How to calculate the motion of this object?

[LasTSurvivoR]

Ok here guys, I have poor english so , underestimate my mistakes :)
I have a question which bothers my mind ...

We have a object let's say that Its a cube 4cm/4cm/4cm lenghts.Density of it 8g/cm(3)
Its on a apartment roof ( 45m lenght) We will let it fall from that distance.
I know the basic calculation ; h=1/2gt² But air friction isn't considered in this.

I want a example real fall , CONSIDERING Air Friction , how can I calculate it ?Which way should I use ?Whats the formuLa?When does it falls ?How much velocity will it have when it reached the ground ?

Thanks for ur answers.

 

[Galileo]

The resistive force can quite generally be modeled by:

\vec F_{res}=-(k_1v+k_2v^2)\hat v,
where k_1, k_2 are constants which depend on the medium (air in this case) and on the size and shape of the object which is falling.

It's quite a difficult problem, I wouldn't know what values to use.

If v is not too big and the object small enough, the first term dominates, so the force is reasonably well modeled by:

\vec F_{res}=-k_1v\hat v
So for the cube falling, the equation of motion is:

mg-k_1v=m\frac{dv}{dt}

which you can solve for the height as a function of time.

 

[krab]

In the case you are considering, the flow around the object is turbulent and so the resistance force is proportional to v^2. Usually k_2 is hard to calculate from first principles, but experimentally the terminal velocity v_\infty is accessible. In that case,
k_2=mg/v_\infty^2

 

[dextercioby]

I'm sorry,Galileo,but since the maximum force of resistance of air (dynamic pressure times surface facing the Earth) is round 1.5N (i'm assuming the square does not rotate in the air) and the weight is 5N,i guess it should be included.It would be far a greater error if u included only Stokes force...

Daniel.

 

[reilly]

Note that with friction having both v and v*v dependence with constant coefficients, the dynamical equation becomes a particularly simple form of the Ricatti Eq.

The form is dv/dt = a + b*v +c*v*v. This can be integrated from

dv/(a + b*v + c*v*v) = dt.

regards,
Reilly Atkinson

 

[vincentchan]

yes, but, then v(t) can't be find explicitly...

 

[dextercioby]

I think it can be found explicitely.By the looks of it,through integration it woud yield something like
C_{1}\arctan C_{2}v =t+C_{3}
,in the case,the constants are positive.If they are negative,then it would be a ratio of natural logarithms.Which can be explicitated.
Given initial conditions and dividing through the constant C_{1} [/tex] and applying the inverse function [itex \tan [/tex] can be found the explicit dependence v=v(t).<br /> <br /> Daniel.

 

[vincentchan]

read carefully, b and c are negative...

 

[dextercioby]

read carefully, b and c are negative...

So what??
I=-cv^{2}-bv+a=-c(v^{2}+\frac{b}{c}v-\frac{a}{c})(1)
You can redefine constants,or just use them like that:
I=-c[(v+\frac{b}{2c})^{2}-(\frac{b^{2}}{4c^{2}}+\frac{a}{c})] (2)
,which could lead to a natural logarithm,since the last term is positive.
I think u can explicitate the natural logarithm.

Daniel.

 

[vincentchan]

it will lead you to TWO natural log, intead of one,
if you have difficulty doing integration, the answer is something like this:
A (ln(B+v) +ln(C+v)) + D=t
which i have no ideal how to find v(t) explicitly

 

[dextercioby]

it will lead you to TWO natural log, intead of one,
if you have difficulty doing integration, the answer is something like this:
A (ln(B+v) +ln(C+v)) + D=t
which i have no ideal how to find v(t) explicitly

\ln[(B+v)(C+v)]=\frac{t-D}{A}(1)

(B+v)(C+v)=e^{\frac{t-D}{A}} (2)

Which is a second order in "v".You can find "v" as a function of "t" by using the quadratic formula.

Daniel.

 

[vincentchan]

sry, it is a minus signs in between, the right form is...
A (ln(B+v) -ln(C+v)) + D=t
sry wasting your time ...

 

[vincentchan]

www.calc101.com
a very good site solving integraion...again, sry for my mistake...but again, this cannot solved explicitly as i said b4...

edit:
dex,
can i know a little bit background about you, ie. age, education..etc, you are just so smart that i don't believe you came from my planet...

 

[dextercioby]

sry, it is a minus signs in between, the right form is...
A (ln(B+v) -ln(C+v)) + D=t
sry wasting your time ...

****! I had solved for a minus there and then i saw a plus in your formula and decided to redo calculations.

v(t)=\frac{Ce^{\frac{t-D}{A}}-B}{1-e^{\frac{t-D}{A}}}

Anyway,somthing like that.I knew it all along it can be solved.
I may have messed up the pendulum,but integration is one of my strengths...

Daniel.

 

[vincentchan]

yes, you are right, it is indeed solvable... my bad this time