Classical MechanicsMechanics is the study of how things move: how planets move around the sun, how a skier moves down the slope, or how an electron moves around the nucleus of an atom. So far as we know, the Greeks were the first to think seriously about mechanics, more than two thousand years ago, and the Greeks' mechanics represents a tremendous step in the evolution of modern science. Nevertheless, the Greek ideas were, by modern standards, seriously flawed and need not concern us here. The development of the mechanics that we know today began with the work of Galileo (1564-1642) and Newton (1642-1727), and it is the formulation of Newton, with his three laws of motion, that will be our starting point of our course. In the late eighteenth and early nineteenth centuries, two alternative formulations of mechanics were developed, named for their inventors, the French mathematician and astronomer Lagrange (1736-1813) and the Irish mathematician Hamilton (1805- 1865). The Lagrangian and Hamiltonian formulations of mechanics are completely equivalent to that of Newton, but they provide dramatically simpler solutions to many complicated problems and are also the taking-off point for various modern developments. Unfortunately these formulations will be out of our discussion because they are not part of the high school physics at this moment. The term classical mechanics is somewhat vague, but it is generally understood to mean these three equivalent formulations of mechanics. Until the beginning of the twentieth century, it seemed that classical mechanics was the only kind of mechanics, correctly describing all possible kinds of motion. Then, in the twenty years from 1905 to 1925, it became clear that classical mechanics did not correctly describe the motion of objects moving at speeds close to the speed of light, nor that of the microscopic particles inside atoms and molecules. The result was the development of two completely new forms of mechanics: relativistic mechanics to describe very high-speed motions and quantum mechanics to describe the motion of microscopic particles. I wish to include an introduction to relativity in the "optional" Chapter later in this course. Quantum mechanics requires a whole separate course, and I have made no attempt to give even a brief introduction to quantum mechanics. Newton's Laws of Motion Although classical mechanics has been replaced by relativistic mechanics and by quantum mechanics in their respective domains, there is still a vast range of interesting and topical problems in which classical mechanics gives a complete and accurate description of the possible motions. I shall discuss problems in the framework of the Newtonian formulation. Space and Time Newton's three laws of motion are formulated in terms of four crucial underlying concepts: the notions of space, time, mass, and force. This section reviews the first two of these, space and time. In addition to a brief description of the classical view of space and time, I give a quick review of the machinery of vectors, with which we label the points of space.

VectorsVectors can be approached from three points of view-geometric, analytic, and axiomatic. Although all three points of view are useful, we shall need only the geometric and analytic approaches in our discussion of mechanics. From the geometric point of view, a vector is a directed line segment. In writing, we can represent a vector by an arrow and label it with a letter capped by a symbolic arrow. In print, bold- faced letters are traditionally used.In order to describe a vector we must specify both its length and its direction. Unless indicated otherwise, we shall assume that parallel translation does not change a vector. Thus the arrows at left all represent the same vector.If two vectors have the same length and the same direction they are equal. The vectors B and C are equal:B = C.The length of a vector is called its magnitude. The magnitude of a vector is indicated by vertical bars or, if no confusion will occur, by using italics. For example, the magnitude of A is written , or simply A. If the length of A is , then = A = .If the length of a vector is one unit, we call it a unit vector. A unit vector is labeled by a caret; the vector of unit length parallel to A is . It follows that =,and converselyA = . The Algebra of VectorsMultiplication of a Vector by a Scalar If we multiply A by a positive scalar b, the result is a new vector C = bA. The vector C is parallel to A, and its length is b times greater. Thus = , and =b .The result of multiplying a vector by -1 is a new vector opposite in direction (anti parallel) to the original vector.Multiplication of a vector by a negative scalar evidently can change both the magnitude and the direction sense. Addition of two vectors Addition of vectors has the simple geometrical interpretation shown by the drawing. The rule is: To add B to A, place the tail of B at the head of A. The sum is a vector from the tail of A to the head of B.Subtraction of Two Vectors Since A - B = A + (- B), in order to subtract B from A we can simply multiply it by -1 and then add.The sketches below show how.

An equivalent way to construct A - B is to place the head of B at the head of A. Then A - B extends from the tail of A to the tail of B, as shown in the right hand drawing above.It is not difficult to prove the following laws. We give a geometrical proof of the commutative law; try to cook up your own proofs of the others.A+B=B+A Commutative lawA + (B + C) = (A + B) + C Associative lawc(dA) = (cd)A(c + d)A = cA + dAc(A + B) = cA + cB Distributive law

Proof of the Commutative law of vector addition

Although there is no great mystery to addition, subtraction, and multiplication of a vector by a scalar, the result of "multiplying" one vector by another is somewhat less apparent. Does multiplication yield a vector, a scalar, or some other quantity? The choice is up to us, and we shall define two types of products which are useful in our applications to physics.Scalar Product ("Dot" Product) The first type of product is called the scalar product, since it represents a way of combining two vectors to form a scalar. The scalar product of A and B is denotedby A B and is often called the dot product. A. B is defined byA. B = cosHere is the angle between A and B when they are drawn tail to tail.Since cos is the projection of B along the direction of A,A B = (projection of B on A).

Similarly,A B = (projection of A on B).If A. B = 0, then = 0 or = 0, or A is perpendicular to B (that is, cos = 0). Scalar multiplication is unusual in that the dot product of two nonzero vectors can be 0.Note that A A = IAI2.By way of demonstrating the usefulness of the dot product, here is an almost trivial proof of the law of cosines. Example 1 Law of CosinesC= A+BC.C=(A+B).(A+B)C2 =2 + 2 +2 cos This result is generally expressed in terms of the angle :C2=A2+B2-2AB cos(We have used cos =cos (-) = - cos ) Note: Work and the Dot ProductThe dot product finds its most important application in the discussion of work and energy. As you may already know, the work W done by a force F on an object is the displacement d of the object times thecomponent of F along the direction of d. If the force is applied at an angle to the displacement,W = (F cos )d.Granting for the time being that force and displacement are vectors,W = F. d. Vector Product ("Cross" Product) The second type of product we need is the vector product. In this case, two vectors A and B are combined to form a third vector C. The symbol for vector product is a cross:C = A X B. An alternative name is the cross product.The vector product is more complicated than the scalar product because we have to specify both the magnitude and direction of A X B. The magnitude is defined as follows: ifC = A X B,then.where is the angle between A and B when they are drawn tail to tail. (To eliminate ambiguity, is always taken as the angle smaller than .) Note that the vector product is zero when = 0 or even if IAI and I BI are not zero.When we draw A and B tail to tail, they determine a plane. We define the direction of C to be perpendicular to the plane of A and B. A, B, and C form what is called a right hand triple. Imagine a right hand coordinate system with A and B in the xy plane as shown in the sketch. A lies on the x axis and B lies toward the y axis. If A, B, and C form a right hand triple, then C lies on the z axis. We shall always use right hand coordinate systems such as the one shown at left. Here is another way to determine the direction of the cross product. Think of a right hand screw with the axis perpendicular to A and B. Rotate it in the direction which swings A into B. C lies in the direction the screw advances.(Warning: Be sure not to use a left hand screw. Fortunately, they are rare. Hot water faucets are among the chief offenders; your honest everyday wood screw is right handed.)A result of our definition of the cross product is thatB X A = -A X B.Here we have a case in which the order of multiplication is important. The vector product is not com- mutative.. (In fact, since reversing the order reverses the sign, it is anticommutative.)We see thatA A=0for any vector A.Area as a VectorWe can use the cross product to describe an area. Usually one thinks of area in terms of magnitude only. However, many applications in physics require that we also specify the orientation of the area. For example, if we wish to calculate the rate at which water in a stream flows through a wire loop of given area, it obviously makes a difference whether the plane of the loop is perpendicular or parallel to the flow. (In the latter case the flow through the loop is zero.) Here is how the vector product acomplishes this:Consider the area of a quadrilateral formed by two vectors, C and D.The area of the parallelogram A is given byA = base X height= CD sin (Discussions are left at this point. Ask the questions that arises naturally)= IC DI.If we think of A as a vector, we haveA = C D.

Components of a Vector The fact that we have discussed vectors without introducing a particular coordinate system shows why vectors are so useful; vector operations are defined without reference to coordinate systems. However, eventually we have to translate our results from the abstract to the concrete, and at this point we have to choose a coordinate system in which to work.AyyA

Axx

For simplicity, let us restrict ourselves to a two-dimensional system, the familiar xy plane. The diagram shows a vector A in the xy plane. The projections of A along the two coordinate axes are called the components of A. The components of A along the x and y axes are, respectively, Ax and Ay. The magnitude of A is | A | = ( + , and the direction of A is such that it makes an angle = arctan (Ay/ Ax) with the x axis. Since the components of a vector define it, we can specify a vector entirely by its components. ThusA = (Ax, Ay) or, more generally, in three dimensions,A = (Ax,Ay,Az). Prove for yourself that | A | = (Ax2 + Ay2 + Az2. The vector A has a meaning independent of any coordinate system. However, the components of A depend on the coordinate system being used. Unless noted otherwise, we shall restrict ourselves to a single coordinate system, so that if A= B,

then Ax = Bx Ay = By Az = Bz.

The single vector equation A = B symbolically represents three scalar equations. All vector operations can be written as equations for components. For instance, multiplication by a scalar gives cA = (cAx,cAy). The law for vector addition is A + B = (Ax + Bx, Ay + By, Az + Bz). By writing A and B as the sums of vectors along each of the coordinate axes, you can verify that A B = AxBx + AyB y + AzBz. We shall defer evaluating the cross product until the next section. Example :Vector Algebra

Let A = (3,5, 7) B = (2,7,1).

Find A + B, A - B, | A |, | B |, A. B, and the cosine of the angle between A and B.

A + B = (3 + 2, 5 + 7, 7 + 1) = (5,12, -6) A B = (3 2, 5 7, 7 1) = (1, 2, 8) | A | = (32 + 52 + 72 = = 9.11 | B | = (22 + 72 + 12 = = 7.35 A.B = 3 2 + 5 7 7 1 = 34cos (A, B) = = = 0.507

Example :Construction of a Perpendicular Vector Find a unit vector in the xy plane which is perpendicular to A = (3,5,1). We denote the vector by B = (Bx, By, Bz). Since B is in the xy plane, Bz = 0. For B to be perpendicular to A, we have A B= 0.

A. B = 3Bx + 5By = 0

Hence By = Bx. However, B is a unit vector, which means that Bx2 + By2 = 1. Combining these gives Bx2 + Bx2 = 1, or Bx == 0.857 and By = Bx = 0.514. The ambiguity in sign of Bx and By indicates that B can point along a line perpendicular to A in either of two directions.

Base VectorsBase vectors are a set of orthogonal (perpendicular) unit vectors, one for each dimension. For example, if we are dealing with the familiar cartesian coordinate system of three dimensions, the base vectors lie along the x, y, and z axes. The x unit vector is denoted by , the y unit vector by and the z unit vector by.The base vectors have the following properties, as you can readily verify:k

. = . = . = 1 . = . = . = 0 = = = . We can write any vector in terms of the base vectors. A = Ax + Ay + Az The sketch illustrates these two representations of a vector. To find the component of a vector in any direction, take the dot product with a unit vector in that direction. For instance,Az = . A . Az

AyAxzAyx

It is easy to evaluate the vector product A B with the aid of the base vectors.

A B= )Consider the first term:

B = + + .

(We have assumed the associative law here.) Since = 0, = , and = ,we find

B = ). The same argument applied to the y and z components gives

B = )

B = ).

A quick way to derive these relations is to work out the first and then to obtain the others by cyclically permuting x, y, z, and ,, (that is, x y, y z, z x, and , .) A simple way to remember the result is to use the following device: write the base vectors and the components of A and B as three rows of a determinant like thisA B =

= + (For instance, if A = + 3 and B = 4 + + 3, then,

A B = = 10 7 11.

Displacement and the Position VectorSo far we have discussed only abstract vectors. However, the reason for introducing vectors here is concretethey are just right for describing kinematical laws, the laws governing the geometrical properties of motion, which we need to begin our discussion of mechanics. Our first application of vectors will be to the description of position and motion in familiar three dimensional space. Although our first application of vectors is to the motion of a point in space, don't conclude that this is the only application, or even an unusually important one. Many physical quantities besides displacements are vectors. Among these are velocity, force, momentum, and the gravitational and electric fields.

Although vectors define displacements rather than positions, it is in fact possible to describe the position of a point with respect to the origin of a given coordinate system by a special vector, known as the position vector, which extends from the origin to the point of interest. We shall use the symbol r to denote the position vector. The position of an arbitrary point P at (x,y,z) is written aszrxyP( x, y, z)

r = (x, y, z) = x + y + z. Unlike ordinary vectors, r depends on the coordinate system. The sketch to the right shows position vectors r and r' indicating the position of the same point in space but drawn in different coordinate systems. If R is the vector from the origin of the unprimed coordinate system to the origin of the primed coordinate system, we havey'Pr'rRzz'xyx'

r' = r R. In contrast, a true vector, such as a displacement S, is independent of coordinate system. As the bottom sketch indicates, zxyRx'y'z'

Sr1r2

S = r2 r1 = ( + R) ( + R) = .

Velocity and Acceleration Motion in One Dimension Before applying vectors to velocity and acceleration in three dimensions, it may be helpful to review briefly the case of one dimension, motion along a straight line.Let x be the value of the coordinate of a particle moving along a line. x is measured in some convenient unit, such as meters, and we assume that we have a continuous record of position versus time.The average velocity of the point between two times, t l and t 2, is defined by = (We shall often use a bar to indicate an average of a quantity.)The instantaneous velocity v is the limit of the average velocity as the time interval approaches zero. = The limit we have introduced in defining is precisely that involved in the definition of a derivative. In fact, we have = . In a similar fashion, the instantaneous acceleration is a = = .

The concept of speed is sometimes useful. Speed s is simply the magnitude of the velocity: s = |v|.

Motion in Several Dimensions Our task now is to extend the ideas of velocity and acceleration to several dimensions. Consider a particle moving in a plane. As time goes on, the particle traces out a path, and we suppose that we know the particle's coordinates as a function of time. The instantaneous position of the particle at some time t1 is r(t1) = [x (t1), y(t1)] or r1 = (x1y1) Position at time t2Position at time t1r1r2(x2 , y2)(x1 , y1)r2 r1r1r2

where x1 is the value of x at t = t1 , and so forth. At time t2 the position is r2 = (x2,y2).The displacement of the particle between times t1 and t2 isr2 r1 = (x2 x1, y2 y1) We can generalize our example by considering the position at some time t, and at some later time t + .t. The displacement of the particle between these times isr(t + t)rr(t)

r = r(t + t) r(t). This vector equation is equivalent to the two scalar equations x = x(t + t) x(t) y = y(t + t) y(t). The velocity v of the particle as it moves along the path is defined to bev = = . yy(t + t)try(t)xx(t)x(t + t)

which is equivalent to the two scalar equationsx = = y = = Extension of the argument to three dimensions is trivial. The third component of velocity is z = = . Our definition of velocity as a vector is a straightforward generalization of the familiar concept of motion in a straight line. Vector notation allows us to describe motion in three dimensions with a single equation, a great economy compared with the three equations we would need otherwise. The equation v = dr/dt expresses the results we have just found.

Alternatively, since r - x + y + z, we obtain by simple differentiation = + + as before. Let the particle undergo a displacement r in time t. in the limit t 0, r becomes tangent to the trajectory, as the sketch indicates. However, the relation r t = v t, Which become exact in the limit t 0, shows that v is parallel to r; the instantaneous velocity v of a particle is everywhere tangent to the trajectory.

Alternative Approach (without using vectors)

Rectilinear Kinematics:The kinematics of a particle is characterized by specifying, at any given instant, the particles position, velocity, and acceleration.

Position:The straight-line path of a particle will be defined using a single coordinate axis s, Fig K-1a. The origin O on the path is a fixed point, and from this point the position coordinate s is the distance from O to the particle, usually measured in meters (m) and the sense of direction is defined by the algebraic sign on s. although the choice is arbitrary, in this case s is positive since the coordinate axis is positive to the right of the origin. Likewise, it is negative if the particle is located to the left of O. Realize that position is a vector quantity since it has both magnitude and direction. Here, however, it is being represented by the algebraic scalar s since the direction always remains along the coordinate axis.sOPosition (a)s

Displacement:The displacement of the particle is defined as the change in its position. For example, if the particle moves from one point to another, Fig. K-1b, the displacement issODisplacement (b)Fig K-1ss's

s = s' s

In this case s is positive since the particles final position is to the right of its initial position i.e.. s' > s. Likewise, if the final position were to the left of its initial position. s would be negative.The displacement of a particle is also a vector quantity, and is should be distinguished from the distance the particle travels. Specifically, the distance traveled is a positive scalar that represents the total length of path over which the particle travels. Another important feature of the distance traveled is that it never decreases.Velocity:If the particle moves through a displacement s during the time interval t, the average velocity of the particle during this time interval is

vavg =

If we take smaller and smaller values of t, the magnitude of t, the magnitude of s becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as v = , or

v = (1)

Since t or dt is always positive, the sign used to define the sense of the velocity is the same as that of s or ds. For example, if the particle is moving to the right, Fig K-1c, the velocity is positive; whereas if it is moving to the left, the velocity is negative. The magnitude of the velocity is known as the speed, and it is generally expressed in units of m/s.OVelocity (c)

ssv

Occasionally, the term average speed is used. The average speed is always a positive scalar and is defined as the total distance traveled by a particle, sT, divided by the elapsed time t, i.e.,

(vsp)avg = OAverage velocity and Average speed (d) Fig. K-1(Cont.)

ssP'P

For example, the particle in Fig K-1d travels along the path of length sT in time t, so its average speed is (vsp)avg = sT / t, but its average velocity is vavg = s/t.

Acceleration.Provided the velocity of the particle is known at two points, the average acceleration of the particle during the time interval t is defined as

aavg = Here v represents the difference in the velocity during the time interval t, i.e., v = v' v, Fig K-1e.OAcceleration (c)

savv'

The instantaneous acceleration at time t is a vector that is found by taking smaller and smaller values of t and corresponding smaller and smaller values of v, so that a = , ora = (2)

Substituting Eq. (1) into this result, we can also write

a = ODeceleration (f)Fig K-1(Cont.)

savv'P'P

Both the average and instantaneous acceleration can be either positive or negative. In particular, when the particle is slowing down, or its speed is decreasing, the particle is said to be decelerating. In this case, v' in Fig K-1f is less then v, and so v = v' v will be negative. Consequently, a will also be negative, and therefore it will act to the left, in the opposite sense to v. Also, notice that if the particle is originally at rest, then it can have an acceleration if a moment later it has a velocity v'; and, if the velocity is constant, then the acceleration is zero since v = v v = 0. Units commonly used to express the magnitude of acceleration are m/s2 orFinally, an important differential relation involving the displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between Eqs. (1) and (2), which gives

a ds = vdv(3)

Although we have now produced three important kinematic equations, realize that the above equation is not independent of Eqs. (1) and (2)

Constant Acceleration.a = ac.When the acceleration is constant, each of the three kinematic equations ac = dv/dt, v = ds/dt, and acds = vdv can be integrated to obtain formulas that relate ac, v, s, and t.

Velocity as a Function of Time.Integrate ac = dv/dt, assuming that initially v = v0 when t = 0.

=

v = v0 + act (Constant Acceleration)(4)

Position as a Function of Time.Integrate v = ds/dt = v0 + act, assuming that initially s = s0 when t = 0.

= dt

s = s0 + v0t + act2 (Constant Acceleration)(5)

Velocity as a Function of Position.Either solve for t in Eq. (4) and substitute Eq. (5) or integrate v dv = acds, assuming that initially v = v0 at s = s0.

=

v2 = +2ac (s s0) (Constant Acceleration)(6)

The algebraic sings of s0, v0, and ac used in the above three equation are determined from the positive direction of the s axis an indicated by the arrow written at the left of each equation. Remember that these equations are useful only when the acceleration is constant and when t = 0, s = s0, v = v0. A typical example of constant accelerated motion occurs when a body falls freely toward the earth. If air resistance is neglected and the distance of fall is short, then the downward acceleration of the body when it is close to the earth is constant and approximately 9.81 m/s2 or 32.2 ft/s2.

More about the Derivative of a VectorIn Sec. 1.6 we demonstrated how to describe velocity and acceleration by vectors. In particular, we showed how to differentiate the vector r to obtain a new vector v = dr / dt. We will want to differentiate other vectors with respect to time on occasion, and so it is worthwhile generalizing our discussion.A(t + t)(A)A(t)

Consider some vector A(t) which is a function of time. The change in A during the interval from t to t + t isA + AAACase 1

A = A(t + t) A(t). In complete analogy to the procedure we followed in differentiating r, we define the time derivative of A by =

It is important to appreciate that dA/dt is a new vector which can be large or small, and can point in any direction, depending on the behavior of A.A + AAACase 2

There is one important respect in which dA/dt differs from the derivative of a simple scalar function. A can change in both magnitude and direction-a scalar function can change only in magnitude. This difference is important. The figure illustrates the addition of a small increment A to A. In the first case A is parallel to A; this leaves the direction unaltered but changes the magnitude to | A | + | A |. In the second, A is perpendicular to A. This causes a change of direction but leaves the magnitude practically unaltered.In general, A will change in both magnitude and direction. Even so, it is useful to visualize both types of change taking place simultaneously. In the sketch to the left we show a small increment A resolved into a component vector .A|| parallel to A and a component vector .A perpendicular to A. In the limit where we take the derivative, .A|| changes the magnitude of A but not its direction, while .A changes the direction of A but not its magnitude.AAA||A

Students who do not have a clear understanding of the two ways a vector can change sometimes make an error by neglecting one of them. For instance, if dA/dt is always perpendicular to A, A must rotate, since its magnitude cannot change; its time dependence arises solely from change in direction. The illustrations below show how rotation occurs when .A is always perpendicular to A. The rotational motion is made more apparent by drawing AAAAAAAAA

the successive vectors at a common origin.AAAAAAA

Contrast this with the case where A is always parallel to A. A

A

A

A

A

A

A

A

A

Drawn from a common origin, the vectors look like this:

A

A

A

A

The following example relates the idea of rotating vectors to circular motion.

Example :Uniform Circular Motion Circular motion plays an important role in physics. Here we look at the simplest and most important case-uniform circular motion, which is circular motion at constant speed. Consider a particle moving in the xy plane according to r = r (cos t + sin t), where r and are constants. Find the trajectory, the velocity, and the acceleration.|r | = [r2 cos2 t + r2 sin2 tUsing the familiar identity sin2 + cos2 = 1.yrtx

|r | = [r2 cos2 t + r2 sin2 t = = constant. The trajectory is a circle. The particle moves counterclockwise around the circle, starting from (r, 0) at t = 0. It traverses the circle in a time T such that T = 2. is called the angular velocity of the motion and is measured in radians per second. T, the time required to execute one complete cycle, is called the period.yrtxv

v = = r( sin t + cos t ) We can show that v is tangent to the trajectory by calculating v r: v. r = r2 ( sin t cos t + cos t sin t) = 0 Since v is, perpendicular to r, it is tangent to the circle as we expect. Incidentally, it is easy to show that |v| = r = constant. xtray

a = = r2 [ cos t sin t] = 2r The acceleration is directed radially inward, and is known as the centripetal acceleration. We shall have more to say about it shortly.

Example : Circular Motion and Rotating Vectors the motion given by position vectorr = r (cos t + sin t). The velocity is v = r ( sin t + cos t). Since r. v = r2( cos t sin t + sin t cos t) = 0, we see that dr /dt is perpendicular to r. We conclude that the magnitude of r is constant, so that the only possible change in r is due to rotation. Since the trajectory is a circle, this is precisely the case: r rotates about the origin.We showed earlier that a = 2 r. Since r. v = 0, it follows that a v = 2r v = 0 and dv / dt is perpendicular to v. This means that the velocity vector has constant magnitude, so that it too must rotate if it is to change in time.That v indeed rotates is readily seen from the sketch, which shows v at various positions along the trajectory. In the second sketch the same

c

b

v

a

hgfeda

bcdefghv

velocity vectors are drawn from a common origin. It is apparent that each time the particle completes a traversal, the velocity vector has swung around through a full circle.Perhaps you can show that the acceleration vector also undergoes uniform rotation Suppose a vector A(t) has constant magnitude A. The only way A(t) can change in time is by rotating, and we shall now develop a useful expression for the time derivative dA/ dt of such a rotating vector. The direction of dA/dt is always perpendicular to A. The magnitude of dA/ dt can be found by the following geometrical argument.A(t). A = constant

The change in A in the time interval t to t + t isA = A (t + t) A(t). Using the angle defined in the sketch, |A| = 2A sin .v

= tr

For