A bicyclist makes a circular turn with a radius R=120 m at a speed of v=18 m/s.

Question

A bicyclist makes a circular turn with a radius R=120 m at a speed of v=18 m/s. See the drawing below. The combined mass of the bicycle and the cyclist is m= 92 kg. Draw a free body diagram for the system (bicycle plus bicyclist). Calculate the magnitude of the normal force F_N exerted by the road on the bicycle plus bicyclist system. Find the magnitude of the frictional force f_s exerted by the road on the bicycle plus bicyclist system. The coefficient of static friction between the bicycle tires and the road is = 0.35. Determine the maximum value of the static friction that the road can exert on the bicycle tires without making the bicycle plus cyclist slide out of the curve. Will the bicyclist be able to take the turn without sliding out of the curve? Please explain your reasoning. of rotation.

Explanation / Answer

a) free body diagram consists of weight down, centrifeugal force towards right and friction towards left from the bicyclist

b) Normal force N = weight of both man and bicycle

N = 92*9.8 = 901.6 N

c) if he doesnot slide then

centrifeugal force = friction

f = mv^2/r

f = 92*18^2/120 = 248.4 N

d) us = 0.35

static friction = us*N

fs = 0.35*901.6 = 315.56 N

e) He will not slide, for the bicycle to slide the centrifeugal force should be greater than the static friction. so the bicycle will not slide.

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