect to the axis of rotation

86 Angular Momentum of a Bowling Ball The momentum of inertia of the ball is 2/5MR 2 The angular momentum of the ball is L z = I The direction of the angular momentum is in the positive z direction

87 Impulse and Momentum Recall for linear motion the linear impulse is equal to the change in linear momentum: F t mv mv Using angular analogies, we find angular impulse to be equal to the change in angular momentum: f 0 t t I I f 0

88 Example 9: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for s. What is the final angular velocity? I = mr 2 = (2 kg)(0.4 m) 2 I = 0.32 kg m 2 Applied torque t FR t = s R F 2 kg o 0 rad/s R = 0.40 m F = 200 N Impulse = change in angular momentum t t = I f I o 0 FR t = I f f = 0.5 rad/s

89 Conservation of Momentum In the absence of external torque the rotational momentum of a system is conserved (constant). 0 I f f I o o = t t I f f I o o I o = 2 kg m 2 ; o = 600 rpm I f = 6 kg m 2 ; o =? f I00 I f 2 (2 kg m )(600 rpm) 6 kg m 2 f = 200 rpm

90 Summary Rotational Analogies Quantity Linear Rotational Displacement Displacement x Radians q Inertia Mass (kg) I (kgm 2 ) Force Newtons N Torque N m Velocity v m/s Rad/s Acceleration a m/s 2 a Rad/s 2 Momentum mv (kg m/s) I (kgm 2 rad/s)

91 The Vector Product Torque can also be found by using the vector product of force F and position vector r. For example, consider the figure below. Torque r F Sin q q F The effect of the force F at angle q (torque) is to advance the bolt out of the page. Magnitude: (F Sin q)r Direction = Out of page (+).

92 Definition of a Vector Product The magnitude of the vector (cross) product of two vectors A and B is defined as follows: A x B = l A l l B l Sin q In our example, the cross product of F and r is: F x r = l F l l r l Sin q Magnitude only q F Sin q r F In effect, this becomes simply: (F Sin q) r or F (r Sin q)

93 Example: Find the magnitude of the cross product of the vectors r and F drawn below: Torque 12 lb r x F = l r l l F l Sin q 6 in r x F = (6 in.)(12 lb) Sin 60 0 r x F = 62.4 lb in. 6 in. r x F = l r l l F l Sin q 60 0 Torque 12 lb r x F = (6 in.)(12 lb) Sin r x F = 62.4 lb in. Explain difference. Also, what about F x r?

94 Direction of the Vector Product. The direction of a vector product is determined by the right hand rule. A x B = C (up) B x A = -C (Down) What is direction of A x C? A C B A -C Curl fingers of right hand in direction of cross pro-duct (A to B) or (B to A). Thumb will point in the direction of product C. B

95 Torque Example: What are the magnitude and direction of the cross product, r x F? 10 lb 50 0 r x F = l r l l F l Sin q r x F = (6 in.)(10 lb) Sin in. r x F = 38.3 lb in. Magnitude F r Direction by right hand rule: Out of paper (thumb) or +k Out r x F = (38.3 lb in.) k What are magnitude and direction of F x r?

96 Cross Products Using (i,j,k) y j k z Consider 3D axes (x, y, z) i x Define unit vectors, i, j, k Consider cross product: i x i i i i x i = (1)(1) Sin 0 0 = 0 Magnitudes are zero for parallel vector products. j x j = (1)(1) Sin 0 0 = 0 k x k = (1)(1)Sin 0 0 = 0

97 Vector Products Using (i,j,k) z k y j i j i x Consider 3D axes (x, y, z) Define unit vectors, i, j, k Consider dot product: i x j i x j = (1)(1) Sin 90 0 = 1 Magnitudes are 1 for perpendicular vector products. j x k = (1)(1) Sin 90 0 = 1 k x i = (1)(1) Sin 90 0 = 1

98 Vector Product (Directions) k y j i x Directions are given by the right hand rule. Rotating first vector into second. z j i x j = (1)(1) Sin 90 0 = +1 k k i j x k = (1)(1) Sin 90 0 = +1 i k x i = (1)(1) Sin 90 0 = +1 j

99 Vector Products Practice (i,j,k) z k y j j i x Directions are given by the right hand rule. Rotating first vector into second. i x k =? k x j =? - j (down) - i (left) j x -i =? + k (out) k i 2 i x -3 k =? + 6 j (up)

100 Using i,j Notation - Vector Products Consider: A = 2 i - 4 j and B = 3 i + 5 j A x B = (2 i - 4 j) x (3 i + 5 j) = 0 k -k 0 (2)(3) ixi + (2)(5) ixj + (-4)(3) jxi + (-4)(5) jxj A x B = (2)(5) k + (-4)(3)(-k) = +22 k Alternative: A = 2 i - 4 j B = 3 i + 5 j Evaluate determinant A x B = 10 - (-12) = +22 k

101 Conservation of Angular Momentum The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero Net torque = 0 -> means that the system is isolated L tot = constant or L i = L f For a system of particles, L tot = SL n = constant

102 Conservation of Angular Momentum, cont If the mass of an isolated system undergoes redistribution, the moment of inertia changes The conservation of angular momentum requires a compensating change in the angular velocity I i i = I f f This holds for rotation about a fixed axis and for rotation about an axis through the center of mass of a moving system The net torque must be zero in any case

103 Conservation Law Summary For an isolated system - (1) Conservation of Energy: E i = E f (2) Conservation of Linear Momentum: p i = p f (3) Conservation of Angular Momentum: L i = L f

104 Conservation of Angular Momentum: The Merry-Go-Round The moment of inertia of the system is the moment of inertia of the platform plus the moment of inertia of the person Assume the person can be treated as a particle As the person moves toward the center of the rotating platform, the angular speed will increase To keep L constant

105 Motion of a Top The only external forces acting on the top are the normal force n and the gravitational force M g The direction of the angular momentum L is along the axis of symmetry The right-hand rule indicates that t = r F = r M g is in the xy plane

106 Motion of a Top, cont The direction of d L is parallel to that of t in part. The fact that L f = d L + L i indicates that the top precesses about the z axis. The precessional motion is the motion of the symmetry axis about the vertical The precession is usually slow relative to the spinning motion of the top

107 Gyroscope A gyroscope can be used to illustrate precessional motion The gravitational force Mg produces a torque about the pivot, and this torque is perpendicular to the axle The normal force produces no torque

108 Gyroscope, cont The torque results in a change in angular momentum d L in a direction perpendicular to the axle. The axle sweeps out an angle df in a time interval dt. The direction, not the magnitude, of L is changing The gyroscope experiences precessional motion

109 Gyroscope, final To simplify, assume the angular momentum due to the motion of the center of mass about the pivot is zero Therefore, the total angular momentum is L = I due to its spin This is a good approximation when is large

110 Precessional Frequency Analyzing the previous vector triangle, the rate at which the axle rotates about the vertical axis can be found p df dt Mgh I p is the precessional frequency

111 Gyroscope in a Spacecraft The angular momentum of the spacecraft about its center of mass is zero A gyroscope is set into rotation, giving it a nonzero angular momentum The spacecraft rotates in the direction opposite to that of the gyroscope So the total momentum of the system remains zero