AN
INTRODUCTION TO NEWTONIAN MECHANICS by Edward Kluk Dickinson State University, Dickinson ND |
m Dvx / Dt = Fx
(0a), m Dvy
/ Dt = Fy
(0b)
when transformed to a uniformly rotating (x',y') reference frame, takes a
new form that includes Coriolis and centrifugal pseudoforces
m Dv'x / Dt = F'x +
2mv'y +
m2x'
(1a)
m Dv'y / Dt = F'y -
2mv'x +
m2y'
(1b).
These pseudoforces are formally describing effects of body's inertia as seen
from a rotating reference frame. With their help we have explained a phenomenon
of stationary satellites.
Now we will try to learn how an ordinary pendulum behaves
in inertial and noninertial reference frames. We are considering a reference
frame as inertial if its axes have directions toward fixed distant stars.
Pendulum in an inertial
reference frame
In Fig. 1 there is presented a pendulum swinging along
the circular arc of radius L. The gravity force mg acting upon
the pendulum is resolved into two components. The component along the pendulum
string is neutralized by a strength of the string, so it does not influence
the pendulum motion. The other component mg
sin pulls the pendulum
along the circular arc toward its bottom (equilibrium) point. This component
is a net force F acting on the pendulum. It can be resolved
into its components Fx and
Fy along x and y axis respectively. These components
are
Fx = - mg
sincos and
Fy = mg
sinsin .
Inserting these components into Newton second law for
inertial frames of reference (0a), (0b) we have
m Dvx / Dt = - mg
sincos (2a),
m Dvy / Dt = mg
sinsin .
(2b)
Canceling masses on both sides of each equation we can instantly conclude
that the pendulum motion is independent of the pendulum mass. Generally,
however, this set of equations is too difficult to be discussed here without
a substantial simplification. Limiting pendulum swings to small angles
(less than
11.5o or 0.2
radians) we notice that a maximum of the
force in the equation (2b) is much smaller than in the equation
(2b). Check it on your calculator finding values of
cos and
sin. This
means that accelerations, velocities and displacements in
y direction are very small and can be disregarded. Also
cosin the equation
(2a) can be replaced by 1. Then, the simplified
equation (2a) can be rewritten as follows
Dvx / Dt = - g
sin= -(g/L) L
sin = - (g/L) x .
(3)
But this looks like the equation for a harmonic
oscillator with the spring constant k = g/L and mass
m = 1 kg. Thus, the period T for
such pendulum is given by
T =
2(L/g)½
, (3a)
and x displacement from equilibrium point is
x(t) = xo
cos(t) , where
=
(g/L)½ .
(3b)
Now it is a time for a first real experiment. Make at
least 1 m long pendulum using thin strong string and rather
heavy machine screw nut. Arrange a point like fixed suspension with the freedom
of swinging in any vertical plane. Measure the pendulum length
L from the suspension point to the center of the
nut. Measure with a stopwatch elapsed time for ten full swings counting swings
very carefully. Make two more such measurements. The results should
be almost identical. If they differ substantially, you probably have miscounted
swings. Find from your data an average period T
for a single swing. Knowing L and
T calculate the gravity acceleration
g. You should be off its value of 9.8 m/s2
no more than 2%. This means that our theory works.
So far we were assuming that the pendulum swings only
in a single vertical plane. But if realising the pendulum you give it a small
side push it will be moving along an oval path on the spherical surface with
the radius L . Let us assume that the plane x,y in Fig.
2 is tangential to the mentioned above sphere at the pendulum equilibrium
point O. Notice, we have changed the meaning of
y-axis !!! Now the unmarked force in Fig. 2 pulling the pendulum
toward its equilibrium position, if we are regarding only a small swings,
has the magnitude F = k (x2 +
y2)½ ( the force is proportional
to the pendulum distance from the equilibrium point) and its components are
Fx = - k x and
Fy = - k y . Here
x and y are coordinates
of the pendulum and k = g/L. Applying to
this situation Newton second law (0a,b) we have
Dvx / Dt = - kx
(4a), Dvy / Dt
= - ky
(4b).
It means that each coordinate of the pendulum behaves like a harmonic oscillator.
We have already found a solution of (4a) . This solution is
x(t) = xo
cost . Thus (4b) has
a solution y(t) = yo
cost . Using exactly
the same method the additional solutions
x(t) = xo
sint and
y(t) = yo
sint
can be found.
Now imagine yourself starting the pendulum from the
intersection of x-axis and the oval trajectory in Fig.
2. If you just let it go without any additional push, then its y coordinate
will be always zero and x coordinate will be described by the function
x(t) = xo
cost . But if you induce
an initial velocity vo in y direction,
then the y coordinate will not stay equal to zero. Then the only possible
choice is to take y(t) = yo
sint because we
must also have y(0) = 0. To find a shape of trajectory
let us make the following simple calculation
[x(t)/xo]2 + [y(t)/yo]2
=
[cost]2 +
[sint]2 =
1.
Thus this trajectory is an ellipse with the semiaxes
xo and
yo. If yo =
xo the ellipse becomes a circle.
In our particular case
yo depends on
vo . A relation between them follows from
the y component of pendulum velocity. This component
vy(t) = Dy(t)/Dt = D[yo
sint]/Dt =
yo
cost ,
thus yo = vo
/. Then this particular
motion of the pendulum is described as follows
x(t) = xo
cost and
y(t) = ( vo
/)
sint
(5a).
If the pendulum starts from y axis with an initial velocity
in opposite direction to the direction of x axis, then its
motion is described by
x(t) = - (vo
/)
sint and
y(t) = yo
cost
(5b).
Pendulum in a
rotating reference frame
Suppose that the pendulum is fixed on a slowly rotating
turntable and the pendulum suspension is placed exactly on the rotation axis
of the turntable. In Fig.3 there are shown two reference frames. The lab
frame fixed in the lab and rotating frame fixed on the turntable. The origins
of both frames are exactly on the rotation axis of the turntable. Initially
the pendulum bob is at the point P in the rotating frame. The pendulum
is released from the turntable without any push
at the instance t = 0 when both reference frames overlap.
If we disregard the earth rotation, the subsequent motion of the pendulum
as seen from the lab frame is described by the equations
(4a,b) and (5b). The initial speed
vo in the negative x direction as seen
from the lab frame is related to yo (the distance
of point P from the origin of rotating reference frame) and
T
(the angular speed of the turntable), and vo =
yo
T
. Therefore an observer placed in the lab frame will see a trajectory
of the pendulum as an ellipse with the semiaxis
yo along y axis and
the semiaxis yo
T
/ along
x axis.
But can we disregard the earth rotation? Studying a
relatively slow motion in a time interval of a few minutes we can because
centrifugal and Coriolis forces related to the earth rotation are very small.
They are proportional to the angular velocity of the earth, and the square
of this velocity, respectively. This velocity is only 7.27 x
10-5 rad/s. Therefore its effects in a short time interval
are hardly measurable. In a long time interval, however, they cumulate and
become well visible.
The same motion seen by an observer in the rotating
frame is described by prohibitively looking equations
(1a,b). Luckily enough we do not need to solve them. We already
know how coordinates x,y of the pendulum evolve in time in the lab frame,
and we also discovered lately how they can be transformed from the lab frame
to the rotating frame. The transformation
formulae as applied to the present situation have the following form
x' = x cos
T
t + y
sinT
t (6a),
y' = - x
sinT
t + y
cosT
t
(6b).
Inserting into them x and y as described by the
equations (5b) with vo = yo
T
we obtain
x'(t) = yo[-
(T
/)
sint
cosT t
+ cost
sinT t
] (7a),
y'(t) =
yo[(T
/)
sint
sinT
t + cost
cosT t
] (7b).
This result looks still very complicated and practically impossible to visualize
without graphing. Some conclusions are easier to reach if it is reshaped
with help of trigonometric relations like cos a cos b =
(1/2)[cos(a+b) + cos(a-b)] etc. Then
x'(t) = (yo/2)[1 -
(T
/)]
sin(+T)t - (yo/2)[1
+(T
/)]
sin(-T)t (8a),
y'(t) = (yo/2)[1 -
(T
/)]
cos(+T)t + (yo/2)[1
+(T
/)]
cos(-T)t (8b).
According to this result the pendulum motion seen from the rotating frame
is a superposition of two simple motions, clockwise circular motion with
the radius (yo/2)[1 -
(T
/)] and angular frequency
+T, and counterclockwise
circular motion with the radius (yo/2)[1 +
(T
/)] and angular frequency
-T . Here we are observing a
rotational Doppler shift. The pendulum seen from the lab frame exhibits a
single angular frequency
, whereas in the
rotational frame it exhibits two shifted frequencies
+T and
-T . Its original
angular frequency
is modulated by
the angular frequency
T
of the turntable.
In a particular case when
=
T
the result is very simple because (8a,b)
are reduced to x'(t) = 0 and y'(t)
= (yo/2)[1 +
(T
/)] = yo
. So, in the rotating frame the pendulum stays at rest because
a centrifugal force acting upon it equilibrates the component of gravity
force that pulls the pendulum toward the turntable rotation axis. This is
a very similar case to the stationary satellite
case.
Now you are ready to simulate with help of included
applet some pendulum trajectories as they are seen from lab and rotating
frames. For the simulation the formulae (8a,b) are used. Do not forget that
this pendulum always starts from a rest in the rotating frame.
Set the pendulum period Tp
to 10 s and the turntable rotation period
Tt also to 10s. This means that
=
T
, and in the rotating frame the pendulum should stay at rest whereas in the
lab frame it should have a circular trajectory. After running this case in
the rotating frame click "reset" and select the inertial (in this case lab)
frame. A change of the frame resets other parameters, so you have to restore
them to desired values. Run it to see what you are expecting to see.
Now clear the view window and make two new experiments taking
Tp=5s , Tt = 10s
and Tp=5s , Tt =
50s . Conclusions from the obtained patterns are obvious. A smaller
ratio of Tp / T t leads to
a narrower elliptical trajectory in the inertial reference frame and a trajectory
with more spikes in the rotating frame. Imagine replacing the turntable by
our planet Earth, placing a very long pendulum on the North Pole and directing
x and y-axes of an inertial frame toward two fixed stars. Well, we have a
Foucault pendulum on North Pole. If our pendulum is 100 m long its period
would be only 20 s whereas earth rotational period is 86400 s. Thus, its
trajectory viewed from the inertial frame would be an exteremely elongated
ellipse with its longer axis/shorter axis ratio equal to 4320. The same
trajectory viewed from the earth would have 8640 spikes per 24 hours. Hopefully
you understand now how Foucault pendulum works but you must admit that the
understanding process is neither simple nor short. And this one is probably
the simplest and shortest if you start from scratch.
So far we were investigating pendulum motions with
Tp / T t ratio smaller or
equal one. Evidently for these cases initial centrifugal force is smaller
than initial force pulling the pendulum toward the rotation axis. This is
why, the pendulum initially "falls" toward the rotation axis but misses it
being redirected by Coriolis and centrifugal forces. For
Tp / T t ratio greater than
one pendulum trajectories in the inertial frame are still elliptical but
their views in the rotating frame is quite different. Examine two such cases
taking Tp = 20s , Tt =
10s with two turns and
Tp=50s , Tt = 10s
with three turns. This time initial centrifugal forces are greater than initial
forces pulling the pendulum toward the rotation axis. It results in initial
motion away from the rotation axis.
The most important conclusion following from our
investigation of pendulum motion is that it provides information about reference
frames. It tells us how much inertial or noninertial a given frame of reference
is or, to what extend we may apply in it Newton second law without taking
corrections for pseudoforces. You have to be aware that we really do not
know if ideally inertial frames exist. We only know that some reference frames
are more inertial than the others. In these more inertial frames Newton second
law correctly predicts motions for longer time intervals. For most engineering
problems reference frames fixed on the earth surface are enough inertial.
But long range artillery shells are influenced by Coriolis force due to their
high speed and relatively long time of flight.
Pendulum in vertically
accelerating reference frame
Motion of pendulum is influenced not only by a
rotation of the reference frame but also by a linear acceleration of such
frame. Let us assume that (x,y)-frame in Fig.4 is an inertial frame. The
pendulum is fixed in (x',y')-frame accelerating vertically with a constant
acceleration a . A real case of such situation could be arranged
in an accelerating elevator. But practically elevators cannot accelerate
for too long. In (x,y)-frame the pendulum motion will be described
with help of the equations (0a,b) with properly chosen
net force components Fx and
Fy.
Introduction
Discovering existence
of inertial and noninertial reference frames
we have learned that Newton second law describing planar motion in an inertial
(x,y) reference frame
The relations between the pendulum coordinates in both reference frames are following (see Fig.4)
x(t) = x'(t), y(t) = y'(t) + yo(t) where yo(t) = yo(0) + (a/2) t2 .
Here yo(0) describes position of (x',y') frame with respect to (x,y) frame at t = 0. Then
vx(t) = Dx(t)/Dt = Dx'(t)/Dt = v'x(t) and vy(t) = Dy(t)/Dt = Dy'(t)/Dt + a t = v'y(t) + a t .
Consequently
Dvx(t)/Dt = Dv'x(t)/Dt and Dvy(t)/Dt = Dv'y(t)/Dt + a . (9)
The net force components acting on the pendulum contain components Tx and Ty of the force exerted by the string and gravity force. Therefore they can be presented as follows
Fx = Tx and Fy = Ty - mg . (10)
Inserting the relations (9) and (10) into (0a,b) we obtain the equations describing pendulum motion in (x',y')-frame
m Dv'x / Dt = Tx (11a), m Dv'y / Dt = Ty - m(g + a) . (11b)
Inserting the relations (10) into (0a,b) the equations describing pendulum motion in (x,y)-frame are obtained
m Dvx / Dt = Tx (12a), m Dvy / Dt = Ty - mg. (12b) .
But the equations (11a,b) and (12a,b) differ only by the term - ma in (11b). Therefore motion of pendulum fixed in (x',y')-frame looks almost like motion of the same pendulum fixed in (x,y)-frame except that in (x',y')-frame on a top of the gravity force - mg there is an additional pseudoforce - m a. This new pseudoforce describes a well known inertial effect. Put on a smooth horizontal surface a sheet of smooth paper and on this paper a heavy smooth object. Pull quickly this sheet in a horizontal direction. If you pull it fast enough, the object will not move whereas the sheet of paper will. If an actual acceleration of the sheet of paper is a , with respect to a reference frame placed on this paper the object moves with an acceleration - a . Then, it looks like a force of - ma is acting on the object in this reference frame.
We have already solved the equations (0a,b) for a pendulum with small swings. But the equations (12a,b) are the same as (0a,b) for a pendulum. Then for small swings the pendulum period is given by the formula (3a) and x coordinate of the pendulum varies in time according to the formula (3b). Almost the same solution apply also to the equations (11a,b) . The only difference is that we have to replace g by g + a . This replacement is a clear consequence of comparison of (11b) and (12b). Therefore for a pendulum in the accelerating reference frame
T = 2{L/(g + a)}½ , (13a)
and
x'(t) = x'o cos(t) , where = {(g + a)/L}½ . (13b)
Therefore the pendulum in the accelerating reference frame (x',y') feels this acceleration as an additional "gravitational" field. This field can be positive or negative. If the (x',y')-frame is freely falling down, then a = - g or g + a = 0 , consequently = 0 and x'(t) stays constant. So the pendulum appears to be weightless in this frame. The same kind of weightlessness are experiencing astronauts on the board of orbiting Shuttle or Mir station. But there the earth gravity acceleration is neutralized by another pseudoforce which is a centrifugal force. Notice, the reference frames like freely falling elevator, Shuttle or Mir station (if they are not rotating) look remarkably like inertial frames where Newton second law without inclusion of pseudoforces is valid. The last statement is a simplified form of principle of equivalence which assumes identity of physics laws in inertial frames and mentioned above frames.
Start the incorporated applet, select vertically accelerating frame with the tracer off and measure ten pendulum periods for the following accelerations: - 4.9, 4.9, 9.8 and 19.6 m/s2. Compare your results with predictions given by the formula (13a).
Evaluation
If at this point you :
the objectives of this lesson are fully achieved. If you have doubts try to read it once more concentrating on them, but do not try to memorize this text. Physics is not about memorizing, it is about understanding.
Last update: May 7, 1997 | E - mail to Edward Kluk |
Copyright (c) 1996 Edward Kluk |