{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture-10

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Remarks: Terminal speed is defined as the speed when t
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern