Galileo’s Twist
Climb up the Eiffel tower, a rod made of one metre iron and one metre wood. Holding it horizontal drop it. It will reach the ground with the heavier iron side touching down first. Is this a violation of Galileo’s: time of fall or acceleration is independent of mass?
Labels: funphysics
posted by Rajesh Lal at
2:07 AM
36 Comments:
The wooden side, being lighter, is more affected by wind resistance than the heavier iron side and so will fall more slowly.
Arrows and darts similarly work on the principle that wind resistance will drag the feathered end more than the metal end which will then arrive first.
I think it is related to the fact that the center of mass is closer to the iron end. The torque on the center of mass caused by the air resistance is not zero because the length at the wood side is longer that at the iron side (taking the cm as a reference). So, the bar turns.
PS: sry 4 my Engrish :P
Galileo's theory only applies to individual objects, not a single object made of two different materials. The theory relates to the mean acceleration applicable to the whole object, not the acceleration rates of individual parts of that object.
If separate rods of iron and wood were dropped they would hit the ground simultaneously.
A single object will naturally orientate itself during freefall according to its centre of gravity.
The two objects will fall at the same time if air resistance won't interfere . But if air resistance affects the two objects. then, the heavier one will touch the ground 1st
I think the rod will stay horizontal. Only wind resistance could cause one end to fall slower.
Someone spoke of a dart... but a dart or arrow have fins that increase air resistance directionally. This rod seems to have equal size on both ends, so they will fall at the same speed.
The rod will not stay horizontal, Ragknot, because the torque on the center of mass will be non zero.
Suppose the iron is on the left side, and the wood on the right side. The center of mass will be between the left side and the middle point of the bar, that is, the surface of the bar at the left of the center os mass will be less than the surface at its right.
The force of the air friction is dependent of this surface, and thus the force applied on the right part (referenced to the center of mass) is higher than that applied on the left, and the bar will rotate. The rotation is centered on the center of mass and is couter-clockwise, until it is vertical (iron side down, wood up), position in which the torque is zero.
I hope I made myself more or less clear :)
Incidentally - a video demonstration of feather and hammer falling on the moon is at:
http://video.google.co.uk/videoplay?docid=6926891572259784994#
Not only do they hit the ground at the same time, but you can also see that they are falling significantly more slowly than they would on the earth.
I think Miguel has it. The centre of mass (= centre of gravitational force) is nearly at the centre of mass of the iron half of the rod. The wind resistance will act near or at the centre of the rod, hence there will be a torque acting until the rod is vertical, with the iron side down.
There will no torque on the horizontal bar unless air resistance creates it. For air resistance to be a factor the area faceing down would have to be greater on one end or the other. It the iron end had fins, the rod would go wood end first.
Since it is supposedly a rod of constantant radius, each end would have same air resistance, thus it will stay horizontal. The same principals of a wood ball and a metal ball of the same size, would fall equally.
Hi Ragknot. Miguel and I have assumed that the rod is geometrically uniform. e.g. a right circular cylinder with a perfectly smooth surface. The air resistance is uniformly distributed over the entire rod and so effectively acts at the geometrical centre. i.e each end end of the rod experiences the same air resistance. The air resistance doesn't cause a torque by itself.
Remember the old-fashioned phrase for torque is "a [turning] couple". In this case, the second force is the gravitational force. I assume that the phrase came about because you need (at least) a couple of forces (acting at different places) to cause a torque. This is stronger at the iron end than the wood end. That only happens because the rod is no longer accelerating at g (because of the air resistance). In the limit that the rod is falling at it's terminal velocity, the air resistance will equal mg, where m is the mass of the rod.
If the rod is instantaneously accelerating at g' and the horizontal separation between the centre of mass and the centre of the geometry is r, then the torque is m(g-g')r. This is so because the air resistance and the nett gravitational force must both be m(g-g'). The two forces are pointing in opposite directions.
If the rod were falling in a vacuum, there would be no air resistance and hence no torque.
I'm not to pleased with my explanation, but It's better than nothing. I've got to stop now, sorry.
Hi Chris, I have not seen your name here for a short while, I missed your input on several ToM's.
I understand that the center of mass of the rod, is somewhere within the iron section, while the center of the air resistance would be the center of the two meter rod. This distance between the two different forces could potentically cause the rod to rotate.
But for this potential issue to cause a reaction there would have to be something unknown.
First the air resistance is equal to both halfs of the rod. Second the g force is causeing equal accerlation on both the center of the wood and the center of the iron. So any unknown is still unknown to me.
Imagine the traditional wooden ball and the iron ball connected via a taunt rubber band, which is just taunt enough to NOT pull the two together. They are dropped together. The band would not be streched any more as the ball dropped even though their masses are different. The air resistance is the same - the acceleration is equal.
I still say that if fins were added to the iron end, the wood end would hit the ground first, but as long as the rod is a rod, it will remain horizontal until some unknown force causes the rotation. Perhaps one end hits a fixed object.... then it would begin rotate.
One problem is bugging me since I gave a thought about the one posted here; maybe you can help me to solve it.
If we have to identically sized perfect spheres, one made of wood and the other made of iron, and drop them at the same time, nobody doubts they'll hit the floor simultaneously, right? Even if they are droped side by side, touching each other in just one point (as good perfect spheres do), their relative position wouln't change in perfect air conditions, that is, they'll hit the floor side by side, touching each other in the same point, right? Theory says so. But...
If we place a tinny spot of super-hyper glue in the point of contact, we get a ridig body and, using the same reasoning as in the bar problem, it will rotate and the iron sphere will hit the ground first!! how can this be??? I think I know the solution, but I want to hear from you guys first :)
One variation from my last post:
If we split the iron-wood bar in two, and drop them separate, they hit the floor at the same time and still in the horizontal position. This is true even if we drop them side by side, touching each other as if the bar is complete: they'll act as separate bodies as they are separate bodies, and will hit the floor side by side and horizontal. But if we glue them, we have one rigid body, and they will rotate!!
How can we explain this strange behaviour????
Climb up the Eiffel tower, a rod made of one metre iron and one metre wood. Holding it horizontal drop it. It will reach the ground with the heavier iron side touching down first.
Is this a violation of Galileo’s: time of fall or acceleration is independent of mass?
Yes it is a violation if the ground is level.
If the metal end hits first, then ground must be higher at that end of the rod, or the rod hit some fixed object on the way down.
To: Miguel Tato
What is the stange force your thinking of?
Ok, Suppose your on the moon and drop the rod from a height of 1000 feet over a cliff.
Which end would hit the earth first, the wood or the iron?
I say neither, they would both hit at the same time.
Can't give moe than a minute for now. If had two identical size spheres (connected by a very thing strong, rigid rod) one of iron and the other of wood, the iron end would hit first.
When talikng about dropping two different masses, from a relatively low height, the air ressitance doesn't have any significant effect.
Ragknot, you're right in that if you fitted suficiently large fins to the iron end, then the air resistance could be made much larger on that end and you could cause the wooden end to hit first.
PS I'll post a clearer explanation later; as there seems to be some confusion seting in.
When an object is in free fall, the mass or weight has no effect on the acceleration. Some how you guys think a spot of glue will take acceleration from one object and give to another because one's mass is more than the other.
I can tell you from experience, I've done that many times when working on a roof. I take a hammer and drop it to get ground just before climbing down. The heavy end does not cause torque to rotate the hammer. It hits the ground at the same angle I drop it, level, wood end down or steel end down. If is give a slight spin, it continues that spin. My coworkers usually their bet, and guys wait to see a new guy fall for it. Many have tried it for themselves thinking it a trick.
Go try it for your self.
I go back to my first post: the lighter the object the greater the effect of the air resistance, and hence it will fall more slowly. That is why the rod in question will turn so that the metal end lands first.
What if it was rigid, hollow cardboard tubing instead of wood? I'm sure it would turn even more.
I believe that if a metal rod, a wooden rod and a cardboard rod were dropped simultaneously from a great height then they would land in that order.
Ragknot's hammer trick is a sure bet from the roof of a single storey house. Air resistance would not have time to work from that height. Try it from the top of a skyscraper instead.
Hi Ragknot. I've not been posting so much as my leg is much better and I've had to return to the "real world".
I'd already covered the observation that you made about the hammer in my 10:35 pm post. Until the object is at a substantial fraction of its terminal velocity, the air resistance effect will be neglible, so the torque will be negligible and further, the torque has to be applied for some time before a significant rotational speed is acquired and then some more delay as an actual change in angle occurs. Everything I've said in my posts is correct, I'm very sure of that. I also understand why some people aren't getting their heads around what's going on.
I think Wiz's 4:24 pm post puts it very nicely. In essence he's saying that the terminal velocity of the iron end of the rod is higher than the wooden end of the rod - that should be enough to let the result be understood.
Galileo put forward a nice argument about why objects should fall at the same rate. He noted that if you [doing the experimants in a vacuum] and assumed that a heavy object would fall more rapidly than a light one, then if the two objects were merged together, then they should fall faster than either one alone. But if they were tied together with a piece of string (of negligible mass etc.), then the light one would retard the heavy one and so the combination should fall at an intermediate rate. He considered that to be ridiculous. The simplest way out is to assume that they must both fall at the same rate.
They both hit the ground at the same time. Galilieo once said that the lightest fether and a bowling ball will hit the ground at the same time (or something of that effect)
Being tied together it would mostlikly be different. But when you look back when you were taught about things like this you remember how and when he does this stuff. Every simple indeed.
... that would be true in a vacuum or with a very short fall (1 cm ay).
By considering things like Galileo's alledged Tower of Pisa experiment, Einstein was able to develop his General Theory of Relativity.
It's to do with the relative position of the 'centre of pressure' and the 'centre of mass'. If we drop the rod with it in a horiontal orientation, the centre of pressure - ie the point at which the air resistance appears to act - is in the centre of the rod as it starts to falls. The centre of mass - the point at which gravity appears to act - is within the iron part of the rod. This creates the torque - the turning effect - as the centre of mass (the effective point at which the downward force due to gravity appears to act on the rod) is to one side of the centre of pressure (the effective point at which the force due to air resistance appears to act). As the rod falls, the centre of pressure ends up - as a result of this turning effect - behind the centre of gravity, and this (eventually) leads to a stable configuration in which there is no further turning effect, and movement of the rod from this 'dart-like' configuration results in a net turning effect (a torque) which acts to restore that stable configuration. This is also how a dart and an arrow work. Without air (and hence air resistance) - which is what galileo was talking about: objects falling in a vacuum - there is no air resistance, no force acting through the centre of pressure (as there's no pressure!), and hence no turning effect as the object falls. Incidentally, if you drop a hammer even through a short distance, the torque is so great that you can be absolutely sure that the hammer will hit 'heavy end down'.
Hi AccordionPlayer. I agree with the bulk of what you said, although I'm turning a blind eye to the use of the word "pressure" - but I empathise with you, it does seem surprisingly hard to explain this non-trivial problem.
I cannot agree with your statements regarding the hammer being dropped. The torque will be negligible for a short fall. I don't know what the terminal velocity would be, but wouldn't be surprised if it were around 500 kph (km/hour). g is approx 10 m/s² => 36 kph/s. So it would take very roughly 15 seconds to approach terminal velocity (in a vacuum) and that would correspond to a height of about 1 km. Until terminal velocity is approached, the air resistance is fairly unimportant. I accept that these numbers are very rough and ready, but the give an idea about the magnitudes. In the case of a hammer, the air resistance might be significant after a mere 100 m fall (as the centre of mass is very close the the hammer's head). Even so, the associated torque only produces a rotational acceleration - the rotational acceleration will cause the rotational speed to build up (very roughly) uniformly with time. The change in angle is the integral of the rotational speed. Altogether a change in angle is not going to happen rapidly. Also, as the final vertical orientation is approached, the torque will be reduced towards zero, as you have discussed, and as you also have discussed, it will undergo a damped oscillation around the vertical orientation.
I'm using rough and ready calculations as I don't want to get involved with full blown solutions to differential equations and the inevitable exponential (or worse) time dependencies. I pretty sure that a pair of (simultaneous) second order non-linear differential equations is the minimum that is required to do a useful treatment - and even that would require the use of some fairly gross approximations to the problem.
Again I say that I think Wiz has given the most easily understandable way of getting the gist of what happens.
PS I defer to Ragknot for the experimental confirmation. See his post dated Nov 20, 11:25 am.
Galileo realized, even during his earliest studies (published in his book On motion) that the speed of a falling body is independent of its weight . He argued as follows: suppose, as Aristotle did, that the manner in which a body falls does depend on it weight (or on some other quality, such as its ``fiery'' or ``earthy'' character), then, for example, a two pound rock should fall faster than a one pound rock. But if we take a two pound rock, split it in half and join the halves by a light string then one the one hand this contraption should fall as fast as a two pound rock, but on the other hand it should fall as fast as a one-pound rock (see Fig. 4.3). Since any object should have a definite speed as it falls, this argument shows that the Aristotle's assumption that the speed of falling bodies is determined by their weight is inconsistent; it is simply wrong. Two bodies released from a given height will reach the ground (in general) at different times not because they have different ``earthliness'' and ``fiery'' characteristics, but merely because they are affected by air friction differently. If the experiment is tried in vacuum any two objects when released from a given height, will reach the ground simultaneously (this was verified by the Apollo astronauts on the Moon using a feather and a wrench).
This result is peculiar to gravity, other forces do not behave like this at all. For example, if you kick two objects (thus applying a force to them) the heavier one will move more slowly than the lighter one. In contrast, objects being affected by gravity (and starting with the same speed) will have the same speed at all times. This unique property of gravity was one of the motivations for Einstein's general theory of relativity (Chap. 7).
Hi Ragknot. I see you've been doing some copying and pasting - the (Chap 7.) is a giveaway, as is the abscence of typos :)
Before Einstein set his mind to work, it was assumed that there were two kinds of mass; gravitational and inertial. That they had the same value was regarded as a bit of luck. Einstein realised that they were in fact identical.
If Mg is gravitational mass and Mi is inertial mass, then the acceleration, g', of a mass in a gravitational field of strength g would be, g' = (Mg/Mi)g. Obviously this assumes no air resistance clouding the issue.
Einstein realised that g' was necessarily equal to g and so it immediately follows that Mg was necessarily identical Mi. That was/is a very deep observation.
The article was a little confusing to me. It seems that if you follow Aristole's theories you would say that the metal end of the rod goes first. If you cut a two pound stone in half and connect with a light string.... sounds a lot like out ToM.
I tried to find the Chap 7 reference but I figured it was refering to Einstin's book.
If the two ends of the rod have equal cross section, and the mass does not effect the acceleration.
then their will be no torque on the rod.
http://phyun5.ucr.edu/~wudka/Physics7/Notes_www/node49.html#SECTION02422100000000000000
Hi Ragknot. Your last comment is right, if their is no air resistance.
Wow... lot's of debate on this topic.
Here goes what I believe is a simple explanation.
We all know that, a seperate iron rod and wooden rod would reach the ground at the same time if they were in a vacuum (not subject to drag).
This happens, because the force imparted on an object (on it's centre of gravity) is proportional to its mass. i.e. F=M*g
and the acceleration of a body is proportional to the force applied and inversely proptional to its mass i.e. F=M*a OR a= F/M
hence if gravity is the only acting force F=M*g and a= F/M OR a=g
Now let's assume that the iron bar, and wooden bar have the same surface texture (one is no rougher than the other) and that they both have the same surface area. If they are now subject to drag (from the air), the time that a seperate iron bar and a seperate wooden bar would take to reach the ground would change. The iron bar would hit the ground first.
Why ?
Because the new force added into the picture, drag force, is proportional to surface area, (and other parameters including drag coeeficient and fluid density, which are also the same for both bars) but not mass.
So if we look at the the forces on each bar they are the drag force and the force due to gravity.
F_total= F_g -F_d
( minus sign for F_d since it points in the opposite direction)
and the acceleration on each bar is:
a= F_g -F_d = g - F_d/M
now the mass of the bar matters, as it doesn't simply cancel out. The smaller the mass, the smaller the net downward acceleration will be. Hence a wooden bar would fall slower than an iron bar, due to drag.
Now if we decide to connect the two bars, effectively the same thing happens, but since they are joined, the composite bar will rotate such that the iron side hits first. (F_g will act on the centre of gravity, which is located on the iron portion of the bar, and the integrated moment about the centre of gravity will impart an angular acceleration according to M=I*alpha, where M is the moment , I is the caculated moment of inertia of the structure, and alpha is the angular acceleration)
If the height is small, or the surfaces are smooth, or the surface areas are small, the drag force may be small, such that its effect would be imperceptible.
Hope this explanation is clear.
Cam
oops
"a= F_g -F_d = g - F_d/M"
should be
"a= (F_g -F_d)/M = g - F_d/M"
Cam
Cam,
Thanks for include drag. That was the unknown I didn't include. The terminal velocity is determined by the weight and air resistance. Air resistance would be the same for the iron have and wood half, so the acceleration of gravity would be the same but as soon as drag becomes significant the wood end would slow it's acceleration.
I final see that.
I see that the proposition I posted
November 20, 2009 10:23 AM
about dropping over a 1000 foot cliff is still true, though.
Both ends would hit at the same time.
Hi Ragknot. LOL. Drag is just another name for air resistance or wind resistance. ("a rose by any other name...").
It hadn't occurred to me that you didn't know that drag, a force that typically increases with speed, is the reason that falling objects have a terminal velocity. i.e. that object will stop accelerating when the drag is equal to the weight.
I'll add the terminal velocity of a horizontal rod would be lower than for a vertical rod.
Cam, I think Wiz's explanation was far simpler. i.e. that, in principle, each half of the rod has a different termnal velocity. It avoids the concept of torque.
You can sign in using name/url (you can leave url blank). It's slightly less work than being Anonymous and adding Cam at the end. It's the method that Wizard of Oz uses (I guess).
chickennn!!!
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