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Unformatted text preview: Fall 2003 Math 308/501–502 3 Mathematical Models 3.4 Newtonian Mechanics Fri, 19/Sep c 2003, Art Belmonte Summary We’ll use Newton’s theory of motion from physics to model linear motion. Our starting point is Newton’s second law, F = ma which says that force = mass × acceleration. We take into account gravitational force and possibly air resistance. Hand Example Example A A mass of 0 . 2 = 1 5 kg is released from rest. [This means that its initial velocity is zero; i.e., v( ) = 0.] As the object falls, air provides a resistance proportional to the velocity. Specifically, the resistance force is R (v) = - . 1 v = - 1 10 v , where velocity is measured in m / s. If the mass is dropped from a height of 50 m, what is its velocity when it hits the ground? Solution Velocity Let x be the distance of the object above the ground, so that the upward direction is positive. There are two forces acting on the object. 1. The gravitational force,- mg , is pulling the mass downward. Here g = 9 . 8 m / s 2 . Below we’ll use 9 . 8 = 98 10 = 49 5 (exact rationals). 2. Air resistance acts in the direction opposite to motion. Since the mass is falling downward, air resistance acts upward . Since a downward falling mass has negative velocity, the force of air resistance is- 1 10 v , which is positive , as required. Apply Newton’s second law to obtain- mg- 1 10 v = F = ma = m d v dt , whence (recalling m = 1 5 ) d v dt =- g- 1 10 m v d v dt =- 49 5- 1 2 v v + 1 2 v =- 49 5 This is both a separable and a linear equation, so we may employ any of the techniques from §2 . 2 or §2 . 4 that we prefer. Let’s go through the steps of the Conventional Procedure (CP) for linear equations for practice since we’ll actually use the separable techique in the second part of this problem! 1. The DE in SLF is v + 1 2 v = - 49 5 . Here p ( t ) = 1 2 , the coefficient of v in the SLF. 2. Construct an integrating factor: μ( t ) = exp ( R p ( t ) dt ) or exp ( R 1 2 dt ) = e t / 2 ....
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