An icicle with mass 13lb that has the shape of a perfect cone is dropped from 25 feet above the ground. The icicle is 40 inches in length, and 5 inches in diameter at its base. With a 15 mile per hour wind, how much will the wind deviate the icicle from its straight, vertical path?
I have calculated the icicle to fall at a final velocity of 27 miles per hour, with a flight time of 1.25 seconds.
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That should read, the distance the icicle is diverted along the ground would approximately be
2.7 x 10^(-5)m = 0.027mm
That is the distance along the ground is only approximately 1/3rd of a millimetre.
The initial acceleration of the icicle under the force of the sideways wind (= 0.00156N) is equal to 0.00156/m = 0.00156/5.897 = 0.000265 m`s^-2` (F =ma and since the mass of the icicle is constant).
Since this is so small compared to the downward acceleration due to gravity, and since the height the icicle falls from is only 25ft (= 7.62m), the effect of the sideways wind will be negligible. Assuming linearity of the path of the icicle (when the distance is so small this is reasonable), the distance would be approximately the ratio of the two component velocities times the height.
The time the icicle takes to reach the ground is approximately `sqrt(7.62/(9.8/2) = 1.247s` , so the two component velocities will be 1.247 x 9.8 = 12.2`ms^-1` downwards and 1.247 x 0.000265 = 0.00033``
`<span style="font-family: Serif;" data-mce-style="font-family: Serif;" face="Serif">The distance the icicle is diverted along the ground then would approximately be<br data-mce-bogus="1"></span>`
`0.00033/12.2 x 7.62 = 2.7 x 10^(-5)m`
`<font face="Serif">That is, approximately only 1/3rd of a millimetre.<br data-mce-bogus="1"></font>`
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.
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