How Do You Calculate Speeds on a Banked Race Track Without Friction?

AI Thread Summary
To calculate speeds on a banked race track without friction, the radius of the smallest circular path is 111 m and the largest is 157 m, with an outer wall height of 15.7 m. The angle of the bank can be determined using the height and the difference in radius, resulting in an angle of approximately 18.85 degrees. The correct formula for calculating speed involves using the square root of the product of radius, gravitational acceleration, and the tangent of the angle. A participant initially miscalculated the speed due to forgetting to take the square root, but upon correction, found the right answer. Understanding the components of net force and proper calculations is crucial for solving such problems.
Miss1nik2
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On a banked race track, the smallest circular path on which cars can move has a radius of 111 m, while the largest has a radius of 157 m, as the drawing illustrates. The height of the outer wall is 15.7 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.


I am having trouble finding an answer because I wasnt given any angles. Please help me in any way that you can. Thank you VERY much!
 
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Miss1nik2 said:
On a banked race track, the smallest circular path on which cars can move has a radius of 111 m, while the largest has a radius of 157 m, as the drawing illustrates. The height of the outer wall is 15.7 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.


I am having trouble finding an answer because I wasnt given any angles. Please help me in any way that you can. Thank you VERY much!
You didn't attach a picture, but I believe that the track is sloped at an angle that has a 15.7m rise over the track width of (157m -111m) = 46m. From that, you should be able to detremine the angle.
 
Using that information I found the angle to be 18.85 degrees. And I found the answer to part (a) 371.38. But that answer is not correct.

Do you know what I did wrong? Any additional help?? Thank you...
 
And I found the answer to part (a) 371.38. But that answer is not correct.

Did you remember about the square root?
 
Miss1nik2 said:
Using that information I found the angle to be 18.85 degrees. And I found the answer to part (a) 371.38. But that answer is not correct.

Do you know what I did wrong? Any additional help?? Thank you...
your angle looks good. I didn't see your work or check out the math on my own, but you may have used the wrong component of net force in the horizontal direction. What did you use, and why?
 
I used the equation v= sq.rt. r*g*tan (theta)

So, my work for part a is...

sq. rt. (111)(9.8)tan(18.85) = 371.38.

The computer tells me that is the wrong answer...
 
hage567 said:
Did you remember about the square root?
oh, yeah, should've done the math...take the sq root...
 
It worked! THANK YOU! I guess I did forget to take the sq rt. Thank you both.
 

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