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Finding the angle of a BANKED curve

by rocky811
Tags: angle, banked, curve
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rocky811
#1
Sep18-04, 10:23 PM
P: 15
hi. I'm new here and I just have a quick question. I have tried to figure out this problem, but I am just not sure where to go since there is not a lot given. If someone could give me some more direction, that would be great.

PROBLEM: " A car can negotiate an unbanked curve safely at a certain maximum speed when the coefficient of static friction between the tires and the ground is 0.965. At what angle should the same curve be banked for the car to negotiate the curve safely at the same maximum speed without relying on friction?

** Now I know that the equation tan (theta)= v ^2/rg gives the angle, but without the velocity or radius of the circle, how am i supposed to answer this? I also know that I need to do something with the coefficient of static friction, but i just don't know where to start **
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Sirus
#2
Sep18-04, 10:49 PM
P: 579
Mmm, my favorite kind of question. You have the correct formula for tan (theta), but try thinking a little more about the case of the unbanked curve, and develop an equation for something that is also a part of the equation you have. An interesting result.
rocky811
#3
Sep18-04, 10:50 PM
P: 15
NEVERMIND!!! I actually figured out the question myself because I couldn't give up just yet. So if anyone ever has a question like this...here is how I answered it:

I know that the coefficient of static friction (i'll call it Us) is * gravity= v^2/r.....so i did .965*9.8 m/s^2= v^2/ (1) *I just made the radius equal to one*
I got the velocity to equal 3.075 m/s.

From there I used the equation TAN theta= v^2/rg....and I got the angle to be 43.97 which is correct! :)

rocky811
#4
Sep18-04, 10:52 PM
P: 15
Finding the angle of a BANKED curve

Thanks for your idea...that is exactly what I did! :)
Sirus
#5
Sep18-04, 10:53 PM
P: 579
Nicely done. That is basically what I did, except you don't even have to sub for r because it cancels when you sub back into the first equation. What fun!


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