Question

A proud new Jaguar owner drives her car at a speed of 35 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic). What is the minimum radius of curvature for the corner in order for the car not to skid

Answer 1

Answer:

178.6 m

Explanation:

Since the car moves in a circular path, it experiences a centripetal force, F = mv²/r where m = mass of car, v = speed of car = 35 m/s and r = radius of curvature of path.

Now, for the car not to skid, this centripetal force must be equal to the frictional force, F' acting in the opposite direction.

So, F' = μN where μ = coefficient of static friction(since the car does not move in this direction) and N = normal force = mg where m = mass of car and g = acceleration due to gravity = 9.8m/s²

F' = μmg

Since F = F'

mv²/r = μmg

dividing both sides by m, we have

v²/r = μg

multiplying both sides by r, we have

v² = μgr

dividing both sides by μg, we have

r = v²/μg

Here we use μ = coefficient of static friction(since the car does not move in this direction) = 0.70. Substituting the other variables into the equation, we have

r = v²/μg

r = (35 m/s)²/(0.70 × 9.8m/s²)  

r = 1225 m²/s²/6.86m/s²)  

r = 178.6 m

So, the minimum radius of curvature of the corner is 178.6 m