Re: How fast would this object be falling?
- From: "Cwatters" <colin.wattersNOSPAM@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 8 Sep 2008 22:09:42 +0100
<john_redman@xxxxxxxxxxx> wrote in message
news:05e8450f-c9aa-4a99-86ac-7a0ad737a810@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Folks
I have a couple of questions about gravity and I have not been able to
find an online calculator able to work this out. I flopped Physics O-
Level 30 years ago so even if I knew how to do this then, I don't know
now.
1/ If I drop a lead bar weighing 1kg from a height of 15m, how fast
would it be travelling when it hits the ground? The lead bar bar is in
the form of a cylinder 18cm long and 2.5cm in diameter - this I think
gives a weight of 1 kg (1.25^2 x pi x 20 = 88cm^3 at 11.6g/cm^3), and
I assuming it is dropped vertically along its long axis.
Ignoring air resistance..
V^2 = U^2 + 2gh
V = final velocity
U = Initial velocity = 0
g = 9.8 m/s^2
h = 15 m
so
V = SQRT(2gh)
and
V = 17.15 m/s
2/ From what minimum height would it need to be dropped to achieve its
terminal velocity and how long would it take to fall this far?
I assume this is easy but I found this hard way back when and it
hasn't got any easier :-)
Nope.
In free fall at terminal velocity the force due to drag will balance the
force due to gravity.
The force due to gravity is F = mass x g
The drag force is given by the equation here
http://en.wikipedia.org/wiki/Drag_(physics)
In theory equating the two will allow you to solve for the terminal
velocity.
The problem is that Cd (the drag coefficient) for a cylinder varies with
Reynolds Number. See the curve here..
http://scienceworld.wolfram.com/physics/CylinderDrag.html).
The Reynolds Number is a scaling factor that corrects for viscosity and
scale factors. (aside: It means model aircraft don't allways behave like
full size aircraft). Unfortunately the Reynolds number also depends on
velocity.... making it all horribly horribly complicated.
.
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