6. [Euler Leaf - ongoing problem] There are many, many problems in engineering t

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6. [Euler Leaf - ongoing problem] There are many, many problems in engineering that can't be solved by calculating a single equation. Instead, they're solved by simulating what happens over time. Imagine a leaf in a stream. You may know its position and velocity right now, but the only way to find out where it will be ten seconds from now is to follow it. This is because, at every point in time, the stream is applying forces to the leaf to move it in a different direction, and those forces change depending on where the leaf is This problem extends of the entire quarter - we'll be adding a piece to it every week. For this week we're going to take one step in the simulation, and we're going to work in one dimension (ie, the leaf moves along a line). Specifically, we're given the position of the leaf. x = 1.1 meter, and its velocity, v 0.5 meters/second. Represent the force acting on the leaf as an equation that depends on the position, r: F(x) = 0.2 sin(x)(x2+x) The goal is to calculate the new position x' and velocity v' of the leaf at a time h seconds from now. There are a variety of ways to do this; the simplest is called Euler integration x, = x + hv a = F(x)/m where a is the acceleration created by the force F acting on the mass m = 0.3 kg of the ea Calculate the following (format output to 2 decimal places) a) Calculate the new position and velocity after taking one step with h-0.1 b) Calculate the position and velocity after taking another step (h-0.1), but this time tarting with the position and velocity calculated 1n part a) c) Calculate the new position and velocity after taking one step with h - 0.2. (Use the same starting position and velocity as in part a).) Self check: a) x. x.x5. b) v, = x.x9. c) v, x.x7 Answer the following (in the homework template document) a) Does taking two steps with h-0.1 give the same answer as taking one step with h-0.2? b) Hypothesize which is more accurate (two steps with h-0.1 or one with h- 0.2) Justify your answer.

Explanation / Answer

A) x'= x+(vh) = 1.1+(0.5*0.1) = 1.15 m

v' = v+(h*a) = 0.5+(0.1*a)

a = f(x)/m = 0.2*sin(1.1)*(1.1^2+1.1)/0.3 = 1.372 m/s^2

then v' = 0.5+(0.1*1.372) = 0.6372 m/s

B) x'' = x'+(v'h) = 1.15+(0.6372*0.1) = 1.213 m

v''= v'+(a*h)

a = 0.2*sin(1.15)*(1.15^2+1.15)/0.3 = 1.504 m/s^2

v'' = (0.6372)+(1.504*0.1) = 0.7876 m/s

C) x' =x+(v*h) = (1.1)+(0.5*0.2) = 1.2 m

v' = v+(a*h)

a = 0.2*sin(1.1)*(1.1^2+1.1)/0.3 = 1.372 m/s^2

v' = (0.5)+(1.372*0.2) = 0.7744 m/s

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a) no slightly changes are there taking two steps with h= 0.1 and taking single step with h = 0.2

B) Since force is not constant

it is better to take two steps with h =0.1

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