Road Bank Top Speed Calculation | G & \mu

[dlingo]

Advanced Road Bank question

Homework Statement

There is a road bank designed for 60km/hr (16.7 m/s) with a radius of 70m and no friction. What is the top speed with a mew of 0.8$$\mu$$ ?
G=gravitational constant
V=16.7m/s

Homework Equations

Tan$$\theta$$=V^2/(R*G)
Fc=(M*V^2)/R

The Attempt at a Solution

First I calculated the angle at which the car won't move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)) The component of gravity added to a component of Centripetal force.
then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.

 

[gneill]

The Attempt at a Solution

First I calculated the angle at which the car won't move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)) The component of gravity added to a component of Centripetal force.

You might want to recheck where the angle is. I suspect you want $$M g cos(\theta) + F_c sin(\theta)$$

then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.

See if you can find the components of -centripetal force (well, I suppose we could call it centrifugal force if we whisper softly) and gravitational force that lie along the road surface.

 

[dlingo]

There is one thing I don't understand, Is the Fc parallel to the ground or the bank?

 

[gneill]

There is one thing I don't understand, Is the Fc parallel to the ground or the bank?

Fc is in the same direction as the radius vector for the moving object describing the circle. In this case, it's parallel to the ground.

 

[rwisz]

Thank you for sharing your work and thought process in solving this advanced road bank question. It appears that you are on the right track in your approach to solving this problem. However, I would like to offer some suggestions and clarifications to help you reach the correct solution.

First, let's define some variables to make the solution clearer. Let V be the velocity of the car, R be the radius of the road bank, and μ be the coefficient of friction. We can also define the angle of the road bank as θ.

Next, let's review the equations you have used. The equation Tanθ = V^2/(R*G) is correct, and it represents the relationship between the angle of the road bank, velocity, radius, and gravitational constant. This equation can also be rewritten as θ = arctan(V^2/(R*G)).

The equation Fc = (M*V^2)/R is also correct and represents the centripetal force required for circular motion. However, the equation Fc = Ffriction + the component of Fnormal pushing back against Fc is not necessary in this problem and may lead to confusion.

Instead, we can use the equation Ffriction = μ*Fnormal, where Fnormal is the normal force acting on the car due to the road bank. This normal force is equal to the sum of the component of gravity acting down the slope (G*sinθ) and the component of centripetal force acting perpendicular to the road bank (Fc*cosθ).

Putting all of this together, we can set up the following equations:

Tanθ = V^2/(R*G)
θ = arctan(V^2/(R*G))
Fc = (M*V^2)/R
Ffriction = μ*(G*sinθ + Fc*cosθ)

We can solve for the unknown variable, V, by substituting the second equation into the first equation and then solving for V. This will give us the maximum velocity of the car on the road bank.

I hope this helps clarify the solution to this problem. Keep up the good work in your studies!