Banked Frictionless Curve, and Flat Curve with Friction

Part A = r = 65.4 m

Part B = μ_{min} = 0.364

Solutions Below:

^{2}throughout this problem.

Part A

Express your answer in meters.

This problem requires some algebraic manipulation to solve. Since the radius is part of the formula for centripetal acceleration (a_{c} = v^{2}/r), we can solve for r using that formula. You should have learned in your class about dimensional translation/rotation and how mg = F_{N}cosθ and F_{N}sinθ equals centripetal force(ma_{c}). Since we know mass and g, if we can relate the above expressions, we can find r. Solve like this:

F_{N}sinθ = ma_{c}

Since a_{c} = v^{2}/r:

F_{N}sinθ = mv^{2}/r:

And we know (from the explanation above) that F_{N}cosθ = mg:

If we divide the first formula (for F_{N}sinθ) by the cos(for F_{N}cosθ), we end up with:

((F_{N}sinθ) / (F_{N}cosθ)) = (mv^{2}/r) / (mg)

Which gives us:

tanθ = (mv^{2}/r) / (mg)

simplified to:

tanθ = v^{2} / (rg)

We get the tanθ because the F_{N}‘s on the left cancel and sin/cos = tan. So to solve now:

First, find velocity in terms of m/s:

55km/h = 55000m/h = 15.277m/s

Next use the formula we solved for:

tanθ = v^{2} / (rg)

tan(20) = 15.277^{2}/r) / (9.8r)

0.364 = (23.81/r)

r = 65.4m

Part B

_{min}, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car’s speed is still 55.0 km/hour and that the radius of the curve is given by the value you found for r in Part A.

In order for the car to not slide around, there needs to be centripetal force. Since there isn’t a slope, gravity won’t cause the car to slide down towards the center of the circle (F_{N}sinθ). So the centripetal force will be coming entirely from the frictional force. Centripetal force is given by: F_{c} = ma_{c} and frictional force is given by F_{fr} = F_{N}μ.

Centripetal acceleration is given by: a_{c} = v^{2}/r

From Part A we can plug in values

a_{c} = 15.277^{2}/65.4

a_{c} = 3.57m/s^{2}

Now equate friction with centripetal force:

F_{fr} = F_{c}

F_{N}μ = ma_{c}

mgμ = ma_{c}

gμ = a_{c}

9.8μ = 3.57

μ = 0.364

I was wondering why you can equate F_fr to F_c in part B?

There needs to be some force acting on the car so that it stays in the circle instead of continuing straight ahead. If the car was on ice, or some other very smooth surface, there wouldn’t be any friction, and the car would just keep going straight and there wouldn’t be any centripetal force at all. In other words, the frictional force is by definition the same as centripetal force in this part of the problem, because it’s what keeps the car going around in a circle. Kind of how the gravitational force in Part A is related to the centripetal force.

Remember that centripetal force is just mv^2 / r. If something is moving around in a circle, then there is a centripetal force. The centripetal force is going to have to be equal to whatever is keeping that object moving in a circle in the first place.