Question

A car is safely negotiating an unbanked circular turn at a speed of 18 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 dirver slow the car?

Answer 1

so initially

u m g = mv^2/r

u = v0^2/( g r)

now

u/3 = v^2/( g r)

so

1/3 v0 ^2/(gr) = v^2

v = v0/sqrt(3) = 18/sqrt(3)= 10.4 m/s

Answer 2

Answer 3

The centripetal acceleration is provided by the force of friction

(mu) N = mv*v/R
mu is the coefficient of friction
N is the normal reaction,
R is the radius of the turn
v is the velocity
m is the mass of the car

So, v is proportional to the square root of the coefficient of friction. So, use when the mu becomes 1/3 times, the maximum velocity possible becomes 1/square root(3) times.

So, the velocity in this case = 18/Sqrt(3) = 10.39 m/s