Circular Motion - Roadway Bank Angle

[JeffNYC]

Homework Statement

A) 3000 pound car is negotiating a circular interchange or radius 300ft at 30mph. Assuming the road is level, find the force between the tires and the road such that the car stays on the circular path without skidding.

B) Next, Find the angle at which the roadway should be banked so that no lateral force is exerted on the tires of the automobile.

Homework Equations

F = ma
m = 3000/32
tanθ = v^2 / rg
θ = tan-1(v^2 / rg ) .
v = velocity of the vehicle = 30
r = radius of banking = 300
g = accleration due top gravity

The Attempt at a Solution

I really don't know - this is a question from a calculus course and is one of the more challenging problems at the chapter's end. Having never taken even introductory physics, if anyone could help me out with the calculations + brief explanation that would be wonderful.

Jeff

 

[LowlyPion]

Homework Statement

A) 3000 pound car is negotiating a circular interchange or radius 300ft at 30mph. Assuming the road is level, find the force between the tires and the road such that the car stays on the circular path without skidding.

B) Next, Find the angle at which the roadway should be banked so that no lateral force is exerted on the tires of the automobile.

Homework Equations

F = ma
m = 3000/32
tanθ = v^2 / rg
θ = tan-1(v^2 / rg ) .
v = velocity of the vehicle = 30
r = radius of banking = 300
g = accleration due top gravity

The Attempt at a Solution

I really don't know - this is a question from a calculus course and is one of the more challenging problems at the chapter's end. Having never taken even introductory physics, if anyone could help me out with the calculations + brief explanation that would be wonderful.

Jeff

The first question is asking you what the centripetal acceleration on the car is. You converted your weight to mass, but you need to convert your speed to seconds to make this calculation correctly. (Once you have the acceleration you use the mass to determine the force.

The second part is asking where the lateral forces on the tires (similar to what you just calculated but taking the angle into account) will balance with the inward component of weight down the incline. Since they are supposed to balance you need not be concerned with friction.

 

[JeffNYC]

So,

1 mph = 0.44704 meters per second
30mph = 13.4112 meters per second

Centripetal Acceleration = (13.4112`^2)/300

= 0.599534*mass = .5999534*93.75 = 56.2063

So I have my acceleration now.

So, θ = tan-1(v^2 / rg )

= arctan(179.86/300g)

what do I use for "g"?

 

[LowlyPion]

So,

1 mph = 0.44704 meters per second
30mph = 13.4112 meters per second

Centripetal Acceleration = (13.4112`^2)/300

= 0.599534*mass = .5999534*93.75 = 56.2063

So I have my acceleration now.

So, θ = tan-1(v^2 / rg )

= arctan(179.86/300g)

what do I use for "g"?

If you are using meters per second then 9.8m/s² is the acceleration you need.

But you also need to convert 300 feet to meters too.