# Car turning on a bank (circular motion problem)

• theneedtoknow
In summary, the problem lacks sufficient information to calculate the radius of curvature and the coefficient of static friction. The equations provided by the individual in the conversation are valid, but they cannot be solved without additional information. Additionally, the assumption that the force of static friction is equal to the coefficient of static friction times the normal force is not necessarily justified. The direction of the force of static friction is also not necessarily down slope, as it depends on the speed of the car.
theneedtoknow

## Homework Statement

A racecar is turning on a bank inclined at 15%. It's speed is 225 km/h and it's mass is 1450kg.
1. Calculate the radius of curvature
2. Calculate the centripetal acceleration.
3. If the car mantains circular traction, what is the magnitude of the force of static friction
4. What is the coefficient of static friction.

## The Attempt at a Solution

Is it just me or is there too many unknowns to be able to solve this question? (The answers are supposed to be actual numbers, not expressions in terms of unknowns)

So , I will have the force of gravity pointing straight down, I will have the Normal Force pointing perpendicular to the bank, and I will have the static friction force (fs) pointing parallel to the bank towards the centre.

So my Fy's are supposed to balance so we have:
Ny - fy - mg = 0
Ncosθ - μNsinθ - mg = 0
N (cosθ - μsinθ) = mg

And for Fx we have
fx + Nx = mv^2/R
μNcosθ + Nsinθ = mv^2/R

N(μcosθ + sinθ) = mv^2/R
plug in the N i got from the Fy equations:

g(μcosθ + sinθ) / (cosθ - μsinθ) = v^2/R

Now, I'm somehow supposed to provide the radius of curvature, and μ, (centripetl acceleration and magnitude of friction will then just come from those 2), but this is a single equation with 2 unknowns, how can i possibly be able to figure out both?

I would agree with you that more information is needed to solve this problem.

In addition, I should point out that you are not justified in saying that the force of static friction is equal to the coefficient of static friction times the normal force. This is the case only if the two surfaces in contact are on the verge of sliding past each other. There is no language in the problem saying that this is the case.

Finally, the direction of the force of static friction is not necessarily down slope. If the speed is low enough (lower than what is necessary for the car to make it round the curve on a friction-free road) then static friction up slope is needed to maintain the car on track.

## 1. How does a car turn on a bank?

When a car turns on a bank, it is essentially moving in a circular motion. The car's tires exert a force on the road, causing the car to accelerate towards the center of the turn. This centripetal force allows the car to maintain its circular motion. Additionally, the car's inertia causes it to resist any changes in its direction, helping it to stay on the curved path.

## 2. How does the angle of the bank affect a car's turning?

The angle of the bank, also known as the banking angle or incline angle, plays a crucial role in a car's turning ability. The steeper the bank, the greater the centripetal force acting on the car. This means that the car can turn at a higher speed without slipping or skidding. However, if the bank angle is too steep, it can also cause the car to roll over, so it must be carefully designed and maintained.

## 3. What is the role of friction in a car turning on a bank?

Friction between the car's tires and the road is necessary for a car to turn on a bank. The frictional force between the tires and the road provides the centripetal force needed to keep the car on its curved path. Without enough friction, the car would slip or skid, making it difficult to maintain control and stay on the road.

## 4. How does the speed of the car affect its turning on a bank?

The speed of the car plays a significant role in its turning ability on a bank. As the car's speed increases, the centripetal force needed to maintain its circular motion also increases. Therefore, a faster-moving car will require a steeper bank angle to prevent it from slipping or skidding. On the other hand, if the car is moving too slowly, it may not have enough centripetal force to stay on the bank and may roll off the turn.

## 5. How do different factors, such as weight and size, impact a car's turning on a bank?

Different factors, such as weight and size, can impact a car's turning ability on a bank. Generally, a heavier car will require a higher centripetal force to turn at the same speed as a lighter car. Furthermore, a larger car will have a larger turning radius, meaning it will require a wider banked turn to maintain its circular motion. These factors must be taken into account when designing and driving on banked turns.

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