Banked Turn: Navigating Car Dynamics on Curves

In summary, the problem involves finding the minimum and maximum speeds of a car entering a turn with radius ##R## and banked at angle ##\theta##, without skidding sideways. Using Newton's laws and drawing a free body diagram, the correct equations can be derived to solve for the minimum speed, which is given by ##\displaystyle v_{min} = \sqrt{\frac{gR(\sin \theta - \mu \cos \theta)}{\mu \sin \theta + \cos \theta}}##.
  • #1
Mr Davis 97
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44

Homework Statement


A car enters a turn whose radius is ##R##. The road is banked at angle ##\theta##, and the coefficient of friction between the wheels and the road is ##\mu_s##. Find the minimum and maximum speeds of the car to stay on the road without skidding sideways

Homework Equations


Polar coordinates
Newton's laws

The Attempt at a Solution



So I am confused about how to get started on this problem. I know that there are three forces acting on the car (friction, gravity, and normal force). But I am confused about how to put this into equations that involve the radius and the centripetal force. I have tried using polar coordinates, but since the vehicle is on a banked turn, the forces don't really align in any way. How should I get started? (Once I have the equations it seems that the problem is not too hard).
 
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  • #2
You don't need to use polar coordinates, Cartesian is easier.
Draw a free body diagram of the road cross section, it should look very similar to the block on a ramp problems you've done on the past.
 
  • #3
All I get is that ##F_{friction} = mg \sin \theta## and ##N = mg \cos \theta##. None of these relate to the velocity of the car, so I don't see how I can get a minimum or maximum.
 
  • #4
Mr Davis 97 said:
All I get is that ##F_{friction} = mg \sin \theta## and ##N = mg \cos \theta##. None of these relate to the velocity of the car, so I don't see how I can get a minimum or maximum.
Neither of those equations is correct.
What are the force components in the horizontal direction? What must the net force be in that direction?
 
  • #5
haruspex said:
Neither of those equations is correct.
What are the force components in the horizontal direction? What must the net force be in that direction?
So we're situating our coordinate system not perpendicular to the ramp but perpendicular to the ground?
 
  • #6
Mr Davis 97 said:
So we're situating our coordinate system not perpendicular to the ramp but perpendicular to the ground?
I feel that will be less confusing in this case because it relates directly to the known accelerations.
 
  • #7
But if I do that won't I end up with 2 equations with four forces, since each force has a component in each direction?
 
  • #8
Mr Davis 97 said:
But if I do that won't I end up with 2 equations with four forces, since each force has a component in each direction?
Have you drawn a free body diagram yet?

The math is very similar regardless of your choice of coordinates. You will need to find vector components either way.
Don't get hung up on which coordinates to choose.
Finding the correct answer to the questions in #4 is much more important.
 
  • #9
Mr Davis 97 said:
But if I do that won't I end up with 2 equations with four forces, since each force has a component in each direction?
There are three forces, one of which is known, so two unknowns. You know the angle, so even if you take horizontal and vertical components there are still only two unknowns.
I see billy sees no advantage in using horizontal and vertical axes. Maybe.
 
  • #10
haruspex said:
I see billy sees no advantage in using horizontal and vertical axes. Maybe.
No I do agree with your suggestion in #6, just trying to point out that either coordinates will work fine.
 
  • #11
So I do as you say haruspex, but I keep getting the answer ##\displaystyle v_{min} = \sqrt{\frac{gR(1 - \mu)}{\mu \sin \theta + \cos \theta}}##, which is not the correct answer. The correct answer is ##\displaystyle v_{min} = \sqrt{\frac{gR(\sin \theta - \mu \cos \theta)}{\mu \sin \theta + \cos \theta}}##. I cannot for the life of me figure out what I am doing wrong...
 
  • #12
Wait, nevermind. I figured out what I was doing wrong. I now have the right answer.
 
  • #13
Mr Davis 97 said:
Wait, nevermind. I figured out what I was doing wrong. I now have the right answer.
Well done.
 

Related to Banked Turn: Navigating Car Dynamics on Curves

1. What is a banked turn?

A banked turn refers to the curve or bend in a road or racetrack that is angled or sloped, allowing vehicles to navigate the turn at higher speeds while maintaining control.

2. How does a banked turn affect a car's dynamics?

A banked turn affects a car's dynamics by providing a centripetal force that helps keep the car on the path of the curve. This allows the car to maintain a higher speed without sliding or drifting off the road.

3. What is the purpose of a banked turn in a racetrack?

The purpose of a banked turn in a racetrack is to allow cars to maintain high speeds while cornering, ultimately reducing lap times. The angled surface of the turn helps cars generate more downforce, providing better grip and stability.

4. How do different factors affect a car's performance on a banked turn?

Factors such as speed, weight distribution, tire grip, and aerodynamics can all affect a car's performance on a banked turn. Higher speeds can generate more downforce, while a heavier car may require more centripetal force to navigate the turn. Additionally, better tire grip and aerodynamics can provide better traction and stability.

5. What is the difference between a positive and negative banked turn?

A positive banked turn is angled upwards, with the outer edge of the turn being higher than the inner edge. This is typically used for higher speeds. In contrast, a negative banked turn is angled downwards, with the inner edge being higher than the outer edge. This type of turn is used for lower speeds and can be more challenging to navigate.

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