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Circular motion vertically

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    this question is about a pendulum's speed. the string is 1.0m long, in which r=1m. it says the tension is twice of the bob's weight when the string is vertical, which means when its at the bottom point of the circular motion.


    2. Relevant equations
    since [tex]F_T[/tex] should overcome [tex]F_g[/tex], so [tex]F_c=F_T-F_g[/tex].
    here is how i solve this question:
    [tex]F_c=F_T-F_g=\frac{mv^2}{r}[/tex]
    [tex]{2F}_g-F_g=\frac{mv^2}{r}[/tex]
    [tex]F_g=\frac{mv^2}{r}[/tex]
    [tex]g=\frac{v^2}{r}[/tex]
    sub in, then v should be equal to about 3.13m/s
    3. The attempt at a solution
    i got this answer twice, but the teacher said im wrong.
     
  2. jcsd
  3. Jan 27, 2007 #2
    I believe the only thing that matters is the force which keeps it in circular motion: namely, the tension in the string.

    So 2mg = mv^2/r.

    I believe.
     
  4. Jan 27, 2007 #3
    y is there no gravity then?
     
  5. Jan 27, 2007 #4
    maybe because at that point its tangential velocity is perpendicular to gravity?
     
  6. Jan 28, 2007 #5
    The tension force is what gives the centripetal acceleration, not gravity.
     
  7. Jan 28, 2007 #6
    if you draw the freebody diagram, tension up, gravity down. the cetripetal force is up, so tension must overcome gravity in order to give an upward force. so shouldnt it be tension minus gravity to get the net force?
     
  8. Jan 28, 2007 #7
    Hmm.. You're right, I forgot that centripetal force comes from a NET force, in your case this is [itex]mg[/itex], I don't know how this is wrong then...
     
  9. Jan 28, 2007 #8

    arildno

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    Dearly Missed

    Your teacher is wrong, or the problem has been stated poorly.
     
  10. Jan 28, 2007 #9
    oh...really...ok...tthx
     
  11. Jan 28, 2007 #10

    Chi Meson

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    Oh no.

    Your teacher is wrong. You are right. Your solution is perfectly correct. IF your teacher still insists that you are wrong, come back here to this thread and describe what the teacher thinks is correct.

    As a teacher myself, I get very annoyed with other Physics teachers who don't understand physics.
     
  12. Jan 28, 2007 #11

    Moonbear

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    Just a moderation note in case anyone is confused on what happened: the question got posted 3 times, and I've merged all three discussions of it. This puts some of the replies in a strange order.
     
  13. Jan 28, 2007 #12

    arildno

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    There is one other option than that the teacher is wrong:

    The problem has been worded poorly, so that with "vertical" is meant the direction anti-parallell to the direction of the force of gravity, commonly called "upwards". In this case, the pendulum is making full rotations, rather than just oscillate in the lower half-plane.
     
  14. Jan 28, 2007 #13

    Chi Meson

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    Oh, good call arildno! That might just be the case.
     
  15. Jan 29, 2007 #14
    well, i think the teacher is wrong then... coz i asked her wat does vertical mean. yeah, well, the test is over anyways. mayb she just made a careless mistake on my homework...
     
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