Finding the angle of a car moving on a circular track

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Homework Help Overview

The problem involves determining the banking angle of a circular racetrack where a car travels at a specific speed without friction. The original poster expresses confusion about setting up the problem and relating the forces acting on the car to the incline of the track.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss drawing force diagrams and identifying forces acting on the car, including normal force and weight. There are attempts to apply Newton's second law and to split forces into components. Questions arise about the role of friction and the correct components of the forces in both the x and y directions.

Discussion Status

The discussion is ongoing, with participants providing guidance on setting up equations based on the forces identified. There is recognition of the need to clarify the components of the forces and how they relate to the acceleration of the car. Multiple interpretations of the force components are being explored.

Contextual Notes

Participants note the absence of friction in the scenario and the implications of this on the force analysis. There is also mention of the need for mass in some calculations, although it is suggested that it may not be necessary for the solution.

Singdasorrow
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Homework Statement


A circular automobile racetrack is banked at an angle X, to the horizontal, such that no friction between the road and tires are required when a car travels at 30.0 m/s. The radius of the track is 400m. What is the angle?

The problem i have is setting up the problem. I have drawn a force diagram for the car moving on the incline, but i am very confused relating it to the slope. The only thing i can think of is to take the components of the radius of the track, but in all honesty i am having issues doing this.

Tips?
 
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Singdasorrow said:
I have drawn a force diagram for the car moving on the incline,
What forces act on the car? In what directions?

Once you get the forces correct, apply Newton's 2nd law. Hint: In what direction is the car accelerating?
 
my force diagram has the normal force perpendicular to the incline, the weight perpendicular to the ground, and going in a forward motion with friction in the opposite. The acceleration though is towards the center of the circle I split it into components of X and Y.
Sum of the forces in the Y direction= 0,
sum of the forces in the x direction = (mass of the car x radial acceleration which was 2.25 m/s^2)- friction (which here is 0)

i feel like i am forgetting a force tho
 
Singdasorrow said:
my force diagram has the normal force perpendicular to the incline, the weight perpendicular to the ground, and going in a forward motion with friction in the opposite. The acceleration though is towards the center of the circle I split it into components of X and Y.
Good, except for the bit about friction--there's no friction here.

Sum of the forces in the Y direction= 0,
Good. So what are the force components in the y direction?
sum of the forces in the x direction = (mass of the car x radial acceleration which was 2.25 m/s^2)- friction (which here is 0)
Good. So what are the forces in the x direction?

You need to set up those two equations, putting in the forces. When you take the components, you'll get functions of the angle.
 
In the y direction we have Normal force and weight perpendicular to the group (weight times sinX)

In the X direction we have just radial acceleration of the car, i cannot think of anything else

I can relate the two because of neuton's second law that the sum of the forces = mass x acceleration. so N -wsinX=2.25. Here tho this feels wrong...
 
Singdasorrow said:
In the y direction we have Normal force and weight perpendicular to the group (weight times sinX)
This is a bit mixed up. The weight is vertical and thus is totally in the y-direction. But what's the y-component of the normal force?
In the X direction we have just radial acceleration of the car, i cannot think of anything else
That's the acceleration, but what about the force? Which of the two forces acting on the car has an x-component.

(Note: I am assuming that x and y refer to horizontal and vertical.)
 
the y component of the normal force then would be Normal force x sin(X)

I'm not quite sure i understand how to get the force from the acceleration without knowing the mass of the vehicle.

And yes ur note is correct
 
Singdasorrow said:
the y component of the normal force then would be Normal force x sin(X)
No, not exactly. You're mixing up your sine and cosine.

I'm not quite sure i understand how to get the force from the acceleration without knowing the mass of the vehicle.
Don't worry about the 'ma' part of the equation. First figure out ƩFx. (You'll find that you won't need the actual mass. Just call it 'm' and see what happens.)
 
ƩFx = 2.25 m/s^2 times mass = mass x acceleration. It simplifies to 2.25 = a
ƩFy = N (the normal force) x cos(X) - mass x gravity = mass times acceleration.

I plugged in the acceleration i found from ƩFx, but i couldn't simplify it enough to solve it
 
  • #10
Singdasorrow said:
ƩFx = 2.25 m/s^2 times mass = mass x acceleration. It simplifies to 2.25 = a
All you've done here is give the acceleration. You have not evaluated ƩFx. (You need to do something similar to what you did for ƩFy.)

ƩFy = N (the normal force) x cos(X) - mass x gravity = mass times acceleration.
Good. But what's the acceleration in the y direction?

I plugged in the acceleration i found from ƩFx, but i couldn't simplify it enough to solve it
Think like this. There are only two forces, the weight and the normal force. List the x and y components of each. To find ƩFx just add up the x-components; To find ƩFy just add up the y-components (you did that one).
 
  • #11
the only force that i know for ƩFx is the acceleration times the mass bc friction is 0

the acceleration in the y direction should b 0 because there is no movement up or down
 
  • #12
oooo i have the x component of the normal force. so Ncos(X) = 2.25 m/s^2
 
  • #13
woot thank you very much for your help! I've figured it out to the end <3
 

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