First convert the measurements into standard units.
If the perfect cone shaped icicle weighs 13lb then it weighs
(13 x 0.454) kg = 5.897kg
If it is dropped from a height of 25ft, in metres this is
(25 x 0.305 ) m = 7.620 m
If the icicle is 40 inches in length and 5 inches in diameter then it is
1.016m long and has diameter of 0.127m.
To calculate the net force on the cone as it falls we need to use the equation
`F = mg - 1/2 rho v^2 A C_d`
where `m` is the cone's mass, g is the acceleration due to gravity (9.8ms^-2), `` is the current velocity of the cone as it falls, `A` is the area of the base of the cone (we use this whether the cone is facing point downwards, or base downwards) and `C_d` is the drag coefficient. A typical quoted drag coefficient for cones is 0.5. However, that is for a cone with half-angle of `30^o``60^o`.Check it's not particularly different to that: half-angle of this cone is (measurement units don't matter here, so use the original whole numbers in inches)
`90 - tan^(-1)(25/5) = 90 - tan^(-1)(5) = 90 - 78.69^o = ` `11.31^0`
Well it is quite different to `30^o` really. The website quoted below suggests that `C_d` for a cone with half-angle ```epsilon` is given by
`C_d = 0.0112*epsilon + 0.162`
which gives `C_d = 0.289` in this case
So, given this value for `C_d` , air density ```rho = 1.275` , and `A = pi (0.127/2)^2 = 0.013m^2` , the current net force as the cone falls is
`F = 5.897g - 1/2(1.275)v^2(0.013)(0.289) = 57.79 - 0.0024v^2`
The terminal velocity is found at equilibrium when F=0. This implies a terminal velocity of 155m`s^-1` `<span style="font-family: Serif;" data-mce-style="font-family: Serif;" face="Serif">s-1</span>`which is 347 miles per hour, so we differ there I think!
Assuming the wind is coming in perpendicular to the downward flight of the cone, then the result will be that the cone is blown to the side. 15mph is 6.71`ms^(-1)` . To calculate the sideways force of the wind from this we need the formula
`F_w = 0.00256Av_w^2 C_d` where `A = 0.127/2 ` = 0.00645 `m^2` (half base of cone times height as the shape the cone presents to the wind is now effectively a triangle), `v_w` is the wind speed in mph and `C_d` is the drag coefficient of the cone sideways on. Assume the drag is like that of a half-sphere, ie 0.42, then we would have
`F_w = (0.00256)(0.00645)(225)(0.42)` = 0.00156N
As the icicle falls, the cone is pushed sideways according to the force diagram with base `F_w` and upward strut `F` given above. For example, after 1s when `F = 57.79 - 0.0024(9.8^2) = 57.56N`. The resultant force is then (using Pythagoras)
`sqrt(F^2 + F_w^2) = sqrt(57.56^2+0.00156^2) = 57.56N` (essentially no change)
Over bigger distances however, the force increases and hence the acceleration of the cone (since F = ma, and the mass stays constant). Successive application of this rule will give the displacement of the cone at ground level.