Need help with fluid mechanics problem set by lecturer :[

[Keval]

Note: Have been taught nothing about fluid mechanics yet, this task was set to help develop self learning skills, yet with some research I've carried out I'm still at a brick wall so i hope you don't mind helping me!

Homework Statement

Explain the phenomenon that causes a noticeable difference in the freefall time of different sixed balls.

Determine the value of engineering co-efficients

Formulate a mathematical moel that takes in account gravity and this other phenomenon to predict the fall time for a spherical object from a height of 6m (I'm supposed to do an experiment next week to put my equations to test, for now just use the letters)

2. What I've done + Relevant equations

Obviously the phenomenon is drag.

Researched about drag and found out that

\mathbf{F}_d= -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v}

Where F_d: Force of drag, p=density of fluid( 1.1877 @298K/25C),
v=speed of object
a=reference area ( \frac{\pi.d^2}{4} for a sphere),
C_d=drag coefficient(0.47 for a smooth sphere),
v^=is the http://www.mathhelpforum.com/wiki/Unit_vector" indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).

 

[djeitnstine]

So what exactly is your problem or what you need help with? It would be nice if you stated the issue you're having.

So far you're on the right track. This does not seem like an extremely difficult project at your level to perform.

 

[Keval]

What do i do now to develop the mathematical model I'm just confused and lost now

 

[djeitnstine]

Well in any physical problem such as this one, the first thing you want to do is draw a diagram of the sphere and check the forces acting on it. This is called a free body diagram.

Ideally, there will only be two major forces acting on your sphere as it falls, can you identify them?

 

[Keval]

Gravity pulling the sphere downwards, then upwards drag right? :s

 

[djeitnstine]

Exactly. Now since you have one of the two required equations, you now need to find an equation for free fall for constant acceleration.

 

[Keval]

so then using f=ma

mass.gravity - drag = mass.acceleration due to gravity

then putting in the drag eqn i get

mg - -{1 \over 2} \rho v^2 A C_d \mathbf{\hat v} = mg

this right?, then just substitute in all the values? doesn't seem right to me argh

 

[djeitnstine]

That's night quite right. You're trying to get the time that it will take for the object to fall. The idea is to find some equation that relates acceleration and time due to gravity.

You need to do a little research, I do believe that equation is somewhere in the forum (hint you can search the PF library).

What you want to do is construct some equation of the form x = f(t) (x: displacement) that is, displacement as a function of time.

 

[djeitnstine]

The reason why I said to look for the forces at play here is so that you can get an idea of what type of equations you will need to predict its motion.

 

[Keval]

<br /> <br /> <br /> &amp; s &amp;&amp; = \tfrac12(u+v)t<br />
s= ut + \tfrac12 at^2 <br /> <br />

is it the top one i should be using?

 

[djeitnstine]

Good. The bottom one is a better approximation to what you want.

 

[Keval]

so then i put

6 = (0)t + 0.5(9.8)t^2
6 = 4.9t^2
t = \sqrt\frac{6}{4.9}

 

[djeitnstine]

Looks good to me. 6m is not that high and this is a good first approximation of the time it will take to land.

 

[djeitnstine]

Just as a pointer, this equation assumes no drag and constant acceleration.

 

[Keval]

yeah i figured but i need one that takes into consideration the drag :\

 

[djeitnstine]

Are you familiar with basic calculus? You can derive the equation yourself.

 

[Keval]

Im good with calculus but I've never applied calculus to mechanics

all I've heard is that vdot = dV/dt or something lol

 

[djeitnstine]

Ok good. So remember the equation you placed. Let's assume down to be positive and up to be negative. The correct form of the equation you placed earlier would be

mg-\frac{1}{2}\rho v^2 A C_d = ma -don't worry about the vector for now we are in 1D so its not needed.

now let some constant k = \frac{1}{2}\rho A C_d

Now, show me what the resulting equation looks like and how can we get an explicit equation in displacement (x or s)from this.

 

[Keval]

1stly wasnt the equation a double minus as k= -0.5pv^2ACd and mg - Fd = ma => mg - -k = ma

so mg + k = ma?

and I am confused by your last line, how do i get an explicit equn in displacemnt sorry I am so terrible atm

 

[djeitnstine]

The drag equation you showed assumed that F_d acts opposite to velocity correct?

Now in my statement I said down is positive and up is negative, if the ball is falling down what direction does drag act? and what is its sign according to my convention in the equation.

(look at your free body diagram again)

Also in the constant k I did not include the v^2 term look carefully and try again.

How to get the equation as a function of displacement? isn't it true that a=\frac{dv}{dt} and that v= \frac{ds}{dt} ?

 

[Keval]

so you want

mg-k=m\frac{dv}{dt}

 

[djeitnstine]

not quite, I said that k = 0.5pACd, v is not constant.

 

[Keval]

Huh i don't follow

mg- \frac{1}{2}\rho{A}{C_d} = m\frac{dV}{dT}?

 

[djeitnstine]

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

now let some constant k = \frac{1}{2}\rho A C_d <--- there is no v^2 in this constant k

Thus mg-kv^2 = m\frac{dv}{dt}

Now since we don't need to be dealing with non-linear differential equations we can make these approximations

\Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v

Now do you think you can turn that into an equation in terms of s from there?

 

[Keval]

ahh i understand the constant bit now and why the v^2 is separate

your last latex is messed up :s

 

[djeitnstine]

Oops, fixed it. Its simply transforming the differential into a delta.

 

[Keval]

I understood upto here

what assumption did you make to get this mg-kv^2 = m\frac{dv}{dt}

to this
\Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v

also the above is same as this right?

\frac{dt}{dt} \left(gt-\frac{k}{m}v^2 \right)= \frac{dv}{dt}

the greek delta confuses me

Now do you think you can turn that into an equation in terms of s from there?

no >_>

 

[djeitnstine]

If we don't let \Delta \rightarrow \infty then we can descretize any differential this is common knowledge from calculus you should know this. If you do not, restudy calculus 1.

In other words we will not get an exact equation for velocity v, unless you want to impress your professor by solving a first order inhomogeneous non-linear differential equation, this is always an option.

also how can you not do this? You are simply letting me do the work don't be lazy...

How do you not know that s=\frac{v}{t} or that v=st?

 

[Keval]

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

 

[djeitnstine]

\Delta v = v-v_0 Or simply put, change in velocity. I hope this seems familiar...