《物理双语教学课件》Chapter 7 Rolling Torque and Angular Momentum 力矩与角动量Word文档格式.docx

1、 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 of the center of the wheel is, and the angular speed of the wheel about its cente

2、r is. So differentiating the equation with respect to time gives us.（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 translatio

3、nal 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 roll

4、ing motion to be pure rotation about an axis passing 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 assign to a rolling bicyc

5、le wheel is the same angular speed that the rider assigns to the wheel as he or she observes it in pure rotation about an axis through its center of mass. （3） To verify this answer, lets use it to calculate the linear speed of the top of the wheel from the point of view of a stationary observer, as

6、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, in which is the angular speed of the wheel and is the rotation

7、al inertia of the wheel about the axis through P. （2） From the parallel-axis theorem, we have, in which M is the mass of the wheel and is its rotational inertia about an axis through its center of mass. （3） Substituting the relation about its rotational inertia, we obtain. （4） We can interpret the f

8、irst of these terms as the kinetic energy associated with the rotation of the wheel about an axis through its center of mass, and the second term as the kinetic energy associated with the translational motion of the wheel.4. Friction and rolling（1） If the wheel rolls at constant speed, it has no ten

9、dency 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 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

10、 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-Yo1. A yo-yo, as shown in the figure, is a physics lab

11、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 moti

12、on of the yo-yo directly with Newtons second law. The above figure shows its free-body diagram, in which only the yo-yo axle is shown. （1） Applying Newtons second law in its linear form yields, Here M is the mass of the yo-yo, and T is the tension in the yo-yos string. （2） Applying Newtons second la

13、w in angular form about an axis through the center of mass yields, Where 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. So After eliminating T in both equations we obtain. Thus a

14、n ideal yo-yo rolls down its string with constant acceleration.7.3 Torque revisitedIn 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

15、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 particles position relativ

16、e 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 as2. Discuss the direction and magnitude of （）.7.4 Angular MomentumLike all other linear quantities, linear momentum has its angular counterpart. The figure s

17、hows 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, where is the position vector of the particle with respect to O.2. The SI unit of angular momentum is the kilogram-met

18、er-square per second （）, equivalent to the joule-second （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 is the angle between r and p when these two vectors are tail to tail.7.5 Newtons second law

19、 in angular form1. 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 is2. The vector sum of all torques acting on a particle is equal to the time rate of change of the angular mom

20、entum 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: Differentiating each side with respect to time yields7.6 The Angular Momentum of A System of ParticlesNow we turn our attention

21、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: labels the particles.2. With time, the angula

22、r 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 equati

23、on. Thus3. Some torques are internal, associated with forces that the particles within the system exert on one another; other torques are external, associated with forces that act from outside the system. The internal forces, because of Newtons law of action and reaction, cancel in pair （give a litt

24、le more explanation）. So, to add the torques, we need consider only those associated with external forces. Then above equation becomesThis is Newtons second law for rotation in angular form, express for a system of particles. The equation has meaning only if the torque and angular momentum vectors a

25、re 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 mass of the system. 7.7 The Angular Momentum of a Rigid Body Rotating about a Fixed AxisWe next evaluat

26、e 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

27、 of rotation.1. We can find the angular momenta by summing the z components of the angular momenta of the mass elements in the body. In figure （a）, a typical mass element of the body moves around the z axis in a circular path. The position of the mass elements is located relative to the origin O by

28、position vector. The radius of the mass elements circular path is, the perpendicular distance between the element and z axis.2. The magnitude of the angular momentum of this mass element, with respect to O, is and are the linear momentum and linear speed of the mass element, and is the angle between and3. We are interested in the component of that parallel to the rotation axis, here the z axis. That z component is4. The z component of the angular momentum for the rotating rigid body as a whole is found by adding up the contributions of all