Friction between tires and pavement supplies the centripetal acceleration necessary for a car to turn. Calculate the maximum speed at which a car can round a turn of radius 30m (a) when the road is dry; (b) when the road is wet. (μk = 0.7 - rubber on dry concrete; μk = 0.5 - rubber on wet concrete)
To calculate the maximum speed at which a car can round a turn, we will use the concept of centripetal force and apply Newton's laws of motion. The centripetal force required to keep an object moving in a circular path is given by the formula:
F = m * v^2 / r
Where:
F is the centripetal force
m is the mass of the car
v is the velocity of the car
r is the radius of the turn
In this case, the friction between the tires and the pavement supplies the centripetal force. The maximum friction force is given by:
Ff = μ * m * g
Where:
Ff is the maximum friction force
μ is the coefficient of friction between the tires and the pavement
m is the mass of the car
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the maximum friction force provides the centripetal force, we can set Ff equal to F:
Ff = F
This gives us:
μ * m * g = m * v^2 / r
Now we can solve for the velocity v:
v^2 = μ * g * r
v = sqrt(μ * g * r)
For the road being dry (a), with a coefficient of friction μk = 0.7:
v(a) = sqrt(0.7 * 9.8 * 30) = 17.48 m/s
For the road being wet (b), with a coefficient of friction μk = 0.5:
v(b) = sqrt(0.5 * 9.8 * 30) = 13.23 m/s
Therefore, the maximum speed at which the car can round the turn is 17.48 m/s when the road is dry and 13.23 m/s when the road is wet.