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Acceleration of Coasting Vehicle Down Inclined Plane 
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#1
Jun2709, 03:45 PM

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This caught my interest after someone posted the question on a cycling forum of would a heavy and light cyclist have different accelerations down an inclined plane when coasting from a stop. Many people had observed that heavier cyclists do indeed have a greater acceleration, but I wasn't satisfied with any of the explanations. I was trying to figure out why with a diagram and equations but I've failed. I'm thinking that the heavier rider applies a greater torque to the wheel thereby causing a greater acceleration. Is that correct? Anyone able to work out the equation for the linear acceleration given the mass, plane angle, and radius of the wheel? Neglect air resistance. Thanks!
PS What's a good explanation for why a wheel rolls downhill, in terms of forces? 


#2
Jun2709, 04:33 PM

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#3
Jun2709, 05:15 PM

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So you don't think the heavier rider will have a greater acceleration from the start, assuming they both start with zero velocity?



#4
Jun2709, 05:32 PM

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Acceleration of Coasting Vehicle Down Inclined Plane
If by "from the start" you mean the first few seconds, that would be correct  wind resistance wouldn't be a factor right away.
But there are other forms of resistance that are probably not directly proportional to weight either. The rolling friction of the tires will vary with weight, but probably not linearly and viscous friction in the bearings of the wheels probably won't vary that much with the weight of the rider. In addition, the moment of inertia of the wheels (resistance to angular acceleration) will not be affected by the weight of the rider. 


#5
Jun2709, 06:24 PM

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What's wrong with this reasoning: The cyclist begins moving forward because the wheels begin to rotate. The wheels rotate because the rider's weight translates to a torque on the wheels. The heavier the rider the greater the torque, therefore the greater the acceleration. So even at very slow speeds the heavier cyclist should have a greater acceleration.



#6
Jun2709, 06:35 PM

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The air drag force for a cyclist is proportional to the velocity squared (not Stoke's Law drag), to the frontal area (of the cyclist), and the drag coefficient. But the mass of the cyclist scales roughly as the frontal area to the 3/2 power. So a heavier cyclist will have less air drag, pound for pound. Other than this, and the rotational inertia of the wheels, both a light and heavy cyclist, if coasting, will accelerate at the same rate, dv/dt =g sin(theta).



#7
Jun2709, 10:01 PM

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#8
Jun2709, 10:44 PM

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What doesn't change is the angular kinetic energy of the wheels, which reduces the rate of acceleration. If all of the mass of the bicycle were in the rims and tires, then the rate of acceleration down the inclined plane with the tires rolling would be 1/2 that if the inclined plane were frictionless and no angular energy was added to the wheels. As the ratio of mass of the bicycle and rider increases compared to the weight of the rolling rims and tires, the smaller the overall impact of the angular energy in the wheels versus the linear kinetic energy of the bicycle and rider, and the faster the rate of acceleration. This is ingoring factors such as the tires would deform more under a heavier load, which would create more rolling resistance.



#9
Jun2809, 12:18 AM

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#10
Jul1509, 03:31 PM

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They will both accelerate at the same speed, assuming they both have the same coefficient of drag and frontal area. As previously stated: everything falls at the same rate.



#11
Jul1509, 05:41 PM

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