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Tension Circular Motion Physics Problem?

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data

    A 0.35 kg ball on a string is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the tension is 18 N. What is the tension in the string at 12 o'clock and 6 o'clock position?

    2. Relevant equations
    centripetal force= net force=tension


    3. The attempt at a solution
    I can do most of the problems except this one. Would the tension act as the radius? I wouldn't think so since tension is a force. I drew a free body diagram, but I still can't get the correct answer. I only really need to know how to solve for velocity because centripetal acceleration= v^2/r. then, you just multiply that by mass and add it to Fg, right?
     
    Last edited: Nov 5, 2009
  2. jcsd
  3. Nov 5, 2009 #2

    tiny-tim

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    Hi Penguin10! :smile:
    Tension can only act along the string, so it always lies along the radius.

    Your free body diagram is only useful if you're prepared to include the tangential acceleration. :wink:

    Forget it, and just take components along the radius. :smile:
     
  4. Nov 5, 2009 #3
    What do you mean by taking components along the radius?
     
  5. Nov 5, 2009 #4
    Nevermind, I got the answers. For the 6 o'clock, you add Fg and the given tension. For the 12 o'clock, you subtract the force of gravity from the tension. Could anyone explain why this is? Because I thought that net force was equal to tension which is equal to centripetal force. If centripetal force is always toward the center and gravity is down, then why do you add the tension and Fg for 6 o'clock? Wouldn't you subtract them?
     
  6. Nov 6, 2009 #5

    tiny-tim

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    I mean do Fnet = ma in the direction of the string …

    the same way you do Fnet = ma (or in that case, Fnet = 0) along the normal direction to find the normal reaction force on a slope. :wink:
    eugh! :yuck:

    the tension isn't given, the speed (and therefore the centripetal acceleration ) is constant, but the tension isn't.
    if you mean "centripetal force", that's just another name for the tension.

    if you mean "centripetal acceleration times mass", then yes, that's ma = -mv2/r,

    and it equals Fnet, which for 6 o'clock is mg minus T …

    so yes, you do subtract them (and T = mg + mv2/r).
     
  7. Nov 6, 2009 #6
    Well, it's one of those online homework things...when I added the tension for the 3 o'clock and the force of gravity, it said my answer was correct. The same for subtracting......I drew a picture and gravity was pointing down, as always, so why would you add them? It doesn't make any sense to me personally.........
     
  8. Nov 6, 2009 #7

    tiny-tim

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    You're completely confusing the tension and the centripetal acceleration …

    what you call "the tension for the 3 o'clock" is equal to the centripetal acceleration, mv2/r = T3

    and it's constant, so in the formula I just mentioned, T6 = mg + mv2/r, you get the tension at 6 o'clock by adding what you call "the tension for the 3 o'clock".

    We usually subtract T from mg, to get the centripetal acceleration,

    but in this problem the centripetal acceleration, mv2/r, is given (and you keep calling it the tension), and so you shift the equation around …

    T6 - mg = mv2/r (subtracting) becomes T6 = mg + mv2/r (adding)
     
  9. Nov 9, 2009 #8
    Nevermind, I've got it. I wasn't confusing the tension and centripetal acceleration. I just forgot that you have to add and subtract in order to isolate tension
     
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