Why does a heavier sledder go faster?

[jpdodd]

In my experience, a heavier sled rider always goes faster than a lighter one (given the same sled design). When looking at this in terms of energy, it doesn't make sense. If we disregard air resistance and friction, they should have the same KE at the bottom as given by the equations below:

ME = KE + PE

KEi + PEi = KEf + PEf

KEi = 0 at the start, PEf = 0 at the end, so we get: PEi = KEf

Or mgh = 1/2mv^2

When rearranged, the vf = sqrt(2gh)

Using these energy equations, the final velocity has nothing to do with mass... So why does a heavier sledder always go faster?

The same scenario can be applied to soapbox cars and pine derby cars. You almost always want to go heavier. Can someone explain this in terms of energy?

Thanks,

jon

 

[AJ Bentley]

If we disregard air resistance and friction, they should have the same KE at the bottom

You answered your own question.

 

[jpdodd]

I've seen several places as I've been searching that have noted as such... I guess what I'm having trouble with is how that ties in with energy?

I know that friction creates heat which would take energy away from a system... so the system with the greater friction would have less energy at the end (bottom of the hill). Wouldn't a person that has a heavier weight also have more friction?

 

[cjl]

Because in the real world, you can't disregard air resistance and friction.

 

[Matterwave]

I've seen several places as I've been searching that have noted as such... I guess what I'm having trouble with is how that ties in with energy?

I know that friction creates heat which would take energy away from a system... so the system with the greater friction would have less energy at the end (bottom of the hill). Wouldn't a person that has a heavier weight also have more friction?

More friction, but not more air resistance.

 

[cosmik debris]

I've heard this heavier equals faster thing touted by quite a few people, even olympic sports persons on the radio. I think you are on the right track, if friction and air-resistance are the same in both cases then mass doesn't make any difference. The variable of interest is probably friction. One would initially think that more weight means more friction but in the case of runners on ice or snow things may be different. Perhaps more weight melts the ice or snow more and gives a more slippery layer between the runners and the surface.

I don't know the answer but I'd be interested also.

 

[AJ Bentley]

Playing Devil's Advocate: I can give you one example where mass(weight) gives a speed advantage from personal experience.

Speed-sailing with a Windsurfer. The wind force on the sail is countered by the sailor hanging out over the side in a harness attached to the sail.
A heavier sailor is able to counter more wind force and therefore use a larger sail, giving a significant increase in propulsive force. The extra weight adds little to frictional forces and the speed advantage is so significant that lighter sailors need to wear water-filled jackets to add weight.

But that's a special case.

 

[jpdodd]

https://www.physicsforums.com/showthread.php?t=458789

I found this explanation on how the air resistance affects the acceleration. I get that. What I'm having trouble with is how to connect the "energy" aspect of an objects motion with the "acceleration" part.

For example: I can solve problems using constant acceleration equations or I can use Energy equations.

Can anyone connect the two for me?

 

[cjl]

To expand a little bit here...

Yes, the magnitude of the frictional force will be larger on a heavier object. However, that's not what matters - what matters is the ratio of the frictional force to the mass. If an object which is twice as massive has twice the friction, the end result will be a tie, with both objects traveling at the same speed.

If the only resistance encountered by the object was a traditional friction force, which scales in linear proportion to the normal force, then this would be the case - all objects would still go the same speed, just like if there weren't any friction at all. However, not all frictional forces scale the same way. I'm not sure how the frictional force of a sled on snow scales, but I would guess that due to the way the snow behaves, the frictional force is less than twice as great with twice as much normal force. In addition, there is air resistance to consider - the air resistance is dependent on the shape of the object, not its mass, so two objects with identical shape but different mass will have the same aerodynamic drag, allowing the heavier one to go faster.Finally, to correct an error in your initial post: No, the two objects should not have the same KE at the end. Even disregarding all drag, a heavier object should have more KE at the bottom, due to its higher mass. Your derivation is correct, showing that with no drag, the final velocity is the same (implying that an object twice as massive should have twice the KE).

 

[Borek]

Ice melts under pressure, creating a thin layer of water that makes friction much lower. Perhaps you should search along these lines, for sure that's how it works in at least some cases.

 

[D H]

Ice melts under pressure, creating a thin layer of water that makes friction much lower. Perhaps you should search along these lines, for sure that's how it works in at least some cases.

That is a widely used explanation for why ice is so slippery. You can find this meme expressed all over the internet, and even in some physics classes and physics texts. It is however completely wrong. The math just doesn't add up with the pressure melting explanation for why ice is slippery.

• Consider an 80 kg hockey player who has all of his weight on a 10 mm by 3 mm section of a single hockey skate blade. This would lower the melting point of the ice under that blade by a couple of degrees. Yet the best ice for hockey is about -9°C, well below the melting point under that single blade. The pressure explanation doesn't work.
  
  
• Consider a petite 40 kg figure skater who has her weight on both blades. Figure skates are considerably flatter and a bit wider than hockey skates; I'll use 100 mm×5mm as the contact area for one blade. She reduces the melting point under her skates by a mere 0.03°C. The best ice for figures is about -5°C, well below this -0.03°C figure. Once again the numbers don't add up.
  
  
• Consider a 20 kg young boy who walks on ice at -10°C in his Sunday dress shoes. He reduces the melting point of the ice by an imperceptibly small amount. That the melting point of the ice under the boy's shoes is well above the ice temperature doesn't stop the boy from slipping and falling.
  
  
• Pressure melting cannot be responsible for the low friction of ice. See "resources" below.

So if it isn't pressure, what is it that makes ice so slippery? There is indeed a layer of water between the blade and the solid ice, but it isn't pressure that makes this happen. Three factors come into play.

• There is a thin water-like layer at the ice/air boundary, even at temperatures down to -157°C. This layer is just a molecule thick at this extreme cold. It's significantly thicker at warmer temperatures. Near 0°C, that water-like layer becomes a bit too thick for proper skating. Instead of gliding over the ice, the skate has to slosh through that now thick water-like layer.
  
  
• The surface in contact with the ice has to be somewhat flat so that water-like layer can be present between the surface and the ice. Ice isn't very slippery at all to a needle dragged across ice. This also explains why ice cleats work. Note that the pressure melting explanation fails completely here; a pinpoint surface maximizes pressure.
  
  
• Frictional heating also appears to play some role. The water-like layer in ice below -35°C is thinner than the microscopic roughness of even the most finely ground ice skate blade. Extremely cold ice or snow isn't all that slippery if one isn't moving. Start moving, however, and friction warms the ice below the blade to the point where a water-like layer thick enough to mask that microscopic roughness can form.

Some resources:

S.C. Colbeck, Pressure melting and ice skating, American Journal of Physics, 63:10 (1995)

Pressure melting cannot be responsible for the low friction of ice. The pressure needed to reach the melting temperature is above the compressive failure stress and, if it did occur, high squeeze losses would result in very thin films. Pure liquid water cannot coexist with ice much below -20 °C at any pressure and friction does not increase suddenly in that range. If frictional heating and pressure melting contribute equally, the length of the wetted contact could not exceed 15 μm at a speed of 5 m/s, which seems much too short. If pressure melting is the dominant process, the water films are less than 0.08 μm thick because of the high pressures.​

R. Rosenberg, Why is ice slippery?, Physics Today 58:12 (2005)

Everyday experience suggests why ice surfaces should be slippery: Water spilled on a kitchen floor or rainwater on asphalt or concrete can create the same kinds of hazards for walkers and drivers that ice can. Presumably, the liquid makes the surface slippery because liquids are mobile, whereas solid surfaces are relatively rigid. Asking why ice is slippery is thus roughly equivalent to asking how a liquid or liquid-like layer can occur on the ice surface in the first place.​

You can find this article on the internet, but given the strict rules about copyright at this site, I'm not going to post the link. Search and yea shall find.

Y. Li and G. A. Somorjai, Surface Premelting of Ice, J. Phys. Chem. C 111:27 (2007)

In this review, we summarize the available experimental data from recently developed molecular level techniques on the surface structure, surface premelting layer thickness, and friction of ice. We conclude that surface premelting of ice is responsible for the unique surface properties of the important substance.​

Also see
Science of Hockey: Why is Ice Slippery at http://www.exploratorium.edu/hockey/ice2.html
Explaining Ice - The Answers are Slippery at http://www.nytimes.com/2006/02/21/science/21ice.html?pagewanted=all&_r=0

 

[mikeph]

DH, I don't think I understand your examples. They seem to be based on analysis of the temperature change of the ice, but temperature being an intensive variable, I don't know what volume (or depth) to which this temperature change is being attributed. If a constant amount of heat is being added to the surface through compression, couldn't this produce a very large temperature change if the boundary layer is made very thin?

(I do like the J.Phys.Chem article!)

 

[jpdodd]

Finally, to correct an error in your initial post: No, the two objects should not have the same KE at the end. Even disregarding all drag, a heavier object should have more KE at the bottom, due to its higher mass. Your derivation is correct, showing that with no drag, the final velocity is the same (implying that an object twice as massive should have twice the KE).

Yes. I realize now that I stated that in error. The KE would be larger for a larger mass.


But what I'm still stuck on is how to tie "energy" equations into "force" equations... If the air resistance/friction is really what causes heavier objects to go faster down the slope, how can I connect the two "types" of equations?

By the way, thank you all for the insight into the snow aspect... but that's not really what I need clarified at this point :)

 

[D H]

DH, I don't think I understand your examples. They seem to be based on analysis of the temperature change of the ice ...

Maybe I worded my examples a bit poorly. I'll try again. I wasn't talking about a temperature change. I was talking about the extent to which pressure can reduce the melting point. Water freezes into ice I_h at temperatures just below 0°C and at one atmosphere of pressure. Ice (ice I_h to be specific) is a weird substance. With most substances, increasing the pressure increases the melting point. Ice I_h is backwards: Increasing the pressure decreases the melting point of ice I_h. The question at hand is whether this decrease in melting point due to increased pressure explains why ice is slippery. The answer is that it doesn't.

I used as examples a hefty hockey player on a single blade, a petite figure skater on two blades, and a kid in his Sunday-best shoes. The melting point of the ice immediately below the hockey player's blade is perhaps -2°C. This slight change in the melting point can't explain how hockey players are able to skate on ice that is -9°C, or even colder if they're playing outside in Canada. The pressure explanation doesn't work because the ice is colder than the melting point, even after accounting for the reduction in melting point due to pressure.

The pressure explanation is even worse for the figure skater. She weighs half as much as the hockey player and the skate blades have a much greater surface area in contact with the ice due to the flatter nature of the figure skate blades versus hockey skate blades. The pressure explanation is even worse yet for the little kid on the ice. He weighs half as much as the figure skater and has an even greater surface area in contact with the ice. The pressure explanation for why ice is slippery is flat out wrong.

 

[Borek]

That is a widely used explanation for why ice is so slippery. You can find this meme expressed all over the internet, and even in some physics classes and physics texts. It is however completely wrong. The math just doesn't add up with the pressure melting explanation for why ice is slippery.

Thanks for the correction and explanation.

Now, if I only could unlearn what has sticked to my brain at the time Americans were still flying to the Moon...

 

[Jespermj]

But what I'm still stuck on is how to tie "energy" equations into "force" equations... If the air resistance/friction is really what causes heavier objects to go faster down the slope, how can I connect the two "types" of equations?

By the way, thank you all for the insight into the snow aspect... but that's not really what I need clarified at this point :)

I think what you are looking for in terms of "tying energy and force equations" is the equation of work W=F*d. Force times distance yields the unit of Joule, the unit of energy. Is this along the lines of what you are looking for?