Yes, for spherical rain drops.
Consider the sum of the forces and examine the drag component.
downward: mg
upward (drag): (1/2)*rho*Vmax^2 *Cd*A, where A = pi*r^2
At terminal Vmax, the force balance is:
(1/2)*rho*Vmax^2*Cd*A = mg
Use m = rho * pi*r^3 :
(1/2)*rho*Vmax^2*Cd*A = rho*pi*g*r^3
Solve for Vmax = sqrt [( (2g) / Cd) * r ].
The terminal velocity then increases as the square root of the radius of the raindrop. A raindrop twice as large in diameter will fall roughly 41% faster at terminal velocity.
2007-08-01 12:55:43
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answer #1
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answered by Mick 3
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Love your questions, Alex.
Let's work out the answer....
F = Ma = W - D(V); where F is net force on a large raindrop of mass M, and W is the weight of the drop and D is the drag force on it. W = Mg; and D(V) = 1/2 rho Cd A V^2. D(V) indicates the drag on the large raindrop is dependent on its velocity.
So the issue is if the terminal velocity of the large drop V > v the terminal velocity for the small drop.
I will presume Cd is the same for each size drop. This is based on my understanding that the shape of a drop is fairly consistent no matter what the size. And since both drops are made of the same material, shape would be the greater determinant of the coefficient of drag.
However, clearly the cross sectional area A for a large drop is bigger than A' for a small drop. That is one way to define the size of each drop. The volume of the large drop V = AL; where L is some length. For the small drop it's v = A'L'.
Also m = Rho v < M = Rho V because we assume the density of the rainwater (Rho) is the same and, because the volume is smaller, the mass contained in the smaller drop will be less than that of the larger one. Then is also follows w = mg < Mg = W; so the weight of the smaller drop is also smaller than that of the larger drop.
Finally, terminal velocity for each drop size happens when F = Ma = 0 and f = ma' = 0 for the large and small drops respectively.
Then, putting this all together, we have:
F = Ma = 0 = (W - D(v)); and Mg = 1/2 rho Cd A V^2
f = ma' = 0 = (w - d(v)); and mg = 1/2 rho Cd A' v^2
Taking the ratio M/m = (A/A')(V^2/v^2) and mixing these factors all around, we have V^2 = (M/m) (A'/A) v^2; so that V = v sqrt(M/m * A'/A), which is inconclusive until we note M = Rho V = Rho AL; where Rho is the density of water and m = Rho v = Rho A'L'.
Thus, on substitution, V = v sqrt(Rho AL/Rho A'L' * A'/A) = v sqrt(L/L'); where L > L' the length of the larger drop is greater than the smaller drop. This results because we presumed the shape of the large and small drops were similar. So L/L' > 1.00.
Finally, V = v sqrt(L/L' > 1.00); so that V > v and the terminal velocity of the big drop is greater than that of the smaller drop.
The answer to your question is yes; big rain drops do fall faster than smaller ones all other things equal. And this results from the fact that the weight of raindrops increases with size faster than the drag force does.
2007-08-01 13:02:03
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answer #2
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answered by oldprof 7
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Yes. Larger raindrops fall faster than smaller raindrops.
See:
http://www.grow.arizona.edu/Grow--GrowResources.php?ResourceId=146
created by the University of Arizona's Department of Civil Engineering.
In a vacuum all objects fall with the same acceleration and their velocity continually increases at that rate until SPLAT! they hit the ground; however in the real world there is AIR and air causes drag (friction) and slows down objects that are falling towards the earth. After awhile the drag on an object equals the force of gravity and the object stops accelerating and is said to have reached it's TERMINAL VELOCITY.
This terminal velocity varies with the object under consideration. A person falling out of an airplane and who doesn't use a parachute will reach a terminal velocity somewhere between 120 and 150 mph. If that same person opens a parachute, then the terminal velocity becomes much lower and the person should land without serious injury.
Small raindrops have a terminal velocity of about 1 mph and large ones of abour 20 mph.
See the above website which has an interactive experiment to calculate terminal velocity of various sized raindrops.
VOILA!
2007-08-01 12:15:19
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answer #3
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answered by Mario 3
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Probably. Smaller raindrops would be more subject to wind resistance.
Only in a vacuum would both drops fall at the same speed.
2007-08-01 12:21:19
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answer #4
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answered by Anonymous
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Vincent G and Mario were so far the only persons giving the right answer. Small drops fall at a slower speed. The ratio of surface per volume is greater in small drops. You just think of floating fog that consists of very small droplets. Then think of the "wet fog" when the fog looses water and the slightly larger droplets drift down and wet the ground. Then think of "spry rain" that is easily drifted horizontally in weak and moderate wind, because the small drops fall slowly. Then think of heavy rain with big drops that is not affected by weak winds, because the big drops fall with a high speed.
I added all this to help people who lack the gift of observing natural phenomena, to believe Vincent G's And Mario's excellent answers.
2007-08-01 12:15:38
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answer #5
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answered by Ernst S 5
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No. The speed at which things fall is not at all dependent on mass. Think of the experiment done by Galileo on the Leaning Tower of Pisa. Whether this is myth or not, the fact remains that both a bowling ball and a bouncy ball would fall with the same acceleration: with an acceleration of 9.81 m/s^2. What would decide which one fell faster would be how high up they started their fall. The drop starting higher up would hit with a faster velocity.
2007-08-01 11:45:08
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answer #6
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answered by Broadway Baby 2
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Yes, as aerodynamic drag is more likely to reduce the termial velocity of smaller drops to a smaller value than bigger ones.
If it was in a vacuum, they would fall at the same rate.
2007-08-01 11:48:52
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answer #7
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answered by Vincent G 7
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Ideally, no, because both have same acceleration by gravity. In reality, yes, because large rain drop has larger surface area thus larger air resistance.
2007-08-01 13:10:36
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answer #8
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answered by BenL 2
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No in an experiment preformed by idont know who but he was the person who dropped the big ball and the little ball at the same time and they both landed at the same time.
2007-08-01 11:45:52
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answer #9
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answered by Anonymous
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That is an interesting question and I hope you find some reasonable answers
2016-08-24 10:37:56
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answer #10
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answered by Anonymous
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