Consider a wet roadway banked as in the following figure, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between the tires and the roadway.
The radius of the curve is R = 50 m. (a) If the banking angle is 25o, what is the maximum speed the car can have before sliding up the banking?
(b) What is the minimum speed the car can have before sliding down the banking?
To find the maximum speed the car can have before sliding up the banking, we need to consider the forces acting on the car. Let's break it down step by step:
Step 1: Determine the normal force on the car
The normal force acts perpendicular to the surface of the roadway and helps to balance the car's weight. In this case, the normal force is split into two components: one perpendicular to the surface (N⊥) and one parallel to the surface (N║).
Step 2: Determine the frictional force
The frictional force acts parallel to the surface of the roadway and opposes the motion of the car. In the case of a wet roadway, we need to consider both the coefficient of static friction (μs) and the coefficient of kinetic friction (μk).
Step 3: Equate the forces
To determine the maximum speed the car can have before sliding up the banking, we need to equate the centrifugal force with the static frictional force.
Now let's calculate the maximum speed the car can have before sliding up the banking (part a):
Step 1: Determine the normal force on the car
N⊥ = mgcosθ
where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the banking angle (θ) is 25°, we can calculate the normal force.
N⊥ = mgcos25°
Step 2: Determine the frictional force
The maximum static frictional force (fstatic) is given by:
fstatic = μsN⊥
Given that the coefficient of static friction (μs) is 0.30, we can calculate the maximum static frictional force.
fstatic = 0.30 × N⊥
Step 3: Equate the forces
The centrifugal force (Fcentrifugal) is given by:
Fcentrifugal = mRω²
where R is the radius of the curve and ω is the angular velocity of the car.
The maximum speed the car can have before sliding up the banking occurs when the centrifugal force is equal to the maximum static frictional force.
fstatic = Fcentrifugal
0.30 × N⊥ = mRω²
We can simplify this further by substituting the value of N⊥:
0.30 × mgcos25° = mRω²
Now we can solve for ω (angular velocity). We know that ω = v/R (where v is the linear velocity of the car), so substitute this into the equation:
0.30 × mgcos25° = mR(v/R)²
Simplify further:
0.30gcos25° = v²
Finally, solve for v (linear velocity):
v = √(0.30gcos25°)
Plug in the values of g and cos25°, and calculate the square root to find the maximum speed the car can have before sliding up the banking.
For part b (the minimum speed the car can have before sliding down the banking), the process is similar. The only difference is that instead of equating the forces, we compare the gravitational force component parallel to the surface of the roadway to the kinetic frictional force.
Kinetic frictional force (fkinetic) = μkN║
The minimum speed the car can have before sliding down the banking occurs when the gravitational force component parallel to the surface of the roadway equals the kinetic frictional force.
mgsinθ = μkN║
Substitute N║:
mgsinθ = μkN⊥sinθ
Simplify:
mg = μkgcosθ
Now we can solve for v (linear velocity):
v = √(μkgcosθ)
Plug in the values of μk, g, and cosθ, and calculate the square root to find the minimum speed the car can have before sliding down the banking.