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Help on circular motion problem

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data

    A ball of mass m is attached to a rigid vertical rod by means of two massless strings each a length L. Both strings are attached on the rod a vertical distance D apart. The rod, strings, and ball are rotated at some speed and both strings are taught. If the tension in the upper string is given as FT, find the speed of the ball and the tension in the lower string.
    2. Relevant equations

    F=ma
    a=v^2/r

    3. The attempt at a solution

    look at my pictures in the attachments... kinda hard to do physics problems online in my opnion. The problem is I feel like the teacher didnt make the question clear enough. can we assume that both strings are at the same angle? can we assume that the y-acceleration is 0?
    also, I didnt even use the value D at all....so i feel like i MUST be going wrong somewhere. the first attachment is the entire problem, but if you cant read it, then i've attached 3 more close ups. THANKS! image.jpeg
     

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  2. jcsd
  3. Apr 29, 2015 #2

    BvU

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    The two strings are taut and equal length. What does that say about ##\theta## and ##\alpha## ?
     
  4. Apr 29, 2015 #3

    haruspex

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    After using BvU's tip, note that you invented those unknowns for the angles, so they cannot feature in your answer.
     
  5. Apr 29, 2015 #4
  6. Apr 29, 2015 #5

    BvU

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    In physics taught is the past tense of teach. :smile:

    A string isn't taught, a string can be wet, dry, curled up or wound up or it can be taut. Which google.
     
  7. Apr 29, 2015 #6
    oh so it means both angles are the same.
     
  8. Apr 29, 2015 #7

    BvU

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    Makes life a lot easier doesn't it :rolleyes: ! And you get a chance to use D
     
  9. Apr 29, 2015 #8
    if they are the same angle then the y acceleration would not be 0 right? the y components of the tension forces cancel out and the y acceleration would be 9.8?
     
  10. Apr 29, 2015 #9

    BvU

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    Ah, just because they are both taut doesn't mean the tensions are equal ! (In fact they are asking for the tension in the lower string, so perhaps that's already Obvious to you).

    You do have me wondering if there are more givens: the angular speed ##\omega## for instance.
    [edit] but perhaps having ##F_T## is enough.
     
    Last edited: Apr 29, 2015
  11. Apr 29, 2015 #10

    haruspex

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    We're given the geometry, the mass, and the upper string tension. The lower string tension can be deduced from those. One can go on to find the rotation rate if desired.
     
  12. Apr 29, 2015 #11
    oh yeah, so the y acceleration is not 0 right?
     
  13. Apr 29, 2015 #12

    haruspex

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    No, if the strings remain taut then the geometry doesn't change, so there can be no vertical movement.
    What three forces act on the mass? Which way is the mass accelerating?
     
  14. Apr 29, 2015 #13
    the 3 forces are the 2 tension forces and the gravitatioanl force. there would only be vertical acceleration if there is air resistance right? it is acceleration horizontally towards the center as of now.
     
  15. Apr 29, 2015 #14

    haruspex

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    Yes, except that I don't see how air resistance would lead to vertical acceleration.
    Are you able to solve it now?
     
  16. Apr 29, 2015 #15
    yes i think i can solve it ... ill post a pic later so you can check my work. its just that as the ball slows down (due to air resistance) would the ball not be in the air anymore? it would go down due to gravity and would lay along the pole. (wouldnt that be a small vertical acceleration) this is ofcourse if air resistance was introduced into the problem.
     
  17. Apr 29, 2015 #16

    haruspex

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    Sure, but you are asked to find the tension while the strings are both taut.
     
  18. Apr 29, 2015 #17
    hmm i got an answer but it looks really complex, and i dont know if i wanna simplify it. do you mind checking it out to see if I made any glaring errors? Thanks! image (7).jpg glaring errors?
     
  19. Apr 29, 2015 #18

    haruspex

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    Your expression for the tension is correct.
    Not sure if your final answer for velocity is right, but it certainly doesn't need to be that complicated.
    Go back to where you had ##r\sin(\theta)## terms. Express either r or ##\sin(\theta)## in terms of the other. What do you notice?
     
  20. Apr 29, 2015 #19
    hmm, well i took theta to be the angle at the pole... not the ball and I think its correct. the angle at the ball is 90-tehta and cos(90-tehta) is the same as sin theta. i agree its really complicated, but i think its because of all the squeare roots. i do notice however, that I can factor out the sin(theta) in the square root and put it outside the radical, but is the logic itself wrong?
     
  21. Apr 29, 2015 #20

    haruspex

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    Maybe you didn't understand my post. I'm suggesting you can make the final steps much easier (and get a simpler answer) if you first look at the relationship between r and theta. You can express r in terms of theta (or sin theta in terms of r) and the given dimensions D, L. Do that in your expression for v2.
     
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