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OŸ “vector” which we have created by œ x b is artificial, or slightly diferent ïn its
character from ø and b, because it was made up with a special rule. lf œ and b
are called ordinary vectors, we have a special name for them, we call them polar
0ectors. Examples of such vectors are the coordinate ?, force #'", momentum 7ø,
velocity , electric fñeld #, etc.; these are ordinary polar vectors. Vectors which
involve just one cross product in their defnition are called a#al 0ectors or pseudo
uectors. Examples of pseudo vectors are, of course, torque 7 and the angular
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mmomentum E. It also turns out that the angular velocity œ is a pseudo vector,
as is the magnetic field Ö.
In order to complete the mathematical properties of vectors, we should know
all the rules for their multiplication, using dot and cross products. In our
applications at the moment, we will need very little of this, but for the sake of
completeness we shall write down all of the rules for vector multiplication so that
we can use the results later. These are
(a) œ<(b+c)=aœaxb+axe,
(b) (œa) x b= œ(œ x b),
e œ-(bxe)—=(axb)-c,
() (b xe) = (a x b) 6010)
(đ) œ < (b x e) = b(œ - c) — c(œ - b),
(e) axœ=0,
( œ-(œ x b) =0.
20-2 The rotation equations using cross products
Now let us ask whether any equations in physics can be written using the
cross product. The answer, of course, is that a great many equations can be so
written. For instance, we see immediately that the torque is equal to the position
vector cross the Íorce:
T—=rx Œ. (20.11)
This is a vector summary of the three equations 7x = 1; — zF¿y, etc. By the
same token, the angular momentum vector, if there is only one particle present,
1s the distanece from the origin multiplied by the vector momentum:
TL —rxp. (20.12)
For three-dimensional space rotation, the dynamical law analogous to the law #' =
dp/dt of NÑewton, is that the torque vector is the rate of change with time of the
angular momentum vector:
T = dL/dt. (20.13)
TÝ we sum (20.13) over many particles, the external torque on a system is the
rate of change of the total angular momentum:
Text — dL:oị /dt. (20.14)
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Another theorem: I the total external torque is zero, then the total vector
angular momentum of the system is a constant. Thịis is called the law of conser-
0ation oƒ angular momentum. TÝ there is no torque on a given system, its angular
mmomentum cannot change.
What about angular velocity? ls i a vector? We have already discussed
turning a solid object about a fñxed axis, but for a moment suppose that we are
turning i% simultaneously about #uo axes. It might be turning about an axis
inside a box, while the box is turning about some other axis. 'Phe net result of
such combined motions is that the object simply turns about some new axisl
The wonderful thing about this new axis is that it can be fgured out this way.
T the rate of turning in the z-plane is written as a vector in the z-direction
whose length is equal to the rate of rotation in the plane, and ïf another vector is
drawn in the -direction, say, which is the rate oŸ rotation in the zz-plane, then
1ƒ we add these together as a vector, the magnitude of the result tells us how
fast the object is turning, and the direction tells us in what plane, by the rule of
the parallelopgram. 'Phat is to say, simply, angular velocity is a vector, where we
draw the magnitudes of the rotations in the three planes as projections at right
angles to those planes.*
As a simple application of the use of the angular velocity vector, we may
evaluate the power being expended by the torque acting on a rigid body. The
pOwer, Of course, is the rate of change of work with time; in three dimensions,
the power turns out to be P =7 -ứ.
AII the formulas that we wrote for plane rotation can be generalized to three
dimensions. For example, If a rigid body is turning about a certain axis with
angular velocity œ, we might ask, “What is the velocity of a poïint at a certain
radial position r?” We shall leave it as a problem for the student to show that
the velocity of a particle in a rigid body is given by 0 = œ x?, where œ is
the angular velocity and z is the position. Also, as another example of cross
products, we had a formula for Coriolis force, which can also be written using
cross products: #2 = 2w x œ. That is, if a particle is moving with velocity 0
in a coordinate system which is, in fact, rotating with angular velocity œ, and
we want to think in terms of the rotating coordinate system, then we have to
add the pseudo force #,.
— * That this is true can be đerived by compounding the displacements of the particles of
the body during an infinitesimal time Af. It is not self-evident, and is left to those who are
interested to try to fgure it out.
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20-3 The gyroscope
Let us now return to the law of conservation of angular momentum. 'Phis law
may be demonstrated with a rapidly spinning wheel, or gyroscope, as follows
(see Eig. 20-1). TỶ we sit on a swivel chair and hold the spinning wheel with
1ts axis horizontal, the wheel has an angular momentum about the horizontal
axis. Angular momentum around a 0erfical axis cannot change because of the
(frictionless) pivot of the chair, so iƒ we turn the axis of the wheel into the vertical,
then the wheel would have angular momentum about the vertical axis, because it
is now spinning about this axis. But the ss¿em (wheel, ourself, and chair) canwnof
have a vertical component, so we and the chaïr have to turn in the direction
opposite to the spin of the wheel, to balance ït.
` JÐ_k h 1)
BEFORE AFTER
Fig. 20-1. Before: axis is horlzontal; moment about vertical axis = 0.
After: axis Is vertical; momentum about vertical axis Is still zero; man
and chair spin in direction opposite to spin of the wheel.
First let us analyze in more detail the thing we have just described. What is
surprising, and what we must understand, is the origin of the forces which turn
us and the chaïr around as we turn the axis of the gyroscope toward the vertical.
Jigure 20-2 shows the wheel spinning rapidly about the -axis. 'Pherefore is
angular velocity is about that axis and, it turns out, its angular momentum is
likewise in that direction. NÑow suppose that we wish to rotate the wheel about
the z-axis at a small angular velocity ©; what forces are required? After a short