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been replaced by relativistic mechanics and by quantum mechanics in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