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Small question about speed of the ball

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello everyone,

    I have this very quick and simple question:
    This is the drawing of the problem and I have to find the speed of the ball.
    uDKmQfo.jpg

    2. Relevant equations


    3. The attempt at a solution
    I found ac to be 0.83m/s2
    and R = 1.2855

    do I simply use this formula to find the speed? ac = v2/R
    ?
    thank you
     
  2. jcsd
  3. Apr 18, 2015 #2

    Simon Bridge

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    The relationship between the (magnitude of) tangential velocity and the centripetal acceleration is ##a_c=v_t^2/r## ... you know this already (it should be in your notes next to the equation you wrote). So the only remaining issue must be whether or not the velocity you are asked to find is the tangential velocity.
    How could you go about figuring that out?
     
  4. Apr 18, 2015 #3
    Well it only says to find the speed of the ball, nothing else. I assumed its the tangential velocity. what other velocity is there to be calculated ?
     
  5. Apr 18, 2015 #4

    Simon Bridge

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    Exactly ... so you have answered your own question.
     
  6. Apr 18, 2015 #5
    But I dont understand because I dont get the correct answer.
    answer is 3.25 and I get
    0.83 = v2/1.2855
    v=1.03294
     
  7. Apr 18, 2015 #6

    Simon Bridge

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    From your value of ##a_c## I get their answer - so it is down to how you did the math.
    What did you use for your value of "r"?
     
  8. Apr 18, 2015 #7

    ehild

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    Your ac is wrong. How did you get it?
     
  9. Apr 19, 2015 #8
    1.2855 = sin(40)*2
    Forces in X = m*ac = -T*sin(40)
    1kg*ac = -T*sin(40)
    Forces in Y = 0 = -w+T*cos(40)
    0 = -1kg*9.81+T*cos(40)

    then when I do a solve with both equations I get T = 12.8 N and ac = 0.839
     
  10. Apr 19, 2015 #9

    ehild

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    ac=T sin(40), how is 12.8 sin(40) equal to 0.839?
     
  11. Apr 19, 2015 #10
    oh wow I really feel stupid now. I misstyped it in my calculator, I assumed it was correct since I was getting the correct value for T.

    thank you very much :P
     
  12. Apr 19, 2015 #11

    ehild

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    You are welcome.
    The tension and weight as vectors ad up to the centripetal force and from the right triangle you can see at once that Fcp=W tan(theta)
     
  13. Apr 19, 2015 #12
    thank you very much for helping me, appreciate it ;P
     
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