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Acceleration of Coasting Vehicle Down Inclined Plane

by ellerbro
Tags: acceleration, coasting, inclined, plane, vehicle
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ellerbro
#1
Jun27-09, 03:45 PM
P: 4
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?
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ideasrule
#2
Jun27-09, 04:33 PM
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Quote Quote by ellerbro View Post
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?
I think the most important reason for the difference is air resistance. When a cyclist goes downhill, he accelerates until air resistance matches the force of gravity, at which point his speed stops increasing (the concept of terminal velocity). Since gravity exerts a stronger force on heavier bikes, terminal velocity is higher for heavier bikes; it takes more air resistance to resist gravity.
ellerbro
#3
Jun27-09, 05:15 PM
P: 4
So you don't think the heavier rider will have a greater acceleration from the start, assuming they both start with zero velocity?

russ_watters
#4
Jun27-09, 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.
ellerbro
#5
Jun27-09, 06:24 PM
P: 4
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.
Bob S
#6
Jun27-09, 06:35 PM
P: 4,663
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).
russ_watters
#7
Jun27-09, 10:01 PM
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Quote Quote by ellerbro View Post
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.
The force is due to his mass, but he resists acceleration due to his mass, so it cancels out. This is why all objects, when dropped, fall at the same speed regardless of their mass.
rcgldr
#8
Jun27-09, 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.
ellerbro
#9
Jun28-09, 12:18 AM
P: 4
Quote Quote by russ_watters View Post
The force is due to his mass, but he resists acceleration due to his mass, so it cancels out.
That definitely clears things up. Thanks for all your input folks.
gearhead
#10
Jul15-09, 03:31 PM
P: 31
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.
rcgldr
#11
Jul15-09, 05:41 PM
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Quote Quote by gearhead View Post
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.
Except that mass in the rims of a bicycle accelerate at half the rate of the non-rotating mass. The higher the ratio of non-rotating mass to rotating mass, the faster the rate of acceleration.


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