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Chapter 6 Rotation and Angular Momentum. Chapter 6 Rotation and Angular Momentum Basic Requirements： 1. Understand the concept of translation and rotation; 2. Master the kinematic equations for constant angular acceleration; 3. Master the relationship between the linear and angular variables; 4. Master kinetic energy of rotation; 5. Master the calculation of the rotational inertia; 6. Master the parallel-axis theorem; 7. Learn to apply Newton's second law for rotation; 8. Master the work-kinetic energy theorem for rotation. 9. Understand the concept of rolling; 10. Master the kinetic energy of rolling; 11. Master the forces of rolling; 12. Master angular momentum; 13. Learn to apply Newton's second law in angular form; 14. Understand the angular momentum of a system of particles; 15. Understand the angular momentum of a rigid body rotating about a fixed axis; 16. Master the conservation law of angular momentum. Review and Summary Static Equilibrium A rigid body at rest is said to be in static equilibrium. For such a body , the vector sum of the external forces acting on it is zero: Fnet 0 (balance of forces ) (6-3) If all the forces lie in the xy plane, this vector equation is equivalent to two component equations: Fn e, t 0 x and Fnet , y 0 (balance of forces) (6-7,6-8) Static equilibrium also implies that the vector sum of the external torques acting on the - 135 - Chapter 6 Rotation and Angular Momentum. body about any point is zero, or net 0 (balance of torques) (6-5) If the forces lie in the xy plane, all torque vectors are parallel to the z axis, and Eq.6-5 is equivalent to the single component equation net , z 0 (balance of torques). (6-9) Angular Position To describe the rotation of a rigid body about a fixed axis, called the rotation axis, we assume a reference line is fixed in the body, perpendicular to that axis and rotating with the body. We measure the angular position of this line relative to a fixed direction. When is measured in radians, s (radian measure), (6-10) r Where s is the arc length, of a circular path of radius r and angle . Radian measure is related to angle measure in revolutions and degrees by 1r e v 3 6 0 2 r a d . (6-11) Angular Displacement A body that rotates about a rotation axis , changing its angular position from 1 to 2 , undergoes an angular displacement 2 1 , (6-13) Where is positive for counterclockwise rotation and negative for clockwise rotation. Angular Velocity and Speed If a body rotates through an angular displacement in a time interval t , its average angular velocity avg is avg . (6-14) t The (instantaneous) angular velocity of the body is d . (6-15) dt Both avg and are vectors, with directions given by the right hand rule. they - 136 - Chapter 6 Rotation and Angular Momentum. are positive for counterclockwise rotation and negative for clockwise rotation. The magnitude of the body’s angular velocity is the angular speed. Angular Acceleration If the angular velocity of a body changes from 1 to 2 in a time interval t t1 t 2 , the average angular acceleration avg of the body is 2 1 avg (6-16) t2 t1 t The (instantaneous) angular acceleration of a body is d . (6-17) dt Both avg and are vectors. The Kinematic Equations for Constant Angular Acceleration Constant angular acceleration ( =constant) is an important special case of rotational motion .The appropriate kinematic equations are 0 at (6-18) 1 0 0t t 2 (6-19) 2 2 02 2 ( 0 ) (6-20) 1 0 ( 0 )t (6-21) 2 1 0 t t 2 (6-22) 2 Linear and Angular Variables Related A point in a rigid rotating body , at a perpendicular distance r from the rotation axis , moves in a circle with radius r . If the body rotates through an angle , the point moves along an arc with length s given by s r (radian measure), (6-23) Where is in radians . The liner velocity v of the point is tangent to the circle, the point’s liner speed is - 137 - Chapter 6 Rotation and Angular Momentum. given by v r (radian measure), (6-24) Where is the angular speed (in radians pre second) of the body . The liner acceleration a of the point has both tangential and radial components. The tangential component is at r (radian measure), (6-28) Where is the magnitude of the angular acceleration (in radians per second-squared ) of the body . The radial component of a is v2 ar 2r (radian measure), (6-28) r If the point moves in uniform circular motion. the period T of the motion for the point and the body is 2 r 2 T (radian measure) (6-25,6-26) v Rotational Kinetic Energy and Rotational Inertia The kinetic energy K of a rigid body rotating about a fixed axis is given by 1 2 K I (radian measure) (6-40) 2 In which I is the rotational inertia of the body, defined as I mi ri 2 (6-39) for a system of discrete particles and as I r 2 dm (6-43) for a body with continuously distributed mass. The r and ri in these expressions represent the perpendicular distance from the axis of rotation to each mass element in the body. The Parallel-Axis Theorem The parallel-axis theorem relates the rotational inertia I of a body about any axis to that of the same body about a parallel axis through the center of mass: I I com Mh 2 (6-42) - 138 - Chapter 6 Rotation and Angular Momentum. Here h is the perpendicular distance between the two axes. Torque Torque is a turning or twisting action on a body about a rotation axis due to a force F . If F is exerted at a point given by the position vector r relative to the axis, then the magnitude of the torque is (r )( F sin ) Where Ft is the component of F perpendicular to r , and is the angle between r and F . The quantity r is the perpendicular distance between the rotation axis and an extended line running through the F vector. This line is called the of action of F . Similarly, r is the moment arm of Ft . The SI unit of torque is the Newton-meter ( N m ). A torque is positive if it tends to rotate a body at rest counterclockwise and negative if it tends to rotate the body in the clockwise direction. Newton’s Second Law in Angular Form The rotational analog of Newton’s second law is n e t I , (6-45) Where net is the net torque acting on a particle or rigid body, I is the rotational inertia of the particle or body about that rotation axis. and is the resulting angular acceleration about that axis. Work and Rotational Kinetic Energy The equations used for calculating work and power in rotational motion correspond to equations used for translational motion and are f W d (6-56) i When is constant, Eq. 6-56 reduces to W ( f i ) (6-57) dW And P (6-58) dt The form of the work-kinetic energy theorem used for rotating bodies is - 139 - Chapter 6 Rotation and Angular Momentum. 1 2 1 2 K K f Ki I f I i W (6-60) 2 2 Rolling Bodies For a wheel of radius R that is rolling smoothly (no sliding), com R (6-67) Where com is the linear speed of the wheel’s center and is the angular speed of the wheel about its center. The wheel may also be viewed as rotating instantaneously about the point P of the “road” that is in contact with the wheel. The angular speed of the wheel about this point is the same as the angular speed of the wheel about its center. The rolling wheel has kinetic energy 1 1 K I com 2 M 2com (6-70) 2 2 Where I com is the rotational moment of the wheel about its center and M is the mass of the wheel. If the wheel is being accelerated but is still rolling smoothly, the acceleration of the center of mass acom is related to the angular acceleration about the center with acom R (6-71) If the wheel rolls smoothly down a ramp of angle , its acceleration along an x axis extending up the ramp is g sin acom , x (6-75) 1 I com / MR 2 Torque as a Vector In three dimensions, torque is a vector quantity defined relative to a fixed point (usually an origin); it is r F, (6-76) Where F is a force applied to a particle and r is a position vector locating the particle relative to the fixed point (or origin). The magnitude of is given by rF sin rF r F , (6-77,6-78,6-79) - 140 - Chapter 6 Rotation and Angular Momentum. Where is the angle between F and r , F is the component of F perpendicular to r , and r is the moment arm of F . The direction of is given by the right-hand rule for cross products. Angular Momentum of a particle The angular momentum of a particle with linear momentum p , mass m , and linear velocity is a vector quantity defined relative to a fixed point (usually origin); it is r p m(r ) (6-80) The magnitude of is given by rm sin (6-81) rp rm (6-82) r p r m (6-83) Where is the angle between r and p , p and are the components of p and perpendicular to r , and r is the perpendicular distance between the fixed point and the extension of p . The direction of is given by the right-hand rule for cross products. Newton’s Second Law in Angular Form Newton’s second law for a particle can be written in angular form as d net , (6-85) dt Where net is the net torque acting on the particle, and is the angular momentum of the particle. Angular Momentum of a System of Particles The angular momentum L of a system of particles is the vector sum of the angular momentum of the individual particles: n L 1 2 n i (6-88) i 1 The time rate of change of this angular momentum is equal to the net external torque on the system (the vector sum of the torques due to interactions of the particles of the - 141 - Chapter 6 Rotation and Angular Momentum. system with particles external to the system): dL net (system of particles) (6-91) dt Angular Momentum of a Rigid Body For a rigid body rotating about a fixed axis is L I (rigid body, fixed axis). (6-93) Conservation of Angular Momentum The angular momentum L of a system remains constant if the net external torque acting on the system is zero: L acons tan t (isolated system) (6-94) Or Li L f (isolated system) . (6-95) This is the law of conservation of angular momentum. It is one of the fundamental conservation laws of nature, having been verified even in situations (involving high-speed particles or subatomic dimensions) in which Newton’s laws are not applicable. Examples Example 1 .Show that the moment of inertia of a uniform hollow cylinder of inner 1 radius R1 ,outer radius R2 ,and mass M , is I M ( R12 R2 ) ,as stated in the 2 2 figure, if the rotation axis is through the center along the axis of symmetry. 2 Solution：We know that the moment of inertia of a thin ring of radius R is m R . So we divide the cylinder into thin concentric cylindrical rings or hoops of thickness dR , one of which is indicated in Fig.6-1. If the density (mass per unit volume ) is ,then dm dV , Where dV is the volume of the thin ring of radius R , thickness dR , and height h . Since dV (2R)(dR)(h) Fig. 6-1 Example 1 We have - 142 - Chapter 6 Rotation and Angular Momentum. dm 2hR dR Then the moment of inertia is obtained by integrating (summing) over all these hoops : I R 2 dm R1 R 4 R14 2hR 3 dR 2h 2 ， 4 R2 Where we are given that the cylinder has uniform density , constant .(If this were not so ,we would have to know as a function of R before the integration could be carried out) The volume V of this hollow cylinder is V (R2 R1 )h ,so its 2 2 mass M is M v ( R2 R12 )h. 2 Since ( R2 R1 ) ( R2 R1 )( R2 R1 ) ,we have 4 4 2 2 2 2 h 1 I ( R2 R12 )( R2 R12 ) 2 2 M ( R12 R2 ) 2 (Answer) 2 2 As stated in figure. As a check ,note that for a solid cylinder , R1 0 and we obtain, with R2 R0 : 1 I 2 MR0 , 2 Which is that given in Fig.10-21c for a solid cylinder of mass M and radius R 0 . Example 2 what will be the speed of a solid sphere of mass M and radius R 0 when it reaches the bottom of an incline if it starts from rest at a vertical height H and rolls without slipping ? see Fig.6-2. Ignore losses due to dissipative forces ,and compare your result to that for an object sliding down a frictionless incline. - 143 - Chapter 6 Rotation and Angular Momentum. Solution ： We use the law of conservation of energy, and we must now include rotational kinetic energy. The total energy at any point a vertical distance y above the base of the incline is 1 1 Mv 2 I CM 2 Mgy , 2 2 Where v is the speed of the CM . We equate Fig. 6-2 Example 2 the total energy at the top （ y H and v 0) to the total energy at the bottom ( y 0 ); 1 1 0 0 MgH Mv 2 I CM w 2 0 . 2 2 The moment of inertia of a solid sphere about an axis through its COM is 2 ICOM MR0 . 2 5 Since the sphere rolls without slipping , the speed , v , of the center of mass with respect to the point of contact (which is momentarily at rest at any instant) is equal to the speed of a point on the edge relative to the center. We therefore have w v / R . Hence 12 v 2 1 Mv 2 MR02 2 MgH . 2 25 R0 Dividing out the M , s and R , s, we obtain 1 1 2 v gH 2 5 So 10 v gH . (Answer) 7 Note first that v is independent of both the mass M and the radius R of the sphere . Also , we can compare this result for the speed of a rolling sphere to that for an object - 144 - Chapter 6 Rotation and Angular Momentum. 1 sliding down a plane without rotating and without friction ( Mv 2 mgH ), 2 in which case v 2 gH , which is greater. An object sliding without friction transforms its initial potential energy into translational kinetic energy (none into rotational kinetic energy ), so its speed is greater. Example 3 Suppose a 60-kg person stands at the edge of a 6.0-m-diameter circular platform, which is mounted on frictionless bearings and has a moment of 2 inertia of 1800kg. m . The platform is at rest initially, but when the person begins running at a speed of 4.2m/s (with respect to the ground).around its edge the platform begins to rotate in the opposite direction as in figure 6-3. Calculate the angular velocity of the platform. Fig. 6-3 Example 3 Solution： The total angular momentum is zero initially .Since there is no net torque, L is conserved and will remain zero ,as in Fig.6-3.The person’s angular momentum is L per (mR )(v / R) and we take this as positive .the angular 2 momentum of the platform is Lplat I .Thus L Lper Lplat v 0 mR 2 I R So m . m R v ( 6 0 k g) ( 3 . 0 ) ( 4m 2 s / ) 2 . 4 r a d /.s 0 2 (Answer) I 1800g mk . The frequency of rotation is f w / 2 0.067rew / s and the period T 1 / f 15s per revolution. (Answer) - 145 - Chapter 6 Rotation and Angular Momentum. Example 4 A uniform rod of length l and mass m is placed horizontally by putting its left end B on the edge of a table and holding its right end A with your hand . Then you release the end A suddenly . find , at the instant end A is released (a) the acceleration a of the rod’s center of mass ? (b) the force F exerted on the rod at end B ? Solution: Draw the free-body diagram of the rod .set the coordinate system as in the figure. At the moment the end A is released there are two forces acting on the rod: the gravitational force F mg exerted at the g center of the rod , downward. The force F n B A exerted at end B from the table. Apply the Newton’s second law for the center of mass of the rod. Fig. 6-4 Example 4 F ma i c In direction: F mg ma c (1) vc 2 In n direction: Fn mac m (2) l/2 According to the theorem of rotation we write: l mg I z (3) 2 from the relationship between the linear and angular quantities we have l ac (4) 2 1 and I z ml 2 (5) 3 At the moment end A is just released .the speed of the rod’s center of mass v 0 c thus v 0 c Substituting v 0 into Eq. (2), yields F 0 c N Then solving equations (3), (4) and (5) for a yields c - 146 - Chapter 6 Rotation and Angular Momentum. 1 1 ac g .a gˆ 3 3 Substituting ac into Eq. (1) we obtain 1 1 F mg F ˆ mg (Answer) 4 4 Example 5 A uniform solid ball of radius r and mass m rolls down a circular track from rest without sliding .The track, radius R , is in the vertical plane. At the beginning the ball is in the same height as the center of the circular track , Find (a) the ball’s speed as it rolls down to the bottom of the track, (b) the normal force on the track from the ball at the same moment as in (a). Solution：(a) Taking the circular track as the reference frame , the system of the problem includes the ball , FN the track and the Earth. Draw the free-body diagram FS of the ball in the process of rolling down the track as Fg shown in figure 6-5. Three forces , the gravitational force Fg , the normal force F and the static Fig. 6-5 Example 5 N frictional force F are acting on the ball, But only the gravitational force Fg , being S a conservative force, does work in the rolling process .Thus the mechanic energy of the system conserved. When the ball is at the bottom of the track, its center is chosen as the reference of the zero potential energy. From the conservation law of mechanic energy , we have mg ( R r ) (1/ 2)mv2 (1/ 2) I c 2 (1) Where I c is the rotational inertia of the ball about a axis through its center, v is the speed of the center of mass (COM) of the ball , Then we have I c (2 / 5)mr 2 (2) v r (3) Solve these three equations for v yield v (10 / 7) g ( R r ) - 147 - Chapter 6 Rotation and Angular Momentum. (b) When the ball rolls down to the bottom of the circular track, two forces, F and F .exerted on the ball , Write Newton’s second law in the normal direction n Fn mg mv2 / ( R r ) Therefore FN mg mv / ( R r ) (17 / 7)mg This is the normal force on the ball from the track, Write Newton’s third law it is known that the normal force FN on the track from the ball of the bottom FN FN FN ( 1 7 / 7 ) g m (Answer) Example 6 A thin rod of mass m and length l is connected with a small ball of the same mass m at the rod’s one end, and the other end of the rod is pivoted on a frictionless hinge as shown in the figure ,The rigid body is held at rest horizontally and then released. What is (a) the rotational inertia of the rigid body about the hinge ? (b) the angular speed of the rigid body as its rod forms an angle with the vertical line ? (c) the angular acceleration of the rigid body at the same instant as in(b)? (d) the normal acceleration of the center of mass of the rigid body at the same instant as in (b)? Solution: (a) the rotational inertia of the rigid body about the hinge is Fig. 6-6 Example 6 I0 (1/ 3)ml 2 ml 2 (4 / 3)ml 2 (1) (Answer) (b) During the rotation of the rigid body only the gravitational force does work so the process obeys the conservation law of mechanic energy . But first of all we should find out the center of mass of the rigid body by M1 X1 M 2 X 2 3 X COM l (2) M1 M 2 4 When the rod forms an angle with the vertical line its center of mass C com has - 148 - Chapter 6 Rotation and Angular Momentum. descent a distance h (3 / 4)l cos (3) Applying the conservation law of mechanic energy. We write (2m) gh (1/ 2) I 0 2 Substituting (1), (2) and (3) into above equation and solve for , yields (3 / 2) ( g / l ) cos (Answer) (c) To find out the angular acceleration of the rigid body at the same instant as in (b) we use Newton’s second law for rotation I 0 (2mg )( 1 l sin ) 9 g 3 sin (Answer) I 4 3 ml 2 8l (d) The normal acceleration of the center of mass of the rigid body at the same instant as in (b) is 3 v 2t ( l ) 2 3 3 9 27 an 4 3 l 2 g cos g cos (Answer) r 4 l 4 4 4 16 Example 7 As show in the figure, the masses of wheels A and B are m1 and m2 and the radii of them are r1 and r2 respectively, there is a thin rope rolling around the two wheels and connecting the wheels as shown in the figure, where wheel A rotates about the fixed O axis. (1) when wheel B drops what is the acceleration of the center of the wheel? (2) what is pulling force of the rope? Solution: We consider the problem as the combined motion of the two wheels. That is the rotation of wheels A and the plane motion of wheel B . Then we can apply the law of rotation (the Newton’s second law for rotation) and the Newton’s second law to get the answer. As always we begin with a free-body diagram as shown in the figure Wheel A rotates about the fixed axis O (perpendicular to the page). From the law of rotation 0 I 0 A - 149 - Chapter 6 Rotation and Angular Momentum. Fig. 6-7 Example 7 1 FT r1 m1r12 A (1) 2 Wheel B rotates about the instantaneous axis C (perpendicular to the page) while its , center of mass C is moving down translationally. According to Newton s second law for the motion of the center of mass of wheel B m2 g FT m2 ac (2) From the law of rotation for wheel B 1 FT r2 m2 r22 B (3) 2 By the relationship between the linear quantities and the angular quantities we have aA aB A , B (4) r1 rB Where a A and a B are accelerations of a point at the edge of wheel A and B respectively. There are two more relations which help the problem solving a A aC a B (5) FT FT (6) Solving above six equations for a C and FT , obtaining - 150 - Chapter 6 Rotation and Angular Momentum. 2m1 m2 aC g (Answer) 3m1 2m2 m1m2 FT FT, g (Answer) 3m1 2m2 Example 8 A thin rod of mass m1 and length can rotate about a frictionless axis O freely. A small ball of mass m2 is suspended at the end of a massless cord of length on the same axis as shown in the figure . At the beginning the rod rests in vertical position and the ball is pulled with its cord makes an angle of with the rod. Then the ball is released and swing down making a elastic collision with the rod. The rod deflects a maximum angle 60 . Write down enough equations to determine the angle . Solution: We analyze the whole process by dividing it into four successive parts as follows and then use conservation law in each of them (a) During the swing process of the ball, only the gravitational force does work, so the mechanic energy of the ball is conserved. 1 m2 g(1 cos ) m2 v 2 (1) (Answer) 2 Where v is the speed of the ball just before it Fig. 6-8 Example 8 collides with the rod. (b) As the ball collides with the rod, there is no external torque about the reference point O , thus the angular momentum of the system (the ball plus the rod) is conserved 1 m2v m2v m1 2 (2) (Answer) 3 Where v and v are the speed of the ball just before and after the collision, is the angular speed of the rod just after the collision. - 151 - Chapter 6 Rotation and Angular Momentum. (c) Since the collision between the ball and the rod is elastic, the mechanic energy of the system is conserved just before and after the elastic collision. 1 1 11 m2 v 2 m2 v , 2 m1l 2 2 (3) (Answer) 2 2 23 (d) In the consequential swing process of the rod the mechanic energy of the rod also is conserved 11 23 2 2 m1l m1 g 1 cos 60 (4) (Answer) 2 Solving above four equations for we can get the answer. Note that the most common mistake in solving this problem is apply the conservation law of linear momentum during the collision between the ball and the rod m2v m2v m1 instead of equation (2). This is wrong because during the collision the horizontal component of the force on the rod from the axis can not be neglected. This does not satisfy the condition of applying the conservation law of linear momentum. Problem Solving 1 (12) The wheel in Fig.6-9 has eight equally spaced spokes and a radius of 30cm . It is mounted on a fixed axle and is spinning at 2.5rev / s . You want to shoot a 20 cm long arrow parallel to this axle and through the wheel without hitting any of the spokes. Assume that the arrow and the spokes are very thin. (a) What minimum speed must the arrow have? (b) Does it matter where between the axle and rim of the wheel you aim? If so, what is the best location? Fig. 6-9 Problem 1 Solution: The angular speed of the rotating wheel is (2.5rev / s)(2 rad / rev) 5 rad / s The angle between the maximum angular displacement when a arrow passes throug the wheel, is - 152 - Chapter 6 Rotation and Angular Momentum. 2rad rad 8 4 The time required for a spaced spoke turns around the angle is ( / 4)rad 1 t s 5 rad / s 20 During this time interval the 20cm long arrow must pass though the wheel with the required speed l 0.2cm v 4m / s (Answer) t (1/ 20) s (b) Since in equal time interval the wheel turns a equal angle it does not matter where between the axle and rim of the wheel you aim. (Answer) 2 (16) An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through one of the slots at the outside edge of the wheel, as in Fig.6-10, travels to a distant mirror, and returns to the wheel just in time to pass through the next slot in the wheel. One such slotted wheel has a radius of 5.0 ㎝ and 500 slots around its edge. Measurements taken when the mirror is L=500m from the wheel indicate a speed of light of 3.0×105 ㎞／s.(a)What is the (constant) angular speed of the wheel? (b)What is the linear speed of a point on the edge of the wheel? Solution: The distant for light beam traveling through one slot to the reflecting mirror and backing through the next slot in the wheel is d 2L 2 500m 1000m The time for the light bean travel that distance is d 1000m 1 t 105 s v 3.0 10 m / s 3 8 During this time interval the wheel just turns one slot, the angular displacement is 2 rad Fig. 6-10 Problem 2 500 So the angular speed of the wheel is ( 2 / 5 0r0a)d 5 r 3 . 8 13 0 a d / s (Answer) t (1 / 3 ) 1 0 - 153 - Chapter 6 Rotation and Angular Momentum. The linear speed of a point on the edge of the wheel is v R (0.005m)(3.8 103 rad / s) 1.9 102 m / s (Answer) 3 (20) In Fig.6-11, a cylinder having a mass of 2.0 ㎏ can rotate about its central axis through point O. Forces are applied as shown F1 6.0 N , F2 4.0 N , F3 2.0 N and F4 5.0N .Also, r 5.0cm and R 12cm . Find the (a)magnitude and(b)direction of the angular acceleration of the cylinder.(During the rotation, the forces maintain their same angles relative to the cylinder.) Solution: We apply the Newton’s second law for rotation to find out the angular acceleration of the cylinder. Since F4 passes through the rotational axis it produce no torque about the same axis therefore F1 R F2 R F3 r I Taking the counterclockwise direction as positive for torques and the angular Acceleration of the cylinder, we have Fig. 6-11 Problem 3 F1 R F2 R F3 r I F F R Fr 6.0 N 4.0 N 0.12m 2.0 N 0.05m 1 2 1 1 2.0Kg 0.12m 2 mR 2 2 2 9.7rad / s 2 (Answer) The direction of the angular acceleration of the cylinder is counterclockwise. 4 (22) Figure 6-12 shows a rigid assembly of a thin hoop (of mass m and radius R 0.150m ) and a thin radial rod (of mass m and length L 2.00 R ). The assembly is upright, but if we give it a slight nudge, it will rotate around a horizontal axis in the plane of the rod and hoop, through the lower end of the rod. Assuming that the energy given to the assembly in such a nudge is negligible, what would be the assembly’s angular speed Fig. 6-12 Problem 4 - 154 - Chapter 6 Rotation and Angular Momentum. about the rotation axis when it passes through the upside-down (inverted) orientation? Solution: In the rotating process only the gravitational force does work on the rigid assembly, so we can use the conservation law of mechanical energy to find out the answer. But first we must find out the position of the center of mass of the assembly my my mR m3R yc 2R mm mm Next we should work out the rotational inertia of the rigid assembly about a horizontal axis in the plane of the rod and hoop through the low end of the rod 1 2 I0 I0Rod I0Hoop m(2R) 2 mR 2 m 3R 0.244m kgm 2 1 3 2 Choose point c as the reference level for U 0 . From the conservation law of mechanical energy, we have 1 Mgyc I 0 0 2 2 Solving for 0 yields 2 m m g 4R 2 2 9.8 m 2 4 0.15 s 9.82 rad (Answer) I0 0.224kg m 2 s 5 (23) A tall, cylindrical chimney falls over when its base is ruptures. Treat the chimney as a thin rod of length 55.0m . At the instant it makes an angle of 35.0 with the vertical as it falls, what are (a) its angular speed, (b) the radial acceleration of the top, and (c) the tangential acceleration of the top.(Hint: Use energy considerations, not a torque.) (d) At what angle is the tangential acceleration equal to g ? Solution: In the system under consideration ( the chimney plus the earth ),only the gravitational force does work. So we can apply the conservation law to get the answer. Supposing the mass of the chimney is M and taking the position of the center of mass of the chimney when it makes an angle of 35.0 with the vertical as it falls to be the zero potential energy level.(a) Form the principle of conservation of mechanic energy we have L 1 1 Mg (1 cos ) ( ML2 ) 2 2 2 3 - 155 - Chapter 6 Rotation and Angular Momentum. 3g (1 cos ) 3(9.8m / s)(1 cos 35.0 ) L 55.0m = 0.311 rad s （Answer） （b）The linear speed of the top of the chimney is v L 55.0m 0.311rad s 17.1 m s So the radial acceleration of the top is v 2 (17.1 m s)2 an 5.32 m s 2 (Answer) L 55.0m (c) The tangential acceleration of the top can be obtained by Newton’s second law for rotation 0 I 0 l 1 Mg sin ( ML2 ) 2 3 3g The angular acceleration of the rotationally falling chimney is sin 2L when 35.0 3(9.8 m s 2 ) sin 35.0 0.153 rad s 2 2(55.0m) So the tangential acceleration of the top is a L (35.0m)(0.153 rad s 2 ) 8.43 m s 2 (Answer) 3g 3 (d) From equation a L L sin g sin when a g 2L 2 3 we have sin 1 ，Then 2 41.8 (Answer) 6 (31) Four particles, each of mass 0.20 ㎏, are placed at the vertices of a square - 156 - Chapter 6 Rotation and Angular Momentum. with sides of length 0.50ｍ.The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal, as shown in Fig.6-13. (a)What is the rotational inertia of the body about axis A? (b)What is the angular speed of the body about axis at the instant rod AB swings through the vertical position? Solution: (a) The rotation inertia of the body about axis A is I A ml 2 ml 2 m( l 2 l 2 ) 2 4ml 2 4(0.2kg )(0.50m) 2 0.20 kg m 2 (answer) Note that the particle at vertex A does not have rotational inertia about axis A. Fig. 6-13 Problem 6 (b) During the rotation of the body only the gravitational force does work, so the body is mechanic energy is conserved. The center of the mass C of the body is located at the center of the square, at the instant rod AB swings through the vertical position the center of mass C is a distance l below its initial position. Thus the potential energy of the body in the initial position is U 4mgl . At the vertical position the kinetic energy of the body is 1 2 k I . From the conservation law of mechanic energy we have 2 1 2 I 4mgl 2 Solving for yields 8mgl 8(0.20kg )(9.8 m s 2 )(0.50m) 6.26 rad s I 0.20kg m 2 (Answer) 7 (40) Figure 6-14 shows a rigid structure consisting of a circular hoop of radius R and mass m , and square made of four thin bars, each of length R and mass m . The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 2.5s . Assuming R 0.50m and m 2.0kg ,calculate (a) The structure’s rotational inertia about the axis of rotation and (b) its angular momentum about that - 157 - Chapter 6 Rotation and Angular Momentum. axis. Solution: (a) The structure’s rotational inertia about the axis of rotation is 1 1 1 I I s I n ( mR 2 mR 2 mR 2 0) ( mR 2 mR 2 ) 3 3 2 3.17mR 2 3.17(2.0kg )(0.50m 2 ) 1.6 2 kg m 2 (Answer) (b) The angular speed of the rotating rigid structure is 2 rad 2.51rad / s T 2.5s Thus the angular momentum of the rigid structure about the rotational axis is L I (1.6kg m2 )(2.51 rad s) 4.0kg m2 s (Answer) Fig. 6-14 Problem 7 8 (43) In Fig.6-15, a small 50g block slides down a frictionless surface through height h 20cm and then sticks to a uniform rod of mass 100g and length 40cm .The rod pivots about point O through angle before momentarily stopping. Find . Solution : The whole process can be divided into three parts : (1), The small block slides down the frictionless surface through height h , In this part only the gravitational force , being a conservitive force ,does work, so the law of conservation of mechanic energy holds 1 m1 gh 2 m1v1 (1) 2 Where v1 is the speed of the block before it collides with Fig. 6-15 Problem 8 the rod. (2) The small block collides with the rod and sticks to it. During this interaction there is no net torque acting on the block –rod system relative to the point O , the angular momentum of the system is conserved (Note: since there is a net force acting on the rod at point O by the pivot, the law of conservation of linear momentum does not - 158 - Chapter 6 Rotation and Angular Momentum. hold! ) m1v1 L I (2) Where is the angular speed of the system about point O just after the collision. I is the rotational inertia of the block-rod system about point O ,which is 1 I m2 L2 m1L2 (3) 3 (3) The block-rod system swings up until it momentarily stops , During this process the mechanic energy of the system is conserved , we thus write 1 2 I (m1 m2 ) ghcom (4) 2 Where hcom is the height change of the center of mass of the block-rod system in the swing up process. In the vertical position the center of mass of the system is below point O at m1 L m2 (0.5L) (0.05kg )(0.4m) (0.1kg )(0.2m) Lcom 0.2667m m1 m2 0.05kg 0.10kg So hcom Lcom (1 cos ) (5) Substituting all the known values into equations (1)~(5), we can then get the answer as the following steps: Form Eq.(1) v1 2 gh 2(9.8m / s 2 )(0.2m) 1.98m / s From Eq.(3) 1 0.1kg I ( m2 m1 ) L2 ( 0.05kg )(0.4m) 2 0.0133kgm2 3 3 From Eq.2 we have m1v1L I (0.05kg )(1.98m / s)(0.4m) /(0.0133kgm2 ) 2.97rad / s From Eqs. (4) and (5), we have 1 2 I (m1 m2 ) gLcom (1 cos ) 2 - 159 - Chapter 6 Rotation and Angular Momentum. cos 0.85 Thus the angle we look for is 31.8 (Answer) 9 (44) A certain gyroscope consists of a uniform disk with a 50cm radius mounted at the center of an axle that is 11cm long and of negligible mass. The axle is horizontal and supported at one end. If the disk is spinning around the axle at 1000rev / min , what is the precession rate? Solution: According to formula of the procession rate of a Gyroscope Mgr I Where the rotational inertia of the disk about the horizontal axle is 1 I MR 2 2 The angular speed of the spinning dist is 1000rev 2rad 1 min 104.7rad / s 1 min 1rev 60 s Therefore the precession rate of the gyroscope is Mgr 2 gr 2 9.8m / s 2 0.11m 0.08rad / s (Answer) 1 R 2 0.5m 2 104.7rad / s MR 2 2 10 (45) Figure6-16 shows an overhead view of a ring that can rotate about its center like a merry-go-round. Its outer radius R2 is 0.800m , its inner radius R1 is R2 / 2.00 , its mass M is 0.800kg , and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of 8.00rad / s with a cat of mass m M / 4.00 on its outer edge, at radius R2 . By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius R1 ? Solution: During the process as the cat crawls to the inner edge from the out edge of the rotating ring there is not torque acting along the rotating axis, so the angular Fig. 6-16 Problem 10 - 160 - Chapter 6 Rotation and Angular Momentum. momentum of the cat-ring system is conserved. From the conservation law of angular momentum of a rigid body rotating about a fixed axis we have I 22 I11 (1) Where I 2 and I 1 are the rotational inertia of the cat-ring system with the cat on the ring , s inner edge and on the out edge, respectively. Work out for them I 2 I ring I cat2 1 2 M R12 R2 mR12 2 1 2 8.00 Kg 2 0.400m2 0.800m2 2.00 Kg 0.400m2 3.2 Kgm 2 0.32 Kgm 2 3.52 Kgm 2 I1 I ring I cat1 1 2 2 2 M R12 R2 mR2 4.48Kgm 2 (3) From the problem, the initial angular speed of the cat-ring system 1 8.00rad / s Substituting 1 , I1 and I 2 into equation (1) and solving for 2 lead to I1 4.48Kgm 2 2 1 8.00rad / s 10.18rad / s I2 3.53Kgm 2 Next let us calculate the kinetic energy of the car-ring system both at the initial stage K1 and at the final stage K 2 . The initial kinetic energy of the system is I112 4.48Kgm2 8.00rad / s 143.4 J 1 1 K1 2 2 2 The final kinetic energy of the system is K2 1 2 I 2 2 3.52 Kgm 2 10.18rad / s 182.4 J 2 1 2 2 Thus the amount of the increased kinetic energy of the cat-ring system as the cat crawls to the inner edge from the outer edge of the ring is K K 2 K1 182.4J 143.4J 39 J (Answer) - 161 - Chapter 6 Rotation and Angular Momentum. 11 (46) A uniform wheel of mass 10.0kg and radius 0.400m is mounted rigidly on an axle through its center (Fig.6-17). The radius of the axle is 0.200m , and the rotational inertia of the wheel-axle combination about its central axis is 0.600kg m2 . The wheel is initially at rest at the top of a surface that is inclined at angle 30.0 with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down the surface by 2.00m , what are (a) its rotational kinetic energy and (b) its translational kinetic energy? Solution: The key idea here is that as the wheel-axle combination rolls down the inclined surface only the gravitational force does work. So this process obey the conservation law of mechanic energy. First we should find out the mass of the axle m. Let M the mass of the wheel then the rotational inertia of the wheel-axle combination is Fig. 6-17 Problem 11 I I wheel I axle 1 2 1 M R 2 r 2 mra2 2 Where R and r are the radius of the wheel and the axle, respectively. Substituting the known values and solving for m . We get m 2 1 I 2 M Rw ra 2 2 ra2 2 0.200m 2 1.1Kgm 10 Kg 0.400m 0.200m 2 1 2 2 2 5.0 Kg The potential energy of the rigid body before releasing is U M m gh sin 30 10.0 Kg 5.0 Kg 9.8m / s 2 2.0m sin 30 147J From the conservation law of mechanic energy we have K rot K tran U - 162 - Chapter 6 Rotation and Angular Momentum. 2 I M m r 2 U 1 2 1 2 Substituting the known values and solving for , the angular speed of the combination when it moves down the surface by 2.00m , lead to 13.2rad / s Therefore, the rotational kinetic energy of the wheel-axle combination is K rot 2 I 1.1Kgm 2 13.0rad / s 95.8J 1 2 1 2 2 (Answer) The translational kinetic energy of the wheel-axle combination is K tra 1 M m V 2 1 M m r 2 2 2 1 10.0 Kg 5.0kg 0.2m 2 13.2rad / s 2 (Answer) 2 52.3 J 12 (47) In Fig.6-18, a constant horizontal force Fapp of magnitude 12N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 0.10m , and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? （b）What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder? Solution: this is the plane notion of a rigid body. There are two forces exerted on the cylinder: the applyed force Fapp and the frictional force F . Consider first the translation motion of the center of mass of the cylinder. From Newton’s second law, we have Fapp F ma (1) Fig. 6-18 Problem 12 Where a c is the magnitude of the acceleration of the center of mass of the cylinder. Next consider the rotational motion of the cylinder. From Newton’s second law for rotation we write FappR FR I (2) Where R is the radius of cylinder, I is the rotational inertia of the cylinder. - 163 - Chapter 6 Rotation and Angular Momentum. 1 I MR 2 2 (3) There is a relation between the linear acceleration of the center of mass of the cylinder a c and the angular acceleration of the rolling cylinder a c R (4) Solving above four equations. We obtain ac 1.6m / s 2 (Answer) 16rad / s 2 (Answer) 4 N The frictional force F Fapp mac 12 N 10 Kg 1.6m / s 2 Choose the positive direction of x axis to the right. In unit-vector vector notation the frictional force acting on the cylinder can be written as F 4i N (Answer) 13 (49) In Fig.6-19, a small 0.50kg block has a horizontal velocity v0 of magnitude 3.0m / s when it slides off a table of height h 1.2m . Answer the following in unit-vector notation for a coordinate system in which the origin is at the edge of the table (at point O ), the positive x direction is horizontally away from the table, and the positive y direction is up. What are the angular momentum of the block about point A (a) just after the block leaves the table and (b) just before the block strikes the floor? What are the torques on the block about point A (c) just after the block leaves the table and (d) just before strikes the floor? Solution: Taking point A as the reference point, rA and rB are the position vectors, when the block just leaves the table and before it strikes the floor. (a) The angular momentum of the block about point A just after block leaves the table is L0 rA mV0 Fig. 6-19 Problem 13 - 164 - Chapter 6 Rotation and Angular Momentum. Its magnitude is L0 r0 mv0 sin 90 1.2m1.5kg 3.0m / s 1.8kgm2 / s ( Answer ) Its direction can be determined by the right hand rule, which is perpendicular to the plane formed by rA and v 0 , entering into this page ( ). (Answer ) (b) The angular momentum of the block about point A , just before the block strikes the floor is L rB mv We should find out the velocity v of the block in that moment in horizontal direction v x v0 3.0m / s . In vertical direction the block is in free fall motion, thus 1 2 h gt 2 2h 21.2m t 0.5s g 9.8m / s 2 v y gt 9.8m / s 2 0.5s 4.9m / s The horizontal displacement of the block is R v0 t 3.0m / s 0.5s 1.5m So the magnitude of the velocity of block just before it strikes the floor is v vx v y 2 2 3.0m / s 2 4.9m / s 2 5.75m / s The angle made by velocity v with the x axis is vy 5.75m / s tg 1 tg 1 58.5 vx 3.0m / s Thus the magnitude of angular momentum of the block in (b) is L rB mv sin 1.5m 0.5kg 5.75m / s sin58.5 3.7kgm2 / s (Answer) - 165 - Chapter 6 Rotation and Angular Momentum. The direction of L is perpendicular to the plane formed by rB and v , entering into this page ( ), also by applying the right hand rule. (Answer) (c) The torque on the block about point A just after the block leaves the table is 0 rA Fg rA mg Note that, at this moment, rA and g are in the same vertical line but opposite in direction. so the torque 0 is zero. (Answer) (d) Just before the block strikes the floor the torque on the block about point A is rB mg Since now rB g the magnitude of the torque acting on the block is rB mg 1.5m0.5kg 9.8m / s 2 7.4kgm 2 / s 2 (Answer) The direction of the torque is perpendicular to the plane formed by rB and g which is entering into this page ( ). (Answer) 14 (50) An impulsive force F (t ) acts for a short time t on a rotating rigid body of rotational inertial I . Show that dt RF avg t I ( f i ) Where is the torque due to the force, R is the moment arm of the force, Favg is the average value of the force during the time it acts on the body, and i and f are the angular velocities of the body just before and just after the forces acts (The quantity dt RF avg t is called the angular impulse, analogous to Favg t , the linear impulse). Solution: The angular impulse, the time accumulation effect of a torque, is defined as - 166 - Chapter 6 Rotation and Angular Momentum. dt RF avg t (1) For a variable force F t , the torque produced by it is a function of time t RF t (2) Substituting Eq.(1) and set t t 2 t1 and avg RFavg , we obtain RF t dt RFavg t avg t t2 t1 (3) While according to the Newton’s Second law for rotation and the definition of angular acceleration, we have 2 1 avg t I I (4) t Recast above equation obtaining avg I 2 1 (5) Combining equation (2)、(3) and (5) we get RF t dt RFavg t I 2 1 t2 t1 (Answer) - 167 -

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