## (23) Frames of Reference: The Centrifugal ForceThe other case of a moving frame of reference studied here is:
## (2) The local frame rotates around some fixed point.
An earlier section described motion in a circle and introduced the -
F
_{centripetal} = mV^{ 2}/ R
To anyone who has not studied motions and accelerations, "
As will be seen, the centrifugal force is not a "real" force, in the sense that in any motion calculated "in the frame of the universe" (or in one moving uniformly with respect to the universe) it does not appear at all. In such a frame, if an object moves around a circle, a centripetal force is needed to maintain that motion--or else it flies off at a tangent, with constant velocity along a straight line.
Unfortunately, if you sit (for instance) in a roller coaster car going around a vertical loop (as in the picture above, from an amusement park in Japan), it is a bit difficult to visualize your motion with respect to the fixed Earth. (To demonstrate that difficulty, we later solve the motion from that viewpoint.) It is
## EquilibriumEquilibrium is the state at which all forces balance each other, so that no net motion exists. Under usual conditions, this happens when all forces on the object in equilibrium add up to zero:F_{1} + F_{2} + ... = 0centrifugal force F with magnitude _{c}-
F
_{c} = m V^{2}/ Rthe direction of F. The "equilibrium" condition, required to avoid any motion in the rotating frame, is now
_{c} is AWAY from the centerF_{1} + F_{2} + ... + F_{c} = 0(If the above condition does not hold, the object will move inside the rotating frame and the situation is more complicated. For now let it just be said that yet another force then enters the equations, the Coriolis force, discussed qualitatively in the next section.) F from both sides of the equality gives_{c}F_{1} + F_{2} + ... = -F_{c}-F) has the same magnitude as _{c}F but the opposite direction, _{c}towards the center of rotation. It therefore equals the centripetal force needed to keep the object rotating at a fixed distance R from the axis. Viewed in the non-rotating frame of the universe, the above equation
F_{1} + F_{2} + ... = -F_{c}That result was alread derived from circular motions two sections back. The difference between the equations governing such motions in the outside frame F_{1} + F_{2} + ... = -F_{c}F_{1} + F_{2} + ... + F_{c} = 0## ExamplesWhen calculating the motions of the oceans and the atmosphere, it is much easier to use reference points on the rotating Earth and add a centrifugal force to all equations. That is one reason why the observed accelerationg due to gravity departs from the average value of 9.81: at the equator, the centrifugal force must be subtracted from the force of gravity, while at the pole no centrifugal force exists. Observations of g give values from 9.78 at the equators to 9.83 at the poles, but the centrifugal force is responsible for only part of that difference. The rest arises because the Earth is not a perfect sphere: the centrifugal force of its rotation causes its equator to bulge out, making the surface there more distant from the Earth´s center and thus weakening the gravitational pull.
- (You may also note that the direction of the centrifugal force is away from the
axis of rotation, and therefore it points away from the Earth´s center only at the equator. One therefore expects at all other points (except for the poles) a very slight difference between the vertical as defined by a plumb line and the direction to the center of the Earth.)
Another example is the "loop the loop" feature found on some roller coasters in amusement parks. There the track descends on a long slope and then, at the bottom, turns in a complete circle (
If you try to solve this problem from the point of view of the outside world, the situation can be confusing. At point A, on the top of the loop,
Let us try solve that motion, using the concept of the centripetal force. A car going around a loop, with radius R and velocity V, is accelerating at a rate of V
The centripetal force is provided by two sources: the weight -
mg + F
_{R} = + mV^{ 2}/ R _{R} pushes the car down, a negative one pulls it up. Hence -
F
_{R} = + mV^{ 2}/ R - mg above the car and therefore it can only push up against them. The rails then, reacting to the force, must push it down, somewhat similar to the situation in "Objects at Rest", in section #18 on Newton's second law. Thus F_{R} must be positive: if it were negative it would mean that the rails were pulling the car upwards, which they cannot do.
We thus require F -
mV
^{ 2}/ R - mg > 0 -
mV
^{ 2}/ R > mg-
--If
all forces on the car ceased at point A, it would continue along a straight line to point B, in accordance with Newton's first law. --If only gravity acted, it would follow a parabolato point C. --For the rails to exert a positive pressure, they must constrain the car to a tighter curvaturethan gravity alone, forcing it to move to point D. can be solved in the outside frame of reference--but at the price of a few complications. Viewed in the frame of the passengers, it boils down to the much simpler requirement that the centrifugal force at A is equal to or greater than gravity, which is exactly the last of the equations obtained above. From that it is easy to show that S must be at least one loop-radius higher than A. |

About a classroom demonstration of "loop the loop" with a "hot wheels" toy car--click here.

Site about **roller coasters**--the biggest, highest, the one with the most loops (8) and the one with the biggest loop ("Moonsault Scramble" in Japan, pictured at the beginning of this page)--click here.

**Next Stop: #24 Rotating Frames of Reference**

Author and curator: David P. Stern, u5dps@lepvax.gsfc.nasa.gov

Last updated6 March 1999