Question

A car is safely negotiating an unbanked circular turn at a speed of 25 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

Answer 1

Answer:

The velocity must be reduced to one third to stay on the road

Explanation:

The sideways force that friction must resist comes from the centrifugal acceleration due to the turn.

fc=mv2Rfc=mv2R

the frictional force is given by

ff=μmgff=μmg where μμ is the static friction coefficient

if the car is not to skid

fc≤fffc≤ff so    

mv2R≤μmgmv2R≤μmg

v≤μgR−−−−√v≤μgR

thus vv varies as the square root of μμ

so if μμ is reduced by 9, vv must be reduced by 9–√=39=3

and thus the speed must be reduced to 26 m/s

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