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Rotation - lifting buckets

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    You are to design a rotating cylindrical axle to lift 800N buckets of cement from the ground to a rooftop 78.0m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise.

    What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm cm/s when it is turning at 7.5rpm?

    If instead the axle must give the buckets an upward acceleration of 0.400 m/s^2, what should the angular acceleration of the axle be?

    2. Relevant equations

    KE=0.5 I w^2
    GPE = mgh or force x height
    I=0.5 m r^2 (for a disc/cylinder)

    3. The attempt at a solution

    GPE=800x78=62400

    Then i would have though KE=GPE but as theres no mass for the cylinder that can't be right

    thanks in advance
     
  2. jcsd
  3. Dec 9, 2008 #2

    Redbelly98

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    Don't worry about energy or moments of inertia, you won't need those for this problem.

    Instead, how about the equation that relates radius, speed, and angular velocity?
     
  4. Dec 9, 2008 #3
    For the steady speed question
    1.What does 7.5 rpm translate to in revolutions per second?
    2.For the rope to rise at 2cm per second, the 2cm traced out on the circumference of the axle must correspond to the answer of my first question. Now find the axle radius (and hence diameter) which will satisfy this.

    For the acceleration question:
    You keep the axle diameter you have just calculated then relate the linear acceleration of the bucket to the angular acceleration of the axle.
     
  5. Dec 9, 2008 #4
    v=rw
    0.02m/s=r x 0.785 rad/s
    r = 0.0255 m
    d=0.051 m
    (correct)

    a = v^2 / r = w^2 r

    a = 0.016 m/s^2

    but thats only that radial actually... is that what i want?
     
    Last edited: Dec 9, 2008
  6. Dec 9, 2008 #5

    Redbelly98

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    Nope. Try differentiating with respect to time:

    v = rw
     
  7. Dec 9, 2008 #6
    ummm

    a=dw / dt ?

    a(tan)=r (dw/dt) also
    but how do i do dw/dt when i dont have an equation?
     
  8. Dec 9, 2008 #7

    Redbelly98

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    The left hand side is correct, but what happened to the "r" from the original equation?

    This time you got the right hand side correct.

    If you do the derivatives correctly, you'll have the equation. dw/dt will be the only unknown quantitiy.
     
  9. Dec 10, 2008 #8
    ok using:
    a(tan)=r (dw/dt)
    0.4=0.255 (a)
    a= 15.68 rad/s^2

    Sorry i wasnt thinking properly

    thanks lots
     
  10. Dec 10, 2008 #9

    Redbelly98

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    Looks good.

    I didn't realize a(tan) meant tangential acceleration. I was trying to figure out how the tangent function got into this :blushing:
     
  11. Dec 10, 2008 #10
    ha sorry ^^

    thanks for the help
     
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