Acceleration, bicyclists and ski racing

[pheadden]

So I'm a 195# cyclist, coasting down a hill. I pass the 120 # cyclist who is also coasting. I'm not sure why. Ignoring wind resistance and friction - which I believe aren't significant here - shouldn't we accelerate at the same rate? I believe it has to do with the fact that I'm carrying more momentum, but I can't quantify it. It's the same deal as downhill skiers, they tend to be larger because with more weight, they carry more momentum into the flats. Any thoughts?

 

[phyzguy]

Your problem is with "ignoring wind resistance and friction". They certainly are important! If there were no wind resistance and friction, you would both go at the same rate. You go faster because there is wind resistance and friction.

 

[pheadden]

Thanks, Wind resistance and friction definitely have an impact, but with me being the larger rider with a presumably larger cross section, I would think I have more wind resistance than the smaller rider. The friction of my bike's bearings and tires on the road, has got to be negligible. I think it has to do with momentum, P=MV, but I can't reconcile F=MA.

 

[phyzguy]

Thanks, Wind resistance and friction definitely have an impact, but with me being the larger rider with a presumably larger cross section, I would think I have more wind resistance than the smaller rider. The friction of my bike's bearings and tires on the road, has got to be negligible. I think it has to do with momentum, P=MV, but I can't reconcile F=MA.

Yes, being larger you presumably have a larger cross-section. But assume that you have two riders of exactly the same shape, but Rider B is scaled up relative to rider A by a factor X. Rider B's mass will be X^3 times larger than Rider A, so the force pulling him downhill will be X^3 times larger. But his cross-sectional area will be only X^2 times larger, so the wind resistance holding him back will only increase by X^2. So the net acceleration of Rider A will be:
[tex] a_A = \frac{F_G - F_W}{M}[/tex]

while the net acceleration of Rider B will be:

[tex] a_B = \frac{X^3 F_G - X^2 F_W}{X^3M} = \frac{F_G - \frac{F_W}{X}}{M}[/tex]

So Rider B will accelerate faster.

 

[Khashishi]

Keep in mind that if there were no air resistance and friction, you would basically keep accelerating downhill and never hit a top speed.

 

[pheadden]

Thanks, but my experience is that the heavier rider, Rider A, passes rider B while coasting down hill, even if they start at the same velocity, v₀. PHyzguy's analysis indicates the opposite should occur. I believe Newton's 2nd law of motion indicates that in a vacuum (no friction/resistance), a bowling ball and a feather will accelerate at the same rate. Thus if they are both dropped from a building, they will hit the ground at the same time. So I'm still confused.

 

[phyzguy]

No, you've missed it. Rider B is the bigger, heavier rider. He is X times larger in scale and X^3 times heavier than rider A. X is a number greater than 1. For the case you gave with Rider A at 120 pounds, and Rider B at 195 pounds, X would be about 1.17.