Lesson Plan #34         http://www.phy6.org/stargaze/Lframes2.htm

#23    Accelerated Frames of Reference:   Inertial Forces   

#23a   Frames of Reference: The Centrifugal Force   

Uniformly moving frames of reference experience no new forces. Uniformly accelerating frames and rotating frames do so. If we want to express the equations of motion in their own frame of reference, we must always add "inertial forces" to represent the effects of their acceleration. The centrifugal force is one such force, described here and illustrated by examples.

Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern

This lesson plan supplements:

 #23  "Accelerated Frames of Reference:   Inertial Forces" Sframes2.htm,
                              on the web   on   http://www.phy6.org/stargaze/Sframes2.htm

  #23a  "Frames of Reference: The Centrifugal Force" Sframes3.htm,
                              on the web     http://www.phy6.org/stargaze/Sframes3.htm

Related web pages (not used in the lesson)
                         #23b  "Loop-the-Loop" Sframes4.htm,
                         #24a  "The Rotating Earth" Srotfram1.htm

"From Stargazers to Starships" home page: ....stargaze/Sintro.htm
Lesson plan home page and index:             ....stargaze/Lintro.htm

Note: This lesson uses vectors, and some way of denoting them
on the board and in the notebook must be agreed on by the class.
In this lesson plan, all vector quantities will be underlined.


Goals: The student will learn

  • That when applying the laws of motion to an object in an accelerating frame, using coordinates defined in such frame, one must always add an "inertial force", representing forces caused by the frame's acceleration.

  • In a uniformly accelerating frame, the inertial force opposes the acceleration. For instance, in a braking car (acceleration directed backward, opposed to velocity), the inertial force tends to push forward.

  • In a uniformly rotating frame, one must add to the forces a "centrifugal force" mv2/r, directed away from the axis of rotation.

  • To clearly distinguish between the centripetal and centrifugal forces:

    • The centripetal force is used when we calculate motion, using coordinates, velocities and accelerations of a non-rotating frame of reference.

    • The centrifugal force is needed to describe the same motion, using coordinates, velocities and accelerations of a rotating frame of reference.

  • Examples of the inertial forces:
    1. The force on objects inside a braking bus,
    2. The centrifugal force on objects on the rotating Earth
    3. The centrifugal force in a roller-coaster car going around a "loop the loop".

Terms: inertial force, accelerated frame of reference, rotating frame of reference, centrifugal force, bulge of the Earth.

Stories: Variation of g and the bulge of the Earth. (Optional, at the end of this lesson plan: pumping a swing)


Starting the lesson:
Up to now we have dealt with uniformly moving frames of reference. That was easy: moving from one such frame to another, it was only necessary to re-define velocities and coordinates in terms appropriate to the new frame. The forces were the same, because forces, as Newton showed, depend only on acceleration.

Tody we go a step further, to accelerating frames of reference--in particular, two examples:

  • a bus whose driver suddenly hit the brakes, and
  • a rotating frame of reference, e.g. a bus turning a sharp corner.

From experience everyone knows what to expect. In the first case, the passengers are flung forward, in the second, they are pushed to the outside of the curve, by what is known as a centrifugal force.

Both these are a rather special type of force, called inertial forces: they only appear in the accelerating frame.

  • If we treat the motion in the coordinates of the accelerating frame--e.g. if your coordinates are those of the inside of the bus--they must be included.

  • If you solve the problem using the coordinates of an outer frame of reference, which does not accelerate or rotate--e.g. if your coordinates refer to position in the outside world--they do not exist.

For example: the bus which suddenly stops.

  • If we calculate position relative to the outside world, the situation is very simple. The bus is slowed down by its brakes, but the passengers keep going, until something--the seat belt, the seat in front--stops them.

  • If we calculate position relative to the bus, the passengers are pushed forward, to the seat in front, to the limits of the seat belt.

This can be quite confusing. Some teachers and textbooks avoid inertial forces altogether or call them "fictional forces". As we will see, however, they can be quite useful. One thing you should remember: be sure you know to which frame of reference your coordinates belong. With accelerated frames, all frames are not all equivalent.

***********

The best known example, of course, is a rotating frame of reference, where the centripetal force pushes inwards, and the centrifugal force pushes outwards. Which to use? Here is the rule: [on the board--all please copy]

    The centripetal force is only used in a nonrotating frame of reference. It is the force holding the rotating mass to its curved path, and is directed towards the axis of rotation.

    The centrifugal force is only used in a rotating frame. It must be added to other forces in that frame, in order to take into account the frame's rotation, and is directed away from the axis of rotation.

As you will see, both describe the same physics, and either can be used to study motion. However, the centrifugal force seems more intuitive: this is what you feel when you sit in a car going around a sharp curve. You feel yourself pushed to the outside of the curve. Your mind is keyed to the frame of reference of the car that surrounds you, and in that frame you feel pushed outwards.

As a result, almost anyone knows intuitively what a centrifugal force is, while "centripetal force" is only familiar if you learned about it in school.

(Then continue with the material of section #23 in "Stargazers.")


Teaser

Ask students if anyone knows what is a "Thank-you-ma'am"--that is, as a noun-phrase, name of some object. And how is it connected to inertial forces?

    (If no one does, challenge the class to find out by the time of the next lesson, cautioning them that a pretty detailed dictionary may be needed.)

    Answer
        A "Thank-you-ma'am" (=madam) is a sharp turn in the road, or a sharp angle at the bottom of a dip in the road. It's a humorous American expression, probably from the stagecoach travel of the 1800s. When a stagecoach went through such a sudden turn, unprepared passengers were thrown against each other--not exactly pleasant, if you had hefty neighbors! The culprit, of course, was the inertial force caused (in the frame of the stagecoach) by the sudden turn.
Questions

An astronaut lies horizontally on a couch, flat on the back, inside a space shuttle as it takes off. In which direction is the astronaut accelerating?

    --The acceleration is directed forward (=upwards, at least at first).


Does the astronaut feel pushed--and if so, which way?

    The astronaut feels pushed down, towards the couch, opposite to the acceleration


We agree the astronaut's body is pressed towards the couch. Viewed from the outside world, why that pressure?
    Because the shuttle is accelerating, the couch is accelerating with it, and it in turn accelerates the astronaut's mass, through the pressure the couch exerts on it.


Viewed from inside the cabin, why that pressure?
    The cabin is an accelerating frame of reference with acceleration a, and it exerts an inertial force –ma on anything inside it.


When working with vectors in cartesian coordinates (x,y), it is useful to define unit vectors xu and yu.
    These are vectors of unit length in the directions of the coordinate axes


What are unit vectors useful for?
    Vectors combine two properties--direction and magnitude. Sometimes it is convenient so look at each of these separately. A vector V in the x-direction can be written V xu: the factor V contains all the information about the magnitude, while the unit vector tells everything about the direction.


Say you fire a gun at an angle upwards, so that
  • the initial upwards velocity is u, and
  • the initial horizontal velocity is w.
[draw it all on the board].
Let the direction of u be the y axis, and of w the x axis. After t seconds, the upward velocity is (u–gt) while the horizontal one remains w. Express the velocity vector V after time t.
    V = (u–gt) xu + w yu


Is motion around a circle with constant speed ("uniform rotation") an accelerated motion?
    Yes, it is.


Why is it accelerated, if the speed does not change?
    The speed doesn't, but the direction of the motion does. By Newton's laws, only if a body moves with constant speed along a straight path are no forces needed to maintain that motion.


So, for a body with mass m to move with constant speed V around a circle of radius R, a force is needed. What is that force called, what is its magnitude and what is its direction?
  • It is called the centripetal force
  • Its magnitude is m V2/R
  • Its direction is towards the center of the circle.


You are standing on the floor, and your body is subject to two forces:
  • Your weight of F1 newtons,
  • the reaction F2 newtons of the floor, not letting you sink any further.
Your body is in equilibrium--it does not move. What can you say about those two forces?
    They must cancel, that is, add up to zero: F1 + F2 = 0


[The discussion that follow may be helped by a crude drawing on the board].

You are sit on a whirling turntable, part of a carnival ride, at a distance R from the axis, going around at a velocity V. As before, F1 is your weight and F2 is the force exerted on you by the ride.
    You hold tight and keep your position. If ru is the unit vector pointing radially outwards from where you sit, which of these equations correctly describes your force balance?

F1 + F2 = (mV2/R) ru        

or
F1 + F2 = –(mV2/R) ru      ?

The total force F1 + F2 on you--in the first case it is outwards, in the second inwards. Which is it?
    Inwards: F1 + F2 = – mV2/R ru


Why?
    Because... as your body goes around the curve, in the absence of forces it would continue straight--but holding on to the turntable, you are instead pulled into the circle, towards the center. You need an inwards-directed centripetal force to keep you going around the circle.


But wait!
Another way of writing the same equation is

F1 + F2 + (mV2/R) ru = 0
How can you interpret this in the rotating frame?
    "You keep your position in the rotating frame only if all forces on your body add up to zero--including a centrifugal force (mV2/R) ru"
The second equation thus shows how the situation may be interpreted in the rotating frame of reference.


Note: the subject below is discussed in more detail in #24a "The Rotating Earth".

You live on a rotating Earth. How does the centrifugal force affect you on the equator?

    The force on your body is the sum of gravity and the centrifugal force. On the equator, the two are opposed, and therefore the effective value of g is slightly smaller there.


How does the centrifugal force affect you away from the equator?
    It reduces the pull of gravity by a lesser amount than at the equator. In addition, it also slightly shifts the direction in which bodies fall, because while gravity is radial, the direction of the centrifugal force is parallel to the equatorial plane.


In which direction is the shift?
    [sketch on the board the two forces and their head-to-tail vector addition: the gravity arrow goes down towards the center of the Earth, the centrifugal arrow continues from its end parallel to the equator.]

    Falling bodies are slightly displaced towards the equator.


Does the centrifugal force of the Earth's rotation affect the Earth itself?
    Yes, it, makes the earth oval instead of a perfect sphere. The equatorial radius of Earth is 6357 km, while the polar radius is 6378km (average radius, 6371 km)


Originally the meter was defined in terms of the distance from the equator to the pole, which was to be equal to 10,000,000 meters. If that were still the definition,would you expect the length of the equator to be 40,000 km? A little more? A little less?
    A little more.
    Note: the actual meter was defined as the distance between two fine scratches on a "standard meter" metal bar kept in Paris, supposedly based on the above distance. Later it was redefined in terms of wavelengths of light, which can be very precisely measured.


Jupiter and Saturn both rotate around their axes in about 10 hours. Would you expect them to be more oval or more spherical than the Earth?
    More oval; in fact, they look oval in the telescope. For Earth the difference between polar and equatorial radii is 1/297 of the radius; for Jupiter 1/15, for Saturn 1/9.5.


[Optional]
You are sitting on a rotating platform, right on the middle. You want one of the people sitting nearer to the edge of the platform to join you at the middle, so you extend your hand and pull that person in, against the centrifugal force.

What do you think: do you need energy to overcome the centrifugal force? Does your arm perform work?

    If you guessed that you did need energy, you are right.

There exists an interesting application. You are probably aware children can "pump up" a swing and keep it going; you may even have done so yourself. How is it done?

The process is best seen in swings on which you stand rather than sit. The centrifugal force on the swing varies: it is greatest at the bottom of the swing, where motion is fastest, but zero when the swing briefly stops at the end points. What you do is, near the end points you lower your body, then near the bottom of the swing you stand up straight again.

By standing up, you overcome the centrifugal force, which is directed towards your legs, so you invest energy. But energy is conserved and must go to somewhere else in the motion! It actually goes to the energy of the swinging motion, making the swing move more vigorously, or at least making it overcome the energy loss to friction, which would otherwise gradually slow it down.

The "pumping" of a swing by moving legs and body while sitting down is somewhat similar.

    [Incidentally: the "Exploratorium" science museum in San Francisco had--and maybe still has--a swing which can be "pumped" by a rope which controls a weight on it. By pulling up the weight near the bottom of the swing and letting go near the end points, you can increase its swings and keep them going.]


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Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   audavstern("at" symbol)erols.com .

Last updated: 10-24-2004