Answer:
r=20.6m
Explanation:
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29 km/h and the μs between tires and track is 0.32?
the acceleration experienced by the bicycle will be the centripetal acceleration
which is given by a=v^2/r
v=velocity of the bicyclist 29km/h =8.05m/s
the centripetal force is given as
f=mv^2/r
note the maximum value of the static frictional force
fmax=∪N
U=the coefficient of static friction
N=mg=normal reaction of the bicycle which is equal to the the weight of the bicycle
equating fmax=fcentripetal
mv^2/r=Umg
r=v^2/Ug
r=8.05^2/(.32*9.81)
r=20.6m , is the smallest radius of an unbanked track
