Chapter 9 Flashcards
Newton’s Second Law in a Rotating Frame
Derivative in nonrotating frame as function of derivative in rotating frame.
Newton’s Second Law in a Rotating Frame
Centrifugal Force
Coriolis Force
Coriolis Force compared to Centrifugal Force
In the last expression I have replaced RS-2
by V, the speed of a point on the equator as the earth rotates with angular velocity Q.
Since V is approximately 1000 mi/h, shows that for projectiles with v << 1000
mi/h it will be a good starting approximation to ignore the Coriolis force, and this is
what I shall do in this section
The vector Omega x r is the velocity of an object as it is
dragged eastward with speed Omega*rho by the earth’s rotation. Therefore,
the centrifugal force, m (Omega x r) x Omega, points radially outward from
the axis and has magnitude m Omega^2rho.
The earth’s rotation carries the object around
a circle of latitude, and the vector velocity of this circular
motion as seen from a nonrotating frame) is tangent to this circle
The Foucault Pendulum
As a final and striking application of the Coriolis effect, let us consider the Foucault
pendulum, which can be seen in many science museums around the world and is
named for its inventor, the French physicist Jean Foucault (1819-1868).This is a pendulum made of a very heavy mass m suspended by a light wire from a tall ceiling.
This arrangement allows the pendulum to swing freely for a very long time and to
move in both the east-west and north-south directions. As seen in an inertial frame,
there are just two forces on the bob, the tension T in the wire and the weight mgo . In
the rotating frame of the earth, there are also the centrifugal and Coriolis forces, so
the equation of motion in the earth’s frame is
A Foucault pendulum comprises a bob of mass m
suspended by a light wire of length L from the point P on a high
ceiling. The tension force on the bob is shown as T and its x and
y components are Tx and Ty . For small oscillations the angle /3
is very small.
X and Y Equations of Motion
We now need to examine the x and y components of the equation of motion (9.58).
This requires that we identify the x and y components of T. If you look at Figure 9.16,
you will see that, by similar triangles, Tx / T = —x/L and similarly for Ty . Combining
this with (9.59), we find that
Foucault Pendulum Again
These two equations of
motion can be rewritten as?
Hint, Omegacostheta = Omegaz, the z compenent of the earth’s angular velocity.
We can solve the coupled equations using the trick, introduced in Chapter
2, of defining a complex number. Recall that not only does this complex number contain the same information as the
position in the xy plane, but a plot of eta in the complex plane is an actual bird’s eye
view of the pendulum’s projected position (x, y). If we multiply the second equation
of (9.62) by i and add it to the first, we get the single differential equation. This is a second-order, linear, homogeneous differential equation and so has exactly
two independent solutions. Thus if we can find two independent solutions, we shall
know that the most general solution is a linear combination of these two. As often
happens, we can find two independent solutions by inspired guesswork: We guess
that there is a solution of the form
However, eventually the complex exponential e -iQz t begins to change, causing
the complex number ri = x iy to rotate through an angle Q z t. In the Northern
Hemisphere, where Q is positive, this means that the number x + iy continues
to oscillate sinusoidally (due to the factor cos coot), but in a direction that rotates
clockwise. That is, the plane in which the pendulum is swinging rotates slowly
clockwise, with angular velocity Q 2 , as indicated in Figure 9.17(b). In the Southern
Hemisphere, where Q z is negative, the corresponding rotation is counterclockwise. If the Foucault pendulum is located at colatitude 9 (latitude 90° — 9), then the rate
at which its plane of oscillation rotates is
Q, = Q cos O. (9.68)
At the North Pole (0 = 0), Q, = Q and the rate of rotation of the pendulum is the same
as the earth’s angular velocity. This result is easy to understand: As seen in an inertial
(nonrotating) frame, a Foucault pendulum at the North Pole would obviously swing
in a fixed plane; meanwhile, as seen in the same inertial frame, the earth is rotating
counterclockwise (as seen from above) with angular velocity Q. Clearly then, as seen
from the earth, the pendulum’s plane of oscillation has to be rotating clockwise with
angular velocity Q.
At any other latitude, the result is much more complicated from an inertial point of
view, but the rate of rotation of the Foucault pendulum is easily calculated from (9.68).
At the equator (9 = 90°), Q, = 0 and the pendulum does not rotate. At a latitude
around 42° (the approximate latitude of Boston, Chicago, or Rome),
Q, = Q cos 48° ti 3S .
Since Q equals 360°/day, 3 Q = 240°/day, and we see that in the course of 6 hours (a
time for which a long, well-built pendulum will certainly continue to swing without
significant damping), the pendulum’s plane of motion will rotate through 60° — an
easily observable effect.
Inertial Force in an Accelerating but Nonrotating Frame
The motion of a body, as seen in a frame that has acceleration A relative to an inertial
frame, can be found using Newton’s second law in the form m*rdoubledot = F + Finertial, where
F is the net force on the body (as measured in any inertial frame) and Finertiai is an
additional inertial force
Finertial = -mA
