Yeah, I am now realising why we NEVER had to do a problem like this.
On the MCAT, they fall at the same rate.
In graduate level physics, you will tear your hair out, estimate it, then have an engineer test it for you.
On the MCAT, they fall at the same rate.
In graduate level physics, you will tear your hair out, estimate it, then have an engineer test it for you.
But since bouyancy is just the mass of air displaced, it is once again just dependent on volume. So both balls have the same bouyancy.
Edit: The more massive ball is being pulled toward the earth "more" but it is also resisting acceleration. The two balls will hit at the same time.
Now that can be the end of the thread.
Two feathers of same weight and shape falling from the same distance in a confined room with no initial air movement, which will fall faster? 
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?
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?
related discussion I found through Google
Basically, in this case buoyancy has no effect because both objects are already "submerged" in air and have displaced the same volume (and thus the same mass) of air. Looks like jagec is correct and buoyancy is strictly dependent on volume, assuming both objects have a higher density than air.
Basically, in this case buoyancy has no effect because both objects are already "submerged" in air and have displaced the same volume (and thus the same mass) of air. Looks like jagec is correct and buoyancy is strictly dependent on volume, assuming both objects have a higher density than air.
bouyancy does not play a part, air resistance does. look at my post above i gave you the answer. 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.
bottom line: the heavier ball will hit the ground sooner, its overall acceleration is actually larger. the thread would now really be officially over if you numbskullls would "get it" but much like 1=0.9999... this will probably continue to be debated even though i have mathematically proven the answer to you already.
ina vacuum yes, with air resistance, no.
think about how ridiculously stupid it sounds. if you took a balloon, filled it with air to be the same size as a bowling ball, and dropped both from the same height, you're saying they both fall at the same rate. wrong! what if you filled the balloon with helium? now it doesn't even fall at all!
BurnItDwn
Lifer
- Oct 10, 1999
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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.
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.
air resistance is NOT proportional to mass, or the University of Washington College of Engineering needs to fire a LOT of their tenured professors.
So you got your PhD from....?
you don't need a tall building. drop a helium balloon, time it's fall. it doesn't even fall. equal volume, different density, it fits the description of the problem
where did I say air resistance was proportional to mass? So you got your reading comprehension skills from?
In all seriousness, you need to reconsider your arguments and read up on how air friction works. jagec is entirely correct and you're entirely wrong.
Apparently your reading comprehension skills could use some work as well.
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.
The original question did not state both objects having a higher than air density.
But even if they do, I'm still right that the heavier object is less affected by air resistance. I proved it using Newton's Law. Jagec is as clueless as you and has no idea what he's talking about. next he'll say 1 and 0.9999... are not the same. You can debate it till the cows come home but you can't rewrite the laws of physics, They are what they are. Have you read my proof up above? Did you understand it?
OK the heavier ball will fall faster. It will continue to accelerate until it reaches terminal velocity. The lighter ball will reach terminal velocity sooner. Right?
right
but even before terminal velocity, the heavier ball will out accelerate the lighter one depending on how much of a difference in weight they are.
if you noticed my proof up above a= g-r/m. The r/m term will be significant for small values of m or large values of r (air resistance)
When you dropped the medicine ball and beach ball they probalby did not reach enough velocity for the air resistance to affect it enough. the air resistance is proportional to the velocity of the object (among other things but for this problem mainly the velocity)
for smaller values of r you'll need lighter values of m to see a significant difference, so try a lighter object vs. a much heavier one if the distance you're going to drop them will be so small.
You're welcome, and thanks for the acknowledgement :beer:
beer for you, just don't drop it:laugh:Thanks but the proof really means nothing to me.
I do however understand that the difference won't be seen in short distances unless one object is a lot lighter, the more massive ball will accelerate faster, and that the more massive ball will reach terminal velocity later (when the resistance equals it's weight) and would therefore be moving faster.I retract my previous statements, I now think you're right (though not for the reasons you thought, unless I misunderstood your explanations). Nonetheless here is a rigorous explanation of why they shouldn't fall at the same speed. I actually started out this proof to say that they do fall at the same speed
As you have stated there are several forces being applied to the object so we'll consider one at a time
These forces are F_gravity, F_friction, F_buoyancy right?
F_gravity = G*M*m/r^2, we agree here that m differs depending on the weight of the object. For small falls (ie you're not dropping the balls from outer space), r is the radius of the Earth + whatever distance you're dropping the ball. Obviously r_Earth >> drop_distance, so we can approximate r = r_earth for both objects. G and M are also constants (gravitational constant and the mass of the earth, so we get that F_gravity = k*m, where k = a_grav = G*M/r^2. We conclude that the forces are different, but the accelerations are the same.
Next we consider F_friction. Friction is a function of both shape, the viscosity of air, and speed. We have concluded above that the acceleration caused by gravity is the same, thus speeds will always be equivalent, and so if they are in the same medium and have the same shape, they will have the same amount drag regardless of mass.
Finally, F_buyancy. This only matters if the object have a lower density than the medium they are in. I won't go into more detail because I already explained it above. Otherwise the forces are the same regardless of mass and is only dependent on volume (which is assumed to be the same).
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
we can subtract equation 2 from equation 1 to get
F_total_m1 - F_total_m2 = a_grav(m1 - m2)
m1*a_net1 - m2*a_net2 = a_grav(m1 - m2)
rearranging, we obtain
a_net1 = m2/m1*a_net2 + a_grav*(1 - m2/m1)
Again, I apologize, clearly a_net1 != a_net2. It turns out, however, that this is the case *because* air friction and buyancy are independent of mass.
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