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Lecture-10 - à infty v T = mg/k(1 v T =/=\infty(2 v T is...

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AMS 361: Applied Calculus IV (DE & BVP) Outline for Lecture 10 2. 3 Classical Mechanics models Resistance Proportional to Velocity (dropping to earth) Note: My description (particularly the notation) of this problem is different from the textbook, but basic idea is the same. Assuming one object was dropped from an airplane with initial speed v0, we can write its equation of motion by using Newton’s 2 nd law: m v’ = -kv + mg m= mass of object, v = velocity, k=frictional constant, g = gravitational constant. We have made several assumptions: (1) speed pointing down is positive (going up is negative); (2) frictional force is always in the opposed direction of the motion; so the frictional force always slows the movement of the object; (3) this setup only explains the situation to drop an object to earth. We can solve the above equation easily to get V(t) = (v0 - mg/k) e^{-kt} + mg/k
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Remarks: Terminal speed is defined as the speed when t
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Unformatted text preview: à infty, v T = mg/k (1) v T =/= \infty (2) v T is independent of V0 (the initial speed) (3) Even if v0 =0, we can reach the positive speed (4) if k=0 (frictionless), v T à \infty (5) if k=\infty, v T à Resistance Proportional to Square of Velocity (high speed situation…) Equation can be written as m v’ = -kv|v| – mg whose solution is For Upward move: ( |v| = v because v is positive) V’ = -g – r v^2 We found v(t) = sqrt(g/r) tan {C1 – t sqrt(rg) } where C1 = arc tan {v0 sqrt(r/g) } For Downward move: ( |v| = -v because v is negative) V’ = -g + r v^2 We found v(t) = sqrt(g/r) tan h {C2 – t sqrt(rg) } where C2 = arc tanh {v0 sqrt(r/g) } Where tanh (x) = sinh (x)/cosh (x) = ½ {e^x – e^(-x)}/ ½ {e^x + e^(-x)} Remarks: (1) If v0=0, we can easily see C2=0 so v(t) = -sqrt(g/r) tan h {t sqrt(rg) } (2) Terminal speed is t=infty, we have v(t=infty) = sqrt(q/r). This is the terminal speed....
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