Question

Problem: Two bicycles roll down a hill which is 20 m high. Both bicycles have a total mass of 12 kg and 700 mm diameter wheels (r = 0.350 m). The first bicycle has wheels with a mass of 0.60 kg each, and the second bicycle has wheels with a mass of 0.30 kg each. Neglecting air resistance, which bicycle has the faster speed at the bottom of the hill? (Consider the wheels to be thin hoops).

Answer 1

The only friction is static friction, so there are no non-conservative forces. (Static
friction involves no motion and since work is defined as W = Fd, when there is no
distance involved, there is no work/energy used).
Mechanical Energy is Conserved, so:
ΣEi = ΣEf
KEi +PEi = KEf +PEf
(1/2)mvi^2 + (1/2)Iωi^2 + mghi = (1/2)mvf^2 + 2*(1/2)Iωf^2 + mghf
mghi = (1/2)mvf^2 + 2(1/2)Iωf^2 (The two for two wheels).
Now w is given by ω=v/r where v is the velocity of the rim of the wheel and is the
same as the velocity of the bike. I is given by Mr2 where M is the mass of the tire.
mgh = (1/2)mvf^2 + Mr^2vf^2/r2= (1/2)mvf^2 + Mvf^2
vf=sqrt(mgh/((1/2)*m+M))
vf=sqrt((12*9.8*20)/(0.5*12+0.3))
So for the first bike vf = 19.3 m/s
For the second bike, the wheels have a mass of 0.6 kg, and we get
vf = 18.9 m/s
so, first bike will reach the bottom first.
This is because of More of the potential energy goes into rotational kinetic energy
and less into translational kinetic energy when the wheel has more mass