Test on monday (banked road problem)

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In summary, the conversation is about a problem involving a car driving on a banked, frictionless, circular track with a 31-degree angle. The goal is to find the maximum velocity that the car can drive without slipping down or sliding up the incline. The method for finding the velocity is to use centripetal acceleration, but there is confusion about whether the force of N (normal force) should be equal to mgcos(theta) or Ncos(theta) = mg. The person is seeking clarification before their upcoming test on Monday.
  • #1
jrmed13
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URGENT! test on monday! (banked road problem)

Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal acceleration by saying that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta)? My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.
 
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  • #2
hm...

Okay, so maybe I should make my question more clear... here goes...
The problem:

The Daytona International Speedway in Daytona Beach, FL, is famous for its races(blahblahblah)... Both of its courses feature four-story 31.0degree banked curves, with maximum radius of 316m. If the car negotiates the curve too slowly, it tends to slip down the incline of the turn, whereas if it's going too fast, it may begin to slide up the incline.
(a) Find the necessary centripetal acceleration on this banked curve so that the car won't slip up or slide down the incline.
(b) Calculate the speed of the race car.


My problem with the solution:

It says in the book to set Ncos(theta) = mg. However, I was wondering, why can't N be = to mgcos(theta)?? My understanding is that two forces on an object are equal when the object does not accelerate in either direction. The car isn't moving into the road or out of the road, so why doesn't N = mgcos(theta)?? Why does it HAVE to be Ncos(theta) = mg?

I understand that centripetal force is equal to Nsin(theta). And I know that I have to solve for N. My only problem is understanding why it has to be Ncos(theta)=mg and why it can't be N=mgcos(theta).

I have a test on Monday! Help would be much appreciated.
 
  • #3


Hi there,

I understand your urgency in preparing for your test on Monday. Let me try to provide some clarification on your banked road problem.

Firstly, you are correct in your understanding that for an object to stay in uniform circular motion, the centripetal force (mv^2/r) must be equal to the net force in the radial direction (nsin(theta)).

Now, regarding your confusion with the normal force (n), let me explain it in more detail. The normal force is the force exerted by the surface on the object in contact with it. In this case, the car is in contact with the banked road and the normal force is perpendicular to the surface of the road.

When we draw a free body diagram for the car, we can see that there are two forces acting on it in the radial direction: the normal force (n) and the component of the car's weight (mgcos(theta)). These two forces must add up to the net force in the radial direction, which is mv^2/r. Therefore, we can write the equation n + mgcos(theta) = mv^2/r.

In your equation, you have used ncos(theta) = mg, which is not correct. This would only be true if the road was not banked and the car was moving horizontally. However, in this case, the road is banked at an angle (theta), so we have to take into account the component of the weight in the radial direction, which is mgcos(theta).

I hope this explanation helps clear up your confusion. Remember to always draw a clear free body diagram and consider all the forces acting on an object in order to solve problems like this. Good luck on your test!
 

1. What is a "banked road problem"?

A banked road problem is a physics concept that involves a vehicle traveling on a curved road. The road is angled or banked at a certain angle, which affects the vehicle's speed and stability. This problem is often used to test students' understanding of circular motion and forces.

2. What is the purpose of a test on Monday about the banked road problem?

The purpose of the test is to assess students' understanding of the banked road problem and their ability to apply their knowledge of physics concepts to real-world situations. It also helps teachers identify any areas where students may need further instruction or review.

3. How do you solve a banked road problem?

To solve a banked road problem, you must first draw a free-body diagram and identify all the forces acting on the vehicle. Then, you can use Newton's laws of motion and the principles of circular motion to set up and solve equations. Finally, you should check your answer to ensure it makes sense in the context of the problem.

4. What are some common mistakes students make when solving a banked road problem?

Some common mistakes include not properly drawing the free-body diagram, not considering all the forces at play, and using incorrect equations or formulas. Another mistake is not checking the answer for reasonableness or forgetting to include units in the final answer.

5. Are there any real-world applications of the banked road problem?

Yes, the banked road problem has many real-world applications, such as designing banked turns on racetracks, roller coasters, and high-speed highways. It also helps engineers determine the optimal angle of banking for a road to ensure safe and efficient travel for vehicles.

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