Question
Asked Sep 6, 2019
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Quadratic Drag in One Dimension Consider an object that is dropping
quadratic drag force
vertically (positive y direction is vertically down) subject to a
fdrag = -f(v)v with f(v) cu2.
(a) Write down the differential equation from Newton's second law for this object.
(b) Show that the terminal speed is vter =
(c) Solve for vy in the case of vyo<Uter
(d) Solve for v in the case of vyo> Uter
Vmg/c.
Hint: You may find the integrals in Appendix E in Thornton & Marion useful (p613). |
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Quadratic Drag in One Dimension Consider an object that is dropping quadratic drag force vertically (positive y direction is vertically down) subject to a fdrag = -f(v)v with f(v) cu2. (a) Write down the differential equation from Newton's second law for this object. (b) Show that the terminal speed is vter = (c) Solve for vy in the case of vyo<Uter (d) Solve for v in the case of vyo> Uter Vmg/c. Hint: You may find the integrals in Appendix E in Thornton & Marion useful (p613). |

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Expert Answer

Step 1

(a) Using Newton’s law,

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=mg-Fag та - mg -cv? dv m mg-cv dt

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Step 2

(b) The object has zero acceleration when it reaches terminal speed. Therefore,

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mg-cv 0 mg Vi C

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Step 3

(c) Re-arrange the equation ...

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dv -= gdt 1 mg dv --fgat gt arctanh ter v(t)=v tanh

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