Need help with fluid mechanics problem set by lecturer :[

[djeitnstine]

Well here in england pre-uni we were taught parts of calculus in our syllabus which was

Basic Differentiation + Integration, Differentiation + Integration of Exponentions Logs and Trig, Volumes of revolution using int, Integration of partial fractions and further trig, Solving first order linear equations

have learned nothing of limits

and i know that speed=d/t but i don't see how it fits into the last eqn u gave me

This is very unfortunate, when one is taught basic calculus the definition of the derivative should be the first thing you learn. Then the definition of the integral. I suggest you ask for your money back.

 

[Keval]

i understand that Delta[t/v] means change in time or velocity

 

[Keval]

This is very unfortunate, when one is taught basic calculus the definition of the derivative should be the first thing you learn. Then the definition of the integral. I suggest you ask for your money back.

We are taught for free until we start university which i only started on wednesday which we pay £3225 a year

Can you show me the differential way maybe ill undestand that

 

[djeitnstine]

Unfortunately differential equations is at least 3 semesters of math away from you at the moment.

You say you understand what that \frac{\Delta v}{\Delta t} means so why do you have an issue? Mathematically there is nothing wrong with any of my statements in the post containing the equation of velocity for your sphere.

 

[djeitnstine]

Simply use all the relations given. Its simply plug and chug nothing complicated. Then after you get an equation in x you just plug all your values in and get an answer. Very simple.

 

[Keval]

\Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v
If your trying to work out the change in velocity how can you use v^2 on the left when that's wht your trying to work out for this part.

and your saying once i get that sorted i use \frac{\Delta v }{\Delta t }=s

 

[nvn]

Keval: I'm currently not understanding the explanation either. Not to mention several other things, velocity is changing nonlinearly. Therefore, you need to use differentials, not differences.

Anyway, it is good you found an air drag equation, but I currently think you should keep researching and looking for a different equation. In particular, the first equation http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object". Remember, v(t) is dx/dt, so change v(t) in that equation to dx/dt. Then multiply both sides of the equation by dt. Next, integrate both sides. I.e., integrate one side from zero to x, and integrate the other side from zero to t.

By the way, I have not checked your Cd = 0.47 value. Ensure you are using the Cd value your teacher wants you to use, and/or a Cd value roughly applicable to your velocity range.

 

[Keval]

Keval: I'm currently not understanding the explanation either. Not to mention several other things, velocity is changing nonlinearly. Therefore, you need to use differentials, not differences.

Anyway, it is good you found an air drag equation, but I currently think you should keep researching and looking for a different equation. In particular, the first equation http://en.wikipedia.org/wiki/Drag_(physics)#Velocity_of_a_falling_object". Remember, v(t) is dx/dt, so change v(t) in that equation to dx/dt. Then multiply both sides of the equation by dt. Next, integrate both sides. I.e., integrate one side from zero to x, and integrate the other side from zero to t.

By the way, I have not checked your Cd = 0.47 value. Ensure you are using the Cd value your teacher wants you to use, and/or a Cd value roughly applicable to your velocity range.

Hey just got back from uni and saw thiss could you use some latex to show me? I'm sort of confused, am still reading up on it thought

 

[nvn]

You can write the latex, if you wish. Use the hints. Show your work; and then someone might check your math.

 

[Keval]

okay here's trying this new method i get lost pretty early :\

F=ma

mg-F_d=ma using F_d=-\frac{1}{2}\rho A C_d v^2

mg-kv^2=ma using k=-\frac{1}{2}\rho A C_d

mg-k(\frac{ds}{dt})^2=m\frac{dv}{dt}

mg-k\frac{d^2s}{dt^2}=m\frac{dv}{dt}

now what? o.o, I'm not sure if i shoud be using v=dx/dt, or v=ds/dt.

And I've never done 2nd order de's if that's what I've ended up with sofar so helping hand please

 

[viscousflow]

Hello Keval,

I looked at nvn's equations and how he's trying to help you. I think he's trying to say that Drag is equal to some constant times velocity, not velocity squared.

 

[nvn]

Keval: Your post 40 is not what I advised in post 37. Read what I wrote more carefully. viscousflow, I didn't think I was saying what you said. I thought I was saying, velocity is a function of time.