《物理双语教学课件》Chapter 7 Rolling Torque, and Angular Momentum 力矩与角动量_课件适合双语教学

《物理双语教学课件》Chapter 7 Rolling Torque, and Angular Momentum 力矩与角动量由刀豆文库小编整理，希望给你工作、学习、生活带来方便，猜你可能喜欢“课件适合双语教学”。

Chapter 7 Rolling Torque, and Angular Momentum

7.1 Rolling When a bicycle moves along a straight track, the center of each wheel moves forward in pure translation.A point on the rim of the wheel, however, traces out a more complex path, as the figure shows.In what follow, we analyze the motion of a rolling wheel first by viewing it as a combination of a pure translation and a pure rotation, and then by viewing it as rotating along.1.Rolling as rotation and translation combined

(1)Imagine that you are watching the wheel of a bicycle, which paes you at constant speed while rolling smoothly, without slipping, along a street.As shown in the figure, the center of ma O of the wheel moves forward at constant speed vcm.The point P at which the wheel contacts the street also moves forward at speed

vcm, so that it always remains directlybelow O.(2)During a time interval t, you see both O and P move forward by a distance s.The bicycle rider sees the wheel rotate through an angle  about the center of the wheel, with the point of the wheel that was touching the street at the beginning of t moving through arc length s.We have the relation where R is the radius of the wheel.(3)The linear speed

vcm

sR,of the center of the wheel is ds/dt,d/dt.and the angular speed  of the wheel about its center is So differentiating the equation with respect to time gives us vcmR.(4)The figure shows that the rolling motion of a wheel is a combination of purely translational and purely rotational motions.(5)The motion of any round body rolling smoothly over a surface can be separated into purely translational and purely rotational motion.2.Rolling as pure rotation

(1)The figure suggests another way to look at the rolling motion of a wheel, namely, as pure rotation about an axis that always extends through the point where the wheel contacts the street as the wheel moves.That is, we consider the rolling motion to be pure rotation about an axis paing through point P and perpendicular to the plane of the figure.The vectors in then represent the instantaneous velocities of points on the rolling wheel.(2)The angular speed about this new axis that a stationary observer aign to a rolling bicycle wheel is the same angular speed that the rider aigns to the wheel as he or she observes it in pure rotation about an axis through its center of ma.(3)To verify this answer, let’s use it to calculate the linear speed of the top of the wheel from the point of view of a stationary observer, as shown in the figure.3.The Kinetic energy: Let us now calculate the kinetic energy of the rolling wheel as measured by the stationary observer.(1)If we view the rolling as pure rotation about an axis through P in above figure, we have

KIp2/2, in which 

is the angular speed of the wheel and Ip is the rotational inertia of the wheel about the axis through P.(2)From the parallel-axis theorem, we have

IpIcmMR2, in which M is the ma of the wheel and

Icm

is its rotational inertia about an axis through its center of ma.(3)Substituting the relation about its rotational inertia, we

2obtainK1Icm21MR221Icm21Mvcm.2222(4)We can interpret the first of these terms as the kinetic energy aociated with the rotation of the wheel about an axis through its center of ma, and the second term as the kinetic energy aociated with the translational motion of the wheel.4.Friction and rolling

(1)If the wheel rolls at constant speed, it has no tendency to slide at the point of contact P, and thus there is no frictional force acting on the wheel there.(2)However, if a force acting on the wheel, changing the speed vcm of the center of the wheel or the angular speed 

about the center, then there is a tendency for the wheel to slide, the frictional force acts on the wheel at P to oppose that tendency.(3)Until the wheel actually begins to slide, the frictional force is a static frictional force fs.If the wheel begins to slide, then the force is a kinetic frictional force fk.7.2 The Yo-Yo 1.A yo-yo, as shown in the figure, is a physics lab that you can fit in your pocket.If a yo-yo rolls down its string for a distance h, it loses potential energy in amount mgh but gains kinetic energy in both translational and rotational form.When it is climbing back up, it loses kinetic energy and regains potential energy.2.Let us analyze the motion of the yo-yo directly with Newton’s second law.The above figure shows its free-body diagram, in which only the yo-yo axle is shown.(1)Applying Newton’s second law in its linear form yields FTMgMa, Here M is the ma of the yo-yo, and T is the tension in the yo-yo’s string.(2)Applying Newton’s second law in angular form about an axis through the center of ma yields TR0I, Where R0 is the radius of the yo-yo axle and I is the rotational inertial of the yo-yo about its center axis.(3)The linear acceleration and angular acceleration have relation aR0.So After eliminating T in both equations we obtain ag121I/MR0.Thus an ideal yo-yo rolls down its string with constant acceleration.7.3 Torque revisited In chapter 6 we defined torque  for a rigid body that can rotate around a fixed axis, with each particle in the body forced to move in a path that is a circle about that axis.We now expand the definition of torque to apply it to an individual particle that moves along any path relative to a fixed point rather than a fixed axis.The path need no longer be a circle.1.The figure shows such a particle at point P in the xy plane.A single force F in that plane acts on the particle, and the particle’s position relative to the origin O is given by position vector r.The torque  acting on the particle relative to the fixed point O is a vector quantity defined as

rF.2.Discu the direction and magnitude of (rFsin).7.4 Angular Momentum 1.Like all other linear quantities, linear momentum has its angular counterpart.The figure shows a particle with linear momentum p(=mv)located at point P in the xy plane.The angular momentum l of this particle with respect to the origin O is a vector quantity defined as

lrpm(rv), where r is the position vector of the particle with respect to O.2.The SI unit

of

angular

momentum

is

the kilogram-meter-square per second(kgm2/s), equivalent to the joule-second(Js).3.The direction of the angular momentum vector can be found to use right-hand rule, as shown in the figure.4.The magnitude of the angular momentum is where lrmvsin, is the angle between r and p when these two vectors are tail to tail.7.5 Newton’s second law in angular form

1.We have seen enough of the parallelism between linear and angular quantities to be pretty sure that there is also a close relation between torque and angular momentum.It is dldt.2.The vector sum of all torques acting on a particle is equal to the time rate of change of the angular momentum of that particle.The torque and the angular momentum should be defined with respect to the same origin.3.Proof of the equation: angular momentum can be written as:

lm(rv)

Differentiating each side with respect to time yields dvdrdlm(rv)m(ravv)r(ma)r(F)(rF) dtdtdt

7.6 The Angular Momentum of A System of Particles Now we turn our attention to the motion of a system of particles with respect to an origin.Note that “a system of particles” includes a rigid body as a special case.1.The total angular momentum L of a system particles is the vector sum of the angular momenta l of the particles:

nLl1l2l3lnlii1, in which

i(1,2,3,)labels the

particles.2.With time, the angular momenta of individual particle may change, either because of interactions within the system(between the individual particles)or because of influences that may act on the system from the outside.We can find the change in L as these changes take place by taking the time derivative of above equation.Thus

nndlidLi.dti1dti13.Some torques are internal, aociated with forces that the particles within the system exert on one another;other torques are external, aociated with forces that act from outside the system.The internal forces, because of Newton’s law of action and reaction, cancel in pair(give a little more explanation).So, to add the torques, we need consider only those aociated with external forces.Then above equation becomes

extdL dtThis is Newton’s second law for rotation in angular form, expre for a system of particles.The equation has meaning only if the torque and angular momentum vectors are referred to the same origin.In an inertia reference frame, the equation can be applied with respect to any point.In an accelerating frame, it can be applied only with respect to the center of ma of the system.7.7 The Angular Momentum of a Rigid Body Rotating about a Fixed Axis We next evaluate the angular momentum of a system of particles that form a rigid body, which rotates about a fixed axis.Figure(a)shows such a body.The fixed axis of rotation is the z axis, and the body rotates about it with constant angular speed .We wish to find the angular momentum of the body about the axis of rotation.1.We can find the angular momenta by summing the z components of the angular momenta of the ma elements in the body.In figure(a), a typical ma element

mi

of the body moves around the z axis in a circular path.The position of the ma elements is located relative to the origin O by

position vector ri.The radius of the ma element’s circular path is ri, the perpendicular distance between the elementand z axis.2.The magnitude of the angular momentum element, with respect to O, is where pi

li

of this ma

li(ri)(pi)(sin900)(ri)(mivi), and

vi

are the linear momentum and linear speed

900

is the angle between ri and of the ma element, and pi.3.We are interested in the component of li that parallel to the rotation axis, here the z axis.That z component is lizlisin(risin)(mivi)rimivi.4.The z component of the angular momentum for the rotating rigid body as a whole is found by adding up the contributions of all the ma elements that make up the body.Thus, because

nni1vr, we may write

n

ni12Lzlizmivirimi(ri)ri(miri)I

i1i15.The following table extends the list of corresponding linear and angular relations.11 7.8 Conservation of Angular Momentum So far we have discued two powerful conservation laws, the conservation of energy and the conservation of linear momentum.Now we meet the third law of this type, the conservation of angular momentum.1.If no net external torque acts on the system, from Newton’s second law in angular form, we have

dL/dt0, or Laconstant.This result, called the law of conservation of angular momentum, can also be written as

LiLf.2.This means if the net external torque acting on a system is zero, the angular momentum of the system remains constant, no matter what changes take place within the system.3.Above equations are vector equations.They are equivalent to three scalar equations corresponding to the conservation of angular momentum in three mutually perpendicular directions.Depending on the torques acting on a system, the angular momentum of the system might be conserved in only one or two directions but not all directions.This is if any component of the external torque on a system is zero, then that component of the angular momentum of the system along that axis cannot change, no matter what changes take place within the system.4.Like the other two conservation laws that we have discued, It holds beyond the limitation of Newton’s mechanics, It holds for particles whose speeds approach that of light(where the theory of relativity reigns), and it remains true in the world of subatomic particles(where quantum mechanics reigns).No exceptions to the law of conservation of angular momentum have ever been found.5.We now discu four examples involving this law.P283 13

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