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Homework Help: Uniform Circular Motion.

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A car, which weighs 1000 N, travels over a bumpy road with a constant speed. Gravity acts. The road at point B is in the shape of a circle with a radius of 100 meters.
    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam1/sp08/fig21.gif [Broken]

    Which one of these statements correctly relates the force of the car on the road at points A (a valley) and B (the top of a hill)?

    (a) The magnitude of the force of the car on the road is larger at point A than it is at point B.
    (b) The magnitude of the force of the car on the road is larger at point B than it is at point A.
    (c) You need to know the road radius at point A to answer this question.


    2. Relevant equations
    a = v^2/r^2 = w^2*r



    3. The attempt at a solution

    I thought at the point A, it's considered as the lowest point of circular path, while
    the point B is considered as the highest point.

    So, at A Weight is somewhat canceled with the tension(thinking about uniform circular path) but at B weight is added to the tension..

    That's why i thought B has larger force than at point A....

    What's wrong in my attempt..?

    Please help me out here....

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 5, 2010 #2

    rock.freak667

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    At point B, the centripetal force equation is

    [tex]F_c= mg-N_B[/tex]

    where N is the normal reaction. What is the equation at A?
     
  4. Jan 7, 2010 #3
    Isn't it F= (N)a - mg at A?

    But I'm still confused it seems to me that normal force is the same for both points...

    It's not the same, is it?



    Thanks for your reply.
     
  5. Jan 7, 2010 #4

    rock.freak667

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    So at A, NA= mv2/r +mg

    and at B, NB=mg-mv2/r

    which one will be bigger?
     
  6. Jan 9, 2010 #5
    Thanks a lot.
    I've got it !!
     
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