What is the maximum speed with which a 1200-kg car can round a turn of radius 68 m on a flat road if the coefficient of static friction between tires and road is 0.80?
To determine the maximum speed at which a car can round a turn on a flat road, we need to consider the friction between the tires and the road. The maximum speed occurs when the centripetal force required to keep the car moving in a circular path is equal to the maximum static friction force that the tires can provide.
The centripetal force is given by the formula:
F = m * v^2 / r
Where:
F is the centripetal force,
m is the mass of the car (1200 kg),
v is the velocity of the car, and
r is the radius of the turn (68 m).
The maximum static friction force is given by:
f_max = μ * N
Where:
f_max is the maximum static friction force,
μ is the coefficient of static friction (0.80), and
N is the normal force between the car and the road.
The normal force (N) is equal to the weight of the car, which is given by:
N = m * g
Where:
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the expression for N, we can rewrite the maximum static friction force equation as:
f_max = μ * m * g
Now, we have two equations:
F = m * v^2 / r (equation 1),
f_max = μ * m * g (equation 2).
At the maximum speed, the centripetal force (F) is equal to the maximum static friction force (f_max), so we can set equation 1 equal to equation 2:
m * v^2 / r = μ * m * g
Simplifying and solving for v:
v^2 = μ * r * g
v = sqrt(μ * r * g)
Now we can substitute the given values to find the maximum speed:
v = sqrt(0.80 * 68 * 9.8)
Using a calculator, we can evaluate the expression:
v ≈ 24.04 m/s
Therefore, the maximum speed at which the 1200-kg car can round the turn is approximately 24.04 m/s.