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Coaster Dynamics
Physics Primer
Chapter 14
Circular Motion
One of earliest and greatest triumphs of Newton's Laws of Motion was that they led to the understanding of the motion of planets.
In Isaac Newton's time, astronomers had concluded by careful observations that the earth existed in a solar system of planets which revolved in circular (or near-circular) orbits around the sun. Disregarding any theological debates this may have stirred (with the sun being the center of the solar system, and not earth), this puzzled the scientists of the day. Because... how was it that heavenly bodies seemed to move naturally in circular orbits, when objects on earth did not?
The explanation is found in Newton's Second Law of Motion... and it led Newton to the introduction of his Universal Law of Gravitation.
Uniform Circular Motion
Newton's Second Law of Motion states that the acceleration of an object is in the direction of, and proportional to, the applied force. In the case of linear acceleration, this was somewhat intuitive, even at that time. With constant linear acceleration, the velocity changes magnitude, but not in direction (always being in the direction of the applied force).
However, recall that acceleration is a vector quantity, with both a magnitude and direction. Can there be a case of constant acceleration with a velocity change in direction only, and not in magnitude? The answer, of course, is yes... and it is called uniform circular motion.
In pondering the motion of the earth around the sun, Newton realized that there must be a force causing the earth to change direction in such a way that its path was a circle... that is, the earth was undergoing constant acceleration. He further surmised that the earth's orbit was a case of uniform circular motion -- or at least, close to it (the orbit is actually an ellipse, but that detail isn't key to this discussion).
In accordance with his Second Law, Newton proposed a force acting on the earth... which he called gravity. By the nature of the resulting circular motion, he deduced that the force must pull the earth inward, toward the center of the orbit. This is illustrated in Figure 14-A.
Figure 14-A. Force Diagram for Uniform Circular Motion.
The inward pulling force is called the centripetal force, and the resulting acceleration is called the centripetal acceleration (sometimes, also called the radial acceleration). In the case of uniform circular motion, the centripetal force is constant, and the resulting acceleration and orbital speed are constant.
All the mathematics of Newton's Second Law worked... but, what is the magnitude of the centripetal force?
In answering this question, Newton demonstrated yet again why he is considered a genius among history's greatest scientists and mathematicians. The problem was that the mathematics existing at the time did not offer a solution to this problem.
So... what did Isaac Newton do? He independently invented the mathematics of fluxional calculus (ie., derivatives and integrals).
With the aid of calculus, Newton was able to derive an equation for centripetal acceleration. For those with a calculus background, a derivation is included in the ADVANCED STUDY section that follows. Although the derivation below is not the same as Newton's, it provides a means to quickly confirm Newton's result.
For those who haven't studied calculus -- or just plain forgot everything you learned about it -- the equation for centripetal force is given below. Just trust us that it is correct.
The centripetal acceleration, ac, is equal to:
(14-1) ac = v²/R
where v is the velocity, and R is the radius.
And, after applying Newton's Second Law of Motion:
(14-2) Fc = mv²/R
where Fc is the centripetal force.
Physics Trivia: In accordance with Newton's Third Law of Motion, centripetal force has an equal and opposite twin brother... called centrifugal force. For motion in a roller coaster loop, the centripetal force is the force acting on you to pull you through the loop. The centrifugal force is the (equal and opposite) force exerted by you to your car seat. These terms are often interchanged mistakenly... but since they have equal magnitude, it usually doesn't matter very much. However, when performing motion analyses, you are generally interested in the forces acting on your system (ie., as depicted with a force diagram) -- so the correct term in this case is centripetal force.
ADVANCED STUDY
Derivation of Centripetal Acceleration
The motion of a particle undergoing uniform circular motion is illustrated in Figure 14-B.
Figure 14-B. Motion Diagram for Uniform Circular Motion.
The particle moves from point 1 to point 2, and for convenience, it is assumed the particle subtends an arc of a total angle, ß, with the angle from each endpoint to the x-axis equal to ß/2.
For uniform circular motion, the magnitude of the acceleration and velocity are constant. Thus, we have
(14-3) |a| = a = v/
t
(14-4) |v| = v = s/
t.
Inserting equation (14-3) into (14-4) yields
(14-5) v = s/( v/a ),
from which we find that
(14-6) ( v²/a ) = s.
In the diagram of Figure 14-B, the true particle displacement, from P1 to P2, is equal to the arclength, s. We will call the length of the line segment from P1 to P2,
s. Note that the magnitudes of s and
s will approach one another as the arc angle, ß, approaches zero.
We will now replace the displacement, s, in equation (14-6) with the approximate value,
s:
(14-7) ( v²/a ) =
s.
We point out now that equation (14-7) is only approximately correct -- and that it becomes closer and closer to the correct relationship as
s approaches zero.
From the diagram in Figure 14-B, a little trigonometry will show that:
(14-8)
s = (2R)(sin( ß/2 )).
Inserting equation (14-8) into (14-7) now yields
(14-9) ( v²/a ) = (2R)(sin( ß/2 )).
With the change in parameters, from
s to ß, we note again that (14-9) is only approximate -- and that it becomes closer and closer to the correct relationship as ß approaches zero. Thus, the true relationship is achieved for
(14-10) ( v²/a ) = ( lim ß –> 0 )( (2R)(sin( ß/2 )) ).
Calculus students will recognize the right hand side of equation (14-10) as the derivative of the equation with respect to ß.
The derivative of the sine function is equal to
(14-11) d( sin( ß/2 )/dß = (1/2)( cos( ß/2 ) ).
Inserting equation (14-11) into (14-10) gives us
(14-12) ( v²/a ) = (2R)( d( sin( ß/2 )/dß )
= (2R)(1/2)( cos( ß/2 ) ) = (R)( cos( ß/2 ) ).And, at the limit, ß = 0, and cos( ß/2 ) = 1. Thus, equation (14-12) reduces to
(14-13) v²/a = R.
Finally, we simply solve equation (14-13) for a, the centripetal acceleration:
(14-14) a = v²/R.
Centripetal Acceleration and G Forces
Just for fun... let's analyze the motion and forces for one of those funky NASA "G Force Trainers." You know what I mean... it has a big, long, crane-like arm that spins horizontally, and attached to the end is a small cage-like thing where you sit (facing toward the center) as the whole contraption spins around. The purpose of the machine is to subject the victim... err, test pilot, to a sustained high G force.
For the following specifications:
Test pilot's mass = 70 kg
Radius of machine's arm = 25 m
Rotation speed = 10 revolutions/minute,what will the G force on the test pilot be?
To begin, we note that for the spinning arm, the relationship between the rotation and tangential velocity is
(14-15) v = (R)(µ)
where µ is the rotation rate (radians/s). If you're not familiar with equation (14-15), you can learn more about rotational dynamics in any good physics textbook.
First, let's convert the rotation rate from RPM to radians/s, and calculate the tangential velocity:
  v = (R)(µ) = (25 m)(10 rev/min)(6.28 rad/1.0 rev)(1.0 min/60 s)
    = 26.2 m/s.Inserting the above value of v, and the other given parameters, into equation (14-2) yields
  Fc = mv²/R = (70 kg)(26.2 m/s )²/(25 m) = 1920 kg-m/s².
Finally, we convert the force, in kg-m/s² (ie., newtons) to G's:
  Fg = Fc/mg = (1920 kg-m/s²)/( (70 kg)(9.81 m/s²) ) = 2.79 G's.
Physics Trivia: A typical "Top Gun" jet fighter pilot may begin to loss consciousness if subjected to a sustained force of 5 G's, or more.
Nonuniform Circular Motion
If an object is traveling in a circular path, but the net force on it is not a constant centripetal force, then the object will experience nonuniform circular motion, and its velocity will vary. An important example of such motion is found in a vertical roller coaster loop.
Figure 14-B illustrates the motion of a roller coaster car as it travels through a vertical round loop.
Figure 14-C. Motion of a Roller Coaster Car in a Vertical Round Loop.
For purposes of our analysis, the movement in the z-direction is very small, compared to that in the x and y directions. Hence, as depicted in the figure, we can assume that the car travels only in the x-y plane.
Because of the geometric constraint imposed by the track, the car is compelled to move in a circular path -- which creates a centripetal force. If there were no gravity, the car would simply rotate around the loop in uniform circular motion. However, the presence of the gravity force greatly complicates matters.
The situation is depicted in the first diagram in Figure 14-B. Here, we see that in addition to the centripetal force (which is always directed toward the center of the loop), the force due to the weight of the car always acts in the downward (negative y) direction. As the car moves around the loop, the orientation of the two forces changes with time. Therefore, any calculation of forces and accelerations would need to be performed using the instantaneous parameters. Such calculations can get very complex, very quickly.
Nevertheless, one generalized way to analyze such systems is to resolve all the forces acting on the car into a set of only two net forces: a radial force and a tangential force. This is depicted in the second diagram in Figure 14-B.
In this case, the instantaneous acceleration is given by the equation:
(14-16) a = rar + tat
where a is the net acceleration, ar is the radial acceleration in the inward direction, r, and at is the tangential acceleration in the tangential direction, t.
Equation (14-16) can be employed in analyses by the mathematically gifted, or with the aid of a computer. In either case, such calculations are not a trivial matter... so we won't press this topic any further.
It may be of interest to note that in the
Coaster Dynamics program, the motion of the roller coaster car through loops is not performed in this manner. Instead, the analysis uses state space equations of motion in the form of differential equations. Although also extremely complex, this method provides a more systematic, and sanity-preserving, procedure. Furthermore, the computer calculations can be performed using "standard" numerical integration methods... making their solution easier.Nevertheless... despite the difficulties of analyzing systems with nonuniform circular motion, an indispensable principle of physics comes to the rescue: conservation of energy.
By employing the principle of conservation of energy, many useful analyses are possible, with a minimum of mathematical complexity. This is demonstrated in the example below.
Motion of a Roller Coaster Car in a Vertical Round Loop
For a roller coaster car traveling through a vertical round loop, calculate the velocity and centripetal force when the car is at the position corresponding to ß = 45 degrees... as depicted in the first diagram in Figure 14-C.
We are given the following data for the system:
The car's mass = 453.6 kg
Loop radius = 20 m
Loop entrance velocity (at ß = -90 degrees) = 35 m/s.By means of conservation of energy, we know that the change in kinetic energy of the car is equal to the negative change in potential energy:
(14-17) (1/2)(m)(v1² - v0²) = -mgh.
In this case, the origin of the coordinate system is at the loop entrance point (corresponding to the position ß = -90 degrees, as shown in the figure). At ß = 45 degrees, the change in height, h, is equal to
h = R + Rsin(45°) = (20.0)( 1 + sin(45°) ) = 34.1 m.
Solving equation (14-17) for the velocity at the desired point, v1, yields
v1² = ( v0² - 2gh )½
= ( (35.0 m)² - (2)(9.81 m/s²)(34.1 m) )½ = 23.6 m/s.From equation (14-2), the centripetal force is equal to
Fc = mv²/R = (453.6 kg)( 23.6 m/s)² )/(20 m) = 12600 kg-m/².
Finally, we convert the force, in kg-m/s² (ie., newtons) to G's:
  Fg = Fc/mg = (12600 kg-m/²)/((453.6 kg)(9.81 m/s²)) = 2.83.
Creating exciting -- and safe -- loops and hills is an important and challenging goal of all roller coaster builders. Be sure to see
Coaster Lab Lessons #8 and #9 to learn more about the ramifications of circular motion and G forces in roller coaster design.
Copyright © 2001, Cyclone Software, Pleasanton, CA, USA. All rights reserved.