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制作 张昆实 谢 丽 Yangtze University

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 What Is Physics? 6-2 Equilibrium 6-3 The Rotational Variables 6-4 Are Angular Quantities Vectors

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制作 张昆实 谢 丽 Yangtze University

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  1. BilingualMechanics Chapter 6 Rotation and Angular Momentum 制作 张昆实 谢 丽 Yangtze University

  2. Chapter 6Rotation and Angular Momentum 6-1 What Is Physics? 6-2 Equilibrium 6-3 The Rotational Variables 6-4 Are Angular Quantities Vectors 6-5 Relating the Linear and Angular Variables 6-6 Kinetic Energy of Rotation

  3. Chapter 6Rotation 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

  4. Chapter 6Rotation 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

  5. 6-1What 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 rotationof aregid body and the related angular momentum.

  6. Question 1 What is the linear momentumand the angular momentum of each of these four objects? 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 pathat constant speed.

  7. a constant , a constant • For each of these four objects: • The linear momentum of its center of mass is constant. 2. Its angular momentum about its center of mass, or about any other point, is also constant. (6-1) 6-2 Equilibrium ★ two requirements for equilibrium objects satisfying Eq.(13-1) are said to be in equilibrium.

  8. ★two requirements for equilibrium a constant , a constant (6-1) (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. 6-2 Equilibrium ●Static equilibrium ●unstablestatic equilibrium: If a small force can displace the body and end the equilibrium.

  9. The analysis of static equilibrium is very important in engineering practice. 6-2 Equilibrium Landing gear

  10. 6-2 Equilibrium Static equilibrium in Building designing is very important especially in the earthquake area !

  11. 6-2 Equilibrium The Yangtze River bridge at JingZhou(荆州) The analysis of static equilibrium is also very important in designing bridgies

  12. ★The translational motion of a body is governed by Newton's second law in its linear momentum form : (6-2) If the body is in translational equilibrium (6-3) is a constant (balance of forces) ★The rotational motion of a body is governed by Newton's second law in its angular momentum form : (6-4) If the body is in rotational equilibrium (6-5) is a constant (balance of torques) 6-2 Equilibrium

  13. 6-2 Equilibrium • ★ The two Requirements for a body to be in Equilibrium • The vector sum of all the external forces that act on the body must be zero. Vector equation Component equations (6-6) (balance of forces) 2.The vector sum of all the external torquesthat act on the body, measured about an possible point, must also be zero. Vector equation Component equations (6-6) (balance of torques)

  14. ★Equilibrium equations for a body lie in the xy plane: (balance of forces) (6-7) (balance of forces) (6-8) (balance of torques) (6-9) 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 eliminateone force equation and two torque equations from Eqs. 6-6 √ × √ × (6-6) √ ×

  15. cog ★ The gravitational force on a body acts at a specific point, called the center of gravity (cog) of the body. ★ If is the same for all elements of a body, then the body’s center of gravity (cog) is coincident with the body’scenter of mass(com). 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. com

  16. Rotation axis Rigid body 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 circlewhose center lies on the axis of rotation, and every point moves through the same angleduring a particulartime interval.

  17. reference line zero angular position (radian measure) (6-10) (6-11) (6-12) 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)

  18. An angular displacement in the counterclockwise direction is positive “+”, and one in the clockwise directionis negative “” . r t2 O . t1 x ★Angular Velocity average angular velocity (6-14) 6-3 The Rotational Variables ★Angular Displacement (6-13)

  19. The instantaneous angular acceleration The average angular acceleration (6-16) (6-17) 6-3 The Rotational Variables The instantaneous angular velocity (6-15) unit (rad/s) or (rev/s) ★Angular Acceleration unit (rad/s2) or (rev/s2)

  20. z 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. In pure rotation, a vector defines an axis of rotation, not a direction in which something moves !

  21. caution 6-4 Are Angular Quantities Vectors 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 displacementsfail this test !

  22. (2-43) (6-18) (2-47) (6-19) (2-48) (6-20) (2-49) (6-21) (2-50) (5-22) Three replacements: 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

  23. M a s a (radian measure) (6-23) ★The Speed Differentiating Eq6-23 with respect to t r M is the linear speed, is the angular speed (radian measure) (6-24) So we can obtain : Caution: The angle and the angular speed must be measured in radians. 6-5 Relating the Linear and Angular Variables Relate the linear variables and to the angular variables and ★The Position t v r a w 0

  24. ★The Acceleration Tangential acceleration radial acceleration differentiating (6-29) (6-27) (radian measure) (6-28) 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 (6-25) Substituting for v from Eq.6-24 and canceling r, we can find (radian measure) (6-26)

  25. Substituting for v from , we can get (6-38) O ★The rotational inertia ( moment of inertia) (radian measure) (6-39) (6-40) 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 (6-37) : the mass of the ith particle, : Its speed

  26. The SI unit for is kilogram-square meter ( ) 6-6 Kinetic Energy of Rotation ★The rotational inertia ( moment of inertia) (6-39) ★The rotational inertia of a rigid bodydepends on: (1) the mass of the rigid body ; (2)the mass distribution; (3) the rotation axis (a) Parallel-Axis Theorem (b) Perpendicular-Axis Theorem

  27. ( 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 ) 6-7 Calculating the Rotational Inertia If a rigid body consists of a great many adjacent par-ticles, replace the sum in with an integral:

  28. ( : thickness , : density) then the rotational inertia of the ring can be found Through integration we get Since , the rotational inertiaof the disk is mass element 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 dr o

  29. TABLE 6-2 Some Rotational Inertias 6-7 Calculating the Rotational Inertia

  30. z Zc (6-42) M h (parallel-axis theorem) C 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 axiszparalle to zc, it can be proved that the rotational inertia of the rigid body about the z axis is Where M is the mass of the rigid body, h is the distance between the two parallel axes.

  31. y Rotation axis through P dm r y- b • p x-a h b x o a Rotation axis through center of mass 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 0 0 R2 h2

  32. 6-7 Calculating the Rotational Inertia Sample Problem 6-3 A thin, uniformrod 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 axisthrough thecenter? (b) What is the rod’s rotational inertia about a newrotation axis that isperpendicular to the rod and through the left end?

  33. mass element Solution: (b) using parallel-axis theorem 6-7 Calculating the Rotational Inertia Solution: (a) for a continuous object using

  34. (perpendicular- axis theorem) Where is therotational inertia about the axis, respectively. 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 and axes lie in the plate and the axis perpendicular to the plate. The rotational inertia of the rigid body about the axis is

  35. y x Example: What is the rotational inertia of a disk about any diameter? , axesare lie in the plane of the disk, axis is perpendicular to them. C R (from symmetry) 6-7 Calculating the Rotational Inertia ★ Perpendicular-Axis Theorem (perpendicular-axis theorem)

  36. radial component torque Rotation axis Line of action of tangential component o Moment arm of : the moment arm 6-7 Calculating the Rotational Inertia ★ Torque A force is applied on a body (only shown a cross section) Two equivalent ways of computing the torque

  37. torque Rotation axis Line of action of The ability of to rotate the body depends not onlyon the magnitude of its tangential component , but alsoon just how far from o the force is applied. o Moment arm of SI unit : N• m Torques obey the superposition principle : when several torques act on a body, the net torque (or resultant torque) is the sum of individual torques. torque: counterclockwise positive; clockwise negative 6-7 Calculating the Rotational Inertia

  38. (6-45) From 6-8 Newton’s Second Law for Rotation ★Newton’s Second low for rotation y m r A force act on the particle ,only the tangential component of the force can accelerate the particle along the circular path. So with the Newton’s second law we can write Rod o x Rotation axis The torque acting on the particle is (6-47) ( radian measure ) (6-48)

  39. (6-53) (work, one-dimensional) 6-9 Work and Rotational Kinetic Energy ★with the work- kinetic energy theorem (Eq 4-10) ,we relate the change in kinetic energy to the work W, writing (6-52) The rate at which the work is done is the power, which we can find with Eqs 4-41 and 4-46 (power, one-dimensional) (6-54)

  40. We can calculate the work with a rotational equivalent of Eq6-53 (work, rotation about fixed axis) (6-56) When is constant, Eq 6-56 reduces to (work, constant torque) (6-57) The rate at which the work is done is the power (6-58) (power, rotation about fixed axis) 6-9 Work and Rotational Kinetic Energy ★A rotational situation is similar, so we can write a rotational kinetic energy as (work-kinetic energy theorem) (6-55)

  41. Using and we can rewrite the Eq6-59 (6-60) Substituting into Eq 6-60 yields 6-9 Work and Rotational Kinetic Energy ★Proof of Equation 6-55 through 6-58 During the rotation, force does work on the body. Let’s assume that the only energy of the body that changed by is the kinetic energy. Then we can apply the work-kinetic energy theorem (6-59)

  42. From ,we can rewrite Eq 6-61 as (6-62) The work done during a finite angular displacement from to is then We can find power P for rotational motion from Eq 6-62 6-9 Work and Rotational Kinetic Energy We write that work dW as Ftds. However, we can replace ds with . Thus we have (6-61)

  43. 6-3 From Left to Right : Five replacements 6-9 Work and Rotational Kinetic Energy

  44. 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 translationof thecenter of mass and rotation of the rest of the object around that center.

  45. A bicycle wheel rolls along a street. During a time interval , both O and the contact point P move a distance . The wheel rotates through an angle about the center of the wheel, The initial contact point P moves through arc length: P P 6-12 (6-66) (6-67) (smooth rolling motion) • 6-10 Rolling as Translation and Rotation Combined R DifferentiatingEq. 6-66 with respect to time (with R held constant) gives us

  46. + = 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 , faster than any other portion of the wheel. • 6-10 Rolling as Translation and Rotation Combined T P The motion of any round body rolling smoothly over a surface can be separated into purely rotational and purely translational motions.

  47. The same angular speed as that the wheel rotates about its “ COM ” axis ! • 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. 6-24 What is the angular speed of the rolling wheel about the new axis ? 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 !

  48. (6-68) where is the angular speed of the wheel and is the rotational inertia of the wheel about the axis through P. From the parallel-axis theorem, we have (6-69) where M is the mass of the wheel, is its rotational inertia about an axis through its center of mass. 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 we have

  49. Using the relation , yields Konig’s theorem (6-70) A rolling object has two type of kinetic energy : a rotational kinetic energy due to its rotation about its center of mass and a transla-tional kinetic energy due to translation of its center of mass. 6-11 The Kinetic Energy of Rolling

  50. tend to makethe wheel slideat P. If the wheel does not slide, that frictional force is a static frictional force , the motion is smooth rolling. Differentiating with respect to time we have (6-71) (smooth rolling motion) 6-12 The Forces of Rolling ★ Friction and Rolling If a net force acts on the rolling wheel to speed it up cause and . Tendency of slide A frictional force must acts on the wheel at P to oppose that sliding tendency.

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