Physics Question?

Question

A circular curve of highway is designed for traffic moving at 60 km/hr on a stormy day.What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

the radius of the curve is 200m. traffic is moving along the highway at 40km/hr on a stormy

Answer 1

Force of friction = mu mg
= Centrifugal force = mv^2 / r

So mu = v^2 / gr

They give you v. You know g. They are going to need to give you the radius of the turn or something like that....

Answer 2

The car will not slide when C = mv^2/r = kN = kmg cos(theta) = F; where C is centrifugal/centripetal force of a car of mass m going v = 60 kph around a curve with r radius of turn. k is the coefficient of friction, which is what you are looking for. g = 9.81 m/sec^2 on Earth's surface and theta is the angle of the curve embankment wrt to the horizon (theta = 0 is assumed).

C = F means C - F = 0 = ma = f; so that the net force actining along the radius of turn is 0 and the radial acceleration is also a = 0. In other words, the car is neither skidding outward from the center of turn nor sliding inward toward that center.

Solve for k = v^2/gr from the earlier equations. As you failed to specity the turn radius of the curve, we cannot solve for k If you have that available, you can do the math since v and g are already known.

Note the equation to interpret what physics it implies. It says that sharper turns require higher k values to avoid skidding off the road. It also says that doubling the velocity in the turn will increase that minimum k four times. In other words, doubling the kinetic energy (1/2 mv^2) of a turn makes it four times tougher to not skid out of control. This is why we slow down in sharp turns.

Answer 3

It really depends on how fast the car is going and how much banking they have put on the curve, if they designed it for 60 kph then they should have sufficient amount of camber that you don't need any friction to negotiate the curve if you are doing exactly 60 kph, if the car is going faster it will tend to slip off the outside of the curve and slower will slip off the inside - best to have as much friction as you can get though, particularly if anyone needs to brake.

Answer 4

Frictional force between the tires and road will balance the centrifugal force on car (which will try to push the car out of circular path).

frictional force = mu * mg (where mu is coefficient of friction)

centrifugal force = mv^2/R

so mu * mg >= mv^2/R

so minimum value of mu >= v^2/(Rg)
v = 60 kmph = 16.67 m/s
=> mu >= 16.67^2 /(R * 9.81)

min mu = 0.14