Maximum speed of a car on a banked road?

Question

A curve with a radius of 120 m is banked at and angle of 10 degrees. If a car with wheels with a coefficient of static lateral friction is .90 to round this curve without skidding, what is the maximum possible speed?

Answer 1

Radius r = 120 m

Angle of banking θ = 10 deg

Coefficient of friction µ = 0.90

Make free body diagram of the car.

The forces on the car are

1. Weight mg downward , where m = car's mass

2. Normal force Fn by the road perpendicular to the road

3. Friction f = µFn down the incline. At maximum speed, the car is just about to skid. Therefore, f equals µFn. At maximum speed, the car tends to skid up the incline. Therefore, the friction is down the incline.

There is no acceleration in vertical direction.

Therefore, total vertical force = 0

Vertical component of normal force = Fn * cosθ upward

Vertical component of friction = µFn*sinθ downward

Vertical component of weight = mg downward

Therefore, for total vertical force = 0,

Fn * cosθ - µFn*sinθ - mg = 0

Or Fn(cosθ - µsinθ) = mg

Or Fn = mg/(cosθ - µsinθ)----------------(1)

Total horizontal force toward the centre

F = Fn * sinθ + µFn * cosθ ---------------------(2)

Substituting the value of Fn from (1) into (2)

F = {mg/(cosθ - µsinθ)} * sinθ + µ * {mg/(cosθ - µsinθ)} * cosθ

F = {mg/(cosθ - µsinθ)} * (sinθ + µcosθ)

F = mg*(sinθ + µcosθ)/(cosθ - µsinθ)

Divide numerator and denominator on rhs by cosθ

F = mg*(tanθ + µ)/(1 - µtanθ)--------------------------(3)

F is centripetal force

Therefore, F = mv^2/r,------------------------(4)

where v is car's speed.

From (3) and (4)

mg*(tanθ + µ)/(1 - µtanθ) = mv^2/r

Divide by m

g*(tanθ + µ)/(1 - µtanθ) = v^2/r

gr*(tanθ + µ)/(1 - µtanθ) = v^2

v = sqrt[gr*(tanθ + µ)/(1 - µtanθ)] ---------------(5)

v = sqrt[9.8 * 120 * (tan 10 deg + 0.9)/(1 - 0.9 * tan 10 deg)]

v = sqrt[9.8 * 120 * (0.176 + 0.9)/(1 - 0.9 * 0.176)]

v = sqrt(9.8 * 120 * 1.076/0.8416)

v = sqrt(1503.536)

v = 38.8 m/s

Ans: 38.8 m/s

Note: Some people memorize the formula in equation (5) and use that formula directly to solve the problem fast. You can do so if you need to calculate answer quickly and if your teacher is OK with using that formula directly. But I wanted you to understand. That is why I gave the detailed answer.

Answer 2

The formula for "banking of curves" problem of this type is given by

v^2/rg = (tan A + mu)/[1 - mu(tan A)]

where

v = maximum velocity allowed

r = radius of curvature = 120 m (given)

A = angle of bank = 10 degrees (given)

mu = coefficient of friction = 0.90 (given)

Substituting appropriate values,

v^2/(120*9.8) = (tan 10 + 0.9)/(1 - 0.9*tan 10)

v^2 = 120*9.8*(tan 10 + 0.9)/(1 - 0.9*tan 10)

v^2 = 1504.5191

v = 38.79 m/sec.