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I was watching the men's luge ride with my dad. My dad said, the mass of the athlete must be at an optimum level so that he wins. I said, his volume should be minimum, but it has nothing to do with the mass, as the acceleration is independent of mass.

Is it just like any other "block on an incline" problem? Or am I wrong?

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4 Answers 4

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The acceleration equation needs to include force terms for air drag $F_a$, and runner friction $F_f$ in addition to the gravity term $g \sin(\theta)$ where $\theta$ is the slope of the luge run and $g$ is gravitation.

As Singh pointed out, gravitation exerts a force proportional to mass, so the total acceleration is

$$a = g \sin(\theta) - \frac{F_f + F_a}{m}$$

Runner friction is more or less proportional to mass, so we can replace it with a constant, i.e. the mass of the rider is not a consideration, giving

$$a = g \sin(\theta) - K - \frac{F_a}{m} $$

Air drag depends on frontal surface area, $A$ and the square of velocity $v^2$, so

$$F_a = DAv^2$$ where D is a drag coefficent to account for the rider's aerodynamic smoothness (or lack thereof).

User11865's answer points out that the rider's surface area is proportional to $\sqrt{m}$ and this is what gives heavier riders an advantage, especially at higher speeds. Let's wrap the density and shape of the human body into a constant, say, $B$, and the acceleration equation now looks like

$$a = g \sin(\theta) - K - \frac{DBv^2}{\sqrt{m}}$$

The acceleration lost to air drag is the only term that depends on mass and having more mass makes it smaller.

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I'm actually skeptical that the runner friction is independent of the mass -- the fricition is dependent on how deeply the blades dig into the ice, and this is going to be very mass-dependent. –  Jerry Schirmer Feb 13 at 22:00
    
Generally, $F_{fr} \propto \mu F_{N}$ is a horrible model for friction when tested empirically. It's just simple for intro physics class. –  Jerry Schirmer Feb 13 at 22:01
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Uhm, if I am reading the answer from User11865 right, the rider's surface area is proportional to m^(2/3) –  Abel Molina Feb 15 at 7:47

I think you are wrong because volume do not have any relation with acceleration, but the cross-sectional area have an effect on acceleration, as more the cross-sectional area more will be the drag and so the retarding force.

The other thing you said that acceleration is independent of mass is also wrong, you know that $$F=ma$$ $$\implies a={\frac{F}{m}}$$ Now from here it is clear that as mass decreases acceleration increases for the same amount of force.

As you dad said that mass must be optimum, I cannot provide its scientific reason because I have not learned optimization but you can get an idea(verification) of it by watching this video.

Link: https://www.youtube.com/watch?v=nPLx9j-bBHo

Between 21:00 - 22:00 min

Do watch the video.

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@GlenTheUdderboat Have you watched it between 21:00 - 22:00 min ? –  Singh Feb 8 at 19:25
    
You must have seen that an optimum amount of mass was required to bring the body in motion. –  Singh Feb 8 at 19:29
    
A luge run (have a look) isn't like a car or a train. A luge doesn't have wheels, and they aren't powered either. –  Glen The Udderboat Feb 8 at 19:34
    
You are right but that was just a verification –  Singh Feb 8 at 19:38
    
Here in case of luge run you can see that as more the mass of the player, then it will take more time to accelerate and even take more time to come at rest as compared to a light weight player who achieves the same acceleration in less time by applying the same amount of force.Do you now agree with me? Therefore the word optimum is used here. –  Singh Feb 8 at 19:40

This should probably be a comment or an edit to the question, but, for now, I'll post it as a community wiki answer. (Also, you will note there's no physics here.) These are quotations from the "Luge: Nutrition Fact Sheet" by the U.S. Olympic Committee - Sport Performance Division. (I found it through Googling heaviest luge competitor, first hit, but I cannot get the direct link.) I think it is from 2010.

Body Composition:

Power to weight ratio is key. The ideal body composition is a large lean mass with high levels of muscularity in the upper body. Given the explosiveness required at the start of a run, athletes will utilize mostly fast twitch muscle in the upper body. While it is known that a high power to weight ratio is important for the start, there is a debate on what the optimal total mass should be. The debate is whether an athlete should be lighter to benefit during the start or if an athlete should be heavier to benefit from gravitational pull during the run. It has been hypothesized that a .01s advantage on the start will multiply into a .03s advantage in the end. However, there is still a culture that pressures athletes to gain any type (fat and lean mass) of weight to increase total mass.

Body Mass. Our US luge athletes are smaller and lighter than ther international competitors. The debate over lighter mass for the start versus greater mass for the run remains; and athletes have been historically pressured to gain weight for performance. The sport science team should play a role in this decision as to what will maximize performance.

Perhaps it should be noted that the US doesn't have a stellar Olympic record at the luge. :)

This is just to point out, that your question is a good one.

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It has to do with drag of the air. This is related to the surface to mass ratio. The surface of a sphere increases with the square of the radius while mass increases with the cube. So the surface to mass ratio is proportional to $r^2/r^3 = 1/$r. This means that overweight lugers would have a big advantage. They don't want that, so lighter lugers are allowed to wear weights to decrease their surface to mass ratio. A luge official told me this.

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+1, because this must be the dominant effect. On the other hand, I can imagine adjusting your height on turns might be useful, as it would adjust your moment of inertia around the axis you go around on the bank. I'd like to think luge athletes have to have some skill besides just being dense deadweights. –  Chris White Feb 8 at 16:26
    
so basically the air drag factor weighs in here. It has nothing to do with contact friction, or gravity, right? Secondly, you said "sphere"; so how can such a concept be extended to the human body in this practical application? –  Saurabh Raje Feb 8 at 17:12

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