Physics Discussion Thread

ramana · Post by **ramana** » 16 Dec 2008 22:41

P1- No rise in sea level as the ice is already displaced in the water. If ice on land melts and runs off to the sea, the level will rise.

P2- You have to use the Bernoulli effect but at 20 cm its difficult.

P3 need to think.
P4 same as P1. No rise as the metal is included in the displacement of the boat.

negi · Post by **negi** » 16 Dec 2008 23:39

P2- Sip water from the straw , and then try to hit the Pendulum 'BOB' with the water jet.

Amber G. · Post by **Amber G.** » 17 Dec 2008 22:29

ramana wrote:P1- No rise in sea level as the ice is already displaced in the water. If ice on land melts and runs off to the sea, the level will rise.

P2- You have to use the Bernoulli effect but at 20 cm its difficult.

P3 need to think.
P4 same as P1. No rise as the metal is included in the displacement of the boat.

Do other people agree (or Disagree)? Any comments.
Negi - One can bring the straw close to BOB (but not touch it),and blow and bob may get wet by the jet but (practically speaking) would it swing it enough?

Rahul M · Post by **Rahul M** » 17 Dec 2008 22:34

p3 : read genius !

(I did *guess* the answer correctly when I read it, much to my astonishment !)

JwalaMukhi · Post by **JwalaMukhi** » 17 Dec 2008 23:21

P4: I think the water level will decrease. Because when the metal object is in the boat and presumably the boat is still floating, then the water displaced by the metal object will equal to its weight and hence flotation.
However, when the metal is thrown in water unless it is in a shape that will cause it to float on the surface by itself, the metal will sink. when it sinks it will displace less water than its weight. Assuming the metal sinks: So the water level should fall compared to original, based on the comparison of densities of water and metal.

Amber G. · Post by **Amber G.** » 17 Dec 2008 23:42

Rahul - yes but is Gleick correct ?

(You are right about the source of the problem)

Rahul M · Post by **Rahul M** » 17 Dec 2008 23:47

you mean rpf was wrong ?

(in which case me too ! )

SaiK · Post by **SaiK** » 17 Dec 2008 23:56

P3: since the s shaped pipe is immersed, the sucking action should have the reverse force to back into water that is being sucked, hence cancels out. no spin is my thought.

Amber G. · Post by **Amber G.** » 17 Dec 2008 23:57

Rahul - I said Gleick, not rpf.

rpf (in all his writings) was careful, not to reveal the answer. (BTW, the orgin of the problem is from a text book published in 1800's)

Rahul M · Post by **Rahul M** » 18 Dec 2008 00:18

hmmm this gets interesting !

been sometime I read it and presumed that gleick had passed on RPF's words.

added later : my (uncalculated and naive) idea back then was that being sucked in won't produce any resultant torque as opposed to the opposite case.

p.s. just checked it on the net.
it tells me torque will be small enough not to make a difference in practice. but frankly that has nothing to do with the problem itself where we can freely ignore such inconvenient reality !
thanks a lot for the mental jog ma'm !
also, interesting to know that mach started it all.

Tanaji · Post by **Tanaji** » 18 Dec 2008 18:13

This question was posed on another forum I sometimes frequent, there was a lot of discussion but no consensus:

If an aircraft (pick any winged one) was placed on a hypothetically large treadmill, and the treadmill was run to the takeoff speed , but the aircraft had its engines off, will the plane take off?

negi · Post by **negi** » 18 Dec 2008 20:07

^^
No, for the relative velocity of the airframe wrt the surrounding air will remain unchanged . Yes if one would have placed the airframe in a wind tunnel and blown air at speeds high enough to generate lift , and somehow managed to control the lifting surfaces remotely perhaps then AC would have taken to air momentarily

SaiK · Post by **SaiK** » 18 Dec 2008 20:18

simply no.. the engine is off means, and hoping the large treadmill has a size, then it should have a fall in seconds. otoh, if the engine thrust matches the spead of treadmill, then it might take off.

Amber G. · Post by **Amber G.** » 18 Dec 2008 21:32

I will say yes - (That is the air craft will take of ) to do that:
1 Mount the air craft in the right direction (That is same direction as the movement of treadmill)
2. Keep the brakes on (or have large friction at the ball bearings of the wheels - if the treadmill is long enough, it will happen, given enough time, for any aircraft whether the brakes are on or not)
Aircraft will pick up velocity and take off.
(Engines need never be turned on)

Have seen gliders taking off . ( not much difference if the pull comes from a winching rope or a treadmill).

SaiK · Post by **SaiK** » 19 Dec 2008 07:05

wouldn't that be catapulting? yes that is a possibility, if the engines kicks off.. else its a nose dive. i 'd say "no" if the engines were to be off.

SaiK · Post by **SaiK** » 19 Dec 2008 07:16

bunch of questions to be answered. some are answered btw. there are mind boggling ones too.

http://star.tau.ac.il/QUIZ/

SaiK · Post by **SaiK** » 23 Dec 2008 19:31

If the earth were to fall into the sun if it were to stop in its orbit, the time taken would be 0.177 times a year, which is 70 days

The Moon Impact Probe (MIP) was released for the orbiting Chandrayaan spacecraft in mid-November.
Chandrayaan was then in its circular orbit, hundred kilometres above the lunar surface, with an orbital period of two hours. The probe weighing about thirty kilograms took just over twenty minutes to hit the lunar surface, after being released from the orbiting spacecraft.

In this connection, it is of interest to note that the above two time periods, i.e, the orbital period and the time taken to impact (after release from the orbit) are related by a well-known result in celestial mechanics. The latter time is square root of 2 divided by 8, of the orbital period of two hours.

This is also 0.177 of the orbital period, which works out to be just over twenty minutes.
Arthur Clarke apparently knew this result when in his science fiction story Jupiter V, he stated that by a well-known theorem, a probe dropped from a spacecraft orbiting Jupiter, time taken is 0.177 times the orbital period! Rarely do we have such knowledgeable science fiction writers. (This theorem may not be known to most present day students of astronomy or physics!)

A related ‘textbook’ problem is to estimate the time it would take for the earth to “fall into” the sun if it were to be suddenly stopped in its orbit. Again the answer is 0.177 times a year which gives about seventy days. So in case of such a contingency, humanity would have about two months to gear itself for the challenge (to devise means of escape)!

An explanation for the above result relating the two time periods involves the basic physics of orbits in a gravitational field.

An object in orbit has both kinetic and potential energy, but as it is still in a bound state while in orbit, the potential energy is dominant. However, when the object is stopped in circular orbit and it goes plunging into the central mass (around which it is orbiting), its total energy doubles (the energy is now all potential; the path becomes an elongated ellipse with eccentricity approaching unity).

The orbital period squared is related to the orbital radius cubed, but as gravitational potential energy scales inversely as the radius, the period squared is now proportional to the inverse cube of the energy. So when the energy doubles, the period squared decreases by a factor of 8, and the time period decreases by a factor of square root of 8, and as the time taken to hit the surface is half of this (only half the orbit is completed), we get the required result, that is one divided by four times square root of two or 0.177 as it can easily be verified.

This equation between the orbital period of any spacecraft and the time taken for a released object to hit the planet surface is universal and has the status of a theorem, similar to the well-known result that the escape velocity (at any orbit) is square root of two times the orbital velocity.

Amber G. · Post by **Amber G.** » 31 Dec 2008 03:03

Few points:
Wrt to P1-P4 - Any more comments/thoughts?

We have Two different answers for P4.- Which one is correct? comments.

For P1 – We have one answer up till now – Further comments/ disagreements/agreements?

P2 – Is still open.
For P3 – As Rahul pointed out, it’s a fairly famous problem(with answer not so obvious

SaiK

since the s shaped pipe is immersed, the sucking action should have the reverse force to back into water that is being sucked, hence cancels out. no spin is my thought.

Can you elaborate (specifically how it is different when the water is sucked vs when it coming out?

From the similar origin as P3 there two classic questions for those who have not seen them:
Q1 – When one looks at his/her face in a mirror, and suppose the face has a til ( beauty mark) on the left side of face, it shows up as if it is on the right side in the mirror, while a “tiki/tilak” on forehead remains on top (upside) . How does mirror know which side is “up”? (After all "up/down" ought to be symmetrical to "left/right" for the mirror)

Q2 – What keeps a railroad train on the tracks?
Hint: No...it is NOT the flanges. (flanges are for extra safety not the primary reason)

John Snow · Post by **John Snow** » 31 Dec 2008 03:13

The W ( = mg)
The reaction of the rail to the weight (on level train) R upward and through CG cancel out keeping the train on the track.

Tan(a) = h/b = centrifugal force keeps it on the track when negotiating the curve V squared /rg where r = radius of the curve taken by the track H = height of the outer track b = width of the track = guage.

From High School mamory 1972

Amber G. · Post by **Amber G.** » 31 Dec 2008 03:38

Q2 - was, why doesn't the train (even on a straight track) derails? (An Ordinary car - with an ordinary driver - will not be able to drive on, say, two narrow strips)

Amber G. · Post by **Amber G.** » 31 Dec 2008 04:09

Short comment on

SaiK wrote:
If the earth were to fall into the sun if it were to stop in its orbit, the time taken would be 0.177 times a year, which is 70 days

The Moon Impact Probe (MIP) was released for the orbiting Chandrayaan spacecraft in mid-November..Chandrayaan was then in its circular orbit, hundred kilometres above the lunar surface, ....
This equation between the orbital period of any spacecraft and the time taken for a released object to hit the planet surface is universal and has the status of a theorem, similar to the well-known result that the escape velocity (at any orbit) is square root of two times the orbital velocity.

While relationship between escape velocity and orbital velocity is square root of two (but only for circular orbit), but the first part (time to fall to sun "if earth would stop in orbit") specifically applied to MIP is not entirely correct. The ratio (.177) is correct if you neglect the radius of sun (compared with the radius of the orbit of earth), which may be okay for sun/earth but not quite okay for moon/MIP as one can not, to put it mildly, neglect the radius of moon compared to orbital radius - after all, the orbit is only about 100Km above the surface of the moon! In fact free falling body (s=1/2 gt^2) type equation would give a better approximation--

BTW - If R= radius of moon (1738Km) , and r = radius of CY1's orbit (~1840Km) then the time to fall (if it "stops" in orbit) would be:
=(r^(3/2)/sqrt(2GM)) ( (R/r) sqrt (((r-R/R)) + arctan (sqrt((r-R)/R))
Of couse, when r>>R, first term is zero, and second is pi/2 so you get the answer consistent with .177 but in the case of MIP r is almost equal to R)

negi · Post by **negi** » 31 Dec 2008 06:32

Q1- Mirror question: Imo in order to visualize the mirror problem one has to imagine that instead of a mirror one is looking at one's clone facing him/her , Now let us take the three coordinate axis viz. 'x','y' and 'z' For both the person and his clone orientation of the axis 'x' remain unchanged relative to each other . However the 'y' and 'z' axis for each one points in the opposite direction relative to each one.

When one looks at the mirror one is actually visualizing the pattern created by the light rays bounced from the mirror and hence the apparent reversal in the 'y' and 'z' axis . It is the Human eye and Brain which actually create this illusion of the left-right image reversal the reversal in 'z' axis is not apparent as in mirror we cannot perceive depth

Hope it made sense.

negi · Post by **negi** » 31 Dec 2008 06:43

Amber G. wrote: For P1 – We have one answer up till now – Further comments/ disagreements/agreements?

Ramana sir already answered it, however if we consider anomalous behavior of water and account for increase in density at 4 deg C , the water level should actually rise for a brief period

no ?

negi · Post by **negi** » 31 Dec 2008 07:03

Amber G. wrote:Q2 - was, why doesn't the train (even on a straight track) derails? (An Ordinary car - with an ordinary driver - will not be able to drive on, say, two narrow strips)

Perhaps I am missing something, but even a car will not derail as long as it too has a fixed 'wheelset' like a train coach and not a split axle with with a differential. One only needs to ensure that the two axle's (front and rear) are exactly parallel to each other , I don't see any reason why such an arrangement will derail (of course the tracks should be parallel)

-added later after googling--

The tracks have a convex surface and the wheels are concave , more importantly the the diameter of the wheel reduces marginally along the width towards the outward direction. Above design ensures that wheels will never slip unless one or both break free from the axle .

SaiK · Post by **SaiK** » 31 Dec 2008 07:26

amber g, thanks for asking, made me to go back to the question. i was under the understanding it was just an S shaped tube with water flowing into it one way.. correcting my understanding and coming back again to the problem, i would like to know about the question.

A. When you say S shaped, is water flowing thru an S shaped tube that spurts out water on both ends? or
B. if its just an S shaped spinner that spins on a jet of water that is coming out through a separate hose.

if its A, then I wonder how it might spin, because (hoping the water jet gushes out with equal force on either ends) then it should cancel out the spin. i hope this is not the case.

B. if i may imagine the S shaped spinner like a flat fan blade that spins on an axis, water gushing out (pushing action)..(assumption is that the jet is hitting only one end of the S shape), I would assume the answer would be the same direction, since the force acting on the axis is the same coming from the same end of the S blade.

John Snow · Post by **John Snow** » 31 Dec 2008 08:40

Becaue the wheel has lip (inside) and the distance between the tracks is with in the tolerence of axial play of the wheels.
The wheels are locked on the shaft. In case of turning it does not require a differential because the track inside ( that is closer to the center of curvature is already short than the track that is farther (out side), however because of the axial thrust we often hear the squeeling and squeeking (like our pujaya Pradhan mantri speaking).

The wheel and axle design is now changed from jute fibre soaked in oild with brass journal bearings to needle roller bearings whose cage can be locked or the racer be locked for smooth differential (turnings) Also these days the carriage (or buggy) are designed such way that on load (of the BOX car or carriage) wheels (set) move away from each other lowering the CG to aviod toppling on curves. ( Hence caution on curves , you may slide into unkonwn holes)

John Snow · Post by **John Snow** » 31 Dec 2008 08:50

If there is bearing with direction horizontally opposed outlets, then the torque exerted by each is action on the pivot point, the torque vectors add up rather tan nullifying. the momnetum is at the pivot bearing point on which the two s shaped rotate. It is like Kaplan turbine with two lobes

Amber G. · Post by **Amber G.** » 31 Dec 2008 22:05

Saik - this is like:

(Water coming in part in the middle is actually vertical if you consider the plane of "S" as horizontal)

Here is a actual demo (and close up) of the experiment performed in a lab to find the answer...

(You can manipulate the direction of water flow by raising or lowering the bucket)

SaiK · Post by **SaiK** » 31 Dec 2008 23:05

thanks.. i correct again to see that this should when immersed have the same direction as when it was spewing water, cause the direction of the forces acting on the S shape remains the same. In the first case, it's because of the reverse effect of water flow, in the suction mode, its because of the suction itself.

vina · Post by **vina** » 31 Dec 2008 23:23

P1 --> Water level rises

P2 --> Take the straw and blow in the gap between the pendulum and glass wall and start it oscillating. Keep blowing until the amplitude increases (you need to time it well , esp when the pendulum is at max position and need a strong breath), but hopefully make it swing enough to hit the wall.

P3 --> Wont spin. When working as a sprinkler, the torque is set up because of the reaction (Newton's 3rd law) from the water jets exiting (exactly how a tip jet helicopter works .. google /wikipedia for it). However, when it is sucking, the reaction if any will be at the place where water exits ie at the the point where water exits, really no force acting in the arms of the sprinkler to cause it to spin.

P4--> Falls, some one explained it correctly with densities. P1 is the opposite of this case (density of ice is less than that of water , so equivalent wt of displaced water has lower volume than that of melt mater of the ice).

negi · Post by **negi** » 31 Dec 2008 23:35

vina wrote: P2 --> Take the straw and blow in the gap between the pendulum and glass wall and start it oscillating. Keep blowing until the amplitude increases (you need to time it well , esp when the pendulum is at max position and need a strong breath), but hopefully make it swing enough to hit the wall.

Yikes and all this time I thought glass == glass tumbler . Yeah now Bernoulli's principle should be applicable.

JwalaMukhi · Post by **JwalaMukhi** » 01 Jan 2009 00:24

With a Caveat, that the density of water trapped will return to original density of the sea water. If the water densities differ because the ice is fresh water and it melts into salt water, which has different density then the level will rise.
P1: The level will remain same. Because it is all floating ice, it has already displaced enough liquid and is floating. When it melts it will have same density as the rest of the liquid and hence there is no change in the level.

Amber G. · Post by **Amber G.** » 01 Jan 2009 03:40

Vina - good explanations .
Negi -

ikes and all this time I thought glass == glass tumbler . Yeah now Bernoulli's principle should be applicable.

.. It was a glass tumble

.. ( The fact it is a glass tumbler is irreverent - , sorry if it caused confusion...There is no glass wall etc..one has to figure out how one can make (practically speaking ...or in actual experiment) the pendulum swing (with significant amplitude) ....

BTW, if one has not already goggled it, here is video file of P3 (Above demo).
video of ordinary case

And the one for P3 (Will it spin like this - or same as above ..

Problem P3 video

P4 - Answer (level will decrease) has been nicely explained by others. (It is easy to check experimentally, by say, floating a small katori inside a larger pan etc...

P1 - Answer first posted by Ramana would be correct, except as Jwala Mukhi has pointed out, that it so happens that the floating ice in sea is almost pure water (why ? - freezing salt water, unless the salt concentration is very high - produces pure(r) water- ice) while sea water is salty and has higher density (about 2-3% higher)than the water produced by melted ice. So when ice melts it will occupy about 2-3% more volume. Rough calculations for actual earth's floating ice gives rise of about 4-5 cms of rise in sea level due to floating polar ice.

SaiK · Post by **SaiK** » 01 Jan 2009 06:12

wow! p3 does a reverse per the video!

vina · Post by **vina** » 01 Jan 2009 11:13

Amber G. wrote: And the one for P3 (Will it spin like this - or same as above ..
Problem P3 video

Yep Amber G. Sorry, was too lazy to actually put pen on paper and work that out . Yup, by examining the flow of water in the S arms under suction, you can see that there is a momentum change of the water in the elbow/bend and that will exert a force. Draw those force vectors on paper on paper and you can see that the resulting torque will make it spin in a direction that is opposite of when the water is coming out as a jet. In the "normal case", this force due to momentum change is in the radial direction and is balanced by the tension in along the arm , while in the "suction" case, it is in the tangential direction and results in a torque.

added later..-- On thinking deeper during my "cigarette break" aka. BR Browsing

, I think there will be no net torque in the suction case during steady state . However, there will be a initial torque that will start the arms spinning and will fall off when the tangential speed of the wheel is the same as that of the water getting sucked in. The "starting" torque is like the forces a bend in water pipes experience and the "thrust block" calculations based on momentum change principle will apply. It is just that since there is no physical "thrust block" restraining, the S wheel will start spinning.

(why ? - freezing salt water, unless the salt concentration is very high - produces pure(r) water- ice) while sea water is salty and has higher density (about 2-3% higher)than the water produced by melted ice. .

All this was such a long time ago. But isnt the eutectic point of a " very high concentration water " aka. brine something like -22C or so ?. Question is if you take even such a brine solution below that and water freezes out, the phase diagram would be different and the resulting "ice" wouldn't it be different than the case of fresh water ice ?.

My guess is if you freeze brine, you will end up getting two solid mixtures..ice and water , both in solid states. But will that be like a solid block of fresh water ice ?..Dont know. I guess it might end up being an amorphous mixture of ice flakes and salt. Seriously doubt if the ice crystals will be able to line up nicely and "include" or is it "acclude" the salt crystals and form a physically and mechanically compact block .

I think that is what happens with "sea ice" which becomes eventually "fresh water ice". The salt falls off /leaches off when temperatures fall really low on the surface and the freezing water is nearly "salt free" and 'fresh ice' forms in huge blocks.

vina · Post by **vina** » 01 Jan 2009 14:13

Amber G. wrote: From the similar origin as P3 there two classic questions for those who have not seen them:
Q1 – When one looks at his/her face in a mirror, and suppose the face has a til ( beauty mark) on the left side of face, it shows up as if it is on the right side in the mirror, while a “tiki/tilak” on forehead remains on top (upside) . How does mirror know which side is “up”? (After all "up/down" ought to be symmetrical to "left/right" for the mirror)

I am busy right now to take out pen and paper , but will do it when I get some time. However if we do a ray trace, I suspect we will find that what we are seeing in a mirror is a "virtual" image and I also suspect it has something do with the size of he mirror (if I remember from long long long ago, the min you can see a full straight image of yourself is if it is at least 1/2 your height right or something , too lazy to ray trace) , where the rays have the chance to cross left to right (so you see the mirror inversion in horizonal plane) , but maybe since the height of mirror is limited the rays haven't had the chance to cross in the vertical plane (for virtual image, the mirror is before the focal point right?) , so unlike the image on a pinhole camera, where the "real image is inverted" , in a mirror the virtual image is upright. (I think it fundamentally down to we are taller than we are broader! ..so the focal point is decided by our height rather than breadth!)

Say yes, if the answer is right, if not give me time, will put pen on paper and get back.

Q2 – What keeps a railroad train on the tracks?
Hint: No...it is NOT the flanges. (flanges are for extra safety not the primary reason)

This I suspect has to do with the shape of the rail and the wheel. Both are definitely not flat. The rail has a convex top and wheel is an appropriate concave shape .. I would think that if the wheel moves along the rail in a transverse direction (in either way, left or right), I think the way the resultant force from the rails and the centre of gravity of the wagon would interact would be to set up a restoring moment that tends to bring the wagon back to it's original position. Basically, the energy state would be higher (ie. potential energy increase) if the wagon moves left or right and the center position would be the one with least energy.. and the wagon is in a state of neutral stability in the center.

Something analogous to the rolling of a ship, where within certain conditions (ie. the right moment arm remains positive) , the buoyancy force acting upwards through the center of buoyancy and the weight acting downwards through the center of gravity keep the ship in a stable equilibrium ..

Again.. not done after putting pen on paper, but just "mental" mast****** oops work (pardon the French).

Gosh, it sure sucks hard to be working over the weekend in a nearly empty office.. Thanks Amber G and BRF , atleast helps me take the nose off the grinding wheel once a while (BRF is my own cigarette break I guess!). BRF rocks!.

BTW, pardon me if I get the convexity, concave shape business wrong.. It has been a really long time. Now all that my addled brains can recollect is that certain bonds , esp Mortgage Backed Securities have "negative convexity" and we all know that any securitized asset, especially MBS is worthless , while Pamela Anderson's very real "assets" have magnificent "positive convexity" ...

ArmenT · Post by **ArmenT** » 01 Jan 2009 15:15

Problem P3 was mentioned in "Surely You Must Be Joking, Mr. Feynman". IIRC he managed to argue very convincingly, that they would move in the same direction, on one day. The very next day, he argued even more convincingly that they would move in opposite directions. In the end, he decided that the only way to settle it was try it for himself and ended up causing a minor explosion in the lab.

For the record: http://en.wikipedia.org/wiki/Feynman_sprinkler

Nandu · Post by **Nandu** » 01 Jan 2009 23:19

The mirror question. My answer is, it is not physics, it is physiology and maybe a little psychology.

We have two eyes, and they are spaced horizontally and not vertically. Also, the human face has a rough horizontal symmetry. When we see our reflection in a mirror, our brain correlates it with how we see another person, and hence we perceive it as a left/right switch.

A bit of physics is involved also. Gravity still points down, in our world or mirror world, so switching up/down will not work to satisfy laws of physics either. However, I submit that we will see a horizontal flip rather than a vertical one, even if we look in a mirror while under free fall.

ChandraS · Post by **ChandraS** » 02 Jan 2009 00:12

vina wrote:
Q2 – What keeps a railroad train on the tracks?
Hint: No...it is NOT the flanges. (flanges are for extra safety not the primary reason)
This I suspect has to do with the shape of the rail and the wheel. Both are definitely not flat. The rail has a convex top and wheel is an appropriate concave shape .. I would think that if the wheel moves along the rail in a transverse direction (in either way, left or right), I think the way the resultant force from the rails and the centre of gravity of the wagon would interact would be to set up a restoring moment that tends to bring the wagon back to it's original position. Basically, the energy state would be higher (ie. potential energy increase) if the wagon moves left or right and the center position would be the one with least energy.. and the wagon is in a state of neutral stability in the center.

Something analogous to the rolling of a ship, where within certain conditions (ie. the right moment arm remains positive) , the buoyancy force acting upwards through the center of buoyancy and the weight acting downwards through the center of gravity keep the ship in a stable equilibrium ..

Ah finally a question to contribute my gyaan.

The rails have a very, very slight convexity at the top..not enough to produce the higher energy state. The reason is the shape of the wagon wheels. If you observe them closely, you will see that the bearing surface is inclined like a truncated base of a cone. To achieve full bearing/contact between the wheels and the rails, the sleepers are also inclined to the same degree. The sleeper, if you observe, have a concave surface at the top such that the inclination at the point of the rails is the same as the wagon wheels. Clicky will give a general idea. Now if you draw out force diagrams, you will see that the restoring force is due to surface friction, which is why you have a max speed at curves. For Indian Railways, this inclination is 5 degrees. Of course all this is from the classes nearly ten years ago, so happy to be corrected.

SaiK · Post by **SaiK** » 02 Jan 2009 00:46

Q1: cause the mirror image is from an object perpendicular to it, causes reversal only on the x axis (parallel to surface ). since, the bindi or tilak is exactly at ZERO axis point, there is no reversal observed, but if you observed closely even the tilak will have reversal.. check the patterns of a bindi.

Q2:

the tracks have curved surface that grooves in to the wheel pretty much creating a soft lock, thus reducing flanges wear and tear.

the axles are not like car or any road mobile.. its a fixed axle and does not rotate to turn left or right. *(no yaw /not sure here)

for turns the flanges help it to keep on the track. excessive speed on curves can cause derailment. certain coach designs for high speed trains, bank (yaw/roll?)to either left or right to prevent derailment.