Download Lecture 21 April 4, 2017

Physics 3309
April 4, 2017
(21-1)
LAST TIME: Introduced noninertial reference frames and examples.
2
.
Setting up the reference frame on the surface of the earth is
shown on the figure to the right. y points north, x points
east, and z points outward or up. In the rotating reference
frame,  is given by
Ω sin
Ω cos
.
Note that the x-component is zero.  here is the colatitude
(measured down from the north pole) or 90o – latitude. Our
next job is to use our general equation to write the
component equations for x, y, and z in the rotating frame
of the earth. How big is Using the period of the earth’s
rotation as 24 hours, we get about 7.29×10-5 rad/s. You can
see why we get by without these corrections for many
applications.
Here is a figure that more clearly shows the basic
geometry. Let  be the distance from point O to the line
connecting the center of the earth to the north pole.
Comment on selecting an inertial reference frame here
since the earth rotates around the sun and the sun with our
solar system rotates around the center of the Milky Way
galaxy. Consider the omegas for these rotations.
Then
sin .This is important because it is the
radius of the circle that the origin of our coordinate
system has with respect to an inertial frame at the center
of the earth. Now let’s apply our equation to the reference frame on the surface of the earth for a
static situation; i.e., a mass attached to a string near the surface of the earth.
Let
0;
0;
0;
0; sin
.Now
we
assume that a mass m is supported on a string so that the only real forces acting on mass m are T
.Our equation now takes the form
and
0
sin
1 .
We usually identify T with mg, which is what we measure when we hang something on the end of
.This means that the figure
a string near the surface of the earth. We will also identify
that expresses each of these forces is as shown. If we let a be the angle
between mg and mgo, then we may set up the following relationship using
the law of sines.
sin
sin
sin 90
since90
istheanglebetween
and –mA. At the equator, 2R = 0.034 m/s2. max is about 0.1o and this
occurs at a latitude and colatitide of 45 degrees.
For dynamic cases near the surface of the earth, the Coriolis force is by far the most important as
we will see now. r is very small compared to R, so any centrifugal force will be small compared
to the acceleration of the origin of our frame attached to the surface of the earth.
Let’s do the problem in its entirety to first order in . Now our equation for motion near the surface
of the earth is given by
2
.F and –mA have
already been combined into mg, where g varies slightly depending on latitude. Therefore our
equation becomes
2
.
to get the x, y, and z components of the motion.
We need to calculate
Therefore
det
cos
0
sin
sin
cos
sin cos
Now we write everything in component form and try to solve the equations. Here are the three
equations, where m cancels.
2
cos
sin
;
2
cos ; and
2
sin Let’s check the zeroth order where  = 0. It is easy to see that
0
.These are
our usual equations that we got when we neglected the earth’s rotation. It is worth noting that
depend on ,so if we solve for , we can directly substitute that value in and get the
remaining equations solved.
2
cos
sin
;
We substitute the values for 2 tcos cos
2
sin
sin
2
2 ;
2
sin
backinto . We neglect terms that are of order  2 to get
2 cos Integrate to get
2
2 sin
sin
.
.
This may be integrated one more time to get
1
3
sin
cos
sin
.
We use this value of x(t) to get the values for y and z. Again  2 terms are neglected.
Here are the equations.
2
2
cos
; sin
cos
1
2
; sin
Now let’s go back to check on our intuition concerning dropping an object form rest at the equator.
At the equator,  = 90o.
0;
;
0.We want to find the drift
that occurs as a result of the earth’s rotation. First, find the time required for the object to hit the
earth. This set z(t) = 0.
; Our equations become
Therefore
; .
.Using h = 100 m gives a value of x = 2.2 cm.
Suppose that we give the object an initial velocity in the easterly direction so that
0.You will
then notice that the time of flight to the ground is not quite the same as before since the second
term in the equation for z(t) no longer vanishes.
Let’s look at the Coriolis force in different locations and with different velocities as another
qualitative example. (a) Move south near the north pole. (b) Move east on the equator. (c) Move
south on the equator.
(a)
(b)
(c)
so only the z-component of  contributes and
so  = 90 degrees and 2
2m
so  = 90 degrees and 2
3 2
is west (along –y).
Ω
2m Ω ,so up here.
Ω
0.
One of the most interesting applications that clearly
shows the effects of the earth’s rotation is the Foucault
pendulum. Typically, L is quite long, and the
connection to point P so as to allow torque-free
operation. Small ocsillations are considered as usual. In
our notation, the equation of motion is
2
.
g is the local value with the correction for the rotation
made. When  is small, the following approximations
are valid.
≅
;
;
; ≅ 0.
This is equivalent to saying that mass m is moving in a horizontal plane under the action of Tx and
Ty. Since ≅ 0,the equations of motion are given by
2
cos ,
2
sin .
and
Reminder – we have solved a similar set of equations before. Do you remember what they were?
.
Suppose Bx = 0 and look at the equations.
det
0
so you can see the similarities between
and
.
Our equations for the Foucault pendulum are
2Ω
0
and
0.
2Ω
Back in lecture 6, I showed you how to solve the equations for the motion of a charged particle ini
a magnetic field, but these equations have an extra wrinkle. They do not involve only the first and
second derivatives, but they also have the function itself in the equation. Remind about earlier
solution.
These equations seem ideally suited to solve using a complex variable given by
so that
and
.Multiply the equation for y by i and add the two equations to get
4 0.
2Ω
Were it not for the factor of i, this would be a damped harmonic oscillator, but the factor of i
changes things. As usual, we search for a solution of the form
and see that
.
and
Therefore,
2
0
Now we may get the equation for 
2
0
So
2
4
⁄
4
⁄
2
Usually,
≫ soweuse
0.57 .
.
.To see this consider L = 30 m. This gives
This gives a value for of
.
To understand what happens with this equation, let’s give some initial conditions. We need to
remember that  is a complex variable, so we need 4 initial conditions, one set for x and one set
for y. Choose the initial conditions as 0
; 0
0;
0.Remember that C1
and C2 are complex as well to get the four unknowns.
Now
.
We also need the general expression for
.
.
0
0 .
and
Therefore,
Using
0
0
0so
.
0
Now we equate the real and imaginary parts of the left- and right-hand sides to get
0
⇒
0
5 0.
.
Now go back to the equation a + c = A to get
1
and
2
Now a = A – c so
2
.
2
Now we may write the final result for (t).
2
2
2
cos
sin
.
Recall that x(t) = Re (t) so
cos
cos
sin
sin
≅
cos
cos
.
≅
sin
cos
Now we get y(t) in the same way.
Im
cos
sin
sin
cos
The behavior is shown on the next page in graphical form.
6 .
The period of the oscillations in
z is given by 2/z and at 42
degrees latitude is given by
z =  cos 48o, which implies a
period of about 36 hours. At the
north pole, we would get a
period of 24 hours.
NEXT TIME: Begin Chapter 10 on Rotation of Rigid Bodies
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