![]()
![]()
Coaster Dynamics
Physics Primer
Chapter 8
Newton's Second Law of Motion
To develop his laws of motion, Newton first needed to expand upon Gaileo's idea of inertia. As proposed by Galileo, inertia is the property of matter that resists changes in motion. In order to quantify inertia, Newton introduced the property of mass.
In the precise world of physics, mass is the measure of an object's inertia. Mass is not the same as the weight or size of an object (which can easily be confused with the imprecise language of common speech). Stated simply... the more mass an object has, the larger will be the force required to change its motion.
The relationship between mass and acceleration is quantified by Newton's Second Law of Motion... which is the most fundamental of all laws in classical mechanics.
Newton's words can be loosely translated as follows:
Newton's Second Law of Motion:
The acceleration of an object is directly proportional to the net force on the object, is in the direction of the net force, and is inversely proportional to the mass of the object.This somewhat wordy statement of his Second Law of Motion was needed in Newton's time because the mathematical equations we use today weren't invented yet. All those words are summed up in the tidy equation:
(8-1) a ~ F/m.
Note that the acceleration, a, is a vector quantity, the force, F, is also a vector, but the mass, m, is a scalar.
With the assignment of appropriate units (think Metric... not British),
we can replace the proportionality sign with an equality sign -- yielding the famous equation:
(8-2) F = ma.
Mass, Weight, and Force
Intuitively, we generally associate mass with weight. However, they are not the same. Weight is the force exerted on an object by gravity.
From Newton's Second Law of Motion, weight is equal to:
(8-3) W = mg
where g is the acceleration due to gravity (equal to 9.81 m/s² on earth at sea level).
If you are limited to life on the surface of earth, mass and weight are inextricably tied. However, mass is a more fundamental property than weight. For example, on earth, the weight of a 1.0 kg mass is 1.0 newton. But, on the moon, where gravity is only about 1/6 that on earth, a 1.0 Kg mass would weigh 0.165 newtons.
The confusion between mass and weight is one of the most egregious deficiencies of the British Engineering System of measurement. The use of the same term, the "pound", in reference to both mass and weight is not only a matter of verbal confusion, but it greatly complicates calculations relating masses and forces. The key point to remember is that mass and weight are not equal... so a pound-force is not the same as a pound-mass.
Physics Trivia: In the year 1999, the $110 million unmanned spacecraft, Mars Polar Lander, sent to investigate Mars, crash landed during its descent from orbit. After an extensive NASA investigation, it was concluded that "the 'root cause' of the loss of the spacecraft was the failed translation of english units into metric units in a segment of ground-based, navigation-related mission software."[1] The calculation error was caused by the failure to use the correct conversion factor, gc.
In this case, a simple mistake in the calculation relating forces and masses directly caused the catastrophic destruction of a $110 million spacecraft.
[1] Source: Mars Climate Orbiter, Mishap Investigation Board, Phase I Report, NASA Report, November 10, 1999.
Force Diagrams
When analyzing the motion of a system, the first step is to identify the "system" (ie., identify the objects and forces to include, and decide what items not to include). The second step is to draw a force diagram. A force diagram is simply a drawing of the object(s) in questions, and all the forces, geometry, and interactions.
For example, the force diagram for the idealized model of a roller coaster car is shown below.
Figure 8-A. Force Diagram for an Idealized Roller Coaster Car.
Using the information from the force diagram, the next step is to derive equations describing the motion of the system.
Motion of a Sliding Mass on a Frictionless Inclined Plane
Consider the case of a block sliding down an inclined plane with no friction between the block and plane. The force diagram is shown below.
Figure 8-B. Force Diagram for a Sliding Block on a Frictionless Inclined Plane.
For this analysis, we choose a x-y coordinate system which remains fixed to the plane surface, with the positive y-axis pointing upward (opposite the direction of gravity). The force, W is the weight, N is the "normal" force exerted on the bottom of the block by the plane. Note that N is always perpendicular to the surface of the plane. Because we assume no friction, there are no forces in the direction parallel to the plane surface.
From the geometry in Figure 8-B, it can be shown that the net force acting on the block in the y-direction is:
(8-4) Fy = Ncos(ß) - W.
What is the normal force, N? To determine its value, note that since the block must slide along the surface of the plane, the acceleration in the direction perpendicular to the surface must always be zero. From Newton's Second Law of Motion, we then know that the net force in that direction must also be zero. Resolving the normal force components yields:
(8-5) 0 = Wcos(ß) - N
and therefore,
(8-6) N = Wcos(ß).
For the case of a non-zero angle, ß, combining equations (8-2), (8-3), (8-4) and (8-6) yields:
(8-7) ay = Fy/m = ( mgcos²(ß) - mg )/(m) = (g)( cos²(ß) - 1 ).
Notice that if the block is on a flat surface, ie. ß = 0, equations (8-4) and (8-7) reduce to Fy = -W and ay = 0.
From the geometry in Figure 8-B, it can also be shown that a displacement, s, along the surface of the plane will have x and y components related by
(8-8) tan(ß) = sy/sx.
Hence, the accelerations will similarly be related as follows:
(8-9) ax = ( ay )/( tan(ß) ).
Equations (8-7) and (8-9) can be called the equations of motion for the system. In this case, they relate the x and y components of the acceleration -- which is sufficient to fully describe the motion of the block.
ADVANCED STUDY
Differential Form of Newton's Second Law of Motion and State Space Equations
Newton's Second Law of Motion, F = ma, is one of the most well known equations in physics (along with Albert Einstein's famous equation, E = mc²). It follows from the interpretation of Newton's words in Principia Mathematica Philosophiae Naturalis.
However, in Principia, Newton actually expressed his Second Law in terms of an object's momentum, not its mass. The following translation is the most fundamentally correct statement of Newton's Second Law of Motion:
(8-10) F = dp/dt
where p is the momentum... which is equal to the product of mass and velocity:
(8-11) p = mv.
Expressed in words, equation (8-10) says that the force is equal to the time rate of change of the momentum. In mathematical jargon, this is called a differential equation
Why is equation (8-10) more correct than equation (8-2)? Because F = ma is only valid when the mass is constant.
Consider, for example, the case of calculating the propulsion force and acceleration of a liquid fuel rocket. For a rocket engine, the vast majority of the total mass of the system is composed of the fuel. As the rocket begins its launch, the mass will decrease rapidly as the fuel is burned away. Thus, F = ma cannot be applied accurately due to the changing mass.
On the other hand, equation (8-10) and (8-11) are correct, where F is the instantaneous force, p is the instantaneous momentum, m is the instantaneous mass, and v is the instantaneous velocity.
Let us now reconsider the equations of motion for a block sliding down a frictionless inclined plane.
In the section above, we derived equations (8-7) and (8-9), which we called the equations of motion. However, we could have derived other equations to describe the motion (for instance, equations relating the velocity components). There is no particular form for the equations of motion that is more true than any other. The only requirements are that they are sufficient to fully describe the motion, and that they are valid.
Returning to the notion of state determined systems, equations (8-7) and (8-9) are equations of motions for the state variables, ay and ax.
Within the field of Mechanical Engineering, the study of the motion and behavior of physical systems is called System Dynamics. In this specialized field, the behavior of state determined systems are analyzed in specific ways in which equations of motion are derived in a special form called state space equations. These equations are merely equations of motion that use a special set of state variables. And, what's so special about state space equations? Nothing... other than the choice of the state variables, and their representation in the form of first-order differential equations.
When analyzing the motion of objects, the two state variables of choice are force and velocity. Using these state variables will yield the so called state equations. These equations are no more correct than any others... but by using the preferred state variables, it simply turns out that the state equations will be in a form that is convenient for solution -- especially using "standard" computer algorithms for numerical integration.
Returning once again to the case of a block sliding down a frictionless inclined plane, let's now derive the state equations. For this case, our state variables of choice are the force, Fy, and the velocity, ds/dt.
For the force in the y direction, we simply solve equation (8-7) for the force, instead of the acceleration:
(8-12) dpy/dt = d(mvy)/dt = may = (mg)( cos²(ß) - 1 ) = Fy.
For motion along the surface of the plane, the displacement, s, is equal to:
(8-13) s = (
y)/(sin(ß)).
Therefore, noting that the incline angle, ß, is a constant, the velocity is equal to:
(8-14) ds/dt = d( y/sin(ß) )/dt = (dy/dt)/(sin(ß))= (vy)/(sin(ß)).
Equations (8-12) and (8-14) are the state space equations for the system.
As a final note, you may be interested to learn that in the Coaster Dynamics program, all the calculations for the motion of the roller coaster car are performed using state space equations, of the form just described -- except, of course, the geometry of the roller coaster tracks, and the resulting equations, are much more complex than in this example.
So, you can now see that all this arcane math really does have application... in the "real world" of computer simulation.
Copyright © 2001, Cyclone Software, Pleasanton, CA, USA. All rights reserved.