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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Newton’s First Law. A body remains in a state of rest, or in a state of uniform motion in a straight line if there is no net external force acting on it.
But the more interesting situation is when there is a net external force acting on the body.
Sometimes it’s easier to visualize Newton’s second law in terms of the x-, y-, and z-components of the position vector R of the moving body. If we project the acceleration vector a = R" and the forces onto the x-, y-, and z-axes, then for a body of mass m,
mx” = the sum of the forces in the x-direction my' = the sum of the forces in the y-direction mZ' = the sum of the forces in the z-direction
We’ll look at motion in a plane with xmeasuring the horizontal distance and y measuring the vertical distance up from the ground. We don’t need the z-axis for our examples because the motion is entirely along a line or in a plane.
Chapter 5
¦ Dunk Tank
What is the ball doing if
00 = 90° ?
Figure 5.1 was done in the Tool, where $o is in radians.
Imagine your favorite professor seated over a dunk tank. Let’s construct a model that will help you find the secret to hitting the target and giving your teacher a swim!
You hurl a ball at the target from a height of 6 ft with speed v0 ft/sec and with a launch angle of 90 degrees from the horizontal1. The target is centered 10 ft above the ground and 20 ft away. Let’s suppose that air resistance doesn't have much effect on the ball over its short path, so that gravity, acting downwards, is the only force acting on the ball.
Newton’s second law says that
mR" = —mgj
where m is the ball’s mass, R(t) is the position of the ball at time t relative to your hand (which is 6 ft above the ground at the instant t = 0 of launch), and g = 32 ft/sec2 is the acceleration due to gravity. In coordinate terms,
mx" = 0
mf = ~mg
Since X (0) = v0cos â0 and / (0) = v0sin 60, one integration of these second-order ODEs gives us
X(t) = V0 cos 00 y( t) = V0 sin 00 — gt Then because x(0) = 0 and y(0) = 6, a second integration yields
x(t) = (v0 cos00)t
1 2 (2)
y(t) = 6 + (t>0 sin00)t--gt2
To hit the target at some time T we want x(T) = 20 and y(T) = 10. So values of T > 0, 00, and v0 such that
x( T) = 20 = (v0 cos 00) T
y[ T) = 10 = 6 + (uosin0o) T- ^gT2 ^
lead to hitting the target right in the bull’s eye and dunking your professor.
You can try to use system (3), or you can just adjust your launch angle and pitching speed by intuition and experience. The screen shot in Figure 5.1 shows you how to get started with the latter approach. If you play the dunking game on Screen 1.3 you’ll find that you can dunk without hitting the target head-on, but that a little up or a little down from the center works fine.
1 The sin, cos and other trig functions in the ODE Architect Tool expect angles to be measured in
radians. Note that 00 = 1 radian corresponds to 360/2n ^ 57.3 degrees. The multimedia modules, however, will accept angles measured in degrees.
Longer to Rise or to Fall?
Figure 5.1: This ODE Architect screen shows paths of a ball thrown at ten different angles 00. Which paths lead to dunking the prof?
¦ Longer to Rise or to Fall?
Drag forces are usually determined by observation. They differ widely from one body to another.
Throw a ball straight up in the air and ask observers whether the ball takes longer to rise or to fall. You’ll get four answers:
1. Longer to rise
2. Longer to fall
3. Rise-time and fall-time are the same
4. It all depends ...
What’s your answer?
A mathematical model and ODE Architect suggest the answer. The forces acting on the ball of mass m are gravity and air resistance, so Newton’s second law states that
mR" (t) = —mgj + F
where R is the position vector of the ball, and F is the drag on the ball caused by air resistance. In this case R(t) = y(t)j where j is a unit vector pointing upward (the positive y direction). If the drag is negligible, we can set F = 0. For a light ball with an extended surface, like a whiffle ball, the drag, called viscous drag, exerts a force approximately proportional to the ball’s velocity but opposite in direction:
F(v) = -kv = -k7j
Chapter 5
If the ball is solid and dense, like a baseball or a bowling ball, then we have Newtonian drag, which acts opposite to the velocity with magnitude proportional to the square of the speed:
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