What is the maximum speed that the car can turn in a curve of radius 29.0 m without slipping? The coefficient of static friction between a road and a car's tires is 0.500, and the coefficient of kinetic friction is 0.390. Assume that the road is horizontal and flat.
To find the maximum speed at which the car can turn without slipping in a curve, we need to analyze the forces acting on the car.
When the car is turning, two forces are relevant: the normal force (N) acting perpendicular to the road, and the frictional force (Ff) acting parallel to the road. The maximum static frictional force is given by:
Ff(max) = μs * N
where μs is the coefficient of static friction.
Since the car is turning without slipping, the maximum static frictional force is equal to the centripetal force required to keep the car moving in a circle of radius r at a speed v:
Ff(max) = m * v^2 / r
where m is the mass of the car.
Setting these two equations equal to each other, we have:
μs * N = m * v^2 / r
Now, we need to find an expression for N. Since the road is flat and horizontal, N is equal to the weight of the car:
N = m * g
where g is the acceleration due to gravity.
Substituting this expression into the equation, we get:
μs * m * g = m * v^2 / r
Simplifying the equation, we find:
v^2 = μs * g * r
Finally, to find the maximum speed (v) without slipping, we can take the square root of both sides:
v = √(μs * g * r)
Now we can substitute the given values into the equation to find the maximum speed:
μs = 0.500 (coefficient of static friction)
g = 9.8 m/s^2 (acceleration due to gravity)
r = 29.0 m (radius of the curve)
Plugging these values into the equation:
v = √(0.500 * 9.8 * 29.0)
Calculating this expression, we find:
v ≈ 18.7 m/s
Therefore, the maximum speed at which the car can turn without slipping in a curve of radius 29.0 m is approximately 18.7 m/s.