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

[fredtam]

Originally posted by: 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?

Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

 

[element]

Originally posted by: fredtam
Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

Why would it resist acceleration if the force of gravity is what is accelerating it? What other force is acting on it in a vacuum?

 

[fredtam]

Originally posted by: element

Originally posted by: fredtam
Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

Why would it resist acceleration if the force of gravity is what is accelerating it? What other force is acting on it in a vacuum?

Massive objects resist acceleration. I can't explain it mathematically. Google it.

 

[ActuaryTm]

In the infamous words of Ralph Wiggum...

 

[MrDudeMan]

Originally posted by: element

Originally posted by: fredtam
Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

Why would it resist acceleration if the force of gravity is what is accelerating it? What other force is acting on it in a vacuum?

thats a good point, but for the purpose of this thread, the balls will fall at different speeds on earth, and that is what is important to me right now. thanks for your input.

 

[fredtam]

¨Galileo could not explain why objects of different mass accelerate the same in free fall (no air resistance)
¨Heavier objects are pulled to the earth with greater force but their greater mass also resist acceleration more

¨So the ratio of the greater force to the greater mass is the same for all objects in Òfree fallÓ and the acceleration is the same for all objects
¨f/m=f/m=g

 

[dullard]

Here is what Element said:

F=ma...

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)

Lets rearrange a bit for clarity:

1) F = ma
2) F = ma - r
3) a = g
4) F = mg - r
Combining (1) and (4) you got:
5) mg - r = ma
Rearranging (5) you got:
6) a = g - r/m

Notice how #3 and #6 conflict (unless r = 0 they cannot both be true). The conflict is pretty clear where you said that 'a' is without air resistance and then later said 'a' is with air resistance. Then you use both definitions in your math.

True you half admit in your post to ignoring semantics, but it makes your math difficult to follow.

 

[Armitage]

Originally posted by: fredtam

Originally posted by: element

Originally posted by: fredtam
Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

Why would it resist acceleration if the force of gravity is what is accelerating it? What other force is acting on it in a vacuum?

Massive objects resist acceleration. I can't explain it mathematically. Google it.

I can
F=ma

 

[fredtam]

Originally posted by: Armitage

Originally posted by: fredtam

Originally posted by: element

Originally posted by: fredtam
Because the more massive object also resists acceleration at a value equal and opposite of the increased gravity pull. I think.

Why would it resist acceleration if the force of gravity is what is accelerating it? What other force is acting on it in a vacuum?

Massive objects resist acceleration. I can't explain it mathematically. Google it.

I can
F=ma

OK I guess I can.

 

[element]

oh yeah inertia...duh! how did I forget that?

 

[Woodchuck2000]

Originally posted by: 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)

Ah, then your original nomenclature is screwy g is constant(ish) a is a variable. F wasn't well defined either way. I stand by my original comments

 

[jkasmann]

Originally posted by: fredtam
¨Galileo could not explain why objects of different mass accelerate the same in free fall (no air resistance)
¨Heavier objects are pulled to the earth with greater force but their greater mass also resist acceleration more

¨So the ratio of the greater force to the greater mass is the same for all objects in Òfree fallÓ and the acceleration is the same for all objects
¨f/m=f/m=g

Right. Another way to say this is that the gravitational force applied to the more massive object is greater than the gravitational force applied to the less massive object, as shown above (F=GMm/r^2), giving F=km. Also, by F=ma, a=F/m. Combining these equations, a_grav=km/m=k, so acceleration due to gravity in a vacuum is equal for all objects. Because no other forces are applied in a vacuum, they accelerate at the same rate, and will hit at equal moments if started at the same time for an initial velocity of 0.

If you consider air resistance, a_net=(Fg - magnitude(Fr))/m, giving a_net=(Fg)/m - magnitude(Fr)/m. Fg is force of gravity. Fr is force of resistance, which as others have said, does not depend on mass, and shape/volume is constant in our case. Magnitude(Fr) is because I have already taken direction into account in the equation with the minus sign (it is known that Fr opposes the path of the object). As before, Fg/m is constant; the magnitude(Fr)/m term will be smaller when m is larger, causing a larger net acceleration, and when m is smaller, the term will be larger, causing a smaller net acceleration. An object with larger mass will accelerate faster and land first. This part is very similar to what has been said in this thread before.

Right?

 

[RaynorWolfcastle]

Why is this thread still being discussed? Read the 50th post in the thread (mine) to get a mathematically rigorous explanation of the relation between the a_net of 2 different mass objects; the math really isn't that difficult to work out and it shows what's going on pretty clearly.

 

[TuxDave]

Man.. you should've dumped this into the 'highly technical' forum. We would've pumped out the answer in 4 posts.

 

[Heisenberg]

/bangs head on table at this thread

 

[element]

Originally posted by: TuxDave
Man.. you should've dumped this into the 'highly technical' forum. We would've pumped out the answer in 4 posts.

says the man who is trying to use an irrational number as an integer in his sig.

hehe just kiddin.....I know i know don't start that again...0.999...=1

 

[element]

Originally posted by: RaynorWolfcastle
Why is this thread still being discussed? Read the 50th post in the thread (mine) to get a mathematically rigorous explanation of the relation between the a_net of 2 different mass objects; the math really isn't that difficult to work out and it shows what's going on pretty clearly.

You wrote this:

So we are left with
F_total_m1 = m1*a_grav + F_buoyancy + F_friction
F_total_m2 = m2*a_grav + F_buoyancy + F_friction

which can't be right, your friction force is adding to the gravitational force. Since one is opposing the other, this makes no sense. it should be -F_Buoy. - F_Frict. But the buoyancy is negligible so it shouldn't even be there.

 

[jkasmann]

Originally posted by: element

Originally posted by: RaynorWolfcastle
Why is this thread still being discussed? Read the 50th post in the thread (mine) to get a mathematically rigorous explanation of the relation between the a_net of 2 different mass objects; the math really isn't that difficult to work out and it shows what's going on pretty clearly.

You wrote this:

So we are left with
F_total_m1 = m1*a_grav + F_buoyancy + F_friction
F_total_m2 = m2*a_grav + F_buoyancy + F_friction

which can't be right, your friction force is adding to the gravitational force. Since one is opposing the other, this makes no sense. it should be -F_Buoy. - F_Frict. But the buoyancy is negligible so it shouldn't even be there.

Not exactly, it is a notational difference. Force is a vector quantity; in that notation, assuming the gravitational force is taken to be the positive direction, then F_friction will be negative, because it is in the opposite direction relative to the gravitational force, so you will want to add. The only reason you can write the equations with '-' in this case is because you know the forces are opposing each other, though really, you then are saying magnitude(F) and not F.

 

[jagec]

Thinking about this one today, I realized that I was considering objects massive enough and a drop short enough that terminal velocity is never approached (eg the iron+wood cannonballs out of a window). Oops!

 

[MrDudeMan]

i am going to do a test on this tomorrow and video tape it for everyone to see. i will drop two aquafina bottles off of a tall building, one full of water and one completey empty...we will see what happens.

 

[TuxDave]

Originally posted by: MrDudeMan
i am going to do a test on this tomorrow and video tape it for everyone to see. i will drop two aquafina bottles off of a tall building, one full of water and one completey empty...we will see what happens.

.......... make sure you do it lots of times to take care of statistical variations.

 

[ActuaryTm]

Originally posted by: TuxDave
.......... make sure you do it lots of times to take care of statistical variations.

n>=32

 

[silverpig]

Originally posted by: 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.

:beer:

<-- physics

 

[silverpig]

Originally posted by: 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.

Tsk tsk tsk. And you were doing so well. Faster time to terminal velocity usually means the object reaches the ground SLOWER. Calculate how long a soap bubble takes to reach 99% of terminal velocity (it's somewhere around 1/10000 of a second). I can tell you now that a bowling ball will take much longer to reach 99% of terminal velocity.

 

[silverpig]

n/m weird double post