Why do heavier people descend faster on a bike?

The FBD was a nice touch, but it doesn't include the most important variable here: drag. While both riders will have the same force per unit mass pulling them down the incline, they will not have the same force per unit mass opposing their descent in the form of wind resistance. Sure heavier people tend to have a larger frontal area to move through the wind, but not enough so to overcome their added mass.

And I guess the question has already been answered, but hey what's a letsrun question/answer thread without something being repeated a few times?

ah, drag

i guess thats why i still have to take AP physics next year

Wow, I thought letsrun people were smarter than this. We're already on the second page and nobody has talked about moment of inertia.

Let's take the case where there is no wind resitance and both the fat and skinny riders are riding on identical bikes without pedaling. The fat rider will descend faster than the skinny rider. The reason is that the wheels need to accelerate in order for the bike to move.

The wheels will have a finite mass and therefore a finite moment of inertia. The moment of inertia of a rotating object is analogous to the mass of a translating object. Just as a larger force will accelerate a mass more, a larger torque will cause a larger angular acceleration in a rotating object of a given moment of intertia.

The torque is simply the force (m*g*sin(theta)) multiplied by the distance from the center of the rotating object. Since the position of the bike and the riders do not change depending on how fat the the rider is, the only difference is the mass of the rider, which is larger for the fatter rider. Therefore, the fat rider will have a larger angular acceleration and thus a larger translation acceleration since the wheels will not slip on the road.

As for the wind resistance, I can't tell you who that would benefit. It's much more complicated.

cushen, you're almost right, but you forgot the fact that the wheels are not massless. The poster above me brought up rotational motion, and you have to include the energy of rotational motion in your equation. Therefore, the more accurate form of your equation is this:

mgh = 1/2m(v^2) + (2)(1/2)(i)(w^2) + Q

where i is the moment of inertia for the wheels (about equal to 1/2 times the mass times the radius squared), doubled because there are two wheels. w (should be omega) is the angular speed of the wheel, or v/r. Q is just a constant accounting for any losses due to friction or air resistance.

What the equation says is that the potential energy of the rider at the top of the hill is translated into both translational and rotational energy, the sum of which is not to exceed the original potential energy.

With this added term, you can take a look at the above poster's reasoning, which I haven't really read carefully, and put it all together.

Sorry, when I said that i is about 1/2m(r^2), I was wrong; it should be more like m(r^2), but it doesn't make a huge difference here.