Calculating Centripetal Force on a Banked Corner Track

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In summary, the conversation discusses the calculation of friction needed for a speed car racer to make turns on a banked track. After some calculations, the coefficient of static friction is determined to be around 2.97, indicating that friction is used by the racer. Alternatively, the angle at which the track should be banked can also be calculated to require no friction. The conversation also raises the question of why finding the coefficient of friction is not enough, to which the response is that it is more than what was needed for the question at hand.
  • #1
samdiah
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A speed car racer averaged 378.11 km/h in the duration of the race. The ends of the tracks are banked at 18 degrees and have a radius of curvature of 382 m. Did the driver rely on friction for the required centripetal force?

I have a diagram, but it won't paste on here.

I am not sure what to do. So far I just ignored the average speed and calculated another speed...which makes no sense.
and my v ended up being 1216.4m/s ---Overall I don't know how to solve.
 
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  • #2
Do not ignore the given speed. Assume the car is going that fast when it rounds the turns. A force is required to make the car turn. Figure out if the bank of the turn is steep enough to give the car enough force to make the turn, or if it needs help from friction.
 
  • #3
After a lot of thinking i decided to find the coefficient of static friction. If the coefficient = 0 then there is no friction used by racer, but if its not equal to zero then the racer did use friction. Is this a logical analysis? If not then what should I do?

I did a lot of calculations for above and got a coefficient of static friction of about 2.97. Is this resonable.

Ff=us*Fn
...y component's Fn

...then x-component's Fn

...set them equal and get and equation

... then solved like below

cos18-usin18=((m9.8)(382.11(sin18+cos18)/m(378.11)^2)
usin18= 0.032995-cos18/-1
u = 0.918061451/sin18
u=2.97

Can someone please tell me if this is correct?
 
  • #4
I don't see how you got your result, but you need to be careful with your units. There is nothing wrong with calculating the necessary coeficient, but it's more than you need to do to answer this yes or no question. Why not just calculate the contripetal force required to turn the car (it will be in terms of the unkown mass of the car) and compare that to the net force that would act on the car if there were no friction? These are both relatively simple calculations. Draw a free body diagram of the car on the bank, with the only two forces that are acting (if there is no friction) and resolve those into horizontal and vertical components. What must the net vertical force be? How much horizontal force is there when the vertical condition is satisfied?

If you decide friction is needed, you could take the extra step to find the frictional force, but you were not asked to do that.
 
  • #5
is it ok to calculate the angl at which the curve should be banked in order for the car to go without friction. I did this and got 71 deegrees.
 
  • #6
I calculated the two normal forces (one from x and one from y) to be Fn=mg/cos0 and Fn=mv^2/rsin0, and set them equal and then solved for the angle of bank required.

Is this right?
 
  • #7
Another question: why wasn't coefficient calculation enough? If the value of coefficient had been zero then would that not indicate that there's no friction. Or if it is greater than zero then why can't we assume that there is friction.

The question just popped into my mind. Thanks for all the help so far!
 
  • #8
samdiah said:
I calculated the two normal forces (one from x and one from y) to be Fn=mg/cos0 and Fn=mv^2/rsin0, and set them equal and then solved for the angle of bank required.

Is this right?
Finding the angle that requires no friction is a great approach. The equations are good.

Another approach would be to use the given angle and find the speed required to give no friction.
 
  • #9
samdiah said:
Another question: why wasn't coefficient calculation enough? If the value of coefficient had been zero then would that not indicate that there's no friction. Or if it is greater than zero then why can't we assume that there is friction.

The question just popped into my mind. Thanks for all the help so far!
Finding the required coefficient is enough. It's just more than you needed to do.
 
  • #10
Thank you very much! I really appreciate all the help guys provide! Hopefully I can do something for this site someday.
 

FAQ: Calculating Centripetal Force on a Banked Corner Track

What is "Banked Corner Victory"?

"Banked Corner Victory" refers to a phenomenon in racing where a vehicle is able to maintain its speed and traction while navigating a curved track with a banked surface. This allows the vehicle to take the turn at a higher speed and gain an advantage over competitors.

How does a banked corner help achieve victory?

A banked corner allows a vehicle to maintain a higher speed while turning, as the banked surface helps to counteract the centrifugal force that would otherwise pull the vehicle outward. This allows for a smoother and faster turn, giving the vehicle an advantage over competitors who may have to slow down to navigate the turn.

What types of vehicles benefit from banked corners?

Banked corners are most commonly used in racing, particularly in motorsports such as NASCAR and Formula One. However, any vehicle that needs to navigate curved tracks, such as roller coasters or bobsleds, can benefit from banked corners.

What are some examples of famous banked corners?

One famous example of a banked corner is the Indianapolis Motor Speedway, which has banked turns of up to 9 degrees. The Talladega Superspeedway and Daytona International Speedway are also known for their banked corners in NASCAR racing. In Formula One, the Monza circuit in Italy has a famously banked corner known as the "Parabolica."

Are there any downsides to using banked corners?

While banked corners can provide a competitive advantage, they can also be dangerous if not designed and maintained properly. High speeds and sharp turns on a banked surface can lead to accidents if the banking is too steep or if there are issues with the track surface. Additionally, some argue that banked corners take away from the skill and precision required in racing, as the banked surface helps to guide the vehicle through the turn rather than relying solely on the driver's abilities.

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