《物理双语教学课件》Chapter 6 Rotation 定轴转动

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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

1 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

sr is measured relative to the positive direction of the x axis, and  is given by

(radianmeasure).

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

1rev360o2r2rad r4.If the body rotates about the rotation axis as in the right figure, changing the angular position of the reference

2 line from 1 to 2, the body undergoes an angular displacement

given by 21

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

 of a rotating body can be either positive or negative, depending on whether the body is rotating in the direction of increasing  (counterclockwise) or decreasing  (clockwise).5.Angular velocity

(1).Suppose that our rotating body is at angular position

1

at time t1 and at angular position 2 at time t2.We define the average angular velocity of the body in the time interval t from t1 to t2 to be

21t2t1t

In which t.

is the angular displacement that occurs during

(2).The (instantaneous) angular velocity , with which we shall be most concerned, is the limit of the average angular velocity as t

is made to approach zero.Thus

limdt0tdt

If we know

(t), we can find the angular velocity  by

3 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 . (5).We establish a direction for the vector of the angular velocity  by using rule,

a as right-hand 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 the angular velocity at times

2

and

1

and t1, respectively.The average angular acceleration of the rotating body in the interval from t1 to t2 is defined as

In which

21t2t1 t is the change in the angular velocity that occurs

4 during the time interval t.

(2).The (instantaneous) angular acceleration , with which we shall be most concerned, is the limit of this quantity as is made to approach zero.Thus

limdt0tdtt

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/s2) or the revolution per second-squared (rev/s2).

(4).The angular acceleration also is a vector.Its direction depends on the change of the angular velocity. 7.Rotation with constant angular acceleration:

dddt0tdt

d10td(0t)dt0tt2dt2Here we suppose that at time

t0, 00.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: d Angular velocity：



dsdr

vr

5 Angular acceleration：

atr anv

6.2 Kinetic Energy of Rotation 1.To discu kinetic energy of a rigid body, we cannot use the familiar formula

Kmv2/2

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,

K111122m1v12m2v2m3v3mivi2 2222In which mi is the ma of the ith particle and

is its speed.The sum is taken over all the particles in the body.2.The problem with above equation is that

is not the same for all particles.We solve this problem by substituting for v in the equation with

r, so that we have

211Kmi(ri)2(miri2)22

In which  is the same for all particles.3.The quantity in parentheses on the right side of above equation tells us how the ma of the rotating body is distributed about its axis of rotation.

(1).We call that quantity the rotational inertia (or moment of

6 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

Imiri2

(2).The SI unit for I is the kilogram-square meter (kgm2).(3).The rotational inertia of a rotating body depends not only on its ma but also on how that ma is distributed with respect to the rotation axis.

4.We can rewrite the kinetic energy for the rotating object as

K1I22

Which gives the kinetic energy of a rigid body in pure rotation.It’s the angular equivalent of the formula

2KMvcm/2, which gives the kinetic energy of a rigid body in pure translation.

7 6.3 Calculating the Rotational Inertia 1.If a rigid body is made up of discrete particles, we can calculate its rotational inertia from

Imiri2.

2.If the body is continuous, we can replace the sum in the equation with an integral, and the definition of rotational inertia becomes

Ir2dm.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 ma, 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 ma 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

Irrddr2R0rdr320111dR42(R2)R2mR2

422a thin rod (1)I1r2drl2l213r3l2l21l11()32(l)l2ml2 321212(2)

13l03l1I2I1m()2ml2212I2r2drl1(l)l23

341ml2ml2ml2121238 3.The parallel-axis theorem: If you know the rotational inertia of a body about any axis that paes through its center of ma, you can find its rotational inertia about any other axis parallel to that axis with the parallel-axis theorem:

IIcmMh2

Here M is the ma 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 ma of the arbitrarily shaped body shown in cro 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 ma element with coordinates x and y.The rotational inertia of the body about the axis through P is then

Ir2dm[(xa)2(yb)2]dm

Which we can rearrange as

9 I(x2y2)dm2axdm2bydm(a2b2)dmIcm00MhIcmMh22 6.4 Newton’s Second Law for Rotation

1.Torque: The following figure shows a cro section of a body that is free to rotate about an axis paing through O and perpendicular to the cro 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  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  as a vector cro product of the position vector and the force

rF 

Discu the 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 ma m fastened to the end

10 of a male 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 governed by Newton’s second law: acting on the particle is

Ftmatat

.The torque

rFFtrmatrm(r)r(mr2)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

I.

(2) For the situation in which more than one force is applied to the particle, we can extend the equation as I.Where  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) Although 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 aembly of single particles.

11 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 ma m fastened to the end of a male rod.During the rotation, Force F does work on the body.Let us aume 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

KKfKiWK1212IfIiW22

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  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 with dsrdd, ..We would get

dWFdsFtdsFtrddThus the work done during a finite angular displacement from i to f is then

Wd.Above equation holds for

if12 any rigid body rotating about a fixed axis.3.We can find the power P for rotational motion

PdWd dtdt 13

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