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Basic Kinetics question

  1. May 27, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    http://img39.imageshack.us/img39/3778/funkyrider.jpg [Broken]

    So I got this funky rider that starts its movement from the top of the hill A at initial speed (Vo). From that point, the bicycles role on freely (without using pedals) at the circular path.

    Calculate the velocity of the rider at the bottom point B.

    W= (weight of the rider+bicycle) = 650 [N]
    Vo (initial velocity) = 27 km/h
    H = 14m
    R= 10m

    Comment: Ignore friction
    C is point mass


    3. The attempt at a solution

    HERE is my problem. Last year in my notebook I wrote this:

    http://img204.imageshack.us/img204/6220/vfold.jpg [Broken]

    I tried resolving it but I didn't get the same result. This result is the same one in the manual so I reckon it's correct.

    I tried

    a = -9.81
    a = 9.81
    a = v2/R = 7.52/10
    a = g2/R = 9.812/2

    Nothing works. The third one makes slight sense to me because according to the formula of circular motion that's what "a" ought to be, v2/R...problem is that using this I'm not considering "g" and that's impossible, therefor the last a makes most sense to me. None of them gives me my old answer!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 27, 2012 #2
    The equation in your old book is incorrect(the answer is right, somehow :uhh:). You seem to have interchanged the addition and multiplication signs :tongue2:

    From energy conservation it comes out to be,

    [itex]mgh + \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2[/itex]

    cutting out m, you get....

    [itex]\frac{1}{2}v_f^2 = \frac{1}{2}v_i^2 + gh[/itex]

    And of course, g is..........?
     
  4. May 27, 2012 #3

    Femme_physics

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    Hmm, how can you just cut out m? dividing everything by m?

    I'm a bit confused as to why we shouldn't take m into consideration...but maybe it's only for the first clause, since after all it's being defined to us as point mass...

    which is another confusing factor. If the thing is "point mass"-- it shouldn't really have any weight
     
  5. May 27, 2012 #4
    Yes, I divided everything by m. It's non zero, thus legal to divide.

    Mass is taken into consideration for the energy equation, but luckily, it gets divided out. If, somehow, there were some mass joining the original body at an intermediate stage of motion(not effecting the other values), then the final mass would be changed, and you couldn't just 'divide mass out' then.

    A 'point mass' means it has mass(and may have weight) but its dimensions are infinitely small.
     
    Last edited: May 27, 2012
  6. May 27, 2012 #5

    Femme_physics

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    True, m isn't changed through. Thanks, I'll keep it all in check. Much appreciated Infinitum :)
     
  7. May 27, 2012 #6
    Glad to help! :smile:
     
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