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The Hammer and the Feather


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#1 alan2here

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Posted 12 July 2007 - 06:48 PM

I really tried to use the search but frustratingly couldn't find a thread answering my question.

On the moon (About 1/3rd of Earth Gravity (Irrelevant to out discussion so I wont mention it again) and no atmosphere) a hammer apparently reaches the ground when dropped from a set height at the same time as a feather.

I understand how air resistance effects terminal velocity. And how air makes things of different shapes fall differently, so I would expect the 1KG of feathers in a weightless bag with all the air sucked out to fall at the same rate as the one 1KG hammer.

However I am having trouble grasping how the 1 single small light feather can reach the ground at the same time as the much heavier hammer.

1. It takes much effort for me to lift the hammer to a high ledge using my muscles to push up against gravity, however it takes far less effort for me to get the feather up there.

2. It seems counter intuitive based on how gravitational body's effect each other in space when you think of the hammer, feather and moon as all being body's of mass (Like planets are). You would presumably expect two planets that are put next to each other to come hurtling together quickly however you would expect two footballs to take much longer to reach each other.

?
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#2 insane_alien

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Posted 12 July 2007 - 07:12 PM

remember a = F/m

and gravitational fields provide a force per unit mass

so a more massive object has a greater force acting upon it.

the F/m ratio is the same so the acceleration is the same.
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#3 timo

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Posted 12 July 2007 - 07:43 PM

@1: You need more energy to pick up the hammer. That's reflected in the hammer hurting more when you drop it on your foot, because it has a higher momentum at the same speed.
@2: The attraction that the feather and the hammer extert on the moon is very small, ignored in the calculation claiming they both drop within the same time and in fact small enough that there's no significant (measurable) change if it was considered. Ignoring part of the mutual attraction is not valid in your planet/planet or ball/ball example.
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#4 alan2here

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Posted 12 July 2007 - 10:15 PM

Thanks, that makes it a lot clearer, I almost understand now whereas I bairly understood at all before, maybe a thread explaining this should be stickied.
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#5 w=f[z]

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Posted 12 July 2007 - 11:24 PM

This question became quite controversial on another forum.

I recall that a cyber-fistfight broke out.
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#6 J.C.MacSwell

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Posted 13 July 2007 - 06:40 PM

I really tried to use the search but frustratingly couldn't find a thread answering my question.

On the moon (About 1/3rd of Earth Gravity (Irrelevant to out discussion so I wont mention it again) and no atmosphere) a hammer apparently reaches the ground when dropped from a set height at the same time as a feather.

I understand how air resistance effects terminal velocity. And how air makes things of different shapes fall differently, so I would expect the 1KG of feathers in a weightless bag with all the air sucked out to fall at the same rate as the one 1KG hammer.

However I am having trouble grasping how the 1 single small light feather can reach the ground at the same time as the much heavier hammer.

1. It takes much effort for me to lift the hammer to a high ledge using my muscles to push up against gravity, however it takes far less effort for me to get the feather up there.

2. It seems counter intuitive based on how gravitational body's effect each other in space when you think of the hammer, feather and moon as all being body's of mass (Like planets are). You would presumably expect two planets that are put next to each other to come hurtling together quickly however you would expect two footballs to take much longer to reach each other.

?


1. Also takes gravity as much effort to drop the hammer from the same ledge

2. The gravitational effect of two feathers on each other is less than two hammers on each other

@1: You need more energy to pick up the hammer. That's reflected in the hammer hurting more when you drop it on your foot, because it has a higher momentum at the same speed.
@2: The attraction that the feather and the hammer extert on the moon is very small, ignored in the calculation claiming they both drop within the same time and in fact small enough that there's no significant (measurable) change if it was considered. Ignoring part of the mutual attraction is not valid in your planet/planet or ball/ball example.


It is the same as what the moon exerts on the feather or hammer respectively.

Obviously this force will not displace the moon very much though, so it is usually ignored. (though I knew you knew alll that)
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#7 ParanoiA

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Posted 17 July 2007 - 05:00 PM

Funny, I was looking for this very topic - I think. This thread is focused on the moon so I'm not sure it's the same thing. Are you all saying that regardless of an objects weight, all objects fall at the same speed (that is, if air friction is removed)?

Not sure I'm wording that right. I remember in my intro to physics class, that I didn't do very well in by the way, that objects "fall" at the same speed, it's just that friction from the air "catches" one object more than another, seeming to indicate that more weight equals speedier descent, when it's really about momentum and friction.

Am I making any sense? It's like I know what I mean and can't say it right. I poked around in wikipedia looking for the constant we used for gravity, trying to remember this stuff, but I can't find it. Seems like it was 9.15 or something base 10 to the -26 power or something like that...
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#8 someguy

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Posted 17 July 2007 - 09:07 PM

if objects fall at a different rate on the moon than on the earth you can know right away that mass is a factor. any object you drop on earth is so tiny compared to earth that the difference in acceleration is negligible. the moon is so much smaller than the earth you can notice the difference of dropping an object on the moon and dropping the same one on earth. if you dropped the moon on the earth you would notice a difference. you don't notice the gravity of the hammer or the feather it's way too tiny. you notice the gravity of the moon if you drop the moon on the earth the moon will pull the earth and the earth will pull the moon so the moon will fall faster than the hammer. the difference between the gravity of the hammer and the feather is extremely tiny. but we can notice the difference in mass between the two by judging their weight because we are really tiny and such weaklings but if we were the size of the earth they both would be like two grains of sand, and if we were really tiny we could notice the difference in weight of two grains of sand but it is no surprise for us that two grains of sand fall at the same rate.
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#9 ParanoiA

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Posted 17 July 2007 - 09:18 PM

Well, not really what I'm asking, or maybe it is but you're way over my head.

Just trying to remember here...

Seems like when calculating the velocity of an object (on earth), say dropping a baseball from 100 feet up, that we would always use that constant for gravity - I'm thinking g, but I don't remember the particular letter used for its designation. That number would come up over and over again when calculating introduction level physics problems.

In addition, I thought this value was independent of mass and that all objects on earth fall at this rate when friction ( I'm guessing air friction ) is disregarded.

Just wondering if this is right or not, or if I dreamed it....
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#10 Klaynos

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Posted 17 July 2007 - 10:04 PM

Well, not really what I'm asking, or maybe it is but you're way over my head.

Just trying to remember here...

Seems like when calculating the velocity of an object (on earth), say dropping a baseball from 100 feet up, that we would always use that constant for gravity - I'm thinking g, but I don't remember the particular letter used for its designation. That number would come up over and over again when calculating introduction level physics problems.

In addition, I thought this value was independent of mass and that all objects on earth fall at this rate when friction ( I'm guessing air friction ) is disregarded.

Just wondering if this is right or not, or if I dreamed it....


Looks about right, g is correct it's normally taken to be 9.81m/s/s

F_gravity = mg
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#11 timo

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Posted 17 July 2007 - 10:21 PM

I'm gonna leave out the details here for simplification:

Just trying to remember here...

Seems like when calculating the velocity of an object (on earth), say dropping a baseball from 100 feet up, that we would always use that constant for gravity - I'm thinking g, but I don't remember the particular letter used for its designation. That number would come up over and over again when calculating introduction level physics problems.

Probably g = 9.81 m/sē which is the gravitational acceleration for all bodies here on earth (as an approximation at least).

In addition, I thought this value was independent of mass and that all objects on earth fall at this rate when friction ( I'm guessing air friction ) is disregarded.

Yes. Since g is the acceleration all objects dropped from a height H will move according to h(t) = H - gtē/2, where h(t) is the height at time t.

The reason for this behaviour is that the masses cancel out: The gravitational force is Fg = m*g. Acceleration is a=F/m = Fg/m => a = g, when Fg is the only force working. Friction only depends on geometry and velocity, i.e. FF = FF(geometry, v). Then, mass does not cancel out in calculating the acceleration and you instead have a = F/m = (Fg + FF)/m = (mg + FF(geometry,v))/m = g + FF(geometry,v)/m. Note that FF and Fg will have opposite directions, hence the acceleration will be reduced. But the heavier the particle, the less the reduction.

EDIT: Perhaps I should stop posting when Klaynos is online :D.
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#12 ParanoiA

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Posted 17 July 2007 - 11:49 PM

Cool! Thanks fellas that's exactly what I was talking about and very well explained. The way the masses cancel out makes sense too. Now when you add friction, that's when I remember really struggling. Angular velocity ruined my confidence in math. I don't know why I could never really "get it".
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#13 Klaynos

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Posted 18 July 2007 - 08:49 AM

Friction is shape dependent, so roughness and physical dimention affect the amount of resistance, this means you can have more massive things falling slower than less massive ones and vice versa in a sufficiently dense atmosphere, or with enough engineering...
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#14 Externet

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Posted 19 July 2007 - 04:06 AM

I have a problem with all this...

Two identical balloons, inflated to exactly
-the same volume, with the
-same shape, and
-same roughness,
-dropped from the exact same height
-at the same time,
will not touch ground simoultaneously if one is filled with air and the other with water.

No way.
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#15 314159

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Posted 19 July 2007 - 04:29 AM

Quite right. They would both feel the same acceleration except for the effect of air resistance. The balloons both have to push past the air.

The balloons experience a force due to gravity. The heavier balloon feels a greater force as it has more mass, so it feels the effect of gravity more strongly, but it is more massive, and so requires a greater force to accelerate it the same as the lighter balloon. These balance out if you have no air resistance, but air resistance depends on the shape of the object (and its velocity), so both balloons are slowed down by a force. The actual force due to gravity on the heavier object is greater, so the air resistance force has less effect proportionally. The lighter balloon may have almost all of the force negated by the air resistance.

Take an example. One balloon (Balloon A) has air in and a mass of 10 grams. The other (balloon B) has water and a mass of 1 kilo. The force due to gravity on A is mg, which is 0.098 Newtons of force. The force on B is 9.8 Newtons.

That force on A will accelerate it at 0.098/0.01 = 9.8 m/s/s
That force on B will accelerate it at 9.8/1 = 9.8 m/s/s

So they'd fall at the same rate.

Now introduce air resistance of, say, 0.09 newtons, the same for both balloons, as they are the same shape.

The force on A is now 0.008 Newtons, accelerating it at 0.8 m/s/s
The force on B is now 9.701 Newtons, accelerating it at 9.701 m/s/s

You'll see the heavier balloon fall faster, as the air resistance is a smaller fraction of the force acting on it.
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#16 Klaynos

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Posted 19 July 2007 - 09:19 AM

I believe you also need to go some momentum calculations for the air particles relative to the objects to get a full picture of what's going on.
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#17 swansont

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Posted 19 July 2007 - 11:34 AM

There is also the bouyancy force from displacing the air, which will be more noticable for the air-filled balloon. One other reason the comparison needs to occur in vacuum. A balloon that displaces 1L will feel a bouyancy force of ~g* 1/22.4 * 30grams, or IOW, ~1 gram of the balloon's mass (from the balloon itself and the extra air under pressure) will be offset by the bouyancy force. That will be a non-negligible fraction of that balloon's mass.
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#18 danny8522003

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Posted 29 July 2007 - 09:18 PM

Just to be picky...

Two objects of different masses falling in a frictionless environment wold not hit the ground at the same time. The effect is however so tiny when you do it on the moon or the Earth that it's impossible to measure.

My tutor derived an equation that included the 'reduced mass' to calculate the acceleration due to gravity, but i can't remember it off hand. I also can't find the wiki article that contained it.
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#19 Klaynos

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Posted 29 July 2007 - 10:32 PM

http://en.wikipedia...._due_to_gravity

;)
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#20 danny8522003

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Posted 29 July 2007 - 11:23 PM

Thanks, knew it was there somewhere!
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