Coaster Dynamics: Physics Primer Chapter 11



Coaster Dynamics
Physics Primer

Chapter 11

Kinetic and Potential Energy

Following from Newton's Second Law of Motion and the concept of work is the notion of alternate forms of energy. Two forms of energy fundamental to motion dynamics are Kinetic Energy, the energy of motion, and Potential Energy, or "stored" energy.

Kinetic Energy

When work is done on an object to accelerate it, the energy of the object increases. This increase in energy is manifested in the higher velocity. This is an intuitive idea, given our everyday experiences.

To be scientifically precise, the term "kinetic energy" is equal to the amount of work done on an object -- and it has units of force×length (a newton-meter in the Metric System... which is called a joule). That is:

(11-1)         KE = W = Fs,

where KE is the kinetic energy.

To quantify kinetic energy, let's first consider the case of a constant force.

For a constant force, Newton's Second Law of Motion tells us that the acceleration is also constant. For the case of constant acceleration in the y-direction, from equation (5-17) of Chapter 5, the displacement, y, for a constant acceleration, ay, is equal to:

(11-2)         y = y0 + (1/2)(vy0 + vy)(t).

Let's assume that we choose a coordinate system and reference frame such that the initial velocity and position coordinates are all equal to zero. Then, the acceleration is equal to:

(11-2)         y = (1/2)(vy0 + vy)(t) = (1/2)(vy)(t).

Also,

(11-3)           ay = v/t = ( vy - vy0 )/t = (vy)/t.

From Newton's Second Law of Motion,

(11-4)         KE = W = (Fy)(y) = ( may )( y).

Inserting equations (11-2) and (11-3) into (11-4) yields:

(11-5)         KE = ( m )( (vy)/t )( (1/2)(vy)(t) ) = (1/2)(m)(vy²).

If we generalize equation (11-5) for any direction, and drop the "y" subscript, we have:

    (11-6)         KE = (1/2)(mv²).

Equation (11-6) is the definition of Kinetic Energy.

This derivation for kinetic energy assumed that the force is constant. However, with a little calculus, it can easily be found that the same result is obtained for a non-constant force. This is shown in the Advanced Study section below for those who have endured a course in calculus.

ADVANCED STUDY

Derivation of Kinetic Energy for a Non-Constant Force

Consider the work done by a force that is in the x-direction and varies in magnitude only. From equation (9-8) of Chapter 9, the work is equal to the integral:

(11-7)         Wx = §( (F(x) )dx

where F(x) is the instantenous force.

Acceleration is equal to:

(11-8)         a = dv/dt = ( dv/dx )( dx/dt ) = ( dv/dx )( v ).

From equations (11-7) and (11-8), and Newton's Second Law of Motion,

(11-9)         W = §( (F(x) )dx = §( ma )dx = §( (m)( (dv/dx)(v) ) )dx
                        = §( (m)(v)( dv/dx) )dx = §( (mv ) )dv = m§( v )dv.

The indefinite integral of v is equal to (1/2)(v²). Therefore,

(11-10)         KE = W = m§( v )dv = (m)(1/2)(v²) = (1/2)(mv²).

Potential Energy

Potential energy is energy that can be "stored" and "recovered".

For instance, one type of potential energy is associated with a spring. For a spring, a force is required to stretch it. This results in energy being stored in the spring material. If released, the spring will return to its previous length, and the stored energy is released. Another example of potential energy is that associated with gravity... and it, of course, is the potential energy of most interest to roller coaster designers.

How do we calculate the potential energy of gravity?

To begin, consider the case of throwing an object up into the air. When the object leaves your hand, it will have an initial velocity. Then, as the object moves upward, it will decelerate due to the force of gravity. As it slows down, the kinetic energy of the object decreases, since KE = (1/2)(mv²). Where does the kinetic energy go? It becomes potential energy of gravity. At some point, the object will stop and then fall back toward the ground, re-accelerating as it drops. Thus, the potential energy is re-converted back into kinetic energy.

To quantify the gravitational potential energy, recall that the force of gravity is a constant equal to the weight, -mg (assuming the positive y-direction is upward, opposite the force of gravity). Thus, the work done on the object as it moves up is

(11-11)         W = Fs = (-mg)(y1 - y0).

For the case of the object moving upward, v1 < v0 -- which means that the kinetic energy has decreased, and the potential energy has increased. At the same time, y1 > y0 -- which means that an increase in potential energy corresponds to an increase in the height. It follows, therefore, that the change in potential energy is equal to the negative of the work done by gravity. That is,

(11-12)         PE = -W = -Fs = -(-mg)(y1 - y0) = (mg)(y1 - y0).

In general, the gravitational potential energy can be expressed as

    (11-13)         PE = mgh

where h is the difference in height (assumed positive for increasing y).

Center of Gravity

When calculating potential energy, the mass of an object is assumed to be totally concentrated at a theoretical center of gravity... and the force of gravity acts as if it were applied only at that single point in space. This is one of the basic idealizations used to model a roller coaster car.

In all calculations in the Coaster Dynamics program, and in almost every analysis in the Physics Primer, the roller coaster is modeled as a single car. However, what happens if we choose to model a train of cars, instead of a single car?

It can be shown that for an assembly of masses, the laws of motion will be valid when applied to a single center of gravity for the entire assembly. We won't prove this principle here in the Physics Primer, leaving that topic to a comprehensive physics textbook.

The center of gravity for an assembly is located at the "mass-weighted mean" of the center of gravity's for the individual masses. That is, the x, y, and z coordinates of the center of gravity are equal to:

(11-14)         cgx = ( Si=n,1{ (mi)(xi) } )/( Si=n,1{ mi } )

(11-15)         cgy = ( Si=n,1{ (mi)(yi) } )/( Si=n,1{ mi } )

(11-16)         cgz = ( Si=n,1{ (mi)(zi) } )/( Si=n,1{ mi } )

where Si=n,1{ -- }, means the summation of the enclosed quantity for all the masses 1 to n (n = total number of subassemblies), and x1, y1, and z1 are the coordinates of the center of gravity for mass 1, x2, y2, and z2 are the coordinates of the center of gravity for mass 2, etc.

For example, if there are 3 masses in the assembly, the coordinates of the center of gravity will be:

(11-17)         cgx = ( (m1)(x1) + (m2)(x2) + (m3)(x3) )/( m1 + m2 + m3 )

(11-18)         cgy = ( (m1)(y1) + (m2)(y2) + (m3)(y3) )/( m1 + m2 + m3 )

(11-19)         cgz = ( (m1)(z1) + (m2)(z2) + (m3)(z3) )/( m1 + m2 + m3 ).

The concept of the center of gravity will be needed in the next chapter when we analyze the motion of a train of roller coaster cars over a hill.

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