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Coaster Dynamics
Physics Primer
Chapter 5
Motion in 2 Dimensions: Velocity and Acceleration
Newtonian Mechanics
Newtonian Mechanics is the branch of physics devoted to the study of the motion of objects and particles. An "object" is a general term meant to represent a physical body which has mass. In the mathematical analysis, objects are actually conceptualized as "point masses", which have all their mass concentrated at a single point in space. This is called a particle. Thus, Newtonian Mechanics is more properly described as the analysis of the motion of particles.
Regardless of semantics, a key characteristic of particle motion is that the motion can be fully described via translation -- that is, the movement of the particle through space. A particle does not possess other motion properties, such as rotation.
2D Particle Velocity
The velocity of a particle is defined to be the rate of change of its position with time.
Figure 5-A is a graph of the x-y position of a particle for two moments in time, t1 and t2.
Figure 5-A. Y versus X for a moving particle.
When a particle moves in space, its change in position is equal to the displacement, s -- which is a vector quantity indicating the magnitude of the change, and the direction of the movement. Thus, from the definition of velocity,
(5-1) v = (change in position)/(change in time) =
From the graph in Figure 5-A, it is seen that the velocity is equal tos/
t.
(5-2) vx = (x2 - x1)/(t2 - t1)
(5-3) vy = (y2 - y1)/(t2 - t1)
(5-4) v = xvx + yvy.
Note that the components vx and vy are scalars, while v is a vector. Also, "speed" is a scalar, which is equal to the magnitude of velocity.
In Figure 5-B, the x-positions of the particle are plotted versus time.
Figure 5-B. X versus time for a moving particle at 2 time points.
Using the data in Figure 5-B, equation (5-2) gives the magnitude of the x-component of the average velocity. This is the average velocity because we don't know what the positions were in between time t1 and t2. However, Figure 5-C shows the x-positions for two intermediate points in time.
Figure 5-C. X versus time for a moving particle at 4 time points.
By repeated application of equation (5-2), the average velocities for the 3 time segments, t1-to-t2, t2-to-t3, and t3-to-t4 are equal to
(5-5) t1-to-t2: vx = (x2 - x1)/(t2 - t1)
(5-6) t2-to-t3: vx = (x3 - x2)/(t3 - t2)
(5-7) t3-to-t4: vx = (x4 - x3)/(t4 - t3).From the graph in Figure 5-C, and equations (5-5) through (5-7), it is seen that the velocities are equal to the slope of the individual line segments for each time interval of the graph (position versus time).
Through a similar process, we can derive analogous equations for the y-component of the velocities.
The velocities in equations (5-5) through (5-7) might be considered "local" velocities, because they are a better estimate for the different velocities in the smaller intervals of time. As the time interval used to measure the velocity is made smaller and smaller, the value of the calculated velocities will become more and more precise -- and will theoretically approach the true instaneous velocity at any exact moment in time.
2D Particle Acceleration
The acceleration of a particle is defined to be the rate of change of its velocity with time.
Analogous to the derivation of the velocity equations, we find that the acceleration is equal to
(5-8) a = (change in velocity)/(change in time) =
v/
t
(5-9) ax = (vx2 - vx1)/(t2 - t1)
(5-10) ay = (vy2 - vy1)/(t2 - t1)
(5-11) a = xax + yay.
Figure 5-D is a graph of the x-velocity of a particle for 4 points in time.
Figure 5-D. X-Velocity versus time for a moving particle at 4 time points.
By repeated application of equation (5-9), the average accelerations for the 3 time segments, t1-to-t2, t2-to-t3, and t3-to-t4 are equal to
(5-12) t1-to-t2: ax = (vx2 - vx1)/(t2 - t1)
(5-13) t2-to-t3: ax = (vx3 - vx2)/(t3 - t2)
(5-14) t3-to-t4: ax = (vx4 - vx3)/(t4 - t3).The accelerations in equations (5-12) through (5-14) are equal to the slopes of the individual line segments for each of the time segments in the graph of Figure 5-D (velocity versus time).
Similar equations can also be derived for the y-components of the acceleration.
Also, as in the case of velocity, the "local accelerations" above will approach the true instantaneous accelerations as the time interval for measurement is made smaller and smaller.
Constant Acceleration -- Free Fall
An important special case of acceleration occurs when the acceleration is constant. Of particular interest, with respect to roller coaster motion, is free fall due to gravity.
Consider the case of holding a ball in the air, and then releasing it -- letting it drop to the ground.
If we neglect the affect of any air resistance, the ball will drop and accelerate at a constant rate. This fact was discovered in the times of Aristotle, although the reason for it wasn't correctly explained until Isaac Newton published his famous Second Law of Motion in 1686. For the moment, we won't explore that topic -- instead, we'll just accept the fact that the acceleration during free fall is a constant, which we'll call "g" (magnitude equal to 9.81 m/s²).
Since the y-acceleration is constant, it follows from equation (5-10) that the velocity magnitude will increase linearly... and the average velocity will be equal to the midpoint between the velocity at the beginning and end of the time interval. That is, if t0 = 0,
(5-15) vy,ave = (1/2)(vy0 + vy).
For an assumed constant average velocity, the y-position is equal to
(5-16) y = y0 + (vy,ave)(t).
Inserting equation (5-16) into (5-15) yields:(5-17) y = y0 + (1/2)(vy0 + vy)(t).
If we use a coordinate system with the positive y-axis pointing upward (away from the center of the earth), and insert a constant value of -g for the acceleration (negative because it is downward), we get:
(5-18) ax = 0
(5-19) vx = 0
(5-20) x = x0
(5-21) ay = -g
(5-22) vy = vy0 + (ay)(t)
Inserting equation (5-22) into (5-17) yields the following equation for the y position:
(5-23) y = y0 + (vy0)(t) + (1/2)(ay)(t²).
Finally, solving equation (5-22) for t, and substituting that expression into equation (5-17) yields
(5-24) vy² = vy0² + (2ay)(y - y0).
If we now make the additional simplifying assumptions that the ball is motionless when it is released, and that the origin of the coordinate system is located at the point where the ball is released (ie., vy0 = 0, x0 = 0, and
y0 = 0), and replace ay with -g, we then get:(5-25) ax = 0
(5-26) vx = 0
(5-27) x = 0
(5-28) ay = -g
(5-29) vy = -(g)(t)
(5-30) y = -(1/2)(g)(t²).Equations (5-25) through (5-30) will prove quite useful when designing roller coasters.
ADVANCED STUDY
Differential Form of the Velocity and Acceleration Equations
From equation (5-1), the average velocity is equal to
(5-31) v =
s/
t.
It was further shown that as the time interval is made smaller and smaller, the average velocity will approach the theoretical instantaneous value. That is,
(5-32) v = ( lim
t –> 0 )(
s/
t ).
Students of calculus will recognize equation (5-32) as the definition of the derivative. Thus,
(5-33) Instantaneous Velocity = v = ds/dt.
Similarly, from equation (5-8), the average acceleration is equal to
(5-34) a =
v/
t,
and the instantaneous acceleration is equal to
(5-35) Instantaneous Acceleration = a = dv/dt = d²s/dt².
Simply stated in words: velocity is equal to the first time derivative of the displacement, and acceleration is equal to the second time derivative of the displacement.
From the definition of the derivative, you will also recognize that the instantaneous velocity is equal to the instantaneous slope of the line of the displacement versus time graph, and the instantaneous acceleration is equal to the instantaneous slope of the velocity versus time graph.
For the case of free fall, we can simply calculate the integral of equation (5-28) to obtain:
(5-36) vy = § ay = § (-g) = (-g)(t - t0) = -(g)(t)
(5-37) y = § vy = § (-g)(t) = (1/2)(-g)(t -t0)² = -(1/2)(g)(t²),
which are equivalent to equations (5-29) and (5-30).
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