Calculating Centripetal Force on a Banked Corner Track

[samdiah]

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.

 

[OlderDan]

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.

 

[samdiah]

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?

 

[OlderDan]

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.

 

[samdiah]

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.

 

[samdiah]

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?

 

[samdiah]

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!

 

[OlderDan]

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.

 

[OlderDan]

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.

 

[samdiah]

Thank you very much! I really appreciate all the help guys provide! Hopefully I can do something for this site someday.