Doc Al said:
If you model the bike as having a wheel of mass m, with all its mass on its rim, then you should be able to show that the speed of the bike is independent of the mass of the rider. (Thus they reach the bottom simultaneously.)
I may do something wrong, check my derivation, please.
m is the mass of the two wheels. M is the mass of the rider and the other parts of the bike. R is the radius of the wheels, I is the moment of inertia of the two wheels. If all the mass of the wheels is concentrated on the rims then I=mR^2. If the wheels do not slip the angular velocity of the wheels multiplied by R is the same "v" as the velocity of the centre of mass of the whole bike and rider together.
The bike starts from rest at the top of the slope. The angle of incline is \alpha.
According to conservation of mechanical energy, the sum of change of KE after traveling a distance s along the incline and the change of PE equals zero.
\frac{1}{2}[(m+M)+I/R^2]v^2 = (m+M)gs\sin(\alpha)
v=\sqrt{ \frac{2(m+M)gs\sin(\alpha)}{m+M+I/R^2}} = ds/dt
We get s(t) by integration:
s=\frac{(m+M)g\sin(\alpha)}{2(m+M+I/R^2)}t^2
If all the mass m is on the rims then I/R^2=m and
s=\frac{(m+M)g\sin(\alpha)}{2(m+M+m)}t^2=\frac{g\sin(\alpha)}{2(1+\frac{m}{m+M})}t^2
If m is comparable to M the time needed to rich the bottom of the slope does depend on the mass of the rider.
ehild
