Galileo's Experiment (with a twist)
Sam drops two solid spheres (made of aluminum 6061 T6) with different radii simultaneously from the top of a tall building. Before the drop, the bottommost points of the spheres are aligned at the same height from the ground. The spheres fall through the air and hit the ground.
Which sphere will hit the ground first?
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5 solutions
In a vacuum they would both hit at the same time. I believe a vacuum is the absence of atmosphere. So I believe the atmosphere is the wild card.
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If the experiment is conducted in vacuum, then both balls will land at the same time. However, in this problem, the experiment is conducted in air, so air resistance comes into play, and the larger ball lands first.
Two balls falling through a vacuum would not accurately be described as falling "through the air". Therefore, while the density of the air is not defined, we know it is not a perfect vacuum.
The greater force on the larger sphere does not necessarily prove that it will have a larger acceleration. Because of the increased mass, a larger force is required to produce the same acceleration as for the smaller sphere (a = F/m).
Maybe something seems to be wrong: “4/3 pi R^3 ro” has not the same dimension of F. Maybe you forget g?
You're forgetting a bouyancy term... If the sphere is filled with hydrogen, then your model breaks.
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The problem says that spheres are solid and made of aluminum. Aluminium is much more denser than air, so the buoyancy effect will be negligible.
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it didn't say that when I made my comment...
The problem doesn't talk about each object's mass. the small sphere could be filled with water while the one could be filled with the gas H2. Fred is right, in a vacuum, they would both hit at the same time. We were really not given enough information to understand the true nature of the problem
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No - I also think we need more information - however (though it could have been altered since you posted) the Q does state that the spheres are solid and constructed of the same material
Going over this all again, the information I would have liked would have been the density of Al whatever compared to the density of the atmosphere. Would this have made a difference? In which case, the deciding factor is chemistry rather than mechanics.
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In general, the densities of solids is much greater than the densities of gases. For reference, the density of Aluminium 6061 is 2700 kg/m³, whereas the density of air is about 1.3 g/m³, which is a thousand times less.
The sphere used are solid and made of aluminum so both balls have the same density and much greater than the density of air.
Also the balls are thrown in air (which is not vacuum), so the factor of air resistance comes into play causing the smaller ball to land after the larger ball.
AWESOME EXTREME CASE PSYCOLOGY......since it has to be correct in all cases, why not think of the worse...I also thought the same way. Brilliant.
That's falling through a liquid because there is the force of DRAG which is F. Quadratic Drag force isn't the same thing as air resistance. You still need weight in that equation.
one additional note... if one of the spheres is extremely small "like a dust particle" where the size of the object is near the mean free path length, then brownian motion must also be included in the equation. At that size air resistance is a quantized thing and starts acting very strange.
What abt inertia
the larger sphere can't have larger acceleration because it has larger mass.so a=F/m can't be larger. as total F=mg-6πnrv here we can tell that the small sphere should reach the ground first.
Wasn't it Galileo who dropped two different size rocks from the tower of Pisa and they both hit the ground at the same time???
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Yes, he did a similar experiment. But he did not observe the time difference, because the to objects were not very different in size or weight (a factor of 2 or 3 maybe) and they were both heavy.
Imagine you do this experiment with a ball of 1millimeter diameter, and a ball of 1meter diameter, and you will probably agree that the smaller one will take longer time to come down.
Relevant wiki: Fluid Mechanics
Air resistance will be proportional to R 2 (since, the flow will be turbulent as suggested by the range of values that can be assumed by Reynolds Number in this situation) but mass depends on volume which in turn is proportional to R 3 . Acceleration is same for both the spheres ( = g ) but; retardation = mass air resistance , which comes out to be inversely proportional to R . Hence, larger the sphere, smaller the retardation. In other words, increase in mass dominates the increase in the air resistance (caused by increase in radius).
Once again, this question does not provide enough information. A classical physics problem has these in a vacuum. In a vacuum, it is probably close enough to say they impact at the same time. The most accurate answer is that the smaller sphere impacts first, as the CoG is lower, and thus the initial acceleration is greater so the acceleration at all points during the fall will be greater, the distance traveled is the same, so the small ball hits first.
If you would like to include air resistance, you need to provide the equations that you would like to be used, as the more you know, the more difficult this question becomes. In example, with circular airfoils, it is quite common for wake to be the driving force of aerodynamic drag. So what speeds we are dealing with becomes very relevant, and the distances have to be known.
These questions need to be moderated.
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An engineer has no problem solving this :) But you are right, too.
It has air in the margin so the relevant gas and scales of the fall are very explicit.
I agree that the size of the object and height over ground will affect what model of air resistance is relevant but even picking and choosing these parameters within reason can you really produce another result than the large one hitting the ground first? The critique only makes sense if different models produce qualitatively different results but Im not sure they do.
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My critique is that there is uncertainty. A question should have an answer that is clear, with one prevailing set of rational assumptions. I could probably conjure up a rarefied gas theory, extreme conditions setup where when all the assumptions are backed and I can definitively say that the smaller object hits first. That's not what I am getting at.
Let's talk quick and easy assumptions. In mechanics, the force due to gravity is typically considered F=mg. To be more accurate, you may easily go to F = Gmm/r^2. This will tell you (in a vacuum certainly), that the smaller object hits first.
Likewise, in terms of air resistance, for low velocities, the force is proportional to R^2 V^2, but in even lower velocities, proportional to R^2 V is also close enough. To be more accurate than these simple equations immediately becomes more complex.
To an engineer, two aluminum balls above the ground at level height will strike the ground at the same time. Both the difference in acceleration due to gravity, and the acceleration due to air resistance would be negligible. Which is more negligible, and under what scenarios is not a trivial limit. That is why I make my critique.
what if the atmosphere is mercury or a super dense gas. neither ball even fall. thus depends on the atmosphere.
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Provided, Sam survives breathing air, it's quite clear.
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But what if Sam is in a space suit and can breathe?
The question says that the spheres fall through air. Air is taken to be a mixture of nitrogen and oxygen molecules, whose density is much smaller than that of aluminium.
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Then why is there an option that says "depends on the atmosphere?"
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@Siva Budaraju – The question says "The spheres fall through the air and hit the ground.", so the atmosphere cannot be denser than the balls.
Actually, at the relevant speeds in this problem air resistance is proportional to R 2 , but the answer is still correct.
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According to Stokes Equation F = 6 π η R v ; v is instantaneous velocity & η is coefficient of viscosity.
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Stokes Law applies to laminar flow. In turbulent flow the drag force is proportional to R 2 . For the sizes of the objects and velocities expected in the problem the air flow is turbulent.
The nature of flow, laminar vs. turbulent, depends on the Reynolds number, R e = v R / μ , where v is the velocity, R is the size and μ is the kinematic viscosity. As a rule of thumb the flow is laminar for R e < 2 , 0 0 0 and turbulent for larger R e .
For air μ ≈ 1 0 − 5 m 2 / s . If the size is about R = 0 . 1 m and the velocity is v = 1 0 m / s the Reynolds number is R e = 1 0 5 . Therefore the turbulent regime applies to this problem.
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@Laszlo Mihaly – Thank you for sharing this information.
With the term 'atmosphere', shouldn't we assume or think that they could be in vacuum and therefore falling at the same time (i mean in my opinion the correct answer is "it depends")
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Provided, Sam survives breathing air, it's quite clear.
But is vacuum even an atmposphere?
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No it's not, but when you say atmosphere, you can think at like the conditions of the air (pressure, density,etc), and vacuum can be thought as a very special condition of the air (where is practically totally absent).. anyways mine was not a critique, I just wanted to know if other people thought the same as me
The question says that the spheres fall through air, so it is not a vacuum.
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Yeah, but as said by D. moore, usually classical physics problems are usually ideally taken in vacuum, and when the problem says air, it usually means the space between the spheres and the ground (in this case).. anyways mine was not a critique, just wanted to know if other people thought the same as me..
No, the question clearly states they are falling through air.
The puzzle should specifically mention that air resistance should be taken into account
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I guess that's the whole point (not to mention it directly), Sam being there and mentioning that it falls through air shall be sufficient.
There are a lot of implicit hints, which makes it quite clear commonsense-wise. 1. They "fall through the air", so there should be air resistance involved. 2. They are dropped from a top of a tall building, so the atmosphere is more than likely Earth's. 3. And the problem itself is in the Intermediate quiz, it would be in the Beginner if only gravity mattered :)
If the atmosphere is super thin, or non-existant, the spheres will fall at the same rate.
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Yes, but the question says they are falling through air, from the top of a tall building. Do you know of any buildings in non-existent atmospheres (again, not to mention the fact that it says “air”. vacuum is not air).
I agree that in a "standard" atmosphere, the smaller one will experience the greater impact from air friction, however, whether this is significant totally depends on the atmosphere, which was not defined in the problem. I believe that either you have to work the problem as a science problem or an engineering one, and this question presents a problem, because if we make the "assumption" that the atmosphere is near "standard" (engineering approach), then we must also assume that the balls are average or proportional to the image size and that we have an average measuring device and an average building height. In this case, we probably can't measure the difference between the two and so they fall the same. If on the other hand, we take the problem at face value and try to consider the extremes that the spheres can be, (very small and very large), then you also have to consider the extremes on the atmosphere. In this case, in a vacuum, the relative sizes of the balls make no difference and they hit at the same time. Also, in a liquid mercury atmosphere where the particle and massive sphere are buoyant, they then also both, do not fall to the ground. So, in the scientific approach, it is reasonable to say it depends. I think the problem is ambiguous because there are several rational assumption paths to follow.
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Neither a vacuum nor a liquid mercury atmosphere would qualify as “air” in my book. The problems says the balls are falling through air. Does the density of the air (which again is not zero, because that wouldn’t be air) make a difference as to which hits first? I’m pretty sure it doesn’t.
Also, the question states that the balls fall through the air and hit the ground, another reason we know the air is not dense enough to prevent the balls from falling.
What you describe as "retardation" is otherwise known as drag. Drag is not a function of mass but of cross-sectional area so, given a sufficient difference in cross-sectional area between the two spheres and sufficient travel time through a fluid (air, in this case) the object with the smaller area will travel faster. As to other comments below, rather than imagine a micron sized particle and an aluminum sphere, imagine a small aluminum sphere and a large beachball, now the sphere lands first.
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Not true. If the two objects are made of the same material, the larger object will land first, because its mass is so much larger.
See Mr Mihaly's answer: the acceleration, without allowing for air resistance, for each is mg - which tallies with the old puzzle of a pound of lead and a pound of feathers - however, in this case, the 2 spheres do not have the same mass.
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Flawed....the acceleration is g not mg that's the force... so neglecting air resistance as per feathers and lead they will hit ground same time
This is correct. If two balls had the same mass, then the smaller one would reach down first. In this question, both have the same density, so the larger ball reaches down first.
The drawing has written that the atmosphere is air, but it doesn't say how dense it is. If the air is almost absent, they will fall at the same speed! https://www.youtube.com/watch?v=eRNC5kcvINA
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That's right, but the larger ball would still be very slightly faster. It will be difficult to detect the time difference between the two balls though.
People have played with the idea that the atmosphere could be dense, which is not compelling given the descriptions -- but what about the material that the balls are made of? As the density of the balls gets lower and we start to look at the floatiness of the objects, the problem of wind resistance becomes dominated by sailing effects. For a material just barely above air density in still air, the air compression below the balls could be enough to let the smaller ball drop first. That still doesn't make the answer "D" unless you want to take wind into account, in which case the argument gets muddy.
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The problem says that solid spheres made of aluminum are used, and aluminium is much more denser than air.
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Yes, of course. I was proposing a related question to consider, not criticizing the question.
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@Mel Nicholson – Oh okay, I misunderstood. It's a good exercise to think what happens when the density of the balls is slightly greater than density of air.
In a vacuum they would fall at the same rate and strike the ground simultaneously so it does depend on the atmosphere.
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The option "Depends on the atmosphere" has been removed. The problem talks about balls being dropped in air, so it cannot be assumed to be a vacuum.
Air resistance is proportional to R squared and the velocity of the object squared. Also, you too are neglecting bouyancy. Your model breaks if the ball is filled with hydrogen or air, or a lot of stuff. This question does not have a clean answer. In some cases the answer will be dependant on the Reynolds number, and in those cases the smaller sphere will hit the ground first because it will have less turbulence induced drag.
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I have commented on this, including the Reynolds number, very early on, see my comments. Sparsh Singh's solution is not correct, but the answer is the correct one.
At first I thought this was a poor question, it might depend on the radius. But the more I thought about it, even if the radii were large and the building not very tall, then the large one would experience less drag per unit mass and so arrive first. This would always be true even if only infinitesimally.
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However, for certain values of R1 and R2, might we end up with the air passing around the smaller sphere in a laminar flow, while the larger sphere experiences turbulent airflow, which would have more drag, possibly canceling negating the velocity advantage that the first answer suggests? The complete absence of boundaries for R1, R2, and the properties of the atmosphere make this a set of related problems, rather than one homogeneous problem.
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I agree the question is rather loose and has many ill defined boundaries.
Drag per unit mass does not make sense, drag is proportional to v^2 and area that is all.
https://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment
Please refer to this post which seems to demonstrate that most here are way overthinking the problem. As for never making it to a Physics book.... well you will see.... the answer is they will land at the same time. For those who did not read the whole problem, the spheres are solid. With respect!
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If the experiment is conducted in vacuum, then both balls will land at the same time. However, in this problem, the experiment is conducted in air, so air resistance comes into play, and the larger ball lands first.
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so air resistance comes into play, and the larger ball lands first.
Could you explain why?
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@Sandor Gered – There is air in this problem, so the balls experience a resistive drag force which is proportional to radius squared. The retardation is greater for the smaller ball because retardation is ρ R 3 R 2 which is inversely proportional to radius. This causes bigger ball to land first.
The supposed leaning tower of Pisa experiment works because the metal balls (perhaps cast iron?) will not reach a significant fraction of there their respective terminal velocities by falling tens of metres, and so they will not show noticeable of a difference in their drop times.
(The terminal velocity of a 100mm diameter Iron ball in air at low altitude is around 135 m/s and will take about at least a 1 km drop to get there. Whilst the terminal velocity of a 200mm diameter iron ball is trans-sonic, if given about 4 km to drop!).
The question asks about aluminium balls but does not specify how high the drop , or how big the balls are. Even so the smaller ball will always be closer to it terminal velocity then the larger even if only minutely.
what about inertia?
Since the two balls are solid with the same density, the only thing can influence is the volume due to air friction. As the force of air friction is directly proportional with the volume (bigger volume results in a bigger air friction), results the small one hits first.
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Proportional to cross sectional area, not volume. The force down is proportional to volume.
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Proportional to cross sectional area, not volume
In this case it is the same (both are spheres, if increases cross sectional area, the volume increases too). In addition here the force down is not important, due to g. But the bigger one's air friction is bigger => the bigger hits last.
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@Sandor Gered – The force might be bigger, but if you want to calculate acceleration, you must divide that force by the mass, which is even bigger in case of the larger sphere (F~r^2 vs. m~r^3)
The problem with your answer is that if there is no air resistance ( as in a vacuum) both balls will hit at the same time.
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If the experiment is conducted in vacuum, then both balls will land at the same time. However, in this problem, the experiment is conducted in air, so air resistance comes into play, and the larger ball lands first.
It's incomplete questions like this, that would never make it in a physics book! Conditions must be spelled out! Period.
And In a vacuum it is the same time so the answer is actually the last one
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If the experiment is conducted in vacuum, then both balls will land at the same time. However, in this problem, the experiment is conducted in air, so air resistance comes into play, and the larger ball lands first.
The problem states "through the air" which means that there will be at least some air resistance. Yes, as air resistance is reduced the answer approaches "the same time", but it only is "the same time" when there is no air, which the problem statement specifically excludes.
It isn't perfectly stated, as with extremes of size on the balls various aerodynamic forces might come into play reversing the results. But, within a reasonable definition of "the air" and what Sam is physically able to hold so he can drop them, the air resistance increases with the surface area of the sphere while the gravitational force increases with the volume; the ratio of downward force to resistance is thus C (r^3/r^2) or C r; the larger radius sphere will have a greater net downward force at any given velocity.
We know acceleration Due to gravity is equal upon different objects. The parameter of interest here is velocity, which we can get from kinetic energy= 1/2mv^2 which we get from potential energy (U). U=mgh. assuming the spheres are solid and uniform, gravity will act upon the center of the sphere. The larger sphere’s Center is higher, and therefore it has more potential energy. Also, the bottom of the sphere is closer to the ground. .
The larger sphere will hit first simply because it is larger.
The answers lead to misinterpretation. As explained by other members, the big ball will touch the ground first because mass grows faster than air friction area. However, with no atmosphere, there is no air resistance, so every object would fall at the same speed. In conclusion, the answer "depends on the atmosphere" should be the right answer (or we can phrase it "the big ball will touch the ground faster or at best at the same time as the small ball depending on the conditions)".
Do you guys agree ?
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The option "Depends on the atmosphere" has been removed. The problem talks about the balls being dropped in air, and not vacuum. Hence, the larger ball would land before the smaller ball.
I said they would fall at the same rate. The basis for this is Galileo's experiment at the tower of Pisa where he dropped 2 cannon balls of different weights and both landed simultaneously.
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We presume the cannonballs Galileo used in his experiment had the same radii/diameter, that is why they landed at the same time ~ regardless of weight. These spheres in this problem have very different radii/diameters.
There was not enough information to answer this question. The person who wrote this question wrote it poorly and needs some basic physics. You need to change the question and answers to be correct. First, they need to state if this is with or without air resistance. Without any air resistance, the objects will fall at the same time regardless of size, shape or weight. WITH air resistance, the size or the shape also doesn't matter (there is no physics equation that will calculate the free fall of objects according to size larger or smaller). It's WEIGHT due to inertia F=mg is what matters. So, the LARGER sphere could well weigh the same as the SMALLER sphere or vice versa (e.g.: What weighs more? A kilogram of bricks or a kilogram of feathers? They weigh the damn same). Furthermore, the air resistance between the two objects depending on their mass could be so negligibly different once they reach terminal velocity which is when F_net = 0 and and velocity is constant that you would need a really good measuring device to tell.
- Physics Grad
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- The question does say that the ball falls through air, so there will be air resistance.
For air resistance, the shape and size of the object do matter. This is why cars have a streamlined design, and are not just a cuboid. It helps reduce the air drag.
Both spheres are solid spheres made of the same material, so their mass is proportional to the cube of their radius. This means that the larger sphere weighs more.
The terminal velocity also depends on the radius. The bigger sphere would have a greater terminal velocity compared to the smaller sphere. Even if both spheres reach terminal velocity, the bigger sphere will be faster and reach the ground first.
By reaching the ground, the center point of the spheres travel the same distance downwards, so no difference comes from the potential energy calculation (well, if we want to be very acurate, the gravitational acceleration is slightly higher in the case of the small sphere, as its center is closer to the gravitational center). But the main different comes from the air resistance, which force is proportional to R squared, while the gravitational force is proportional to R cubed.
Acceleration due to gravity IS equal. But objects travelling through air achieve terminal velocity due to air resistance. So at some point the forces of gravity and resistance equalize. The philosophical question is which ball is affected by air resistance first? Since the mass of a sphere which depends on its volume (r cubed) varies exponentially with surface area (r squared), as r increases the mass to surface ratio increases and the effects of air resistance are proportionally less. The small sphere reaches terminal velocity first and therefore its rate of acceleration while initially equal to the larger sphere declines more quickly as air resistance builds.
Frank: You are incorrect. The formula for distance is d = 1/2 a0 t^2 + v0 t + s0. a0 = initial acceleration (in this case gravity, g), v0 = initial velocity (in this case zero) and s0 = initial position from the measure point (since we are measuring from the top of the building were the spheres are dropping, we again have a zero value).
d = 1/2 g t^2. Note that mass and volume are NOT part of the equation.
The formula for velocity is v = a0 t + v0. a0 and v0 represent the same values as in the distance equation. Again, mass and volume are NOT part of the equation.
v = g t
That being said, the volume of the spheres matter if h represents a large enough distance. Someone jumping out of a plane at 10,000 feet will reach terminal velocity due to air resistance - they will stop accelerating towards the ground. Someone jumping off of a 25 foot high platform will accelerate until reaching the ground - there's too little time for air resistance to take effect. Francis' mention of Galileo's experiment at the Leaning Tower of Pisa reinforces this point.
The question is totally misleading; the runtime parameters are incomplete.
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I concur... to me, 20m is a "very tall building" which fits the given description... but that's typically not enough height to give terminal velocity restricted by air resistance.
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Vfinal = 2gh therefore does not depend on the mass of the spheres. In the void the two spheres touch the ground at the same speed. but in the presence of air the sphere of smaller radius loses less speed due to the friction of the air by having less area in contact. Therefore the sphere of lesser radius comes first.
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The smaller sphere also has a smaller mass, so even though it experiences smaller drag force, it retards more than the bigger sphere and reaches the ground second.
Equating E=1/2mv^2 to mgh, we get v=(2gh)^1/2; which means velocity is independent of mass, which you refer to largeness.
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Your energy equation is true only if there are no dissipative forces. In this problem the spheres fall through air. There is air resistance, due to which the spheres lose some energy on their way down, so the conservation of energy equation as you have written does not hold true.
I think the problem is ill posed. With no air tlhey would hit the ground at the same time indicating the gravity force is the same on both objects. The only consideration is aerodynamic force. The drag force on a sphere is D = Cd * .5 * rho * V^2 * A. If Cd, rho, are the same for both spheres then F-D=mg - D = mg - Cd * .5 * rho * (dx/dt)^2 * A a differential equation that can be solved for t corresponding to a given x=h and and two values of A. It seems that the smaller shere would hit first or have the smallest t. The supplied solution is incorrect.
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Please note that the mass m is proportional to the radius cubed and the area is proportional to the radius squared. Therefore the larger the radius, the less important is the drag force. That is why the larger ball will hit the ground first.
Ignoring buoyancy and using Stokes’ Law, the force on the sphere is F = mg -6𝜋𝜂 R 𝔳 where 𝜂 is the dynamic viscosity of air and 𝜐 the velocity of the sphere. If the spheres are dropped from a sufficient height, they will reach a constant velocity when F =0. Substituting 3 4 R 3 ρ for m, where ρ is the density of the sphere, the terminal velocity will be v t =\(\frac{2}{9𝜂}\) R 2 ρ g . Since the terminal velocity is the crucial parameter that determines which sphere is first, the larger sphere will be the first one. If buoyancy is included in the analysis, the terminal velocity will be: \(\frac{2}{9𝜂}\) R 2 g ( ρ s p h e r e − ρ a i r ) .
The larger sphere experiences more air resistance compared to the small ball. So technically it will touch the ground last. However, the terminal velocity of the bigger ball is greater than the terminal velocity of the smaller ball, simply because its mass is greater and it requires a greater upward force due to drag to balance the weight. Given that the terminal velocity is greater, the bigger sphere will eventually overtake the smaller ball in a long fall (given by the tall building).
The terminal velocity of BOTH spheres will be identical, if dropped in a vacuum. The larger sphere will have a greater terminal MOMENTUM, not velocity. Because drag is greater on the larger surface area, the small sphere will touch ground first.
"To take this question to the extreme: think of one of the spheres as small as a few microns in diameter, like a dust particle. That will float in air for quite a long time."
I think this limit is misleading. The quadratic drag law applies for an object with a fully turbulent boundary layer. It may not apply for a microscopic object because in such a case the Reynolds number will be very small. In that case the flow around the tiny object will be laminar, or at least not fully turbulent, and the drag will be dominated by molecular viscosity.
Relevant wiki: Fluid Mechanics
The downward force is F t o t = m g − F = g 3 4 π R 3 ρ − A R 2 v 2 , where ρ is the density, F is the air resistance, that is proportional to the cross sectional area and the square of the velocity. It is clear that for larger radius the first term will dominate the second term, so that the larger sphere will have larger force pulling it down. From F t o t = m a it follows that it will have larger acceleration. Therefore the larger sphere will reach the ground first.
Note: To take this question to the extreme: think of one of the spheres as small as a few microns in diameter, like a dust particle. That will float in air for quite a long time.