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Circular motion question - finding rpm

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A 0.60 kg sphere rotates around a vertical shaft supported by two strings, as shown. If the tension in the upper string is 18 N. Calculate
    the tension in the lower string?
    the rotation rate (in rev/min) of the system?

    2. Relevant equations

    v=2(pi)r/T
    Fr= mv^2/r
    ω = v/R

    3. The attempt at a solution

    I believe I have the first part correct, my question is really about the second part. For the second part of the question: I calculated the x components of each tension force to find the centripetal force = 20.96N. I then found the velocity by rearranging the equation: Fr = mv^2/r to get 3.74m/s. I used the equation ω = v/R to get 9.35 rads/sec and converted it to 89.3 rpm by multiplying 9.35 by 60s and dividing it by 2(pi). Am I on the right track?? Just seems like a small answer…
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2014 #2
    The upper string has to supply force for the mass to rotate and to stay afloat.
     
  4. Feb 13, 2014 #3

    BvU

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    [strike]Don't follow the 3.74 m/s. Did you divide by 6 or (0.6 x 9.81) instead of by 0.6 kg ?[/strike]

    (Edit: Jgray is fully correct)
     
    Last edited: Feb 13, 2014
  5. Feb 13, 2014 #4
    I don't see any workings of the first part.
     
  6. Feb 13, 2014 #5

    BvU

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    @aziz: I think the poster is happy with the result of the first part (I support him/her in that) and now asks for help in the second part.

    [strike]If I fill in v = 3.74m/s in mv^2/r I do not get the 21 N...[/strike]

    (Edit: Jgray is fully correct)
     
    Last edited: Feb 13, 2014
  7. Feb 13, 2014 #6
    Hmm, maybe I do have an error in my first part then. I found the centripetal force:
    T1x+T2x =
    sin θ(T1) + sin θ(T2)=
    sin53(18N) + sin53(8.2N) --8.2 N is the answer i found for the second tension force for part 1--
    14.37+6.55
    20.92N

    For the velocity:
    v^2= Fr(r)/m
    (20.92N)(0.4m)/0.6kg=
    square root of 13.95=
    3.74m/s
     
  8. Feb 13, 2014 #7

    BvU

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    Fully correct, it was me that used r=4 (from the 3,4,5 triangle) instead of the 0.4 in the exercise. Sorry.

    So the 3.74 is fine too. And the 9.35 and the 89.2.
     
  9. Feb 13, 2014 #8
    Thanks for the help! :)
     
  10. Feb 13, 2014 #9

    BvU

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