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
