Angle for a banked curve (friction-less surface)

  • Thread starter physgrl
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  • #1
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Homework Statement



At what angle should the roadway on a curve with a 50 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is frictionless?

a. 0°
b. 12.2°
c. 17.1°
d. 35.0°
e. 73.2°


Homework Equations



β=arctan(v^2/r*g)


The Attempt at a Solution



im using β=arctan(v^2/r*g) but the key says the answer is b?
can anyone help me figure out why its not 16.4??
 

Answers and Replies

  • #2
993
13
Why not try to derive the formula?
 
  • #3
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i did...i though of the triangle as if the gravity was pointing downward the centripetal force outward and then the angle should be the same as the angle for the curve right?
 
  • #4
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If your acceleration is towards the centre the force will be too. You need to find the angle of the road that gives a normal force with the horizontal component equal to the force needed to keep the circular motion described by v = 12 m/s and r = 50m

I'm aware that my response is a mouthful :smile:, feel free to ask me for clarification.
 
  • #5
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do you mean like a triangle with the vertical 9.81*m and the horizontal (12^2/50)*m ??
 
  • #6
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do you mean like a triangle with the vertical 9.81*m and the horizontal (12^2/50)*m ??
Yes. But the horizontal component will be a fraction of the normal force, which depends on the angle.
 
  • #7
I like Serena
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Welcome to PF, physgrl! :smile:

I derived the same formula as you have.
When I fill in the values I also get 16.4 degrees.
So I believe your answer is correct (assuming you did not make a typo).
 
  • #8
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Welcome to PF, physgrl! :smile:

I derived the same formula as you have.
When I fill in the values I also get 16.4 degrees.
So I believe your answer is correct (assuming you did not make a typo).
I believe your answer is incorrect, as I got 17.1 as my answer. You should investigate why you are using arctan.
 
  • #9
I like Serena
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I believe your answer is incorrect, as I got 17.1 as my answer. You should investigate why you are using arctan.
Here's my interpretation of the forces.

attachment.php?attachmentid=40391&stc=1&d=1319734802.gif


We have:
[tex]F_{resultant} = {m v^2 \over r}[/tex]
[tex]\tan \beta = {F_{resultant} \over m g }[/tex]
The OP's formula follows from this.
 

Attachments

  • #10
ehild
Homework Helper
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16.4° is correct, the answer key is wrong. It happens quite often.

ehild
 
  • #11
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Oooh I see where I went wrong. My apologies to all. I just always have trouble with the normal force being more than gravity. Its very unintuitive to me.
 
  • #12
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Thanks to everyone!!
 
  • #13
I like Serena
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You're welcome! :smile:
 

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