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Banked Curves

  1. Sep 29, 2005 #1
    Original Problem:

    Code (Text):
    A curve with a 120-m radius on a level road is banked at the correct angle for
    a speed of 20 m/s.
    Now I think you go about finding the banked angle first and to do that I used..

    Code (Text):
    tan^-1([v^2]/[gr])
    Using that I got the bank angle.. 18.78 degrees. And I am stuck there.

    This problem has been posted before and no solution, or even help, was posted.
     
  2. jcsd
  3. Sep 29, 2005 #2

    Andrew Mason

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    What is the problem asking for besides the angle? You seemed to have provided the correct answer.

    AM
     
  4. Sep 29, 2005 #3

    Fermat

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    He's looking for the minimum coefft of friction.

    See here
     
  5. Sep 30, 2005 #4

    Fermat

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    I saw the original post as well as this one but never responded to either of them because I tend to get a bit confused about some aspects of circular motion.
    If an object is moving in a circle, then there is a centripetal force, acting towards the centre, that provides centripetal acceleration keeping the object constantly travelling in a circle.
    But there is not only a centripetal force, there is also a centrifugal force, of the same magnitude, acting outwards on the object, tending to push it away from the centre of rotation.
    But these forces are equal and opposite. If so, then there is no net force on the object, hence no acceleration !!!

    Obviously, I'm missing something.

    This why I tend to avoid questions that require a discussion of centrigugal forces.

    Which is what is happening in this situation.

    There is a centrifugal force, acting radially outwards, and is parallel to the gound.
    The car is on a banked curve.
    By resolving the Centrifugal force, Fc, along the slope of the bank, you were able to equate this to the resolved component of the car down the slope.
    You had Fc = mv²/r and the car's component as mgsin@ and so were able to get the angle, @, of the bank.
    When v = 30 m/s, then Fc is that much greater, so there is a net force up the slope of (Fc.cos@ - mgsin@).
    Let this be the (static) frictional force.
    Now you have to work out the normal reaction of the slope to the car. Remember to take into account the components of both mg, the weight of the car, and Fc, the new centrifugal force.
     
  6. Sep 30, 2005 #5

    Andrew Mason

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    The centrifugal force, which is not a real force, results from the fact that the car is an accelerated (non-inertial) frame of reference. The only real forces acting on the car are gravity and the normal force of the road, neither of which are directed outward.

    The centripetal force - which causes the car to curve its path of travel - is inward toward the centre of the circle and is supplied by the road. The car rider feels a centrifugal effect tending to push him/her outward but it is due to inertia, not an external force. It is similar to a person experiencing linear forward acceleration who feels a push back into the seat. There is no force pushing the person back - that is just the effect of inertia in an accelerated frame.

    Since the apparent 'centrifugal force' phenomenon appears only in accelerated frames of reference, it does not appear where the centripetal force is supplied by gravity (as in an orbiting spacecraft). Such a frame is physically equivalent to an inertial (non-accelerated) frame of reference.

    AM
     
  7. Oct 1, 2005 #6

    Fermat

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    Thanks for the explanation, Andrew. That clears up some things.
    I still have a sticking point though.

    In the attachment I've decribed the setup for the first part of the above problem where the angle of the bank is calculated.
    I have shown a centrifugl force, Fcf , acting outwards on the car. This provides a component up the slope to counter the downwards component of the car's mass. Equating the two components gives the angle of the slope.

    Now for the sticking point.

    Is there actually any force acting outwards on the car ?
    Or is the only external force acting on the car due to road friction ?
    Is this friction then acting down the slope ?
    Is centrifugal force actually a force, or is it mainly a statement of convenience (calling it a force) to describe the effect it has/gives ?
     

    Attached Files:

    • Fcf.jpg
      Fcf.jpg
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    Last edited: Oct 1, 2005
  8. Oct 2, 2005 #7

    Andrew Mason

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    No. There is only an outward inertial effect which we call centrifugal 'force'. It is not a real force.
    The normal force of the road and gravity are the other external forces acting on the car.

    The direction of the friction force depends on the speed of the vehicle. In the absence of any friction, there is only one speed ([itex]v_c=\sqrt{rgtan\alpha}[/itex]) where the horizontal component of the normal force is exactly equal to the centripetal force needed to make the turn. Any more speed, and the car slides up and off the road. Any less, the car slides down to the bottom. So think of the friction as the force that keeps the car from sliding up if it is going faster than the critical speed vc and the force that keeps it from sliding down if it is going slower.

    It is not a force. It is an inertial effect. It should be called centrifugal effect. See: this link, for example. However, some astrophysicists like to refer to it as a force.

    AM
     
    Last edited: Oct 2, 2005
  9. Oct 2, 2005 #8

    Fermat

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    Thanks. I now have a much better idea about centrifugal force. It's only taken me about 40 yrs to find out it's not actually a force, but an inertial effect :smile:

    errm, just one last question ...

    In the 1st attachment is my working for the first part of the above question, where there is no friction involved. Here, I have assumed a centrifugal "force" acting radially outwards and equated its slope component to the slope component of the car's mass. This gave me the critical velocity, ([itex]v_c=\sqrt{rgtan\alpha}[/itex]) .
    In the 2nd attachment, for the same situation, I have a centrifugal force acting inwards, this force being produced by the horizontal component of the normal reaction, mgcos@.
    But here I get [itex]v = \sqrt{\frac{1}{2}grsin(2\alpha)}[/itex]
    Have I missed something ?
     

    Attached Files:

  10. Oct 2, 2005 #9
    Can you use the velocity 30 m/s and use that for tan^-1([v^2]/[gr]) and see the difference in angle between 30m/s and 20m/s and find the percentage difference?
     
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