Unusual Physics: Understanding a Strange Truck Phenomenon in Reality

[munky99999]

This is set in reality.

You have a truck, set with high wheels(railroad wheels). Your driving on the railroadtracks going 50km/h. It's completely dry. You stop easily and quickly.

The weird thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

How is this explained?

 

[vincentchan]

The weird thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

are you saying it speed up when you hit the brake? that is not possible... what if you keep hiting the brake... will your truck keep speeding up? so the truck can principally go from east coast to west coast without using any fuel..

 

[Integral]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

 

[munky99999]

vincentchan no you only gain 8-10km/h then it starts to lower.

Integral that's amazing, at first I denied the possibility, then I thought it might be a friction thing.

University of Toronto concluded they could not understand how it's possible.

 

[dextercioby]

University of Toronto concluded they could not understand how it's possible.

Then University of Toronto should start reading...phyics...

Daniel.

 

[Bystander]

(snip)a net increase in applied force and accelerates.

The problem statement doesn't explicitly include any applied force --- it implies a constant 50km/h (maybe 200 N/m² air resistance) balanced by engine and drive train, transitioning to zero driving force upon application of brakes (wheel lock-up is implied, or must be assumed), and the rotational kinetic energy of the wheels is conserved (some dissipation in the brakes stopping the rotation) to be added to that of the truck (I ain't going to do the integration of the cycloid across a railroad wheel) as it hydroplanes along the rails.

 

[LURCH]

This is set in reality.

You have a truck, set with high wheels(railroad wheels). Your driving on the railroadtracks going 50km/h. It's completely dry. You stop easily and quickly.

The weird thing. If its raining and you try to stop from 50km/h, you gain 8-10km/h on average.

How is this explained?

Do you have a link or reference to this experiment?

 

[physicsisphirst]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

so what you are saying is that when the brakes hit the friction is reduced because of the wet tracks - ie the sliding friction is less than the rolling friction. it isn't that a force is added, but the frictional force is reduced so there is a net increase in force, hence the acceleration. is this the same sort of thing that happens on slippery snow where it is more advantageous to pump the brakes (hence, maintain some of the rolling wheel friction) rather than slamming the brakes and locking the wheels?

 

[munky99999]

physicsisphirst well i guess so. Very interesting stuff eh?

 

[GENIERE]

I'm off to the local pub to make a few bets! If I were fair-minded I'd split the winnings with Integral, that's if I were fair-minded.

 

[munky99999]

but instead your absent-minded?

 

[Fliption]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

Then why not just make a train that slides on a wet track instead of rolling wheels?

 

[EL]

The problem statement doesn't explicitly include any applied force --- it implies a constant 50km/h (maybe 200 N/m2 air resistance) balanced by engine and drive train, transitioning to zero driving force upon application of brakes (wheel lock-up is implied, or must be assumed), and the rotational kinetic energy of the wheels is conserved (some dissipation in the brakes stopping the rotation) to be added to that of the truck (I ain't going to do the integration of the cycloid across a railroad wheel) as it hydroplanes along the rails.

This seems like a nice explanation.


However this does not make sense to me:

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

What force is applied more than friction when the wheels are locked?

 

[DaveC426913]

Things to factor in:

1] A speedometer does not measure actual vehicle speed. It measures the rotation of the tires. If the wheels happen to be spinning faster (for whatever reason and however briefly), the speedometer will show an increase in speed.

2] The engine is still trying to drive the vehicle, even when the brakes are applied. In a small car, the engine quickly drops to a lower rev, but I'll bet those monster diesels don't slow to idle quite so easily.

A combination of one or both these factors will likely yield a sensical answer.



<trite> Everybody's looking at this as an idealized study in Newtonian physics. It's not ideal, it's a locomotive! This is what you get when you send a bunch of physicists out to do a mechanic's job! </trite>

 

[da_willem]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

This only works if a constant force is applied to the truck even when the brakes are on. Else it would violate conservation of energy

 

[The_Thinker]

it's a good thing that u mentioned that, for some time i didn't understand what was going on...

 

[physicsisphirst]

This only works if a constant force is applied to the truck even when the brakes are on. Else it would violate conservation of energy

hey! that's a good point! in fact, that is the point since the force that does exist is that of friction.

even though we have less friction by the application of the brakes there is presumably no force being applied by the locomotive when the brakes are on. so what does that leave us with? it leaves us with the inertia of the locomotive going at say x km/hr and a frictional force that has just been reduced.

so here is a similar set up:

say you have a curling stone that you slide over a carpet that is on a rink. what will happen when the rock hits the edge of the carpet and slides onto the ice? will it speed up or not?

 

[gerben]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

but the static friction provides the force that makes the train go forwards, while the kinetic friction (when braking) is directed against the direction of the trains motion and wil only slow it down...

(it is not the case that the non-braking train has to overcome static friction while the braking train has to overcome kinetic friction)

 

[Gokul43201]

I can't see how the truck will speed up ! The mass of the wheels is negligible compared to that of the truck.

 

[da_willem]

I can't see how the truck will speed up ! The mass of the wheels is negligible compared to that of the truck.

It are railroad wheels, which are pretty heavy I guess. Maybe their rotational energy is about enough to acomplish the acceleration. Let's see:

T=\frac{1}{2}I\omega ^2=\frac{1}{2}(\frac{1}{2}MR^2) (\frac{v}{R}) ^2=\frac{1}{4}Mv^2

So if the mass of the truck is x times that of the total mass of the wheels this can by conservation of energy accomplish a change in velocity given by:

\frac{1}{2}m(v+\Delta v)^2=\frac{1}{4}\frac{m}{x}v^2
\Delta v=v[\sqrt{1+\frac{1}{2x}}-1]

So for v=50km/h this can only account for the observed increase in speed when the mass of the wheels is comparable to the mass of the truck. I guess this is not a very realistic scenario, and how would this energy be converted to translational energy?

 

[physicsisphirst]

I can't see how the truck will speed up ! The mass of the wheels is negligible compared to that of the truck.

but wouldn't the real issue be friction?
say the total energy of the system is given by

<br /> E_{Total} = E_{kinetic} - E_{friction}<br />

the frictional energy is negative because it is going out of the system.

now if by applying the brakes we cause skidding and thereby reduce the frictional force we also would decrease the frictional energy that is lost and thereby increase the total energy and hence the speed of the train.

does this make sense or am i missing something (my physics is rusty from disuse and I'm trying to lubricate it)?

 

[Bystander]

It are railroad wheels, which are pretty heavy I guess. Maybe their rotational energy is about enough to acomplish the acceleration. Let's see:

T=\frac{1}{2}I\omega ^2=\frac{1}{2}(\frac{1}{2}MR^2) (\frac{v}{R}) ^2=\frac{1}{4}Mv^2

So if the mass of the truck is x times that of the total mass of the wheels this can by conservation of energy accomplish a change in velocity given by:

\frac{1}{2}m(v+\Delta v)^2=\frac{1}{4}\frac{m}{x}v^2
\Delta v=v[\sqrt{1+\frac{1}{2x}}-1]

Not too bad --- good for a "C" --- now, account for the conservation of angular momentum to bring your grade up to "B."

So for v=50km/h this can only account for the observed increase in speed when the mass of the wheels is comparable to the mass of the truck. I guess this is not a very realistic scenario, and how would this energy be converted to translational energy?

Not so good --- but, in Delft, you may not be familiar with N. Amer. track inspection vehicles (Detroit, or Japanese, iron with rail car wheels swapped for automotive tires and wheels) --- the wheel mass fraction is higher than one might think.

An 8-10 km/hr jump? It is also necessary to consider that measurement of speed is sometimes accomplished by monitoring rotational speed of a trailing (or leading) pony wheel (unbraked, low moment of inertia) connected to an automotive speedometer. If this happens to be the hysteresis type, it is subject to overshoot during transitions between an indicated steady state speed and a higher or lower new steady state as a result of a step impulse (braking of heavy wheels on a near frictionless surface).

How does one get an "A?" Do the analysis without the hints.

 

[PBRMEASAP]

The key here it that kinetic friction is always lower then static friction. As long as the wheels are rolling the friction is static friction. AS soon as the wheels start to slide kinetic friction is the key. So by sliding the force due do friction is reduced, thus the train sees a net increase in applied force and accelerates.

This is quite remarkable, since it means that on a train track buffered with short wet spots every few feet, one can achieve perpetual motion by slamming on the brakes periodically. The engine can be turned off as soon as you get some speed!

BRILLIANT! *hands Integral a Guinness*

 

[da_willem]

but wouldn't the real issue be friction?
say the total energy of the system is given by

<br /> E_{Total} = E_{kinetic} - E_{friction}<br />

the frictional energy is negative because it is going out of the system.

now if by applying the brakes we cause skidding and thereby reduce the frictional force we also would decrease the frictional energy that is lost and thereby increase the total energy and hence the speed of the train.

does this make sense or am i missing something (my physics is rusty from disuse and I'm trying to lubricate it)?

There is no 'friction energy' part of the system. It is a dissipative power. Once energ is converted in this 'friction energy' (ie heat, sound etc.) it is gone. So having less friction doesn't speed you up, absence of friction doesn't add energy to the motion. As said before, this only works for an applied force, even when the breaks are on.

 

[da_willem]

Not too bad --- good for a "C" --- now, account for the conservation of angular momentum to bring your grade up to "B."

Not so good --- but, in Delft, you may not be familiar with N. Amer. track inspection vehicles (Detroit, or Japanese, iron with rail car wheels swapped for automotive tires and wheels) --- the wheel mass fraction is higher than one might think.

You're right. I've never heard of a vehicle with train wheels. I guess we don't have em in Delft or I've just never noticed them. As I pointed out in the beginning of my post I acknowledged the fact that the mass of the wheels might be higher than you think, but comparable to the mass of a truck? What trucks and how many wheels are we talking about?!

And if I knew this was going to be an exam I would have pushed my analysis some further

 

[Gokul43201]

It are railroad wheels, which are pretty heavy I guess. Maybe their rotational energy is about enough to acomplish the acceleration. Let's see:

T=\frac{1}{2}I\omega ^2=\frac{1}{2}(\frac{1}{2}MR^2) (\frac{v}{R}) ^2=\frac{1}{4}Mv^2

So if the mass of the truck is x times that of the total mass of the wheels this can by conservation of energy accomplish a change in velocity given by:

\frac{1}{2}m(v+\Delta v)^2=\frac{1}{4}\frac{m}{x}v^2
\Delta v=v[\sqrt{1+\frac{1}{2x}}-1]

So for v=50km/h this can only account for the observed increase in speed when the mass of the wheels is comparable to the mass of the truck. I guess this is not a very realistic scenario, and how would this energy be converted to translational energy?

I believe I get a similar result. I have :

\frac{\Delta v}{v} \approx\frac {f}{4}

where f = \frac {mass~of~wheels}{mass~of~rest~of~truck}

 

[Rogerio]

I can't see how the truck will speed up ! The mass of the wheels is negligible compared to that of the truck.

I can't see it, too.

Absence of friction doesn't add energy to the truck.

And by applying the brakes we don't transform the wheel's energy of rotation into truck's energy of translation.

In other words, the wheels' energy plays no role here.

I think the truck won't speed up.

 

[da_willem]

I think the truck won't speed up.

Me too; or we're all missing something here...

 

[Rogerio]

Me too; or we're all missing something here...

And if munky99999 is playing with the words?

Where he said (post#1) "If its raining and you try to stop from 50km/h, you gain 8-10km/h on average", maybe the "average" could mean "the velocities average until the truck stops." Then, it could be possible.

However, he said (post#4) "vincentchan no you only gain 8-10km/h then it starts to lower".

So, he meant "the truck really speeds up about 8-10km/h ".

I think it's impossible.

 

[T.Roc]

1. missing data: what is the exact grade of the track? even slight differences from level will throw off results.

2. never use physics when logic will do.

consider:
1. i can run 10mph on level ground.
(feet making "circles" limits max speed)
2. i can run 13mph down the side of a mountain.
3. i come back to the same mountain when it is covered in snow, and attatch long, smooth things to my feet, and am able to go down the mountain at 20mph.

the only thing that changed was replacing the limit of my gait (wheels) with a low friction slide.
if acceleration and gravity can not be felt to be different, the why not momentum of bigm ass on "level" ground compared to gravity pulling my narrowm ass down a hill?

TRoc