《物理双语教学课件》Chapter 6 Rotation 定轴转动

Chapter 6 Rotation
In this chapter, we deal with the rotation of a rigid body about a fixed axis. The first of these restrictions means that we shall not examine the rotation of such objects as the Sun, because the Sun-a ball of gas-is not a rigid body. Our second restriction rules out objects like a bowling ball rolling down a bowling lane. Such a ball is in rolling motion, rotating about a moving axis.
6.1 The Rotational Variables
1.Translation and Rotation: The motion is the one of pure translation, if the line connecting any two points in the object is always parallel with each other during its motion. Otherwise, the motion is that of rotation. Rotation is the motion of wheels, gears, motors, the hand of clocks, the rotors of jet engines, and the blades of helicopters.
2.The nature of pure rotation: The
right figure shows a rigid body of
arbitrary shape in pure rotation
around a fixed axis, called the axis
of rotation or the rotation axis.
(1). Every point of the body moves in a circle whose center
lies on the axis of the rotation.
(2). Every point moves through the same angle during a particular time interval.
3. Angular position : The above figure shows a reference line, fixed in the body, perpendicular to the axis, and rotating with the body. We can describe the motion
of the rotating body by specifying the
angular position of this line, that is,
the angle of the line relative to a fixed
direction. In the right figure, the
angular position
θ is measured relative to the positive direction of the x axis, and θ is given by ).(measure radian r s
=θ
Here s is the length of the arc (or the arc distance ) along a circle and between the x axis and the reference line, and r is a radius of that circle.
An angle defined in this way is measured in radians (rad) rather than in revolutions (rev) or degree. They have relations
rad r
r rev o ππ223601=== 4. If the body rotates about the rotation
axis as in the right figure, changing
the angular position of the reference
line from 1θ to
2θ, the body undergoes an angular displacement
θ∆ given by 12θθθ-=∆
The definition of angular displacement holds not only for the rigid body as a whole but also for every particle within the body. The angular displacement θ∆ of a rotating body can be either positive or negative, depending on whether the body is rotating in the direction of increasing θ (counterclockwise ) or decreasing θ (clockwise ).
5. Angular velocity
(1). Suppose that our rotating body is at angular position 1θ at time 1t and at angular position 2θ at time 2t . We define
the average angular velocity of the body in the time interval t ∆ from 1t to 2t to be
t t t ∆∆=--=θθθω121
2 In which
θ∆ is the angular displacement that occurs during t ∆.
(2). The (instantaneous) angular velocity ω, with which we shall be most concerned, is the limit of the average angular velocity as t ∆ is made to approach zero. Thus
dt d t t θθω=∆∆=→∆0lim If we know
)(t θ, we can find the angular velocity ω by
differentiation.
(3). The unit of angular velocity is commonly the radian per second (rad/s) or the revolution per second (rev/s).
(4). The magnitude of an angular velocity is called the angular speed , which is also represented with ω.
(5). We establish a
direction for the vector
of the angular velocity
ω by using
a right-hand
rule , as shown in the figure.
Curl your right hand about the rotating record, your fingers pointing in the direction of rotation. Your extended thumb will then point in the direction of the angular velocity vector .
6. Angular acceleration
(1). If the angular velocity of a rotating body is not constant, then the body has an angular acceleration. Let
2ω and 1ω be the angular velocity at times 2t and 1t , respectively. The
average angular acceleration of the rotating body in the interval from 1t to 2t is defined as
t t t ∆∆=--=ωωωα1212 In which
ω∆ is the change in the angular velocity that occurs
during the time interval t ∆.
(2). The (instantaneous) angular acceleration α, with which we shall be most concerned, is the limit of this quantity as t ∆ is made to approach zero. Thus
dt d t t ωωα=∆∆=→∆0lim
above equations hold not only for the rotating rigid body as a whole but also for every particle of that body .
(3). The unit of angular acceleration is commonly the radian per second-squared (rad/s 2) or the revolution per second-squared (rev/s 2).
(4). The angular acceleration also is a vector. Its direction depends on the change of the angular velocity.
7. Rotation with constant angular acceleration:
2
00002
1)(t t dt t d t dt d t dt d dt d αωθαωθαωθαωωαωαω+=→+=→+=+=→=→= Here we suppose that at time ,0=t 00=θ. We also can get a parallel set of equations to those for motion with a constant linear acceleration.
8. Relating the linear and angular variables: They have relations as follow:
Angular displacement: r d s d d ⨯=θθ
Angular velocity ： r v ⨯=ωω
Angular acceleration ：v
a r a n t ⨯=⨯=ωαα
6.2 Kinetic Energy of Rotation
1. To discuss kinetic energy of a rigid body, we cannot use the familiar formula 2/2mv K = directly because it applies only to particles. Instead, we shall treat the object as a collection of particles-all with different speeds. We can then add up the kinetic energies of these particles to find kinetic energy of the body as a whole. In this way we obtain, for the kinetic energy of a rotating body,
∑=+++=22332222112
1212121i i v m v m v m v m K In which i m is the mass of the i th particle and i v is its speed.
The sum is taken over all the particles in the body.
2. The problem with above equation is that i v is not the same
for all particles. We solve this problem by substituting for v in the equation with
r ω, so that we have ∑∑==222)(21)(21ωωi i i i r m r m K
In which ω is the same for all particles.
3. The quantity in parentheses on the right side of above equation tells us how the mass of the rotating body is distributed about its axis of rotation.
(1). We call that quantity the rotational inertia (or moment of
inertia) I of the body with respect to the axis of rotation. It’s a constant for a particular rigid body and for a particular rotation axis. We may now write ∑=2i i r m I
(2). The SI unit for I is the kilogram-square meter (2m kg ⋅).
(3). The rotational inertia of a rotating body depends not only on its mass but also on how that mass is distributed with respect to the rotation axis .
4. We can rewrite the kinetic energy for the rotating object as 221ωI K =
Which gives the kinetic energy of a rigid body in pure rotation. It’s the angular equivalent of the formula 2/2cm Mv K =, which gives the kinetic energy of a rigid body in pure translation.
6.3 Calculating the Rotational Inertia
1. If a rigid body is made up of discrete particles , we can calculate its rotational inertia from ∑=2i i r m I .
2. If the body is continuous , we can replace the sum in the equation with an integral, and the definition of rotational inertia becomes dm r I ⎰=2. In general, the rotational inertia of any rigid body with respect to a rotation axis depends on (1). The shape of the body, (2). The perpendicular distance from the axis to the body’s center of mass, and (3). The orientation of the body with respect to the axis .
The table gives the rotational inertias of several common bodies, about various axes. Note how the distribution of mass relative to the rotational axis affects the value of the rotational inertia I . We would like to give the example of rotational inertia for
a thin circular plate
2224200322
1)(21241mR R R R d dr r dr rd r I R
=====⎰⎰⎰πσπσθσθσπ a thin rod
223223222112
1)(1212)2(3131)1(ml l l l r dr r I l l
l
l ==⨯===--⎰λλλλ 2
222
212230223
1124123121)2()(313
1)
2(m l m l m l m l l m I I l l l dr r I l ==+=+====⎰λλλ
3. The parallel-axis theorem : If you know the rotational inertia of a body about any axis that passes through its center of mass, you can find its rotational inertia about any other axis parallel to that axis with the parallel-axis theorem:
2Mh I I cm +=
Here M is the mass of the body and h is the perpendicular distance between the two parallel axes.
4. Proof of the parallel-axis theorem : Let O be the center of mass of the arbitrarily shaped
body shown in cross section
in the figure. Place the origin
of coordinates at O. Consider
an axis through O
perpendicular to the plane of
the figure, and another axis of P parallel to the first axis, Let the coordinates of P be a and b.
Let dm be a mass element with coordinates x and y . The rotational inertia of the body about the axis through P is then dm b y a x dm r I ])()[(222⎰⎰-+-==
Which we can rearrange as
22222200)(22)(Mh I Mh I dm b a ydm b xdm a dm y x I cm cm +=+--=++--+=⎰⎰⎰⎰
6.4 Newton’s Second Law for Rotation
1. Torque : The following figure shows a cross section of a body that is free to rotate about an axis passing through O and
perpendicular to the cross section. A force F is applied at point P, whose position relative to O is defined by a position vector r . Vector F and r make an angle ϕ with each other. (For simplicity, we consider only forces that have no component parallel to the rotation axis: thus, F is in the plane of the page). We define the torque τ as a vector cross product of the position vector and the force
F r
⨯=τ Discuss t he direction and the magnitude of the torque .
2. Newton’s second law for rotation
(1). The figure shows a simple case of
rotation about a fixed axis. The
rotating rigid body consists of a single
particle of mass m fastened to the end
of a massless rod of length r . A force F acts as shown, causing the particle to move in a circle about the axis. The particle has a tangential component of acceleration t a governed by Newton’s second law:
t t ma F =. The torque acting on the particle is αατ)()(2mr r r m r ma r F F r t t ====⨯= . The quantity in parentheses on the right side of above equation is the rotation inertia of the particle about the rotation axis. So the equation can be reduced to ατI =.
(2) For the situation in which more than one force is applied to the particle, we can extend the equation as ατI =∑. Where ∑τ is the net torque (the sum of all external torques) acting on the particle. The above equation is the angular form of Newton’s second law .
(3) Altho ugh we derive the angular form of Newton’s second law for the special case of a single particle rotating about a fixed axis, it holds for any rigid body rotating about a fixed axis , because any such body can be analyzed as an assembly of single particles.
6.5 Work and Rotational Kinetic Energy
1. Work-kinetic energy theorem : Let’s again consider the situation of the figure, in which
force F rotates a rigid body
consisting of a single particle of
mass m fastened to the end of a
massless rod. During the rotation,
Force F does work on the body. Let us assume that the only energy of the body that changed by F is the kinetic energy. Then we can apply the work-kinetic energy theorem to get W I I K W K K K i f i f =-=∆⇒=-=∆222121ωω
Above equation is the angular equivalent of the work-kinetic energy theorem for translational motion. We derive it for a rigid body with one particle, but it holds for any rigid body rotated about a fixed axis .
2. We next relate the work W done on the body in the figure to the torque τ on the body due to force F. If the particle in Fig. 11-17 were move a differential distance ds along its circular path, the body would rotate through differential angle θd , with θrd ds =. We would get θτθd rd F ds F s d F dW t t ===⋅= . Thus the work done during a finite angular displacement from i θ to f θ is then
⎰=f i d W θθθτ. Above equation holds for
any rigid body rotating about a fixed axis .
3. We can find the power P for rotational motion ωτθτ===dt
d dt dW P。