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Finding number of rotations till arm breaks (circular motion)

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A 400 g steel block rotates on a steel table while attached to a 1.20 m -long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.91 Nperpendicular to the tube. The maximum tension the tube can withstand without breaking is 60.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1)
    knight_Figure_07_55.jpg

    2. Relevant equations
    ∑Ft=thrust-ƒk
    ∑Fr=mv2/r
    at=αr
    v=ωr
    ωf2i2+2αΔΘ
    2π radians=1 revolution

    3. The attempt at a solution
    ∑Ft=mat=4.91N-(.4kg*.6*9.81m/s2
    at=2.5556kg m/s2 /.4kg
    at=6.839m/s2

    α=6.839m/s2/1.2m
    α=5.324s-2

    I assume radians are assumed as units in α..not sure on this point.

    ∑Fr=mv2/r
    Since 60N is the max ∑Fr can be
    60N=.4kgv2/1.2m
    180 m2/s2=v2
    v=13.416 m/s

    v/r=ω
    13.416m/s/1.2m
    ω=11.18s-1

    Im assuming radians are implicit here. Again not sure.


    ωf2i2+2αΔΘ

    We know ωi2 is zero so:

    ωf2=2αΔΘ
    ωf2/2α=ΔΘ
    (11.18s-1)2/(2*5.324s-2)=ΔΘ
    ΔΘ=11.73

    Again I'm assuming ΔΘ is in radians. So

    11.73 * (1 revolution)/(2π radians)

    1.86 revolutions.

    My answer is wrong. That's all I got.
     
  2. jcsd
  3. Feb 9, 2016 #2

    haruspex

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    I make it closer to 1.87, but other than that I agree. I see nothing wrong in your logic.
    But I strongly recommend working entirely symbolically until the final step. This has many advantages. In the present problem, you would have discovered that the radius was irrelevant, and you would have avoided the need to take a square root.
     
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