# Road Bank Top Speed Calculation | G & \mu

• dlingo
In summary, the conversation discusses a homework problem involving a road bank designed for a certain speed and radius with no friction. The question is to find the top speed with a given coefficient of friction. The conversation includes the equations and attempts at a solution, with a focus on finding the components of centripetal force and gravitational force along the road surface.
dlingo

## 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.

Last edited:

dlingo said:

## 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.

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

dlingo said:
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.

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!

## 1. How is Road Bank Top Speed calculated?

Road Bank Top Speed is calculated by taking into account the coefficient of friction (μ) of the road surface and the gravitational force (G) acting on the vehicle. The formula for calculating top speed on a banked road is v = √(GμR), where v is the top speed, R is the radius of the curve, G is the gravitational force, and μ is the coefficient of friction.

## 2. What is the significance of the coefficient of friction (μ) in Road Bank Top Speed calculation?

The coefficient of friction represents the amount of grip or traction between the tires of the vehicle and the road surface. A higher μ means there is more grip and the vehicle can maintain a higher speed on a banked road without slipping. A lower μ means there is less grip and the vehicle will have a lower top speed on a banked road.

## 3. How does the radius of the curve (R) affect Road Bank Top Speed?

The radius of the curve plays a critical role in determining the top speed on a banked road. A larger radius means the curve is more gradual, allowing the vehicle to maintain a higher speed. A smaller radius means the curve is sharper, requiring the vehicle to slow down to prevent slipping.

## 4. Is it safe to drive at the calculated Road Bank Top Speed?

The calculated top speed on a banked road is based on ideal conditions and assumes the driver is experienced and the vehicle is in good condition. It is important to always follow posted speed limits and drive at a safe and comfortable speed for the road conditions.

## 5. How can Road Bank Top Speed calculation be useful in real-world situations?

Road Bank Top Speed calculation can be useful for engineers and designers when planning and constructing roads and curves. It can also be used by race car drivers to determine the maximum speed they can safely maintain on a banked track. Additionally, understanding road bank top speed can help drivers make informed decisions about their speed and driving on banked roads.

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