science buff question: two balls of equal volume and different density...

[crazychicken]

you guys are all confusing force with velocity.

f=ma, a is constant, so the ratio of f to m must be a constant.

however, none of this has to do with velocity.

if you ask any physics professor, they will say that any 2 object fall at the same SPEED in a vacuum, but of course they will cause a different force when they hit.

 

[Shivatron]

I can't believe this -- I thought Anandtech was smarter than this!

Things that are right:

the acceleration is slowed down by air resistance pushing up on the falling object. The amount air resistance affects the acceleration is proportional to the mass. the larger the mass, the less air resistance will affect it.

Things that are wrong:

They will hit at the same time given that they are the same shape and size to rule out wind resistance.

The simplest test for this is the thought experiment described above: imagine a beach ball full of feathers, imagine a beach ball full of lead shot, dropped of a tall building at the same time. Result?

Sorry, you got it backwards...Newtonian physics dictates that if they are both the same size (and since they are assumed to be perfect spheres of the same volume, they fulfill this), they'll fall at the same rate.

If you ignore air resistance, yes. But we aren't doing that, are we?

The wind resistance is the same.
The Force UP (wind resistance) is therefore the same.
The Force DOWN is different.

This isn't bad but needs some clarification. The wind resistance is proportional to velocity (usually v^2, but there are complicated equations dealing with this and I'm not going to get into CFD here). The force up is not the same unless they are moving at the same speed. As for the force down, well on this count the poster is right, the force down is different assuming different masses.

Assume that Fup = constant for a moment (this makes for a simple calculation). The difference between these two forces (Fdown - Fup) gives

Fdown - Fup = Fnet
M*G - Fup = M*Anet
M*G - M*Anet = Fup
M*(G-Anet) = Fup
G-Anet = Fup/m
G-Fup/M = Anet

Thus, large M tends to decrease the Fup/M term causing Anet to approach G for large M or small Fup, as you would expect. Sorry for the quick and dirty notation guys...

BTW the poster above me did a proper calculation comparing two falling masses. I only did one, just to demonstrate how M affects the final Anet term.

 

[SoylentGreen]

They dropped a hammer and feather on the moon. They hit at the same time.

 

[Shivatron]

Originally posted by: SoylentGreen
They dropped a hammer and feather on the moon. They hit at the same time.

And we all know how air resistance is a factor on the moon.

 

[dullard]

The math from Element is just plain awful (for example he uses 'a' to symbolize two different things). However, his results and conclusions are correct.

If you keep his math up you can calculate the terminal velocity - which is a function of mass! I'd rather not type all the math, so here is a link: NASA Glenn Research Center.

If we have two objects with the same area and drag coefficient, like two identically sized spheres, the lighter object falls slower. This seems to contradict the findings of Galileo that all free falling objects fall at the same rate with equal air resistance. But Galileo's principle only applies in a vacuum, where there is NO air resistance and drag is equal to zero.

 

[Armitage]

Originally posted by: element
wow just wow.

mass proportional to wind resistance? yeah you get partial marks....how about -1x10^50/100 for your mark? the mass is not affected by wind resistance, but the acceleration of the object will be as wind resistance is a force that will act in opposition to the acceleration due to gravity.

F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.

If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.

All that's left is the mass of each object being different.

Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)

F=ma-r, a =g, the constant for gravity

so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.

as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?

Wow ... I can't believe it took an hour into the discussion for somebody to get this right. Even at 0100 :roll:

 

[simms]

So was I right or wrong?

 

[simms]

Originally posted by: Armitage

Originally posted by: element
wow just wow.

mass proportional to wind resistance? yeah you get partial marks....how about -1x10^50/100 for your mark? the mass is not affected by wind resistance, but the acceleration of the object will be as wind resistance is a force that will act in opposition to the acceleration due to gravity.

F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.

If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.

All that's left is the mass of each object being different.

Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)

F=ma-r, a =g, the constant for gravity

so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.

as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?

Wow ... I can't believe it took an hour into the discussion for somebody to get this right. Even at 0100 :roll:

Makes sense, but with different accelerations results in different times to reach terminal velocity. So one object will fall faster than the other because of your equatl a=g-r/m (if r/m is significant).

 

[Shivatron]

Originally posted by: simms
So was I right or wrong?

You said:

logically a heavier mass will have less wind resistance, so lead will go down first.

Nope, sorry. Assuming same shape and size, F_airresistance is mass-independant. Having said that, it is because of this the net acceleration of the mass IS mass dependant.

 

[dullard]

Originally posted by: simms
So was I right or wrong?

Your result was right, but for all the wrong reasons. Pretty much everything you said was incorrect, but you are right they will land at different times.

 

[Shivatron]

Originally posted by: simms
Makes sense, but with different accelerations results in different times to reach terminal velocity. So one object will fall faster than the other because of your equatl a=g-r/m (if r/m is significant).

Bigger M = Smaller R/M = Bigger A = Faster time to terminal velocity. The faster to terminal velocity, the sooner the mass gets to the ground (for those of you who can't understand this... well... I don't know what to say!).

As an added advantage to the heavier mass, the heavier mass has a larger terminal velocity. See Dullard's link.

 

[Armitage]

To put it simply, there are 2 forces acting on the objects. Gravity pulling down, drag (wind resistance) pushing up.

drag is only a function of size, shape, and speed. Therefore, at a given speed the drag on the two objects is identical.
The force due to gravity is proportional to mass. Therefore, the force due to gravity is greater on the more massive object. This is also known as "weight".

So, if both objects have the same force acting up (drag), but the more massive object has more force acting down, the more massive object will accelerate faster.

Because drag is proportional to velocity, if they can fall far enough, they may eventually reach a speed where the drag force is equal to the force of gravity on the object. At this point, the object will stop accelerating. Also known as terminal velocity. Again, because the more massive object has a higher gravitational pull on it, the drag needed to balance it out is higher, and therefore its terminal velocity is higher.

 

[rahvin]

I became very very dissapointed reading this thread.

 

[Armitage]

Originally posted by: simms

Originally posted by: Armitage

Originally posted by: element
wow just wow.

mass proportional to wind resistance? yeah you get partial marks....how about -1x10^50/100 for your mark? the mass is not affected by wind resistance, but the acceleration of the object will be as wind resistance is a force that will act in opposition to the acceleration due to gravity.

F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.

If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.

All that's left is the mass of each object being different.

Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)

F=ma-r, a =g, the constant for gravity

so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.

as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?

Wow ... I can't believe it took an hour into the discussion for somebody to get this right. Even at 0100 :roll:

Makes sense, but with different accelerations results in different times to reach terminal velocity. So one object will fall faster than the other because of your equatl a=g-r/m (if r/m is significant).

Terminal velocity really has nothing to do with it. They will accelerate at different rates immediately. No need to reach terminal velocity to have a difference.

 

[RaynorWolfcastle]

Originally posted by: rahvin
I became very very dissapointed reading this thread.

, at least I eventually got it right... I'm gonna blame my erroneous replies on the hour I was posting at

simms, read my last post with all the calculations, it explains everything in pretty good detail, I think

 

[Armitage]

Originally posted by: fredtam

Originally posted by: BurnItDwn
Anyone who said they fall at the same rate do this simple experiment.

Drop a beach ball off of a tall building. Time it's fall.
Drop a Bowling ball off of a tall building. Time it's fall.


Assuming they are both the same size, Which one will hit the ground first?

I assure you, the Bowling ball will hit the ground MUCH sooner.

OK I don't bowl but I just dropped a medicine ball and one of my daghter's blow up ball of approximately the same size out of my window. Guess what? They hit at the same time.

At the low speeds achieved dropping them out of the window (I'm assuming a normal residential window, 2 maybe 3 stories), drag is pretty small. I'm not suprised you didn't notice a difference, but it was there. Drop them off a tall building and you will see it.

 

[MrDudeMan]

so i had this debate with a friend, and i dont care who is right or wrong to shove it in his face, but i want him to understand that, given the circumstances in the original question, the balls will fall at different speeds and he still is saying i am wrong...so what is it? i am definitely saying they fall at different speeds based on the NASA site and the explanations in this thread.

 

[Armitage]

Originally posted by: MrDudeMan
so i had this debate with a friend, and i dont care who is right or wrong to shove it in his face, but i want him to understand that, given the circumstances in the original question, the balls will fall at different speeds and he still is saying i am wrong...so what is it? i am definitely saying they fall at different speeds based on the NASA site and the explanations in this thread.

Yes, they will definitely fall at different speeds.

 

[Woodchuck2000]

In answer to the first post...
The answer's correct, the methodology is wrong. You're incorrect to assume that a=32ft/sec as this isn't true once the object is in motion and there is drag force operating. The only constant other than the shapes and masses is the weight of the object (assuming the height is not too great). Also, the statement F=Ma-r is incorrect, it should be F=ma+r

You formulate your equation as a = (mg - kv^2) where k is a co-efficient of drag for the particular object. This neatly solves as v = sqrt(mg/k)
And thus nicely proves the point.

F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.

If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.

All that's left is the mass of each object being different.

Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)

F=ma-r, a =g, the constant for gravity

so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.

as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?

 

[KillyKillall]

Originally posted by: MrDudeMan
fall from an equal point above the earths surface. if ball 1 is 100000x heavier than ball 2, does ball 1 fall faster? [\q]

Are we talking about descending during puberty?

 

[Woodchuck2000]

Originally posted by: Armitage

Originally posted by: element
wow just wow.

mass proportional to wind resistance? yeah you get partial marks....how about -1x10^50/100 for your mark? the mass is not affected by wind resistance, but the acceleration of the object will be as wind resistance is a force that will act in opposition to the acceleration due to gravity.

F=ma, a=32ft/sec, m= whatever the mass of the object is. The acceleration is always the same IF you ignore air resistance.

If you don't ignore air resistance, then the shape of the object has to be taken into account. The shapes were given as 2 spheres of equal volume. So the shape is out as far as affecting the rate at which they will fall.

All that's left is the mass of each object being different.

Since the force on the objects can be shown as F=ma without air, with air resistance it would be F=ma-r (r stands for resistance (air resistance but I can't write ar or you'll think it's acceleration times r, which it's not)

F=ma-r, a =g, the constant for gravity

so therefore F=mg-r, and mg-r=ma
solve for a, you get a= g-r/m (divided both sides by m)
this gives us what the acceleration will be with wind resistance.

as you can see, if the mass is large, r/m becomes small and wind resistance won't affect it much. but if m is small, the 2nd term r/m starts to become significant. This will "subtract" more from the acceleration. get it?

Wow ... I can't believe it took an hour into the discussion for somebody to get this right. Even at 0100 :roll:

Dammit jim, he's right... But for all the wrong reasons! As explained in my last post....

 

[element]

The answer's correct, the methodology is wrong. You're incorrect to assume that a=32ft/sec as this isn't true once the object is in motion and there is drag force operating. The only constant other than the shapes and masses is the weight of the object (assuming the height is not too great). Also, the statement F=Ma-r is incorrect, it should be F=ma+r

Not quite. if you say F=ma+r, you are saying that the air resistance is acting in the same direction as gravity. It is not, it's acting opposite to it. I think you just misunderstood what I meant by F=ma-r. When I wrote that, I meant the sum of all forces (non negligible) acting on it. The a is gravity, which is a constant. The part that's not a constant is air resistance. Not gravity. I should have probably written F=mg-r so that you would not be confused that the a in there should not be the overall accelration, but the acceleration due to gravity only. edit: I just noticed I did write F=mg-r, so I don't see where I went wrong.

I made only 1 other mistake and that is calling air resistance wind resistance. Not quite the same thing but you know what I meant. (even in zero wind the air resistance is non zero)

 

[element]

Originally posted by: dullard
The math from Element is just plain awful (for example he uses 'a' to symbolize two different things). However, his results and conclusions are correct.

If you keep his math up you can calculate the terminal velocity - which is a function of mass! I'd rather not type all the math, so here is a link: NASA Glenn Research Center.

If we have two objects with the same area and drag coefficient, like two identically sized spheres, the lighter object falls slower. This seems to contradict the findings of Galileo that all free falling objects fall at the same rate with equal air resistance. But Galileo's principle only applies in a vacuum, where there is NO air resistance and drag is equal to zero.

Why is my math so awful? a= the overall acceleration, g= the accelration due to gravity. Please show me where my "a" means something other than the acceleration of the object due to gravity minus the deceleration it experiences due to air resistance.

 

[yukichigai]

As has been mentioned, air resistance is independant of mass, simply a function of surface area. While the force of air resistance on each ball will be identical, the force of gravity on the ball will be greater on the ball with more mass. Therefore there will be more net force on the ball with more mass, therefore the heavier ball will fall faster.

However, keep in mind that the difference in speed, acceleration, etc. is likely not noticeable to the human eye or even some measuring equipment, depending on the differing densities. For all intents and purposes they will fall at the same rate.

 

[element]

Originally posted by: yukichigai
As has been mentioned, air resistance is independant of mass, simply a function of surface area. While the force of air resistance on each ball will be identical, the force of gravity on the ball will be greater on the ball with more mass. Therefore there will be more net force on the ball with more mass, therefore the heavier ball will fall faster.

However, keep in mind that the difference in speed, acceleration, etc. is likely not noticeable to the human eye or even some measuring equipment, depending on the differing densities. For all intents and purposes they will fall at the same rate.

If the force of gravity is greater on the ball with more mass, then why do they fall at the same rate in a perfect vacuum?