Soapbox Racer Physics

The force F acting to move a soapbox cartie down a slope of angle θ is m g sin(θ), where m is the mass of the cartie and g is the acceleration due to gravity.

In order to accelerate the cartie, this force has to overcome drag, rolling friction and wheel inertia, and whatever is left over accelerates the cartie. So;

m g sin(θ) = Ft + Fr + Fd + FR

Where;

Ft = Translational Force (i.e. moving the whole machine down the hill)
Fw = Force rotating the wheels
Fd = Aerodynamic drag force
FR = Rolling friction

Aerodynamic drag F_d is given by;

F_d = -(1/2)ρC_dAv²   (see http://en.wikipedia.org/wiki/Drag_(physics) )

Rolling friction is linear in both velocity and vehicle mass, and has a C_R (rolling resistance coefficient) of about 0.0055. It is proportional to the force (Newtons) normal to the track. See http://en.wikipedia.org/wiki/Rolling_resistance. For all four wheels, we get;

F_R = -C_Rmg cos(θ)

Ft is pretty straightforward - for a cartie of mass m accelerating at a;

Ft = m a

The Fw term is a little more fiddly; the angular acceleration ω of the wheel with moment of inertia I is caused by the torque T

T = I ω

I for a hoop of mass mw is mw r2, and ω is given by a / r, so the above equation becomes;

T = mw r2 (a / r)
T = mw r a

T also equals Fw r, so

Fw r = mw r a
Fw = mw a

i.e. The radius of the wheel has no effect - the extra inertia of the larger wheel is cancelled out by the larger torque applied to rotate it.

Substituting these two terms back into the initial equation we now have;

m g sin(θ) = m a + mw a + Fd + FR
m g sin(θ) = (m + mw) a + Fd + FR

which can be rearranged for a as;

a = (m g sin(θ) - Fd - FR) / (m + mw)

Or, in full (substituting the expressions for Fd and FR

a = (m g sin(θ) - (1/2)ρACdv2 - CRmg cos(θ)) / (m + mw)