How to calculate the motion of this object?

1. Jan 10, 2005

LasTSurvivoR

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

We have a object lets 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 isnt 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 ?

Last edited: Jan 10, 2005
2. Jan 10, 2005

Galileo

The resistive force can quite generally be modelled 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.

3. Jan 10, 2005

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,
$$k_2=mg/v_\infty^2$$

4. Jan 10, 2005

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.

5. Jan 10, 2005

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

6. Jan 10, 2005

vincentchan

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

7. Jan 10, 2005

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 $C_{1} [/tex] and applying the inverse function [itex \tan [/tex] can be found the explicit dependence [itex] v=v(t)$.

Daniel.

8. Jan 11, 2005

vincentchan

read carefully, b and c are negative....

9. Jan 11, 2005

dextercioby

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.

10. Jan 11, 2005

vincentchan

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

11. Jan 11, 2005

dextercioby

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

12. Jan 11, 2005

vincentchan

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

13. Jan 11, 2005

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.......

Last edited: Jan 11, 2005
14. Jan 11, 2005

dextercioby

****!!!!!!!!! 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.

15. Jan 11, 2005

vincentchan

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